Physics - Course Companion - David Homer and Michael Bowen-Jones - Oxford 2014
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OXFORD IB DIPLOM A PROGRAM M E
2 0 1 4 ED I TI O N
PH YSI CS C O U R S E C O M PA N I O N
David Homer Michael Bowen-Jones
Great Clarendon Street, Oxford, OX2 6DP, United Kingdom Oxford University Press is a department of the University of Oxford. It furthers the Universitys objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries Oxford University Press 2014 The moral rights of the authors have been asserted First published in 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above. You must not circulate this work in any other form and you must impose this same condition on any acquirer British Library Cataloguing in Publication Data Data available 978-0-19-839213-2 1 3 5 7 9 10 8 6 4 2 Paper used in the production of this book is a natural, recyclable product made from wood grown in sustainable forests. The manufacturing process conforms to the environmental regulations of the country of origin. Printed in Great Britain
Adrian Hillman/Shutterstock; p420: sciencephotos / Alamy; p427: Zigzag Mountain Art/Shutterstock; p445: Ziga Cetrtic/Shutterstock; p461: Shutterstock; p475: Kamioka Observatory, ICRR (Institute for Cosmic Ray Research), The University of Tokyo; p481: ANDREW LAMBERT PHOTOGRAPHY/SCIENCE PHOTO LIBRARY; p486: Paul Noth/Condenast Cartoons; p499: Kamioka Observatory, ICRR (Institute for Cosmic Ray Research), The University of Tokyo; p507: MIKKEL JUUL JENSEN / SCIENCE PHOTO LIBRARY; p540: NASA, ESA, and STScI; p549: PERY BURGE/SCIENCE PHOTO LIBRARY; p552: Ricardo high speed carbon-fibre flywheel; p570: Shutterstock; p584: iStock; p587: Travelpix Ltd/Getty Images; p593: Allison Herreid/Shutterstock; p594: GIPhotoStock/Science Source; p599a: DAVID PARKER/SCIENCE PHOTO LIBRARY; p599b: DAVID PARKER/SCIENCE PHOTO LIBRARY; p612: Donald Joski/Shutterstock; p614: Image courtesy of Celestron; p616: Israel Pabon/Shutterstock; p617: PETER BASSETT/ SCIENCE PHOTO LIBRARY; p620: Edward Kinsman/Getty Images; p623: US AIR FORCE/SCIENCE PHOTO LIBRARY; p627: iStock; p631: GUSTOIMAGES/ SCIENCE PHOTO LIBRARY; p641: ROBERT GENDLER/SCIENCE PHOTO LIBRARY; p642: ALMA/NAOJ/NRAO/EUROPEAN SOUTHERN OBSERVATORY/ NASA/ESA HUBBLE SPACE TELESCOPE/SCIENCE PHOTO LIBRARY; p643: Alan Dyer, Inc/Visuals Unlimited/Corbis; p644a: CHRIS COOK/SCIENCE PHOTO LIBRARY; p644b: ESO/NASA; p644c: GALEX, JPL-Caltech/NASA; p646: C. CARREAU/EUROPEAN SPACE AGENCY/SCIENCE PHOTO LIBRARY; p654: ROYAL ASTRONOMICAL SOCIETY/SCIENCE PHOTO LIBRARY; p657: NASA/ CXC/ SAO/NASA; p658: ROYAL OBSERVATORY, EDINBURGH/SCIENCE PHOTO LIBRARY; p664: ESA and the Planck Collaboration; p665: GSFC/ NASA; p667: ESO/VISTA/J. Emerson; p676: GSFC/NASA; p677: NASA/WMAP Science Team; p677: NATIONAL OPTICAL ASTRONOMY OBSERVATORIES/ SCIENCE PHOTO LIBRARY; p680: NASA/JPL-Caltech/S.Willner (HarvardSmithsonian CfA); p683: NASA/SCIENCE PHOTO LIBRARY; p689: Dorling Kindersley/Getty Images Artwork by Six Red Marbles and OUP
Acknowledgements
The authors and publisher are grateful for permission to reprint extracts from the following copyright material:
The publishers would like to thank the following for permissions to use their photographs:
P651 Nick Strobel, table Main Sequence Star Properties from www. astronomynotes.com, reprinted by permission.
Cover image: James Brittain/Corbis; p1: Shutterstock; p3: ANDREW BROOKES, NATIONAL PHYSICAL LABORATORY/SCIENCE PHOTO LIBRARY; p4: Victor Habbick/Shutterstock; p9: Shutterstock; p10a: Pavel Mitrofanov/ Shutterstock; p10b: Yury Kosourov/Shutterstock; p10c: ANDREW LAMBERT PHOTOGRAPHY/SCIENCE PHOTO LIBRARY; p27: OUP; p49: Shutterstock; p63: EMILIO SEGRE VISUAL ARCHIVES/AMERICAN INSTITUTE OF PHYSICS/ SCIENCE PHOTO LIBRARY; p66: Georgios Kollidas/Shutterstock; p74: Liviu Ionut Pantelimon/Shutterstock; p80: Helen H. Richardson/The Denver Post/Getty Images; p81: JOHN HESELTINE/SCIENCE PHOTO LIBRARY; p82: MATTEIS/LOOK AT SCIENCES/SCIENCE PHOTO LIBRARY; p85: Autoguide; p91: Rainer Albiez/Shutterstock; p92: MAURICIO ANTON/SCIENCE PHOTO LIBRARY; p101: SCIENCE PHOTO LIBRARY; p106: ANDREW MCCLENAGHAN/SCIENCE PHOTO LIBRARY; p115: BERENICE ABBOTT/ SCIENCE PHOTO LIBRARY; p115: Bill McMullen/Getty Images; p116: Shutterstock; p124a: F1 Online/REX; p124b: sciencephotos/Alamy; p131a: Shutterstock; p131b: FRIEDRICH SAURER/SCIENCE PHOTO LIBRARY; p136: Harvard.Edu; p146: ANDREW LAMBERT PHOTOGRAPHY/SCIENCE PHOTO LIBRARY; p154: GIPHOTOSTOCK/SCIENCE PHOTO LIBRARY; p160: Jerry Di Marco, Montana State Univ. Physics Dept; p169: Volodymyr Krasyuk/ Shutterstock; p195: TREVOR CLIFFORD PHOTOGRAPHY/SCIENCE PHOTO LIBRARY; p229: MARTYN F. CHILLMAID/SCIENCE PHOTO LIBRARY; p245: Shutterstock; p246: Henri Silberman/Getty Images; p252: DAVID DUCROS, CNES/SCIENCE PHOTO LIBRARY; p253: FPG/Hulton Archive/ Getty Images; p254: ASSOCIATED PRESS; p255: Patty Lagera/Getty Images; p258a: Portrait of Nicolaus Copernicus (1473-1543) (oil on canvas), Pomeranian School, (16th century) / Nicolaus Copernicus Museum, Frombork, Poland / Giraudon / The Bridgeman Art Library; p258b: Tycho Brahe, Planella Coromina, Josep or Jose (1804-90) / Private Collection / Look and Learn / The Bridgeman Art Library; p258c: Leemage/Getty Images; p258d: Portrait of Isaac Newton (1642-1727) 1702 (oil on canvas), Kneller, Sir Godfrey (1646-1723) / National Portrait Gallery, London, UK / The Bridgeman Art Library; p258e: DEA PICTURE LIBRARY/Getty Images; p265: Kevin Clogstoun/Getty Images; p267: GORONWY TUDOR JONES, UNIVERSITY OF BIRMINGHAM/SCIENCE PHOTO LIBRARY; p271: SCIENCE PHOTO LIBRARY; p272a: Phil Degginger/Alamy; p272a: Phil Degginger/ Alamy; p272b: Deutsche Bundespost/NobbiP/Wikipedia; p278: Dr Steven Murray/Shutterstock; p279: Bromsgrove School (W. Dainty/C. Shakespear); p280: Bromsgrove School (W. Dainty/C. Shakespear); p296: GORONWY TUDOR JONES, UNIVERSITY OF BIRMINGHAM/SCIENCE PHOTO LIBRARY; p301: Photo courtesy of Berkeley Lab; p303: MissMJ/Wikipedia; p307: Shawn Hempel/Shutterstock; p316: Public Domain/Wikipedia; p322a: Ramon grosso dolarea/Shutterstock; p322b: Worldpics/Shutterstock; p325a: Markuso/Shutterstock; p325b: www.Quebecgetaways.com; p327: Shutterstock; p337: Shutterstock; p348: Public Domain/Wikipedia; p353: Shutterstock; p366: Dietrich Zawischa; p368: GIPHOTOSTOCK/SCIENCE PHOTO LIBRARY; p370: labman.phys.utk.edu; p372: GIPHOTOSTOCK/ SCIENCE PHOTO LIBRARY; p375: CHARLES D. WINTERS/SCIENCE PHOTO LIBRARY; p385: www.astro.cornell.edu; p386: NOAA; p391: Robert
P499 John Updike, Telephone Poles and Other Poems from Cosmic Gall, (Deutsch, 1963), copyright 1959, 1963 by John Updike, reprinted by permission of Alfred A. Knopf, an imprint of the Knopf Doubleday Publishing Group, a division of Random House LLC, and Penguin Books Ltd, all rights reserved. Sources: P472 Albert Einstein Considerations concerning the fundaments of theoretical physics, Science 91:492 (1940) p112 Neutrino faster than light scientist resigns, BBC News 2013 BBC
Contents 1 Measurements and uncertainties Measurements in physics Uncertainties and errors Vectors and scalars
1 8 18
2 Mechanics Motion Forces Work, energy, and power Momentum
27 44 61 73
91 1 00
4 Oscillations and waves O scillations Travelling waves Wave characteristics Wave behaviour S tanding waves
D Astrophysics
S imple harmonic motion S ingle- slit diraction Intererence Resolution The D oppler eect
353 3 64 3 67 3 76 3 81
S tellar quantities S tellar characteristics and stellar evolution C osmology S tellar processes Further cosmology
3 91 405
Internal assessment
10 Fields (AHL) D escribing felds Fields at work
11 Electromagnetic induction (AHL)
3 Thermal physics Temperature and energy changes Modelling a gas
9 Wave phenomena (AHL)
115 1 23 1 34 1 45 1 58
Electromagnetic induction Power generation and transmission C apacitance
The interaction o matter with radiation Nuclear physics
5 Electricity and magnetism
A Relativity The beginnings o relativity Lorentz transormations S pacetime diagrams Relativistic mechanics General relativity
1 92 21 7
Index
693
42 7 43 9 45 5
475 492
5 07 513 522 529 534
B Engineering physics 2 45 257
7 Atomic, nuclear, and particle physics D iscrete energy and radioactivity Nuclear reactions The structure o matter
(with thanks to Mark Headlee for his assistance with this chapter) 687
227
6 Circular motion and gravitation C ircular motion Newtons law o gravitation
65 9 660 666 675
12 Quantum and nuclear physics (AHL)
E lectric felds Heating eect o an electric current E lectric cells Magnetic eects o electric currents
1 69
641
Rigid bodies and rotational dynamics Thermodynamics Fluids and uid dynamics Forced vibrations and resonance
5 49 559 5 70 5 82
C Imaging 2 67 2 82 2 90
Introduction to imaging Imaging instrumentation Fibre optics Imaging the body
5 93 608 62 0 62 6
8 Energy production E nergy sources Thermal energy transer
3 07 329
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Course book defnition
The IB Learner Profle
The IB D iploma Programme course books are resource materials designed to support students throughout their two- year D iploma Programme course o study in a particular subj ect. They will help students gain an understanding o what is expected rom the study o an IB D iploma Programme subj ect while presenting content in a way that illustrates the purpose and aims o the IB . They reect the philosophy and approach o the IB and encourage a deep understanding o each subj ect by making connections to wider issues and providing opportunities or critical thinking.
The aim o all IB programmes to develop internationally minded people who work to create a better and more peaceul world. The aim o the programme is to develop this person through ten learner attributes, as described below.
The books mirror the IB philosophy o viewing the curriculum in terms o a whole- course approach; the use o a wide range o resources, international mindedness, the IB learner profle and the IB D iploma Programme core requirements, theory o knowledge, the extended essay, and creativity, action, service ( C AS ) . Each book can be used in conj unction with other materials and indeed, students o the IB are required and encouraged to draw conclusions rom a variety o resources. Suggestions or additional and urther reading are given in each book and suggestions or how to extend research are provided. In addition, the course books provide advice and guidance on the specifc course assessment requirements and on academic honesty protocol. They are distinctive and authoritative without being prescriptive.
IB mission statement The International B accalaureate aims to develop inquiring, knowledgeable and caring young people who help to create a better and more peaceul world through intercultural understanding and respect. To this end the organization works with schools, governments and international organizations to develop challenging programmes o international education and rigorous assessment. These programmes encourage students across the world to become active, compassionate and lielong learners who understand that other people, with their dierences, can also be right.
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Inquirers: They develop their natural curiosity. They acquire the skills necessary to conduct inquiry and research and snow independence in learning. They actively enj oy learning and this love o learning will be sustained throughout their lives. Knowledgeable: They explore concepts, ideas, and issues that have local and global signifcance. In so doing, they acquire in-depth knowledge and develop understanding across a broad and balanced range o disciplines. Thinkers: They exercise initiative in applying thinking skills critically and creatively to recognize and approach complex problems, and make reasoned, ethical decisions. C ommunicators: They understand and express ideas and inormation confdently and creatively in more than one language and in a variety o modes o communication. They work eectively and willingly in collaboration with others. Princip led: They act with integrity and honesty, with a strong sense o airness, j ustice and respect or the dignity o the individual, groups and communities. They take responsibility or their own action and the consequences that accompany them. O p en-minded: They understand and appreciate their own cultures and personal histories, and are open to the perspectives, values and traditions o other individuals and communities. They are accustomed to seeking and evaluating a range o points o view, and are willing to grow rom the experience. C aring: They show empathy, compassion and respect towards the needs and eelings o others. They have a personal commitment to service, and to act to make a positive dierence to the lives o others and to the environment. Risk-takers: They approach unamiliar situations and uncertainty with courage and orethought, and have the independence o spirit to explore new roles, ideas, and strategies. They are brave and articulate in deending their belies.
B alanced: They understand the importance o intellectual, physical and emotional balance to achieve personal well- being or themselves and others. Refective: They give thoughtul consideration to their own learning and experience. They are able to assess and understand their strengths and limitations in order to support their learning and personal development.
What constitutes malpractice? Malpractice is behaviour that results in, or may result in, you or any student gaining an unair advantage in one or more assessment component. Malpractice includes plagiarism and collusion. Plagiarism is defned as the representation o the ideas or work o another person as your own. The ollowing are some o the ways to avoid plagiarism:
words and ideas o another person to support ones arguments must be acknowledged
passages that are quoted verbatim must be enclosed within quotation marks and acknowledged
C D -Roms, email messages, web sites on the Internet and any other electronic media must be treated in the same way as books and j ournals
the sources o all photographs, maps, illustrations, computer programs, data, graphs, audio- visual and similar material must be acknowledged i they are not your own work
works o art, whether music, flm dance, theatre arts or visual arts and where the creative use o a part o a work takes place, the original artist must be acknowledged.
A note on academic honesty It is o vital importance to acknowledge and appropriately credit the owners o inormation when that inormation is used in your work. Ater all, owners o ideas ( intellectual property) have property rights. To have an authentic piece o work, it must be based on your individual and original ideas with the work o others ully acknowledged. Thereore, all assignments, written or oral, completed or assessment must use your own language and expression. Where sources are used or reerred to, whether in the orm o direct quotation or paraphrase, such sources must be appropriately acknowledged.
How do I acknowledge the work of others? The way that you acknowledge that you have used the ideas o other people is through the use o ootnotes and bibliographies. Footnotes ( placed at the bottom o a page) or endnotes ( placed at the end o a document) are to be provided when you quote or paraphrase rom another document, or closely summarize the inormation provided in another document. You do not need to provide a ootnote or inormation that is part o a body o knowledge. That is, defnitions do not need to be ootnoted as they are part o the assumed knowledge. B ibliograp hies should include a ormal list o the resources that you used in your work. Formal means that you should use one o the several accepted orms o presentation. This usually involves separating the resources that you use into dierent categories ( e.g. books, magazines, newspaper articles, internet-based resources, C D s and works o art) and providing ull inormation as to how a reader or viewer o your work can fnd the same inormation. A bibliography is compulsory in the E xtended E ssay.
C ollusion is defned as supporting malpractice by another student. This includes:
allowing your work to be copied or submitted or assessment by another student
duplicating work or dierent assessment components and/or diploma requirements.
O ther orms o malp ractice include any action that gives you an unair advantage or aects the results o another student. Examples include, taking unauthorized material into an examination room, misconduct during an examination and alsiying a C AS record.
v
Using your IB Physics Online Resources What is Kerboodle? Kerboodle is an online learning platorm. I your school has a subscription to IB Physics Kerboodle O nline Resources you will be able to access a huge bank o resources, assessments, and presentations to guide you through this course.
What is in your Kerboodle Online Resources? There are three main areas or students on the IB Physics Kerboodle: planning, resources, and assessment.
Resources There a hundreds o extra resources available on the IB Physics Kerboodle O nline. You can use these at home or in the classroom to develop your skills and knowledge as you progress through the course. Watch videos and animations o experiments, difcult concepts, and science in action. Hundreds o worksheets read articles, perorm experiments and simulations, practice your skills, or use your knowledge to answer questions. Look at galleries o images rom the book and see their details close up. Find out more by looking at recommended sites on the Internet, answer questions, or do more research.
Planning B e prepared or the practical work and your internal assessment with extra resources on the IB Physics Kerboodle online. Learn about the dierent skills that you need to perorm an investigation. Plan and prepare experiments o your own. Learn how to analyse data and draw conclusions successully and accurately.
One of hundreds of worksheets.
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Practical skills presentation.
Assessment C lick on the assessment tab to check your knowledge or revise or your examinations. Here you will fnd lots o interactive quizzes and examstyle practice questions. Formative tests: use these to check your comprehension, theres one auto-marked quiz or every sub-topic. E valuate how confdent you eel about a sub-topic, then complete the test. You will have two attempts at each question and get eedback ater every question. The marks are automatically reported in the markbook, so you can see how you progress throughout the year. Summative tests: use these to practice or your exams or as revision, theres one auto- marked quiz or every topic. Work through the test as i it were an examination go back and change any questions you arent sure about until you are happy, then submit the test or a fnal mark. The marks are automatically reported in the markbook, so you can see where you may need more practice. Assessment practice: use these to practice answering the longer written questions you will come across when you are examined. These worksheets can be printed out and perormed as a timed test.
Don't forget! You can also fnd extra resources on our ree website www.oxfordsecondary.co.uk/ib-physics Here you can fnd all o the answers and even more practice questions. vii
Introduction Physics is one o the earliest academic disciplines known i you include observational astronomy, possibly the oldest. In physics we analyse the natural world to develop the best understanding we can o how the universe and its constituent parts interrelate. Our aim as physicists is to develop models that correspond to what is observed in the laboratory and beyond. These models come in many orms: some may be quantitative and based on mathematics; some may be qualitative and give a verbal description o the world around us. But, whatever orm the models take, physicists must all agree on their validity beore they can be accepted as part o our physical description o the universe. Models used by physicists are linked by a coherent set o principles known as concepts. These are overarching ideas that link the development o the subject not only within a particular physical topic (or example, orces in mechanics) but also between topics (or example, the common mathematics that links radioactive decay and capacitor discharge) . In studying physics, take every opportunity to understand a new concept when you meet it. When the concept occurs elsewhere your prior knowledge will make the later learning easier. This book is designed to support your learning o physics within group 4 o the IB Diploma Programme. Like all the disciplines represented in this subject group it has a thorough basis in the acts and concepts o science, but it also draws out the nature o science. This is to give you a better understanding o what it means to be a scientist, so that you can, or example, identiy shortcomings in scientifc topics presented to you in the media or elsewhere. Not everyone taking IB Physics will want to go on to be a physicist or engineer, but all citizens need to have an awareness o the importance o science in modern society. The structure o this book needs an explanation; all o the topics include the ollowing elements:
Understanding The specifcs o the content requirements or each sub- topic are covered in detail. C oncepts are presented in ways that will promote enduring understanding.
Investigate! These sections describe practical work you can undertake. You may need to modiy these
viii
experiments slightly to suit the apparatus in your school. These are a valuable opportunity to build the skills that are assessed in IA ( see page 687) .
Nature of science These sections help you to develop your understanding by studying a specifc illustrative example or learning about a signifcant experiment in the history o physics. Here you can explore the methods o science and some o the knowledge issues that are associated with scientifc endeavour. This is done using careully selected examples, including research that led to paradigm shits in our understanding o the natural world.
Theory of Knowledge These short sections have articles on scientifc questions that arise rom Theory o knowledge. We encourage you draw on these examples o knowledge issues in your TOK essays. O course, much o the material elsewhere in the book, particularly in the nature o science sections, can be used to prompt TOK discussions.
Worked example These are step-by-step examples o how to answer questions or how to complete calculations. You should review them careully, preerably ater attempting the question yoursel.
End -of-Topic Questions At the end o each topic you will fnd a range o questions, including both past IB Physics exam questions and new questions. Answers can be ound at www. oxordsecondary.co.uk/ib- physics Authors do not write in isolation. In particular, our ways o describing and explaining physics have been honed by the students we have been privileged to teach over the years, and by colleagues who have challenged our ways o thinking about the subj ect. O ur thanks go to them all. More specifcally, we thank Jean Godin or much sound advice during the preparation o this text. Any errors are, o course, our responsibility. Last but in no sense least, we thank our wives, Adele and B renda, or their ull support during the preparation o this book. We could not have completed it without their understanding and enormous patience. M Bowen-Jones D Homer
1
M EASU REM EN TS AN D U N CERTAI N TI ES
Introduction This topic is different from other topics in the course book. The content discussed here will be used in most aspects of your studies in physics. You will come across many aspects of this work
in the context of other subj ect matter. Although you may wish to do so, you would not be expected to read this topic in one go, rather you would return to it as and when it is relevant.
1.1 Measurements in physics Understanding Fundamental and derived SI units Scientifc notation and metric multipliers Signifcant fgures Orders o magnitude Estimation
Applications and skills Using SI units in the correct ormat or all
required measurements, fnal answers to calculations and presentation o raw and processed data Using scientifc notation and metric multipliers Quoting and comparing ratios, values, and approximations to the nearest order o magnitude Estimating quantities to an appropriate number o signifcant fgures
Nature of science In physics you will deal with the qualitative and the quantitative, that is, descriptions o phenomena using words and descriptions using numbers. When we use words we need to interpret the meaning and one person's interpretation will not necessarily be the same as another's. When we deal with numbers (or equations) , providing we have learned the rules, there is no mistaking someone else's meaning. It is likely that some readers will be more comortable
with words than symbols and vice-versa. It is impossible to avoid either methodology on the IB Diploma course and you must learn to be careul with both your numbers and your words. In examinations you are likely to be penalized by writing contradictory statements or mathematically incorrect ones. At the outset o the course you should make sure that you understand the mathematical skills that will make you into a good physicist.
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1
M E A S U R E M E N TS AN D U N C E R TAI N T I E S
Quantities and units Physicists deal with p hysical quantities, which are those things that are measureable such as mass, length, time, electrical current, etc. Quantities are related to one another by equations such as m which is the symbolic orm o saying that density is the ratio = __ V o the mass o an obj ect to its volume. Note that the symbols in the equation are all written in italic ( sloping) onts this is how we can be sure that the symbols represent quantities. Units are always written in Roman ( upright) ont because they sometimes share the same symbol with a quantity. S o m represents the quantity mass but m represents the unit metre . We will use this convention throughout the course book, and it is also the convention used by the IB .
Nature of science The use of symbols
Greek
The use o Greek letters such as rho () is very common in physics. There are so many quantities that, even using the 5 2 Arabic letters (lower case and capitals) , we soon run out o unique symbols. Sometimes symbols such as d and x have multiple uses, meaning that Greek letters have become just one way o trying to tie a symbol to a quantity uniquely. O course, we must consider what happens when we run out o Greek letters too we then use Russian ones rom the C yrillic alphabet.
alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu
Russian
nu ksi omicron pi rho sigma tau upsilon phi chi psi omega
Fundamental quantities are those quantities that are considered to be so basic that all other quantities need to be expressed in terms o them. m In the density equation = __ only mass is chosen to be undamental V ( volume being the product o three lengths) , density and volume are said to be derived quantities. It is essential that all measurements made by one person are understood by others. To achieve this we use units that are understood to have unambiguous meaning. The worldwide standard or units is known as S I Systme international dunits. This system has been developed rom the metric system o units and means that, when values o scientifc quantities are communicated between people, there should never be any conusion. The SI defnes both units and prefxes letters used to orm decimal multiples or sub- multiples o the units. The units themselves are classifed as being either undamental ( or base) , derived, and supplementary. There are only two supplementary units in S I and you will meet only one o these during the D iploma course, so we might as well mention them frst. The two supplementary units are the radian ( rad) the unit o angular measurement and the steradian ( sr) the unit o solid angle . The radian is a useul alternative to the degree and is defned as the angle subtended by an arc of a circle having the same length as the radius,
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1 . 1 M E A S U R E M E N T S I N P H YS I C S as shown in fgure 1 . We will look at the radian in more detail in S ub- topic 6 . 1 . The steradian is the three- dimensional equivalent o the radian and uses the idea o mapping a circle on to the surace o a sphere.
r r 1 rad r
Fundamental and derived units In S I there are seven fundamental units and you will use six o these on the D iploma course ( the seventh, the candela, is included here or completeness) . The undamental quantities are length, mass, time, electric current, thermodynamic temperature, amount o substance, and luminous intensity. The units or these quantities have exact defnitions and are precisely reproducible, given the right equipment. This means that any quantity can, in theory, be compared with the undamental measurement to ensure that a measurement o that quantity is accurate. In practice, most measurements are made against more easily achieved standards so, or example, length will usually be compared with a standard metre rather than the distance travelled by light in a vacuum. You will not be expected to know the defnitions o the undamental quantities, but they are provided here to allow you to see j ust how precise they are.
Figure 1 Defnition o the radian.
Figure 2 The international prototype kilogram.
metre (m) : the length o the path travelled by light in a vacuum during 1 o a second. a time interval o _________ 2 9 9 7 9 2 45 8 kilogram (kg) : mass equal to the mass o the international prototype o the kilogram kept at the B ureau International des Poids et Mesures at S vres, near Paris. second (s) : the duration o 9 1 92 63 1 770 periods o the radiation corresponding to the transition between the two hyperfne levels o the ground state o the caesium-1 3 3 atom. amp ere (A) : that constant current which, i maintained in two straight parallel conductors o infnite length, negligible circular cross- section, and placed 1 m apart in vacuum, would produce between these conductors a orce equal to 2 1 0 7 newtons per metre o length. 1 o the thermodynamic temperature o the kelvin (K) : the raction _____ 2 73 .1 6 triple point o water.
mole (mol) : the amount o substance o a system that contains as many elementary entities as there are atoms in 0.01 2 kg o carbon1 2 . When the mole is used, the elementary entities must be specifed and may be atoms, molecules, ions, electrons, other particles, or specifed groups o such particles. candela (cd) : the luminous intensity, in a given direction, o a source that emits monochromatic radiation o requency 5 40 1 0 1 2 hertz and 1 that has a radiant intensity in that direction o ___ watt per steradian. 683 All quantities that are not undamental are known as derived and these can always be expressed in terms o the undamental quantities through a relevant equation. For example, speed is the rate o change o distance with s respect to time or in equation orm v = ___ (where s means the change in t distance and t means the change in time) . As both distance (and length) and time are undamental quantities, speed is a derived quantity.
TOK Deciding on what is fundamental Who has made the decision that the undamental quantities are those o mass, length, time, electrical current, temperature, luminous intensity, and amount o substance? In an alternative universe it may be that the undamental quantities are based on orce, volume, requency, potential dierence, specifc heat capacity, and brightness. Would that be a drawback or would it have meant that humanity would have progressed at a aster rate?
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S
Note I you are reading this at the start o the course, it may seem that there are so many things that you might not know; but, take heart, Rome was not built in a day and soon much will come as second nature. When we write units as m s 1 and m s 2 it is a more efective and preerable way to writing what you may have written in the past as m/s and m/s2 ; both orms are still read as metres per second and metres per second squared.
The units used or undamental quantities are unsurprisingly known as undamental units and those or derived quantities are known as derived units. It is a straightorward approach to be able to express the unit o any quantity in terms o its undamental units, provided you know the equation relating the quantities. Nineteen undamental quantities have their own unit but it is also valid, i cumbersome, to express this in terms o undamental units. For example, the S I unit o pressure is the pascal ( Pa) , which is expressed in undamental units as m 1 kg s 2 .
Nature o science Capitals or lower case? Notice that when we write the unit newton in ull, we use a lower case n but we use a capital N or the symbol or the unit unortunately some word processors have deault setting to correct this so take care! All units written in ull should start with a lower case letter, but those that have been derived in honour o a scientist will have a symbol that is a capital letter. In this way there is no conusion between the scientist and the unit: Newton reers to Sir Isaac Newton but newton means the unit. Sometimes units are abbreviations o the scientists surname, so amp (which is a shortened orm o ampre anyway) is named ater Ampre, the volt ater Volta, the arad, Faraday, etc.
Example o how to relate undamental and derived units The unit o orce is the newton ( N) . This is a derived unit and can be expressed in terms o undamental units as kg m s 2 . The reason or this is that orce can be defned as being the product o mass and acceleration or F = ma. Mass is a undamental quantity but acceleration is not. Acceleration is the rate o change o velocity or a = v where v t represents the change in velocity and t the change in time. Although time is a undamental quantity, velocity is not so we need to take another step in defning velocity in undamental quantities. Velocity is the rate o change o displacement ( a quantity that we will discuss later in the topic but, or now, it simply means distance in a given direction) . s So the equation or velocity is v = ___ with s being the change in t displacement and t again being the change in time. D isplacement ( a length) and time are both undamental, so we are now in a position to put N into undamental units. The unit o velocity is m s 1 and these are already undamental there is no shortened orm o this. The units o acceleration will thereore be those o velocity divided by time and so m s will be ____ which is written as m s 2 . So the unit o orce will be the unit s o mass multiplied by the unit o acceleration and, thereore, be kg m s 2 . This is such a common unit that it has its own name, the newton, ( N kg m s 2 a mathematical way o expressing that the two units are identical) . So i you are in an examination and orget the unit o orce you could always write kg m s 2 ( i you have time to work it out! ) .
___
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Figure 3 Choosing fundamental units in an alternative universe.
Signifcant fgures C alculators usually give you many digits in an answer. How do you decide how many digits to write down or the fnal answer?
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1 . 1 M E A S U R E M E N T S I N P H YS I C S
S cientists use a method o rounding to a certain number o signifcant fgures ( oten abbreviated to s. .) . S ignifcant here means meaningul. C onsider the number 84 072 , the 8 is the most signifcant digit, because it tells us that the number is eighty thousand and something. The 4 is the next most signifcant telling us that there are also our thousand and something. Even though it is a zero, the next digit, the 0, is the third most signifcant digit here. When we ace a decimal number such as 0. 002 45 , the 2 is the most signifcant digit because it tells us that the number is two thousandth and something. The 4 is the next most signifcant, showing that there are our ten thousandths and something. I we wish to express this number to two signifcant fgures we need to round the number rom three to two digits. I the last number had been 0.00244 we would have rounded down to 0.0024 and i it had been 0.00246 we would have rounded up to 0.0025 . However, it is a 5 so what do we do? In this case there is equal justifcation or rounding up and down, so all you really need to be is consistent with your choice or a set o fgures you can choose to round up or down. Oten you will have urther digits to help you, so i the number had been 0.002 45 1 and you wanted it rounded to two signifcant fgures it would be rounded up to 0.0025 .
Some rules or using signifcant fgures
A digit that is not a zero will always be signifcant 345 is three signifcant fgures (3 s..) .
Zeros that occur sandwiched between non- zero digits are always signifcant 3 405 ( 4 s.. ) ; 1 0.3 405 ( 6 s..) .
Non- sandwiched zeros that occur to the let o a non- zero digit are not signifcant 0.3 45 ( 3 s.) ; 0. 03 4 ( 2 s.. ) .
Zeros that occur to the right o the decimal point are signifcant, provided that they are to the right o a non- zero digit 1 .03 4 ( 4 s..) ; 1 .00 ( 3 s..) ; 0. 3 45 00 ( 5 s. .) ; 0. 003 ( 1 s..) .
When there is no decimal point, trailing zeros are not signifcant ( to make them signifcant there needs to be a decimal point) 400 ( 1 s..) ; 400. ( 3 s..) but this is rarely written.
Scientifc notation O ne o the ascinations or physicists is dealing with the very large ( e.g. the universe) and the very small ( e.g. electrons) . Many physical constants ( quantities that do not change) are also very large or very small. This presents a problem: how can writing many digits be avoided? The answer is to use scientifc notation. The speed o light has a value o 299 792 45 8 m s 1 . This can be rounded to three signifcant fgures as 300 000 000 m s 1 . There are a lot o zeros in this and it would be easy to miss one out or add another. In scientifc notation this number is written as 3.00 1 0 8 m s 1 (to three signifcant fgures) . Let us analyse writing another large number in scientifc notation. The mass o the Sun to our signifcant fgures is 1 989 000 000 000 000 000 000 000 000 000 kg ( that is 1 989 and twenty- seven zeros) . To convert
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S this into scientifc notation we write it as 1 .989 and then we imagine moving the decimal point 3 0 places to the let ( remember we can write as many trailing zeros as we like to a decimal number without changing it) . This brings our number back to the original number and so it gives the mass o the Sun as 1 . 989 1 0 30 kg. A similar idea is applied to very small numbers such as the charge on the electron, which has an accepted value o approximately 0.000 000 000 000 000 000 1 602 coulombs. Again we write the coefcient as 1 .602 and we must move the decimal point 1 9 places to the right in order to bring 0.000 000 000 000 000 000 1 602 into this orm. The base is always 1 0 and moving our decimal point to the right means the exponent is negative. We can write this number as 1 .602 1 0 1 9 C . Apart rom avoiding making mistakes, there is a second reason why scientifc notation is preerable to writing numbers in longhand. This is when we are dealing with several numbers in an equation. In writing the value o the speed o light as 3 .00 1 0 8 m s 1 , 3 . 00 is called the coefcient o the number and it will always be a number between 1 and 1 0. The 1 0 is called the base and the 8 is the exponent. There are some simple rules to apply:
When adding or subtracting numbers the exponent must be the same or made to be the same.
When multiplying numbers we add the exponents.
When dividing numbers we subtract one exponent rom the other.
When raising a number to a power we raise the coefcient to the power and multiply the exponent by the power.
Worked examples In these examples we are going to evaluate each o the calculations. 1
1 .40 1 0 6 + 3 .5 1 0 5
S o we write this product as: 7.8 1 0 3
Solution
4
These must be written as 1 . 40 1 0 6 + 0.3 5 1 0 6 so that both numbers have the same exponents.
Solution
They can now be added directly to give 1 .75 1 0 6 2
3 .7 1 0 5 2 .1 1 0 8
Solution
So we write this product as: 7. 8 1 0 3
4.8 1 0 5 _ 3 .1 1 0 2
The coefcients are divided and the exponents are subtracted so we have: 4.8 3 . 1 = 1 .5 48 ( which we round to 1 .5 ) And 5 2 = 3
The coefcients are multiplied and the exponents are added, so we have: 3.7 2.1 = 7.77 (which we round to 7.8 to be in line with the data something we will discuss later in this topic) and: 5 + 8 = 1 3 13
3 .7 1 0 5 2 .1 1 0 8
Solution Again the coefcients are multiplied and the exponents are added, so we have: 3 .7 2 . 1 = 7. 8
6
Here the exponents are subtracted ( since the 8 is negative) to give: 5 8 = 3
This makes the result o the division 1 .5 1 0 3 5
( 3 .6 1 0 7 ) 3
Solution We cube 3 .6 and 3 .6 3 = 46.7 And multiply 7 by 3 to give 2 1 This gives 46.7 1 0 21 , which should become 4.7 1 0 22 in scientifc notation.
1 . 1 M E A S U R E M E N T S I N P H YS I C S
Metric multipliers (prefxes)
Factor 10 24 10 21 10 18 10 15 10 12 10 9 10 6 10 3 10 2 10 1 10 -1 10 -2 10 -3 10 -6 10 -9 10 -12 10 -15 10 -18 10 -21 10 -24
S cientists have a second way o abbreviating units: by using metric multipliers ( usually called prefxes ) . An S I prefx is a name or associated symbol that is written beore a unit to indicate the appropriate power o 1 0. S o instead o writing 2 . 5 1 0 1 2 J we could alternatively write this as 2 . 5 TJ ( teraj oule) . Figure 4 gives the 2 0 S I prefxes these are provided or you as part o the data booklet used in examinations.
Orders of magnitude An important skill or physicists is to understand whether or not the physics being considered is sensible. When perorming a calculation in which someones mass was calculated to be 5 000 kg, this should ring alarm bells. S ince average adult masses ( weights) will usually be 6090 kg, a value o 5 000 kg is an impossibility. A number rounded to the nearest power o 1 0 is called an order of magnitude. For example, when considering the average adult human mass: 6080 kg is closer to 1 00 kg than 1 0 kg, making the order o magnitude 1 0 2 and not 1 0 1 . O course, we are not saying that all adult humans have a mass o 1 00 kg, simply that their average mass is closer to 1 00 than 1 0. In a similar way, the mass o a sheet o A4 paper may be 3 . 8 g which, expressed in kg, will be 3 . 8 1 0 3 kg. S ince 3 . 8 is closer to 1 than to 1 0, this makes the order o magnitude o its mass 1 0 3 kg. This suggests that the ratio o adult mass to the mass o a piece o paper (should 10 you wish to make this comparison) = ____ = 1 0 2 ( 3) = 1 0 5 = 1 00 000. In 10 other words, an adult human is 5 orders o magnitude ( 5 powers o 1 0) heavier than a sheet o A4 paper.
Name yotta zetta exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto zepto yocto
Symbol
d c m n p f a z y
Figure 4 SI metric multipliers.
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Estimation E stimation is a skill that is used by scientists and others in order to produce a value that is a useable approximation to a true value. E stimation is closely related to fnding an order o magnitude, but may result in a value that is more precise than the nearest power o 1 0. Whenever you measure a length with a ruler calibrated in millimetres you can usually see the whole number o millimetres but will need to 1 estimate to the next __ mm you may need a magniying glass to help 10 you to do this. The same thing is true with most non- digital measuring instruments. S imilarly, when you need to fnd the area under a non- regular curve, you cannot truly work out the actual area so you will need to fnd the area o a rectangle and estimate how many rectangles there are. Figure 5 shows a graph o how the orce applied to an obj ect varies with time. The area under the graph gives the impulse ( as you will see in Topic 2 ) . There are 2 6 complete or nearly complete yellow squares under the curve and there are urther partial squares totalling about our ull squares in all. This gives about 3 0 ull squares under the curve. E ach curve has an area equivalent to 2 N 1 s = 2 N s. This gives an estimate o about 60 N s or the total impulse.
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S In an examination, estimation questions will always have a tolerance given with the accepted answer, so in this case it might be ( 60 2 ) N s.
12 These two partial squares may be combined to approximate to a whole square, etc.
force/N
10 8 6 4 2 0 0
1
2
3
4
5
6
7
8
time/s
Figure 5
1.2 Uncertainties and errors Understanding Random and systematic errors Absolute, ractional, and percentage uncertainties Error bars Uncertainty o gradient and intercepts
Nature of science In Sub-topic 1.1 we looked at the how we defne the undamental physical quantities. Each o these is measured on a scale by comparing the quantity with something that is precisely reproducible. By precisely reproducible do we mean exact? The answer to this is no. I we think about the defnition o the ampere, we will measure a orce o 2 10 7 N. I we measure it to be 2.1 10 7 it doesnt invalidate the measurement since the defnition is given to just one signifcant fgure. All measurements have their limitations or uncertainties and it is important that both the measurer and the person working with the measurement understand what the limitations are. This is why we must always consider the uncertainty in any measurement o a physical quantity.
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Applications and skills Explaining how random and systematic errors
can be identifed and reduced Collecting data that include absolute and/or ractional uncertainties and stating these as an uncertainty range (using ) Propagating uncertainties through calculations involving addition, subtraction, multiplication, division, and raising to a power Using error bars to calculate the uncertainty in gradients and intercepts
Equations Propagation o uncertainties: I: y = a b then: y = a + b ab I: y = _____ c y
______ ______ ______ a c b then: ______ y = a + b + c I: y = a n y ______ a then: _______ y = n a
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
Uncertainties in measurement Introduction No experimental quantity can be absolutely accurate when measured it is always subj ect to some degree o uncertainty. We will look at the reasons or this in this section. There are two types o error that contribute to our uncertainty about a reading systematic and random.
Systematic errors As the name suggests, these types o errors are due to the system being used to make the measurement. This may be due to aulty apparatus. For example, a scale may be incorrectly calibrated either during manuacture o the equipment, or because it has changed over a period o time. Rulers warp and, as a result, the divisions are no longer symmetrical. A timer can run slowly i its quartz crystal becomes damaged ( not because the battery voltage has allen when the timer simply stops) .
Figure 1 Zero error on digital calliper.
When measuring distances rom sealed radioactive sources or lightdependent resistors ( LD Rs) , it is hard to know where the source is actually positioned or where the active surace o the LD R is. The zero setting on apparatus can drit, due to usage, so that it no longer reads zero when it should this is called a zero error. Figure 1 shows a digital calliper with the j aws closed. This should read 0.000 mm but there is a zero error and it reads 0.01 mm. This means that all readings will be 0.01 mm bigger than they should be. The calliper can be reset to zero or 0.01 mm could be subtracted rom any readings made. O ten it is not possible to spot a systematic error and experimenters have to accept the reading on their instruments, or else spend signifcant eort in making sure that they are re-calibrated by checking the scale against a standard scale. Repeating a reading never removes the systematic error. The real problem with systematic errors is that it is only possible to check them by perorming the same task with another apparatus. I the two sets give the same results, the likelihood is that they are both perorming well; however, i there is disagreement in the results a third set may be needed to resolve any dierence. In general we deal with zero errors as well as we can and then move on with our experimentation. When systematic errors are small, a measurement is said to be accurate.
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Nature of science Systematic errors Uncertainty when using a 300 mm ruler may be quoted to 0.5 mm or 1 mm depending on your view o how precisely you can gauge the reading. To be on the sae side you might wish to use the larger uncertainty and then you will be sure that the reading lies within your bounds.
Figure 2 Millimeter (mm) scale on ruler.
You should make sure you observe the scale rom directly above and at right angles to the plane o the ruler in order to avoid parallax errors.
Figure 4 Analogue scale.
Figure5 Digital scale.
The digital ammeter in fgure 5 gives a value o 0.27 A which should be recorded as (0.27 0.01 ) A. In each o these examples the uncertainty is quoted to the same p recision ( number o decimal places) as the reading it is essential to do this as the number o decimal places is always indicative o precision. When we write an energy value as being 8 J we are implying that it is ( 8 1 ) J and i we write it as 8. 0 J it implies a precision o 0. 1 J.
1 2
3
4 5 6
7 8
9 10
The meter in fgure 4 shows an analogue ammeter with a airly large scale there is justifcation in giving this reading as being (40 5 ) A.
Figure 3 Parallax error.
Random errors Random errors can occur in any measurement, but crop up most requently when the experimenter has to estimate the last signifcant fgure when reading a scale. I an instrument is insensitive then it may be difcult to j udge whether a reading would have changed in dierent circumstances. For a single reading the uncertainty could well be better than the smallest scale division available. B ut, since you are determining the maximum possible range o values, it is a sensible precaution to use this larger precision. D ealing with digital scales is a problem the likelihood is that you have really no idea how precisely the scales are calibrated. C hoosing the least signifcant digit on the scale may severely underestimate the uncertainty but, unless you know the manuacturers data regarding calibration, it is probably the best you can do. When measuring a time manually it is inappropriate to use the precision o the timer as the uncertainty in a reading, since your reaction time is likely to be ar greater than this. For example, i you timed twenty oscillations o a pendulum to take 1 6. 2 7 s this should be recorded as being ( 1 6.3 0.1 ) s. This is because your reaction time dominates the precision o the timer. I you know that your reaction time is greater than 0.1 s then you should quote that value instead o 0.1 s.
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1 . 2 U N CE R TAI N TI E S AN D E R R O R S
The best way o handling random errors is to take a series o repeat readings and fnd the average o each set o data. Hal the range o the values will give a value that is a good approximation to the statistical value that more advanced error analysis provides. The range is the largest value minus the smallest value. Readings with small random errors are said to be p recise ( this does not mean they are accurate, however) .
Worked examples 1
In measuring the angle o reraction at an airglass interace or a constant angle o incidence the ollowing results were obtained ( using a protractor with a precision o 1 o ) :
T/C
Solution The mean o these values is 45 . 4 and the range is ( 47 44) = 3 . Hal the range is 1 .5 . How then do we record our overall value or the angle o reraction? S ince the precision o the protractor is 1 , we should quote our mean to a whole number ( integral) value and it will round down to 45 . We should not minimize our uncertainty unrealistically and so we should round this up to 2 . This means that the angle o reraction should be recorded as 45 2 . 2
The diagram below shows the position o the meniscus o the mercury in a mercury-in- glass thermometer.
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Express the temperature and its uncertainty to an appropriate number o signifcant fgures.
45 , 47, 46, 45 , 44 How should we express the angle o reraction?
0
Solution The scale is calibrated in degrees but they are quite clear here, so it is reasonable to expect a precision o 0.5 C . The meniscus is closer to 6 than to 6.5 ( although that is a j udgement decision) so the values should be recorded as ( 6.0 0. 5 ) C . Remember the measurement and the uncertainty should be to the same number o decimal places. 3
A student takes a series o measurements o a certain quantity. He then averages his measurements. What aspects o systematic and random uncertainties is he addressing by taking repeats and averages?
Solution S ystematic errors are not dealt with by means o repeat readings, but taking repeat readings and averaging them should cause the average value to be closer to the true value than a randomly chosen individual measurement.
Absolute and fractional uncertainties The values o uncertainties that we have been looking at are called absolute uncertainties. These values have the same units as the quantity and should be written to the same number o decimal places. Dividing the uncertainty by the value itsel leaves a dimensionless quantity (one with no units) and gives us the fractional uncertainty. Percentaging the ractional uncertainty gives the percentage uncertainty.
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Worked example C alculate the absolute, ractional, and percentage uncertainties or the ollowing measurements o a orce, F:
range = ( 2 .8 2 .5 ) N = 0. 3 N, giving an absolute uncertainty o 0.1 5 N that rounds up to 0.2 N
2 .5 N, 2 .8 N, 2 .6 N
We would write our value or F as ( 2 .6 0.2 ) N
Solution
0.2 = 0.077 and the the ractional uncertainty is ___ 2 .6 percentage uncertainty will be 0.077 1 00% = 7.7%
2 .5 N + 2 .8 N + 2 .6 N
mean value = ________________ = 2 .63 N, this is 3 rounded down to 2.6 N
Propagation of uncertainties O ten we measure quantities and then use our measurements to calculate other quantities with an equation. The uncertainty in the calculated value will be determined rom a combination o the uncertainties in the quantities that we have used to calculate the value rom. This is known as p rop agation of uncertainties. There are some simple rules that we can apply when we are propagating uncertainties. In more advanced treatment o this topic we would demonstrate how these rules are developed, but we are going to ocus on your application o these rules here ( since you will never be asked to prove them and you can look them up in a text book or on the Internet i you want urther inormation) . In the uncertainty equations discussed next, a, b, c, etc. are the quantities and a, b, c, etc. are the absolute uncertainties in these quantities. 11
Addition and subtraction
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This is the easiest o the rules because when we add or subtract quantities we always add their absolute uncertainties. When a = b + c
or a = b c
then a = b + c
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In order to use these relationships dont orget that the quantities being added or subtracted must have the same units.
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S o i we are combining two masses m 1 and m 2 then the total mass m will be the sum o the other two masses.
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m 1 = ( 2 00 1 0) g and m 2 = ( 1 00 1 0) g so m = 3 00 g and m = 2 0 g meaning we should write this as:
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m = ( 3 00 2 0) g
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We use subtraction more oten than we realise when we are measuring lengths. When we set the zero o our ruler against one end o an object we are making a judgement o where the zero is positioned and this really means that the value is (0.0 0.5) mm.
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A ruler is used to measure a metal rod as shown in fgure 6. The length is ound by subtracting the smaller measurement rom the larger one. The uncertainty or each measurement is 0.5 mm.
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Larger measurement = 1 95 .0 mm S maller measurement = 1 1 8. 5 mm
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Figure 6 Measuring a length.
Length = ( 76.5 1 .0) mm as the uncertainty is 0. 5 mm + 0.5 mm
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
Nature of science Subtracting values When subtraction is involved in a relationship you need to be particularly careful. The resulting quantity becomes smaller in size (because of subtraction) , while the absolute uncertainty becomes larger (because of addition) . Imagine two values that are subtracted: b = 4.0 0.1 and c =3.0 0.1 . We wont concern ourselves with what these quantities actually are here.
If a = b c then a = 1 .0 and since a = b + c then a = 0.2 We have gone from two values in which the percentage uncertainty is 2.5 % and 3.3% respectively to a calculated value with uncertainty of 20%. Now that really is propagation of uncertainties!
Multiplication and division When we multiply or divide quantities we add their fractional or percentage uncertainties, so: when a = bc or a = __bc or a = __bc a b c ___ ___ then ___ a = b + c
There are very few relationships in physics that do not include some form of multiplication or division. m We have seen that density is given by the expression = __ where V m is the mass of a sample of the substance and V is its volume. For a particular sample, the percentage uncertainty in the mass is 5 % and for the volume is 1 2 % .
The percentage uncertainty in the calculated value of the density will therefore be 1 7% . If the sample had been cubical in shape and the uncertainty in each of the sides was 4% we can see how this brings about a volume with uncertainty of 1 2 % : For a cube the volume is the cube of the side length ( V = l 3 = l l l) V so ___ = V
__ + __ + __ = 4% + 4% + 4% = 1 2 % l l
l l
l l
This example leads us to:
Raising a quantity to a power V l From the cube example you might have spotted that ___ = 3 __ V l
This result can be generalized so that when a = b n ( where n can be a positive or negative whole, integral, or decimal number)
a b ___ then ___ a = n b
The modulus sign is included as an alternative way of telling us that the uncertainty can be either positive or negative.
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Worked example The period T o oscillation o a mass __ m on a spring, m having spring constant k is T = 2 __ k Dont worry about what these quantities actually mean at this stage. The uncertainty in k is 1 1 % and the uncertainty in m is 5 % . C alculate the approximate uncertainty in a value or T o 1 .2 0 s.
Solution First lets adj ust the equation a little we can write it as
( )
m T = 2 __ k
1 __ 2
()
which is o the orm a = 2 __bc
n
Although we will truncate , we can really write it to as many signifcant fgures as we wish and so the percentage uncertainty in , as in 2 will be zero.
Using the division and power relationships: a b c m k T 1 ___ 1 ___ ___ ___ ___ ___ __ __ a = n b + n c or here T = 2 m + 2 k
so the percentage uncertainty in T will be hal that in m + hal that in k. This means that the percentage error in T = 0.5 5 % + 0.5 1 1 % = 8% I the measured value o T is 1 .2 0 s then the 8 absolute uncertainty is 1 .20 ___ = 0. 096 This 1 00 rounds up to 0.1 0 and so we quote T as being ( 1 . 2 0 0.1 0) s. Remember that the quantity and the uncertainty must be to the same number o decimal places and so the zeros are important, as they give us the precision in the value.
Drawing graphs speed/m s 1
3.0 2.0 1.0 0 0
0.1
0.2 0.3 time/s
0.4
Figure 7 Error bars.
An important j ustifcation or experimental work is to investigate the relationship between physical quantities. O ne set o values is rarely very revealing even i it can be used to calculate a physical constant, such as dividing the potential dierence across a resistor by the current in the resistor to fnd the resistance. Although the calculation does tell you the resistance or one value o current, it says nothing about whether the resistance depends upon the current. Taking a series o values would tell you i the resistance was constant but, with the expected random errors, it would still not be defnitive. B y plotting a graph and drawing the line o best ft the pattern o results is ar easier to spot, whether it is linear or some other relationship.
Error bars In plotting a point on a graph, uncertainties are recognized by adding error bars. These are vertical and horizontal lines that indicate the possible range o the quantity being measured. S uppose at a time o ( 0. 2 0.05 ) s the speed o an obj ect was ( 1 .2 0.2 ) m s 1 this would be plotted as shown in fgure 7.
The value could lie anywhere inside this rectangle
speed/m s 1
3.0
This means that the value could possibly be within the rectangle that touches the ends o the error bars as shown in fgure 8. This is the zone of uncertainty or the data point. A line o best ft should be one that spreads the points so that they are evenly distributed on both sides o the line and also passes through the error bars.
2.0 1.0 0 0
0.1
0.2 0.3 time/s
Figure 8 Zone of uncertainty.
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Uncertainties with gradients Using a computer application, such as a spreadsheet, can allow you to plot a graph with data points and error bars. You can then read o the gradient and the intercepts rom a linear graph directly. The application
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
will automatically draw the best trend line. You can then add the trend lines with the steepest and shallowest gradients that are j ust possible while still passing through all the error bars. S tudents quite commonly, but incorrectly, use the extremes o the error bars that are urthest apart on the graphs. Although these could be appropriate, it is essential that all the trend lines you draw pass through all o the error bars. In an experiment to measure the electromotive orce (em) and internal resistance o a cell, a series o resistors are connected across the cell. The currents in and potential dierences across the resistors are then measured. A graph o potential dierence, V, against current, I, should give a straight line o negative gradient. As you will see in Topic 5 the em o a cell is related to the internal resistance r by the equation: = I( R + r) = V + Ir This can be rearranged to give V = Ir So a graph o V against I is o gradient r (the internal resistance) and intercept (the em o the cell) . The table on the right shows a set o results rom this experiment. With a milliammeter and voltmeter o low precision the repeat values are identical to the measurements given in the table. The graph o fgure 9 shows the line o best ft together with two lines that are j ust possible. EMF and Internal Resistance 1.8
I 5/mA
V 0.1/V
15
1.5
20
1.4
1.2
25
1.4
1.0
30
1.3
0.8
35
1.2
0.6
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1.1
0.4
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0.9
0.2
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0.8
85
0.6
90
0.5
1.6
V = 0.013I + 1.68 V = 0.0127I + 1.64 V = 0.0153I + 1.78
1.4
V/V
0.0 0
20
40
60
80
100
120
140
I/mA
Figure 9 A graph o potential diference, V, against current, I, or a cell.
Converting rom milliamps to amps, the equations o these lines suggest that the internal resistance (the gradient) is 1 3.0 and the range is rom 1 2.7 to 1 5.3 (= 2.6 ) meaning that hal the range = 1 .3 . This leads to a value or r = ( 1 3 . 0 1 . 3 ) . The intercept on the V axis o the line o best ft = 1 .68 which rounds to 1 .7 V ( since the data is essentially to 2 signifcant fgures) . The range o the j ust possible lines gives 1 .6 to 1 . 8 V ( when rounded to two signifcant fgures) . This means that = ( 1 .6 0. 1 ) V.
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S
Nature of science Drawing graphs manually O ne o the skills expected o physicists is to draw graphs by hand and you may well be tested on this in the data analysis question in Paper 3 o the IB D iploma Programme physics examination. You are also likely to need to draw graphs or your internal assessment.
Try to look at your extreme values so that you have an idea o what scales to use. You will need a minimum o six points to give you a reasonable chance o drawing a valid line.
Use scales that will allow you to spread your points out as much as possible ( you should fll your page, but not overspill onto a second sheet as that would damage your line quality and lose you marks) . You can always calculate an intercept i you need one; when you dont
include the origin, your axes give you a false origin ( which is fne) .
Use sensible scales that will make both plotting and your calculations clear-cut (avoid scales that are multiples o 3, 4, or 7 stick to 2, 5, and 1 0) .
Try to plot your graph as you are doing the experiment i apparently unusual values crop up, you will see them and can check that they are correct.
B eore you draw your line o best ft, you need to consider whether or not it is straight or a curve. There may well be anomalous points ( outliers) that you can ignore, but i there is a defnite trend to the curve then you should opt or a smooth curve drawn with a single line and not sketched artistically!
Figure 1 0 shows some o the key elements o a good hand- drawn graph. C alculating the gradients on the graph is very useul when checking values. best straight line
line that is just possible
gradient of best straight line (59.6 42.0) 10 6 m 3
58
=
Second just possible line should be added for a real investigation it has been missed out here so that you can clearly see the values on the two lines
(398 288) K
= 1.60 10 7 m 3 K1 56 gradient of just possible line (58.6 42.0) 10 6 m 3
54
=
(398 286) K = 1.48 10 7 m 3 K1
V/10 6 m 3
52
All values on V axis have been divided by 10 6 and are in m 3
Use a large gradient triangle to reduce uncertainties
50 48 46 outlier ignored for best straight line
44 False origin neither line has been forced through this point
Uncertainties in temperature are too small to draw error bars
42 40 260
16
Figure 10 Hand-drawn graph.
280
300
320
340 T/K
360
380
400
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
Linearizing graphs
Note
Many relationships between physical quantities are not directly proportional and a straight line cannot be obtained simply by plotting one quantity against the other. There are two approaches to dealing with non- proportional relationships: when we know the orm o the relationship and when we do not.
Relationships can never be indirectly proportional this is a meaningless term since it is too vague. Consequently the term proportional means the same as directly proportional.
1 (or a gas held I we do know the orm o the relationship such as p __ _ V at constant temperature) or T l (or a simple pendulum) we can plot a graph o one quantity against the power o the other quantity to obtain a straight-line origin graph. An alternative or the simple pendulum is to plot a graph o T 2 against l which will give the same result.
This topic is dealt with in more detail and with many further examples on the website.
You should think about the p rop agation of errors_ when you consider the relative merits of plotting T against l or T 2 against l. The following discussion applies to HL examinations, but offers such a useful technique that SL students may wish to utilize it when completing IAs or if they undertake an Extended Essay in a science subject.
By taking logs o this equation we obtain log y = log k + n log x which we can arrange into log y = n log x + log k and is o the orm y = m x + c. This means that a graph o log y against log x will be linear o gradient n and have an intercept on the log y axis o log k.
N/arbitrary units
I we dont know the actual power involved in a relationship, but we suspect that one quantity is related to the other, we can write a general relationship in the orm y = kxn where k and n are constants.
expontential shape 45 30 25 20 15 10 5 0
N = 32e -0.039t
0
N = N0 e - t by taking logs to the base e we get lnN = lnN0 - t ( where lnN is the usual way o writing log e N) .
40 60 t/arbitrary units
80
100
linear form of same data
B y plotting a graph o lnN against t the gradient will be - t and the intercept on the lnN axis will be lnN0 . This linearizes the graph shown in fgure 1 1 producing the graph o fgure 1 2 . A linear graph is easier to analyse than a curve. C apacitors also discharge through resistors using the same general mathematical relationship as that used or radioactive decay.
20
Figure 11
In N
This technique is very useul in carrying out investigations when a relationship between two quantities really is not known. The technique is also a useul way o dealing with exponential relationships by taking logs to base e, instead o base 1 0. For example, radioactive nuclides decay so that either the activity or the number o nuclei remaining alls according to the same general orm. Writing the decay equation or the number o nuclei remaining gives:
4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0
In N = 0.0385t + 3.4657
0 10 20 30 40 50 60 70 80 90 100 t/arbitrary units
Figure 12
17
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S
1.3 Vectors and scalars Understanding
Applications and skills
Vector and scalar quantities
Solving vector problems graphically and
Combination and resolution of vectors
algebraically.
Equations The horizontal and vertical components of vector A: A H = A cos A V = A sin A
AV
u AH
Nature of science All physical quantities that you will meet on the course are classified as being vectors or scalars. It is important to know whether any quantity is a vector or a scalar since this will affect how the quantity is treated mathematically. Although the concept of adding forces is an intuitive application of vectors that has probably been used by sailors for millennia, the analytical aspect of it is a recent development. In the
Philosophi Naturalis Principia Mathematica, published in 1687, Newton used quantities which we now call vectors, but never generalized this to deal with the concepts of vectors. At the start of the 19th century vectors became an indispensible tool for representing three-dimensional space and complex numbers. Vectors are now used as a matter of course by physicists and mathematicians alike.
Vector and scalar quantities S calar quantities are those that have magnitude ( or size) but no direction. We treat scalar quantities as numbers ( albeit with units) and use the rules o algebra when dealing with them. D istance and time are both scalars, as is speed. The average speed is simply the distance divided by the time, so i you travel 80 m in 1 0 s the speed will always be 8 m s 1 . There are no surprises.
18
Vector quantities are those which have both magnitude and direction. We must use vector algebra when dealing with vectors since we must take into account direction. The vector equivalent o distance is called displacement ( i. e., it is a distance in a specifed direction) . The vector equivalent o speed is velocity ( i.e. , it is the speed in a specifed direction) . Time, as we have seen, is a scalar. Average velocity is defned as being displacement divided by time.
1 . 3 VE CTO RS AN D S C AL ARS D ividing a vector by a scalar is the easiest operation that we need to do involving a vector. To continue with the example that we looked at with scalars, suppose the displacement was 80 m due north and the time was, again, 1 0 s. The average velocity would be 8 m s 1 due north. S o, to generalize, when we divide a vector by a scalar we end up with a new vector that has the direction o the original one, but which will be o magnitude equal to that o the vector divided by that o the scalar.
Commonly used vectors and scalars Vectors
Scalars
Comments
orce (F)
mass (m)
displacement (s)
length/distance (s, d, etc.)
velocity (v or u)
time (t)
momentum (p)
volume (V)
acceleration (a)
temperature (T)
gravitational feld strength (g)
speed (v or u)
velocity and speed oten have the same symbol
electric feld strength (E)
density ( )
the symbol or density is the Greek rho not the letter p
magnetic feld strength (B)
pressure (p)
area (A)
energy/work (W, etc.)
F displacement used to be called space now that means something else!
the direction o an area is taken as being at right angles to the surace
power (P) current (I)
with current having direction you might think that it should be a vector but it is not (it is the ratio o two scalars, charge and time, so it cannot be a vector) . In more advanced work you might come across current density which is a vector.
resistance (R) gravitational potential (VG ) electric potential (VE ) magnetic ux ()
the subscripts tell us whether it is gravitational or electrical ux is oten thought as having a direction it doesnt!
Representing vector quantities A vector quantity is represented by a line with an arrow.
The direction the arrow points represents the direction o the vector.
The length o the line represents the magnitude o the vector to a chosen scale.
When we are dealing with vectors that act in one dimension it is a simple matter to assign one direction as being positive and the opposite direction as being negative. Which direction is positive and which negative really doesnt matter as long as you are consistent. S o, i one orce acts upwards on an obj ect and another orce acts downwards, it is a simple matter to fnd the resultant by subtracting one rom the other.
5.0 N this vector can represent a force of 5.0 N in the given direction (using a scale of 1 cm representing 1 N, it will be 5 cm long)
Figure 1 Representing a vector.
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1
M E A S U R E M E N TS AN D U N C E R TAI N T I E S Figure 2 shows an upward tension and downward weight acting on an obj ect the upward line is longer than the downward line since the obj ect is not in equilibrium and has an upward resultant.
tension = 20 N upwards
Adding and subtracting vectors
resultant = 5 N upwards
When adding and subtracting vectors, account has to be taken o their direction. This can be done either by a scale drawing ( graphically) or algebraically.
weight = 15 N downwards
Figure 2 Two vectors acting on an object.
V2
V1
V1
V
V2
Figure 3 Two vectors to be added.
Figure 4 Adding the vectors.
Scale drawing (graphical) approach Adding two vectors V1 and V2 which are not in the same direction can be done by orming a parallelogram to scale.
Make a rough sketch o how the vectors are going to add together to give you an idea o how large your scale needs to be in order to fll the space available to you. This is a good idea when you are adding the vectors mathematically too.
Having chosen a suitable scale, draw the scaled lines in the direction o V1 and V2 ( so that they orm two adj acent sides o the parallelogram) .
C omplete the parallelogram by drawing in the remaining two sides.
The blue diagonal represents the resultant vector in both magnitude and direction.
Worked example
resultant
36
17 0 180 160 0 1 5 0 2 0 10 30
40
30 40 150 14 0
80 90 10 0 1 1 0 1 70 20 6 0 1 1 0 10 0 90 80 7 0 13 60 0 0 0 5 12 50 0 13
0
D ont orget that the vector must have a magnitude and a direction; this means that the angle is j ust as important as the size o the orce.
8.7 N
6N
14
Solution
Scale 10 mm represents 1.0 N
0 10 20 180 17 0 1 60
Two orces o magnitude 4.0 N and 6.0 N act on a single point. The orces make an angle o 60 with each other. Using a scale diagram, determine the resultant orce.
4N length of resultant = 87 mm so the force = 8.7 N angle resultant makes with 4 N force = 36
20
6N
1 . 3 VE CTO RS AN D S C AL ARS
Algebraic approach Vectors can act at any angle to each other but the most common situation that you are going to deal with is when they are at right angles to each other. We will deal with this frst. v2
Adding vector quantities at right angles Pythagoras theorem can be used to calculate a resultant vector when two perpendicular vectors are added ( or subtracted) . Assuming that the two vector quantities are horizontal and vertical but the principle is the same as long as they are perpendicular.
v1 v1
Figure 5 shows two perpendicular velocities v1 and v2 ; they orm a parallelogram that is a rectangle. ______
The magnitude o the resultant velocity = v 21 + v 22
The resultant velocity makes an angle to the horizontal given by
( )
v2
( )
v1 v1 1 _ tan = _ v2 so that = tan v2
Notice that the order o adding the two vectors makes no dierence to the length or the direction o the resultant.
Worked example A walker walks 4.0 km due west rom his starting point. He then stops beore walking 3 .0 km due north. At the end o his j ourney, how ar is the walker rom his starting point?
Solution
Figure 5 Adding two perpendicular vectors.
sketch
3 km
______
2 resultant = 4 2 + 3 = 5 km
angle = tan
1
()
3 = 3 6.9 __ 4
4 km
B eore we look at adding vectors that are not perpendicular, we need to see how to resolve a vector i.e. split it into two components.
Resolving vectors We have seen that adding two vectors together produces a resultant vector. It is sensible, thereore, to imagine that we could split the resultant into the two vectors rom which it was ormed. In act this is true or any vector it can be divided into components which, added together, make the resultant vector. There is no limit to the number o vectors that can be added together and, consequently, there is no limit to the number o components that a vector can be divided into. However, we most commonly divide a vector into two components that are perpendicular to one another. The reason or doing this is that perpendicular vectors have no aect on each other as we will see when we look at proj ectiles in Topic 2 .
Fsin
F
Fcos
Figure 6 Resolving a force.
The orce F in fgure 6 has been resolved into the horizontal component equal to F cos and a vertical component equal to F sin . ( The component opposite to the angle used is always the sine component.)
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S
Worked example An ice- hockey puck is struck at a constant speed o 40 m s 1 at an angle o 60 to the longer side o an ice rink. How ar will the puck have travelled in directions a) parallel and b) perpendicular to the long side ater 0.5 s?
vx = 40 cos 60 = 2 0 m s 1 distance travelled ( x) = vxt = 2 0 0.5 = 10 m b) Resolving parallel to shorter side: vx = v sin 60
Vy = 40 sin 60
Solution
vy = 40 cos 60 = 3 4. 6 m s 1 distance travelled ( y) = vyt = 3 4.6 0.5 = 17 m
40 ms 1
Just to demonstrate that resolving is the reverse o adding the components we can use Pythagoras theorem to add together our two components giving: _________
60
total speed = 2 0 2 + 3 4. 6 2 = 3 9. 96 m s 1 long side of rink
Vx = 40 cos 60
a) Resolving parallel to longer side:
as the value or vy was rounded this gives the expected 40 m s 1
vx = v cos 60
Adding vector quantities that are not at right angles
V1 = V1 sin 1
You are now in a position to add any vectors.
V1
Resolve each o the vectors in two directions at right angles this will oten be horizontally and vertically, but may be parallel and perpendicular to a surace.
Add all the components in one direction to give a single component.
Add all the components in the perpendicular direction to give a second single component.
C ombine the two components using Pythagoras theorem, as or two vector quantities at right angles.
1
V1x = V1 cos 1
V2y = V2 sin 2
V1 and V2 are the vectors to be added.
V2
Total x component Vx = V1 x + V2x = V1 cos 1 - V2 cos 2 V2x = V2 cos
22
Each vector is resolved into components in the x and y directions. Note that since the x component o V2 is to the let it is treated as being negative ( the y component o each vector is in the same direction . . . so upwards is treated as positive) .
2 2
Figure 7 Finding the resultant of two vectors that are not perpendicular.
Total y component Vy = V1 y + V2y = V1 sin 1 + V2 sin 2 Having calculated Vx and Vy we can fnd the resultant by using ______
Pythagoras so V =
V2x +
V2y
and the angle made with the horizontal = tan
( ).
Vy 1 __ Vx
1 . 3 VE CTO RS AN D S C AL ARS
Worked example Magnetic felds have strength 2 00 mT and 1 5 0 mT respectively. The felds act at 2 7 to one another as shown in the diagram.
Solution The 1 5 0 mT feld is horizontal and so has no vertical component. Vertical component o the 2 00 mT feld = 2 00 sin 2 7 = 90.8 mT
200 mT
(not drawn to scale)
This makes the total vertical component o the resultant feld. Horizontal component o the 2 00 mT feld = 2 00 cos 2 7 = 1 78.2 mT
27 150 mT
C alculate the resultant magnetic feld strength.
Total horizontal component o resultant feld = ( 1 5 0.0 + 1 78.2 ) mT = 3 2 8.2 mT ____________
Resultant feld strength 90.8 2 + 32 8.2 2 = 340 mT The resultant feld makes an angle o: tan 1
(_____) = 1 5 with the1 5 0 mT feld. 90.8 3 2 8.2
Subtraction of vectors S ubtracting one vector rom another is very simple you j ust orm the negative o the vector to be subtracted and add this to the other vector. The negative o a vector has the same magnitude but the opposite direction. Lets look at an example to see how this works: S uppose we wish to fnd the dierence between two velocities v1 and v2 shown in fgure 8. V2 V1 V2 V1 V2
V1
V1
Figure 8 Positive and negative vectors.
V1
In fnding the dierence between two values we subtract the frst value rom the second; so we need v1 . We then add v1 to v2 as shown in fgure 9 to give the red resultant. The order o combining the two vectors doesnt matter as can be seen rom the two versions in fgure 9. In each case the resultant is the same it doesnt matter where the resultant is positioned as long as it has the same length and direction it is the same vector.
V2 V1
V2
Figure 9 Subtracting vectors.
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S
Questions 1
E xpress the ollowing units in terms o the S I undamental units.
5
Write down the order o magnitude o the ollowing ( you may need to do some research) .
a) newton ( N)
a) the length o a human oot
b) watt ( W)
b) the mass o a y
c) pascal ( Pa)
c) the charge on a proton
d) coulomb ( C )
d) the age o the universe
e) volt ( V)
e) the speed o electromagnetic waves in a vacuum
( 5 marks)
( 5 marks) 2
Express the ollowing numbers to three signifcant fgures.
6
a) 2 5 7.5 2 b) 0.002 3 47 c) 0.1 783 d) 7873 e) 1 .997 ( 5 marks)
Without using a calculator estimate to one 2 4.9 signifcant fgure the value o _. 480 b) When a wire is stretched, the area under the line o a graph o orce against extension o the wire gives the elastic potential energy stored in the wire. Estimate the energy stored in the wire with the ollowing characteristic:
C omplete the ollowing calculations and express your answers to the most appropriate number o signifcant fgures.
12 10 force/N
3
a)
a) 1 .3 4 3 .2 1 .3 4 1 0 2 b) _ 2 .1 1 0 3 c) 1 .87 1 0 2 + 1 . 97 1 0 3
6 4 2
d) ( 1 .97 1 0 5 ) ( 1 . 0 1 0 4)
0
e) ( 9.47 1 0 2 ) ( 4.0 1 0 3 )
0
( 5 marks)
4
8
Use the appropriate metric multiplier instead o a power o ten in the ollowing. a) 1 .1 1 0 4 V b) 4.2 2 1 0 4 m c) 8.5 1 0 1 0 W
1
2
3
4 5 6 extension/mm
7
8
( 4 marks) 7
The grid below shows one data point and its associated error bar on a graph. The x- axis is not shown. S tate the y- value o the data point together with its absolute and percentage uncertainty.
d) 4.2 2 1 0 7 m 5.0
e) 3 .5 1 0 1 3 C ( 5 marks)
4.0 3.0 2.0 1.0
24
( 3 marks)
QUESTION S 8
A ball alls reely rom rest with an acceleration g. The variation with time t o its displacement s 1 is given by s = __ gt2 . The percentage uncertainty 2 in the value o t is 3 % and that in the value o g is 2 % . C alculate the percentage uncertainty in the value o s. ( 2 marks)
9
The volume V o a cylinder o height h and radius r is given by the expression V = r2 h. In a particular experiment, r is to be determined rom measurements o V and h. The percentage uncertainty in V is 5 % and that in h is 2% . C alculate the percentage uncertainty in r. ( 3 marks)
1 0 ( IB) At high pressures, a real gas does not behave as an ideal gas. For a certain range o pressures, it is suggested that or one mole o a real gas at constant temperature the relation between the pressure p and volume V is given by the equation pV = A + Bp
where A and B are constants.
In an experiment, 1 mole o nitrogen gas was compressed at a constant temperature o 1 5 0 K. The volume V o the gas was measured or dierent values o the pressure p. A graph o the product pV against p is shown in the diagram below.
c) p was measured to an accuracy o 5 % and V was measured to an accuracy o 2 % . D etermine the absolute error in the value o the constant A. ( 6 marks) 1 1 ( IB) An experiment was carried out to measure the extension x o a thread o a spiders web when a load F is applied to it. 9.0 8.0 7.0 6.0 F/10 2 N 5.0 4.0 3.0 2.0 1.0 0.0 0.0
thread breaks at this point
1.0
2.0
3.0 4.0 x/10 2 m
5.0
6.0
a) C opy the graph and draw a best-ft line or the data points. b) The relationship between F and x is o the orm F = kxn S tate and explain the graph you would plot in order to determine the value n. c) When a load is applied to a material, it is said to be under stress. The magnitude p o the stress is given by F p= _ A
13
where A is the cross- sectional area o the sample o the material.
12 pV/10 2 Nm
Use the graph and the data below to deduce that the thread used in the experiment has a greater breaking stress than steel.
11
B reaking stress o steel = 1 .0 1 0 9 N m 2 10 0
5
10 15 p/ 10 6 Pa
20
a) C opy the graph and draw a line o best ft or the data points. b) Use your graph to determine the values o the constants A and B in the equation pV = A + Bp
Radius o spider web thread = 4.5 1 0 6 m d) The uncertainty in the measurement o the radius o the thread is 0. 1 1 0 6 m. D etermine the percentage uncertainty in the value o the area o the thread. ( 9 marks)
25
1
M E A S U R E M E N TS AN D U N C E R TAI N T I E S 1 2 A cyclist travels a distance o 1 2 00 m due north beore going 2 000 m due east ollowed by 5 00 m south-west. D raw a scale diagram to calculate the cyclists fnal displacement rom her initial position. ( 4 marks)
1 3 The diagram shows three orces P, Q, and R in equilibrium. P acts horizontally and Q vertically.
R
a) C alculate the resultant velocity o the boat relative to the bank o the river. b) The river is 5 0 m wide. C alculate the displacement rom its initial position when the boat reaches the opposite bank. ( 7 marks)
1 5 A car o mass 85 0 kg rests on a slope at 2 5 to the horizontal. C alculate the magnitude o the component o the cars weight which acts parallel to the slope.
P
( 3 marks)
Q
When P = 5 .0 N and Q = 3 . 0 N, calculate the magnitude and direction o R. ( 3 marks)
26
1 4 A boat, starting on one bank o a river, heads due south with a speed o 1 .5 m s 1 . The river ows due east at 0.8 m s 1 .
2
M E CH AN I CS
Introduction Everything moves. The Earth revolves on its axis as it travels around the S un. The S un orbits within the Milky Way. Galaxies move apart. Motion and its causes are important to a study o physics. We begin by defning the meaning
o everyday terms such as distance, velocity, acceleration and go on to develop models or motion that will allow us to predict the uture motion o an obj ect.
2.1 Motion Understanding Distance and displacement Speed and velocity Acceleration
Applications and skills Determining instantaneous and average values
Graphs describing motion Equations o motion or uniorm acceleration
Projectile motion
Fluid resistance and terminal speed
Nature of science An understanding o motion lies at the heart o physics. The areas o the subject are linked by the concepts o movement and the orces that produce motion. Links are used in a creative way by scientists to illuminate one part o the subject by reerence to insights developed or other topics. The study o motion also relies on careul observation o the world so that accurate models o motion can be developed.
or velocity, speed, and acceleration Solving problems using equations o motion or uniorm acceleration Sketching and interpreting motion graphs Determining the acceleration o ree-all experimentally Analysing projectile motion, including the resolution o vertical and horizontal components o acceleration, velocity, and displacement Qualitatively describing the efect o uid resistance on alling objects or projectiles, including reaching terminal speed
Equations Kinematic equations o motion: v = u + at 1 2 s = ut + ___ at 2 2 2 v = u + 2as (v + u) t s = _____________ 2
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2
M E C H AN I C S
Distance and displacement Your j ourney to school is unlikely to take a completely straight line from home to classroom. How do we describe j ourneys when we turn a corner and change direction? The map shows the j ourney a student makes to get to school together with the times of arrival at various points on the way. To keep life simple, the j ourney has no hills.
A
home
bus stop
B school bus stop
Figure 1 Journey to school.
journey leg
time
distance for leg / m
leave home
08.10.00
0
walk to bus stop
08.20.15
800
bus arrives at stop
08.24.30
0
bus arrives near school
08.31.10
2400
walk from bus to school
08.34.00
200
The total length of this j ourney is 3 .4 km including all the twists and turns. This is the distance travelled. As we saw in Topic 1 , distance is a scalar quantity; it has no direction. If the student walks home by exactly the same route, then the distance travelled going home is the same as going to school.
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2 .1 M OTI ON
Nature of science How long is a piece of string? Distance can be measured in any appropriate unit o length: metres, miles, and millimetres are all common. Some countries and some proessions use alternatives. Surveyors use chains, the English-speaking world once used measures o length called rods, poles, and perches all related to agricultural measurements. Astronomers use light years (a unit o length, not time) and a measure called simply the astronomical unit. Sometimes in everyday lie it is a question o using a convenient unit rather than the correct metric version. In your exam, however, lengths will be in multiples and sub-multiples o the metre or in a wellrecognised scientifc unit such as the light year.
Journeys can be defned by the starting point ( A) and fnishing point ( B ) without saying anything about the intermediate route. A vector is drawn that starts at A and fnishes at B . This straight line rom start to fnish has, as a vector, magnitude, and direction. This vector measurement is known as the disp lacement. As you saw in Sub- topic 1 . 3 , when you give the details o a displacement you must always give two pieces o inormation: the magnitude ( or size) o the quantity together with its unit, and also the direction o the vector. This direction can be written in a number o ways. O ne way is as a heading such as N3 5 E meaning that the fnishing direction is at an angle o 3 5 clockwise rom north. I you are a sailor or a keen orienteer, you may have other ways o measuring the direction o displacement. The displacement o the students j ourney to school ( AB ) is also shown on the diagram. The displacement is the vector that connects home to school. This time the j ourney rom school to home is not the same as the trip to school. It has the same vector length, but the direction is the opposite to that o the outward route.
Nature of science Moving in 3D D isplacement has been described here in terms o a j ourney in a at landscape. D oes a change in level alter things? O nly one thing changes, and that is the number o pieces o inormation required to speciy the fnal position relative to the start. Three pieces o inormation are now required: the magnitude plus its unit, the heading, and the overall change in height during the j ourney. S peciying motion in three dimensions requires three numbers or coordinates, you are already
amiliar with the idea o a coordinate rom drawing and using graphs. There is exibility in how the three numbers can be chosen. You may have seen three- dimensional graphs with three axes each at 90 to the others, in this case coordinate numbers are given that relate to the distance along each axis. Another option is to use polar coordinates ( fgure 2 ) : where a distance and two angles are required,
29
2
M E C H AN I C S
one angle is the bearing rom North, the other the angle up or down rom the horizontal needed to look directly at the obj ect above (or below) us.
(r, , )
r
In some circumstances, even the distance itsel may not be required. S ailors use latitude and longitude when they are navigating. They stay on the surace o the sea and this is ee ctively a constant distance rom the centre o the E arth.
Worked examples 1
A cyclist travels 1 6 km in 70 minutes. Calculate, in m s 1 , the speed o the cyclist.
Solution 70 minutes is 60 70 = 42 00 s, 1 6 km is 1 6 000 m. The quantities are now in the units required by the question. The speed o the cyclist 1 6 000 is _____ = 3 .8 m s 1 . 42 0 0
2
The speed o light in a vacuum is 3 .0 1 0 8 m s 1 . A star is 2 2 light years rom Earth ( 1 light year is the distance travelled by light in one year) . C alculate the distance o the star rom Earth in kilometres.
Solution Light travels 3 .0 1 0 8 m in 1 s. S o in a year it travels 3 .0 1 0 8 3 65 2 4 60 60 = 9.5 1 0 1 5 m. The distance o the star rom the Earth is 22 9.5 1 0 1 5 = 2 .1 1 0 1 7 m. The question asks or an answer in kilometres, so the distance is 2 .1 1 0 1 4 km.
30
Figure 2 Polar coordinates.
Speed and velocity In j ust the same way that there are scalar and vector measures o the length o a j ourney, so there are two ways o measuring how quickly we cover the ground. The frst o these is a scalar quantity sp eed, which is defned as distan ce trave lle d o n the j o u rn e y _______________________ . Units or speed, amiliar to you already, may tim e take n o r the j o u rne y
include metre per second ( m s 1 ) and kilometres per hour ( km h 1 ) , but any accepted distance unit can be combined with any time unit to speciy speed. Velocity is the vector term and, j ust as or displacement, a magnitude and direction are required. Examples might be 4.2 m s 1 due north or 5 5 km h 1 at N2 2 .5 E.
Nature of science Measuring speed? To measure speed it is necessary to measure the the distance travelled (using a ruler) and time taken (using a clock) . The trick is in a good choice o ruler, clock and method or recording the measurements! A 30 cm ruler and a wrist watch will be fne or a biologist measuring the speed at which an earthworm moves. B ut i we want to measure the speed o a 1 00 m sprinter, then a measured distance on the ground, a good stop watch and a human observer is barely good enough. Even then the observer has to be careul to watch the smoke rom the starting pistol and not to wait or the sound o the gun. I you need to measure the speed o a soccer ball during a penalty kick, then stop watch-plus-human is no longer adequate to make a valid measurement. Perhaps a video camera that takes rames at a known rate (the clock) and a scale near the path o the ball visible on the picture (the ruler) is needed now? Move up to measuring the speed o a jet aircrat and the equipment needs to change again. C hoosing the right equipment or the task in hand is all part o the job o the working science student.
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Describing motion with a graph I The use o two variables, distance and time, to calculate speeds and velocities means that things become more complicated. Figure 1 showed a map o the j ourney the student makes to school. Part o this j ourney is on oot, part is by bus. It is unlikely that the student will travel at the same speed all the time, as the bus will travel aster than the student walks. I we want to display data in a visual way, a graph o distance against time is one o the most common approaches. 3500 distance travelled/m
3000 2500 2000 1500 1000 500 0
500
1000 time elapsed/s
1500
Figure 3 Distancetime graph.
The distance travelled by the student is plotted on the vertical axis while the time since the beginning o the j ourney is plotted horizontally ( the clock times o the j ourney have been translated into elapsed times since the start o the j ourney or this graph) . The dierent regions o the graph are identifed and you should confrm that they are matched correctly. Notice how the gradient o the graph changes or the dierent parts o the j ourney: small values o the gradient or the walking sections o the j ourney, horizontal ( zero gradient) or stationary at the bus stop, and steep or the bus j ourney. What will the graph or the j ourney home rom school look like, assuming that the time or each segment o the j ourney is the same as in the morning?
distance travelled/m
0
distance travelled on bus = 2400 m
time taken or bus journey = 400 s
Inormation can easily be extracted rom this graph. The gradient o the graph is the speed. Add the overall direction to this speed and we have the velocity too. For the frst walk to the bus stop the distance was 800 m and the time taken 800 was 61 5 s. The constant walking speed was thereore ___ , which is 61 5 1 1 .3 m s .
speed = 2400 = 6 ms 1 400
1000 time elapsed/s
Figure 4 Distancetime graph rom fgure 3 with gradients.
2 400 The gradient o the bus journey is ____ (this is marked on the graph) and so 400 1 the speed was 6.0 m s .
O course, even this j ourney with its changes is simplistic. Real j ourneys have ew straight lines, so we must introduce some ways to handle rapidly varying speeds and velocities.
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Instantaneous and average values The bus driver knows how ast the bus in the students j ourney is travelling because it is displayed on the speedometer. This is the instantaneous sp eed, as it gives the value o the speed at the moment in time at which the speed is determined. 3500
distance travelled/m
3000
2500
1500
Nature of science Calculating a gradient of a graph Taking the gradient o a graph is actually a way o averaging results. As discussed in Topic 1 , a straightline graph drawn through points that have some scatter makes this obvious. The value o the gradient ch an ge in the valu e o n the y- axis . is the _______________________ change in th e valu e o n the x- axis The unit associated with the gradient y- axis u nit is the ________ . D ont orget to x- axis u nit quote the unit every time you write down the value o the gradient.
use as large a tangent line as possible, change in distance = (2500 500) m change in time = (1300 900) s = 5.0 m s 1 gradient = 2000 400
2000 a more realistic graph for the bus
1000
tangent to graph at t = 1000 s
500 1000 time elapsed/s
Figure 5 Instantaneous speed.
The instantaneous speed is also the gradient o the distancetime graph at the instant concerned. Figure 5 shows how this is calculated when 1 000 s into the whole journey to school. The original red line or the bus rom fgure has been replaced by a green line that is more realistic or the motion o a real bus the speed varies as the driver negotiates the trafc. You will requently be asked to calculate gradients on physics graphs. Make sure that you can do this accurately by using a transparent ruler. From a mathematical point o view we can describe the instantaneous speed as the rate of change of position with respect to time. ds A mathematician will write this as __ , where s is the distance and t is the dt s time. You may also have seen ___ , where the symbol means change in. t change in distance s So ___ is just shorthand or ____________ . change in tim e t
There is however another useul measure o speed. This is the average speed and is the speed calculated over the whole the journey without regard to variations in speed. So as an equation this is distan ce trave lle d o ve r w ho le j o u rne y average speed = __________________________ tim e take n o r w ho le j o u rne y
In terms o the distancetime graph, the average speed is equal to the gradient o the straight line that j oins the beginning and the end o the time interval concerned. S o, or the part o the students j ourney up to the moment when the bus arrives at the stop, the distance travelled is 800 m, the time taken is 870 s ( including the wait at the stop) so the average speed is 0.92 m s 1 .
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E verything said here about average and instantaneous speeds can also reer to average and instantaneous velocities. Remember, o course, to include the directions when quoting these measurements.
Worked example Solution
distance/m
The graph shows how the distance run by a boy varies with the time since he began to run.
a)
100 90 80 70 60 50 40 30 20 10 0
( i) The boy runs at constant speed so the graph is straight rom 0 1 0 s. The gradient o the straight line is 48 __ = 4.8 m s 1 . 10
S o the instantaneous speed at 5 .0 s is 4.8 m s 1 . ( ii) Again the boy is running at a constant speed, but this time slower than in the frst 1 0 s. 0
5
10
15
20 time/s
25
30
35
C alculate: a) the instantaneous speed at ( i) 5 .0 s ( ii) 2 0 s
The speed rom 1 0 s to 3 0 s is: ( 90 48) 42 _ = 2 .1 m s 1 . = __ 20 ( 3 0 1 0) The instantaneous speed at 2 0 s is 2 .1 m s 1 . b) The total distance travelled in 3 0 s is 90 m, so 90 the average speed is __ = 3 .0 m s 1 . 30
b) the average speed or the whole 3 0 s run.
Acceleration In real j ourneys, instantaneous speeds and velocities change requently. Again we need to develop a mathematical language that will help us to understand the changes. The quantity we use is acceleration. Acceleration is taken to be a vector, but sometimes we are not interested in the direction and then write the magnitude o the acceleration meaning the size o the acceleration ignoring direction. The defnition is: acceleration =
change in velocity ___________________ tim e taken or the change
m s and this means that the units o acceleration are ____ which is usually s 1
written as m s 2 ( or sometimes you will see m/s 2 ) . It is important to understand what acceleration means, not j ust to be able to use it in an equation. I an obj ect has an acceleration o 5 m s 2 then or every second it travels, its velocity increases in magnitude by 5 m s 1 in the direction o the acceleration vector. As an example: the Japanese N700 train has a quoted acceleration o 0.72 m s 2 . Assume that this is a constant value (very unlikely) . One second ater starting rom rest, the speed o the train will be 0.72 m s 1 . One second later (at 2 s rom the start) the speed will be 0.72 + 0.72 = 1 .44 m s 1 . At 3 s it will be 2.1 6 m s 1 and so on. Each second the speed is 0.72 m s 1 more.
Worked example How many seconds will it take the N700 to reach its maximum speed o 3 00 km h 1 on the S anyo Shinkansen route?
Solution 3 00 km h 1 = 83 .3 m s 1 Time taken to reach the maximum speed: 83 .3 _ = 1 1 6 s, 0. 72 just under 2 minutes.
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Nature of science Spreadsheet models One powerful way to think about acceleration (and other quantities that change in a predictable way) is to model them using a spreadsheet. The examples here use a version of Microsoft Excel but any computer spreadsheet can be used for this. This is a spreadsheet model for the N700 train. The value of the acceleration is in cell B 1 . C ells A4 to A2 9 give the time in increments of 5 s; the computed speed at each of these times is in B
C
D
E
F
G
H
I
J
0.72 speed/ms -1 0 3.6 7.2 10.8 14.4 18.0 21.6 25.2 28.8 32.4 36.0 39.6 43.2 46.8 50.4 54.0 57.6 61.2 64.8 68.4 72.0 75.6 79.2 82.8 86.4 90.0
100 90 80 70 60 50 40 30 20 10 0
speed /m s -1
A 1 acceleration/s 2 time/s 3 4 0 5 5 6 10 7 15 8 20 9 25 10 30 11 35 12 40 13 45 14 50 15 55 16 60 17 65 18 70 19 75 20 80 21 85 22 90 23 95 24 100 25 105 26 110 27 115 28 120 29 125
cells B 4 to B 2 9. The speed is calculated by taking the change in time between the present cell and the one above it, and then multiplying by the acceleration (the acceleration is written as $B $1 so that the spreadsheet only uses this cell and does not drop down a cell every time the new speed is calculated) . Finally, the spreadsheet plots speed against time showing that the graph is a straight line and that the acceleration is uniform.
0
20
40
60
80 time/s
100
120
140
Describing motion with a graph II D istancetime plots lead to a convenient display of speed and velocity changes. Plots of speed ( or velocity) against time also help to display and visualize acceleration. The data table and the graph show a j ourney with various stages on a bicycle. From the start until 1 0 s has elapsed, the bicycle accelerates at a uniform rate to a velocity of + 4 m s 1 . The positive sign means that the velocity is directed to the right.
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5 4 3 2 1 velocity /m s -1 0 1 0 2 3 4 5
20
40
60
80
100
120
140
time/s Figure 6 Velocitytime graph
for the bicycle.
From 1 0 s to 45 s the cycle moves at a constant velocity o + 4 m s 1 and at 45 s the cyclist applies the brakes so that the cycle stops in 5 s. The cycle is then stationary or 1 0 s. From then on the velocity is negative meaning that the cycle is travelling in the opposite direction. The pattern is similar, an acceleration to 3 m s 1 , a period o constant velocity and a deceleration to a stop at 1 20 s. As beore, the gradient o the graph has a meaning. The gradient o this velocitytime graph gives the magnitude ( size) o the acceleration and its sign ( direction) as well. From 45 s to 5 0 s the velocity goes rom 4 m s 1 to 0 and so the f nal sp e e d - in itial sp e e d ____ ( 0 4) = 0. 8 m s 2 . From 90 s to 1 2 0 s acceleration is __________________ 5 tim e take n
3 the magnitude o the acceleration is __ = 0.1 m s 2 . We need to take care 30 with the sign o the acceleration here. B ecause the cycle is moving in the negative direction and is slowing down, the acceleration is positive (as is the gradient on the graph) this simply means that there is a orce acting to the right, that is, in the positive direction which is slowing the cycle down. We will discuss how orce leads to acceleration later in this topic.
The area under a velocitytime graph gives yet more inormation. It tells us the total displacement o the moving object. The way to see this is to realize that the product o velocity time is a displacement (and that product o speed time is a distance) . The units tell you this too: when the units are m e tre multiplied the seconds in _____ second cancel to leave metre only. se co nd
velocity/m s -1
In the case o a graph with uniorm acceleration, the areas, and hence the displacements (distances) are straightorward to calculate. Divide the graph into right-angled triangles and rectangles and then work out the areas or each individual part. This is shown in fgure 7. 5 area = 1 10 4 = 20 m 2 4 3 area = 12 5 4 = 10 m 2 1 0 0 40 60 80 100 120 20 1 2 area = 35 4 = 140 m 3 4 time/s
Figure 7
140
Velocity-time graph broken down into areas.
35
2
M E C H AN I C S The individual areas are shown on the diagram and or the motion up to a time o 60 s the area is 1 70 m; the area rom the 60 s time to the end is 1 2 0 m. As usual, the negative sign indicates motion in the opposite direction to the original. As discussed in Topic 1 , when the velocitytime graph is curved, you will need to: ( i) estimate the number o squares ( ii) assess the area ( distance) or one square, and fnally ( iii) multiply the number o squares by the area o one square. This will usually give you an estimate o the overall distance. Figure 8 gives an example o how this is done. There are about 85 squares between the x-axis and the line. ( You may disagree slightly with this estimate, but that is fne there is always an allowance made or this.)
speed/m s -1
E ach o the squares is 2 s along the time axis and 0.5 m along the speed axis. S o the area o one square is equivalent to 1 .0 m o distance. The total distance travelled is 85 m ( or, at least, somewhere between 80 and 90 m) . 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0
10
20 time/s
30
40
Figure 8 A more difcult area to estimate.
The suvat equations of motion
36
symbol
quantity
s
displacement/distance
u
initial (starting) velocity/speed
v
fnal velocity/speed
a
acceleration
t
time taken to travel the distance s
The graphs o distancetime and speedtime lead to a set o equations that can be used to predict the value o the parameters in motion. They also help you to understand the connection between the various quantities in your study o motion. A note about symbols: rom now on we will use a consistent set o symbols or the quantities. The table gives the list. I you read down the list o symbols they spell out suvat and the equations are sometimes known by this name, another name or them is the kinematic equations o motion. The derivation o the equations o uniormly accelerated motion begins rom a simple graph o speed against time or a constant acceleration rom velocity u to velocity v in a time t.
2 .1 M OTI ON
v
speed
gradient = area =
(v u) t
1 (v u) t 2
u area = u t 0
0
t time
Figure 9 Deriving the frst two equations o motion.
The acceleration is the gradient o the graph: change in speed __ time taken or change The change in speed is v u, the time taken is t. Thereore, vu a= _ t and re arranging gives
v = u + at
frst equation o motion
The area under the speedtime graph rom 0 to t is made up o two parts, the lower rectangle, area and the upper right- angled triangle, area . 1 base height = _ 1 t( v u) area = _ 2 2 area = base height = ut 1 ( v u) t = ut + _ 1 ( at) t s = total area = area + area = ut + _ 2 2 So
1 at2 s = ut + _ 2
second equation o motion
The frst equation has no s in it; the second has no v. There are three more equations, one with a missing t and one with a missing a. There is one equation that has a missing u, but this is not oten used. To eliminate t rom the frst and second equations, re arrange the frst in terms o t: vu t= _ a This can be substituted into the second equation:
(
vu vu 1 _ _ s=u_ a + 2 a a
)
2
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2
M E C H AN I C S and 1 1 1 1 as = u( v u) + _ ( v u) 2 = uv u 2 + _ v2 + _ u 2 _ 2 uv 2 2 2 2 which gives 2 as = v2 u 2 or
v2 = u 2 + 2as
third equation of motion
The derivation o the fnal equation is let to you as an exercise:
( )
v+ u s = _________ t 2
fourth equation of motion
There are two ways to approach this proo. O ne way is to think about v+ u the meaning o the speed that corresponds to ____ and then to recognize 2 the time at which this speed occurs in the motion. The second way is to take the third equation and amalgamate it with the frst.
Remember! These equations only apply if the acceleration is uniform. In other words, acceleration must not change during the motion.
You will not be expected to remember these proos or the equations themselves (which appear in the data booklet) , but they do illustrate how useul graphs and equations can be when solving problems in kinematics.
Worked examples 1
A driver o a car travelling at 2 5 m s 1 along a road applies the brakes. The car comes to a stop in 1 5 0 m with a uniorm deceleration. C alculate a) the time the car takes to stop, and b) the deceleration o the car.
Solution a) O ne way to answer kinematic equation questions is to begin by writing down what you do and dont know rom the question. 1
s = 1 5 0 m; u = 2 5 m s ; v = 0 ; a = ? ; t = ? To work out t, the ourth equation is required: v+ u t s = ____ 2
( )
( )
2 which rearranges to t = ____ s v+ u
Substituting the values gives
( )
2 t = __ 1 50 = 2 6 = 1 2 s 25
b) To fnd a the equation v2 = u 2 + 2 as is best. Substituting: 0 = 252 + 2 a 1 50
38
25 25 25 a = _ = _ = 2 .1 m s 2 . 3 00 12 The minus sign shows that the car is decelerating rather than accelerating. 2
A cyclist slows uniormly rom a speed o 7.5 m s 1 to a speed o 2.5 m s 1 in a time o 5.0 s.
C alculate a) the acceleration, and b) the distance moved in the 5 .0 s.
Solution a) s = ?; u = 7.5 m s 1 ; v = 2 .5 m s 1 ; a = ? ; t = 5 .0 s Use v = u + at and thereore 2 .5 = 7.5 + a 5 .0 5 .0 so, a = ___ = 1 .0 m s 2 5 .0
The negative sign shows that this is a deceleration. 1 b) s = ut + __ at2 so 2 1 s = 7.5 5 .0 __ 1 .0 5 .0 2 2 = 3 7.5 1 2 .5 = 25 m
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Projectile motion Falling freely When an obj ect is released close to the Earths surace, it accelerates downwards. We say that the orce o gravity acts on the obj ect, meaning that it is pulled towards the centre o the E arth. E qually the obj ect pulls with the same orce on the Earth in the opposite direction. Not surprisingly, with small obj ects, the eect o the orce on the E arth is so small that we do not notice it. In Topic 6, we shall look in more detail at the eects o gravity but or the moment we assume that there is a constant acceleration that acts on all bodies close to the surace o the Earth. The acceleration due to gravity at the Earths surace is given the symbol g. The accepted value varies rom place to place on the surace, so that at Kuala Lumpur g is 9.776 m s 2 whereas at Stockholm it is 9. 81 8 m s 2 . Reasons or the variation include variations in the shape o the E arth ( it is not a perect sphere, being slightly attened at the poles) and eects that are due to the densities o the rocks in dierent locations. The dierent tangential speeds o the Earth at dierent latitudes also have an eect. It is better to buy gold by the newton at the equator and sell it at the North Pole than the other way round!
Investigate! Measuring g Alternative 1
There are a number o ways to measure g. This method uses a data logger to collect data. One o the problems with measuring g by hand is that the experiment happens quickly. Manual collection o the data is difcult.
An ultrasound sensor is mounted so that it senses obj ects below it.
S et the logging system up to measure the speed o the obj ect over a time o about 1 s. S et a sensible interval between measurements.
S witch on the logger system, and drop an obj ect vertically that is large enough or the sensor to detect.
The output rom the system should be a speedtime graph that is a straight line; it may be that the loggers sotware can calculate the gradient or you.
You could extend this experiment by testing obj ects o dierent mass but similar size and shape to confrm a suggestion by Galileo that such dierences do not aect the drop.
Ultrasound sensor
Figure 10 Ultrasound sensor.
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M E C H AN I C S
Alternative 2
There are other options that do not involve a data logger.
This system measures the time o ight t o the sphere rom the contacts to the trapdoor.
Measure the distance h rom the bottom o the sphere to the top o the trapdoor ( you might think about why these are the appropriate measurement points) .
A possible way to carry out the experiment is to measure t or one value o h with, o course, a ew repeat measurements or the
electromagnet
ruler
height of fall = h
to switch and timer
1 gt2 with u = 0 to same h. Then use h = ut + __ 2
trap-door
calculate g.
Figure 11 Trap door method.
A magnetic feld holds a small steel sphere (such as a ball bearing) between two metal contacts. The magnetic feld is produced by a coil o wire with an electric current in it. When the current is switched o, the feld disappears and the sphere is released to all vertically.
As the sphere leaves the metal contacts, a clock starts. The clock stops when the sphere opens a small trapdoor and breaks the connection between the terminals o a timing clock or computer. ( The exact details o these connections will depend on the equipment you have.)
This is a one- o measurement that is prone to error. C an you think o some reasons why? O ne way to reduce the errors is to change the vertical distance h between the sphere and trapdoor and to plot a graph o h against t2 . g The gradient o the graph is __ . I you observe 2 an intercept on the h- axis, what do you think it represents?
Alternative 3
There is a urther method that involves taking a video or a multiash image o a alling object and analysing the images to measure g. You will see an example o such an analysis later in this sub-topic.
What goes up must come down displacement
Perhaps you have seen a toy rocket flled with water under pressure and fred vertically upwards? O r you may have thrown a ball vertically, high into the air?
time
Figure 12a Displacementtime for ball thrown vertically.
maximum height
distance
0 0
40
time
Figure 12b Distancetime for ball thrown vertically.
Ater the ball has been released, as the pull o gravity takes eect, the ball slows down, eventually stopping at the top o its motion and then alling back to Earth. I there was no air resistance the displacement time graph would look like fgure 1 2 a. Remember that this is a graph o vertical displacement against time, not the shape o the path the ball makes in the air which is called the traj ectory. The ball is going vertically up and then vertically down to land in the same spot rom where it began. A distancetime graph would look dierent ( fgure 1 2 b) , it gives similar inormation but without the direction part o the displacement and velocity vectors. Make sure that you understand the dierence between these graphs. The suvat equations introduced earlier can be used to analyse this motion. The initial vertical speed is u, the time to reach the highest point is t, the highest point is h and the acceleration o the rocket is g. The
2 .1 M OTI ON
sign o g is negative because upwards is the positive direction. As the acceleration due to gravity is downwards, g must have the opposite sign. The kinematic equations are printed again but with dierences to reect the vertical motion to the highest point: 0 = u gt which comes rom v = u + at 1 1 h = ut __ gt2 which comes rom s = ut + __ at2 2 2
0 = u 2 2 gh which comes rom v2 = u 2 + 2 as I you want to fnd out the time or the entire motion ( that is, up to the highest point and then back to E arth again) , it is simply 2 t.
Reminder! Try to remember this crucial point about the signs in the equations when you answer questions on vertical motion: upwards is +ve and downwards is ve. Something else that students forget is that at the top of the motion the vertical speed of the rocket is zero.
Worked examples 1
A student drops a stone rom rest at the top o a well. She hears the stone splash into the water at the bottom o the well 2.3 s ater releasing the stone. Ignore the time taken or the sound to reach the student rom the bottom o the well. a) C alculate the depth o the well. b) C alculate the speed at which the stone hits the water surace. c) Explain why the time taken or the sound to reach the student can be ignored.
Solution a) The acceleration due to gravity g is 9.8 m s 2 . u = 0; t = 2 .3 s. s= s=
1 ut + __ at2 and thereore 2 1 9.8 2 .3 2 0 + __ 2
= 26 m b) v = u + at; v = 0 + 9. 8 2 . 3 = 2 3 m s 1 c) The speed o sound is about 300 m s 1 and so the time to travel about 25 m is about 0.08 s. Only about 4% o the time taken or the stone to all. 2
A hot-air balloon is rising vertically at a constant speed o 5 .0 m s 1 . A small obj ect is released rom rest relative to the balloon when the balloon is 3 0 m above the ground. a) C alculate the maximum height o the obj ect above the ground. b) C alculate the time taken to reach the maximum height. c) C alculate the total time taken or the obj ect to reach the ground.
Solution a) The obj ect is moving upwards at + 5 .0 m s 1 when it is released. The acceleration due to gravity is 9. 8 m s 2 . When the object is released it will continue to travel upwards but this upward speed will decrease under the inuence o gravity. When it reaches its maximum height it will stop moving and then begin to all. 0 5 v2 = u 2 + 2 as and s = _______ = + 1 .3 m 2 9 . 8 2
This shows that the obj ect rises a urther 1 .3 m above its release point, and is thereore 3 1 .3 m above the ground at the maximum height. 0 5 b) v = u + at; t = ____ = + 0.5 1 s ( the plus sign 9 . 8 shows that this is 0.5 1 s ater release)
c) Ater reaching the maximum height ( at which point the speed is zero) the obj ect alls with the acceleration due to gravity. s = 3 1 .3 m ; u = 0; v = ?; a = 9.8 m s 2 ; t = ? 1 at2 , 3 1 . 3 = 0 0.5 9.8 t2 . Using s = ut + __ 2
( Notice that s is 3 1 .3 m as it is in the opposite direction to the upwards + direction.) This gives a value or t o 2 . 5 3 s. The positive value is the one to use. Think about what the negative value stands or. S o the total time is the 0.5 1 s to get to the maximum height together with the 2 .5 3 s to all back to Earth. This gives a total o 3.04 s which rounds to 3.0 s. Notice that, in this example, i you carry the signs through consistently, they give you inormation about the motion o the obj ect.
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M E C H AN I C S
Moving horizontally There is little that is new here. We are going to assume that the surace o the E arth is large enough or its surace to be considered at ( Topic 6 will go into more details o what happens in reality) and that there is no riction. Gravity acts vertically and not in the horizontal direction. This will be important when we combine the horizontal and vertical motions later. B ecause the horizontal acceleration is zero, the suvat equations are simple. For horizontal motion:
the horizontal velocity does not change
the horizontal distance travelled is horizontal speed time for the motion.
Putting it all together
projected to the right
dropped
A student throws a ball horizontally. Figure 1 3 shows multiple images o the ball every 0 . 1 0 s as it moves through the air. The picture also shows, or comparison, the image o a similar ball dropped vertically at the same moment as the ball is thrown. What do you notice about both images? It is obvious which o the two balls was thrown horizontally. C areul examination o the images o this ball should convince you that the horizontal distance between them is constant. Knowing the time interval and the distance scale on the picture means that you can work out the initial ( and unchanging) horizontal speed.
dropped
Figure 13 Multi-fash images o two alling objects.
projected to the right
The images tell us about the vertical speeds too. The clue here is to concentrate on the ball that was dropped vertically. The distance between images (strictly, between the same point on the ball in each image) is increasing. The distance s travelled varies with time t rom release as s t2 . I t doubles then s should increase by actor o 4. D oes it look as though this is what happens? C areul measurements rom this fgure ollowed by a plot o s against t2 could help you to confrm this. The two motions, horizontal and vertical, are comp letely indep endent of each other. The horizontal speed continues unchanged ( we assumed no air resistance) while the vertical speed changes as gravity acts on the ball. This independence allows a straightorward analysis o the motion. The horizontal and the vertical parts o the motion can be split up and treated separately and then recombined to answer questions about the velocity and the displacement or the whole o the motion. The position is summed up in fgure 1 4. At two positions along the traj ectory, the separate components o velocity are shown and the resultant ( the actual velocity including its direction) is drawn.
Figure 14 Horizontal and vertical speed components.
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Real- lie situations do not always begin with horizontal motion. A wellaimed throw will proj ect the ball upwards into the air to achieve the best range ( overall distance travelled) . B ut the general principles above still allow the situation to be analysed.
2 .1 M OTI ON
vertical speed = 0 at maximum height
at maximum height: 0 = u sin - gt
gt 2 and 2 0 = u 2 sin 2 - 2gh h = (u sin ) t -
horizontal speed is constant if air resistance negligible
displacement
vertical acceleration is -g
range: = 2t (u cos ) t is the time to maximum height
h g only acts on vertical speed initial velocity u at to horizontal initial horizontal component = u cos initial vertical component = u sin
time time for whole motion is twice time to maximum height
Figure 15 Projectile motion.
S tudy fgure 1 5 careully and apply the ideas in it to any proj ectile problems you need to solve.
Worked examples 1
An arrow is fred horizontally rom the top o a tower 3 5 m above the ground. The initial horizontal speed is 3 0 m s 1 . Assume that air resistance is negligible. C alculate:
c) To calculate the velocity, the horizontal and vertical components are required. The horizontal component remains at 3 0 m s 1 . The vertical speed is calculated using v = u + at and is 0 + 9.8 2 .67 = 2 6. 2 m s 1 . _________
The speed is 3 0 2 + 2 6.2 2 = 3 9. 8 m s 1 which rounds to 40 m s 1 .
a) the time or which the arrow is in the air b) the distance rom the oot o the tower at which the arrow strikes the ground c) the velocity at which the arrow strikes the ground.
Solution a) The time taken to reach the ground depends on the vertical motion o the arrow. At the instant when the arrow is fred, the vertical speed is zero. The time to reach the ground can be ound 1 using s = ut + __ at2 2 2 35 t2 = _, so t = 2 .67 s or 2 .7 s to 2 s. . 9.8 b) The distance rom the oot o the tower depends only on the horizontal speed.
The angle at which the arrow strikes the 2 6.2 ground is tan ____ = 41 30 2
An obj ect is thrown horizontally rom a ship and strikes the sea 1 .6 s later at a distance o 3 7 m rom the ship. C alculate: a) the initial horizontal speed o the obj ect b) the height o the obj ect above the sea when it was fred thrown.
Solution a) The obj ect travelled 3 7 m in 1 .6 s and the 37 horizontal speed was ___ = 2 3 m s 1 . 1 .6 1 b) Use s = ut + __ at2 to calculate s above the sea. 2
s = 0 + 0. 5 9.8 1 .6 2 = 1 2 .5 m
s = ut = 3 0 2 .67 = 80. 1 m = 80 m.
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2.2 Forces Understanding
Applications and skills
Objects as point particles
Representing orces as vectors
Free-body diagrams
Sketching and interpreting ree-body diagrams
Translational equilibrium
Describing the consequences o Newtons frst
Newtons laws o motion Solid riction
Nature of science The application o mathematics enhances our understanding o orce and motion. Isaac Newton was able to use the work o earlier scientists and to ormalize it through the use o the calculus, a orm o which he developed or this purpose. He made many insights during his scientifc thinking within the topic o orce and motion, but also beyond it. The story o the alling apple, whether true or not, illustrates the importance o serendipity in science and the requirement that the creative scientist can orm links that go beyond what exists already.
law or translational equilibrium Using Newtons second law quantitatively and qualitatively Identiying orce pairs in the context o Newtons third law Solving problems involving orces and determining resultant orce Describing solid riction (static and dynamic) by coecients o riction
Equations Newtons second law: F = ma static riction equation: F s R dynamic riction equation: F = d R
Introduction We depend on forces and their effects for all aspects of our life. Forces are often taught in most elementary physics courses as though they are pushes or pulls, but forces go well beyond this simple description. Forces can change the motion of a body and they can deform the shapes of bodies. Forces can act at a distance so that there is no contact between obj ects or between a system that produces a force and the obj ect on which it acts.
TOK Aristotle and the concept of force Discussions about the concept o what is meant by a orce go back to the dawn o scientifc thought. Aristotle, a Greek philosopher who lived about 2300 years ago, had an overarching view o the world (called an Aristotelian cosmology) and he can be regarded as being an important actor in the development o science. The German philosopher Heidegger wrote that there would have been no Galileo without Aristotle beore him.
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Despite his importance to us, however, we would not have regarded Aristotle as a scientist in any modern sense. For one thing he is not known to have perormed experiments to veriy his ideas; some o his ideas seem very odd to us today. Aristotle believed in the nature o all objects including living things. He believed that all objects had a natural state which was to be motionless on the surace o the Earth and that all
2 . 2 FO R CE S
objects, i let alone, would try to attain this state. Then he distinguished between natural motion in which, or example, heavy objects all downwards and unnatural or orced motion in which the objects need to have a orce continually applied i they are to remain anywhere other than their natural state. Unortunately or those learning physics this is a very persuasive idea because we know intuitively that i we want to hold something in our hand, our muscles have to keep working in order to do this.
There are many other examples o Aristotelian thought and how later scientists moved our thinking orward. But it is important to remember the contribution that Aristotle made to science, even i some o his ideas are now overturned. What do you think students in 50, 100, or even a 1000 years will make o the physics o our century?
Newtons laws of motion Newtons frst law B y the time o Galileo, scientists had begun to realise that things were not as simple as the Greek philosophers such as Aristotle had thought. They were coming to the view that moving obj ects have inertia, meaning a resistance to stopping and that, once in motion, obj ects continue to move. Galileo carried out an experiment with inclined planes and spheres. In act, this may have been a thought experiment this was oten the way orward in those days but in any event it is easy to see what Galileo was trying to suggest.
(a)
(b)
(c)
Figure 1 Galileos thought experiment.
In the frst experiment ( a) , the two arms o the inclined plane are at the same angle and the sphere rolls the same distance up the slope as it rolled down ( assuming no energy losses) . In the second experiment ( b) , the second arm is at a lower angle than beore but the sphere rolls up this plane to the same height as that rom which it was released. Galileo then concluded that i the second plane is horizontal ( c) , the sphere will go on rolling or ever because it will never be able to climb to the original release height. Newton included this idea in his frst law o motion that says: An obj ect continues to remain stationary or to move at a constant velocity unless an external orce acts on it. Galileo had suggested that the sphere on his horizontal plane went on orever, but it was Newton who realised that there is more to say than this. Unless something rom outside applies a orce to change it, the velocity o an obj ect ( both its speed and its direction) must remain the same.
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M E C H AN I C S This was directly opposed to Aristotles view that a force had to keep pushing constantly at a moving obj ect for the speed to remain the same.
Newtons second law The next step is to ask: if a force does act on an obj ect, in what way does the velocity change? Newton proposed that there was a fundamental equation that connected force and rate of change of velocity. This is contained in his second law of motion. This law can be written in two ways, one way is more complex than the other and we will look at the more complicated form of the law later in this topic. Newtons second law, in its simpler form, says that Force = mass acceleration As an equation this can be written F = ma The appropriate S I units are force in newtons ( N) , mass in kilograms ( kg) , and acceleration in metres per second 2 ( m s 2 ) . As discussed in Topic 1 the newton is a derived unit in S I. We can represent it in terms of fundamental units alone as kg m s 2 . Two things arise from this equation:
Mass is a scalar, so there will not be a change in the direction of the acceleration if we multiply acceleration by the mass. The direction of the force and the direction of the acceleration must be the same. So, applying a force to a mass will change the velocity in the same direction as that of the force.
One way to think about the mass in this equation is that it is the ratio of the force required per unit of acceleration for a given object. This helps us to standardize our units of force. If an object of mass 1 kg is observed to accelerate with an acceleration of 1 m s 2 then one unit of force (1 N) must have acted on it.
Nature of science Inertial and gravitational mass W hat e xactly do we m e an b y m ass? The m as s that we u s e in Ne wto ns s e co nd law o f m o tio n is ine rtial m as s . Ine rtial m as s is the p ro p e rty that p e rm its an o b j e ct to re s is t the e ffe cts o f a fo rce that is trying to change its m o tio n. In o the r p arts o f this b o o k, m as s is u s e d in a diffe re nt co nte xt. We talk ab o u t the we ight o f an o b j e ct and we kno w that this we ight aris e s fro m the gravitatio nal attractio n b e twe e n the
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m as s o f the o b j e ct and the m as s o f the E arth. The two m as s e s in this cas e are gravitatio nal m as s e s and are the re s p o ns e o f m atte r to the e ffe cts o f gravity. It has b e e n e xp e rim e ntally ve rifie d that we ight is p ro p o rtio nal to m as s to b e tte r than 3 p arts in 1 0 1 1 . It is a p o s tu late o f E ins te ins ge ne ral the o ry o f re lativity that ine rtial m as s and gravitatio nal m as s are p ro p o rtio nal.
2 . 2 FO R CE S
Investigate! Force, mass, and acceleration card
break the light beam to calculate, frst, the average speed at each gate and then ( using the distance apart o the gates) , the acceleration o the cart. The kinematic equations are used or this. ( This is why you must pull the cart with a constant orce so that the acceleration is uniorm.)
light gate
cart
Repeat the experiment with two, three, and possibly our elastic threads, all identical, all extended by the same amount. This means that you will be using one unit o orce ( with one thread) , two units ( with two threads) and so on.
Plot a graph o calculated acceleration against number o orce units. Is your graph straight? D oes it go through the origin? Remember that this experiment has a number o possible uncertainties in j udging the best- ft line.
Experiment 1: force and acceleration
The idea in this experiment is to measure the acceleration o the cart ( o constant mass) when it is towed by dierent numbers o elastic threads each one extended by the same amount. The timing can be done using light gates and an electronic timer as shown here. Alternatively it can be done using a data logger with, say, an ultrasound sensor and the cart moving away rom the sensor. Practise accelerating the cart with one elastic thread attached to the rear o the cart. The thread( s) will have to be extended by the same amount or each run. A convenient point to j udge is the orward end o the cart. Make sure that your hand clears the light gates i you are using this method. Your hand needs to move at the same speed as the cart so that the thread is the same length throughout the run.
Experiment 2: mass and acceleration
The setup is essentially the same as or Experiment 1 , except this time you will use a constant orce ( possibly two elastic threads is appropriate) .
C hange the mass o your cart (some laboratory carts are specially designed to stack, one on top o another) and measure the acceleration with a constant orce and varying numbers o carts.
This time Newtons second law predicts that mass should be inversely proportional to acceleration. Plot a graph o acceleration 1 against ____ m ass . Is it a straight line?
2
An aircrat o mass 3 .3 1 0 5 kg takes o rom rest in a distance o 1 .7 km. The maximum thrust o the engines is 83 0 kN. C alculate the take- o speed.
The card on the cart needs to be o a known length so that you can use the time taken to
Worked examples 1
A car with a mass o 1 5 00 kg accelerates uniormly rom rest to a speed o 2 8 m s 1 ( about 1 00 km h 1 ) in a time o 1 1 s. C alculate the average orce that acts on the car to produce this acceleration.
Solution
8.3 1 0 The acceleration o the aircrat is _______ = 3.3 1 0 2 2 .5 1 m s .
vu 28 The acceleration = a = _ = _ = 2 .5 4 m s 2 . t 11 Force = ma = 1 5 00 2 . 5 4 = 3 . 8 kN.
v2 = u 2 + 2 as so v = 0 + 2 2 . 5 1 1 700 which leads to v = 92 .4 m s 1 .
5
5
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M E C H AN I C S
TOK
Newtons third law
But are they really laws?
Newtons third law of motion is sometimes written in such a way that the true meaning o the law has to be teased out. The law can be expressed in a number o equivalent ways.
The essential question is: can Newtons laws of motion be proved? The answer is that they cannot; strictly speaking they are assertions as Newton himself recognized. In his famous Principia (written in Latin as was a custom in those days) he writes Axiomata sive leges motus, or the axioms or laws of motion. However, they appear to be an excellent set of rules that allow us to predict most of the motion that we undertake. They remained unchallenged as a theory for about 200 years until the two theories of relativity were formulated by Einstein at the turn of the twentieth century. Essentially (said Einstein) the rules that Newton proposed do not always apply to, for example, motion that is very fast. However, for the modest speeds at which humans travel, the rules are reliable to a high degree and are certainly good enough for our needs most of the time.
O ne common way to write Newtons third law o motion is: E very action has an equal and op p osite reaction. The frst point to make is that the words action and reaction really mean action orce and reaction orce. S o, once again, Newton is reerring directly to the eects o orces. A second point is that the actionreaction pair must be o the same type. S o a gravitational action orce must correspond to a gravitational reaction. It could not, or example, reer to an electrostatic orce. The law suggests that orces must appear in pairs, but in thinking about a particular situation it is important to identiy all possible orce pairs and then to pair them up correctly. Take, as an example, the situation o a rubber ball resting on a table. At frst glance the obvious action orce here is the weight o the ball, that is, the gravitational pull the E arth exerts on the sphere. This orce acts downwards and i the table were not there the ball would accelerate downwards according to Newtons second law. It would all to the oor. What is the reaction orce here? Given that action orce and reaction orce must pair up like or like, the reaction must be the gravitational orce that the ball exerts on the Earth. This is o exactly the same size as the pull o the Earth on the ball, but is in the opposite direction. What prevents the ball accelerating downwards to the oor? A orce must be exerted by the table on the ball and i there are no accelerations happening, then this orce is equal and opposite to the downwards gravitational pull o the E arth on the ball. B ut the upwards table orce is not the reaction orce to the balls weight we have already seen that this is the gravitation pull on the E arth. The origin o the table orce is the electrostatic orces between atoms. As the ball lies on the table, it deorms the horizontal surace very slightly, rather like what happens when you push downwards with a fnger on a metre ruler suspended by supports at its end the ruler bends in a spring- like way to provide a resistance to the orce acting downwards. The dent in the table surace is the response o the atoms in the table to the weight lying on it. Remove the ball and the surace will return to being at. This upwards orce that returns the table to the horizontal is pushing upwards on the ball. There is a corresponding downwards orce rom the deormed ball ( the ball will become slightly attened as a response to the gravitational pull) . S o here is the second actionreaction pair, between two orces that are electrostatic in origin. In summary, there are our orces in this situation: the weight o the ball, the upwards spring- like orce o the table, and the reactions to these orces which are the pull o the ball on the E arth and the downwards push o the deormed ball on the table.
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2 . 2 FO R CE S
To ( literally) get a eeling or this, take a one- metre laboratory ruler and suspend it between two lab stools as suggested earlier. Press down gently with your fnger in the centre o the ruler so that it becomes curved. The ruler will bend, you will be able to eel it resisting your eorts to deorm it too ar. Remove your fnger and the ruler will return to its original shape. In explaining Newtons third law in a particular example, you must remember to emphasise the nature, the size, and the direction o the orce you are describing. Another common example is that o a rocket in space. S tudents sometimes write that .. .by Newtons third law, the rocket pushes on the atmosphere to accelerate but this shows a poor understanding o how the propulsion actually works. First, o course, the rocket does not push on anything. The act that a rocket can accelerate in space where there is no atmosphere proves this. What happens inside the rocket is that chemicals react together producing a gas with a very high temperature and pressure. The rocket has exhaust nozzles through which this gas escapes rom the combustion chamber. At one end o the chamber inside the rocket the gas molecules rebound o the end wall and exert a orce on it, as a result they reverse their direction. In principle, the rebounding molecules could then travel down the rocket and leave through the nozzles. S o there is an action- reaction pair here, the orce orwards that the gas molecules exert on the chamber ( and thereore the rocket) and the orce that the chamber exerts on the gas molecules. It is the frst o these two orces that accelerates the rocket. I the chamber were completely sealed and the gas could exert an equal and opposite orce at the back o the rocket, then the orward orce would be exactly countered by the backwards orce and no acceleration would occur.
Figure 2 Chemical rockets in action.
Later we shall interpret this acceleration in a dierent way. B ut the explanation given here will still be correct at a microscopic level. Think about the ollowing situations and discuss them with ellow students.
A freman has to exert considerable orce on a fre hose to keep it pointing in the direction that will send water to the correct place.
There is a suggestion to power space travel to deep space by ej ecting ions rom a spaceship.
A sailing dinghy moves orward when the wind blows into the sails.
Free-body force diagrams As a vector quantity, a orce can be represented by an arrow that gives both the scaled length and the direction o the orce. In simple cases where ew orces act this works well, but as the situations become more complex, diagrams that show all the arrows can become complicated. O ne way to avoid this problem is to use a free-body force diagram to illustrate what is happening. The rules or a ree- body diagram or a body are simple, as ollows.
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M E C H AN I C S
The diagram is or one body only and the orce vectors are represented as arrows.
O nly the orces acting on the body are considered.
The orce ( vector) arrows are drawn to scale originating at a point that represents the centre o mass o the body.
All orces have a clear label.
Procedure for drawing and using a free-body diagram
B egin by sketching the general situation with all the bodies that interact in the situation.
S elect the body o interest and draw it again removed rom the situation.
D raw, to scale, and label all the orces that act on this body due to the other bodies and orces.
Add the orce vectors together ( either by drawing or calculation) to give the net orce acting on the body. The sum can be used later to draw other conclusions about the motion o the obj ect.
Examples of free-body diagrams 1
A ball falling freely under gravity with no air resistance. situation diagram ball
pull of Earth on ball
pull of ball on Earth
Earth
free-body diagram for the ball ball acceleration of ball pull of Earth on ball
Figure 3 Ball falling freely under gravity.
This straightorward situation speaks or itsel. There are two orces acting: the Earth pulling on the ball and the ball pulling on the Earth. For the ree-body diagram o the ball we are only interested in the frst o these. So the ree-body diagram is a particularly simple one, showing the object and one orce. Notice that the ball is not represented as a real obj ect, but as a point that reers to the centre o mass o the object.
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2 . 2 FO R CE S 2
The same ball resting on the ground.
general situation
free-body diagram for the ball force on ball from ground
acceleration = 0
ball
ball
ground
force of Earth on ball
Figure 4 Ball resting on the ground.
As we saw earlier, our orces act in this case:
the weight o the ball downwards
the reaction o the Earth to this weight
the upwards orce rom the ground
the reaction o the ball to this spring-like orce.
A ree- body diagram ( fgure 4) defnitely helps here because, by restricting ourselves to the orces acting on the ball, the our orces reduce to two: the weight o the ball and the upwards orce on it due to the deormation o the ground. These two orces are equal and opposite. The net resultant ( vector sum) o the orces is zero and there is thereore no acceleration. 3
An object accelerating upwards in a lift.
The weight o the object is downwards and the size o this orce is the same as though the object had been stationary on the Earths surace. However, the upwards orce o the oor o the lit on the object is now larger than the weight and the resultant orce o the two has a net upwards component. The object is accelerated upwards as you would expect. For the lit, there is an upward orce in the lit cable and a downwards weight o the lit is downwards together with the weight o the object. The resultant orce is upwards and is equal to the orce in the cable less the weight orces o lit and obj ect. general situation
free-body diagram for the mass lift cable
free-body diagram for the lift
acceleration of lift upwards force exerted by lift on object
acceleration of object lift
acceleration of object upwards
object
object
force exerted by cable on lift
force exerted by earth on lift (weight of lift)
force exerted by Earth on object force exerted by object on lift
Figure 5 Mass in a lift accelerating upwards.
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Translational equilibrium
orce 1
orce 2
(a)
orce components add vertically
When an obj ect is in translational equilibrium it is either at rest or moving at a constant velocity ( not j ust constant speed in this case) . Translational here means moving in a straight line. ( There is also rotational equilibrium where something is at rest or rotating at a constant angular speed; this is discussed in option B ) . Newtons frst and second laws remind us that i there is no change o velocity then there must be zero orce acting on the object. This zero orce in many cases is the resultant (addition) o more than one orce. In this section we will examine what equilibrium implies when there is more than one orce. The simplest case is that o two orces. I they are equal in size and opposite in direction, they will cancel out and be in equilibrium ( Figure 6( a) ) .
orce components cancel horizontally (b)
Figure 6 Two equal orces in diferent directions.
I the orces are equal in size but not in the same direction then equilibrium is not possible. Horizontally, in fgure 6( b) , the two components o the orce vectors are still equal and opposite and we could expect that there would be no change in this direction. B ut vertically the two vector components point in the same direction so that overall there will be an unbalanced vertical orce and an acceleration acting on the obj ect on which these orces act. This gives a clue as to how we should proceed when there are three or more orces. free-body diagram
situation diagram
T1
gb
a la
nc
e
T2
sp
ri n
mg
string ring
weight
T2 sin T1
T2 cos horizontally
mg
vertically
Figure 7 Three orces acting on a ring.
Figure 7 shows a situation diagram and a ree- body diagram or a ring on which three orces act. For equilibrium, in whatever direction we resolve the orces, all three components must add up to zero in this direction.
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Figure 7 shows that horizontal and vertical are two good directions with which to begin, because two orces are aligned with these directions and one disappears in each direction chosen. Thus, vertical orce mg has no component in the horizontal direction and horizontal orce T1 has no part to play vertically. B ut whichever direction is chosen, i there is to be no resultant orce and consequently no acceleration, then all the orces must cancel. There is one more consequence o this idea. Figure 8 shows the orces drawn, as usual, to scale and in the correct direction ( in red) . The orces can be moved, as shown by the green arrows, into a new arrangement ( shown in black) . What is special here is that the three orces orm a closed triangle where all the arrows meet.
T2 Move the mg vector vertically upwards to begin at the start of T1 .
Move T2 sideways to the end of T1 . T1 mg
The three vectors now form a closed triangle. The three forces are in translational equilibrium.
Figure 8 Making a triangle for forces.
Algebraically, this must be true because we know (ignoring the directions o the orces) that T1 = T2 cos ( horizontally) and mg = T2 sin ( vertically) So T12 = T22 cos 2 and ( mg) 2 = T22 sin 2 Adding these together gives T12 + ( mg) 2 = T22 ( sin 2 + cos 2 ) and thereore as sin 2 + cos 2 = 1 T12 + ( mg) 2 = T22 This is equivalent to Pythagoras theorem where x = T1 2 , y = ( mg) 2 and z = T2 2 so that x2 + y 2 = z2 . S o, because the sum o the squares o two vector lengths equals the square o the third vector length, the three must ft together as a right- angled triangle. We have a right- angled triangle ormed by the scaled lengths and directions o the three orce vectors. This important fgure is known as a triangle of forces. I you can draw the vectors or a system in this way, then the system must be in translational equilibrium.
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Investigate! Three forces in equilibrium T2
Your arrangement or hanging the spring balances will depend on the resources in your school.
S elect a mass to suit the sensitivity o your spring balances.
Place a vertical drawing board with a large sheet o paper pinned to it behind the balances so that you can mark the position o the strings. When these are marked, use a protractor to measure the angles.
B egin with a simple case, say when T1 is horizontal so that 1 = 90.
The knot must be stationary and when this is true:
T1
2
1
string
mg
I a point is in equilibrium under the infuence o three orces, then the three orces must orm a closed triangle. To veriy this, use a mass and two spring balances with a knot halway along the string between the two spring balances to act as the point obj ect.
T1 /N
1 /
T2 /N
2 /
mg/N
T1 cos 1 = T2 cos 2 ( resolving horizontally)
T1 sin 1 + T2 sin 2 = mg ( resolving vertically)
Construct this table to enable you to veriy that both equations are true or every case you set up.
T1 cos 1 /N
T1 sin 1 /N
T2 cos 2 /N
T2 sin 2 /N
Case 1 For each case check that the equations are correct within experimental error.
Solid friction Friction is the orce that occurs between two suraces in contact. I you live in a part o the world where there is snow and ice then you will know that when the riction between your shoes and the ice disappears it can be a good thing ( or skiing) or a bad thing ( or alling over) .
Investigate! Solid friction
spring balance
In this experiment you use a spring balance to measure the orce that acts between two suraces. It gives some unexpected results. weight
Set up the platorm on rollers with a suitable weight and a spring balance. A string connects the spring balance to the weight. Ensure that the weight is attached securely to the block.
Pull the platorm using the winch system with an increasing orce observing, at the same time, the reading on the balance.
rollers
winch system
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platform
2 . 2 FO R CE S
Note the maximum value o the orce as measured on the balance.
Note the way the size o the orce changes when the platorm begins to move.
Is this orce constant? O r does it uctuate?
C hange the size o the weights on the platorm. How does the total weight aect the orce measured on the balance?
C hange the type o surace at the bottom o the weights. You might use abrasive ( emery) paper or cloth in dierent abrasive grades pinned to the platorm.
Investigate other changes: does lubrication at the bottom o the weights aect the riction?
I you carried out the Investigate! S olid riction experiment you may have ound that:
The orce on the balance increases as you pull harder and harder, but the platorm does not begin to move relative to the weights immediately.
Eventually the platorm suddenly begins to move at a particular value o orce and at this instant the orce, shown by the balance, drops to a new lower value.
This new value is then maintained as the platorm moves steadily.
You may observe stick-slip behaviour where the platorm alternately sticks and then j umps to a new sticking position. This behaviour is associated with two values o riction, but this may be too difcult to observe unless you get very suitable suraces.
The riction orces depend on the magnitude o the weights.
The rictional orces that occur in this experiment are described empirically as:
static riction ( when there is no relative movement between the suraces)
dynamic riction ( when there is relative movement) .
As the pulling orce increases but without any slip happening, the riction is said to be static (because there is no motion) . Eventually however, the pulling orce will exceed the value o the static riction and the suraces will start to move. As the movement continues, the riction orce drops to a new value, lower than the maximum static value. This new lower value is known as the dynamic riction. B oth static and dynamic riction orces are highly dependent on the two suraces concerned.
Static friction The static riction orce F is ound to be given empirically by F s R where F is the rictional orce exerted by the surace on the block. R is the normal reaction o the surace on the block, this equals the weight o the block as there is no vertical acceleration. The symbol s reers to the coefcient o static riction. The less than or equal to symbol indicates that the static riction orce can vary rom zero up to a maximum value. B etween these limits F is
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2
M E C H AN I C S equal to the pull on the block. O nce the pull on the block is equal to F, then the block is j ust about to move. O nce the pull exceeds F then the block begins to slide ( fgure 9) . For these larger orces, the riction operating is in the dynamic regime.
pull on block
pull of Earth on block
friction force exerted by surface on block
reaction force of surface on block
Figure 9 Friction force acting on an object.
Dynamic friction Dynamic riction only applies when the suraces move relative to each other. The riction drops rom its maximum static value (there is an explanation or this below) and remains at a constant value. This value depends on the total reaction orce acting on the surace but (according to simple theory) is not thought to depend on the relative speed between the two suraces. So, or dynamic riction F = dR where d is the coefcient o dynamic riction. The values o s and d vary greatly depending on the pair o suraces being used and also the condition o the suraces ( or example, whether lubricated or not) . A ew typical values are given in the table. I you want to investigate a wider range o suraces, there are many sources o the coefcient values on the Internet search or C oefcients o riction . Each riction coefcient is a ratio o two orces (F and R) and so has no units.
Surface 1
Surface 2
s
d
glass
metal
0.7
0.6
rubber
concrete
1.0
0.8
rubber
wet tarmac
0.6
0.4
rubber
ice
0.3
0.2
metal
metal (lubricated)
0.15
0.06
Values o s and d or p airs o suraces It is possible or the coefcients to be greater than 1 or some surace pairs. This reects the act that or these suraces the riction is very strong and greater than the weight o the block. Remember that the suraces are being pulled sideways by a horizontal orce whereas the reaction orce is vertical so we are not really comparing like with like in these empirical rules or riction.
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2 . 2 FO R CE S
Investigate! Friction between a block and a ramp One way to measure the static coefcient o riction between two suraces in a school laboratory is to use the two suraces as part o a ramp system.
O ne surace is the top o the ramp, the other is the base o the block.
Resolving at 90 to plane, N = mg cos ; resolving along the plane, F = mg sin F S o tan = _ = s N S tart with the ramp horizontal, and then gradually raise one end until the block starts to slip.
free-body diagram of just the block
a block on a ramp
N
Ff
N
Ff
m
At this moment o slip, measure the angle o the ramp. The tangent o this angle is equal to the coefcient o static riction.
2
A skier places a pair o skis on a snow slope that is at an angle o 1 . 7 to the horizontal. The coefcient o static riction between the skis and the snow is 0.02 5 .
mg
mg
Worked examples 1
A box is pushed across a level oor at a constant speed with a orce o 2 80 N at 45 to the oor. The mass o the box is 5 0 kg.
F 45
D etermine whether the skis will slide away by themselves.
Solution reaction force of surface on ski
C alculate:
weight component down slope
friction force
1.7
a) the vertical component o the orce
weight of ski
b) the weight o the box c) the horizontal component o the orce d) the coefcient o dynamic riction between the box and the oor.
Solution a) the vertical component is 2 80 sin 45 = 1 98 N b) the weight o the box = mg = 5 0 9.8 = 490 N c) the horizontal component o the orce = 2 80 cos 45 = 1 98 N d) the vertical component o the orce exerted by the oor on the box = 490 + 1 98 N = 688 N the riction orce = the horizontal component ( the box is travelling at a steady speed) , so 1 98 d = _ = 0. 2 9 688
C all the weight o the skis W. The component o weight down the slope = W sin 1 .7 The reaction orce o the surace on the ski = W cos 1 .7. Thereore, the maximum riction orce up slope = S W cos 1 .7. The skis will slide i S W cos 1 .7 < W sin 1 .7 in other words, i S < tan 1 .7. The value o tan 1 .7 is 0.02 96 and this is greater than the value o S , which is 0.02 5 , so the skis will slide away.
57
2
M E C H AN I C S
Origins of friction Friction originates at the interace between the two materials, in other words at the surace where they meet. The actual causes o riction are still being investigated today and the explanation given here is highly simplifed and a historical one. Leonardo da Vinci mentions riction in his notebooks and some o the next scientifc writings about riction appear in works by Guillaume Amontons rom around 1 700. moving to right
(a)
moving to left
liquid lubricant layer
(b)
Figure 10 How solid friction arises.
One model or solid riction suggests that what are very smooth suraces to us are not smooth at all. At the atomic level, they are actually ull o peaks and troughs o atoms (fgure 1 0(a) ) . When static riction occurs and two suraces are at rest relative to each other, then the atomic peaks rest in the troughs, and it needs a certain level o orce to deorm or break the peaks sufciently or sliding to begin. Once relative motion has started (so that dynamic riction occurs) , then the top surace rises a little above the deormed peaks. Less orce is now required to keep the motion going. B ecause the irregularities on the surace are very small and o atomic size, even the small orces applied in our lab experiments cause large stresses to act on the peaks. The peaks then deorm like a sot plastic irrespective o whether the material is hard steel or something much soter. Moving suraces are oten coated with a lubricant to reduce wear due to riction ( fgure 1 0( b) ) . The lubricant flls the space between the two suraces and either prevents the peaks and troughs o atoms rom touching or reduces the amount o contact. In either event, the atoms rom the suraces do not interact as much as beore and the riction orce and the coefcient are reduced. The origins o these riction orces are bound up in the complex electronic properties o the materials that make up the suraces. However, this simple theory should give you some understanding o riction as well as an awareness that the bulk materials we perceive on the macroscopic scale arise rom microscopic properties that are operating at the atomic level. What is important to realize is that the two equations or static and dynamic riction do not arise rom a study o the interatomic orces between the suraces; they are derived purely rom experiments with bulk materials. The results are empirical not theoretical.
58
2 . 2 FO R CE S
Fluid resistance and terminal speed The assumption that air resistance is negligible is oten unrealistic. An obj ect that travels through a uid ( a liquid or a gas) is subj ect to a complex process in which, as the obj ect travels through the material, the uid is stirred up and becomes subj ect to a drag force. The process is complex, so even ater introducing some simple assumptions about the resistance o a uid, we will still not be able to give a complete analysis. In energy terms, the action o air resistance is to transer some o the energy o the moving body into the uid through which it is moving. S ome uids absorb this energy better than others: swimming through water is much more tiring than running in the air. For your IB D iploma Programme physics exams, you only need to describe the eects o uid resistance without going into the mathematics.
Skydiving In 2 01 2 , Felix B aumgartner j umped saely rom a height o 3 9 km above New Mexico to reach a top speed o 1 3 42 km h 1 aster than the speed o sound. A skydive rom more usual heights will not take place at such high speeds, usually up to about 2 00 km h 1 . The dierence is due to the variation in the resistance o the air at dierent heights above the Earth.
drag force mg drag force
mg
mg body released from rest
forces on body during acceleration
forces on body at terminal speed
Figure 11 Forces acting on skydiver.
The diagram shows the orces acting on a skydiver. The weight o the diver acts vertically downwards and is eectively constant ( because there is little change in the Earths gravitational feld strength at the height o the dive) . The air resistance orce acts in the opposite direction to the motion o the diver and or a diver alling vertically, this will be vertical too. O ther orces acting on the diver include the upwards buoyancy caused by the displacement o air by the diver.
59
2
M E C H AN I C S When the skydiver initially leaves the aircrat, the divers weight acts downwards and because the vertical speed is almost zero, there is almost no air resistance. Air resistance increases as the speed increases so that as the diver goes aster and aster the resistance orce becomes larger and larger. The net orce thereore decreases and consequently the acceleration o the diver downwards also decreases. E ventually the weight orce downwards and the resistance orce upwards are equal in magnitude and, o course, opposite in direction. At this point there is no longer any acceleration and the diver has reached a constant rate o all known as the terminal sp eed. The graph o vertical speed against time is as shown in fgure 1 2 . parachute opens
vertical speed
accelerates to terminal speed
greater drag force so terminal speed smaller
(not to scale)
lands
time
Figure 12 Speedtime for a parachute jump.
Eventually the skydiver opens the parachute. Now, because o the large surace area o the parachute envelope, the upwards resistive orce is much larger than beore and is greater than the weight. As a result, the directions o the net orce and acceleration are also upwards; the vertical velocity decreases in magnitude. Once again a balance will be reached where the upward and downward orces are equal and opposite but at a much lower speed than beore (about 1 2 m s 1 or a landing) and the diver reaches the ground saely.
Maximum speed of a car The top ( maximum) speed o a motor car is determined by a number o actors. The most obvious o these is the maximum orce that the engine can exert through the tyres on the road surace. B ut, as with the skydiver, this is not the only orce acting. There is a considerable drag on the vehicle due to the air and this drag orce increases markedly as the speed o the car becomes larger. Typically, when the speed doubles the drag orce will increase by at least a actor o our. There is a maximum power that the car engine can produce. When the car accelerates and the speed increases towards the maximum, the power dissipated in riction also increases. When the maximum energy output o the engine every second is completely used in overcoming the energy losses, then the car cannot accelerate urther and has reached its maximum speed.
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2 . 3 W O R K , E N E R G Y, A N D P O W E R
Worked example 1
C alculate the upward force acting on a skydiver of mass 80 kg who is falling at a constant speed.
horizontal force of 3 0 N. The mass of P is 2 .0 kg and the mass of Q is 4.0 kg.
Solution
Q
The weight of the skydiver is 80 9. 8 = 784 N. B ecause the skydiver is falling at constant speed ( i.e. terminal speed) the upwards drag force is equal to the downwards weight. The upward force is 784 N. 2
P and Q are two boxes that are pushed across a rough surface at a constant velocity with a
P
F
S tate the resultant force on box Q.
Solution Q is moving at constant speed. S o the resultant force on this box must be 0 N.
2.3 Work, energy, and power Understanding Principle o conservation o energy Kinetic energy Gravitational potential energy
Applications and skills Discussing the conservation o total energy
Elastic potential energy Work done as energy transer
Power as rate o energy transer Eciency
Nature of science The theory o conservation o energy allows many areas o science to be understood at a undamental level. It allows the explanation o natural phenomena but also means that scientists can predict the outcome o a previously unknown efect. The conservation o energy also demonstrates that paradigm shits occur in science: the interchangeability o mass and energy as predicted by Einstein is an example o this.
within energy transormations Sketching and interpreting orce distance graphs Determining work done including cases where a resistive orce acts Solving problems involving power Quantitatively describing eciency in energy transers
Equations work: W = Fs cos 1 kinetic energy: EK = ___ mv2 2
1 k(x) 2 elastic potential energy: Ep = ___ 2
change in gravitational potential energy:
Ep = mgh power = Fv useul power out useul workout = _________________________ eciency = ______________________ total work in total power in
61
2
M E C H AN I C S
Introduction This sub- topic looks at the physics of energy transfers. The importance of energy is best appreciated when it moves or transfers between different forms. Then we can make it do a useful j ob for us. What is important is that you learn to look below the surface of the bald statement: electrical energy is converted into internal energy and ensure that you can talk about the detailed changes that are taking place.
Energy forms and transfers Energy can be stored in many different forms. S ome of the important ones are listed in the table below.
Energy
Nature of energy associated with...
kinetic (gravitational) potential electric/magnetic chemical nuclear elastic (potential) thermal (heat)
the motion o a mass the position o a mass in a gravitational feld charge owing atoms and their molecular arrangements the nucleus o an atom an object being deormed a change in temperature or a change o state
mass
conversion to binding (nuclear) energy when nuclear changes occur mechanical waves in solids, liquids, or gases
vibration (sound)
light
photons o light
Notes sometimes the word gravitational is not used
related to a mass change by E = mc2 The word potential is not always used A change o state is a change o a substance between phases, i.e. solid to liquid, or liquid to gas. This is reerred to as energy transerred as a result o temperature dierence in line with the IB Guide. The colloquial term heat is usually acceptable when reerring to situations involving conservation o energy situations.
the amount o sound energy transerred is almost always negligible when compared with other energy orms sometimes called radiant energy another orm o electric/magnetic
Energy can be transferred between any of its forms and it is during such transfers we see the effects of energy. For example, water can fall vertically to turn the turbine of a hydroelectric power station and drive a generator. Lots of things are happening here. The water molecules are attracted by the Earth and accelerate downwards through the pipe. Their momentum is transferred to the blades of the turbine which rotates and turns the coils in the generator. As a result of the coils turning, electrons are forced to move and there is an electric current. Overall, this chain of physical processes can be summed up as the conversion of gravitational potential energy of the water into an electrical form. Another example of an energy transfer is that of an animal converting stored chemical energy in the muscles leading to kinetic and gravitational potential energy forms with some of the chemical energy also appearing as frictional losses.
62
2 . 3 W O R K , E N E R G Y, A N D P O W E R Learn to recognise the physical (and sometimes chemical) processes that are going on in the system. It is easy to describe the changes in airly broad energy transer terms. Always try, however, to explain the eects in terms o microscopic or macroscopic interactions. Whatever the orm o the energy transer, we use a unit o energy called the joule (J) . This is in honour o James Joule (1 81 81 889) the English scientist who devoted his scientifc eorts to studying energy and its transers. O ne j oule is the energy required when a force of one newton acts through a distance of one metre. When energy changes rom one orm to another we fnd that nothing is lost ( providing that we take care to include every single orm o energy that is included in the changes) . This is known as the p rincip le of conservation of energy which says that energy cannot be created or destroyed. S ince E insteins work at the turn o the twentieth century, we now recognise that mass must be included in our table o energy orms. For most changes the mass dierence is insignifcant, but in nuclear changes it makes a maj or contribution. In some applications, such as when discussing the output o power stations, the joule is too small a unit so you will requently see energies expressed in megajoules (MJ or 1 0 6 J) or even gigajoules (GJ or 1 0 9 J) . Get used to working in large powers o ten and with prefxes when dealing with energies.
Worked example D escribe the mechanisms associated with the energy changes that occur when a balloon is blown up.
Solution Air molecules gain kinetic energy that is used to store elastic potential energy in the skin o the balloon and to make changes in the energy o the air. The air inside is able to exert more orce ( pressure) outwards on the skin o the balloon until a new equilibrium is established between the tension in the skin and the atmospheric pressure.
In some parts o physics, dierent energy units are used. These have usually arisen historically. An example o this is the electronvolt ( eV) , which is the energy gained by an electron when it is accelerated through a potential dierence o one volt. Another example is the calorie used by dieticians; this is an old unit that has lingered in the public domain or perhaps longer than it should! O ne calorie ( cal) is 4.2 J. You will learn the detail about any special units used in the course in the appropriate place in this book.
Doing work In 1 82 6, Gaspard- Gustave C oriolis was studying the engineering involved in raising water rom a ooded underground mine. He realized that energy was being transerred when the steam engines were pumping the water vertically. He described this energy transer as work done, and he recognized that the energy transerred when the pumping engines exerted a orce on a particular mass o water and lited it rom the bottom o the mine to the surace. In other words work done ( in J) = force exerted ( in N) distance moved in the direction of the force ( in m) So, when a weight o 5 N o water is lited vertically through a height o 1 5 0 m, then the work done by the engine on the water is 5 1 50 = 75 0 J. In the example o the mine, the orce and the distance moved are in the same direction ( vertically upwards) , but in many cases this will not
Figure 1 Mine steam engine.
63
2
M E C H AN I C S be the case. A good example of this is a sand yacht, where the force F from the wind acts in one direction and the sail is set so that the yacht moves through a displacement s that is at an angle to the wind. In this case, the distances moved in the direction of the force and by the yacht are not the same. You must use the component of force in the direction of movement. sail distance moved by sand yacht, s force exerted by wind, F plan view
Figure 2 Force on sail.
In this case, work done = F cos s
Work done against a resistive force Work is done when a resistive force is operating too. C onsider a box being pushed at a constant speed in a horizontal straight line. For the speed to be constant, friction forces must be overcome. The force that overcomes the friction may not act in the direction of movement. Again, the work that will be done by the force (or the force provider) is force acting distance travelled cos (figure 3) . force F horizontal component of force = F cos
box
direction of motion
Figure 3 Resistive force.
Worked examples 1
The thrust ( driving force) of a microlight aircraft engine is 3 .5 1 0 3 N. C alculate the work done by the thrust when the aircraft travels a distance of 1 5 km.
that acts at 5 0 to the horizontal. C alculate the work done in moving the box 8.5 m. 55 N 8.5 m
Solution Work done = force distance = 3 5 00 1 5 000 = 5 .3 1 0 7 J 5 3 MJ. 2
64
A large box is pulled a distance of 8. 5 m along a rough horizontal surface by a force of 5 5 N
50
2 . 3 W O R K , E N E R G Y, A N D P O W E R
Solution
distance in which the obj ect will come to rest i a net orce o 4. 0 N opposes the motion.
The component o orce in the direction o travel is 55 cos 50 = 35.4 N.
Solution
The work done = this orce component distance travelled = 3 5 .4 8.5 = 3 01 J.
There must be 2 4 J o work done to stop the motion o the obj ect. The orce acting is 4.0 so 24 __ = 6 m. 4
An obj ect moving in a straight line has an initial kinetic energy o 2 4 J. C alculate the
3
Forcedistance graphs In practice, it is rare or the orce acting on a moving obj ect to be constant. Real railway trucks or sand yachts have air resistance and other orces that lead to energy losses that vary with the speed o the obj ect or the surace over which it runs. We can deal with this i we know how the orce varies with distance. S ome examples are shown in fgure 4. total area = (number of squares) (energy represented by one square)
area = F s 0
distance moved (a)
s s
0
0
F
Worked example
distance moved (b)
The graph shows the variation with displacement d o a orce F that is applied to a toy car. C alculate the work done by F in moving the toy through a distance o 4.0 cm.
Figure 4 Forcedistance graphs.
For a constant orce ( fgure 4( a) ) , the graph o orce against distance will be a straight line parallel to the x-axis. The work done is the product o force distance ( we are assuming that = 90 in this case) ; this corresponds to the area under the graph o orce against distance. When the orce is not constant with distance moved ( fgure 4( b) ) the work done is still the area under the line, but this time you have to work a little harder by estimating the number o squares under the graph and equating each square to the energy that it represents. The product o ( energy for one square) ( number of squares) will then give you the overall work done. There is a urther example o this type o calculation later in the topic on elastic potential energy.
Power Imagine two boys, Jean and Phillipe with the same weight ( lets say 65 0 N) who climb the same hill ( 70 m high) . B ecause they have the same weight and climb the same vertical distance they both gain the same amount o gravitational potential energy. This is at the expense o the chemical energy reserves in their bodies. However, suppose Jean climbs the hill in 1 5 0 s whereas Phillipe takes 3 00 s.
6 5 4 F/N
0
one square = F s
force
force
F
3 2 1 0
0
2
4 d/10 - 2 m
6
Solution The work done is equal to the area o the triangle enclosed by the graph and 1 the axes. This is __ base 2 o the triangle height o 1 4. 0 1 0 - 2 triangle = __ 2 5 .0 = 0.1 0 J
65
2
M E C H AN I C S
James Watt was a mechanical engineer who worked at the end o the 18th century and the beginning o the 19th. He did major work in improving steam engine design and even developed a way o making copies o paper documents that was used in ofces until early in the 20th century.
The obvious dierence he re is that Jean is gaining potential energy twice as ast as Phillipe because Jeans time to make the climb is hal that o Phillipe . This die rence is important when we want to compare two machines taking dierent times to carry out the same amount o work. The quantity p ower is used to measure the rate of doing work, in other words it is the number o j oules that can be converted every second. S o, power is defned as: energy transferred power = __ time taken for transfer Using the correct quantities is important here. I the energy change is in j oules, and the time or the transer measured in seconds then the power is in watts ( W) . 1 W 1 J s 1 In the example o Jean and Phillipe above, both did 45 . 5 kJ o work, but 650 70 Jeans power in climbing the hill was _______ = 3 03 W and Phillipes was 1 50 1 5 2 W because he took twice as long in the climb. The equation work done = force distance can be rearranged to give another useul expression or power. distance moved by force work done power = _ = force __ time time It is easy to see that this is the same as power = force speed. S o the power required to move an obj ect travelling at a speed v with a orce F is Fv.
Nature of science You will see other units or power used in various areas o everyday lie. As an example, the horsepower is one measure used by car manuacturers to provide customers with data about car engines. This is a unit that dates back to the beginning o the 1 8th century and was used to compare the power o an average horse with steam engines. (1 horsepower is equal to about 75 0 W in modern units.)
66
Kinetic energy, KE Kinetic energy is the energy an obj ect has because o its motion; kinetic energy ( sometimes abbreviated to KE) has its own symbol, EK. O bj ects gain kinetic energy when their speed increases.
mass m
KE =
1 2
u
mu 2
KE = change in KE = 12 m(v2 u 2 )
mass m
Figure 5 Kinetic energy of mass.
1 2
mv2
v
2 . 3 W O R K , E N E R G Y, A N D P O W E R Imagine an obj ect of mass m which is at rest at time t = 0 and which is accelerated by a force F for a time T. The kinematic equations and Newtons second law allow us to work out the speed of the obj ect v at time t = T. F The acceleration a is __ m ( using Newtons second law of motion) . mv F ___ Therefore v = 0 + __ m T ( because the initial speed is zero) . S o F = T
The work done on the mass is the gain in its kinetic energy, EK , and is F s where s is the distance travelled and therefore mv vT EK = F s = _ _ T 2 (v + 0) T because s = _ 2
Worked example A car is travelling at a constant speed of 2 5 m s 1 and its engine is producing a useful power output of 2 0 kW. C alculate the driving force required to maintain this speed.
Solution power driving force = _ speed 2 0 000 _ = = 800 N 25
The work done by the force is equal to the gain in kinetic energy and is 1 EK = __ mv2 2
Remember that this is the case where the initial speed was 0. If the obj ect is already moving at an initial speed u then the change in kinetic 1 energy EK will be __ m( v2 u 2 ) . 2 There is a subtle piece of notation here. When we talk about a value of kinetic energy, we write EK, but when we are talking about a change in kinetic energy from one value to another then we should write EK where , as usual, means the change in.
Examiners tip This equation for EK is one that needs a little care. Notice where the squares are; they are attached to each individual speed. This equation is not the same as ___12 m (v u) 2 .
Worked examples 1
A vehicle is being designed to capture the world land speed record. It has a maximum design speed of 1 700 km h 1 and a fully fuelled mass of 7800 kg. C alculate the maximum kinetic energy of the vehicle.
Solution 1 700 km h 1 470 m s 1 ( this is greater than the speed of sound in air! ) 1 EK = __ mv2 = 0.5 7800 470 2 = 8.6 1 0 8 J 2
0.86 GJ 2
A car of mass 1 .3 1 0 3 kg accelerates from a speed of 1 2 m s 1 to a speed of 20 m s 1 . Calculate the change in kinetic energy of the car.
Solution 1 EK = __ m( v2 u 2 ) = 0. 5 1 3 00 ( 2 0 2 1 2 2 ) 2
= 65 0 ( 400 1 44) = 1 . 7 1 0 5 J
67
2
M E C H AN I C S
Gravitational potential energy, GPE mass, m
g
h
gain in gpe( E P) = mgh
Gravitational p otential energy is the energy an obj ect has because o its position in a gravitational feld. When a mass is moved vertically up or down in the gravity feld o the E arth, it gains or loses gravitational potential energy ( GPE) . The symbol assigned to GPE is EP. O nly the initial and fnal positions relative to the surace determine the change o GPE ( assuming that there is no air resistance on the way) . For this reason, gravitational orce is among the group o orces said to be conservative, because they conserve energy ( see the Nature o S cience box below) . The work done when an obj ect is raised at constant speed through a change in height h is, as usual, equal to force distance moved. In this case the orce required is mg and work done, EP = mg h.
Figure 6 Gravitational potential energy.
As usual, the value o g is 9.8 m s 2 close to the Earths surace, but this value becomes smaller when we move away rom the E arth.
Nature of science Conservative forces There is an important dierence between orces such as gravity and rictional orces. When only gravity acts, the energy change depends on the start height and end height o the motion but not on the route taken by an obj ect to get rom start to fnish. Horizontal movement does not have to be counted i there is no riction acting. This type o orce is said to be conservative; in other words, it conserves energy. We could recover all the energy by moving the obj ect back to the start. C ontrast this with the riction orce that acts between a book and a table as the book is moved around the tables surace rom one point to another. I the book goes directly rom start to fnish a certain amount o energy will be used up to overcome the riction. B ut i the book goes by a longer route, more energy is needed ( work done = friction force distance travelled) . When it is necessary to know the exact route beore we can calculate the total energy conversion, the orce is said to be non-conservative. I we move the book back again, we cannot recover the energy in the way that we could when only gravity acted. You will not need to write about the meaning o the term conservative orce in the IB Diploma Programme physics examination.
Energy moving between GPE and KE S ometimes the use o gravitational potential energy and kinetic energy together provides a neat way to solve a problem. A snowboarder is moving down a curved slope starting rom rest ( u = 0) . The vertical change in height o the slope is h = 5 0 m. What is the speed o the snowboarder at the bottom o the slope? Assume we can ignore riction at the base o the board and the air resistance. What we must not do in this example is to use the kinematic ( suvat) equations. They cannot be applied in this case because the acceleration
68
2 . 3 W O R K , E N E R G Y, A N D P O W E R o the snowboarder will not be constant suvat only works when we are certain that a does not change. Although the suvat equations give the correct answer here, they should not be used because the physics is incorrect. The fnal answer happens to be correct only because we use the start and end points and also as a consequence o the conservative nature o gravity. We are also using an average value or acceleration down the slope by assuming that the angle to the horizontal is constant. The equations would not give the correct answer i we brought riction orces into the calculation. C onservation o energy comes to our aid because ( as the riction losses are negligible) we know that the loss o gravitational potential energy as the boarder goes down the slope is equal to the gain in kinetic energy over the length o the slope. B ecause we know that the GPE change only depends on the initial and fnal positions then we do not need to worry at all about what is going on at the base o a board during the ski run.
h = 50 m
Figure 7 Mechanics of snowboarding.
_____
1 So EP = mgh = __ mv2 and v = 2gh , 2
__________
in this case v = 2 9.8 5 0 = 3 1 m s 1 ( This is a speed o about 1 1 0 km h 1 which tells you that the assumption about no air resistance and no riction is a poor one, as any snowboarder will tell you! ) Notice that the answer does not depend on the mass o the boarder; the mass term cancels out in the equations.
Investigate! Converting GPE to KE cart
D evise a way to measure the speed o the cart. You could use a smart pulley that can measure the speed as the string turns the pulley wheel, or an ultrasound sensor, or a data logger with light gates.
When the mass is released, the cart will gain speed, as the gravitational potential energy o the weight changes.
Make measurements to assess the gravitational potential energy lost ( you will need to know the vertical height through which the mass alls) and the kinetic energy gained ( you will need the fnal speed) . Notice that only the alling mass is losing GPE but both the mass and the cart are gaining kinetic energy.
C ompare the two energies in the light o the likely errors in the experiment. Is the energy conserved? Where do you expect energy losses in the experiment to occur?
smart pulley to interface clamp mass
This experiment will help you to understand the conversion between KE and GPE.
Arrange a cart on a track and compensate the track or riction. This is done by raising the let- hand end o the track through a small distance so that the cart neither gains nor loses speed when travelling down the track without the string attached.
Have a string passing over a pulley at the end o the track and tie a weight o known mass to the other end o the track.
Measure the mass o the cart too.
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Worked examples 1
A ball of mass 0.3 5 kg is thrown vertically upwards at a speed of 8. 0 m s 1 . C alculate a) the initial kinetic energy b) the maximum gravitational potential energy c) the maximum height reached.
Solution 1 1 mv2 = __ a) EK = __ 0.3 5 8 2 = 1 1 .2 J 2 2
b) At the maximum height all the initial kinetic energy will have been converted to gravitational potential energy so the maximum value of the GPE is also 1 1 . 2 J. c) The maximum GPE is 1 1 . 2 J and this is equal to mg h, so 1 1 .2 1 1 .2 ________ h = ____ m g = 0 . 3 5 9 . 8 = 3 .3 m. 2
A pendulum bob is released from rest 0.1 5 m above its rest position. C alculate the speed as it passes through the rest position.
0.15 m
Solution EK at the rest position = EP at the release position; 1 __ mv2 = mg h which rearranges to 2
___
____________
v = 2 gh = 2 9.8 0.1 5 = 1 .7 m s 1 .
Elastic potential energy The shape of a solid can be changed by applying a force to it. D ifferent materials will respond to a given force in different ways; some materials will be able to return the energy that has been stored in them when the force is removed. A metal spring is a good example of this; most springs are designed to store energy in this way in many different contexts. The materials that can return energy in this way have stored elastic p otential energy. To discuss elastic potential energy we need to know something about the properties of springs. This is an area of physics that has been studied for a long time. Robert Hooke, a contemporary of Newton, published a rule about springs that has become known as Hookes law.
70
2 . 3 W O R K , E N E R G Y, A N D P O W E R
Investigate! Investigating Hookes law
support
Robert Hooke realized that there was a relationship between the load on a spring and the extension of the spring. spring
Arrange a spring of known unstretched length with a weight hanging on the end of the spring.
You will need to devise a way to measure the extension ( change in length from the original unstretched length) of the string for each of a number of different increasing weights hanging on the end of the spring.
0 5
pointer
10
scale
15 20
pan
Repeat the measurements as you remove the weights as a check.
Plot a graph of force ( weight) acting on the spring ( y- axis) against the extension ( x-axis) . This is not the obvious way to draw the graph, but it is the way normally used. Normally, we would plot the dependent variable ( the extension in this case) on the y-axis.
25 30
weight
table
For small loads acting on the spring, Hooke showed that the extension of the spring is directly proportional to the load. (For larger loads, this relationship breaks down as the microscopic arrangement of atoms in the spring changes. We shall not consider this part of the deformation, however.) Hookes rule means that the graph of force F against extension ( x) is a straight line going through the origin. In symbols the rule is F x, or F = kx where k is a constant known as the spring constant, which has units of N m 1 . The gradient of the graph is equal to k ( this is why the graph is plotted the wrong way round) . We can now relate this graph to the work done in stretching the spring. The force is not constant ( the bigger the extension, the bigger the force
force
Fmax
0
1
Area = 2 Fmax x = work done in extending spring = stored elastic potential energy
0
extension
x
Figure 8 Work done in stretching spring.
71
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M E C H AN I C S
Worked examples
required) but we now know how to deal with this. The work done on the spring is the area under the graph o orce against extension.
1
The area is a right-angled triangle and is equal to
A spring, o spring constant 48 N m 1 , is extended by 0.40 m. C alculate the elastic potential energy stored in the spring.
Solution 1 Energy stored = __ kx2 = 2 2 0.5 48 0.40 = 0.38 J
2
An object o mass 0.78 kg is attached to a spring o unstretched length 5 60 mm. When the obj ect has come to rest the new length o the spring is 62 0 mm. C alculate the energy stored in the spring as a result o this extension.
Solution The change in length o the spring x = 62 0 5 60 = 60 mm The tension in the spring will be equal to the weight o the obj ect = mg = 0.78 9.8 = 7.64 N The energy stored in the 1 spring = __ Fx = 0.5 2 7.64 0.06 = 0.2 3 J
1 __ force to stretch spring extension, 2 1 in symbols EP = __ Fmax x. 2 F We know rom Hookes law that F = k x, so k = ___ and thereore EP is x 1 2 also equal to __ k( x) . 2
Efciency In the Investigate! experiment where gravitational potential energy was transerred to kinetic energy, it is likely that some gravitational potential energy did not appear in the motion o the cart. Some energy will have been lost to internal energy as a consequence o riction and to elastic potential energy stored when the string changes its shape. We need to have a way to quantiy these losses. One way to do this is to compare the total energy put into a system with the useul energy that can be taken out. This is known as the efciency o the transer and can be applied to all energy transers, whether carried out in a mechanical system, electrical system or other type o transer. You can expect to meet efciency calculations in any area o physics where energy transers occur. The defnition can also be applied to power transers, because the energy change in these cases takes place in the same time or the total energy in and the useul work out. useul power out useul work out efciency = __ = __ total energy in total power
Worked example 1
An electric motor raises a weight o 1 5 0 N through a height o 7.2 m. The energy supplied to the motor during this process is 3 .5 1 0 4 J. C alculate: a) the increase in gravitational potential energy b) the efciency o the process.
Solution a) EP = 1 5 0 7. 2 = 1 080 J u se u l w o rk o u t 1 080 b) Efciency = ___________ = ____ = 0.3 1 or 3 1 % 3500 e ne rgy in
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2.4 M OM EN TUM
2.4 Momentum Understanding Newtons second law expressed as a rate of
change of momentum Impulse and forcetime graphs Conservation of linear momentum Elastic collisions, inelastic collisions, and explosions
Applications and skills Applying conservation of momentum in simple
Nature of science The concept of momentum has arisen as a result of evidence from observations carried out over many centuries. The principle of conservation of momentum is an example of a law that has universal applicability. It allows the prediction of the outcomes of physical interactions at both the macroscopic and the microscopic level. Many areas of physics are informed by it, from the kinetic theory of gases to nuclear interactions.
isolated systems including (but not limited to) collisions, explosions, or water jets Using Newtons second law quantitatively and qualitatively in cases where mass is not constant Sketching and interpreting forcetime graphs Determining impulse in various contexts including (but not limited to) car safety and sports Qualitatively and quantitatively comparing situations involving elastic collisions, inelastic collisions, and explosions
Equations momentum: p = mv Newton's second law (momentum version) : p F = ______ t
p2 kinetic energy: EK = ______ 2m
impulse: = Ft = p
Introduction Many sports involve throwing or catching a ball. C ompare catching a table-tennis ( ping-pong) ball with catching a baseball travelling at the same speed. O ne o these may be a more painul experience than the other! What is dierent in these two cases is the mass o the obj ect. The velocity may be the same in both cases but the combination o velocity and mass makes a substantial dierence. Equally, comparing the experience o catching a baseball when gently tossed rom one person to another with catching a frm hit rom a good player should tell you that changes in velocity make a dierence too.
Momentum We call the product o the mass m o an obj ect and its instantaneous velocity v the momentum p of the obj ect ( p = m v) and this quantity turns out to have ar-reaching importance in physics. First, here are some basics ideas about momentum.
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mass m
Momentum is mass velocity never mass speed.
Momentum has direction. The mass is a scalar but velocity is a vector. When mass and velocity are multiplied together, the momentum is also a vector with the same direction as the velocity. Think o the mass as scaling the velocity in other words, j ust making it bigger by a actor equal to the mass o the obj ect.
It ollows that the unit o momentum is the product o the units o mass and velocity, in other words kg m s 1 . There is a shorter alternative to this that we shall see later.
I velocity or mass is changing then the momentum must also be changing. We shall be looking at the two cases where one quantity changes while the other is held constant.
When a net resultant orce acts on an obj ect, the obj ect accelerates and the velocity must change. This means a change in momentum too. S o a net orce leads to a change in momentum.
velocity v
momentum = mv
Figure 1 Momentum and velocity.
Worked examples 1
A ball o mass 0.2 5 kg is moving to the right at a speed o 7.4 m s 1 . C alculate the momentum o the ball.
Solution p = mv = 0.2 5 7.4 = 1 .85 kg m s 2
1
to the right.
A ball o mass 0.2 5 kg is moving to the right at a speed o 7.4 m s 1 . It strikes a wall at 90 and rebounds rom the wall leaving it with a speed o 5 .8 m s 1 moving to the let. C alculate the change in momentum.
Solution From the previous example the initial momentum is 1 .85 kg m s 1 to the right. The fnal momentum is 0.25 5.8 = 1 .45 kg m s 1 to the let. So taking the direction to the right as being positive, the change in momentum = 1 .45 ( + 1 .85 ) = 3 .3 kg m s 1 to the right ( or alternatively + 3 .3 kg m s 1 to the let) .
Collisions and changing momentum You may have seen a Newtons cradle. Newton did not invent this device (it was developed in the twentieth century as an executive toy) , but it helps us to visualize some important rules relating to his laws o motion, and it certainly seems appropriate to associate his name with it.
Figure 2 Newtons cradle.
O ne o the balls ( the right- hand one in fgure 2 ) is moved up and to the right, away rom the remaining our. When released the ball alls back and hits the second ball rom the right. The right-hand ball then stops moving and the let-most ball moves o to the let. It is as though the motion o the original ball transers through the middle three which remain stationary and appears at the let-hand end. You can analyse this in terms o physics that you already know. The right-hand ball gains gravitational potential energy when it is raised. When it is released the potential energy is converted into kinetic energy as the ball gains speed. Eventually the ball strikes the next stationary one along, and a pair o actionreaction orces acts at the suraces o the two balls (you might like to consider what they are) . The stationary ball is compressed slightly by the orce acting on it and this compression moves as a wave through the middle three balls. When the compression reaches the
74
2.4 M OM EN TUM
let-hand ball the elastic potential energy associated with the compression wave is converted into kinetic energy (because there is a net orce on this sphere) and work begins to be done against the Earths gravity. The lethand ball starts to move to the let. When the speed o the ball becomes zero, it is at its highest point on the let-hand side and the motion repeats in the reverse direction. The overall result is that the end spheres appear to perorm an alternate series o hal oscillations. Another way to view this sequence is as a transer o momentum. Think about a simpler case where only two spheres are in contact (in the toy, three balls can be lited out o the way) . The right-hand sphere gains momentum as it alls rom the top o its swing. When it collides with the other sphere, the momentum appears to be transerred to the second sphere. The frst sphere now has zero momentum (it is stationary) and the second has gained momentum. What rules govern this transer o momentum? These interactions between the balls in the Newtons cradle are called collisions. This is the term given to any interaction where momentum transers. E xamples o collisions include fring a gun, hitting a ball with a bat in sport, two toy cars running into each other, and a pile driver sinking vertical cylinders into the ground on a construction site. There are many more.
Investigate! Is momentum conserved? The exact details o this experiment will depend on the apparatus you have in your school. light gate stationary cart
timing card moving cart raised to compensate for friction (see page 68)
The experiment consists o measuring the speed o a cart o known mass and then launching it at another cart also o known mass that is initially stationary. O ten there will be carts available to you o almost identical mass, as this is a particularly easy case to begin with. You will need a way to measure the velocity o the carts just beore and just ater the collision. This could be done in various ways:
using a data logger with motion sensors
using a paper tape system where a tape attached to the cart is pulled through a device that makes dots at regular time intervals on the tape
using a video camera and computer sotware
using a stop watch to measure the time taken to cover a short, known distance beore and ater the collision.
I your carts run on a track, you can allow or the riction at the cart axles and the air resistance. This is done by raising the end o the track a little so that, when pushed, a cart runs at a constant speed. The riction at the bearings and the air resistance will be exactly compensated by the component o the cart weight down the track.
For the frst part o the experiment, begin by arranging that the two carts will stick together ater colliding. This can be done in a number o ways, including using modelling clay, or a pin attached to one cart entering a piece o cork in the other, or two magnets ( one on each cart) that attract.
Make the frst ( moving) cart collide with and stick to the second ( stationary) cart.
Measure the speed o the frst cart beore the collision and the combined speed o the carts ater the collision.
Repeat the experiment a number o times and think careully about the likely errors in the results.
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The initial momentum is mass o frst cart velocity o frst cart. The fnal momentum is (mass o frst cart + mass o second cart) combined velocity o both carts.
and the momentum ater the collision are approximately equal.
What can you say about the total momentum o the system beore the collision compared with the total momentum ater the collision? You should consider the experimental errors in the experiment beore making your j udgement. I you have done the experiment careully, you should fnd that the momentum beore
Now extend your experiment to dierent cases:
where the carts do not stick together
where they are both moving beore the collision
where the masses are not the same, and so on.
You may need to alter how you measure the velocity to cope with the dierent cases.
I you carry out an experiment that compares the momentum beore with the momentum ater a collision then you should fnd ( within the experimental uncertainty o your measurements) that the total momentum in the system does not change. An important point here is that there must be no orce rom outside the system acting on the obj ects taking part in the collision. As we saw earlier, external orces produce accelerations, and this changes the velocity and hence the momentum. You might argue that gravitational orce is acting and gravity certainly is acting on the carts in the experiment but because the gravitational orce is not being allowed to do any work, the orce does not contribute to the interaction because the carts are not moving vertically. Momentum is always constant if no external force acts on the system. This is known as the p rincip le of conservation of linear momentum, the word linear is here because this is momentum when the obj ects concerned are moving in straight lines. This conservation rule has never been observed to be broken, and is one o the important conservation rules that ( as ar as we know) are true throughout the universe. In nuclear physics, particles have been proposed in order to conserve momentum in cases where it was apparently not being conserved. These particles were subsequently ound to exist. Momentum conservation is such an important rule that it is worth us considering a ew dierent situations to see how momentum conservation works. In each o these cases we will assume that the centres o the obj ects lie on a straight line so that the collision happens in one dimension.
Two objects with the same mass, one initially stationary, when no energy is lost This is known as an elastic collision and or it to happen no permanent deormation must occur in the colliding obj ects and no energy can be released as internal energy ( through riction) , sound, or any other way. The spheres in the Newtons cradle lose only a little energy every time they collide and this is why this is a reasonable demonstration o momentum eects.
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2.4 M OM EN TUM u
beore:
mass m 1
0 m s 1
TOK
mass m 2
Popper and alsifability
mass m 1
ater:
mass m 2
0 m s 1 v= u
Figure 3 Elastic collision between two identical masses.
The frst obj ect collides with the second stationary obj ect. The frst obj ect stops and remains at rest while the second moves o at the speed that the frst obj ect had beore the collision. ( Try icking a coin across a smooth table to hit an identical coin head-on.) In this case momentum is conserved because ( using an obvious set o symbols or the mass m o the obj ects and their velocities u and v) m 1 u = m2v B ecause m 1 = m 2 then u = v so the velocity o one mass beore the collision is equal to the velocity o the second mass aterwards. The 1 kinetic energy o the moving mass ( whichever mass is moving) is __ mu 2 2 and it does not change either.
Two objects with diferent masses when no energy is lost This time ( as you may have seen in an experiment or demonstration) the situation is more complicated. u1 u2 beore:
m1 m2
v1 v2 ater:
m1 m2
No-one has yet observed a case where the momentum in an isolated system is not conserved, but we should continue to look! Karl Popper, a philosopher o the twentieth century, argued that the test o whether a theory was truly scientifc was that it was capable o being alsifed. By this he meant that there has to be an experiment that could in principle contradict the hypothesis being tested. Popper argued that psychoanalysis was not a science because it could not be alsifed by experiment. Popper also applied his ideas to scientifc induction. He said that, although we cannot prove that the Sun will rise tomorrow, because it always has we can use the theory that the Sun rises in the morning until the day when it ails to do so. At that point we must revise our theory. What should happen i momentum appears not to be conserved in an experiment? Is it more sensible to look or a new theory or to suggest that something has been overlooked?
Figure 4 Two moving objects with diferent mass in an elastic collision.
Again m 1 u 1 + m 2 u 2 = m 1 v1 + m 2 v2 but this time we cannot eliminate the mass terms so easily. What we do know, though, is that kinetic energy is conserved. S o the kinetic energy beore the collision must equal the kinetic energy ater the collision ( because no energy is lost) . This means that ( summing the kinetic energies beore and ater the collision) 1 1 1 1 __ m 1 u 21 + __ m 2 u 22 = __ m 1 v21 + __ m 2 v22 2 2 2 2
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2
M E C H AN I C S The momentum and kinetic energy equations can be solved to show that
( (
) ( ) (
) )
m1 m2 2 m2 v1 = _ u 1 + _ u 2 m1 + m2 m1 + m2 and
m2 m1 2 m1 v2 = _ u 2 + _ u 1 m1 + m2 m1 + m2 ( You will not be expected to prove these in, or memorize them or, an examination. ) There are some interesting cases when mass m 2 is stationary and is struck by m 1 :
C ase 1 : m 1 is much smaller than m 2 . Look at the v1 equation. I m 1 m 2 m then the frst term becomes roughly _____ u 1 which is u 1 ; the second m term is zero because u 2 = 0. So the small mass bounces o the large mass, reversing its direction ( shown by the minus sign) . The large mass gains speed in the orward direction (the original direction o the small mass) . The magnitude o the speed o the larger mass is roughly 2m ___ u 1 and this is a small raction o the original speed o the small m 2m mass because ___ is much smaller than 1 . m
( ) 2
2
( ) 1
2
( ) 1
2
C ase 2 : m 1 is much greater than m 2 . Here the original mass loses hardly any speed ( though it must lose a little) . The momentum lost by m 1 is given to m 2 which moves o in the same direction, but at about twice the original speed o m 1 . Look at the v1 and v2 equations and satisy yoursel that this is true.
Two objects colliding when energy is lost When a moving object collides with a stationary one and the two objects stick together, then some o the initial kinetic energy is lost. Ater the collision, there is a single object with an increased (combined) mass and a single common velocity. This is known as an inelastic collision (fgure 5) . This is a case you may have studied experimentally in the Investigate! The momentum equation this time is m 1 u 1 = ( m 1 + m 2 ) v1 m A rearrangement shows that v1 = _______ u 1 and, as we might expect, (m + m ) 1
1
2
the fnal velocity is in the same direction as beore but is always smaller than the initial velocity. u1 before:
0
m1
m2 v1
after:
78
m1 + m2
Figure 5 Inelastic collision between two masses.
As or energy loss, the incoming kinetic energy is kinetic energy is ( substituting or v1 ) m 21 1 __ __ ( m + m ) u2 1 2 2 ( m1 + m 2) 2 1 This is m 12 1 _ __ u 21 . 2 (m + m ) 1 2 ( m1 + m2) initial kinetic energy The ratio __ is _ m1 fnal kinetic energy
__1 m u 2
1
2 1
and the fnal
2.4 M OM EN TUM
Two objects when energy is gained There are many occasions when two initially stationary masses gain energy in some way. S ome laboratory dynamics carts have a way to show this. An easy way is to attach two small strong magnets to the ront o the carts so that when the carts are released ater being held together with like poles o the magnets acing each other, the magnets repel and drive the carts apart. The analysis is quite straightorward in this case. The initial momentum is zero as neither obj ect is moving. Ater the collision the momentum is m 1 v1 + m 2 v2 = 0.
before:
0
0
m1
m2
S o m 1 v1 = m 2 v2 The obj ects move apart in opposite directions. I the masses are equal the speeds are the same, with one velocity the negative o the other. I the masses are not equal then v2 m _ m 2 = v1
1 _
after:
v1
v2
m1
m2
Figure 6 Energy gained in a collision.
TOK Helpful or not? Are conservation laws helpul to scientists? On the one hand they allow the prediction o as yet untested cases, but on the other hand they may restrict the progress o science. This can happen i scientists are not prepared to challenge the status quo. In 2012 the results o an experiment suggested that neutrinos could travel aster than the speed o light.
This few in the ace o the accepted science originally proposed by Einstein. Later investigations showed that small errors in the timings had occurred in the experiments, and that there was no evidence or aster-than-light travel. Were the scientists right to publish their results so that others could test the new proposals?
Worked examples 1
A rail truck o mass 45 00 kg moving at a speed o 1 .8 m s 1 collides with a stationary truck o mass 1 5 00 kg. The two trucks couple together. C alculate the speed o the trucks immediately ater the collision.
Solution The intial momentum = 45 00 1 .8 kg m s 1 . The fnal momentum = ( 45 00 + 1 5 00) v, where v is the fnal speed. 45 0 0 1 . 8 Momentum is conserved and so v = __________ ( 45 0 0 + 1 5 0 0 ) 4. 5 1 . 8 _______ 1 = = 1 .3(5 ) = 1 .4 m s . 6.0
2
S tone A o mass 0. 5 kg travelling at 3 . 8 m s 1 across the surace o a rozen pond collides
with stationary stone B o mass 3 .0 kg. S tone B moves o at a speed o 0.65 m s 1 in the same original direction as stone A. C alculate the fnal velocity o stone A.
Solution The initial momentum is 0.5 3 .8 = 1 .9 kg m s 1 The fnal momentum is 3 .0 0.65 + 0.5 vA and this is equal to 1 .9 kg m s 1 1 .9 ( 3 .0 0.65 ) 0.05 vA = __ = _ = 0.1 kg m s 1 0.5 0.5 The fnal velocity o stone A is 0.1 kg m s 1 in the opposite direction to its original motion.
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3 before: after:
2.5 m s 1
0 m s 1
v m s 1
1.9 m s 1
6000 kg
3000 kg
A railway truck o mass 6000 kg collides with a stationary truck o mass 3000 kg. The frst truck moves with an initial speed o 2.5 m s 1 and the second truck moves o with a speed o 1 .9 m s 1 in the same direction. C alculate:
Solution a) The initial momentum is 6000 2 .5 = 1 5 000 kg m s 1 . The fnal momentum is 6000v + 3 000 1 .9 Equating these values: 1 5 000 = 6000v + 5 700, so 6000v = 93 00 and v = 1 .5 (5 ) kg m s 1 . This is positive so the frst truck continues to move to the right ater the collision at a speed o 1 .6 kg m s 1 . 1 6000 2 .5 2 b) The initial kinetic energy = __ 2 = 1 8 75 0 J.
Ater the collision the total kinetic energy =
a) the velocity o the frst truck immediately ater the collision b) the loss in kinetic energy as a result o the collision.
1 1 __ 6000 1 .5 5 2 + __ 3 000 1 .9 2 2 2
= 72 08 + 5 41 5 = 1 2 62 3 or to 2 s. . 1 3 000 J. S o the loss in kinetic energy is 1 8 75 0 1 3 000 = 5 800 J.
Energy and momentum There is a convenient link between kinetic energy and momentum. 1 Kinetic energy, EK= __ mu 2 ; momentum, p = mu and thereore p 2 = m 2 u 2 . 2 So p2 EK = _ 2m
This is oten o use in calculations.
Applications of momentum conservation Recoil of a gun Figure 7 shows a gun being fred to trigger a snow all. This prevents a more dangerous avalanche. When the gun fres its shell, the gun moves backwards in the opposite direction to that in which the shell goes. You should be able to explain this in terms o momentum conservation. Initially, both gun and shell are stationary; the initial total momentum is zero. The shell is propelled in the orwards direction through the gun by the expansion o gas ollowing the detonation o the explosive in the shell. S o this is one o the cases discussed earlier, one in which energy is gained. The explosion is a orce internal to the system. Gas at high pressure is generated by the explosion in the chamber behind the shell. The gas exerts a orce on the interior o the shell chamber and hence a orce on the gun as well. The explosive releases energy and this is transerred into kinetic energies o both the shell and the gun.
80
Figure 7 Firing a gun to cause an avalanche.
The initial linear momentum was zero and no external orce has acted on the system. The momentum must continue to be zero and this can only be so i the gun and the shell move in opposite directions with the same magnitude o momentum. The shell will go ast because it has a small mass compared to the gun; the gun moves relatively slowly.
2.4 M OM EN TUM
Water hoses Watch a fre being extinguished by fremen using a high- pressure hose and you will see the eect o water leaving the system. O ten two or more fremen are needed to keep the hose on target because there is a large orce on the hose in the opposite direction to that o the water. This can be seen in a more modest orm when a garden hose connected to a tap starts to shoot backwards in unpredictable directions i it is not held when the tap is turned on. The cross- sectional area o the hose is greater than that o the nozzle through which the water emerges. The mass o water owing past a point in the hose every second is the same as the mass that emerges rom the nozzle every second. S o the speed o the water emerging rom the nozzle must be greater than the water speed along the hose itsel. The water gains momentum as it leaves the hose because o this increase in exit speed compared with the ow speed in the hose.
Figure 8 Fire fghting.
The momentum o the system has to be constant and so there must be a orce backwards on the end o the hose which needs to be countered by the eorts o the fremen. The kinetic energy and the momentum are being supplied by the water pump that eeds water to the fre hose or whatever originally created the pressure in the supply to the garden tap. The momentum lost by the system per second is (mass of water leaving per second) (speed at which water leaves the nozzle speed in the hose). m The mass o water lost per second is ___ and so the momentum lost per t m ___ second p is t ( v u) where v is the speed o water as it leaves the nozzle and u is the speed o the water in the hose.
v A, cross-sectional area o nozzle
v
u
one second o water
Figure 9 Water leaving the hose.
I we know the cross- sectional area A o the nozzle o the hose and the m density o the water, , then ___ can be determined. Figure 9 shows what t happens inside the hose during a one-second time interval. Every second you can imagine a cylinder o water leaving the hose; this cylinder is v long and has an area A. The volume leaving per second is thereore Av m The mass o the water leaving in one second ___ t
= (density of water) (volume of cylinder) = Av. B ecause the mass entering and leaving the nozzle per second must be the same, this means that the change o momentum in one second m = ___ ( v u) = Av( v u) . I u v then the expression simplifes t to Av2 . The hose is an example o where care is needed when looking at the whole system. C onsider what happens i the water is directed at a vertical wall. The water strikes the wall, loses all its horizontal momentum, and trickles vertically down the wall. The momentum must
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Worked example A mass o 0.48 kg o water leaves a garden hose every second. The nozzle o the hose has a cross- sectional area o 8.4 1 0 5 m 2 . The water ows in the hose at a speed o 0.71 m s 1 . The density o water is 1 000 kg m 3 . C alculate: a) the speed at which water leaves the hose b) the orce on the hose.
Solution a) 0.48 kg o water in one second corresponds to 0 . 48 m 3 a volume o ____ 1 000 leaving the hose per second. This leaves through a nozzle o area 8.4 1 0 5 m 2 so the speed v must be 0 . 48 _____________ = 1 0 0 0 8 . 4 1 0 5
5.71 m s 1 . b) The orce on the hose = mass lost per second change in speed = 0.48 (5 .71 0.71 ) = 2.4 N
have gone into the wall, its oundations and, thereore, the ground. S o we might conclude that the E arth itsel has gained momentum and that we can speed up the Earths rotation by using a garden hose. B ut this is not true, because the water had to be given momentum originally by a pump. This gain in momentum at the pump must have given some momentum to the Earth too. The amount o momentum the E arth gained at the pump is equal and opposite to the momentum gained by the Earth when the water strikes the wall.
Rocketry Earlier in this topic we discussed the acceleration o a rocket and looked at the situation rom the perspective o Newtons second and third laws. A similar analysis is possible in terms o momentum conservation. Rockets operate eectively in the absence o an atmosphere because momentum is conserved. All rockets release a liquid or gas at high speed. The uid can be a very hot gas generated in the combustion o a solid chemical ( as in a domestic rocket) or rom the chemical reaction when two gases are mixed and burnt. O r it can be a uid stored inside the rocket under pressure. In each case, the uid escapes rom the combustion/storage chamber through nozzles at the base o the rocket. As a result the rocket accelerates in the opposite direction to the direction in which the uid is ej ected. Momentum is conserved. The rate o loss o momentum rom the rocket in the orm o high- speed uid must be equal to the rate o gain in momentum o the rocket. We shall look again at the mathematics o the rocket later in this topic.
Helicopters Helicopters are aircrat that can take o and land vertically and also can hover motionless above a point on the ground. Vertical ight is said to have been invented in C hina about 2 5 00 years beore the present, but anyone who has seen the seeds rom certain trees spiral down to the ground will realize where the original design came rom. There were many attempts to build ying machines on the helicopter principle over the centuries, but the frst commercial aircrat ew in the 1 93 0s. A helicopter uses the principle o conservation o linear momentum in order to hover. The rotating blades exert a orce on originally stationary air causing it to move downwards towards the ground gaining momentum in the process. No external orce acts and as a result there is an upward orce on the helicopter through the rotors.
Impulse Earlier we used Newtons second law o motion: F = ma where the symbols have their usual meaning. We can rearrange this equation using one o the kinematic equations: (v u) a = _____ t
Figure 10 Helicopter.
m( v u) Eliminating a gives F = _ t change in momentum which means that force = __ time taken for change or force time = change in momentum
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2.4 M OM EN TUM
This equation gives a relationship between orce and momentum and provides a urther clue to the real meaning o momentum itsel. The equation shows that we can change momentum ( in other words, accelerate an obj ect) by exerting a large orce or a short time or by exerting a small orce or a long time. A small number o people can get a heavy vehicle moving at a reasonable speed, but they have to push or a much longer time than the vehicle itsel would take i powered by its own engine ( which produces a larger orce) .
Worked examples 1
The product o force and time is given the name imp ulse, its units are newton seconds ( N s) . Impulse is the same as change in momentum, and N s gives us an alternative to kg m s 1 as a unit or momentum. You can use whichever you preer.
An impulse o 85 N s acts on a body o mass 5 .0 kg that is initially at rest. C alculate the distance moved by the body in 2 .0 s ater the impulse has been delivered.
Solution The change in momentum is 85 kg m s 1 so that the fnal 85 speed is __ = 1 7 m s 1 . In 5 2.0 s the distance travelled is 34 m.
ch an ge in m om en tu m
There is a short-hand way to write the force = ______________ equation. tim e We have already used the convention that means change in. S o in symbols the equation becomes p F= _ t
2
where p is the symbol or momentum and t ( as usual) means time. ( I you are amiliar with dierential calculus you may also come across dp this expression in the orm F = __ , but using the equation in this orm dt will take us too ar away rom IB D iploma Programme physics. )
An impulse I acts on an object o mass m initially at rest. D etermine the kinetic energy gained by the object.
Solution
Forcetime graphs
The change in speed o the I obj ect is __ m . The obj ect is initially at rest and the gain 1 in kinetic energy is __ mv2 2 2 I 1 I ___ m _ which is __ 2 m = 2m
Up till now we have usually assumed that orces are constant and do not change with time. This is rarely the case in real lie and we need a way to cope with changes in momentum when the orce is not constant. p The equation F = ___ helps here because it suggests that we use a orce t time graph.
( )
2
force
force
area = F T
force
Fmax
F
area = 12 Fmax T 0
0
time (a)
T
0
0
time
T
0
0
(b)
time
T
(C)
Figure 11 Forcetime graphs.
I a constant orce acts on a mass, then the graph o orce, F, against time, t, will look like fgure 1 1 ( a) . The change in momentum is F T and this is the shaded area below the line. The area below the line in a orcetime graph is equal to the change in momentum. Another straightorward case that is more plausible than a constant orce is the graph shown in fgure 1 1 ( b) where the orce rises to a maximum Fmax and then alls back to zero in a total time T. The area under the 1 graph this time is __ F T and, again, this is the change in momentum. 2 max
(
)
The fnal case ( fgure 1 1 (c) ) is one where there is no obvious mathematical relationship between F and t, but nevertheless we have
83
2
M E C H AN I C S a graph o how F varies with t. This time you will need to estimate the number o squares under the graph and use the area o one square to evaluate the momentum change.
Worked examples 1
b) the change in kinetic energy over the 1 0 s o the motion.
The sketch graph shows how the orce acting on an obj ect varies with time.
200 180
15 force/N
160
0
0
5
20
25
time/s
p/kg m s 1
140 120 100 80
The mass o the obj ect is 5 0 kg and its initial speed is zero.
60 40
C alculate the fnal speed o the obj ect. 20
Solution The total area under the F t graph is 1 2 __ 1 5 5 + ( 1 5 1 5 ) = 75 + 2 2 5 2 = 3 00 N s
(
)
This is the change in momentum.
0
1
2
3
4
5 t/s
Solution a) The gradient o the graph is equal to the orce.
300 So the fnal speed is ___ = 6.0 m s 1 50
2
0
The graph shows how the momentum o an obj ect o mass 40 kg varies with time.
6
7
8
9
10
p ___ and this is t
200 The magnitude o the gradient is ___ and this 10 gives a value or orce o 2 0 N. 2
p 200 b) EK = ___ = _____ = 5 00 J 2m 2 40
C alculate, or the obj ect:
2
a) the orce acting on it
Revisiting Newtons second law p In the last sub- topic we used F = ma to show that F also equals ___ . Using t the ull expression or momentum p gives ( m v) F = _____ t
This can be written as ( using the product rule) v m F= m_+ v_ t t You may have to take this algebra on trust but, by thinking through what the two terms stand or, you should be able to understand the physics that they represent. ch an ge in velocity The frst term on the let-hand side is j ust mass ___________ which we ch an ge in tim e
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know as mass acceleration our original orm o Newtons second law o motion. The second term on the right-hand side is something new, it is the change in m ass . instantaneous velocity __________ change in tim e
2.4 M OM EN TUM
Our frst version o Newtons second law was a simpler orm o the law than the ull change in momentum version. The new extra term takes account o what happens when the mass o the accelerating object is changing.
Rockets and helicopters again Rockets We showed that rockets are an excellent example o momentum conservation in action. The difculty in analysing the rocket example is that the rocket is always losing mass ( in the orm o gas or liquid propellant) , so m is not constant. mv v m F = ____ + ____ is our new version o Newtons second law o motion t t and, in this case, F = 0 because there is no external orce acting on the system. mv v m mv S o ____ = ____ . The two terms have quite separate meanings: ____ t
t
t
reers to the instantaneous mass o the rocket ( including the remaining v uel) and to the acceleration o this total mass ___ . The other term t is the ej ection speed o the uel and the rate at which mass is lost m rom the system ___ . S o at one instant in time, the acceleration o t the rocket
Worked example A small frework rocket has a mass o 3 5 g. The initial rate at which hot gas is lost rom the frework has been lit is 3.5 g s 1 and the speed o release o this gas rom the rear o the rocket is 1 3 0 m s 1 . C alculate the initial acceleration o the rocket.
Solution v v m a = ___ = ____ where v t mt is the release speed o the m gas, ___ is the rate o loss t o gas, and m is the mass o the rocket 1 30 3.5 a = _______ = 7. 0 m s 2 65
v v m a = ___ = ____ . t mt
The negative sign reminds us that the rocket is losing mass while gaining speed.
Helicopters With the helicopter hovering, there is a weight orce downwards and so the equation now becomes mv v m Mg = ____ + ____ where M is the mass o the helicopter t t
There is no change in the speed o the helicopter (it is hovering) so v = 0, there is however momentum gained by the air that moves downwards. This is the speed, v, the air gains multiplied by the mass o the air accelerated m every second so Mg = v___ or t weight of helicopter ( in N) = mass of air pushed downwards per second ( in kg s 1 ) speed of air downwards (in m s 1 )
Momentum and safety Many countries have made it compulsory to wear seat belts in a moving vehicle. Likewise, many modern cars have airbags that inate very quickly i the car is involved in a collision. O bviously, both these devices restrain the occupants o the car, preventing them rom striking the windscreen or the hard areas around it. B ut there is more to the physics o the air bag and the seat belt than this. O n the ace o it, someone in a car has to lose the same amount o kinetic energy and momentum whether they are stopped by the windscreen or restrained by the seat belt. What diers in these two
Figure 12 Car safety: seat belts and air bags.
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2
M E C H AN I C S cases is the time during which the loss o energy and momentum occur. Unrestrained, the time to stop will be very short and the deceleration will thereore be very large. A large deceleration means a large orce and it is the magnitude o the orce that determines the amount o damage. S eat belts and air bags dramatically increase the time taken by the occupants o the car to stop and as force time = momentum change, or a constant change in momentum, a long stopping time will imply a smaller, and less damaging, orce.
Momentum and sport This sub- topic began with a suggestion that it was less painul to catch a table-tennis ball than a baseball. You should now be able to understand the reason or the dierence. You should also realize why good technique in many sports hinges on the application o momentum change. Many sports in which an obj ect usually a ball is struck by hand, oot or bat rely on the efcient transer o momentum. This transer is oten enhanced by a ollow through, which increases the contact time between bat and ball. The player maintains the same orce but or a longer time, so the impulse on the ball will increase, increasing the momentum change as well. Think about your sport and how eective use o momentum change can help you.
Investigate! Estimating the force on a soccer ball This experiment will allow you to estimate the orce used to kick a soccer ball. It uses many o the ideas contained in this topic and is a good place to conclude our look at IB mechanics!
To measure the contact time: stick some aluminium oil to the shoe o the kicker and to the soccer ball. S et up a data logger or asttimer so that it will measure the time T or which the two pieces o oil are in contact.
To measure the change in momentum: the ball starts rom rest so all you need is the magnitude o the fnal momentum. Get the kicker to kick the ball horizontally rom a lab bench. Measure the distance s rom where the ball is kicked to where it lands. Measure the distance h rom the bottom o the ball on the bench to the oor. Use the ideas o projectile motion to calculate (i) the time t taken or the ball __ to reach the
h
s aluminum foil foot ball
[
to timer
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The basis o the method is to measure the contact time between the oot and the ball and the subsequent change in momentum o the ball. The use o force contact time = change in momentum allows the orce to be calculated.
]
2h 1 oor h = __ gt2 and so t = __ g . Then using 2 this t, the initial speed u o the ball can be estimated as _st . Measure the mass o the ball M and thereore the change in momentum is Mu which is equal to the force on the ball T.
This method can be modifed or many sports including hockey, baseball, and gol.
QUESTION S
Questions 1
3
(IB) Christina stands close to the edge o a vertical cli and throws a stone at 1 5 m s 1 at an angle o 45 to the horizontal. Air resistance is negligible.
A marble is projected horizontally rom the edge o a wall 1 .8 m high with an initial speed V. A series o fash photographs are taken o the marble and combined as shown below. The images o the marble are superimposed on a grid that shows the horizontal distance x and vertical distance y travelled by the marble.
P
15 m s 1 O
(IB)
The time interval between each image o the marble is 0.1 0 s.
Q
0
0.50
x/m 1.0
1.5
2.0
0 25 m
0.50
y/m
sea
Point P on the diagram is the highest point reached by the stone and point Q is at the same height above sea level as point O . C hristinas hand is at a height o 2 5 m above sea level.
1.50
a) At point P on a copy o the diagram above draw arrows to represent:
2 .0
( i) the acceleration o the stone ( label this A)
Use data rom the photograph to calculate a value o the acceleration o ree all. ( 3 marks)
( ii) the velocity o the stone ( label this V) . b) D etermine the speed with which the stone hits the sea. ( 8 marks) 2
1.0
4
( IB)
(IB)
A cyclist and his bicycle travel at a constant velocity along a horizontal road.
Antonia stands at the edge o a vertical cli and throws a stone vertically upwards.
a) ( i) S tate the value o the resultant orce acting on the cyclist.
The stone leaves Antonias hand with a speed v = 8.0 m s 1 . The time between the stone leaving Antonias hand and hitting the sea is 3 . 0 s. Assume air resistance is negligible.
( ii) C opy the diagram and draw labelled arrows to represent the vertical orces acting on the bicycle.
a) C alculate: ( i) the maximum height reached by the stone ( ii) the time taken by the stone to reach its maximum height. b) D etermine the height o the cli. ( 6 marks)
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2
M E C H AN I C S ( iii) E xplain why the cyclist and bicycle are travelling at constant velocity.
1.4 1.2
b) The total mass o the cyclist and bicycle is 70 kg and the total resistive orce acting on them is 40 N. The initial speed o the cycle is 8.0 m s 1 . The cyclist stops pedalling and the bicycle comes to rest.
P/kW
1
0.4 0 0
( ii) Estimate the distance taken by the bicycle to come to rest rom the time the cyclist stops pedalling.
0.5
1
1.5
2 v/m s -1
2.5
3
3.5
For a steady speed o 2 .0 m s 1 : a) use the graph to determine the power o the boats engine
( iii) State and explain one reason why your answer to b) ( ii) is an estimate.
b) calculate the rictional (resistive) orce acting on the boat. (3 marks)
( 1 3 marks)
A car o mass 1 000 kg accelerates on a straight fat horizontal road with an acceleration a = 0.30 m s 2 . The driving orce T on the car is opposed by a resistive orce o 5 00 N. C alculate T. (3 marks)
0.6 0.2
( i) C alculate the magnitude o the initial acceleration o the bicycle and rider.
5
0.8
8
(IB) The graph shows the variation with time t o the speed v o a ball o mass 0.5 0 kg that has been released rom rest above the Earths surace. 25
A crane hook is in equilibirium under the action o three orces as shown in the diagram. 3.8 kN
20 v/m s -1
6
15 10 5
T2
60
30
0
T1
0
2
4
6
8
10
t/s
The orce o air resistance is not negligible. a) S tate, without any calculations, how the graph could be used to determine the distance allen. C alculate T1 and T2 .
7
( 4 marks)
b) ( i)
C opy the diagram and draw and label arrows to represent the orces on the ball at 2 .0 s.
(IB) A small boat is powered by an outboard motor o variable power P. The graph below shows the variation with speed v o the power P or a particular load.
ball at t = 2.0s
Earths surface
( ii) Use the graph to show that the acceleration o the ball at 2 .0 s is approximately 4 m s 2 .
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QUESTION S
( iii) C alculate the magnitude o the orce o air resistance on the ball at 2 . 0 s.
5.0 m s 1
( iv) State and explain whether the air resistance on the ball at t = 5 .0 s is smaller than, equal to, or greater than the air resistance at t = 2 . 0 s.
B
A immediately before collision
c) Ater 1 0 s the ball has allen 1 90 m. ( i)
v
S how that the sum o the potential and kinetic energies o the ball has decreased by about 800 J. ( 1 4 marks)
B
A immediately after collision
9
(IB) A bus is travelling at a constant speed o 6.2 m s 1 along a section o road that is inclined at an angle o 6.0 to the horizontal. 6.2m s 1
The mass o truck A is 800 kg and the mass o truck B is 1 2 00 kg. a) ( i) C alculate v. ( ii) C alculate the total kinetic energy lost during the collision. b) S uggest where the lost kinetic energy has gone. ( 6 marks)
6.0
a) ( i)
1 1 (IB) D raw a labelled sketch to represent the orces acting on the bus.
( ii) S tate the value o net orce acting on the bus. b) The total output power o the engine o the bus is 70 kW and the efciency o the engine is 3 5 % .
Large metal bars are driven into the ground using a heavy alling obj ect. object mass = 2.0 10 3 kg
C alculate the input power to the engine. c) The mass o the bus is 8.5 1 0 3 kg. D etermine the rate o increase o gravitational potential energy o the bus.
bar mass = 400 kg
d) Using your answer to c ( and the data in b) , estimate the magnitude o the resistive orces acting on the bus. ( 1 2 marks) The alling obj ect has a mass 2 000 kg and the metal bar has a mass o 400 kg. 1 0 (IB) Railway truck A moves along a horizontal track and collides with a stationary truck B . The two j oin together in the collision. Immediately beore the collision, truck A has a speed o 5 .0 m s 1 . Immediately ater collision, the speed o the trucks is v.
The obj ect strikes the bar at a speed o 6.0 m s 1 . It comes to rest on the bar without bouncing. As a result o the collision, the bar is driven into the ground to a depth o 0.75 m. a) D etermine the speed o the bar immediately ater the obj ect strikes it. b) D etermine the average rictional orce exerted by the ground on the bar. ( 7 marks)
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2
M E C H AN I C S 1 2 (IB) crane
An engine for a spacecraft uses solar power to ionize and then accelerate atoms of xenon. After acceleration, the ions are ej ected from the spaceship with a speed of 3 .0 1 0 4 m s 1 . xenon ions = 3.0 10 4 m s 1 speed
5.8 m
wall
spaceship mass = 5.4 10 2 kg
The mass of one ion of xenon is 2 .2 1 0
25
kg metal ball
a) The original mass of the fuel is 81 kg. D etermine how long the fuel will last if the engine ej ects 77 1 0 1 8 xenon ions every second.
To knock a wall down, the metal ball of mass 3 5 0 kg is pulled away from the wall and then released. The crane does not move. The graph below shows the variation with time t of the speed v of the ball after release.
2
b) The mass of the spaceship is 5 .4 1 0 kg. D etermine the initial acceleration of the spaceship. The graph below shows the variation with time t of the acceleration a of the spaceship. The solar propulsion engine is switched on at time t = 0 when the speed of the spaceship is 1 .2 1 0 3 m s 1 .
The ball makes contact with the wall when the cable from the crane is vertical. a) For the ball j ust before it hits the wall use the graph, to estimate the tension in the cable. The acceleration of free fall is 9.8 m s 2 .
10.0
b) D etermine the distance moved by the ball after coming into contact with the wall.
a/10 5 m s 2
9.5
c) C alculate the total change in momentum of the ball during the collision of the ball with the wall. ( 7 marks)
9.0 8.5
3.0
1.0
2.0
3.0 4.0 t/ 10 7 s
5.0
6.0
c) Explain why the acceleration of the spaceship increases with time. d) Using data from the graph, calculate the speed of the spaceship at the time when the xenon fuel has all been used. ( 1 5 marks)
1 3 (IB) A large metal ball is hung from a crane by means of a cable of length 5 .8 m as shown below.
90
2.0 v/m s 1
8.0 0.0
cable
1.0
0.0 0.0
0.2
0.4
0.6
0.8 t/s
1.0
1.2
1.4
3
T H E R M A L P H YS I C S
Introduction In this topic we look at at thermal processes resulting in energy transfer between obj ects at different temperatures. We consider how the energy transfer brings about further temperature changes and/or changes of state or phase.
We then go on to look at the effect of energy changes in gases and use the kinetic theory to explain macroscopic properties of gases in terms of the behaviour of gas molecules.
3.1 Temperature and energy changes Understanding Temperature and absolute temperature Internal energy Specifc heat capacity
Applications and skills The three states o matter are solid, liquid,
Phase change Specifc latent heat
Nature of science Evidence through experimentation Since early humans began to control the use o fre, energy transer because o temperature dierences has had signifcant impact on society. By controlling the energy ow by using insulators and conductors we stay warm or cool o, and we prepare lie-sustaining ood to provide our energy needs (see fgure 2) . Despite the importance o energy and temperature to our everyday lives, conusion regarding the dierence between thermal energy or heat and temperature is commonplace. The word heat is used colloquially to mean energy transerred because o a temperature dierence, but this is a throwback to the days in which scientists thought that heat was a substance and dierent rom energy.
and gas Solids have fxed shape and volume and comprise o particles that vibrate with respect to each other Liquids have no fxed shape but a fxed volume and comprise o particles that both vibrate and move in straight lines beore colliding with other particles Gases (or vapours) have no fxed volume or shape and move in straight lines beore colliding with other particles this is an ideal gas Thermal energy is oten misnamed heat or heat energy and is the energy that is transerred rom an object at a higher temperature by conduction, convection, and thermal radiation
Equations Conversion rom Celsius to kelvin:
T(K) = (C) + 273 Specifc heat capacity relationship: Q = mcT Specifc latent heat relationship: Q = mL
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3
T H E R M AL P H YS I C S
Introduction From the 1 7th century until the end o the 1 9th century scientists believed that heat was a substance that owed between hot and cold obj ects. This substance travelling between hot and cold obj ects was known as phlogiston or caloric and there were even advocates o a substance called rigoric that owed rom cold bodies to hot ones. In the 1 840s James Joule showed that the temperature o a substance could be increased by doing work on that substance and that doing work was equivalent to heating. In his paddle wheel experiment ( see fgure 1 ) he dropped masses attached to a mechanism connected to a paddle wheel; the wheel churned water in a container and the temperature o the water was ound to increase. Although the caloric theory continued to have its supporters it was eventually universally abandoned. The unit calorie which is sometimes used relating to ood energy is a residual o the caloric theory, as is the use o the word heat as a noun.
2 kg
when masses fall they turn the axle and cause the paddle wheels to churn up the water this raises the water temperature
2 kg
thermometer paddles
water
Figure 1 Joules paddle wheel experiment.
Temperature and energy transfer You may have come across temperature described as the degree o hotness o an obj ect. This is a good starting point since it relates to our senses. A pot o boiling water eels very hot to the touch and we know instinctively that the water and the pot are at a higher temperature than the cold water taken rom a rerigerator. The relative temperature o two obj ects determines the direction in which energy passes rom one obj ect to the other; energy will tend to pass rom the hotter obj ect to the colder obj ect until they are both at the same temperature ( or in thermal equilibrium) . The energy owing as a result o conduction, convection, and thermal radiation is what is oten called heat . Figure 2 Dinner is served!
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Temp erature is a scalar quantity and is measured in units o degrees C elsius ( C ) or kelvin ( K) using a thermometer.
3 . 1 T E M P E R AT U R E A N D E N E R G Y C H A N G E S
Nature of science Thermometers Most people are amiliar with liquid- in- glass thermometers in which the movement o a column o liquid along a scale is used to measure temperature. Thermometers are not j ust restricted to this liquid- in- glass type; others use the expansion o a gas, the change in electrical resistance o a metal wire, or the change in em ( electromotive orce) at the j unction o two metal wires o dierent materials. A thermometer can be constructed rom any obj ect that has a property that varies with temperature ( a therm om etric p rop erty) . In the case o a liquid- in- glass thermometer the thermometric property is the expansion o the liquid along a glass capillary tube. The liquid is contained in the bulb which is a reservoir; when the bulb is heated the liquid expands, travelling along the capillary. S ince the bore o the capillary is assumed to be constant, as the volume o the liquid increases with temperature so does the length o the liquid. S uch thermometers are simple but not particularly accurate. Reasons or the inaccuracy could be that the capillary may not be uniorm, or its cross- sectional area may vary with temperature, or it is difcult to make sure that all the liquid is at the temperature o the obj ect being investigated. Glass is also a relatively good insulator and it takes time or thermal energy to conduct through the glass to the liquid, making
the thermometer slow to respond to rapid changes in temperature. Liquid- in- glass thermometers are oten calibrated in degrees C elsius, a scale based on the scale reading at two fxed p oints, the ice point and the steam point. These two temperatures are defned to be 0 C and 1 00 C respectively ( although C elsius actually used the steam point or 0 and ice point or 1 00) . The manuacturer o the thermometer assumes that the length o the liquid in the capillary changes linearly with temperature between these two points, even though it may not actually do so. This is a undamental assumption made or all thermometers, so that thermometers only agree with each other at the fxed points between these points they could well give dierent values or the same actual temperature. D igital thermometers or temperature sensors have signifcant advantages over liquid- in- glass thermometers and have largely taken over rom liquid- in- glass thermometers in many walks o lie. The heart o such devices is usually a thermistor. The resistance o most thermistors alls with temperature ( they are known as ntc or negative temperature coefcient o resistance thermistors) . S ince the thermistor is usually quite small, it responds very quickly to temperature changes. Thermistor thermometers are usually ar more robust than liquid- in- glass ones. steam point
ice point melting ice
ask steam boiling water
unnel beaker
fnding the ice point
fnding the steam point
Figure 3 Calibration o a liquid-in glass thermometer at the ice point and steam point.
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3
T H E R M AL P H YS I C S
Investigate! Calibrating a thermistor against an alcohol-in-glass thermometer This investigation can be perormed by taking readings manually or by using a data logger. Here we describe the manual way o perorming the experiment. The temperature can be adjusted either using a heating coil or adding water at dierent temperatures (including some iced water, perhaps) .
Plot a graph o resistance/ against temperature/C ( since this is a calibration curve there can be no systematic uncertainties and any uncertainties will be random being sufciently small to be ignored in the context o this investigation) .
A multimeter set to ohms or an ohm-meter is connected across the thermistor (an ammeter/ voltmeter method would be a suitable alternative but resistance would need to be calculated) .
The graph shown is typical or a ntc thermistor.
S uggest why, on a calibration curve, systematic uncertainties are not appropriate.
Would the designers o a digital thermometer assume a linear relationship between resistance and temperature?
Would you expect the reading on a digital thermometer to correspond to that on a liquid- in- glass thermometer?
The thermistor is clamped so that it lies below the water surace in a S tyrooam ( expanded polystyrene) container next to an alcohol-inglass thermometer. O btain pairs o values o readings on the multimeter and the thermometer and record these in a table.
thermistor thermometer rubber seals cardboard square Styrofoam cups
resistance of themistor/
calibration of thermistor
2.0 10 4 1.5 10 4 1.0 10 4 0.5 10 4 0
water
0 multimeter
50
100
temperature/C
Absolute temperature The C elsius temperature scale is based on the ice point and steam point o water; by defnition all thermometers using the C elsius scale agree at these two temperatures. B etween these fxed points, thermometers with dierent thermometric properties do not all agree, although dierences may be small. The absolute temp erature scale is the standard S I temperature scale with its unit the kelvin ( K) being one o the seven S I base units. Absolute temp erature is defned to be zero kelvin at absolute zero ( the temperature at which all matter has minimum kinetic energy) and 2 73 . 1 6 K at the triple point o water ( the unique temperature and pressure at which water can exist as liquid water, ice, and water vapour) . D ierences in absolute temperatures exactly correspond to those in C elsius temperatures ( with a temperature dierence o 1 C being identical to 1 K) . For this reason
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it is usual to write the units o temperature dierence as K but not C ( although you are unlikely to lose marks in an examination or making this slip) . To convert rom temperatures in degrees C elsius to absolute temperatures the ollowing relationship is used: T( K) = ( C ) + 2 73 where T represents the absolute temperatures and the temperature in degrees C elsius. Other aspects o the absolute temperature scale will be considered in Sub-topic 3.2.
Internal energy S ubstances consist o particles ( e. g., molecules) in constant random motion. As energy is transerred to a substance the separation o the particles could increase and they could move aster. When particles move urther apart ( or closer to other neighbouring particles) the potential energy o the particles increases. As they move aster their random kinetic energy increases. The internal energy o a substance is the total o the p otential energy and the random kinetic energy o all the p articles in the substance. For a solid, these two orms o energy are present in roughly equal amounts; however in a gas the orces between the particles are so small that the internal energy is almost totally kinetic. We will discuss this urther in the S ub-topic 3 .2 .
Worked example A temperature o 73 K is equivalent to a temperature o A. 3 46 C . C . + 73 C .
B . 2 00 C . D . + 2 00 C .
Solution S ubstituting values into the conversion equation: 73 = + 2 73 so = 73 2 73 = 200 making the correct option B.
Note: Students oten conuse the terms internal energy and thermal energy. Avoid using the term thermal energy talk about internal energy and the energy transerred because o temperature dierences, which is what most people mean by thermal energy.
Specifc heat capacity When two dierent obj ects receive the same amount o energy they are most unlikely to undergo the same temperature change. For example, when 1 000 J o energy is transerred to 2 kg o water or to 1 kg o copper the temperature o each mass changes by dierent amounts. The temperature o the water would be expected to increase by about 0. 1 2 K while that o copper by 2 . 6 K. The two masses have dierent heat cap acities. You may think that it is hardly a air comparison since there is 2 kg o water and 1 kg o copper; and you would be right to think this! I we chose equal masses o the two substances we would discover that 1 kg o water would increase by 0. 2 4 K under the same conditions.
solid
liquid
In order to be able to compare substances more closely we defne the sp ecifc heat cap acity (c) o a substance as the energy transerred to 1 kg o the substance causing its temp erature to increase by 1 K. The defning equation or this is Q c= _ m T where Q is the amount o energy supplied to the obj ect o mass m and causing its temperature to rise by T. When Q is in J, m in kg, and T in K, c will be in units o J kg 1 K 1 . You should check that the data provided show that water has a specifc heat capacity o approximately 4.2 1 03 J kg 1 K1 and copper a value o approximately 380 J kg1 K1 .
gas
Figure 4 Motion of molecules in solids, liquids, and gases.
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Worked example A piece o iron o mass 0. 1 3 3 kg is placed in a kiln until it reaches the temperature o the kiln. The iron is then quickly transerred to 0. 476 kg o water held in a thermally insulated container. The water is stirred until it reaches a steady temperature. The ollowing data are available. Specifc heat capacity o iron = 45 0 J kg 1 K 1 Specifc heat capacity o water
= 4.2 1 0 3 J kg 1 K1
Initial temperature o the water = 1 6 C
b) C alculate the increase in internal energy o the water as the iron cools in the water. c) Use your answers to b) and c) to determine .
Solution a) Using Q = mc T or the iron Q = 0.1 33 45 0 ( 45 ) = 60 ( 45 ) b) Using Q = mc T or the water Q = 0.476 42 00 (45 1 6) = 5 .8 1 0 4 J c) Equating these and assuming no energy is transerred to the surroundings 60 ( 45 ) = 5 .8 1 0 4
Final temperature o the water = 45 C The specifc heat capacity o the container and insulation is negligible.
60 2 700 = 5 . 8 1 0 4
a) S tate an expression, in terms o and the above data, or the energy transer o the iron in cooling rom the temperature o the kiln to the fnal temperature o the water.
60 = 6.1 1 0 4 6. 1 1 0 4 = _ 60 = 1 000 C ( or 1 01 0 C to 3 s.. )
Investigate! Estimating the specifc heat capacity o a metal This is another investigation that can be perormed manually or by using a data logger. Here we describe the manual way o perorming the experiment. A metal block o known mass is heated directly by a heating coil connected to a low-voltage electrical supply. Channels in the block allow the coil to be inserted and also a thermometer or temperature probe. The power supplied to the block is calculated rom the product o the current (I) in the heating coil and the potential dierence (V) across it. The supply is switched on or a measured time (t) so that the temperature changes by at least 1 0 K. When the temperature has risen sufciently the power is switched o, but the temperature continues to be monitored to fnd the maximum temperature rise (it takes time or the thermometer to reach the same temperature as the surrounding block) .
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Thermal energy transerred to the block in time t = VIt
The heater itsel, the temperature probe and the insulation will all have heat capacities that will mean that there is a urther unknown term to the relationship: which should become VIt = ( mc + C) T where C is the heat capacity o everything except the block which will undergo the same temperature change as the block since they are all in good thermal contact. This term may be ignored i the mass o the block is high ( say, 1 kg or so) .
As the block is heated it will also be losing thermal energy to the surroundings and so make the actual rise in temperature lower
3 . 1 T E M P E R AT U R E A N D E N E R G Y C H A N G E S
that might be predicted. You are not expected to calculate cooling corrections in your IB calculations but it is important to recognize that this happens. A simple way o compensating or this is to cool the apparatus by, say 5 K, below room temperature beore switching on the power supply. I you then allow its temperature to rise to 5 K above room temperature, then you can assume that the thermal energy gained rom the room in reaching room temperature is equal to that lost to the room when the apparatus is above room temperature.
This method can be adapted to measure the specifc heat capacity o a liquid, but you must remember that the temperature o the container will also rise.
B to immersion heater circuit A 0100 C thermometer
aluminium block (calorimeter)
TOK
immersion heater
The development o understanding o energy transer due to temperature diferences insulating board
Specifc latent heat The temperature o a block o ice in a reezer is likely to be well below 0 C . I we measure the temperature o the ice rom the time it is taken rom the reezer until it has all melted and reached room temperature we would note a number o things. Initially the temperature o the ice would rise; it would then stay constant until all o it had melted then it would rise again but at a dierent rate rom beore. I we now put the water into a pan and heat it we would see its temperature rise quickly until it was boiling and then stay constant until all o the water had boiled away. These observations are typical o most substances which are heated sufciently to make them melt and then boil. Figure 5 shows how the temperature changes with time or energy being supplied. Energy is continually being supplied to the ice but there is no temperature change occurring during melting or boiling. The energy required to achieve the change o phase is called latent heat; the word latent meaning hidden. In a similar way to how specifc heat capacity was defned in order to compare equal masses o substances, physicists defne:
sp ecifc latent heat o usion ( melting) as the energy required to change the phase o 1 kg o substance rom a solid to a liquid without any temperature change
The term specifc heat capacity is a throwback to the caloric days when energy was thought to be a substance. A more appropriate term would be to call this quantity specifc energy capacity but specifc heat capacity is a term that has stuck even though scientists now recognize its inappropriateness. Are there other cases where we still use an old-ashioned term or concept because it is simpler to do so than to change all the books? Has the paradigm shit believing that heating was a transer o substance into being a transer o energy been taken on by society or is there still conusion regarding this?
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sp ecifc latent heat o vap orization ( boiling) as the energy required to change the phase o 1 kg o liquid into a gas without any temperature change.
The equations or these take the orm Q L= _ m
co
ol s
where Q is the energy supplied to the obj ect o mass which causes its phase to change without any temperature change. When Q is in J, m in kg, L will be in units o J kg 1 .
molecules
boils
wa
s
gas
rm
ols
freezes wa
co
temperature
co
ol s
rm
s
condenses
liquid
solid
wa
rm
s
melts
time
Figure 5 Change of phase (not to scale) . The graph shows how the temperature o a substance changes with time. The portions o the graph that are at indicate when there is no temperature change and so the substance is changing phase.
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This graph assumes that energy is supplied by a constant power source and so, since energy is the product o power and time, a graph o temperature against energy transerred to the substance will take the same shape as this.
I you are told the rate at which energy is supplied ( i.e. the power supplied) then rom the gradients o the temperaturetime graph you can work out the heat capacity o the liquid, the solid and the gas.
From the time o each horizontal section o the graph you can also calculate the specifc heat capacity o the solid and liquid phases.
When a gas transers energy to another obj ect at a constant rate we see the temperature o the gas drop until it reaches the boiling point ( condensing point is the same temperature as this) at which time the graph becomes at. Again as the liquid transers energy away the temperature will all until the melting ( or solidiying) temperature is reached when the graph again becomes at until all the substance has solidifed. Ater this it will cool again.
C ooling is the term generally used to mean alling temperature rather than loss o energy, so it is important not to say the substance is cooling when the temperature is constant ( at parts o the graph) .
3 . 1 T E M P E R AT U R E A N D E N E R G Y C H A N G E S
Worked example A heater is used to boil a liquid in a pan or a measured time. The ollowing data are available. Power rating o heater = 2 5 W Time or which liquid is boiled = 6. 2 1 0 2 s Mass o liquid boiled away = 4.1 1 0 2 kg Use the data to determine the specifc latent heat o vaporization o the liquid.
Solution Using Q = Pt or heater: Q = 2 5 6.2 1 0 2 = 1 5 5 00 J Using Q = mL or the liquid: 1 5 5 00 = 4. 1 1 0 2 L 1 5 5 00 L=_ = 3 . 8 1 0 5 J kg 1 4.1 1 0 2
Molecular explanation of phase change The transer o energy to a solid increases its internal energy this means that the mean random kinetic energy o the molecules increases ( as the vibrations and speeds increase) and the intermolecular potential energy increases ( as the molecules move urther apart) . Eventually some groups o molecules move ar enough away rom their neighbours or the inuence o their neighbours to be reduced chemists would oten use the model o the intermolecular bonds being broken. When this is happening the energy supplied does not increase the mean random kinetic energy o the molecules but instead increases the potential energy o the molecules. E ventually the groups o molecules are sufciently ree so that the solid has melted. The mean speed o the groups o molecules now increases and the temperature rises once more the molecules are breaking away rom each other and j oining together at a constant rate. The potential energy is not changing on average. As the liquid reaches its boiling point the molecules start moving away rom each other within their groups. Individual molecules break away and the potential energy once more increases as energy is supplied. Again a stage is reached in which the mean kinetic energy remains constant until all the molecules are separated rom each other. We see a constant temperature ( although much higher than during melting) . The energy is vaporizing the liquid and only when the liquid has all vaporized is the energy re-applied to raising the gas temperature.
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3.2 Modelling a gas Understanding
Applications and skills
Pressure
Equation o state or an ideal gas
Equation o state or an ideal gas Kinetic model o an ideal gas
Kinetic model o an ideal gas Boltzmann equation
Mole, molar mass, and the Avogadro constant Diferences between real and ideal gases
Mole, molar mass, and the Avogadro constant Diferences between real and ideal gases
Equations Pressure: p = ___AF Number o moles o a gas as the ratio o number
_____
o molecules to Avogadros number: n = NN A Equation o state or an ideal gas: pV = nRT Pressure and mean square velocity ___ o ___ 1 2 molecules o an ideal gas: p = 3 c The mean kinetic energy o ideal gas molecules: R E K = ___32 kB T = ___32 ____ T N mean
A
Nature of science Progress towards understanding by collaboration and modelling Much o this sub-topic relates to dealing with how scientists collaborated with each other and gradually revised their ideas in the light o the work o others. Repeating and oten improving the original experiment using more reliable instrumentation meant that generalizations could be made. Modelling
the behaviour o a real gas by an ideal gas and the use o statistical methods was groundbreaking in science and has had a major impact on modern approaches to science including the quantum theory where certainty must be replaced by probability.
Introduction O ne of the triumphs of the use of mathematics in physics was the successful modelling of the microscopic behaviour of atoms to give an understanding of the macroscopic properties of a gas. Although the S wiss mathematician D aniel B ernoulli had suggested that the motion of atoms was responsible for gas pressure in the mid-1 700s, the kinetic theory of gases was not widely accepted until the work of the S cottish physicist, James C lerk-Maxwell and the Austrian, Ludwig B oltzmann working independently over a hundred years later.
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The gas laws The gas laws were developed independently experimentally between the mid-seventeenth century and the start o the nineteenth century. An ideal gas can be defned as one that obeys the gas laws under all conditions. Real gases do not do this but they approximate well to an ideal gas so long as the pressure is little more than normal atmospheric pressure. With the modern apparatus available to us it is not difcult to veriy the gas laws experimentally. As a consequence o developing the vacuum pump and thus the hermetically- sealed thermometer, the Irish physicist Robert B oyle was able to show in 1 662 that the pressure o a gas was inversely proportional to its volume. B oyles experiment is now easily repeatable with modern apparatus. B oyles law states that or a fxed mass o gas at constant temp erature the p ressure is inversely p rop ortional to the volume. The French experimenter Edm Mariotte independently perormed a similar experiment to that o B oyle and was responsible or recognizing that the relationship only holds at a constant temperature so with modern statements o the law, there is much j ustifcation or calling this law Mariottes law. 1 ( at constant temperature) or Mathematically this can be written as p __ V pV = constant ( at constant temperature) .
A graph o pressure against volume at constant temperature ( i.e. , a B oyles law graph) is known as an isothermal curve ( iso meaning the same and thermo relating to temperature) . Isothermal curves are shown in fgure 2 ( b) on p1 04.
TOK Boyles impact on scientifc method Robert Boyle was a scientifc giant and experimental science has much to thank him or in addition to his eponymous law. In the mid-1600s where philosophical reasoning was preerred to experimentation, Boyle championed perorming experiments. Boyle was most careul to describe his experimental techniques to allow them to be reproduced by others this gave reliability to experimental results and their interpretation. At this time, many experimenters were working independently. The conusion in correct attribution o gas laws to their discoverer may have been less involved had others ollowed Boyle's lead. His rapid and clear reporting o his experimental work and data avoided the secrecy that was common at the time; this was advantageous to the progress o other workers. Although Boyle was apparently not involved in the ensuing quarrels, there was great debate regarding whether Boyle or the German chemist Henning Brand discovered the element phosphorus. This is largely because Brand kept his discovery secret while he
pursued the philosophers stone in an attempt to convert base metals such as lead into gold. Is rapid or requent publication usually a case o a scientist's selpromotion or is the practice more oten altruistic?
Figure 1 Robert Boyle.
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T H E R M AL P H YS I C S The French physicist Jacques C harles around 1 787 repeated the experiments o his compatriot Guillaume Amontons to show that all gases expanded by equal amounts when subj ected to equal temperature 1 changes. C harles showed that the volume changed by ___ o the volume 273 at 0 C or each 1 K temperature change. This implied that at 2 73 C , the volume o a gas becomes zero. C harless work went unpublished and was repeated in 1 802 by another French experimenter, Joseph GayLussac. The law that is usually attributed to C harles is now stated as or a fxed mass o gas at constant p ressure the volume is directly p rop ortional to the absolute temp erature. Mathematically this law can be written as V T ( at constant pressure) V constant ( at constant pressure) . or __ T = Amontons investigated the relationship between pressure and temperature but used relatively insensitive equipment. He was able to show that the pressure increases when temperature increases but ailed to quantiy this completely. The third gas law is stated as or a gas o fxed mass and volume, the pressure is directly proportional to the absolute temperature. This law is sometimes attributed to Amontons and oten (incorrectly) to Gay-Lussac or Avogadro and sometimes it is simply called the pressure law. Maybe it is saest to reer to this as the third gas law! Mathematically this law can be written as p T ( at constant volume) p or __ = constant( at constant volume) . T The Italian physicist, C ount Amadeo Avogadro used the discovery that gases expand by equal amounts or equal temperature rises to support a hypothesis that all gases at the same temperature and pressure contain equal numbers o particles per unit volume. This was published in a paper in 1 81 1 . We now state this as the number o p articles in a gas at constant temp erature and p ressure is directly p rop ortional to the volume o the gas. Mathematically this law can be written as n V ( at constant n pressure and temperature) or __ = constant ( at constant pressure and V temperature) . S ince each o the gas laws applies under dierent conditions the constants are not the same in the our relationships. The our equations can be combined to give a single constant, the ideal gas constant, R. C ombining the our equations gives what is known as the equation o state o an ideal gas: pV _ = R or pV = nRT nT When pressure is measured in pascal ( Pa) , volume in cubic metre ( m 3 ) , temperature in kelvin ( K) , and n is the number o moles in the gas, then R has the value 8. 3 1 J K 1 mol 1 .
The mole and the Avogadro constant The mole, which is given the symbol mol, is a measure o the amount o substance that something has. It is one o the seven S I base units and is defned as being the amount o substance having the same
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number of p articles as there are neutral atoms in 1 2 grams of carbon-1 2 . O ne mole o a gas contains 6.02 1 0 23 atoms or molecules. This number is the Avogadro constant NA. In the same way that you can reer to a dozen roses ( and everyone agrees that a dozen is the name or 1 2 ) you can talk about three moles o nitrogen gas which will mean that you have 1 8.06 1 0 23 gas molecules. The mole is used as an alternative to expressing quantities in volumes or masses.
Molar mass As we have j ust seen, the mole is simply a number that can be used to count atoms, molecules, ions, electrons, or roses ( i you wanted) . In order to calculate the molar mass o a substance ( which diers rom substance to substance) we need to know the chemical ormula o a compound ( or whether a molecule o an element is made up rom one or more atoms) . Nitrogen gas is normally diatomic ( its molecules have two atoms) , so we write it as N 2 . O ne mole o nitrogen gas will contain 6.02 1 0 2 3 molecules but 1 2 .04 1 0 23 atoms. As one mole o nitrogen atoms has a mass o approximately 1 4.01 g then a mole o nitrogen molecules will have a mass o 2 8.02 g this is its molar mass. The chemical ormula or water is, o course, H 2 O , so one mole o water molecules contains two moles o hydrogen atoms and one mole o oxygen atoms. 1 mol o hydrogen atoms has a mass o 1 .00 g and 1 mol o oxygen atoms has a mass o 1 6.00 g, so 1 mol o water has a mass o ( 2 1 .00 g) + 1 6. 00 g = 1 8.00 g. The mass o water is 1 8.00 g mol 1 .
Worked example C alculate the percentage change in volume o a fxed mass o an ideal gas when its pressure is increased by a actor o 2 and its temperature increases rom 3 0 C to 1 2 0 C .
Solution S ince n is constant the equation o state can be written p 1 V1 p 2 V2 _ =_ T1 T2
V and so the We are trying to obtain the ratio __ V 2 1
V p T equation can be rearranged into the orm __ = ___ V p T 2
1
2
1
2
1
V2 p 2 = 2 p 1 and T1 = 3 03 K and T2 = 3 93 K, so __ V 1
393 = ______ = 0.65 or 65 % 2 303
This means that there is a 3 5 % reduction in the volume o the gas.
Investigate! Verifying the gas laws experimentally Boyles law
Pressure is changed using the pump.
This pushes dierent amounts o oil into the vertical tube, which changes the pressure on the gas ( air) .
C hanges are carried out slowly to ensure there is no temperature change.
A graph o pressure against volume gives a curve known as an isothermal ( line at constant temperature) .
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Isothermals can be plotted over a range o dierent temperatures to give a series o curves such as those in fgure 2 ( b) .
A graph o pressure against the reciprocal o 1 volume __ should give a straight line passing V through the origin.
This investigation could be monitored automatically using a data logger with a pressure sensor.
Charless law
Pressure is kept constant because the capillary tube is open to the surrounding atmosphere ( provided the atmospheric pressure does not change) .
C hange the temperature by heating ( or cooling/icing) the water.
Assuming that the capillary is totally uniorm ( which is unlikely) the length o the column o air trapped in the tube will be proportional to its volume.
S tir the water thoroughly to ensure that it is at a constant temperature throughout.
0 5 10
column o trapped gas
15 20 25
thermometer
30 35 40
oil column
capillary tube
Bourdon gauge
45
valve
L
rubber bands
50
air trapped by liquid index
55 60
ootball pump
(a)
pressure
increasing temperature This is your value or absolute zero
(b) volume
Figure 2 (a) Boyles law apparatus and (b) graph.
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Your results are going to be in the region rom 0 C up to 50 C, but i you draw your graph on the scale shown here it will not give a very reliable value or absolute zero it is better to use the page rom 0 C up to 50 C and to use the method o similar triangles to fnd absolute zero
L/mm
200 temperature/
0
Figure 3 Charles law apparatus and graph.
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3 . 2 M O D E LLI N G A G AS
Leave a reasonable length o time in between readings to allow energy to conduct to the air in the capillary when this happens the length o the capillary no longer changes.
A graph o length o the air column against the temperature in C should give a straight line, which can be extrapolated back to absolute zero.
To choose larger ( and thereore better) scales you can use similar triangles to calculate absolute zero.
line, which can be extrapolated back to absolute zero.
To choose larger ( and thereore better) scales you can use similar triangles to calculate absolute zero.
Bourdon gauge
thermometer 250 ml round-bottomed fask containing air water bath
The third gas law
There are many variants o this apparatus which could, yet again, be perormed using a data logger with temperature and pressure sensors.
The gas ( air) is trapped in the glass bulb and the temperature o the water bath is changed.
The pressure is read rom the B ourdon gauge ( or manometer or pressure sensor, etc.) . Leave a reasonable length o time in between readings to allow energy to conduct to the air in the glass bulb when this happens the pressure no longer changes. A graph o pressure against the temperature in o C should give a straight
pressure This is your value or absolute zero
200 temperature/C
0
50
Figure 4 Third gas law apparatus and graph.
Nature of science Evidence for atoms The concept o atoms is not a new one but it has only been accepted universally by scientists over the past century. Around 400 B C E the Greek philosopher D emocritus theorized the existence o atoms named rom ancient Greek: meaning without division. He suggested that matter consisted o tiny indivisible but discrete particles that determined the nature o the matter o which they comprised. E xperimental science did not become ashionable until the 1 7th century and so D emocritus theory remained unproven. S ir Isaac Newton had a limited view o the atomic nature o matter but believed that all matter was ultimately made o the same substance. In the 1 9th century the chemist, John D alton, was the frst to suggest that the individual elements were made o dierent atoms and could be combined in fxed ratios. Yet even D alton and his successors were only able to iner the presence o atoms rom chemical reactions. Less explicit evidence (yet still indirect) came rom experiments with diusion and B rownian motion.
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The microscopic interpretation of gases Difusion The smell o cooked ood wating rom a barbecue is something that triggers the ow o stomach j uices in many people. This can happen on a windless day and is an example o diusion o gases. The atomic vapours o the cooking ood are being bombarded by air molecules causing them to move through the air randomly. The bromine vapour experiment shown in fgure 5 is a classic demonstration o diusion. B romine ( a brown vapour) is denser than air and sinks to the bottom o the lower right- hand gas j ar. This is initially separated rom the upper gas j ar by a cover slide. As the slide is removed the gas is gradually seen to fll the upper gas j ar ( as seen in the j ars on the let) ; this is because the air molecules collide with the bromine atoms. I this was not the case we would expect the bromine to remain in the lower gas j ar.
Brownian motion
Figure 5 Bromine difusion.
In 1 82 7, E nglish botanist, Robert B rown, frst observed the motion o pollen grains suspended in water. Today we oten demonstrate this motion using a microscope to see the motion o smoke particles suspended in air. The smoke particles are seen to move around in a haphazard way. This is because the relatively big and heavy smoke particles are being bombarded by air molecules. The air molecules have momentum, some o which is transerred to the smoke particles. At any instant there will be an imbalance o orces acting on each smoke particle giving it the random motion observed. The experiment o fgure 6( a) shows how smoke in an air cell is illuminated rom the side. E ach smoke particle scatters light in all directions and so some reaches the microscope. The observer sees the motion as tiny specks o bright light that wobble around unpredictably as shown in fgure 6( b) .
these specks o light are the smoke particles scattering light
microscope
convex lens
window
the random motion o a smoke particle showing how it moves linearly in between collisions with air molecules
smoke
lamp air cell (a)
Figure 6 The smoke cell.
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(b)
3 . 2 M O D E LLI N G A G AS
Kinetic model of an ideal gas
z
The kinetic theory of gases is a statistical treatment o the movement o gas molecules in which macroscopic properties such as pressure are interpreted by considering molecular movement. The key assumptions o the kinetic theory are:
c
A gas consists o a large number o identical tiny particles called molecules; they are in constant random motion.
L
m cx
The number is large enough or statistical averages to be made.
Each molecule has negligible volume when compared with the volume o the gas as a whole.
At any instant as many molecules are moving in one direction as in any other direction.
The molecules undergo perectly elastic collisions between themselves and also with the walls o their containing vessel; during collisions each momentum o each molecule is reversed.
There are no intermolecular orces between the molecules between collisions ( energy is entirely kinetic) .
The duration o a collision is negligible compared with the time between collisions.
Each molecule produces a orce on the wall o the container.
The orces o individual molecules will average out to produce a uniorm pressure throughout the gas ( ignoring the eect o gravity) .
y
x L
L
Figure 7 Focusing on the x-component of the velocity.
Using these assumptions and a little algebra it is possible to derive the ideal gas equation: In fgure 7 we see one o N molecules each o mass m moving in a box o volume V. The box is a cube with edges o length L. We consider the collision o one molecule moving with a velocity c towards the righthand wall o the box. The components o the molecules velocity in the x, y, and z directions are cx, cy, and cz respectively. As the molecule collides with the wall elastically, its x component o velocity is reversed, while its y and z components remain unchanged. The x component o the momentum o the molecule is mcx beore the collision and mcx ater the collision. The change in momentum o the molecule is thereore mcx (mcx) = 2mcx As orce is the rate o change o momentum, the orce Fx that the 2mc right-hand wall o the box exerts on our molecule will be _____ where t t is the time taken by the molecule to travel rom the right- hand wall o the box to the opposite side and back again ( in other words it is the time between collisions with the right- hand wall o the box) . x
2 m cx _ m c 2x 2L _ Thus t = _ = = and so F m cx 2L __ L cx
What quantity is equivalent to the reciprocal of t?
Using Newtons third law o motion we see that the molecule must exert mc a orce o ___ on the right-hand wall o the box ( i.e. a orce equal in L magnitude but opposite in direction to the orce exerted by the righthand wall on the molecule) . 2 x
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T H E R M AL P H YS I C S S ince there are N molecules in total and they will have a range o speeds, the total orce exerted on the right- hand wall o the box m Fx = _ ( c 2x + c 2x + c 2x + ... + c 2x ) L Where c x is the x component o velocity o the frst molecule, c x that o the second molecule, etc. 1
2
3
N
1
Which assumption is related to molecules having a range o speeds?
2
With so many molecules in even a small volume o gas, the orces average out to give a constant orce. The mean value o the square o the velocities is given by
z
( c 2x + c 2x + c 2x + .. . + c 2x ) ___ c = N This means that the __ total orce on the right- hand wall o the box is Nm 2 given by Fx = ___ c x. L __ 2 x
c cz cx
1
2
3
N
Figure 8 shows a velocity vector c being resolved into components cx cy, and cz.
cy
Using Pythagoras theorem we see that c2 = c 2x + c 2y + c 2z x
It ollows that the mean values o velocity can be resolved into the mean values o its components such that __
y
Figure 8 Resolving the velocity of a molecule.
__
__
__
c 2 = c 2x + c 2y + c 2z __
c 2 is called the mean square speed o the molecules.
Which assumption is being used here?
On average there is an equal likelihood o a molecule moving in any direction as in any other direction so the magnitude o the mean components o the velocity will be the same, i.e. , __
__
__
__
__
__
__
1 2 c 2x = c 2y = c 2z , so c 2 = 3 c 2x or c 2x = __ c 3
S o our equation or the total orce on the right- hand wall becomes __ Nm 2 1 ___ F = __ c 3 L F The pressure on this wall ( which has an area A) is given by p x = __ and A A = L 2 so that X
Nm __2 Nm __2 1_ 1_ px = _ c or p x = _ c 3 3 L 3 V S ince pressure at a point in a uid ( gas or liquid) acts equally in all directions we can write this equation as Nm __2 1_ p= _ c 3 V As Nm is the total mass o the gas and the density ( Greek, rho) is the total mass per unit volume, our equation simplifes to 1 c__2 p= _ 3
Do collisions between molecules afect this result?
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B e careful not to confuse p, the pressure, with , the density. You should note that the guidance in the IB D iploma Programme physics guide says that you should understand this proo that does not mean that you need to learn it line by line but that each o the steps should make sense to you. You should then be able to answer any question asked in examinations.
3 . 2 M O D E LLI N G A G AS
Molecular interpretation of temperature Returning to the equation o an ideal gas in the orm Nm __2 1 _ p = __ c 3 V we get Nm __ pV = _c2 3 3 Multiplying each side by __ gives 2
__
__
Nm 2 3 3 ___ 1 __ pV = ( __ c ) = N __ m c2 2 2 3 2
C omparing this with the equation o state or an ideal gas pV = nRT __
3 1 nRT = N __ m c2 we can see that __ 2 2
This may look a little conusing with both n and N in the equation so lets recap: n is the number o moles o the gas and N is the number o molecules so lets combine them to give a simpler equation: __
n RT 3 ___ 1 __ = __ m c2 2 N 2
N N B ut ___ = n so __ n is the number o molecules per mole ( the Avogadro N number, NA) making the equation: A
__
3 ___ RT 1 __ = __ m c2 2 N 2 A
R Things get even easier when we defne a new constant to be ___ ; this N constant is given the symbol kB and is called the B oltzmann constant. A
S o the equation now becomes: __
3 1 __ k T = __ m c2 2 B 2
__
1 m c2 should remind you o the equation or kinetic energy and, Now __ 2 indeed, it represents the mean translational kinetic energy o the gas molecules. We can see rom this equation that the mean kinetic energy gives a measure o the absolute temperature o the gas molecules.
The kinetic theory has linked the temperature ( a macroscopic property) to the microscopic energies o the gas molecules. In an ideal gas there are no long- range intermolecular orces and thereore no potential energy components; the internal energy o an ideal gas is entirely kinetic. This means that the total internal energy o an ideal gas is ound by multiplying the number o molecules by the mean kinetic energy o the molecules 3 total internal energy of an ideal gas = __ NkB T 2
We have only considered the translational aspects o our gas molecules in this derivation and this is fne or atomic gases ( gases with only one atom in their molecules) ; when more complex molecules are considered this equation is slightly adapted using a principle called the equipartition o energies ( something not included in the IB D iploma Programme Physics guide) .
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T H E R M AL P H YS I C S
Alternative equation o state o an ideal gas
Note When you look at the unit or Boltzmann's constant you should see a similarity with specifc heat capacity. This is like the specifc heat capacity or one mole o every monatomic gas
The gas laws produced the equation o state in the orm o pV = nRT. Now we have met the B oltzmann constant we can use an alternative orm o this equation. As we have seen n represents the number o moles o our ideal gas and R is the universal molar gas constant. I we want to work with the number o molecules in the gas ( as physicists oten do) the equation o state can be written in the orm o pV = Nk B T where N represents the number o molecules and kB is the B oltzmann constant. Thus nR = NkB and i we consider 1 mol o gas then n = 1 R and N = NA = 6.02 1 0 23 . We see rom this why kB = __ as was used N previously. A
1
1
8.3 J mol K With R = 8. 3 J K 1 mol 1 kB must = ____________ = 1 .3 8 1 0 23 J K 1 6.02 1 0 mol 23
1
Worked example Nitrogen gas is sealed in a container at a temperature o 3 2 0 K and a pressure o 1 .01 1 0 5 Pa. a) C alculate the mean square speed o the molecules. b) C alculate the temperature at which the mean square speed o the molecules reduces to 5 0% o that in a) .
Solution
__
__
3p 3 1 .01 1 0 1 ___________ a) p = __ c2 so c2 = __ = 3 1 .2 5
= 2 .5 3 1 0 5 m 2 s 2 ( notice the unit here?) __
b) c2 T so the temperature would need to be 5 0% o the original value ( i. e. it would be 1 60 K) .
mean density o nitrogen gas over the temperatures considered = 1 .2 kg m 3
Nature of science MaxwellBoltzmann distribution In the 1 85 0s James C lerk Maxwell realized that a gas had too many molecules to have any chance o being analysed using Newtons laws ( even though this could be done in principle) . With no real necessity to consider the motion o individual molecules he realized that averaging techniques could be used to link the motion o the molecules with their macroscopic properties. He recognized that he needed to know the distribution o molecules having dierent speeds. Ludwig B oltzmann had proposed a general idea about how energy was distributed among systems consisting o many particles and Maxwell developed B oltzmanns distribution to show how many particles have a particular speed in the gas. Figure 9 shows the three typical distributions
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or the same number o gas molecules at three dierent temperatures. At higher temperatures the most probable speed increases but overall there are less molecules travelling at this speed since there are more molecules travelling at higher speeds. The speeds o molecules at these temperatures can be greater than 1 km s 1 ( but they dont travel very ar beore colliding with another molecule! ) . The MaxwellB oltzmann distribution was frst verifed experimentally between 1 93 0 and 1 934 by the American physicist I F Zartman using methods devised by the German-American Otto Stern in the 1 92 0s. Zartman measured the speed o molecules emitted rom an oven by collecting them on the inner surace o a rotating cylindrical drum.
3 . 2 M O D E LLI N G A G AS
MaxwellBoltzmann speed distribution 12.00% 273 K 373 K 473 K
relative frequency
10.00% 8.00% 6.00% 4.00% 2.00% 0.00% 0
200
400
600 800 speed/m s 1
1000
1200
1400
Figure 9 MaxwellBoltzmann distribution for the speed of gas molecules.
Nature of science Real gases An ideal gas is one that would obey the gas laws and the ideal gas equation under all conditions so ideal gases cannot be liquefed. In 1 863, the Irish physician and chemist Thomas Andrews succeeded in plotting a series o pV curves or carbon dioxide; these curves deviated rom the Boyles law curves at high pressures and low temperatures. Until this time it had been believed that certain gases could never be liquefed. Andrews showed that there was a critical temperature above which the gas could not be liquefed by simply increasing the pressure. He demonstrated that or carbon dioxide the critical temperature is approximately 31 C.
The diference between ideal gases and real gases The 1 9th century experimenters showed that ideal gases are j ust that ideal and that real gases do not behave as ideal gases. Under high pressures and low temperatures all gases can be liquefed and thus become almost incompressible. This can pV be seen i we plot a graph o ___ against p or one RT mole o a real gas we would expect this plot to give a horizontal straight line o value o 1 .00 or
an ideal gas. Figure 1 0 is such a plot or nitrogen. The pressure axis is in atmospheres this is a non-SI unit that is appropriate or high pressures and quite commonly used by experimenters (e.g. 900 atm converts to 900 1 .01 1 0 5 Pa or 9.1 1 0 7 Pa) . The ideal gas line is a better ft to the real gas line at 1 000 K and better still at low pressure. 3
200 K 500 K
2 PV RT
1000 K ideal gas
1
0 300
600 P (atm)
900
Figure 10 Deviation of a real gas from an ideal gas situation.
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Thus we can see that the best approximation o a real gas to an ideal gas is at high temperature and low pressures. In deriving the ideal gas equation we made assumptions that are not true or a real gas. In particular we said that there are no intermolecular orces between the molecules in between collisions. This is not true or real gases on two accounts:
Short-range repulsive orces act between gas molecules when they approach each other thereby reducing the eective distance in which they can move reely (and so actually reducing the useul gas volume below the value o V that we considered) . At slightly greater distances the molecules will exert an attractive orce on each other this causes the ormation o small groups
o loosely attached molecules. This in turn reduces the eective number o particles in the gas. With the groups o molecules at the same temperature as the rest o the gas, they have the same translational kinetic energy and momentum as other groups o molecules or individual molecules. The overall eect o intermolecular attraction is to reduce the pressure slightly. The D utch physicist Johannes van der Waals was awarded the 1 91 0 Nobel Physics Prize or his work in developing a real gas equation, which was a modifcation o the ideal gas equation. In this equation two additional constants were included and each o these diered rom gas to gas. In the ensuing years many have tried to ormulate a single simple equation o state which applies to all gases, no one has been successul to date.
TOK Empirical versus theoretical models
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Theoretical models such as the kinetic theory o gas are based on certain assumptions. I these assumptions are not met in the real world, the model may ail to predict a correct outcome as is the case with real gases at high pressure and low temperature.
modifed when objects are moving at speeds approaching that o light. Theoretical models are not inherently better than empirical models, or vice versa. The best model, whether it is theoretical or empirical, is the model that gives the best predictions or a particular set o circumstances.
Despite the best eorts o van der Waals and others, there is no theorectical model that gives the right answer or all gases in all states. This, however, does not mean that we cannot make correct predictions about a particular gas we can use an empirical model that is based on how the gas has behaved in these circumstances previously. We can sensibly expect that i the conditions are repeated then the gas will behave in the same way again. In the 17th century, with no understanding o the nature o gas molecules, a theoretical model o gases was not possible and so Boyle, Charles, and others could only ormulate empirical laws that were repeatable with the precision that their apparatus allowed. In the 21st century we still rely on empirical methods when gases are not behaving ideally; even when the van der Waals modifcation o the equation o state or an ideal gas is used, the constants that need to be included are dierent rom gas to gas and determined experimentally. So even though it is, perhaps, aesthetically pleasing to have a one size fts all equation, the reality is that theoretical models rarely account or all circumstances. Even Newtons laws o motion must be
When a set o experimental results are not in line with theory does this mean that the theory must be abandoned?
In March 2012 the ollowing report appeared on the BBC website: The head o an experiment that appeared to show subatomic particles travelling aster than the speed o light has resigned rom his post .. Earlier in March, a repeat experiment ound that the particles, known as neutrinos, did not exceed light speed. When the results rom the Opera group at the Gran Sasso underground laboratory in Italy were frst published last year, they shocked the world, threatening to up-end a century o physics as well as relativity theory which holds the speed o light to be the Universes absolute speed limit.
The media were very quick to applaud the experiment when the initial results were published, but the scientifc community was less quick to abandon a well-established law. Which o the two groups was showing bias in this case?
QUESTION S
Questions 1
7
Two obj ects are in thermal contact. S tate and explain which o the ollowing quantities will determine the direction o the transer o energy between these obj ects? a) The mass o each obj ect.
S pecifc latent heat o usion o water = 3 .3 4 1 0 5 J kg 1
b) The area o contact between the obj ects. c) The specifc heat capacity o each obj ect. d) The temperature o each obj ect.
2
3
S pecifc latent heat o vaporization o water = 2 .2 6 1 0 6 J kg 1
( 4 marks)
a) C alculate the amount o energy needed to be removed rom the water to reeze it.
Two obj ects are at the same temperature. Explain why they must have the same internal energy. ( 2 marks)
b) C alculate the amount o energy required by the water to vaporize it. c) Explain the dierence between the values calculated in a) and b) . (6 marks)
The internal energy o a piece o copper is increased by heating. 8
a) Explain what is meant, in this context, by internal energy and heating. b) The piece o copper has mass 0. 2 5 kg. The increase in internal energy o the copper is 1 .2 1 0 3 J and its increase in temperature is 2 0 K. Estimate the specifc heat capacity o copper. ( 4 marks)
5
C alculate the amount o energy needed to raise the temperature o 3 . 0 kg o steel rom 2 0 C to 1 2 0 C . The specifc heat capacity o steel is 490 J kg 1 K 1 . ( 2 marks)
Specifc heat capacity o water = 42 00 J kg K
1
Specifc heat capacity o copper = 3 90 J kg 1 K 1 ( 2 marks)
6
A fxed mass o an ideal gas is heated at constant volume. S ketch a graph to show the variation with C elsius temperature t with pressure p o the gas? ( 3 marks)
1 0 Under what conditions does the equation o state or an ideal gas, pV = nRT, apply to a real gas? ( 2 marks)
C alculate the energy supplied to 0. 070 kg o water contained in a copper cup o mass 0.080 kg. The temperature o the water and the cup increases rom 1 7 C to 2 5 C . 1
( IB) A container holds 2 0 g o neon and also 8 g o helium. The molar mass o neon is 2 0 g and that o helium is 4 g. C alculate the ratio o the number o atoms o neon to the number o atoms o helium. ( 2 marks)
9 4
2 .0 kg o water at 0 C is to be changed into ice at this temperature. The same mass o water now at 1 00 C is to be changed into steam at this temperature.
The temperature dierence between the inlet and the outlet o an air- cooled engine is 3 0.0 K. The engine generates 7.0 kW o waste power that the air extracts rom the engine. C alculate the rate o ow o air ( in kg s 1 ) needed to extract this power.
1 1 a) ( i) Explain the dierence between an ideal gas and a real gas. (ii) Explain why the internal energy o an ideal gas comprises o kinetic energy only. b) A fxed mass o an ideal gas has a volume o 870 cm 3 at a pressure o 1 .00 1 0 5 Pa and a temperature o 20.0 C . The gas is heated at constant pressure to a temperature o 21 .0 C . C alculate the change in volume o the gas. ( 6 marks)
Specifc heat capacity o air (at constant pressure) = 1 .01 1 0 3 J kg1 K1 (3 marks)
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T H E R M AL P H YS I C S 1 2 (IB)
S tate the relation between
a) The pressure p o a fxed mass o an ideal gas is directly proportional to the kelvin temperature T o the gas, that is, p T. ( i) State the relation between the pressure p and the volume V or a change at constant temperature. ( ii) State the relation between the volume V and kelvin temperature T or a change at a constant pressure. b) The ideal gas is held in a cylinder by a moveable piston. The pressure o the gas is p , its volume is V and its kelvin 1 1 temperature is T . 1
The pressure, volume and temperature are changed to p , V and T respectively. The 2 2 2 change is brought about as illustrated below.
( i)
p 1 , p 2 , T1 and T
( ii) V1 , V2 , T and T2 c) Use your answers to b) to deduce, that or an ideal gas pV = KT, where K is a constant.
( 6 marks)
1 3 A helium balloon has a volume o 0.2 5 m 3 when it is released at ground level. The temperature is 3 0 C and the pressure 1 .01 1 0 5 Pa. The balloon reaches a height such that its temperature has allen to 1 0 C and its pressure to 0.65 1 0 5 Pa. a) C alculate the new volume o the balloon. b) S tate two assumptions that must be made about the helium in the balloon. c) C alculate the number o moles o helium in the balloon. ( 6 marks)
p 1 , V1 , T1
p 2 , V1 , T'
heated at constant volume to pressure p 2 and temperature T'
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p 2 , V2 , T2
heated at constant pressure to volume V2 and temperature T2
4 O S C I L L AT I O N S A N D WAV E S Introduction All motion is either periodic or non-periodic. In p eriodic motion an obj ect repeats its pattern o motion at fxed intervals o time: it is regular and repeated. Wave motion is also periodic and
there are many similarities between oscillations and waves; in this topic we will consider the common eatures but also see that there are dierences.
4.1 2.1 Oscillations Motion Understanding Simple harmonic oscillations Time period, requency, amplitude,
displacement, and phase diference Conditions or simple harmonic motion
Applications and skills Qualitatively describing the energy changes
taking place during one cycle o an oscillation Sketching and interpreting graphs o simple harmonic motion examples
Equations 1 Period-requency relationship: T = ___ f
Proportionality between acceleration and
displacement: a x
Nature of science Oscillations in nature Naturally occurring oscillations are very common, although they are oten enormously complex. When analysed in detail, using a slow-motion camera, a hummingbird can be seen to ap its wings at a requency o around 20 beats per second as it hovers, drinking nectar. Electrocardiographs are used to monitor heartbeats as hearts pulsate, pushing blood around our bodies at about one per second when we are resting and maybe two or three times this rate as we exert ourselves. Stroboscopes can be used to reeze the motion in engines and motors where periodic motion is essential, but too strong vibrations can be potentially very destructive. The practical techniques that have been developed combined with the mathematical modelling that is used to interpret oscillations
Figure 1 Hummingbird hovering over fower.
are very powerul tools; they can help us to understand and make predictions about many natural phenomena.
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Isochronous oscillations
Figure 2 Pocket watch under strobe light.
A very common type o oscillation is known as isochronous (taking the same time) . These are oscillations that repeat in the same time period, maintaining this constant time property no matter what amplitude changes due to damping occur. It is the isochronicity o oscillations such as that o a simple pendulum that has made the pendulum such an important element o the clock. I we use a stroboscope (or strobe) we can reeze an isochronous oscillation so that it appears to be stationary; the strobe is made to ash at a regular interval, which can be matched to the oscillation. I the requency o the strobe matches that o the oscillating object then, when the object is in a certain position, the strobe ashes and illuminates it. It is then dark while the object completes an oscillation and returns to the same position when the strobe ashes again. In this way the object appears motionless. The lowest requency o the strobe that gives the object the appearance o being static will be the requency o the object. I you now double the strobe requency the object appears to be in two places. Figure 2 shows a swinging pocket watch which is illuminated with strobe light at 4 Hz, so the time interval between the images is 0.25 s. The watch speeds up in the middle and slows down at the edges o the oscillation.
Describing periodic motion The graph in fgure 3 is an electrocardiograph display showing the rhythm o the heart o a healthy 48-year-old male pulsing at 65 beats per minute. The pattern repeats regularly with the repeated pattern being called the cycle o the motion. The time duration o the cycle is called the time period or the period (T) o the motion. Thus, a person with a pulse o 65 beats per minute has an average heartbeat period o 0.92 s.
A
Figure 3 Normal adult male heart rhythm. Let us now compare the pattern o fgure 3 with that o fgure 4, which is a data-logged graph o a loaded spring. The apparatus used or the datalogging is shown in fgure 5 and the techniques discussed in the Investigate! section on p1 1 7. The graph or the loaded spring is a more straightorward example o periodic motion, obtained as the mass suspended on a spring oscillates up and down, above and below its normal rest position. There are just over 1 2 complete oscillations occurring in 2 0 seconds giving a period o approximately 1 .7 s. Although the two graphs are very dierent the period is calculated in the same way. As the mass passes its rest or equilibrium p osition its displacement ( x) is zero. We have seen in Topics 1 and 2 that displacement is a vector
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4 . 1 O S C I L L AT I O N S
distance/m
x vs t centre zero 0.25 0.20 0.15 0.10 0.05 0 0.05 0.10 0.15 0.20 0.25
10 seconds
20 seconds
retort stand time
Figure 4 Loaded spring. quantity and, thereore, must have a direction. In this case, since the motion is linear or one dimensional, it is sufcient to speciy the direction as simply being positive or negative. The choice o whether above or below the rest position is positive is arbitrary but, once decided, this must be used consistently. In the graph o fgure 4 the data logger has been triggered to start as the mass passes through the rest position. In this case positive displacements are above the rest position so the mass is moving downwards when timing is started. The maximum value o the displacement is called the amp litude ( x0 ) . In the case o the loaded spring the amplitude is a little smaller than 0.2 5 m. It is much more difcult to measure the amplitude or the heart rhythm ( fgure 3 ) because o determining the rest position additionally with no calibration o the scale it is impossible to give any absolute values. The amplitude is marked as A which is approximately 3 cm on this scale. The frequency ( f ) o an oscillation is the number o oscillations completed per unit time. When the time is in seconds the requency will be in hertz ( Hz) . Scientists and engineers regularly deal with high requencies and so kHz, MHz, and GHz are commonly seen. The vibrations producing sound waves range rom about 2 0 Hz to 2 0 kHz while radio waves range rom about 1 00 kHz to 1 00 MHz. With frequency being the number o oscillations per second and p eriod the time or one oscillation you may have already spotted the relationship between these two quantities: 1 T= _ f
spring
clamp rigid card always more than 25 cm above motion sensor motion sensor mesh guard
Figure 5 Data logger arrangement.
Worked example A childs swing oscillates simple harmonically with a period o 3 .2 s. What is the requency o the swing?
Solution 1 so f = _ 1 T= _ T f 1 = _ = 0. 3 1 Hz 3 .2
Investigate! In this investigation, a motion sensor is used to monitor the position o a mass suspended rom the end o a long spring. The data logger sotware processes the data to produce a graph showing the variation o displacement with time. The apparatus is arranged as shown in fgure 5 with the spring having a period o at least 1 second.
The apparatus is set up as shown above.
The mass is put into oscillation by displacing and releasing it.
The details o the data logger will determine the setting values but data loggers can usually be triggered to start reading at a given displacement value.
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O S C I LL AT I O N S AN D WAVE S
Motion sensors use ultrasound, so reections rom surroundings should be avoided.
The sotware can normally be used to plot graphs o velocity and acceleration ( in addition to displacement) against time.
S imilar investigations can be perormed with other oscillations.
Nature of Science Simple for physicists Simple does not mean easy in the context o physics! In act, mathematically (as discussed in Sub-topic 9.1 ) simple harmonic motion (SHM) is not simple. Simple may be better thought o as simplifed in that we make lie easier or ourselves by either limiting or ignoring some o the orces that act on a body. A simple pendulum consists o a mass suspended rom a string. In dealing with the simple
pendulum the rictional orces acting on it (air resistance and riction between the string and the suspension) and the act that the mass will slightly stretch the string are ignored without having any disastrous eect on the equations produced. It also means that smooth graphs are produced, such as or the loaded spring.
Simple harmonic motion We saw that the graph produced by the mass oscillating on the spring was much less complex than the output o a human heart. The mass is said to be undergoing simple harmonic motion or SHM. In order to perorm SHM an object must have a restoring orce acting on it.
The magnitude o the orce ( and thereore the acceleration) is proportional to the displacement o the body rom a fxed point.
The direction o the orce (and thereore the acceleration) is always towards that fxed point.
Focusing on the loaded spring, the fxed point in the above defnition is the equilibrium position o the mass where it was beore it was pulled down. The orces acting on the mass are the tension in the spring and the pull o gravity (the weight) . In the equilibrium position the tension will equal the weight but above the equilibrium the tension will be less and the weight will pull the mass downward; below the equilibrium position the tension will be greater than the weight and this will tend to pull the mass upwards. The dierence between the tension and the weight provides the restoring orce the one that tends to return the mass to its equilibrium position. We can express the relationship between acceleration a and displacement x as: a x which is equivalent to a = kx This equation makes sense i we think about the loaded spring. When the spring is stretched urther the displacement increases and the tension
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4 . 1 O S C I L L AT I O N S
a
increases. B ecause force = mass acceleration, increasing the orce increases the acceleration. The same thing applies when the mass is raised but, in this case, the tension decreases, meaning that the weight dominates and again the acceleration will increase. So although we cannot prove the proportionality (which we leave or HL) the relationship makes sense. The minus sign is explained by the second bullet point the acceleration is always in the opposite direction to the displacement. S o choosing upwards as positive when the mass is above the equilibrium position, the acceleration will be downwards ( negative) . When the mass is below the equilibrium position ( negative) , the ( net) orce and acceleration are upwards ( positive) . The orce decelerates the mass as it goes up and then accelerates it downwards ater it stopped. Figure 6 is a graph o a against x. We can see that a = kx. takes the orm y = mx + c with the gradient being a negative constant.
0
x 0
Figure 6 Accelerationdisplacement graph.
Graphing SHM From Topic 2 we know that the gradient o a displacementtime graph gives the velocity and the gradient o the velocitytime graph gives the acceleration at any given time. This remains true or any motion. Lets look at the displacement graph or a typical S HM. In fgure 7 the graph is a sine curve ( but i we chose to start measuring the displacement rom any other time it could j ust as easily be a cosine, or a negative sine, or any other sinusoidally shaped graph) . I we want to fnd the velocity at any particular time we simply need to fnd the gradient o the displacementtime graph at that time. With a curved graph we must draw a tangent to the curve (at our chosen time) in order to fnd a gradient. The blue tangent at 0 s has a gradient o + 2.0 cm s 1 , which gives the maximum velocity ( this is the steepest tangent and so we see the velocity is a maximum at this time) . Looking at the gradient at around 1 .6 s or 4.7 s and you will see rom the symmetry that it will give a velocity o 2 .0 cm s 1 , in other words it will be the minimum velocity ( i.e. the biggest negative velocity) . At around 0.8, 2 .4, 3 .9, and 5 .5 s the gradient is zero so the velocity is zero. 2 1.5 1
x/cm
0.5 0 0
0.5
1
2
3
4
5
6
7
t/s
1 1.5 2
Figure 7 The displacement graph for a typical SHM.
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In many situations it is helpful to think about the magnitude of the velocity i. e. speed. For SHM organize a table for an object oscillating between +A and -A. In your table show where the magnitudes of the position, velocity, acceleration and force are (a) zero and (b) maximum. Once you have done that mark in the directions of these these vectors at +A, -A and zero. By doing this you should be thinking physics before algebra.
Using these values you will see that, overall, the velocity is going to change as shown in fgure 8(b) ; it is a cosine graph in this instance. You may remember that the gradient o the velocitytime graph gives the acceleration and so i we repeat what we did or displacement to velocity, we get fgure 8( c) or the acceleration graph. You will notice that this is a reection o the displacementtime graph in the time axis i. e. a negative sine curve. This should be no surprise, since rom our defnition o simple harmonic motion we expect the acceleration to be a negative constant multiplied by the displacement. (a)
1.0 0.5 x/cm
Directions with SHM
0
0
1
2
3
4
5
6
7
4
5
6
7
4
5
6
7
t/s
0.5 1.0 (b)
2.0
v/m s 1
1.0 0
0
1
2
3 t/s
1.0 2.0 (c)
4.0
a/m s 2
2.0 0
0
1
2
3 t/s
2.0 4.0
Figure 8 The variation with time of displacement, velocity, and acceleration.
Worked examples 1
120
On a sheet o graph paper sketch two cycles o the displacementtime (x - t) relationship or a simple pendulum. Assume that its displacement is a maximum at time 0 seconds. Mark on the graph a time or which the velocity is maximum (labeled A) , a time or which the velocity is zero (labeled B) and a time or which the acceleration is a maximum (labeled C) .
Solution x
A
B
C t
4 . 1 O S C I L L AT I O N S
Note
2
This is a sketch graph so no units are needed they are arbitrary.
A, B, and C are each labelled on the time axis as this is what the question asks.
There are just two cycles (two complete periods) marked since this is what the question asks.
A is where the gradient o the displacementtime graph is a maximum.
B is where the gradient o the displacement time graph is a zero.
C is where the displacement is a minimum because a = kx this means that the acceleration will be a maximum. The equation defning simple harmonic motion is a = kx.
b) Two similar systems oscillate with simple harmonic motion. The constant or system S 1 is k, while that or system S 2 is 4k. Explain the dierence between the oscillations o the two systems.
Solution a) Rearranging the equation we obtain a k = _ x S ubstituting the units or the quantities gives ms 2 unit o k = ____ m = s 2
S o the units o k are s 2 or per second squared this is the same as requency squared ( and could be written as Hz 2 ) . b) Reerring to the solution to ( a) we can see that 4 or S 2 the square o the requency would be __ times that o S 1 . This means that S 2 has 4 or twice the requency ( or hal the period) o S. That is the dierence.
a) What are the units o the constant k?
Phase and phase diference Reerring back to fgures 8 a, b, and c we can see that there is a big similarity in how the shapes o the displacement, velocity, and acceleration graphs change with time. The three graphs are all sinusoidal they take the same shape as a sine curve. The dierence between them is that the graphs all start at dierent points on the sine curve and continue like this. The graphs are said to have a phase difference. When timing an oscillation it really doesnt matter when we start timing we could choose to start at the extremes o the oscillation or the middle. In doing this the shape o the displacement graph would not change but would look like the velocity graph (quarter o a period later) or acceleration graph (hal a period later) . From this we can see that the phase dierence between the displacementtime graph and the velocitytime graph is equivalent to quarter T o a period or __ (we could say that velocity leads displacement by quarter o 4 a period) . The phase dierence between the displacementtime graph and T the accelerationtime graph is equivalent to hal a period or __ (we could say 2 that acceleration leads displacement by hal a period or that displacement leads acceleration by hal a period it makes no dierence which) . Although we have discussed phase dierence in term o periods here, it is more common to use angles. However, transerring between period and angle is not difcult: Period T is equivalent to 3 60 or 2 radians, T so __ is equivalent to 1 80 or radians 2 T and __ is equivalent to 90 or __ radians. 4 2
When the phase dierence is 0 or T then two systems are said to be oscillating in p hase.
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Worked example 1
b) The period ( 2 . 0 s) is equivalent to 2 radians, so taking ratios:
C alculate the phase dierence between the two displacementtime graphs shown in fgure 9. Give your answers in a) seconds b) radians c) degrees.
0.2 5 2 .0 ___ = ____ ( where is the phase dierence in 2
radians)
2.0
This gives a value or o 0.79 radian.
1.5
c) The period ( 2 .0 s) is equivalent to 3 60, so taking ratios:
1.0 0.5
2 .0 0.25 ___ = ____ ( where is the phase dierence in 360
0.0 0.5
0.0
1.0
2.0 time/s
3.0
4.0
This gives a value or o 45 .
1.0
2.5 2.0
2.5
1.5 displacement/cm
1.5 2.0
2.5 2.0 1.5 displacement/cm
degrees)
5.0
1.0 0.5 0.5
1.0 0.5 time/s
0.0 0.5
0.0
1.0
2.0
3.0
4.0
5.0
1.0 1.5
0.0 0.0
1.0
2.0
3.0
4.0
2.0
5.0
2.5
time/s
0.25 s
1.0 1.5
2.5
2.0
2.0
2.5
1.5
Figure 9
Solution a) It can be seen rom either graph that the period is 2 .0 s. D rawing vertical lines through the peaks o the two curves shows the phase dierence. This appears to be between 0.2 5 0.2 6 s so we will go with ( 0.2 5 0.01 ) s.
displacement/cm
displacement/cm
2.5
1.0 0.5 time/s
0.0 0.5
0.0
1.0
2.0
3.0
4.0
5.0
1.0 1.5 2.0 2.5
Figure 10
Energy changes in SHM Lets think o the motion o the simple pendulum o fgure 1 1 as an example o a system which undergoes simple harmonic motion. For a pendulum to oscillate simple harmonically the string needs to be long and to make small angle swings (less than 1 0) . The diagram is, thereore, not drawn to scale. Lets imagine that the bob just brushes along the ground when it is at its lowest position. When the bob is pulled to position A, it is at its highest point and has a maximum gravitational potential energy
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4 . 2 T R AV E L L I N G W AV E S ( GPE) . As the bob passes through the rest position B , it loses all the GPE and gains a maximum kinetic energy ( KE) . The bob now starts to slow down and move towards position C when it briefy stops, having regained all its GPE. In between A and B and between B and C the bob has a combination o KE and GPE. In a damped system over a long period o time the maximum height o the bob and its maximum speed will gradually decay. The energy gradually transers into the internal energy o the bob and the air around it. D amping will be discussed in O ption B . 4.
O point of suspension string GPE(max) = mgh KE = 0
GPE(max) = mgh KE = 0 bob C h
A h ground
B GPE = 0 KE = maximum
Figure 11 simple pendulum.
4.2 Travelling waves Understanding Travelling waves Wavelength, requency, period, and wave speed Transverse and longitudinal waves The nature o electromagnetic waves The nature o sound waves
Applications and skills Explaining the motion o particles o a medium
when a wave passes through it or both transverse and longitudinal cases Sketching and interpreting displacement distance graphs and displacement time graphs or transverse and longitudinal waves Solving problems involving wave speed, requency, and wavelength Investigating the speed o sound experimentally
Equation The wave equation: c = f
Nature of science Patterns and trends in physics One o the aspects o waves that makes it an interesting topic is that there are so many similarities with just enough twists to make a physicist think. Transverse waves have many similarities to longitudinal waves, but there are equally many dierences. An electromagnetic wave requires no medium through which to travel, but a mechanical
wave such a sound does need a medium to carry it; yet the intensity o each depends upon the square o the amplitude and the two waves use the same wave equation. Physicists enjoy patterns but at the same time they enjoy the places where patterns and trends change the similarities give confdence but the dierences can be thought-provoking.
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Introduction Whe n yo u co nside r wave s yo u p ro b ab ly imme diate ly think o the rip p le s o n the surace o a lake o r the se a. It may b e diicult to think o many e xamp le s o wave s, b u t this to p ic is inte gral to o u r ab ility to co mmu nicate and, since lie o n E arth is de p e nde nt o n the radiatio n that arrive s ro m the S u n, hu mankind co u ld no t e xist witho u t wave s. Waves are o two undamental types:
Figure 1 Ripples spreading out rom a stone thrown into a pond.
mechanical waves, which require a material medium through which to travel
electromagnetic waves, which can travel through a vacuum.
B oth types o wave motion can be treated analytically by equations o the same orm. Wave motion occurs in several branches o physics and an understanding o the general principles underlying their behaviour is very important. Modelling waves can help us to understand the properties o light, radio, sound . . . even aspects o the behaviour o electrons.
Travelling waves Figure 2 shows a slinky spring being used to demonstrate some o the properties o travelling waves.
Figure 2 Slinky spring being used to demonstrate travelling waves.
fxed end
end ree to move
With one end o the stretched spring fxed ( as in fgure 3 ( a) ) , moving the other end sharply upwards will send a pulse along the spring. The pulse can be seen to travel along the spring until it reaches the fxed end; then it reects and returns along the spring. O n reection the pulse changes sides and what was an upward pulse becomes a downwards pulse. The pulse has undergone a phase change o 1 80 or radians on reection. When the end that was fxed is allowed to move there is no phase change on reection as shown in fgure 3 ( b) and the pulse travels back on the same side that it went out.
Note You should discuss with fellow students how Newtons second and third laws of motion apply to the motion of the far end of the spring.
a
b
Figure 3 Reection o wave pulses.
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O bserving the motion o a slinky can give much insight into the movement o travelling waves. To help ocus on the motion o the spring, it is useul i one o the slinky coils is painted to make it stand out rom the others. The coils are being used to model the particles o the medium through which the wave travels they could be air or water molecules. I a table tennis ball is placed beside the spring, it shoots o at right angles to the direction o the wave pulse. This shows that this is a transverse wave pulse the direction o the pulse being perpendicular to the direction in which the pulse travels along the slinky. When the slinky is vibrated continuously rom side to side a transverse wave is sent along the spring as shown in fgure 4 overlea.
4 . 2 T R AV E L L I N G W AV E S
Figure 4 Transverse waves on a slinky spring. From observing the wave on the spring and water waves on a pond we can draw the ollowing conclusions:
A wave is initiated by a vibrating obj ect ( the source) and it travels away rom the obj ect.
The particles o the medium vibrate about their rest position at the same requency as the source.
The wave transers energy rom one place to another.
A slinky can generate a dierent type o wave called a longitudinal wave. In this case the ree end o a slinky must be vibrated back and orth, rather than rom side to side. As a result o this the coils o the spring will vibrate about their rest position and energy will travel along the spring in a direction parallel to that o the springs vibration as shown in fgure 5 .
rarefaction
compression
Figure 5 Longitudinal waves on a slinky spring.
Describing waves When we describe wave properties we need specialist vocabulary to help us:
Wavelength is the shortest distance between two points that are in phase on a wave, i. e. two consecutive crests or two consecutive troughs.
Frequency f is the number o vibrations per second perormed by the source o the waves and so is equivalent to the number o crests passing a fxed point per second.
Period T is the time that it takes or one complete wavelength to pass a fxed point or or a particle to undergo one complete oscillation.
Amp litude A is the maximum displacement o a wave rom its rest position.
These defnitions must be learned, but it is oten easier to describe waves graphically. There are two types o graph that are generally used when describing waves: displacementdistance and displacementtime graphs. S uch graphs are applicable to every type o wave.
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4
O S C I LL AT I O N S AN D WAVE S For example displacement could represent:
the displacement o the water surace rom its normal at position or water waves
the displacement o air molecules rom their normal position or sound waves
the value o the electric feld strength vector or electromagnetic waves.
Displacementdistance graphs This type o graph is sometimes called a wave profle and represents the displacement o many wave particles at a particular instant o time. O n fgure 6 the two axes are perpendicular but, in reality, this is only the case when we describe transverse waves ( when the graph looks like a photograph o the wave taken at a particular instant) . For longitudinal waves the actual displacement and distance are parallel. Figure 7 shows how the particles o the medium are displaced rom the equilibrium position when the longitudinal wave travels through the medium orming a series o compressions (where the particles are more bunched up than normal) and rarefactions (where the particles are more spread out than normal) . In this case the particles that are displaced to the let o their equilibrium position are given a negative displacement, while those to the right are allocated a positive displacement.
displacement /cm 0.5
wavelength amplitude
0 2
4
8 dist ance/cm
6
0.5 trough
crest
trough
crest
Figure 6 Transverse wave profle. 1 cm equilibrium position with no wave position o particles with wave displacement /cm
t= 0
0.5
wavelength amplitude
0 2
4
6
centre o rareaction
centre o compression
centre o rareaction
8 dist ance/cm
0.5 centre o compression
Figure 7 Longitudinal wave profle.
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centre o compression
4 . 2 T R AV E L L I N G W AV E S
It is very easy to read the amplitude and wavelength directly rom a displacementtime graph. For the longitudinal wave the wavelength is both the crest- to-crest distance and the distance between two consecutive compressions or rareactions. Using a sequence o displacementdistance graphs can provide a good understanding o how the position o an individual particle changes with time in both transverse and longitudinal waves ( fgure 8) . O n this T diagram the wave profle is shown at a quarter period ( __ intervals; 4) diagrams like this are very useul in spotting the phase dierence between the particles. P and Q are in anti- phase here (i.e. 1 80 or radians. You may wish radians out o phase) and Q leading R by 90 or __ 2 to consider the phase dierence between P and R. displacement/cm 0.5 t= 0
Q
P
0 2
4
6
8 distance/cm
6
8 distance/cm
6
8 distance/cm
6
8 distance/cm
6
8 distance/cm
R
0.5 displacement/cm 0.5 t=
T 4
Q R
0 2
4 P
0.5 displacement/cm
R
0.5 t=
T 2
P
0
2
Q
4
0.5 displacement/cm 0.5 P
t=
3T 4
0 2
4 Q
0.5
R
displacement/cm 0.5 t= T
2 0.5
Q
P
0
4 R
Figure 8 Sequence of displacement-distance graphs at ___4T time intervals.
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Displacementtime graphs A displacementtime graph describes the displacement of one particle at a certain position during a continuous range of times. Figure 9 shows the variation with time of the displacement of a single particle. Each particle along the wave will undergo this change ( although with phase difference between individual particles) . displacement/cm 0.5
Note It is common to confuse displacementdistance graphs (where crest-to-crest gives the wavelength) with displacementtime graphs (where crest-to-crest gives the period) .
0 T 2
T
2T time/s
3T 2
0.5
Figure 9 Displacementtime graphs. This graph makes it is very easy to spot the period and the amplitude of the wave.
The wave equation When a source of a wave undergoes one complete oscillation the wave it produces moves forward by one wavelength ( ) . S ince there are f oscillations per second, the wave progresses by f during this time and, therefore, the velocity ( c) of the wave is given by c = f. With f in hertz and in metres, c will have units of hertz metres or ( more usually) metres per second. You do need to learn this derivation and it is probably the easiest that you will come across in IB Physics.
Worked example
directions of wave travel
The diagram below represents the direction of oscillation of a disturbance that gives rise to a wave. M
a) Draw two copies of the diagram and add arrows to show the direction of wave energy transfer to illustrate the difference between (i) a transverse wave and (ii) a longitudinal wave. b) A wave travels along a stretched string. The diagram to the right shows the variation with distance along the string of the displacement of the string at a particular instant in time. A small marker is attached to the string at the point labelled M. The undisturbed position of the string is shown as a dotted line.
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O n a copy of the diagram: ( i) D raw an arrow to indicate the direction in which the marker is moving. ( ii) Indicate, with the letter A, the amplitude of the wave.
4 . 2 T R AV E L L I N G W AV E S
( iii) Indicate, with the letter , the wavelength o the wave.
b)
direction of wave
crest
( iv) D raw the displacement o the string a time T/4 later, where T is the period o oscillation o the wave.
A M
( v) Indicate, with the letter N, the new position o the marker.
A
c) The wavelength o the wave is 5 .0 cm and its speed is 1 0 cm s 1 . N
D etermine: ( i) the requency o the wave
( i) With the wave travelling to the let, the trough shown will move to the let and that must mean that M moves downwards.
( ii) how ar the wave has moved in a quarter o a period.
Solution
( ii) The amplitude ( A) is the height o a crest or depth o a trough.
a) ( i) The energy in a transverse wave travels in a direction perpendicular to the direction o vibration o the medium:
( iii) The wavelength () is the distance equivalent to crest to next crest or trough to next trough. ( iv) In quarter o a period the wave will have moved quarter o a wavelength to the let ( broken line curve) .
direction of energy direction of vibration
( ii) The energy in a longitudinal wave travels in a direction parallel to the direction o vibration o the medium:
trough
( v) Ater quarter o a period the marker is now at the position o the trough ( N) . c)
direction of energy
c 10 ( i) f = __ = __ = 2 . 0 Hz ( since both 5 wavelength and speed are in cm, there is no need to convert the units) 1 ( ii) T = __ = 0.5 s f
direction of vibration
B ecause the wave moves at a constant speed T s = ct = c _ = 1 0 0.1 2 5 = 1 .2 5 cm 4
Investigate! Measuring the speed of sound Here are two o the many ways o measuring the speed o sound in ree air ( i.e. not trapped in a tube) .
Method one using a fast timer
The frst microphone triggers the timer to start.
The second microphone triggers the time to stop.
When the hammer is made to strike the plate the sound wave travels to the two microphones triggering the nearer microphone frst and the urther microphone second.
B y separating the microphones by 1 m, the time delay is around 3 .2 ms.
This is a very simple method o measuring the speed o sound:
Two microphones are connected to a ast timer ( one which can measure the nearest millisecond or even microsecond) .
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2nd microphone
fast timer 1st microphone
hammer
metre ruler plate
Figure 10 Fast timer method for the speed of sound. This gives a value or the speed o sound to be 1 .0 c = _st = _______ 3 1 0 m s 1 3 .2 1 0
Connect a signal generator to a loudspeaker and set the requency to between 500 Hz and 2.0 kHz.
This should be repeated a ew times and an average value obtained.
You might think about how you could develop this experiment to measure the speed o sound in dierent media. S ound travels aster in solids and liquids than in gases can you think why this should be the case ? I the temperature in your country varie s signifcantly over the year you might try to perorm the experiment in a hot or cold corridor and compare your results.
O ne o the microphones needs to be close to the loudspeaker, with the second a metre or so urther away.
C ompare the two traces as you move the second microphone back and orth in line with the frst microphone and the speaker.
Use a ruler to measure the distance that you need to move the second microphone or the traces to change rom being in phase to changing to antiphase and then back in phase again.
The distance moved between the microphones being in consecutive phases will be the wavelength o the wave.
The speed is then ound by multiplying the wavelength by the requency shown on the signal generator.
3
Method two using a double beam oscilloscope
C onnect two microphones to the inputs o a double-beam oscilloscope ( fgure 1 1 ) .
oscilloscope signal generator
loudspeaker microphones
upper trace shifted horizontally to align with lower trace
Figure 11 Double beam oscilloscope method for the speed of sound.
Nature of Science Analogies can slow down progress You may have seen a demonstration, called the bell j ar experiment, to show that sound needs a material medium through which to travel.
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An electric bell is suspended rom the mouth o a sealed bell jar and set ringing (fgure 1 2 ) . A vacuum pump is used to evacuate the jar. When
4 . 2 T R AV E L L I N G W AV E S
there is no longer any air present in the jar, no sound can be heard. This experiment was frst perormed by Robert B oyle in 1 660 although, long beore this time, the Ancient Greeks understood that sound needed something to travel through. The frst measurements o the speed o sound were made over our hundred years ago. These experiments were based on measuring the time delay either between a sound being produced and its echo reecting rom a distant surace or between seeing the ash o a cannon when fred and hearing the bang. From the middle to the end o the seventeenth century physicists such as Robert Hooke and C hristiaan Huygens proposed a wave theory o light. In line with the model or sound, Huygens suggested that light was carried by
a medium called the luminierous ether. It was not until 1 887, when Michelson and Morley devised an experiment in an attempt to detect the ether wind, that people began to realize that electromagnetic waves were very dierent rom sound waves.
electric bell
to vacuum pump
Figure 12 The bell jar experiment.
Electromagnetic waves You may have seen the experiment shown in fgure 1 3 in which a beam o white light is dispersed into the colours o the visible spectrum by passing it through a prism. The act that what appears to be a single colour actually consists o multiple colours triggers the question what happens beyond the red and blue ends o the spectrum? The two colours represent the limit o vision o the human eye but not the limit o detection o electromagnetic radiation by the human body. Holding the back o your hand towards the S un allows you to eel the warmth o the S uns inra- red radiation. In the longer term, your hand will be subj ected to sunburn and even skin cancer caused by the higher energy ultraviolet radiation. Figure 1 4 shows the ull electromagnetic spectrum with the atmospheric windows; these are the ranges o electromagnetic waves that can pass through the layers o the atmosphere. All electromagnetic waves are transverse, carry energy, and exhibit the ull range o wave properties. They travel at 3 00 million metres per second ( 3 .00 1 0 8 m s 1 ) in a vacuum. All electromagnetic waves ( except gamma rays) are produced when electrons undergo an energy change, even though the mechanisms might dier. For example, radio waves are emitted when electrons are accelerated in an aerial or antenna. Gamma rays are dierent, they are emitted by a nucleus or by means o other particle decays or annihilation events. All electromagnetic waves consist o a time-varying electric feld with an associated time-varying magnetic feld. As the human eye is sensitive to the electric component, the amplitude o an electromagnetic wave is usually taken as the waves maximum electric feld strength. B y graphing the electric feld strength on the y-axes, we can use displacementdistance and displacementtime graphs to represent electromagnetic waves. In the ollowing discussion o the dierent areas o the electromagnetic spectrum the range o the wavelength is given but these values are not hard and ast. There are overlaps o wavelength when radiation o the same wavelength is emitted by dierent
Figure 13 Dispersion of white light using a glass prism.
Figure 14 The electromagnetic spectrum.
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O S C I LL AT I O N S AN D WAVE S mechanisms ( most notably X- rays and gamma rays) . In addition to comparing the wavelengths o the radiations it is sometimes more appropriate to compare their requencies ( or example, radio waves) or the energy o the wave photon ( or example, with X- rays and gamma rays) . The requency range is easy to calculate rom wavelengths using the equation c = f with c = 3 . 00 1 0 8 m s 1 . Photon energies will be discussed in Topic 7. Those electromagnetic waves with requencies higher than that o visible light ionize atoms and are thus harmul to people. Those with lower requencies are generally believed to be sae.
Nature of Science Infra-red radiation ( ir~1 1000 m) William Herschel discovered inra-red in 1 800 by placing a thermometer j ust beyond the red end o the spectrum ormed by a prism. Today we might do the same experiment but using an inra-red detector as a modern thermometer. Obj ects that are hot but not glowing, i.e. below 5 00 C , emit inra-red only. At this temperature obj ects become red-hot and emit red light in addition to inra-red. At around 1 000 C objects become white hot and emit the ull visible spectrum colours. Remote controllers or multimedia devices utilize inra-red as do thermal imagers used or night vision. Inrared astronomy is used to see through dense regions o gas and dust in space with less scattering and absorption than is exhibited by visible light.
Ultraviolet radiation ( uv ~ 100 400 nm) In 1 801 , Johann Ritter detected ultraviolet by positioning a photographic plate beyond the violet part o the spectrum ormed by a prism. It can also be detected using uorescent paints and inks these absorb the ultraviolet ( and shorter wavelengths) and re- emit the radiation as visible light. Absorption o ultraviolet produces important vitamins in the skin but an overdose can be harmul, especially to the eyes. The Sun emits the ull range o ultraviolet: UV-A, UVB , and UV- C . These classifcations are made in terms o the range o the wavelengths emitted ( UV- A ~3 1 5 400 nm, UV-B ~2 803 1 5 nm, and UV- C ~ 1 002 80 nm) . UV- C rays, having the highest requency, are the most harmul but, ortunately, they are almost completely absorbed by the atmosphere. UV- B rays cause sunburn and increase the risk o D NA and other cellular damage in living organisms; luckily only about 5 % o this radiation passes through the ionosphere ( this consists o layers o electrically
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charged gases between 80 and 400 km above the Earth) . Fluorescent tubes, used or lighting, contain mercury vapour and their inner suraces are coated with powders. The mercury vapour and powders uoresce when radiated with ultraviolet light. With the atmosphere absorbing much o the ultraviolet spectrum, satellites must be positioned above the atmosphere in order to utilize ultraviolet astronomy, which is very useul in observing the structure and evolution o galaxies.
Radio waves ( radio ~1 mm 100 km) James C lerk Maxwell predicted the existence o radio waves in 1 864. B etween 1 885 and 1 889 Heinrich Hertz produced electromagnetic waves in the laboratory, confrming that light waves are electromagnetic radiation obeying Maxwells equations. Radio waves are used to transmit radio and television signals. VHF ( or FM) radio waves have shorter wavelengths than those o AM. VHF and television signals have wavelengths o a ew metres and travel in straight lines ( or rays) rom the transmitter to the receiver. Long- wave radio relies on reections rom the ionosphere and diracts around obstacles on the Earths surace. S atellite communication requires signals with wavelengths less than 1 0 m in order to penetrate the ionosphere. Radio telescopes are used by astronomers to observe the composition, structure and motion o astronomic bodies. S uch telescopes are physically large, or example the Very Large Array ( VLA) radio telescope in New Mexico which consists o 2 7 antennas arranged in a Y pattern up to 3 6 km across.
Microwaves ( micro ~ 1 mm 30 cm) Microwaves are short wavelength radio waves that have been used so extensively with radar, microwave cooking, global navigation satellite
4 . 2 T R AV E L L I N G W AV E S
systems and astronomy that they deserve their own category. In 1 940, S ir John Randall and D r H A B oot invented the magnetron which produced microwaves that could be used in radar ( an acronym or radio detection and ranging) to locate aircrat on bombing missions. The microwave oven is now a common kitchen appliance; the waves are tuned to requencies that can be absorbed by the water and at molecules in ood, causing these molecules to vibrate. This increases the internal energy o the ood the container holding the ood absorbs an insignifcant amount o energy and stays much cooler. Longer wavelength microwaves pass through the Earths atmosphere more eectively than those o shorter wavelength. The C osmic B ackground Radiation is the elemental radiation feld that flls the universe, having been created in the orm o gamma rays at the time o the B ig B ang. With the universe now cooled to a temperature o 2 .73 K the peak wavelength is approximately 1 .1 mm ( in the microwave region o the spectrum) .
X-rays ( X~30 pm 3 nm) X- rays were frst produced and detected in 1 895 by the German physicist Wilhelm C onrad Roentgen. An X- ray tube works by fring a beam o electrons at a metal target. I the electrons
have sufcient energy, X- rays will emitted by the target. X-rays are well known or obtaining images o broken bones. They are also used in hospitals to destroy cancer cells. S ince they can also damage healthy cells, using lead shielding in an X- ray tube is imperative. Less energetic and less invasive X-rays have longer wavelengths and penetrate esh but not bone: such X- rays are used in dental surgery. In industry they are used to examine welded metal j oints and castings or aults. X- rays are emitted by astronomical obj ects having temperatures o millions o kelvin, including pulsars, galactic supernovae remnants, and the accretion disk o black holes. The measurement o X- rays can provide inormation about the composition, temperature, and density o distant galaxies.
Gamma rays ( < 1 pm) Gamma rays were discovered by the French scientist Paul Villard in 1 900. Gamma rays have the shortest wavelength and the highest requency o all electromagnetic radiation. They are generated, amongst other mechanisms, by naturally occurring radioactive nuclei in nuclear explosions and by neutron stars and pulsars. Only extra-terrestrial gamma rays o the very highest energies can reach the surace o the Earth the rest being absorbed by ozone in the Earths upper atmosphere.
Nature of Science Night vision Humans eyes are sensitive to the electric component o a portion o the electromagnetic spectrum known as the visible spectrum; this is what we call sight. S ome animals, in particular insects and birds, are able to see using the ultraviolet part o spectrum. However, ultraviolet consists o relatively short wavelength radiation that can damage animal tissue, yet these animals appear to be immune to these dangers. It has been speculated whether any animal eye might be adapted to be able to use the inra- red part
o the electromagnetic spectrum. As these long wavelengths have low energy it might be aretched to believe that they can be detected visually. There are animals that have evolved ways o sensing inra-red that are similar to the processes occurring in the eye. For example, the brains o some snakes are able to interpret the inra-red radiation in a way that can be combined with other sensory inormation to enable them to have a better understanding o surrounding danger or ood sources.
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4.3 Wave characteristics Understanding
Applications and skills
Waveronts and rays
Sketching and interpreting diagrams involving
Amplitude and intensity Superposition
Polarization
waveronts and rays Solving problems involving amplitude and intensity Sketching and interpreting the superposition o pulses and waves Describing methods o polarization Sketching and interpreting diagrams illustrating polarized refected and transmitted beams Solving problems involving Maluss law
Equations Relationship between intensity and amplitude:
I A2 Maluss law: I = I0 cos 2
Nature of science Imagination and physics Imagination ... is more important than knowledge. Knowledge is limited. Imagination encircles the world 1 . Einstein was amous or his gedanken thought experiments and one o the qualities that makes a great physicist is surely a hunger to ask the question what i ... ?. Mathematics is a crucial tool or the physicist and it is central to what a physicist does, to be able to quantiy an argument. However,
imagination can also play a major role in interpreting the results. Without imagination it is hard to believe that we would have Huygens principle, Newtons law o gravitation, or Einsteins theory o special relativity. To visualize the abstract and apply theory to a practical situation is a fair that is undamental to being an extraordinary physicist. Viereck, George Sylvester (October 26, 1929) . What lie means to Einstein: an interview. The Saturday Evening Post.
1
Introduction Wavefronts and rays are visualizations that help our understanding of how waves behave in different circumstances. B y drawing ray or wave diagrams using simple rules we can predict how waves will behave when they encounter obstacles or a different material medium.
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4 . 3 W AV E C H A R A C T E R I S T I C S
D efnitions o these two quantities are:
A wavefront is a surace that travels with a wave and is perpendicular to the direction in which the wave travels the ray.
A ray is a line showing the direction in which a wave transers energy and is, o course, perpendicular to a waveront.
The distance between two consecutive waveronts is one wavelength ( ) .
The motion of wavefronts O ne o the simplest ways to demonstrate the motion o waveronts is to use a ripple tank such as that shown in fgure 1 . This is simply a glass- bottomed tank that contains water, illuminated rom above. Any waves on the water ocus the light onto a screen oten placed below the tank. The bright patches result rom the crests ocusing the light and the dark patches rom the troughs deocusing the light. Using a vibrating dipper, plane or circular waves can be produced allowing us to see what happens to waveronts in dierent situations.
lamp
to power supply
elastic bands
vibrator water surface
dipper shallow water tray
wave pattern on screen
white screen
Figure 1 A ripple tank. The images shown in fgure 2 show the eect on the waveronts as they meet ( a) a plane barrier, ( b) a shallower region over a prism-shaped glass and ( c) a single narrow slit. Using waveront or ray diagrams we can illustrate how waves behave when they are reected, reracted, and diracted. We will reer to these diagrams in the sections o this topic ollowing on rom this.
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4
O S C I LL AT I O N S AN D WAVE S (a)
(b)
(c)
Figure 2 Reection, reraction, and difraction o waves in a ripple tank. incident wave ront incident ray
reected wave ront reected ray
i r
metal strip placed in water
Figure 3a Waveront and ray diagrams or reection o waves.
deep water
shallow water
Figure 3b Waveront and ray diagrams or reraction o waves.
Figure 3c Waveront and ray diagrams or difraction o waves by a single slit.
In fgure 3 a we can see that there is no change o wavelength and that the angle o incidence ( i) is equal to the angle o reection ( r) . In fgure 3 b we see the wave slowing down and bending as it enters the denser medium. The wavelength o the wave in the denser medium is shorter than in the less dense medium but the requency remains unchanged ( although it is not possible to tell this rom the the pattern shown in the ripple tank) . Figure 3 c shows diraction where the wave spreads out on passing through the slit but there is no change in the wavelength.
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4 . 3 W AV E C H A R A C T E R I S T I C S
Nature of Science Huygens principle O ne o the ways to predict what will happen to waveronts under dierent circumstances is to use Huygens principle. This was suggested in 1 6 7 8 by the D utch physicist C hristiaan Huygens: the waveront o a travelling wave at a particular instant consists o the tangent to circular wavelets given out by each point on the previous waveront as shown in fgure 4. In this way the waveront travels orward with a velocity c. Huygens was able to derive the laws o reection and reraction rom his principle. What the principle does not explain is why an expanding circular ( but really spherical) wave continues to expand outwards rom its source rather than travel back and ocus on the source. The French physicist Augustin Fresnel ( 1 7881 82 7) adapted Huygens principle to explain diraction by proposing the principle o superposition, discussed in S ub-topic 4.4. The HuygensFresnel principle
( as the overall principle should more accurately be called) is useul in explaining many wave phenomena.
ct secondary wavelets
ct
primary source
secondary sources
plane wavefronts
spherical wavefronts
Figure 4 Huygens principle.
The intensity of waves The loudness o a sound wave or the brightness o a light depends on the amount o energy that is received by an observer. For example, when a guitar string is plucked more orceully the string does more work on the air and so there will be more energy in the sound wave. In a similar way, to make a flament lamp glow more brightly requires more electrical energy. The energy E is ound to be proportional to the square o the amplitude A: E A2 S o doubling the amplitude increases the energy by a actor o our; tripling the amplitude increases the energy by a actor o nine, etc. Loudness is the observers perception o the intensity o a sound and brightness that o light; loudness and brightness are each aected by requency.
I = P/4r2 r
I we picture waves being emitted by a point source, S , they will spread out in all directions. This will mean that the total energy emitted will be spread increasingly thinly the urther we go rom the source. Figure 5 shows how the energy spreads out over the surace area o a sphere. In order to make intensity comparisons more straightorward it is usual to use the idea o the energy transerred per second this is the power ( P) o the source. This means that the intensity ( I) at a distance ( r)
S
Figure 5 Energy spreading out from a point source.
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4
O S C I LL AT I O N S AN D WAVE S I /9 I /4 I
r 2r 3r
Figure 6 Inverse square law.
rom a point source is given by the power divided by the surace area o the sphere at that radius: P I = _2 4r This equation shows that intensity has an inverse-square relationship with distance rom the point source; this means that, as the distance doubles, the intensity alls to a quarter o the previous value and when the distance the is tripled, the intensity alls to one nineth as shown in fgure 6. The S I unit or intensity is W m 2 .
Worked example At a distance o 1 5 m rom the source, the intensity o a loud sound is 2 .0 1 0 4 W m 2 . a) Show that the intensity at 1 2 0 m rom the source is approximately 3 1 0 6 W m 2 . b) D educe how the amplitude o the wave changes.
Solution P P a) Using the equation I = ___ , __ remains constant so I1 r12 = I2 r22 this 4 4 r 2
gives 2 .0 1 0 4 1 5 2 = I 2 1 2 0 2 1 5 2 = 3 .1 1 0 6 W m 2 I2 = 2.0 1 0 4 _ 2 1 20
3 10
6
Wm
2
Note In show that questions there is an expectation that you will give a detailed answer, showing all your working and that you will give a fnal answer to more signifcant fgures than the data in the question actually this is good practice or any answer! b) With the intensity changing there must be a change o amplitude. The intensity is proportional to the square o the amplitude so: __ _________ 2 I A I A 2 .0 1 0 4 _2 = _1 or _1 = _2 = _ = 8.0 2 I2 A2 I2 3 . 1 1 0 6 A2
Thus the amplitude at 1 5 m rom the source is 8.0 times that at 1 2 0 m rom the source. Another way o looking at this is to say 1 that A is proportional to __ r.
Note Because the previous part was a show that question you had all the data needed to answer this question. In questions where you need to calculate data you will never be penalized or using incorrect data that you have calculated previously. In a question that has several parts you might ail to gain a sensible answer to one o the parts. When a subsequent part requires the use o your answer as data dont give up, invent a sensible value (and say that is what you are doing) . You should then gain any marks available or the correct method.
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4 . 3 W AV E C H A R A C T E R I S T I C S
The principle of superposition Unlike when solid obj ects collide, when two or more waves meet the total displacement is the vector sum o their individual displacements. Having interacted, the waves continue on their way as i they had never met at all. This principle is used to explain intererence and standing waves in S ub- topics 4.4 and 4.5 respectively. For a consistent pattern, waves need to be o the same type and have the same requency and speed; the best patterns are achieved when the waves have the same or very similar amplitudes. Figure 7 shows two pulses approaching and passing through each other. When they meet, the resultant amplitude is the algebraic sum o the two amplitudes o the individual pulses.
Figure 7 Superposition of pulses. This principle applies equally well to complete waves as to pulses this is shown in fgure 8. You should convince yoursel that the green wave is the vector sum o the red and blue waves at every instant. It is equally valid to use the principle o superposition with displacementdistance graphs. 0.3
displacement/m
0.2 0.1 0 0
0.1
0.2
0.3
0.4
0.5
0.1 time/s 0.2 0.3
Figure 8 Displacementtime graph showing the superposition of waves.
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O S C I LL AT I O N S AN D WAVE S
Worked examples 1
Two identical triangular pulses of amplitude A travel towards each other along a rubber cord. At the instant shown on the diagram below, point M is midway between the two pulses.
A
C
b) ( i)
E 1 F medium R
medium I
A 2
M
A
What is the amplitude of the disturbance in the string as the pulses move through M?
B
O n entering the new medium the waves refract by the same amount to give parallel wavefronts.
Solution The pulses are symmetrical, so when they meet they completely cancel out giving zero amplitude at M. 2
a)
( ii) As the wavefronts are closer in the medium R the waves are travelling more slowly.
For a travelling wave, distinguish between a ray and a wavefront.
(iii) The frequency of a wave does not change when the wave moves from one medium to another. As c = f c then __ = constant and so:
The diagram below shows three wavefronts incident on a boundary between medium I and medium R. Wavefront C D is shown crossing the boundary. Wavefront EF is incomplete.
c1 c2 1 c1 __ __ = __ __ c = 1
3 A
C
2
2
2
The graphs below show the variation with time of the individual displacements of two waves as they pass through the same point.
E F
medium I
D
medium R displacement
A1 x1
0
0
time
T
D B A 1
( i) O n a copy of the diagram above, draw a line to complete the wavefront EF. ( ii) E xplain in which medium, I or R, the wave has the higher speed. ( iii) B y taking appropriate measurements from the diagram, determine the ratio of the speeds of the wave travelling from medium I to medium R.
Solution a) A ray is a line that shows the direction of propagation of a wave. Wavefronts are lines connecting points on the wave that are in phase, such as a crest or a trough. The distance between wavefronts is one wavelength and wavefronts are always perpendicular to rays.
140
A2 displacement
b)
0 x2 0
T
time
A 2
What is the total displacement of the resultant wave at the point at time T?
Solution The total displacement will be the vector sum of the individual displacements. As they are in opposite directions the vector sum will be their difference, i.e. x1 x2
4 . 3 W AV E C H A R A C T E R I S T I C S
Polarization Although transverse and longitudinal waves have common properties they reect, reract, diract and superpose the dierence between them can be seen by the property o polarization. Polarization o a transverse wave restricts the direction o oscillation to a plane perpendicular to the direction o propagation. Longitudinal waves, such as sound waves, do not exhibit polarization because, or these waves, the direction o oscillation is parallel to the direction o propagation. Figure 9 shows a demonstration o the polarization o a transverse wave on a rubber tube. vibrations go through
same plane as ence
ence with vertical rails acting as slits
vibrations stopped
ence with horizontal rails acting as slits
diferent plane
vibrations stopped at second ence
two planes at right angles
Figure 9 Polarization demonstration. Most naturally occurring electromagnetic waves are completely unpolarized; this means the electric feld vector (and thereore the magnetic feld vector perpendicular to it) vibrate in random directions but in a plane always at right angles to the direction o propagation o the wave. When the direction o vibration stays constant over time, the wave is said to be plane polarized in the direction o vibration this is the case with many radio waves which are polarized as a result o the orientation o the transmitting aerial (antenna) . Partial polarization is when there is some restriction to direction o vibration but not 1 00%. There is a urther type o polarization when the direction o vibration rotates at the same
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4
O S C I LL AT I O N S AN D WAVE S
unpolarized incident beam
p
plane polarized refected beam
p
90 2
requency as the wave this called circular or elliptical polarization and is caused when a wave is in a strong magnetic feld but you will not be examined on this. Figure 1 0 shows how we represent polarized and unpolarized light diagramatically; the double-headed arrow represents a polarized wave, showing the plane o polarization o the wave. The crossed arrows show that the vibration has an electric feld vector in all planes and these are resolved into the two perpendicular planes shown (you may see this marked as our double-headed arrows in some texts) . These diagrams can become a little conusing when rays are added to show the direction in which the waves are travelling. Figure 1 1 shows some examples o this.
or unpolarized light
partially polarized reracted beam
Figure 11 Polarization o refected light.
polarized light
Figure 10 Representing polarized and unpolarized waves.
Polarization of light In 1 809 the French experimenter tienne-Louis Malus showed that when unpolarized light reected o a glass plate it could be polarized depending upon the angle o incidence the plane o polarization being that o the at surace reecting the light. In 1 81 2 the Scottish physicist S ir D avid B rewster showed that when unpolarized light incident on the surace o an optically denser material ( such as glass) , at an angle called the polarizing angle, the reected ray would be completely plane polarized. At this angle the reected ray and reracted ray are at right angles as shown in fgure 1 1 . Today the most common method o producing polarized light is to use a polarizing flter ( usually called Polaroid) . These flters are made rom chains o microcrystals o iodoquinine sulate embedded in a transparent cellulose nitrate flm. The crystals are aligned during manuacture and electric feld vibration components, parallel to the direction o alignment, become absorbed. The electric feld vector causes the electrons in the crystal chains to oscillate and thus removes energy rom the wave. The direction perpendicular to the chains allows the electric feld to pass through. The reason or this is that the limited width o the molecules restricts the motion o the electrons, meaning that they cannot absorb the wave energy. When a pair o Polaroids are oriented to be at 90 to each other, or crossed, no light is able to pass through. The frst Polaroid restricts the electric feld to the direction perpendicular to the crystal chains; the second Polaroid has its crystals aligned in this direction and so absorbs the remaining energy. The frst o the two Polaroids is called the p olarizer and the second is called the analyser.
Maluss law When totally plane- polarized light ( rom a polarizer) is incident on an analyser, the intensity I o the light transmitted by the analyser is directly proportional to the square o the cosine o angle between the transmission axes o the analyser and the polarizer.
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4 . 3 W AV E C H A R A C T E R I S T I C S
Figure 1 2 shows polarized light with the electric feld vector o amplitude E0 incident on an analyser. The axis o transmission o the analyser makes an angle with the incident light. The electric feld vector E0 can be resolved into two perpendicular components E0 cos and E0 sin . The analyser transmits the component that is parallel to its transmission axis, which is E0 cos . We have seen that intensity is proportional to the square o the amplitude o a wave so I0 E02 the transmitted intensity will be I which is proportional to ( E0 cos ) 2 or I E02 cos 2 2
2
E co s I = _______ cancelling E 0 2 gives Taking ratios we have __ I E
orientation of analyser E0
E 0 cos
orientation of light from polarizer
0
2
0
0
2
I = I0 cos
Figure 12 Analysing polarized light.
When = 0 ( or 1 80) I = I0 ( since cos 0 = 1 ) ; this means that the intensity o light transmitted by the analyser is maximum when the transmission axes o the two Polaroids are parallel. When = 90, I = I0 cos 2 90 = 0; this means that no light is transmitted by the analyser when the Polaroids are crossed. Figure 1 3 shows a pair o Polaroids. The let-hand image shows that, when their transmission axes are aligned, the same proportion o light passing through one Polaroid passes through both. The central image shows that when one o the Polaroids is rotated slightly, less light passes through their region o overlap. The right- hand image shows that where the Polaroids are crossed no light is transmitted.
Figure 13 Two Polaroids.
Nature of science Uses of polarization of light Polaroid sunglasses are used to reduce the glare coming rom the light scattered by suraces such as the sea or a swimming pool. In industry, stress analysis can be perormed on models made o transparent plastic by placing the model in between a pair o crossed Polaroids. As white light passes through a plastic, each colour o the spectrum is polarized with a unique orientation. There is high stress in the regions where the colours are most concentrated this is where the model ( or real obj ect being modelled) is most likely to break when it is put under stress. C ertain asymmetric molecules ( called chiral molecules) are op tically active this is the ability to rotate the plane o plane-polarized light. The angle that the light is rotated through is measured using a p olarimeter, which consists o a light source, and a pair o Polaroids. The light passes through the frst Polaroid ( the polarizer) and, initially with no sample present, the second Polaroid ( the analyser) is aligned so that no light passes through. With a sample placed between
the Polaroids, light does pass through because the sample has rotated the plane o polarization. The analyser is now rotated so that again no light passes through. The angle o rotation is measured and rom this the concentration o a solution, or example, can be ound. An easily produced, optically active substance is a sugar solution. Polarization can also be used in the recording and proj ection o 3 D flms. These consist o two flms proj ected at the same time. Each o the flms is recorded rom a slightly dierent camera angle and the proj ectors are also set up in this way. The two flms are proj ected through polarizing flters one with its axis o transmission horizontal and one with it vertical. B y wearing polarized eye glasses with one lens horizontally polarized and one vertically polarized, the viewers let eye only sees the light rom the let proj ector and the right eye light rom the right proj ector thus giving the viewer the perception o depth. In cinemas the screen needs to be metallic as non- metallic at suraces have a polarizing eect.
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Worked examples 1
Unpolarized light o intensity I0 is incident on a polarizer. The transmitted light is then incident on an analyser. The axis o the analyser makes an angle o 60 to the axis o the polarizer.
position shown, the transmission axis o the analyser is parallel to the plane o polarization o the light ( = 0) . S ketch a graph to show how the intensity I o the transmitted light varies with as the analyser is rotated through 1 80.
analyser
Solution a) In unpolarized light the electric feld vector vibrates randomly in any plane ( perpendicular to the direction o propagation) . In polarized light this vector is restricted to j ust one plane.
polarizer
unpolarized light
b) This is an application o Maluss law I = I0 cos 2
C alculate the intensity emitted by the analyser.
Solution I The frst polarizer restricts the intensity to __ . 2 Using Maluss law I = I0 cos 2 0
2
cos 60 = 0.5 so cos 2 60 = 0.2 5 I0 thus I = 0.2 5 _ = 0. 1 2 5 I0 2 a) D istinguish between polarized and unpolarized light. b) A beam o plane- polarized light o intensity I0 is incident on an analyser.
When = 0 or 1 80, cos = 1 and so cos 2 = 1 and I = I0 When = 90, cos = 0 and so cos 2 = 0 and I = 0 These are the key points to ocus on. Note that cos 2 will never become negative. There is no need to include a unit or intensity as this is a sketch graph. I0
transmission axis plane of polarization
I
incident beam
analyser
The angle between the transmission axis o the analyser and the plane o polarization o the light can be varied by rotating the analyser about an axis parallel to the direction o the incident beam. In the
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0 0
90 /
180
4 . 4 W AV E B E H AV I O U R
4.4 Wave behaviour Understanding Reection and reraction
Applications and Skills Sketching and interpreting incident, reected
Snells law, critical angle, and total
internal reection Difraction through a single-slit and around objects Intererence patterns Double-slit intererence Path diference
and transmitted waves at boundaries between media Solving problems involving reection at a plane interace Solving problems involving Snells law, critical angle, and total internal reection Determining reractive index experimentally Qualitatively describing the difraction pattern ormed when plane waves are incident normally on a single-slit Quantitatively describing double-slit intererence intensity patterns
Equations n
sin sin1
v
_________2 = ____1 1 Snells law: ____ n = v 2
2
D Intererence at a double slit: s = ______ d
Nature of science Wave or particle? In the late seventeenth century two rival theories o the nature o light were proposed by Newton and Huygens. Newton believed light to be particulate and supported his view by the acts that it apparently travels in straight lines and can travel through a vacuum; at this time it was a strongly held belie that waves needed a medium through which to travel. Huygens wave model was supported by the work o Grimaldi who had shown that light difracts around small objects and through narrow openings. He
was also able to argue that when a wave meets a boundary the total incident energy is shared by the reected and transmitted waves; Newtons argument or this was based on the particles themselves deciding whether or not to reect or transmit this was not a strong argument and the wave theory o light became predominant. In the 21st century light is treated as both a wave and a particle in order to explain the ull range o its properties.
Introduction Now we have looked at how to describe waves, we are in the position to look at wave properties or, as it is described in IB Physics, wave behaviour. You are likely to have come across some of this topic if you
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4
O S C I LL AT I O N S AN D WAVE S normal
incident ray
refected ray 1 1
medium 1: reractive index = n 1 medium 2: reractive index = n 2 2
have studied physics beore starting the IB D iploma Programme and we have used some o the ideas in the previous sub- topic. In S ub- topic 4.3 we studied polarization a property that is restricted to transverse waves alone. The ideas examined in this sub- topic apply to transverse waves ( both mechanical and electromagnetic) and longitudinal waves. There are many demonstrations o wave properties utilizing sound and light; however, microwaves are commonly used or demonstrating these wave properties too.
Refection and reraction o waves reracted ra
Figure 1 Refection and reraction o waves.
We have looked at much o the content o fgure 1 when we considered polarization. We will now ocus on what is happening to the rays remember we could always add waveronts at right angles to the rays drawn on these diagrams. What the ray diagrams do not show is what is happening to the wavelength o the waves we will return to this in due course. The laws o reection and reraction can be summarized in three laws as ollows: 1
The refected and reracted rays are in the same p lane as the incident ray and the normal. This means that the event o reection or reraction does not alter the plane in which the light ray travels this is not obvious because we draw ray diagrams in two dimensions but, when we use ray boxes to perorm experiments with light beams, we can confrm that this is the case.
2
The angle o incidence equals the angle o refection. The angle o incidence is the angle between the incident ray and the normal and the angle o reection is the angle between the reected ray and the normal. The normal is a line perpendicular to a surace at any chosen point. The angle o incidence and reection are both labelled as 1 on fgure 1 .
3
For waves o a p articular requency and or a chosen p air o media the ratio o the sine o the angle o incidence to the sine o the angle o reraction is a constant called the (relative) reractive index. This is called Snells law ( or D escartess law in the French speaking world) . The angle o reraction is the angle between the reracted ray and the normal. Snells law can be written as sin 1 _ = 1n2 sin 2 For light going rom medium 1 to medium 2 this way o writing Snells law has several variants and the IB course uses one that we will look at soon.
Figure 2 Use o ray box to demonstrate the laws o refection and reraction.
146
When light is normal on a surace S nells law breaks down because the light passes directly through the surace.
4 . 4 W AV E B E H AV I O U R
TOK
Nature of science
Conservation of energy and waves
What happens to light at an interace between two media?
The wave and ray diagram or refection and reraction tells just part o the story. As with many areas o physics, returning to the conservation o energy is important. For the total energy incident on the interace between the two media the energy is shared between the refected wave, the transmitted wave and the energy that is absorbed the urther the wave passes through the second medium, the more o the energy is likely to be absorbed. The conservation o mass/energy is a principle in physics which, to date, has not let physicists down. Does this mean that we have proved that the principle o conservation o energy is inallible?
This is a complex process but in general terms, when charges are accelerated, or example when they are vibrated, they can emit energy as an electromagnetic wave. In moving through a vacuum the electromagnetic wave travels with a velocity o 3 .00 1 0 8 m s 1 . When the wave reaches an atom, energy is absorbed and causes electrons within the atom to vibrate. All particles have requencies at which they tend to vibrate most efciently called the natural frequency. When the requency o the electromagnetic wave does not match the natural requency o vibration o the electron, then the energy will be re- emitted as an electromagnetic wave. This new electromagnetic wave has the same requency as the original wave and will travel at the usual speed in the vacuum between atoms. This process continues to be repeated as the new wave comes into contact with urther atoms o unmatched natural requency. With the wave travelling at 3 .00 1 0 8 m s 1 in space but being delayed by the absorptionre- emission process, the overall speed o the wave will be reduced. In general, the more atoms per unit volume in the material, the slower the radiation will travel. When the requency o the light does match that o the atoms electrons the re- emission process is occurs in all directions and the atom gains energy, increasing the internal energy o the material.
Refractive index and Snells law The absolute refractive index (n) o a medium is defned in terms o the speed o electromagnetic waves as: speed o electromagnetic waves in a vacuum c n = ____ = _ v speed o electromagnetic waves in the medium
The reractive index depends on the requency o the electromagnetic radiation and, since the speed o light in a vacuum is the limit o speed, the absolute reractive index is always greater than 1 ( although there are circumstances when this is not true but that is well beyond the expectation o your IB Physics course) . For all practical purposes the absolute reractive index o air is 1 so it is not necessary to perorm reractive index experiments in a vacuum.
Worked example C alculate the angle o reraction when the angle o incidence at a glass surace is 5 5 ( reractive index o the glass = 1 .48) .
Solution As we are dealing with air and glass there is no dierence between absolute reractive index and relative reractive index.
sin = n glass S nells law gives ____ sin 1
2
sin5 5 = _ = 1 .48 sin 2 sin5 5 0.81 9 sin 2 = _ = _ = 0.5 5 3 1 .48 1 . 48 = sin 1 ( 0.5 5 3 ) = 3 3 .6 3 4
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Reversibility of light You should be able to prove to yourself that rays are reversible. Place a ray box and a glass block on a piece of paper. Mark, on the paper, the path of the beam of light emitted by the ray box as it approaches and leaves the glass block. Then place the ray box on the other side of the block and you will see that the light travels along the same path in the opposite direction. We have seen that for light travelling from medium 1 to medium 2 S nells law can be written as sin 1 _ = 1n2 sin 2 here 1 n 2 means the relative refractive index going from medium 1 to medium 2 . For light travelling in the opposite direction and since light is reversible we have sin 2 _ = 2n1 sin 1 1 It should be clear from this that 1 n 2 = ___ n 2
1
Worked example The ( absolute) refractive index of water is 1 .3 and that of glass is 1 .5 .
c)
water water
a) C alculate the relative refractive index from glass to water. b) Explain what this implies regarding the refraction of light rays. c) D raw a wavefront diagram to show how light travels through a plane interface from glass to water.
n
vac water
= 1 . 3 and n glass =
We are calculating sin va c
______ = sin w a te r
n
vac water
glass
and _____ = sin gla s s
w a te r
This means that
w ate r
glass
n
vac glass
= 1 .5
n water
sin v ac
sin gla s s sin gla s s sin v a c _____ ______ = ______ = sin sin sin va c
glass glass
Solution a) n water =
interface
vac
glass
n glass
n water
1 n water = ___ 1 .3 = 0.87 1 .5
b) With a relative refractive index less than one this means that the light travels faster in the water than the glass and therefore bends away from the normal.
As the refractive index of water is lower than that of glass the light wave speeds up on entering the water. The frequency is constant and, as = f, the wavelength will be greater in water meaning that the wavefronts are further apart. The fact that the frequency remains constant is a consequence of Maxwells electromagnetic equations something not covered in IB Physics.
The critical angle and total internal refection When a light wave reaches an interface travelling from a higher optically dense medium to a lower one the wave speeds up. This means that the wavelength of the wave increases ( frequency being constant) and
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4 . 4 W AV E B E H AV I O U R
the direction o the wave moves away rom the normal the angle o reraction being greater than the angle o incidence so as the angle o incidence increases the angle o reraction will approach 90. O p tical density is not the same as physical density, i.e. mass per unit volume, it is measured in terms o reractive index higher reractive index material having higher optical density. Light incident at any angle > c is totally refected.
Though not bent, part o the normal ray is refected. c 1
2
3
4
critical angle c
5
n1 high index material
light source
Figure 3 Light passing rom more optically dense medium to less optically dense medium. Ray 1 in fgure 3 shows a ray passing rom a more optically dense medium to a less optically dense medium normal to an interace. Most o the light passes though the interace but a portion is reected back into the original medium. Increasing the angle o incidence ( as or rays 2 and 3 ) will increase the angle o reraction and ray 4 shows an angle o incidence when the angle o reraction is 90. The angle o incidence at this value is called the critical angle ( c) . Ray 5 shows that, when the angle o incidence is larger than the critical angle, the light wave does not move into the new medium at all but is reected back into the original medium. This process is called total internal refection. It should be noted that or angles smaller than the critical angle there will always be a reected ray; although this will carry only a small portion o the incident energy. A ray box and semicircular glass block can be used to measure the critical angle or glass as shown in fgure 4. When the beam is incident on the curved ace o the block making it travel towards the centre o the at ace, it is acting along a radius and so enters the block normally and, thereore, without bending. B y moving the ray box, dierent angles o incidence can be obtained. B oth the critical angle and total internal reection can be seen. r1
r = 90 c
i1 semicircular glass slab
critical angle
i2
r2
i2 = r2
ray box
Figure 4 Use o ray box to investigate critical angle and total internal refection.
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4
O S C I LL AT I O N S AN D WAVE S
Worked example C alculate the ( average) critical angle or a material o ( average) absolute reractive index 1 .2 .
Solution The word average is included because the reractive index and critical angle would each be dierent or dierent colours. D ont be surprised i it is let out in some questions it is implied by talking about single values. 1 1 _ sin c = _ n 1 = 1 .2 = 0.83 3 c = sin1 0.833 = 56.4 56
When a ray box emitting white light is used, the light emerging through the glass block is seen to disp erse into the colours o the rainbow. This is due to each o the colours, o which white light is comprised, having a dierent requency. The reractive index or each o the colours is dierent.
Calculating the critical angle sin S nells law gives ____ = 1n2 . sin 1
2
In order to obtain a critical angle, medium 1 must be more optically dense than medium 2 . When 1 = c then 2 = 90 so sin 2 = 1 This gives sin c = 1 n 2 1
n2 n2 = _ n1
When the less dense medium ( medium 2 ) is a vacuum or air then n 2 = 1 1 S o sin c = _ n1
Investigate! Measuring the refractive index There are several possible experiments that you could do to measure the reractive index depending on whether a substance is a liquid or a solid ( we assume the reractive index o gases is 1 although mirages are a good example o how variations in air density can aect the reraction o light) . You could research how to use real and apparent depth measurements to measure the reractive index o a liquid. The investigation outlined below will be to trace some rays ( or beams o light, really) through a glass block o rectangular cross-section.
The arrangement is similar to that shown in fgure 2 on p1 46.
Place the block on a piece o white paper and mark its position by drawing around the edges.
D irect a beam o light to enter the block near the centre o a longer side and to leave by the opposite side.
150
Mark the path o the beam entering and leaving the block you will need at least two points on each beam to do this. Remove the block and use a ruler to mark the path o the beam on either side o the
block and then inside. Add arrows to these to remind you that they represent rays and to indicate which is the incident beam and which is the reracted beam.
Using a protractor, mark in and draw normals or the beam entering and leaving the block.
Remembering that light is reversible and the beam is symmetrical, measure two values or each o 1 and 2 .
C alculate the reractive index o the block using S nells law.
Estimate the experimental uncertainty on your measurements o 1 and 2 .
Note that the uncertainty in 1 and 2 is not the same as that in sin 1 and sin 2 but you sin can calculate the uncertainty in ____ ( and sin 1
2
hence n) by calculating hal the dierence sin sin and ______ between ______ sin sin 1 m ax
1 m in
2 m in
2 m ax
Repeat the experiment or a range o values o the angle o incidence.
Which o your values is likely to be the most reliable?
4 . 4 W AV E B E H AV I O U R
Difraction The frst detailed observation and description o the phenomenon that was named diffraction was made by the Italian priest Francesco Grimaldi. His work was published in 1 665 , two years ater his death. He ound that when waves pass through a narrow gap or slit ( called an aperture) , or when their path is partly blocked by an obj ect, the waves spread out into what we would expect to be the shadow region. This is illustrated by fgure 5 and can be demonstrated in a ripple tank. He noted that close to the edges the shadows were bordered by alternating bright and dark ringes. Given the limited apparatus available to Grimaldi his observations were quite extraordinary.
aperture
obstacle
obstacle
Figure 5 Difraction. Further observation o diraction has shown:
the requency, wavelength, and speed o the waves each remains the same ater diraction
the direction o propagation and the pattern o the waves change
the eect o diraction is most obvious when the aperture width is approximately equal to the wavelength o the waves.
the amplitude o the diracted wave is less than that o the incident wave because the energy is distributed over a larger area.
E xplaining diraction by a single slit is complex and you will only be asked or a qualitative description o single- slit diraction at S L however, as mentioned in Sub-topic 4. 3 , the HuygensFresnel principle gives a good insight into how the single-slit diraction pattern comes about ( see fgure 6) .
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O S C I LL AT I O N S AN D WAVE S
Worked example C omplete the ollowing diagrams to show the waveronts ater they have passed through the gaps.
Figure 6 HuygensFresnel explanation o difraction.
Solution
When the width o the slit is less than or equal to the wavelength o the wave, the waves emerge rom the slit as circular waveronts. As the slit width is increased, the spreading o the waves only occurs at the edges and the diraction is less noticeable.
Plane waves travelling towards the slit behave as i they were sources o secondary wavelets. The orange dots in fgure 6 show these secondary sources within the slit. These sources each spread out as circular waves. The tangents to these waves will now become the new waveront. The central image is bright and wide, beyond it are urther narrower bright images separated by darkness. S ingle-slit diraction is urther explored or those studying HL Physics in Topic 9.
Figure 7 Single-slit difraction pattern.
Double-slit interference We briey discussed the principle o superposition in S ub- topic 4.3 . Intererence is one application o this principle. When two or more waves meet they combine to produce a new wave this is called interference. When the resultant wave has larger amplitude than any o the individual waves the intererence is said to be constructive; when the resultant has smaller amplitude the intererence is destructive. Intererence can be achieved by using two similar sources o all types o wave. It is usually only observable i the two sources have a constant phase relationship this means that although they need not emit the two sets o waves in phase, the phase o the waves cannot alter relative to one another. S uch sources will have the same requency and are said to be coherent.
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4 . 4 W AV E B E H AV I O U R
audio signal generator
speakers a
D
x L
S
L
S
S
L
S
L
S
L
microphone
S
L
S
L
S
L
L loud sound S sot sound
Figure 8 Intererence o sound waves. Intererence o sound waves is easy to demonstrate using two loudspeakers connected to the same audio requency oscillator as shown in fgure 8. Moving the microphone ( connected to an oscilloscope) in a line perpendicular to the direction in which the waves are travelling allows an equally spaced loudsot sequence to be detected. More simply, the eect can be demonstrated by the observer walking along the loudsot line while listening to the loudness. When a coherent beam o light is incident on two narrow slits very close together the beam is diracted at each slit and, in the region difracted beam rom top slit destructive intererence (trough meets trough) S constructive intererence (crest meets crest) lamp
double slit difracted beam rom bottom slit
Figure 9 Intererence o light waves.
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4
O S C I LL AT I O N S AN D WAVE S where the two diracted beams cross, intererence occurs as seen in fgure 9. A pattern o equally spaced bright and dark ringes ( shown in fgure 1 0) is obtained on a screen positioned in the region where the diracted beams overlap. When a crest meets a crest ( or a trough meets a trough) constructive intererence occurs. When a crest meets a trough destructive intererence occurs. A similar experiment to this was perormed by the talented English physicist ( and later physician) Thomas Young in 1 801 . The coherent beam is achieved by placing a single slit close to the source o light this means that the waveronts spreading rom the single slit each reach the double slit with the same phase relationship and so the secondary waves coming rom the double slit retain their constant phase relationship.
Figure 10 Fringes produced by a double-slit.
Path diference and the double-slit equation You will never be asked to derive this equation but the ideas regarding path dierence are vital to your understanding o intererence. B A2 S
d
C
A1
O
P D
Figure 11 The double-slit geometry. Figure 1 1 shows two slits ( apertures) A 1 and A 2 distance d apart. The double slit is at distance D rom a screen. O is the position o the central bright ringe ( arising rom constructive intererence) . B is the position o the next bright ringe above O ; the distance O B is the ringe spacing s. There will be another bright ringe distance s below O . The beams rom A 1 and A 2 to O will travel equal distances and so will meet with the same phase relationship that they had at A 1 and A 2 they have zero path dierence. At B the beam rom A 1 will travel an extra wavelength compared with the beam rom A 2 the path distance ( = A 1 P) equals . B ecause o the short wavelength o light and the act that D is very much larger than d, the line A 2 P is eectively perpendicular to lines A 1 B and C B ( C being the midpoint o A 1 A 2 ) . This means that the triangles A 1 A 2 P and C B O are similar triangles. A1 P BO s Taking ratios _ = _ or _ = _ D CO A1 A2 d D Rearranging gives s = _ d This gives the separation o successive bright ringes ( or bands o loud sound or a sound experiment) .
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4 . 4 W AV E B E H AV I O U R
In general or two coherent beams starting in phase, i the path dierence is a whole number o wavelengths we get constructive intererence and i it is an odd number o hal wavelengths we get destructive intererence. It must be an odd number o hal wavelengths or destructive intererence because an even number would give a whole number ( integer) and that is constructive intererence! S ummarizing this: For constructive intererence the path dierence must = n where n = 0, 1 , 2 1 where For destructive intererence the path dierence must = ( n + __ 2 ) n = 0, 1 , 2
n is known as the order o the ringe, n = 0 being the zeroth order, n = 1 the frst order, etc.
Worked examples 1
In a double-slit experiment using coherent light o wavelength , the central bright ringe is observed on a screen at point O, as shown below.
5 ___ 9 ___ 1 3 ___ , 32 , ___ , 72 , ___ , 1 12 , ___ giving a dierence is __ 2 2 2 2 total o 7 dark ringes.
2
Two coherent point sources S 1 and S 2 oscillate in a ripple tank and send out a series o coherent waveronts as shown in the diagram.
P coherent light
P O
Q
wavelength double slit screen (not to scale)
At point P, the path dierence between light arriving at P rom the two slits is 7.
S1
S2
S tate and explain the intensity o the waves at P and Q?
a) Explain the nature o the ringe at P.
Solution
b) S tate and explain the number o dark ringes between O and P.
C onsidering each waveront to be a crest; at point P two crests meet and so superpose constructively giving an amplitude which is twice that o one wave (assuming the waveronts each have the same amplitude) the intensity is proportional to the square o this so will be our times the intensity o either o the waves alone. At Q a blue crest meets a red trough and so there is cancellation occurring and there will be zero intensity.
Solution a) As the path dierence is an integral number o wavelengths there will be a bright ringe at P. b) For destructive intererence the path dierence must be an odd number o hal wavelengths, so there will be dark ringes when the path
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Investigate! Measuring the wavelength of laser light using a double slit
bright spots
laser double slit
screen
Figure 12 Measuring the wavelength of laser light using a double slit. This experiment, using a gas laser ( or laser pointer) , is a modern version o the one perormed by Young. A laser emits a highly coherent beam o light ideal or perorming this experiment. C are should be taken not to shine the laser beam or its refection into your eye should this happen, look away immediately to avoid the risk o permanent damage to your eye. O ne way to minimise eye damage is to keep sufcient light in the room so that you can still do the experiment. This means that the iris o your eye will not be ully open.
Set up the apparatus as shown in fgure 1 2 the screen should be a ew metres rom the double slit.
The double slit can be homemade by scratching a pair o narrow lines on a piece o glass painted with a blackened material or it could be a ready prepared slide.
Light rom the laser beam diracts through the slits and emerges as two separate coherent waves. B oth slits must be illuminated by the narrow laser beam; sometimes it may be necessary to use a diverging ( concave) lens to achieve this.
The intererence pattern is then proj ected onto the screen and the separation o the spots ( images o the laser aperture) is measured as accurately as possible using a metre ruler or tape measure. This is best done by measuring the distance between the urthest spots, remembering that nine spots would have a separation o 8S.
The distance rom the double slit to the screen ( D) should also be measured using metre rulers or a tape measure.
O nce your readings are taken the wavelength Sd o the light can be calculated rom = __ . D
The slit separation ( d) can be measured using a travelling microscope ( however, d is likely to be provided by a manuacturer) .
When red light is used as the source, the bright ringes are all red. When blue light is used as a source, the bright ringes are all blue. As blue light has a shorter wavelength than red light the blue ringes are closer together than the red ringes. When the source is white light the zeroth order ( central) bright ringe is white but the other ringes are coloured with the blue edges closer to the centre and the red edges urthest rom the centre. C an you explain this?
156
4 . 4 W AV E B E H AV I O U R
Investigate! Measuring the wavelength of microwaves using a double slit receiver
mA milliammeter
transmitter
double slit
Figure 13 Microwave arrangement.
The double slit is adj usted so that the slits are around 3 cm apart for maximum diffraction.
Arrange both transmitter and receiver about half a metre from the slit.
Alter the position of the receiver until the received signal is at its strongest.
occur as in Young' s double-slit interference experiment with light. Look for fringes on both sides of the maximum.
Measure as accurately as you can the values for D, d, and S.
C alculate the wavelength of the microwaves. Repeat for other distances of transmitter and receiver from the slits.
Now slowly rotate the receiver until the signal is weakest.
C over up one of the slits with a book and explain the result.
Remove the book so that two slits are again available and attempt to discover if fringes
These investigations can be repeated with sound waves or radio waves given appropriate transmitters, receivers and slits of the correct dimension.
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4.5 Standing waves Understanding
Applications and Skills
The nature of standing waves
Describing the nature and formation of standing
Boundary conditions
waves in terms of superposition Distinguishing between standing and travelling waves Observing, sketching, and interpreting standing wave patterns in strings and pipes Solving problems involving the frequency of a harmonic, length of the standing wave, and the speed of the wave
Nodes and antinodes
Nature of science Fourier synthesis Synthesizers are used to generate a copy of the sounds naturally produced by a wide range of musical instruments. Such devices use the principle of superposition to join together a range of harmonics that are able to emulate the sound of the natural instrument. Fourier synthesis works by combining a sine-wave signal with sine-wave or cosine-wave
harmonics of correctly chosen amplitude. The process is named after the French mathematician and physicist Jean Baptiste Joseph, Baron de Fourier who, in the early part of the nineteenth century, developed the mathematical principles on which synthesis of music is based. In many ways Fourier synthesis is a visualization of music.
Introduction We have seen in Sub-topic 4. 2 how travelling waves transfer energy from the source to the surroundings. In a travelling wave the position of the crests and troughs changes with time. Under the right circumstances, waves can be formed in which the positions of the crests and troughs do not change in such a case the wave is called a standing wave. When two travelling waves of equal amplitude and equal frequency travelling with the same speed in opposite directions are superposed, a standing wave is formed.
Standing waves on strings Figure 1 shows two travelling waves ( coloured green and blue) moving towards each other at four consecutive times t1 , t2 , t3 , and t4. The green and blue waves superpose to give the red standing wave. As the green and blue waves move forward there are points where the total displacement ( seen on the red wave) always remains zero these are called nodes. At other places the displacement varies between a
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4 . 5 S T A N D I N G W AV E S
maximum in one direction and a maximum in the other direction these are called antinodes. travelling waves standing wave
time t 1
time t2
time t 3
time t 4
2
standing wave
node
node
Figure 1 The formation of a standing wave. At any instant the displacement o the standing wave will vary at all positions other than the nodes. Thus a single rame shot o a standing wave would look like a progressive wave as can be seen rom the red wave in fgure 1 . When representing the standing wave graphically it is usual to show the extremes o standing waves, but over a complete time period o the oscillation the wave will occupy a variety o positions as shown by the arrow in the loop o fgure 2 . extreme positions (antinodes)
nodes
range of motion at antinode
Figure 2 Nodes and antinodes on a string.
Meldes string The apparatus shown in fgure 3 is useul in demonstrating standing waves on a string. A variant o this apparatus was frst used in the late nineteenth century by the German physicist Franz Melde. A string is strung between a vibration generator and a fxed end. When the vibration generator is
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O S C I LL AT I O N S AN D WAVE S
stroboscope
fxed end o string
node
antinode
vibration generator signal generator
Figure 3 Meldes string. connected to an audio requency oscillator, the end o the string attached to the vibration generator oscillates vertically. A wave travels down the string beore undergoing a phase change o 1 80 when it reects at the fxed end. The reected wave superposes with the incident wave and (at certain requencies) a standing wave is ormed. The requency o the audio requency generator is slowly increased rom zero and eventually a requency is reached at which the string vibrates with large amplitude in the orm o a single loop the frst harmonic. I the requency is urther increased, the amplitude o the vibrations dies away until a requency o twice the frst harmonic requency is achieved in this case two loops are ormed and we have the second harmonic requency shown in fgure 3. This is an example o resonance the string vibrates with large amplitude only when the applied requency is an integral multiple o the natural requency o the string. We return to resonance in more detail in Option B. Using a stroboscope to reeze the string reveals detail about a standing wave. For example, when the ash requency is slightly out o synch with the vibration requency, it is possible to see the variation with time o the strings displacement; this will be zero at a node but a maximum at an antinode.
Note Although we oten treat the point o attachment o the string to the vibration generator as being a node, this is not really correct. The generator vibrates the string to set up the wave and thereore the nearest node to the generator will be a short distance rom the vibrator. We call this inaccuracy an end correction but oten draw a diagram showing the node at the vibrator. Because some o the travelling wave energy is transmitted or absorbed by whatever is clamping the fxed end, the reected waves will be slightly weaker than the incident waves. This means that cancellation is not complete and there will be some slight displacement at the nodes. Within each loop all parts o the string vibrate together in phase, but loops next to each other are constistently 180 o out o phase. Although strings are oten used to produce sound waves, it is not a sound wave travelling along the string it is a transverse wave. This wave travels at a speed which is determined by the characteristics o the string.
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4 . 5 S T A N D I N G W AV E S
Harmonics on strings
N=1
We have seen rom Meldes string that a string has a number o requencies at which it will naturally vibrate. These natural requencies are known as the harmonics of string. The natural requency at which a string vibrates depends upon the tension o the string, the mass per unit length and the length o the string. With a stringed musical instrument each end o a string is fxed, meaning there will be a node at either end. The frst harmonic is the lowest requency by which a standing wave will be set up and consists o a single loop. Doubling the requency o vibration halves the wavelength and means that two loops are ormed this is called the second harmonic; three times the undamental requency gives the third harmonic (see fgure 4) .
N=2
N=3
N=4
You will see rom fgure 1 that the distance between two consecutive nodes is equal to hal a wavelength in other words each loop in a standing . Figure 4 shows the frst fve harmonics o a wave wave is equivalent to __ 2 on a string o a fxed length. With the speed o the wave along the string being constant, halving the wavelength doubles the requency; reducing the wavelength by a actor o three triples the requency, etc. (the wave equation c = f applies to all travelling waves) . The dierent harmonics can be achieved either by vibrating the string at the appropriate requency or by plucking, striking or bowing the string at a dierent position although plucking at the centre is likely to produce the frst, third and fth harmonics. B y pinching the string at dierent places a node is produced so, or example, pinching at the centre o the string would produce the even harmonics. In a musical instrument several harmonics occur at the same time, giving the instrument its rich sound.
N=5
Figure 4 Harmonics on a string.
Worked example A string is attached between two rigid supports and is made to vibrate at its frst harmonic requency f.
__
1 so a quarter This means that t = T4 = __ 4f o a period has elapsed and the string has gone through quarter o a cycle to give:
The diagram shows the displacement o the string at t = 0. ( ii) In this case the wave has gone through T 1 = __ and so will have hal a period t = __ 2 2f
(
(a) D raw the displacement o the string at time 1 1 ( i) t = _ ( ii) t = _ 4f 2f (b) The distance between the supports is 1 .0 m. A wave in the string travels at a speed o 2 40 m s 1 . C alculate the requency o the vibration o the string.
Solution (a) ( i) We must remember the relationship 1 between period T and requency f is T = __ . f
)
moved rom a crest to a trough:
(b) The string is vibrating in frst harmonic mode and so the distance between the fxed ends is hal a wavelength ( __ . 2 ) S o __ = 1 . 0 m and = 2 .0 m. 2
Using c = f
c 2 40 f = __ = ___ 2 .0
= 1 2 0 Hz
Standing waves in pipes S tanding waves in a pipe dier rom standing waves on a string. In pipes the wave medium is ( usually) air and the waves themselves are longitudinal. Pipes can have two closed ends, two open ends, or one
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4
O S C I LL AT I O N S AN D WAVE S open and one closed; the latter two being shown in fgure 5 . The sound waves are reected at both ends o the pipe irrespective o whether they are open or closed. The variation o displacement o the air molecules in the pipe determines how we graphically represent standing waves in pipes. There is a displacement antinode at an open end ( since the molecules can be displaced by the largest amount here) and there will be a displacement node at the closed end ( because the molecules next to a closed end are unable to be displaced) . S imilar ideas to strings apply to the harmonics in pipes. S trictly speaking, the displacement antinode orms j ust beyond an open end o the pipe but this end eect can be ignored or most purposes. 1 4
1 2
1st harmonic
1st harmonic 3 4
1 2nd harmonic
3rd harmonic 5 4
3 2
3rd harmonic
5th harmonic 7 4
2 4th harmonic
7th harmonic 9 4
9th harmonic
5 2
5th harmonic
Figure 5 Harmonics in a pipe.
Harmonics in pipes The harmonic in a pipe depends on whether or not the ends o a pipe are open or closed. For a pipe o fxed length with one open and one closed end there must always be a node at the closed end and an antinode at the open end. This means that only odd harmonics are available ( the number o the harmonic is the number o hal loops in this type o pipe) . For a pipe with two open ends there must always be an antinode at each end and this means that all harmonics are achieveable ( in this case the number o loops gives the number o the harmonic) . Lets compare the requencies o the harmonics or the one open end pipe. Suppose the pipe has a length L. The wavelength ( ) o the frst c harmonic would be 4L and, rom c = f, the requency would be __ . 4L c is the speed o sound in the pipe. 3 3c 4 For the third harmonic L = __ , so = __ L and the requency = __ ( or 4 3 4L three times that o the frst harmonic) .
A harmonic is named by the ratio o its requency to that o the frst harmonic.
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4 . 5 S T A N D I N G W AV E S
Worked examples 1
The frst harmonic requency or a particular organ pipe is 3 3 0 Hz. The pipe is closed at one end but open at the other. What is the requency o its second harmonic?
Solution A named harmonic is the ratio o its requency to that o the frst harmonic so in this case the second harmonic will be 660 Hz since the second harmonic has twice the requency o the frst harmonic. 2
The frst harmonic requency o the note emitted by an organ pipe which is closed
at one end is f. What is the frst harmonic requency o the note emitted by an organ pipe o the same length that is open at both ends?
Solution The length o a pipe closed at one end in frst harmonic mode is __ ( = L ) so = 4L 4 This length will be hal the wavelength o a pipe open at both ends so L = __ so = 2 L 2 S ince the wavelength has halved the requency must double so the new requency will be 2 f.
Boundary conditions In considering both pipes and strings we have assumed reections at the ends or boundaries. In meeting the boundary o a string the wave reects ( or at least partially reects) this is known as a fxed boundary; there will be the usual phase change o 1 80 at a fxed boundary meaning that the reected wave cancels the incident wave and so orms a node. The closed ends o pipes and edges o a drumhead also have fxed boundaries. In the case o an open-ended pipe there is still a reection o the wave at the boundary but no phase change, so the reected wave does not cancel the incident wave and there is an antinode ormed the same idea applies to strips o metal vibrated at the centre, xylophones, and vibrating tuning orks. This type o boundary is called a ree boundary.
Comparison o travelling waves and stationary waves The ollowing table summarizes the similarities and dierences between travelling waves and standing waves:
Property energy transer amplitude phase
wave profle (shape) wavelength
requency
Travelling wave energy is transerred in the direction o propagation all particles have the same amplitude within a wavelength the phase is dierent or each particle propagates in the direction o the wave at the speed o the wave the distance between adjacent particles which are in phase all particles vibrate with same requency.
Standing wave no energy is transerred by the wave although there is interchange o kinetic and potential energy within the standing wave amplitude varies within a loop maximum occurs at an antinode and zero at a node all particles within a loop are in phase and are antiphase (180 out o phase) with the particles in adjacent loops stays in the same position
twice the distance between adjacent nodes (or adjacent antinodes) all particles vibrate with same requency except at nodes (which are stationary)
TOK Pitch and frequency Musical pitch is closely linked to requency but also has a psychological component in relation to music. We think o pitch as being someones perception o requency. Musical notes o certain pitches, when heard together, will produce a pleasant sensation and are said to be consonant or harmonic. These sound waves orm the basis o a musical interval. For example, any two musical notes o requency ratio 2:1 are said to be separated by an octave and result in a particularly pleasing sensation when heard. Similarly, two notes o requency ratio o 5:4 are said to be separated by an interval o a third (or strictly a pure third) ; again this interval sounds pleasing. Has music always been thought o in this way? Is the concept o consonance accepted by all societies?
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Questions 1
3
(IB) a)
(IB) a)
A pendulum consists of a bob suspended by a light inextensible string from a rigid support. The pendulum bob is moved to one side and then released. The sketch graph shows how the displacement of the pendulum bob undergoing simple harmonic motion varies with time over one time period.
In terms of the acceleration, state two conditions necessary for a system to perform simple harmonic motion.
b) A tuning fork is sounded and it is assumed that each tip vibrates with simple harmonic motion.
d
displacement
0
0
time
O n a copy of the sketch graph:
The extreme positions of the oscillating tip of one fork are separated by a distance d.
( i) Label, with the letter A, a point at which the acceleration of the pendulum bob is a maximum.
( i)
( ii) Label, with the letter V, a point at which the speed of the pendulum bob is a maximum.
( ii) Sketch a graph to show how the displacement of one tip of the tuning fork varies with time.
b) Explain why the magnitude of the tension in the string at the midpoint of the oscillation is greater than the weight of the pendulum bob.
(iii) On your graph, label the time period T and the amplitude A. ( 8 marks) 4
2
State, in terms of d, the amplitude of vibration.
(IB) The graph below shows how the displacement x of a particle undergoing simple harmonic motion varies with time t. The motion is undamped.
( IB) a) Graph 1 below shows the variation with time t of the displacement d of a travelling ( progressive) wave. Graph 2 shows the variation with distance x along the same wave of its displacement d.
x
d/mm
0
t
a) S ketch a graph showing how the velocity v of the particle varies with time. b) Explain why the graph takes this form. ( 4 marks)
164
Graph 2 4 2 0 0.0 2 4 d/mm
0
Graph 1 4 2 0 0.0 2 4
a)
0.1
0.2
0.3
0.4
0.5
0.6 t/s
0.4
0.8
1.2
1.6
2.0
2.4 x/cm
S tate what is meant by a travelling wave.
QUESTION S
b)
Use the graphs to determine the amplitude, wavelength, frequency and speed of the wave.
wave A 3.0 2.0
( 5 marks) xA /mm
5
1.0
( IB)
0.0 0.0
2.0
4.0
6.0
8.0
t/ms 10.0
1.0
a) With reference to the direction of energy transfer through a medium, distinguish between a transverse wave and a longitudinal wave.
2.0 3.0
The graph below shows the variation with time t of the displacement xB of wave B as it passes through point P. The waves have equal frequencies.
b) A wave is travelling along the surface of some shallow water in the x- direction. The graph shows the variation with time t of the displacement d of a particle of water.
wave B 2.0
10 1.0 xB /mm
8 6
0.0
2.0
4.0
6.0
8.0
t/ms 10.0
1.0
4
2.0
2 d/mm
0.0
( i) C alculate the frequency of the waves.
0 0.0
0.05
0.1
2
0.15
0.2
0.25
0.3 t/s
( ii) The waves pass simultaneously through point P. Use the graphs to determine the resultant displacement at point P of the two waves at time t = 1 . 0 ms and at time t = 8.0 ms.
4 6 8 10
( 6 marks)
Use the graph to determine the frequency and the amplitude of the wave. c) The speed of the wave in b) is 1 5 cm s 1 . D educe that the wavelength of this wave is 2 .0 cm. d) The graph in b) shows the displacement of a particle at the position x = 0. D raw a graph to show the variation with distance x along the water surface of the displacement d of the water surface at time t = 0.070 s. ( 1 1 marks)
7
( IB) a) With reference to the direction of energy transfer through a medium, distinguish between a transverse wave and a longitudinal wave. b) The graph shows the variation with time t of the displacement d of a particular water particle as a surface water wave travels through it. 10 8 6
(IB)
4
a)
B y referring to the energy of a travelling wave, explain what is meant by: ( i) a ray
2 d/mm
6
0 0 2
0.05
0.1
0.15
0.2
0.25
0.3 t/s
( ii) wave speed. 4
b) The following graph shows the variation with time t of the displacement xA of wave A as it passes through a point P.
6 8 10
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4
O S C I LL AT I O N S AN D WAVE S Use the graph to determine or the wave:
The polarizer is then rotated by 1 80 in the direction shown. S ketch a graph to show the variation with the rotation angle , o the transmitted light intensity I, as varies rom 0 to 1 80. Label your sketch- graph with the letter U.
( i) the requency ( ii) the amplitude. c) The speed o the water wave is 1 2 cm s 1 . C alculate the wavelength o the wave. d) The graph in b) shows the displacement o a particle at the position x = 0.
b) The beam in a) is now replaced with a polarized beam o light o the same intensity.
S ketch a graph to show the variation with distance x along the water surace o the displacement d o the water surace at time t = 0.2 0 s.
The plane o polarization o the light is initially parallel to the polarization axis o the polarizer.
e) The wave meets a shel that reduces the depth o the water.
polarizer
wave fronts shelf direction of rotation
direction of travel of wave 30
shallow water
deep water polarization axis polarized light
The angle between the waveronts in the shallow water and the shel is 3 0. The speed o the wave in the shallow water is 1 2 cm s 1 and in the deeper water is 1 8 cm s 1 . For the wave in the deeper water, determine the angle between the normal to the waveronts and the shel. ( 1 2 marks) 8
(IB) a) A beam o unpolarized light o intensity I0 is incident on a polarizer. The polarization axis o the polarizer is initially vertical as shown.
The polarizer is then rotated by 1 80 in the direction shown. O n the same axes in a) , sketch a graph to show the variation with the rotation angle , o the transmitted light intensity I, as varies rom 0 to 1 80. ( 5 marks) 9
(IB) An orchestra playing on boat X can be heard by tourists on boat Y, which is situated out o sight o boat X around a headland. ocean X
polarizer headland Y direction of rotation
The sound rom X can be heard on Y due to A. reraction B . refection polarization axis
166
unpolarized light
C . diraction D . transmission.
QUESTION S 1 0 (IB)
( i)
A small sphere, mounted at the end of a vertical rod, dips below the surface of shallow water in a tray. The sphere is driven vertically up and down by a motor attached to the rod.
( ii) The angle between the wavefronts and the interface in region A is 60. The refractive index An B is 1 .4. D etermine the angle between the wavefronts and the interface in region B .
The oscillations of the sphere produce travelling waves on the surface of the water. rod
( iii) O n the diagram above, construct three lines to show the position of three wavefronts in region B .
water surface
sphere
a) The diagram shows how the displacement of the water surface at a particular instant in time varies with distance from the sphere. The period of oscillation of the sphere is 0.02 7 s.
With reference to a wave, distinguish between a ray and a wavefront.
c) Another sphere is dipped into the water. The spheres oscillate in phase. The diagram shows some lines in region A along which the disturbance of the water surface is a minimum. lines of minimum disturbance
1.5
displacement/mm
1.0 wavefront
0.5 distance/mm
0 0
5
10
15
0.5 1.0
( i)
1.5
Use the diagram to calculate, for the wave: ( i)
O utline how the regions of minimum disturbance occur on the surface.
( ii) The frequency of oscillation of the spheres is increased. S tate and explain how this will affect the positions of minimum disturbance.
the amplitude
( ii) the wavelength ( iii) the frequency
( 1 5 marks)
( iv) the speed. b) The wave moves from region A into a region B of shallower water. The waves move more slowly in region B . The diagram ( not to scale) shows some of the wavefronts in region A.
1 1 (IB) a) D escribe two ways in which standing waves differ from travelling waves. b) An experiment is carried out to measure the speed of sound in air, using the apparatus shown below.
60
direction of motion region A
region B
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4
O S C I LL AT I O N S AN D WAVE S ( i) O n a copy o the diagram, label with the letter P the position along the pipe where the amplitude o oscillation o the air molecules is the largest.
tuning fork, frequency 440 Hz tube
( ii) The speed o sound in the air in the pipe is 3 3 0 m s 1 . C alculate the length l.
tank of water
A tube that is open at both ends is placed vertically in a tank o water until the top o the tube is j ust at the surace o the water. A tuning ork o requency 440 Hz is sounded above the tube. The tube is slowly raised out o the water until the loudness o the sound reaches a maximum or the frst time, due to the ormation o a standing wave. ( i)
Explain the ormation o a standing wave in the tube.
c) Use your answer to b) ( ii) to suggest why it is better to use organ pipes that are closed at one end or producing low requency notes rather than pipes that are open at both ends. ( 8 marks)
1 3 (IB) A microwave transmitter emits radiation o a single wavelength towards a metal plate along a line normal to the plate. The radiation is reected back towards the transmitter. metal plate
( ii) S tate the position where a node will always be produced. ( iii) The tube is raised a little urther. Explain why the loudness o the sound is no longer at a maximum.
microwave transmitter microwave detector
c) The tube is raised until the loudness o the sound reaches a maximum or a second time. Between the two positions o maximum loudness the tube has been raised by 36.8 cm. The requency o the sound is 440 Hz. Estimate the speed o sound in air. ( 1 0 marks)
A microwave detector is moved along a line normal to the microwave transmitter and the metal plate. The detector records a sequence o equally spaced maxima and minima o intensity. a) Explain how these maxima and minima are ormed.
1 2 (IB) a) State two properties o a standing ( stationary) wave. b) The diagram shows an organ pipe that is open at one end.
b) The microwave detector is moved through 1 3 0 mm rom one point o minimum intensity to another point o minimum intensity. O n the way it passes through nine points o maximum intensity. C alculate the ( i) wavelength o the microwaves. ( ii) requency o the microwaves.
l
The length o the pipe is l. The requency o the undamental ( frst harmonic) note emitted by the pipe is 1 6 Hz.
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c) D escribe and explain how it could be demonstrated that the microwaves are polarized. ( 1 1 marks)
5
E LE CTRI CI TY AN D M AG N ETI SM
Introduction Modern society uses a whole range o electrical devices rom the simplest heated metal flaments that provide light, through to the most sophisticated medical instruments and computers. D evices o increasing technical complexity are developed every day.
In this topic we look at the phenomenon o electricity, and what is meant by charge and electric current. We consider the three eects that can be observed when charge ows in an electric circuit.
5.1 Electric felds Understanding Charge Electric feld Coulombs law
Applications and skills Identiying two species o charge and the
Electric current Direct current (dc)
Potential dierence (pd)
Nature of science Electrical theory resembles the kinetic theory o gases in that a theory o the microscopic was developed to explain the macroscopic observations that had been made over centuries. The development o this subject and some o the byways that were taken make this a ascinating study. We should remember the many scientists who were involved. It is a tribute to them that they could make so much progress when the details o the microscopic nature o electronic charge were unknown to them.
direction o the orces between them Solving problems involving electric felds and Coulombs law Calculating work done when charge moves in an electric feld in both joules and electronvolts Identiying sign and nature o charge carriers in a metal Identiying drit speed o charge carriers Solving problems using the drit-speed equation Solving problems involving current, potential dierence, and charge
Equations q
current-charge relationship: I = ______ t q1 q2 ________ Coulomb's law: F = k r2
1 the coulomb constant: k = ________ 4 0
W potential dierence defnition: V = ____ q
conversion o energy in joule to electron-volt: W(eV) W(J) __________ e
electric ield strength: E = ___qF drit speed: I = nAvq
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Charge and feld Simple beginnings Take a plastic comb and pull it through your hair. Aterwards, the comb may be able to pick up small pieces o paper. Look closely and you may see the paper being thrown o shortly ater touching the comb. S imilar observations were made early in the history o science. The discovery that obj ects can be charged by riction ( you were doing this when you drew the comb through your hair) is attributed to the Greek scientist Thales who lived about 2 600 years ago. In those days, silk was spun on amber spindles and as the amber rotated and rubbed against its bearings, the silk was attracted to the amber. The ancient Greek word or amber is ( electron) . In the 1 700s, du Fay ound that both conductors and insulators could be electrifed ( the term used then) and that there were two opposite kinds o electrifcation . However, he was unable to provide any explanation or these eects. Gradually, scientists developed the idea that there were two separate types o charge: positive and negative. The American physicist, B enj amin Franklin, carrying out a amous series o experiments ying kites during thunderstorms, named the charge on a glass rod rubbed with silk as positive electricity . The charge on materials similar to ebonite ( a very hard orm o rubber) rubbed with animal ur was called negative . O ne o Franklins other discoveries was that a charged conducting sphere has no electric feld inside it ( the feld and the charges always being outside the sphere) . Joseph Priestley was able to deduce rom this that the orce between two charges is inversely proportional to the square o the distance between the charges. At the end o the nineteenth century J. J. Thomson detected the presence o a small particle that he called the electron. Experiments showed that all electrons have the same small quantity o charge and that electrons are present in all atoms. Atoms were ound to have protons that have the same magnitude o electronic charge as, but have opposite charge to, the electron. In a neutral atom or material the number o electrons and the number o protons are equal. We now assign a negative charge to the electron and positive to the proton. There are only these two species o charge.
Explaining electrostatics E xperiments also show that positively charged obj ects are attracted to negatively charged obj ects but repelled by any other positively charged obj ect. The possible cases are summed up in fgure 1 . There can also be an attraction between charged and uncharged obj ects due to the separation o charge in the uncharged obj ect. In fgure 1 ( c) the electrons in the uncharged sphere are attracted to the positively charged sphere ( sphere A) and move towards it. The electrons in sphere B are now closer to the positives in sphere A than the fxed positive charges on B . S o the overall orce is towards sphere A as the orce between two charges increases as the distance between them decreases.
170
5.1 ELECTRI C FI ELD S +
+ +
+
+
+
+
+
+
+
+
+
+ +
+ +
+
opposite charges attract (a)
like charges repel (b) +
+ + +
+
+
+
+
+
+
+
-
+
neutral
+ +
+
charge separation means that attraction occurs (c)
Figure 1 Attractions between charges.
We now know that the simple electrostatic eects early scientists observed are due only to the movement o the negatively charged electrons. An object with no observed charge has an exact balance between the electrons and the positively charged protons; it is said to be neutral. Some electrons in conducting materials are loosely attached to their respective atoms and can leave the atoms to move rom one object to another. This leaves the object that lost electrons with an overall positive charge. The electrons transerred give the second object an overall negative charge. Notice that electrons are not lost in these transers. I 1 000 electrons are removed rom a rod when the rod is charged by a cloth, the cloth will be let with 1 000 extra electrons at the end o the process. Charge is conserved; the law o conservation o charge states that in a closed system the amount o charge is constant. When explaining the eects described here, always use the idea o surplus o electrons or a negative charge, and describe positive charge in terms o a lack ( or defcit) o electrons. Figure 2 shows how an experiment to charge a metal sphere by induction is explained.
+ + +
+ + + + + +
connected + to Earth + +
+ +
+ +
+
+
conducting sphere insulating stand
6 electrons travel to Earth
charged rod separates charge
remove charged rod
sphere is earthed and electrons are repelled to Earth
Earth connection broken, rod removed, and electrons re-distributed to leave sphere positive after rearrangement
TOK Inverse-square laws Forces between charged objects is one o several examples o inverse-square laws that you meet in this course. They are o great importance in physics. Inverse-square laws model a characteristic property o some felds, which is that as distance doubles, observed eects go down by one quarter. Mathematics helps you to learn and conceptualize your ideas about the subject. When you have learnt the physics o one situation (here, electrostatics) then you will be able to apply the same rules to new situations (or example, gravitation, in the next topic) . Is the idea o feld a human construct or does it reect the reality o the universe?
Figure 2 Charging by induction.
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Measuring and defning charge The unit o charge is the coulomb ( abbreviated to C ) . C harge is a scalar quantity. The coulomb is defned as the charge transp orted by a current o one amp ere in one second. Measurements show that all electrons are identical, with each one having a charge equal to 1 .6 1 0 1 9 C ; this undamental amount o charge is known as the electronic (or elementary) charge and given the symbol e. C harges smaller than the electronic charge are not observed in nature. 1 2 ( Quarks have ractional charges that appear as __ e or __ e; however, 3 3 they are never observed outside their nucleons.) In terms o the experiments described here, the coulomb is a very large unit. When a comb runs through your hair, there might be a charge o somewhere between 1 pC and 1 nC transerred to it.
Forces between charged objects In 1 785 , C oulomb published the frst o several Mmoires in which he reported a series o experiments that he had carried out to investigate the eects o orces arising rom charges. He ound, experimentally, that the orce between two point charges 1 separated by distance r is proportional to __ thus confrming the earlier r theory o Priestley. Such a relationship is known as an inverse-square law. 2
Investigate! Forces between charges
These are sensitive experiments that need care and a dry atmosphere to achieve a result.
Take two small polystyrene spheres and paint them with a metal paint or colloidal graphite, or cover with aluminum oil. S uspend one rom an insulating rod using an insulating ( perhaps nylon) thread. Mount the other on top o a sensitive top- pan balance, again using an insulating rod.
172
insulating rod
+
insulating support
C harge both spheres by induction when they are apart rom each other. An alternative charging method is to use a laboratory high voltage power supply. Take care, your teacher will want to give you instructions about this. B ring the spheres together as shown in fgure 3 and observe changes in the reading on the balance.
sensitive top-pan balance
Figure 3
5.1 ELECTRI C FI ELD S
The distance d moved by the sphere depends on the orce between the charged spheres. The distance r is the distance between the centres o the spheres.
Vary d and r making careul measurements o them both.
d sideways force on ball r = distance between balls
1 Plot a graph o d against __ . An experiment r perormed with care can give a straight- line graph. 2
d r
Figure 4
Another method is to bring both charged spheres together as shown in fgure 4.
For small deections, d is a measure o the orce between the spheres ( the larger the orce the greater the distance that the sphere is moved) whereas r is the distance between sphere centres.
Nature of science Scientists in C oulombs day published their work in a very dierent way rom scientists today. C oulomb wrote his results in a series o books called Mmoires. Part o C oulombs original Mmoire in which he states the result is shown in fgure 5 .
Figure 5
Later experiments confrmed that the orce is proportional to the product o the size o the point charges q 1 and q 2 . C ombining C oulombs results together with these gives q1 q2 F _ r2 where the symbol means is proportional to. The magnitude o the orce F between two point charges o charge q 1 and q 2 separated by distance r in a vacuum is given by kq 1 q 2 F= _ r2 where k is the constant o proportionality and is known as C oulombs constant. In act we do not always quote the law in quite this mathematical orm. The constant is requently quoted dierently as 1 k=_ 4 0 The new constant 0 is called the permittivity o ree space ( ree space is an older term or vacuum . The 4 is added to rationalize electric and
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5
E L E C T R I C I T Y AN D M AG N E T I S M magnetic equations in other words, to give them a similar shape and to retain an important relationship between them ( see the TO K section on page 1 77) . S o the equation becomes 1 q1 q2 F= __ 4 0 r2 When using charge measured in coulombs and distance measured in metres, the value o k is 9 1 0 9 N m 2 C 2 . This means that 0 takes a value o 8.85 4 1 0 1 2 C 2 N 1 m 2 or, in undamental units, m 3 kg 1 s 4 A 2 . The equation as it stands applies only or charges that are in a vacuum. I the charges are immersed in a dierent medium ( say, air or water) then the value o the permittivity is dierent. It is usual to 1 as the 0 subscript amend the equation slightly too, k becomes ___ 4 in 0 should only be used or the vacuum case. For example, the permittivity o water is 7. 8 1 0 1 0 C 2 N 1 m 2 and the permittivity o air is 8. 85 49 1 0 1 2 C 2 N 1 m 2 . The value or air is so close to the ree- space value that we normally use 8. 85 1 0 1 2 C 2 N 1 m 2 or both. The table gives a number o permittivity values or dierent materials.
Material paper rubber water graphite diamond
Permittivity / 10 12 C2 N 1 m 2 34 62 779 106 71
This equation appears to say nothing about the direction o the orce between the charged obj ects. Forces are vectors, but charge and ( distance) 2 are scalars. There are mathematical ways to cope with this, but or point charges the equation gives an excellent clue when the signs o the charges are included. force on A due to B
charge A
charge B
force on B due to A
r positive direction
Figure 6 Force directions.
We take the positive direction to be rom charge A to charge B ; in fgure 6 that is rom let to right. Lets begin with both charge A and charge B being positive. When two positive charges are multiplied q q together in ____ , the resulting sign o the orce acting on charge B due to r charge A is also positive. This means that the direction o the orce will be assigned the positive direction ( rom charge A to charge B ) : in other words, let to right. This agrees with the physics because the charges are repelled. I both charges are negative then the answer is the same, the charges are repelled and the orce is to the right. 1
2
2
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5.1 ELECTRI C FI ELD S
I, however, one o the charges is positive and the other negative, then the product o the charges is negative and the orce direction will be opposite to the let- to- right positive direction. S o the orce on charge 2 is now to the let. Again, this agrees with what we expect, that the charges attract because they have opposite signs.
Worked examples 1
Two point charges o + 1 0 nC and 1 0 nC in air are separated by a distance o 1 5 mm.
The charges are attracted along the line j oining them. ( D o not orget that orce is a vector and needs both magnitude and direction or a complete answer.)
a) C alculate the orce acting between the two charges. b) C omment on whether this orce can lit a small piece o paper about 2 mm 2 mm in area.
Solution a) It is important to take great care with the prefxes and the powers o ten in electrostatic calculations. The charges are: +1 0 1 0 9 C and 1 0 1 0 9 C. The separation distance is 1 .5 1 0 2 m (notice how the distance is converted right at the outset into consistent units) . 8
8
(+1 1 0 ) (1 1 0 ) So F = ___ 4 0 (1 .5 1 0 2 ) 2 = 4.0 1 0 3 N
b) A sheet o thin A4 paper o dimensions 2 1 0 mm by 2 97 mm has a mass o about 2 g. S o the small area o paper has a mass o about 1 .3 1 0 7 kg and thereore a weight o 1 .3 1 0 6 N. The electrostatic orce could lit this paper easily. 2
Two point charges o magnitude + 5 C and + 3 C are 1 .5 m apart in a liquid that has a permittivity o 2 .3 1 0 1 1 C 2 N 1 m 2 . C alculate the orce between the point charges.
Solution 6
6
(+5 1 0 ) (+3 1 0 ) F = __________________ = 2 3 mN; 4 2 . 3 1 0 ( 1 . 5 ) 1 1
2
a repulsive orce acting along the line j oining the charges.
Electric felds Sometimes the origin o a orce between two objects is obvious, an example is the riction pad in a brake rubbing on the rim o a bicycle wheel to slow the cycle down. In other cases there is no physical contact between two objects yet a orce exists between them. Examples o this include the magnetic orce between two magnets and the electrostatic orce between two charged objects. Such orces are said to act at a distance. The term feld is used in physics or cases where two separated obj ects exert orces on each other. We say that in the case o the comb picking up the paper, the paper is sitting in the electric feld due to the comb. The concept o the feld is an extremely powerul one in physics not least because there are many ideas common to all felds. As well as the magnetic and electrostatic felds already mentioned, gravity felds also obey the same rules. Learn the underlying ideas or one type o feld and you have learnt them all.
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Mapping felds
Investigate! Plotting electric felds the method) . This experiment allows patterns to be observed or electric felds.
+
Put some castor oil in a Petri dish and sprinkle some grains o semolina ( or grits) onto the oil. Alternatives or the semolina include grass seed and hairs cut about 1 mm long rom an artist paint brush.
Take two copper wires and bend one o them to orm a circle just a little smaller than the internal diameter o the Petri dish. Place the end o the other wire in the centre o the Petri dish.
C onnect a 5 kV power supply to the wires take care with the power supply!
O bserve the grains slowly lining up in the electric feld.
S ketch the pattern o the grains that is produced.
Repeat with other wire shapes such as the our shown in fgure 7( c) .
(a)
semolina castor oil
(b)
(c)
Figure 7 At an earlier stage in your school career you may have plotted magnetic feld patterns using iron flings ( i you have not done this there is an Investigate! in S ub- topic 5 .4 to illustrate
In the plotting experiment, the grains line up in the feld that is produced between the wires. The patterns observed resemble those in fgure 8. The experiment cannot easily show the patterns or charges with the same sign.
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+ + + + + + + +
+
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+
The idea o feld lines was frst introduced by Michael Faraday ( his original idea was o an elastic tube that repelled other tubes) . Although feld lines are imaginary, they are useul or illustrating and understanding the nature o a particular feld. There are some conventions or drawing these electric feld patterns.
+
Figure 8 Electric feld patterns.
The lines start and end on charges o opposite sign.
An arrow is essential to show the direction in which a positive charge would move ( i. e. away rom the positive charge and towards the negative charge) .
Where the feld is strong the lines are close together. The lines act to repel each other.
The lines never cross.
The lines meet a conducting surace at 90.
5.1 ELECTRI C FI ELD S
Electric feld strength
TOK
As well as understanding the feld pattern, we need to be able to measure the strength o the electric feld. The electric feld strength is defned using the concept o a p ositive test charge. Imagine an isolated charge Q sitting in space. We wish to know what the strength o the feld is at a point P, a distance r away rom the isolated charge. We put another charge, a positive test charge o size q, at P and measure the orce F that acts on the test charge due to Q. Then the magnitude o the electric feld strength is defned to be F E= _ q
So why not use k? James Maxwell, working in the middle o the nineteenth century, realized that there was an important connection between electricity, magnetism and the speed o light. In particular he was able to show that the permittivity o ree space 0 (which relates to electrostatics) and the permeability o ree space 0 (which relates to electromagnetism) are themselves connected to the speed o light c:
electric orce +q r
test charge, P
+Q
________ 1 2 = c 0
Figure 9 Defnition o electric feld strength.
0
This proves to be an important equation. So much so that, in the set o equations that arise rom the SI units we use, we choose to use 0 in all the electrostatics equations and 0 in all the magnetic equations.
The units o electric feld strength are N C 1 . ( Alternative units, that have the same meaning, are V m 1 and will be discussed in Topic 1 1 .) E lectric feld strength is a vector, it has the same direction as the orce F ( this is because the charge is a scalar which only scales the value o F up or down) . A ormal defnition or electric feld strength at a p oint is the orce p er unit charge exp erienced by a small p ositive p oint charge p laced at that p oint.
However, not all unit systems choose to do this. There is another common system, the cgs system (based on the centimetre, the gram and the second, rather than the metre, kilogram and second) . In cgs, the value o the constant k in Coulombs law is chosen to be 1 and the equation q q . I the appears as F = ________ r numbers are dierent, is the physics the same?
C oulombs law can be used to fnd how the electric feld strength varies with distance or a point charge. Q is the isolated point charge and q is the test charge, so 1 Qq F= __ 4 0 r2 F E=_ q Thereore 1 Qq 1 _ E= __ q 4 0 r2
1
so
2
2
1 Q E = _ _2 4 0 r The electric feld strength o the charge at a point is proportional to the charge and inversely proportional to the square o the distance rom the charge. I Q is a positive charge then E is also positive. Applying the rule that r is measured rom the charge to the test charge, then i E is positive it acts outwards away rom the charge. This is what we expect as both charge and test charge are positive. When Q is negative, E acts towards the charge Q. The feld shape or a point charge is known as a radial feld. The feld lines radiate away ( positive) or towards ( negative) the point charge as shown in fgure 1 0.
+
Q
Figure 10 Radial felds or positive and negative point charges.
Q
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E L E C T R I C I T Y AN D M AG N E T I S M
Worked examples 1
8. 5 1 0 4 = 1 .2 1 0 6 N C 1 b) E = __ 4 0 ( 2 .5 ) 2
C alculate the electric feld strengths in a vacuum
The feld direction is towards the negative charge.
a) 1 .5 cm rom a + 1 0 C charge 2
b) 2 .5 m rom a 0.85 mC charge
Solution a) B egin by putting the quantities into consistent units: r = 1 .5 1 0 2 m and Q = 1 . 0 1 0 5 C .
An oxygen nucleus has a charge o + 8e. C alculate the electric feld strength at a distance o 0.68 nm rom the nucleus.
Solution
1 .0 1 0 = + 4. 0 1 0 8 N C 1 . Then E = ____________ 4 ( 1 . 5 1 0 )
The charge on the oxygen nucleus is 8 1 .6 1 0 1 9 C ; the distance is 6.8 1 0 1 0 m.
The feld direction is away rom the positive charge.
1 .3 1 0 E = _______________ = 2.5 1 0 1 0 N C 1 away rom 4 ( 6 . 8 1 0 )
5
2 2
0
1 8
1 0 2
0
the nucleus. The electric feld strengths can be added using either a calculation or a scale diagram as outlined in Topic 1 . feld due to -q
+q feld
-q feld
-q
+Q
gives test charge
net electric feld
feld due to +Q
feld due to 1 and 2 feld due to 1
+2Q charge 3 feld due to 2 test charge
charge 2 +Q
gives
gives
=
net electric feld
charge 1 +Q feld due to 3
feld due to 3
Figure 11 Vector addition o electric felds.
This vector addition o feld strengths ( fgure 1 1 ) can give us an insight into electric felds that arise rom charge confgurations that are more complex than a single point charge.
Close to a conductor Imagine going very close to the surace o a conductor. Figure 1 2 shows what you might see. First o all, i we are close enough then the surace will appear at ( in j ust the same way that we are not aware o the curvature o the Earth until we go up in an aircrat) . S econdly we would see that all the ree electrons are equally spaced. There is a good reason or this: any one electron has orces acting on it rom the other electrons. The electron will accelerate until all these orces balance out and it is in equilibrium, or this to happen they must be equally spaced.
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Now look at the feld strength vectors radiating out rom these individual electrons. Parallel to the surace, these all cancel out with each other so there is no electric feld in this direction. ( This is the same as saying that any one electron will not accelerate as the horizontal feld strength is
5.1 ELECTRI C FI ELD S
TOK
surace
But does the test charge aect the original feld? I we wanted to use the defnition o the strength o an electric feld practically by using a test charge, we would have to take care. Just as a thermometer alters the temperature o the object it measures, so the test charge will exert a orce on the original charge (the one with the feld we are trying to measure) and may accelerate it, or disturb the feld lines. This means in practice that a test charge is really an imaginary construct that we use to help our understanding. In practice we preer to measure electric feld strengths using the idea that it is the potential gradient, in other words the change in the voltage divided by the change in distance: the larger this value, the greater the feld strength. We will leave urther discussion o this point to Topic 11 but it is ood or thought rom a theory o knowledge perspective: a measurement that we can think about but not, in practice, carry out. The German language has a name or it: gedankenexperiment thought experiment. How can a practical subject such as a science have such a thing?
perpendicular to surace add
surace
parallel to surace cancel
leaving electric feld only perpendicular to surace
Figure 12 Close to conducting suraces.
zero) . Perpendicular to the surace, however, things are dierent. Now the feld vectors all add up, and because there is no feld component parallel to the surace, the local feld must act at 90 to it. S o, close to a conducting surace, the electric feld is at 90 to the surace.
Conducting sphere This can be taken a step urther or a conducting sphere, whether it is hollow or solid. Again, the ree electrons at the surace are equally spaced and all the feld lines at the surace o the sphere are at 90 to it. The consequence is that the feld must be radial, just like the feld o an isolated point charge. So, to a test charge outside the sphere, the feld o the sphere appears exactly the same as that o the point charge. Mathematical analysis shows that outside a sphere the feld indeed behaves as though it came rom a point charge placed at the centre o the sphere with a charge equal to the total charge spread over the sphere. ( Inside the sphere is a dierent matter, it turns out that there is no electric feld inside a sphere, hollow or solid, a result that was experimentally determined by Franklin.)
Worked example Two point charges, a +25 nC charge X and a +1 5 nC charge Y are separated by a distance o 0.5 m. a) C alculate the resultant electric feld strength at the midpoint between the charges.
c) Calculate the magnitude o the electric feld strength at the point P on the diagram. X and Y are 0.4 m and 0.3 m rom P respectively. P
b) C alculate the distance rom X at which the electric feld strength is zero.
0.4 m X
0.3 m Y
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E L E C T R I C I T Y AN D M AG N E T I S M
2 .5 1 0 c) PX = 0.4 m so EX at P is ________ 4 0 . 4 8
Solution
2
0
2 .5 1 0 8 a) EA = __2 = 3 600 N C 1 4 0 0. 2 5
= 1 400 N C 1 along XP in the direction away rom X.
1 .5 1 0 8 EB = __2 = 2 2 00 N C 1 4 0 0. 2 5
1 .5 1 0 PY = 0.3 m so EY at P is ________ 4 0 . 3 8
2
The feld strengths act in opposite directions, so the net electric feld strength is ( 3 600 2 2 00) = 1 400 N C 1 ; this is directed away rom X towards Y.
0
= 1 5 00 N C 1 along PY in the direction towards Y.
b) For E to be zero, EA = EB and so 2 .5 1 0 8 1 . 5 1 0 8 _ = __2 2 4 0 d 4 0 ( 0. 5 d)
P net e
thus
le c t ric
f e ld
2
d 2 .5 _ = _ 1 .5 ( 0.5 d) 2 or d _ = ( 0.5 d)
X
____
2 .5 = 1 .3 _ 1 .5
Y
d = 0.65 1 .3 d The magnitude o the resultant electric feld ___________ strength is 1 400 2 + 1 5 00 2 = 2 1 00 N C 1
2 .3 d = 0.65 d = 0.2 8 m
( The calculation o the angles was not required in the question and is let or the reader.)
Moving charge Investigate! Moving charge around pong) ball with a painted metal surace rom a long insulating thread so that it is midway between the plates.
insulating thread metal plate
C onnect the plates to a high- voltage power supply with a sensitive ammeter or light- beam galvanometer in the circuit.
When the supply is turned on, the ball should shuttle between the plates (it may need to be kickstarted by making it touch one o the plates) .
Notice what happens on the scale o the galvanometer. E very time the ball touches a plate there is a deection on the meter. Look at the direction and size o the deections do they vary?
insulating handle coated ball
A ammeter
+
high voltage (approx. 5 kV) supply
180
Figure 13 Arrange two metal plates vertically and about 2 0 cm apart. S uspend a table tennis ( ping-
5.1 ELECTRI C FI ELD S
In the previous Investigate! the power supply is connected to the plates with conducting leads. Electrons move easily along these leads. When the supply is turned on, the electrons soon distribute themselves so that the plate connected to the negative supply has surplus electrons, and the other plate has a defcit o electrons, becoming positive. As the ball touches one o the plates it loses or gains some o the electrons. The charge gained by the ball will have the same sign as the plate and as a result the ball is almost immediately repelled. A orce acts on the ball because it is in the electric feld that is acting between the plates. The ball accelerates towards the other plate where it transers all its charge to the new plate and gains more charge. This time the charge gained has the sign o the new plate. The process repeats itsel with the ball transerring charge rom plate to plate. The meter is a sensitive ammeter, so when it deects it shows that there is current in the wires leading to the plates. C harge is moving along these wires, so this is evidence that:
an electric current results when charge moves
the charge is moved by the presence o an electric feld.
A mechanism for electric current The shuttling ball and its charge show clearly what is moving in the space between the plates. However, the microscopic mechanisms that are operating in wires and cells are not so obvious. This was one o the maj or historical problems in explaining the physics o electricity. E lectrical conduction is possible in gases, liquids, solids, and a vacuum. O particular importance to us is the electrical conduction that takes place in metals.
Conduction in metals The metal atoms in a solid are bound together by the metallic bond. The ull details o the bonding are complex, but a simple model o what happens is as ollows. When a metal solidifes rom a liquid, its atoms orm a regular lattice arrangement. The shape o the lattice varies rom metal to metal but the common eature o metals is that as the bonding happens, electrons are donated rom the outer shells o the atoms to a common sea o electrons that occupies the entire volume o the metal. positive ions +
+ + +
electrons entering metal
+ +
+
+
+ +
+
+
electrons leaving metal
+ metal rod
Figure 14 Conduction by free electrons in a metal.
Figure 1 4 shows the model. The positive ions sit in fxed positions on the lattice. There are ions at each lattice site because each atom has now lost an electron. O course, at all temperatures above absolute zero
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E L E C T R I C I T Y AN D M AG N E T I S M they vibrate in these positions. Most o the electrons are still bound to them but around the ions is the sea o ree electrons or conduction electrons; these are responsible or the electrical conduction. Although the conduction electrons have been released rom the atoms, this does not mean that there is no interaction between ions and electrons. The electrons interact with the vibrating ions and transer their kinetic energy to them. It is this transer o energy rom electrons to ions that accounts or the phenomenon that we will call resistance in the next sub-topic. The energy transer in a conductor arises as ollows: electric feld
low potential ()
(+) high potential
drit direction
Figure 15
In a metal in the absence o an electric feld, the ree electrons are moving and interacting with the ions in the lattice, but they do so at random and at average speeds close to the speed o sound in the material. Nothing in the material makes an electron move in any particular direction. However, when an electric feld is present, then an electric orce will act on the electrons with their negative charge. The defnition o electric feld direction reminds us that the electric feld is the direction in which a positive charge moves, so the orce on the electrons will be in the opposite direction to the electric feld in the metal ( fgure 1 5 ) . In the presence o an electric feld, the negatively charged electrons drit along the conductor. The electrons are known as charge carriers. Their movement is like the random motion o a colony o ants carried along a moving walkway.
Conduction in gases and liquids Electrical conduction is possible in other materials too. S ome gases and liquids contain ree ions as a consequence o their chemistry. When an electric feld is applied to these materials the ions will move, positive in the direction o the feld, negative the opposite way. When this happens an electric current is observed. I the ele ctric ield is strong e nough it can, itsel, lead to the creation o ions in a gas or liquid. This is known as electrical b reakd o w n. It is a common eect during electrical storms when lightning moves be twe en a charge d cloud and the E arth. You will have see n such conduction in neon display tub es or luorescent tub es use or lighting.
182
5.1 ELECTRI C FI ELD S
Nature of science Models o conduction The model here is o a simple ow o ree electrons through a solid, a liquid, or a gas. B ut this is not the end o the story. There are other, more sophisticated models o conduction in solids that can explain the dierences between conductors, semiconductors and insulators better than the ow model here. These involve the electronic band theory which arises rom the interactions between the electrons within individual atoms and between the atoms themselves. Essentially, this band model proposes that electrons have to adopt dierent energies within the substance and that some groups o energy levels ( called band gaps) are not permitted to the electrons. Where there are wide band gaps,
electrons cannot easily move rom one set o levels to another and this makes the substance an insulator. Where the band gap is narrow, adding energy to the atomic structure allows electrons to j ump across the band gap and conduct more reely this makes a semiconductor, and you will later see that one o the semiconductor properties is that adding internal energy allows them to conduct better. In conductors the band gap is o less relevance because the electrons have many available energy states and so conduction happens very easily indeed. Full details o this theory are beyond IB Physics, but i you have an interest in taking this urther, you can fnd many reerences to the theory on the Internet.
Electric current When charge ows in a conductor we say that there is an electric current in the conductor. C urrent is measured in amp res, the symbol or the unit is A. O ten, in the E nglish speaking world, the accent is omitted. C urrent is linked to ow o charge in a simple way. point P
Tip It is not good practice to write or say that current fows, what is fowing in the circuit is the electric charge. The movement o this charge is what we call current and it is best to write that there is a current in the circuit, or in a component as appropriate.
one coulomb o electron charge
Figure 16 Charge fow leading to current.
Imagine a block o electrons with a total charge o one coulomb moving along a conductor. An observer at point P is watching these electrons move along the conductor. I all the electrons in the block move past the point in one second then, the current is one ampere. I it takes twice as long ( 2 s) or the block to pass, then the current is hal and is 0.5 A. I the block takes 0.1 s to pass the observer, then the current is 1 0 A. Mathematically total charge that moved past a point electric current, I = ____ time taken for charge to move past the point
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5
E L E C T R I C I T Y AN D M AG N E T I S M or in symbols Q I= _ t The ampere is a fundamental unit defned as part o the S I. Although it is explained here in terms o the ow o charge, this is not how it is defned. The SI defnition is based on ideas rom magnetism and is covered in Sub- topics 1 .1 and 5 .4.
Nature of science Another physics link This link between owing charge and current is a crucial one. E lectrical current is a macroscopic quantity, transer o charge by electrons is a microscopic phenomenon in every sense o the word. This is another example o a link in physics between macroscopic observations and inerences about what is happening on the smallest scales.
It was the lack o knowledge o what happens inside conductors at the atomic scale that orced scientists, up to the end o the nineteenth century, to develop concepts such as current and feld to explain the eects they observed. It also, as we shall see, led to a crucial mistake.
Worked examples 1
In the shuttling ball experiment, the ball moves between the two charged plates at a requency o 0. 67 Hz. The ball carries a charge o magnitude 72 nC each time it crosses rom one plate to the other.
b) The charge transerred is 72 nC = 7.2 1 0 8 C Each electron has a charge o 1 . 6 1 0 1 9 C , so the number o electrons involved in the 7.2 1 0 transer is ________ = 4.5 1 0 1 1 1 .6 1 0 8
1 9
2
C alculate: a) the average current in the circuit b) the number o electrons transerred each time the ball touches one o the plates.
a) C alculate the current in a wire through which a charge o 2 5 C passes in 1 5 00 s. b) The current in a wire is 3 6 mA. C alculate the charge that ows along the wire in one minute.
Solution
Solution
a) The time between the ball being at the same plate 1 1 = __ = ____ = 1 . 5 s. The time to transer 72 nC 0.67 is thereore 0.75 s.
Q 25 a) I = ___ , so the current = ____ = 1 7 mA 1 5 00 t
7.2 1 0 C urrent = _______ = 96 nA 0.75 8
b) Q = I t and t = 60 s. Thus charge that ows = 3 .6 1 0 2 60 = 2 .2 C
Charge carrier drift speed Turn on a lighting circuit at home and the lamp lights almost immediately. D oes this give us a clue to the speed at which the electrons in the wires move? In the Investigate! experiment on page 1 85 , the stain indicating the position o the ions moves at no more than a ew millimetres per second. The lower the value o the current, the slower the rate at which the ions move. This slow speed at which the ions move along the conductor is known as the drift sp eed. We need a mathematical model to confrm this observation.
184
5.1 ELECTRI C FI ELD S
Imagine a cylindrical conductor that is carrying an electric current I. The cross- sectional area o the conductor is A and it contains charge carriers each with charge q. We assume that each carrier has a speed v and that there are n charge carriers in 1 m 3 o conductor this quantity is known as the charge density. length o volume swept out in one second v cross-sectional area, A
n charge carriers per unit volume
P charge carrier q
Figure 18 A model or conduction.
Figure 1 8 shows charge carriers, each o charge q, moving past point P at a speed v. In one second, a volume Av o charge carriers passes P. The total number o charge carriers in this volume is nAv and thereore the total charge in the volume is nAvq. However, this is the charge that passes point P in one second, which is what we mean by the electric current. So I = nAvq
Investigate! potassium manganate( VII ) crocodile clip crystal
clips should be attached to leads that are connected to a low voltage power supply ( no more than 2 5 V is required in this experiment) .
+
Wet the flter paper with aqueous ammonia solution.
Take a small crystal o potassium manganate( VII) and place it in the centre o the flter paper. Ensure that the slide is horizontal.
Turn on the current and watch the crystal. You should see a stain on the paper moving away rom the crystal.
Reverse the current direction to check that the eect is not due to the slide not being horizontal.
How ast is the stain moving?
flter paper soaked in microscope slide aqueous ammonia solution
Figure 17 The speeds with which electrons move in a metal conductor during conduction are difcult to observe, but the progress o conducting ions in a liquid can be inerred by the trace they leave.
You should wear eye protection during this experiment.
Fold a piece o flter paper around a microscope slide and fx it with two crocodile clips at the ends o the slide. The crocodile
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Worked example
Solution
1
Rearranging the equation I v= _ nAQ 2 0.65 1 0 and the area A o the wire is ( ________ ) 2 = 3 .3 1 0 7 m 2 .
A copper wire o diameter 0.65 mm carries a current o 0.2 5 A. There are 8.5 1 0 2 8 charge carriers in each cubic metre o copper; the charge on each charge carrier ( electron) is 1 .6 1 0 1 9 C . C alculate the drit speed o the charge carriers.
3
0.25 So v = _________________________ 8.5 1 0 3 .3 1 0 1 .6 1 0 28
7
1 9
= 0. 05 5 mm s 1
The example shows that the drit speed o each charge is less than onetenth o a millimetre each second. You may well be surprised by this result but it is probably o the same order as the speed you observed in the experiment with the potassium manganate( VII) . Although the electron charge is very small, the speed can also be small because there are very large numbers o ree electrons available or conduction in the metal. To see how sensitive the drit speed is to changes in the charge carrier density, we can compare the drit speed in copper with the drit speed in a semiconductor called germanium. The number o charge carriers in one cubic metre o germanium is about 1 0 9 less than in copper. So to sustain the same current in a germanium sample would require a drit speed 1 0 9 times greater than in the copper or a cross-sectional area 1 0 9 as large. The slow drit speed in conductors or substantial currents poses the question o how a lamp can turn on when there may be a signifcant run o cable between switch and lamp. The charge carriers in the cable are driting slowly around the cable. However, the inormation that the charge carriers are to begin to move when the switch is closed travels much more quickly close to the speed o light in act. The inormation is transerred when an electromagnetic wave propagates around the cable and produces a drit in all the ree electrons virtually simultaneously. So the lamp can turn on almost instantaneously, even though, or direct current, it may take an individual electron many minutes or even hours to reach the lamp itsel.
Potential diference Free electrons move in a conductor when an electric feld acts on the conductor. Later we shall see how devices such as electric cells and power supplies provide this electric feld. At the same time the power supplies transer energy to the electrons. As the electrons move through the conductors, they collide with the positive ions in the lattice and transer the energy gained rom the feld to the ions. In situations where felds act, physicists use two quantities called potential and potential difference when dealing with energy transers. Potential dierence (oten abbreviated to pd) is a measure o the electrical potential energy transerred rom an electron when it is moving between two points in a circuit. However, given the very small amount o charge possessed by each electron this amount o energy is also very small. It is better to use the much larger quantity represented by one coulomb o charge.
186
5.1 ELECTRI C FI ELD S
Potential dierence between two points is defned as the work done (energy transerred) W when one unit o charge Q moves between the points. W potential dierence = _ Q The symbol given to potential dierence is V; its unit is the J C 1 and is named the volt (symbol: V) ater the Italian scientist Alessandro Volta who was born in the middle o the eighteenth century and who worked on the development o electricity. The potential dierence between two points is one volt i one joule o energy is transerred per coulomb o charge passing between the two points. A simple circuit will illustrate these ideas: An electric cell is connected to a lamp via a switch and three leads. Figure 1 9 shows a picture o the circuit as it would look set up on the bench. +
lead
cell
conventional current
electronic current
electronic current
conventional current
lamp
Figure 19 Conventional and electronic current in a circuit.
When the switch is closed, electrons ow round the circuit. Notice the direction in which the electrons move and also that the diagram shows the direction o a conventional current. The two directions are opposite; in this case, clockwise or the electron ow and anti-clockwise or the conventional current. The reason or this dierence is explained in a later Nature o science section. You need to take care with this dierence, particularly when using some o the direction rules that are introduced later in this topic. What happens to an electron as it goes round the circuit once? The electron gains electric potential energy as it moves through the cell (this will be covered in Sub-topic 5.3) . The electron then leaves the cell and begins to move through the connecting lead. Leads are designed so that they do not require much energy transer to allow the electrons through (we say that they have a low electrical resistance) and so the potential dierence rom one end o the lead to the other is small. The electron moves through the switch which also gains little energy rom the charge carrier. Ater moving through another lead the electron reaches the lamp. This component is dierent rom others in the circuit, it is deliberately designed so that it can gain much o the electrical potential energy rom
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E L E C T R I C I T Y AN D M AG N E T I S M the electrons as they pass through it. The metal lattice in the flament gains energy and as a result the ions vibrate at greater speeds and with greater amplitudes. At these high temperatures the flament in the lamp will glow brightly; the lamp will be lit. In potential dierence terms, the pds across the leads and the switch are small because the passage o one coulomb o charge through them will not result in much energy transer to the lattice. The pd across the lamp will be large because, or each coulomb going through it, large amounts o energy are transerred rom electrons to the lattice ions in the flament raising its temperature.
Nature of science Conventional and electron currents In early studies o current electricity, the idea emerged that there was a ow o electrical uid in wires and that this ow was responsible or the observed eects o electricity. At frst the suggestion was that there were two types o uid known as vitreous and resinous. B enjamin Franklin (the same man who helped drat the US D eclaration o Independence) proposed that there was only one uid but that it behaved dierently depending on the circumstances. He was also the frst scientist to use the terms positive and negative. What then happened was that scientists assigned a positive charge to the uid thought to be moving in the wires. This positive charge was said to ow out o the positive terminal o a power supply ( because the charge was repelled) and went around the circuit re- entering the power
supply through the negative terminal. This is what we now term the conventional current. In act, we now know that in a metal the charge carriers are electrons and that they move in the opposite direction, leaving a power supply at the negative terminal. This is termed the electronic current. You should take care with these two currents and not conuse them. You may ask: why do we now not simply drop the conventional current and talk only about the electronic current? The answer is that other rules in electricity and magnetism were set up on the assumption that charge carriers are positive. All these rules would need to be reversed to take account o our later knowledge. It is better to leave things as they are.
Worked examples 1
A high efciency LED lamp is lit or 2 hours. C alculate the energy transer to the lamp when the pd across it is 2 40 V and the current in it is 5 0 mA.
Solution 2 hours is 2 60 60 = 72 00 s. The charge transerred is It = 7.2 1 0 3 5 0 1 0 3 = 3 60 C Work done = charge pd = 3 60 2 40 = 86 400 J 2
188
A cell has a terminal voltage o 1 .5 V and can deliver a charge o 460 C beore it becomes discharged.
a)
C alculate the maximum energy the cell can deliver.
b) The current in the cell never exceeds 5 mA. Estimate the lietime o the cell.
Solution W a) Potential dierence, V = __ q so W = qV = 460 1 .5 = 690 J
b) The current o 5 mA means that no more than 5 mC ows through the cell at any time. 46 0 S o _____ = 92 000 s ( which is about 2 5 hours) 0.005
5.1 ELECTRI C FI ELD S
Electromotive force (emf) Another important term used in electric theory is electromotive force ( usually written as em or brevity) . This term seems to imply that there is a orce involved in the movement o charge, but the real meaning o em is connected to the energy changes in the circuit. When charge ows electrical energy can go into another orm such as internal energy ( through the heating, or Joule, eect) , or it can be converted from another orm ( or example, light ( radiant energy) in solar ( photovoltaic) cells) . The term em will be used in this course when energy is transerred to the electrons in, or example, a battery. ( O ther devices can also convert energy into an electrical orm. Examples include microphones and dynamos.) The term potential difference will be used when the energy is transerred from the electrical orm. So, examples o this would be electrical into heat and light, or electrical into motion energy. The table shows some o the devices that transer electrical energy and it gives the term that is most appropriate to use or each one.
Device Cell Resistor Microphone Loudspeaker Lamp Photovoltaic cell Dynamo Electric motor
chemical electrical sound converts electrical energy electrical from light kinetic electrical
electrical internal electrical sound into light (and internal) electrical electrical kinetic
pd or emf? emf pd emf pd pd emf emf pd
Power, current, and pd We can now answer the question o how much energy is delivered to a conductor by the electrons as they move through it. S uppose there is a conductor with a potential dierence V between its ends when a current I is in the conductor. In time t the charge Q that moves through the conductor is equal to It. The energy W transerred to the conductor rom the electrons is QV which is ( I t) V. S o the energy transerred in time t is W = IV t e n e rgy W The electrical power being supplied to the conductor is _____ = __ and t tim e thereore
electrical power P = IV Alternative orms o this expression that you will fnd useul are I = P and V = __ I
__P V
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E L E C T R I C I T Y AN D M AG N E T I S M The unit of p ower is the watt ( W) 1 watt ( 1 W) is the power developed when 1 J is converted in 1 s, the same in both mechanics and electricity. So another way to think o the volt is as the power transerred per unit current in a conductor.
Worked examples 1
A 3 V, 1 .5 W flament lamp is connected to a 3 V battery. C alculate:
a distance o 1 .5 m in 7 s. The motor is 40% efcient. C alculate the average current in the motor while the load is being raised.
a) the current in the lamp b) the energy transerred in 2400 s.
Solution 1 .5 P a) Electrical power, P = IV, so I = __ = ___ = 0.5 A V 3
b) The energy transerred every second is 1 .5 J so in 2 400 s, 3 600 J. 2
An electric motor that is connected to a 1 2 V supply is able to raise a 0. 1 0 kg load through
Solution The energy gained is mgh = 0.01 g 1 .5 = 0.1 47 J The power output o the motor must be 0 . 1 47 = _____ = 0.02 1 W 7 0.021 P The current in the motor = __ = _____ = 1 .8 mA. V 12 Since the motor is 40% efcient the current will be 4.5 mA.
Nature of science A word about potential The use o the term potential dierence implies that there is something called potential which can dier rom point to point. This is indeed the case.
energy; this is described as a loss o potential. Positive charges move rom points o high potential to low potential i they are ree to do so.
An isolated positive point charge will have feld lines that radiate away rom it. A small positive test charge in this feld will have a orce exerted on it in the feld line direction and, i ree to do so, will accelerate away rom the original charge. When the test charge is close, there is energy stored in the system and we say that the system ( the two charges interacting) has a high potential. When the test charge is urther away, there is still energy stored, but it is smaller because the system has converted energy into the kinetic energy o the charge it has done work on the charge. So to move the test charge away rom the original charge transers some o the original stored
Negative charges on the other hand move rom points o low potential to high. You can work this through yoursel by imagining a negative test charge near a positive original charge. This time the two charges are attracted and to move the negative charge away we have to do work on the system. This increases the potential o the system. I charges are ree they will all towards each other losing potential energy. In what orm does this energy re- appear? We shall return to a discussion o potential in Topic 1 0. From now on Topic 5 only reers to potential dierences.
The electronvolt Earlier we said that the energy possessed by individual electrons is very small. I a single electron is moved through a potential dierence o W equal to 1 5 V then as V = __ , so W = QV and the energy gained by this Q 1 9 electron is 1 5 1 . 6 1 0 J = 2 . 4 1 0 1 8 J. This is a very small amount and involves us in large negative powers o ten. It is more convenient to defne a new energy unit called the electronvolt ( symbol eV) .
190
5.1 ELECTRI C FI ELD S
This is defned as the energy gained by an electron when it moves through a potential dierence o one volt. An energy o 1 eV is equivalent to 1 .6 1 0 1 9 J. The electronvolt is used extensively in nuclear and particle physics. B e careul, although the unit sounds as though it might be connected to potential dierence, like the j oule it is a unit o energy.
Worked examples 1
An electron, initially at rest, is accelerated through a potential dierence o 1 80 V. C alculate, or the electron: a) the gain in kinetic energy
Solution (a) The electron gains 1 80 eV o energy during its acceleration. 1 eV 1 .6 1 0 1 9 J so 1 80 eV 2 .9 1 0 1 7 J 1 (b) The kinetic energy o the election = __ mv2 and 2 31 the mass ___ o the electron is 9.1 1 0 kg.
_________
___ = ___________ = 8.0 1 0 2 E ke me
2 2 . 9 1 0 1 7 9 . 1 1 0 3 1
In a nuclear accelerator a proton is accelerated rom rest gaining an energy o 2 5 0 MeV. E stimate the fnal speed o the particle and comment on the result.
Solution
b) the fnal speed.
So v =
2
6
m s 1 .
The energy gained by the proton, in j oules, is 4.0 1 0 1 1 J. ___
2 E ke
_________
2 4. 0 1 0 As beore, v = ___ = ___________ , but using a m 1 .7 1 0 value or the mass o the proton this time. p
1 2
2 7
The numerical answer or v = 2 .2 1 0 8 m s 1 . This is a large speed, 70% o the speed o light. In act the speed will be less than this as some o the energy goes into increasing the mass o the proton through relativistic eects rather than into the speed o the proton.
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5.2 Heating efect o an electric current Understanding
Applications and skills
Circuit diagrams
Drawing and interpreting circuit diagrams
Kirchhos laws
Indentiying ohmic and non-ohmic
Heating eect o an electric current and its
consequences Resistance Ohms law Resistivity Power dissipation
Nature o science Peer review a process in which scientists repeat and criticize the work o other scientists is an important part o the modern scientifc method. It was not always so. The work o Ohm was neglected in England at frst as Barlow, a much-respected fgure in his day, had published contradictory material to that o Ohm. In present-day science the need or repeatability in data collection is paramount. I experiments or other fndings cannot be repeated, or i they contradict other scientists work, then a close look is paid to them beore they are generally accepted.
conductors through a consideration o the VI characteristic graph Investigating combinations o resistors in parallel and series circuits Describing ideal and non-ideal ammeters and voltmeters Describing practical uses o potential divider circuits, including the advantages o a potential divider over a series variable resistor in controlling a simple circuit Investigating one or more o the actors that aect resistivity Solving problems involving current, charge, potential dierence, Kirchhos laws, power, resistance and resistivity
Equations resistance defnition: R = ___VI
2
V electrical power: P = VI = I2 R = _______ R
combining resistors:
in series Rto ta l = R1 + R2 + R3 ... 1 1 1 1 in parallel ___________ = _______ + _______ + _______ + ... R R R R total
1
2
3
RA resisitivity defnition: = ________ l
Efects o electric current Introduction This is the frst o three sub- topics that discusses some o the eects that occur when charge ows in a circuit. The three eects are:
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heating effect, when energy is transerred to a resistor as internal energy
chemical effect, when chemicals react together to alter the energy o electrons and to cause them to move, or when electric current in a material causes chemical changes ( Sub- topic 5 . 3 )
magnetic effect, when a current produces a magnetic feld, or when magnetic felds change near conductors and induce an em in the conductor ( S ub-topic 5 .4) .
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
This sub-topic deals with the heating eect o a current ater giving you some advice on setting up and drawing electric circuits.
Drawing and using circuit diagrams At some stage in your study o electricity you need to learn how to construct electrical circuits to carry out practical tasks. This section deals with drawing, interpreting, and using circuit diagrams and is designed to stand alone so that you can reer to it whenever you are working with diagrams and real circuits. You may not have met all the components discussed here yet. They will be introduced as they are needed.
Circuit symbols A set o agreed electrical symbols has been devised so that all physicists understand what is represented in a circuit diagram. The agreed symbols that are used in the IB D iploma Programme are shown in fgure 1 . Most o these are straightorward and obvious; some may be amiliar to you already. Ensure that you can draw and identiy all o them accurately.
joined wires
wires crossing (not joined)
cell
battery
lamp
ac supply
switch
ammeter
voltmeter
A
V
galvanometer
resistor
variable resistor
potentiometer
heating element
fuse
thermistor
diode
variable power supply
transformer
ac supply
capacitor
Figure 1 Circuit symbols.
193
5
E L E C T R I C I T Y AN D M AG N E T I S M There are points to make about the symbols.
06 V
01 A
S ome o the symbols here are intended or direct current (dc) circuits ( cells and batteries, or example) . D irect current reers to a circuit in which the charge ows in one direction. Typical examples o this in use would be a low- voltage ashlight or a mobile phone. O ther types o electrical circuits use alternating current (ac) in which the current direction is frst one way around the circuit and then the opposite. The time between changes is typically about 1 /1 00th o a second. C ommon standards or the requencies around the world include 5 0 Hz and 60 Hz. Alternating current is used in high- voltage devices ( typically in the home and industry) , where large amounts o energy transer are required: kettles, washing machines, powerul electric motors, and so on. Alternating supplies can be easily transormed rom one pd to another, whereas this is more difcult ( though not impossible) or dc.
There are separate symbols or cells and batteries. Most people use these two terms interchangeably, but there is a dierence: a battery is a collection o cells arranged positive terminal to negative the diagram or the battery shows how they are connected. A cell only contains one source o em. S ub- topic 5 .3 goes into more detail about cells and batteries.
Circuit conventions
It is not usual to write the name o the component in addition to giving its symbol, unless there is some chance o ambiguity or the symbol is unusual. However, i the value o a particular component is important in the operation o the circuit it is usual to write its value alongside it.
Particular care needs to be taken when it is necessary to draw one connecting lead over another. The convention is that i two leads cross and are j oined to each other, then a dot is placed at the j unction. I there is no dot, then the leads are not connected to each other. In the circuit in fgure 2 it is important to know the em o the cell, the data or the lamp at its working temperature, and the ull scale defections (sd) o the two meters in the circuit.
A 6 V, 300 mA
V 010 V
Figure 2 Circuit diagram.
Practical measurements o current and potential diference We oten need to measure the current in a circuit and the pd across components in the circuit. This can be achieved with the use o meters or sensors connected to computers ( data loggers) . You may well use both during the course. S chools use many types and varieties o meters and it is impossible to discuss them all here, but an essential distinction to make is between analogue and digital meters. Again, you may well use both as you work through the course. Analogue meters have a mechanical system o a coil and a magnet. When charge ows through the coil, a magnetic feld is produced that interacts with the feld o the magnet and the coil swings round against a spring. The position reached by the pointer attached to the coil is a measure o the current in the meter.
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D igital meters sample the potential dierence across the terminals o the meter ( or, or current, the pd across a known resistor inside the meter) and then convert the answer into a orm suitable or display on the meter. Ammeters measure the current in the circuit. As we want to know the size o the current in a component it is clear that the ammeter must have the same current. The ammeter needs to be in series with the circuit or component. An ideal ammeter will not take any energy rom the electrons as they ow through it, otherwise it would disturb the circuit it is trying to measure. Figure 2 shows where the ammeter is placed to measure the current. Voltmeters measure the energy converted per unit charge that ows in a component or components. You can think o a voltmeter as needing to compare the energy in the electrons beore they enter a component to when they leave it, rather like the turnstiles (bae gates) to a rail station that count the number o people (charges) going through as they give a set amount o money (energy) to the rail company. To do this the voltmeter must be placed across the terminals o the component or components whose pd is being measured. This arrangement is called parallel. Again, fgure 2 shows this or the voltmeter.
Constructing practical circuits from a diagram Wiring a circuit is an important skill or anyone studying physics. I you are careul and work in an organized way then you should have no problems with any circuit no matter how complex. As an example, this is how you might set up one o the more difcult circuits in this course.
step 1
V
A
A
B
step 2 (a) A A'
B'
step 3 V
step 4 (b)
link A A'; B B' (c)
Figure 3 Variable resistors.
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E L E C T R I C I T Y AN D M AG N E T I S M This is a potential divider circuit that it is used to vary the pd across a component, in this case a lamp. The most awkward component to use here is the potential divider itsel ( a orm o variable resistor, that is sometimes known as a p otentiometer) . In one orm it has three terminals ( fgure 3 ( b) ) , in another type it has three in a rotary ormat. The three- terminal linear device has a terminal at one end o a rod with a wiper that touches the resistance windings, and another two terminals one at each end o the resistance winding itsel. B egin by looking careully at the diagram fgure 3 ( a) . Notice that it is really two smaller circuits that are linked together: the top sub- circuit with the cell, and the bottom sub- circuit with the lamp and the two meters. The bottom circuit itsel consists o two parts: the lamp/ammeter link together with the voltmeter loop. The rules or setting up a circuit like this are:
I you do not already have one, draw a circuit diagram. Get your teacher to check it i you are not sure that it is correct.
B eore starting to plug leads in, lay out the circuit components on the bench in the same position as they appear on the diagram.
C onnect up one loop o the circuit at a time.
Ensure that components are set to give minimum or zero current when the circuit is switched on.
D o not switch the circuit on until you have checked everything.
Figure 3 ( c) shows a sequence or setting up the circuit step-by- step. Another skill you will need is that o troubleshooting circuits this is an art in itsel and comes with experience. A possible sequence is:
C heck the circuit is it really set up as in your diagram?
C heck the power supply ( try it with another single component such as a lamp that you know is working properly) .
C heck that all the leads are correctly inserted and that there are no loose wires inside the connectors.
C heck that the individual components are working b y substituting them into an alternative circuit known to be working.
Resistance We saw in S ub- topic 5 .1 that, as electrons move through a metal, they interact with the positive ions and transer energy to them. This energy appears as kinetic energy o the lattice, in other words, as internal energy: the metal wire carrying the current heats up. However, simple comparison between dierent conductors shows that the amount o energy transerred can vary greatly rom metal to metal. When there is the same current in wires o similar size made o tungsten or copper, the tungsten wire will heat up more than the
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
copper. We need to take account o the act that some conductors can achieve the energy transer better than others. The concept o electrical resistance is used or this. The resistance o a component is defned as potential dierence across the component ____ current in the component The symbol or resistance is R and the defnition leads to a well-known equation V R=_ I The unit o resistance is the ohm ( symbol ; named ater Georg Simon O hm, a German physicist) . In terms o its undamental units, 1 is 1 kg m 2 s 3 A 2 ; using the ohm as a unit is much more convenient! Alternative orms o the equation are: V = IR and I =
__V R
Investigate! Resistance of a metal wire
Take a piece o metal wire ( an alloy called constantan is a good one to choose) and
A
connect it in the circuit shown. Use a power supply with a variable output so that you can alter the pd across the wire easily.
I your wire is long, coil it around an insulator ( perhaps a pencil) and ensure that the coils do not touch.
Take readings o the current in the wire and the pd across it or a range o currents. Your teacher will tell you an appropriate range to use to avoid changing the temperature o the wire.
For each pair o readings divide the pd by the current to obtain the resistance o the wire in ohms.
V
Figure 4
current /A 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70
A table o results is shown in fgure 5 or a metal wire 1 m in length with a diameter o 0. 5 0 mm. For each pair o readings the resistance o the wire has been calculated by dividing the pd by the current. Although the resistance values are not identical ( it is an experiment with real errors, ater all) , they do give an average value or the resistance o 2 . 5 4 . This should be rounded to 2 .5 given the signfcant fgures in the data.
pd/V 0.13 0.26 0.50 0.76 1.01 1.27 1.53 1.78
resistance/ 2.55 2.60 2.50 2.53 2.53 2.54 2.55 2.54
Figure 5 Variation of pd with current for a conductor.
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Nature of science Edison and his lamp The conversion o electrical energy into internal energy was one o the frst uses o distributed electricity. Thomas Edison an inventor and entrepreneur, who worked in the US at around the end o the nineteenth century, was a pioneer o electric lighting. The earliest orms o light were provided by producing a current in a metal or carbon flament. These flaments heated up
until they glowed. Early lamps were primitive but produced a revolution in the way that homes and public spaces were lit. The development continues today as inventors and manuacturers strive to fnd more and more efcient electric lamps such as the light- emitting diodes ( LED ) . More developments will undoubtedly occur during the lietime o this book.
Ohms law Figure 6 shows the results rom the metal wire when they are plotted as a graph o V against I. 2.00 1.80 1.60 1.40 pd/V
1.20 1.00 0.80 0.60 0.40 0.20 0.00 0.00
Worked example The current in a component is 5 .0 mA when the pd across it is 6.0 V. C alculate: a) the resistance o the component b) the pd across the component when the current in it is 1 5 0 A.
Solution V 6 a) R = __ = ______ = 1 .2 k I 5 10
0.10
0.20
0.30
0.40 current/A
0.50
0.60
0.70
0.80
Figure 6 pd against current from the table.
A best straight line has been drawn through the data points. For this wire, the resistance is the same or all values o current measured. Such a resistor is known as ohmic. An equivalent way to say this is that the potential dierence and the current are proportional ( the line is straight and goes through the origin) . In the experiment carried out to obtain these data, the temperature o the wire did not change. This behaviour o metallic wires was frst observed by Georg S imon O hm in 1 82 6. It leads to a rule known as O hms law. O hms law states that the potential dierence across a metallic conductor is directly proportional to the current in the conductor providing that the physical conditions o the conductor do not change.
3
b) V = IR = 1 .5 1 0 4 1 .2 1 0 3 = 0.1 8 A
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B y physical conditions we mean the temperature ( the most important actor as we shall see) and all other actors about the wire. B ut the temperature actor is so important that the law is sometimes stated replacing the term physical conditions with the word temperature.
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Nature of science Ohm and Barlow O hms law has its limitations because it only tells us about a material when the physical conditions do not change. However, it was a remarkable piece o work that did not fnd immediate avour. B arlow was an English scientist who was held in high respect or his earlier work and had recently published an alternative theory on conduction. People simply did not believe that B arlow could be wrong. This immediate acceptance o one scientists work over another would not necessarily happen today. Scientists use a system o peer review. Work published by one scientist or scientifc group must be set out in such a way that other scientists can repeat the experiments or collect the same data to check that there are no errors in the original work. O nly i the scientifc community as a whole can veriy the data is the new work accepted as scientifc act. Towards the end o the 2 0th century a research group thought that it had ound evidence that nuclear usion could occur at low temperatures ( so-called cold usion) . Repeated attempts by other research groups to replicate the original results ailed, and the cold usion ideas were discarded.
Investigate! Variation o resistance o a lamp flament
Use the circuit you used in the investigation on resistance o a metal wire, but repeat the experiment with a flament lamp instead o the wire.
two ways to achieve this: the frst is to reverse the connections to the power supply, also reversing the connections to the ammeter and voltmeter (i the meters are analogue) . The second way is somewhat easier, simply reverse the lamp and call all the readings negative because they are in the opposite direction through the lamp.
Your teacher will advise you o the range o currents and pds to use.
This time do the experiment twice, the second time with the charge owing through the lamp in the opposite direction to the frst. There are
Plot a graph o V (y-axis) against I (x-axis) with the origin (0,0) in the centre o the paper. Figure 7 shows an example o a VI graph or a lamp.
6 pd/V 4 2 0 60
40
0
20 2
20
40
60 current, I/mA
4 6
Figure 7
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TOK But is it a law? This rule o Ohm is always called a law but is it? In reality it is an experimental description o how a group o materials behave under rather restricted conditions. Does that make it a law? You decide.
The graph is not straight ( although it goes through the origin) so V and I are not proportional to each other. The lamp does not obey O hms law and it is said to be non-ohmic. However, this is not a air test o ohmic behaviour because the flament is not held at a constant temperature. I it were then, as a metal, it would probably obey the law. When the resistance is calculated or some o the data points it is not constant either. The table shows the resistance values at each o the positive current points.
Current/mA 20 34 41 47 52 55
There is also another aspect to the law that is oten misunderstood. Our defnition o resistance is that R = ___VI or V = IR. Ohms law states that
Resistance/ 50 59 73 85 96 109
Tip Notice that these resistances were calculated or each individual V data point using __ . They were not I evaluated rom the tangent to the graph at the current concerned. The defnition o resistance is in V V terms o __ not in terms o ___ which I I is what the tangent would give.
V I and, including the constant o proportionality k, V = kI We thereore defne R to be the same as k, but the defnition o resistance does not correspond to Ohms law (which talks only about a proportionality) . V = IR is emphatically not a statement o Ohms law and i you write this in an examination as a statement o the law you will lose marks.
The data show that resistance o the lamp increases as the current increases. At large currents, it takes greater changes in pd to change the current by a fxed amount. This is exactly what you might have predicted. As the current increases, more energy is transerred rom the electrons every second because more electrons ow at higher currents. The energy goes into increasing the kinetic energy o the lattice ions and thereore the temperature o the bulk material. B ut the more the ions vibrate in the lattice, the more the electrons can collide with them so at higher temperatures even more energy is transerred to the lattice by the moving charges. O ther non- ohmic conductors include semiconducting diodes and thermistors; these are devices made rom a group o materials known as semiconductors.
Investigate! Diodes and thermistors
200
You can easily extend the Investigate! lamp experiment to include these two types o device. There are some extra practical points however:
The diode will require a protecting resistor in series with it ( 1 00 is usually appropriate) as the current can become very large at even quite small potential dierences; this will cause the diode to melt as large amounts o energy are transerred to it. The resistor limits the current so that the diode is not damaged.
The thermistor also needs care, because as it heats up its resistance decreases and i too much current is used, the thermistor can also be destroyed.
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
The results o these experiments and some others are summarized in the IV graphs in fgure 8. I
I
0
0
V
0
I
ohmic conductor
0
flament lamp
I
V
semiconducting diode
I
0 0
0
V
0
V neon tube
0 0
V thermistor
Figure 8 IV graphs or various conductors.
Semiconducting diodes S emiconducting diodes are designed only to allow charges to ow through them in one direction. This is seen clearly in the graph. For negative values o V there is actually a very small current owing in the negative direction but it is ar less than the orward current. The nature o semiconductor material also means that there is no signifcant current in the orward direction until a certain orward pd is exceeded.
Thermistors Thermistors are made rom one o the two elements that are electrical semiconductors: silicon and germanium. There are several types o thermistor, but we will only consider the negative temperature coefcient type ( ntc) . As the temperature o an ntc thermistor increases, its resistance alls. This is the opposite behaviour to that o a metal. S emiconductors have many ewer ree electrons per cubic metre compared with metals. Their resistances are typically 1 0 5 times greater than similar samples o metals. However, unlike in a metal, the charge density in semiconductors depends strongly on the temperature. The higher the temperature o the semiconductor, the more charge carriers are made available in the material. As the temperature rises in the germanium:
The lattice ions vibrate more and impede the movement o the charge carriers. This is exactly the same as in metals and also leads to an increase in resistance.
More and more charge carriers become available to conduct because the increase in temperature provides them with enough energy to break away rom their atoms. This leads to a large decrease in resistance.
The second eect is much greater than the frst and so the net eect is that conduction increases ( resistance alls) as the temperature o the semiconductor rises.
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Investigate! How resistance depends on size and shape
You will need a set o wires made rom the same metal or metal alloy. The wires should have a circular cross- section and be available in a range o dierent diameters. You will also need to devise a way to vary the length o one o the wires in the circuit.
Use the circuit in fgure 4 on p1 97 to answer the ollowing questions:
How does the resistance R o one o the wires vary with length l?
How does the resistance R o the wires vary with diameter d when the wires all have the same length?
Try to make things easy or your analysis. D ouble and halve the values o length to see what dierence this makes to the resistance is there an obvious relationship? The diameter o the wire may be more difcult to test in this way, but a graph o resistance against diameter may give you a clue.
O nce you have an idea what is going on, you may decide that the best way to answer these questions is to plot graphs o R against l, and 1 R against __ d 2
Resistivity The resistance o a sample o a material depends not only on what it is made o, but also on the physical dimensions ( the size and shape) o the sample itsel. The graphs suggested in the Investigate! should give straight lines that go through the origin and can be summed up in the ollowing rule. The resistance o a conductor is:
proportional to its length l
inversely proportional to its cross-sectional area A ( which is itsel proportional to d2 ) .
So l R _ A This leads to a defnition o a new quantity called resistivity. Resistivity is defned by RA =_ l The unit o resistivity is the ohm-metre ( symbol m) . Take care here: this is ohm metre. It is not ohm metre 1 a mistake requently made by students in examinations. The meaning o ohm metre 1 is the resistance o one metre length o a particular conductor, which is a relevant quantity to know, but is not the same as resistivity. Resistivity is a quantity o considerable use. The resistance o a material depends on the shape o the sample as well as what it is made rom. Even a constant volume o a material will have values o resistance that depend on the shape. However, the value o the resistivity is the same or all samples o the material. Resistivity is independent o shape or size j ust like quantities such as density ( where the size o a material has been
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
removed by the use o volume) or specifc latent heat ( where the value is related to unit mass o the material) .
Worked examples 1
A uniorm wire has a radius o 0.1 6 mm and a length o 7.5 m. C alculate the resistance o the wire i its resistivity is 7.0 1 0 7 m.
2
Solution Unless told otherwise, assume that the wire has a circular cross-section. area o wire = ( 1 .6 1 0 4) 2 = 8. 04 1 0 8 m 2 RA =_ l l 7.0 1 0 7 7.5 and R = _ = __ = 65 A 8.04 1 0 8
C alculate the resistance o a block o copper that has a length o 0.01 2 m with a width o 0.75 mm and a thickness o 1 2 mm. The resistivity o copper is 1 .7 1 0 8 m.
Solution The cross-sectional area o the block is 7.5 1 0 4 1 . 2 1 0 2 = 9.0 1 0 6 m 2 The relevant dimension or the length is 0.01 2 m, so l 1 .7 1 0 0.01 2 R = _ = _____________ = 0.02 3 m 9.0 1 0 A 8
6
Investigate! Resistivity of pencil lead pencil
D etermine the resistance o the lead.
Measure the length o pencil lead between the crocodile clips.
Measure the diameter o the lead using a micrometer screw gauge or digital callipers and calculate the area o the lead.
Using the data, calculate the resistivity o the lead. The accepted value o the resistivity o graphite is about 3 1 0 5 m but you will not expect to get this value given the presence o clay in the lead as well.
Take the experiment one step urther with a challenge. Use your pencil to uniormly shade a 1 0 cm by 2 cm area on a piece o graph paper. This will make a graphite resistance flm on the paper. Attach the graphite flm to a suitable circuit and measure the resistance o the flm. Knowing the resistance and the dimensions o your shaded area should enable you to work out how thick the flm is. ( Hint: in the resistivity equation, the length is the distance across the flm, and the area is width o the flm thickness o the flm. )
V
A
Figure 9 Graphite is a semi- metallic conductor and is a constituent o the lead in a pencil. Another constituent in the pencil lead is clay. It is the ratio o graphite to clay that determines the hardness o the pencil. This experiment enables you to estimate the resistivity o the graphite. Take a B grade pencil ( sometimes known as #1 grade in the US ) and remove about 1 . 5 cm o the wood rom each end leaving a cylinder o the lead exposed. Attach a crocodile ( alligator) clip frmly to each end. Use leads attached to the crocodile clips to connect the pencil into a circuit in order to measure the resistance o the lead. Expect the resistance o the pencil lead to be about 1 or the purpose o choosing the power supply and meters.
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Practical resistors Resistors are o great importance in the electronics and electrical industries. They can be single value ( fxed) devices or they can be variable. They can be manuactured in bulk and are readily and cheaply available. Resistors come in dierent sizes. Small resistors can have a large resistance but only be able to dissipate ( lose) a modest amount o energy every second. I the power that is being generated in the resistor is too large, then its temperature will increase and in a worst case, it could become a thermal use! Resistors are rated by their manuacturers so that, or example, a resistor could have a resistance o 2 70 with a power rating o 0.5 W. This would ___mean that the maximum current that 0.5 = 43 mA. the resistor can saely carry is ___ 270
Combining resistors Electrical components can be linked together in two ways in an electrical circuit: in series, where the components are j oined one ater another as the ammeter, the cell and the resistor in fgure 1 0, or in p arallel as are the resistor and the voltmeter in the same fgure. A
V
Figure 10
Two components connected in series have the same current in each. The number o ree electrons leaving the frst component must equal the number entering the second component; i electrons were to stay in the frst component then it would become negatively charged and would repel urther electrons and prevent them rom entering it. The ow o charges would rapidly grind to a halt. This is an important rule to understand. In series the potential dierences ( pds) add. To see this, think o charge as it travels through two components, the total energy lost is equal to the sum o the two separate amounts o energy in the components. B ecause the charge is the same in both cases, thereore, the sum o the pds is equal to the total pd dropped across them. C omponents in parallel, on the other hand, have the same pd across them, but the currents in the components dier when the resistances o the components dier. C onsider two resistors o dierent resistance values, in parallel with each other and connected to a cell. I one o the resistors is temporarily disconnected then the current in the remaining resistor is given by the em o the cell divided by the resistance. This will also be true or the other resistor i it is connected alone. I both resistors are now connected in parallel with each other, both resistors have the same pd across them because a terminal o each resistor is connected to one o the terminals o the cell. The cell will have to supply more current than i either resistor were there alone to be precise, it has to supply the sum o the separate currents. To sum up
Currents ... In series In parallel
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...are the same ...add
Potential diferences ... ...add ...are the same
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Investigate! Resistors in series and parallel
For this experiment you need six resistors, each one with a tolerance o 5 % . Two o these resistors should be the same. The tolerance fgure means that the manuacturer only guarantees the value to be within 5 % o the nominal value nominal means the printed value. Your resistors may have the nominal value written on them or you may have to use the colour code printed on them. The code is easy to decipher. 2
You also need a multimeter set to measure resistance directly and a way to j oin the resistors together and to connect them to the multimeter.
First, measure the resistance o each resistor and record this in a table.
Take the two resistors that have the same nominal value and connect them in series. Measure the resistance o the combination. C an you see a rule straight away or the combined resistance o two resistors?
Repeat with fve o the possible combinations or connecting resistors in series.
Now measure the combined resistance o the two resistors with the same nominal value when they are in parallel. Is there an obvious rule this time?
O ne way to express the rule or combining
7 1k (5%) = 27 k (5%)
multiplier tolerance silver gold black brown red orange yellow green blue violet grey white
0 1 2 3 4 5 6 7 8 9
0.01 0.1 1 10 100 1k 10 k 100 k 1M 10 M
10% 5%
R R two resistors R 1 and R 2 in parallel is as ______ . R + R 1
1% 2%
1
2
2
Test this relationship or fve combinations o parallel resistors.
0.5%
Test your two rules together by orming combinations o three resistors such as:
These ideas provide a set o rules or the combination o resistors in various arrangements:
Resistors in series Suppose that there are three resistors in series: R1 , R2 and R3 (fgure 1 1 (a) ) . What is the resistance o the single resistor that could replace them so that the resistance o the single resistor is equivalent to the combination o three? The resistors are in series ( fgure 1 1 ( a) ) and thereore the current I is the same in each resistor. Using the defnition o resistance, the pd across each resistor, V1 , V2 and V3 is V1 = IR 1
V2 = IR 2
and
V3 = IR 3
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E L E C T R I C I T Y AN D M AG N E T I S M The single resistor with a resistance R has to be indistinguishable rom the three in series. In other words, when the current through this single resistor is the same as that through the three, then it must have a pd V across it such that V = IR R1
R2
R3 I
V1
V2
V3
V (a) series
R1 I1 R2 I I2 R3 I3 V (b) parallel
Figure 11 Resistors in series and parallel.
As this is a series combination, the potential dierences add, so V = V1 + V2 + V3 Thereore IR = IR 1 + IR 2 + IR 3 and cancelling I leads to, in series R = R1 + R2 + R3 When resistors are combined in series, the resistances add to give the total resistance.
Resistors in parallel Three resistors in parallel ( fgure 1 1 ( b) ) have the same pd V across them ( fgure 1 1 ( b) ) . The currents in the three separate resistors add to give the current in the connecting leads to and rom all three resistors, thereore I = I1 + I2 + I3 Each current can be written in terms o V and R using the defnition o resistance: V _ V V V _ = + _+ _ R R1 R2 R3
206
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T This time V cancels out and, in p arallel,
step 1 combine series resistors
1 1 1 _1 = _ + _+ _ R
R1
R2
R3
In parallel combinations o resistors, the reciprocal o the total resistance is equal to the sum o the reciprocals o the individual resistances.
step 2 combine parallel resistor separately
The parallel equation needs some care in calculations. The steps are:
calculate the reciprocals o each individual resistor
add these reciprocals together
take the reciprocal o the answer.
It is common to see the last step ignored so that the answer is incorrect; a worked example below shows the correct approach.
More complicated networks When the networks o resistors are more complicated, then the individual parts o the network need to be broken down into the simplest orm.
step 3 combine series resistors together
Figure 1 2 shows the order in which a complex resistor network could be worked out.
Figu re 12
Worked examples 1
Three resistors o resistance, 2 .0 , 4. 0 , and 6.0 are connected. C alculate the total resistance o the three resistors when they are connected: a) in series b) in parallel.
Solution a) In series, the resistances are added together, so 2 + 4 + 6 = 12
Solution The two resistors in parallel have a combined 3 1 1 1 1 1 resistance o __ = __ + __ = __ + __ = __ R R R 4 8 8 1
2
8 R = _ = 2 .67 3 This 2 . 67 resistor is in series with 2 .0 , so the total combined resistance is 2 .67 + 2 .0 = 4.7 . 3
Four resistors each o resistance 1 .5 are connected as shown. C alculate the combined resistance o these resistors.
b) In parallel, the reciprocals are used: 6 + 3 + 2 1 1 1 1 1 1 1 11 __ = __ + __ + __ = __ + __ + __ = ________ = __ R R R R 2 4 12 6 12 1
2
3
The fnal step is to take the reciprocal o the 12 sum, R = __ = 1 .1 11 2
2 .0 , 4.0 , and 8. 0 resistors are connected as shown. C alculate the total resistance o this combination. 4.0 2.0
Solution Two 1 .5 resistors in parallel have a resistance 1 1 1 2 given by __ = ___ + ___ = ___ . S o R = 0.75 . R 1 .5 1 .5 1 .5 Two 0. 75 resistors in series have a combined resistance o 0.75 + 0. 75 = 1 .5
8.0
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E L E C T R I C I T Y AN D M AG N E T I S M
Potential divider In the section on circuit diagrams, a potential divider circuit was shown. This is a circuit commonly used with sensors and also to produce variable potential dierences. It has some advantages over the simpler variable resistor circuit even though it is more complicated to set up.
V1
R1
Rtop
R
Vin
Vin R2
Vin thermistor
V2
Vout
LDR
Vout
V (a)
(b)
(c)
Figure 13 The potential divider.
The most basic potential divider consists o two resistors with resistances R 1 and R 2 in series with a power supply. This arrangement is used to provide a fxed pd at a value somewhere between zero and the em o the power supply. Figure 1 3 ( a) shows the arrangement.
Worked example 1
A potential divider consists o two resistors in series with a battery o 1 8 V. The resistors have resistances 3 .0 and 6.0 . C alculate, or each resistor:
The two resistors have the same current in them, and the sum o the pds across the resistors is equal to the source em. So The current I in both resistors is total pd across both resistors V I = ___ = _ R total resistance As usual,
a) the pd across it b) the current in it.
Solution
The resistors are in series so
a) The pd across the 3 .0 V R 18 3 resistors _______ = _____ 3 + 6 (R + R ) in
1
1
2
= 6.0 V. The pd across the 6.0 is then 1 8 6.0 V = 1 2 V. b) The current is the same in both resistors because they are in series. The total resistance is 9. 0 and the em o the battery is 1 8 V. The current = V 18 __ = __ = 2 .0 A. R 9
208
V1 = IR 1 and V2 = IR 2
R = R1 + R2 This leads to equations or the pd across each resistor R1 R2 V1 = _ Vin and V2 = _Vin ( R1 + R2 ) ( R1 + R2 )
Using a potential divider with sensors It is a simple matter to extend the fxed pd arrangement to a circuit that will respond to changes in the external conditions. S uch an arrangement might be used by a computer that can sense changes in pd and respond accordingly, or example, by turning on a warning siren i a rerigerator becomes too warm, or turning on a light when it becomes dark. Typical circuits are shown in fgures 1 3 ( b) and1 3 ( c) . In fgure 1 3 ( b) a thermistor is used instead o one o the fxed resistors.
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Recall that when a thermistor is at a high temperature its resistance is small, and that the resistance increases when the temperature alls. Rather than calculating the values, or this example we will use our knowledge o how pd and current are related to work out the behaviour o the circuit rom frst principles. S uppose the temperature is low and that the thermistor resistance is high relative to that o the fxed-value resistor. Most o the pd will be dropped across the thermistor and very little across the fxed resistor. I you cannot see this straight away, remember the equations rom the previous section: VinR thermistor VinR resistor Vthermistor = __ and Vresistor = __ ( R thermistor + R resistor ) ( R thermistor + R resistor ) I Rthermistor R resistor ( means much greater than) then R thermistor ( R thermistor + R resistor) and Vthermistor Vin. S o the larger resistance ( the thermistor at low temperatures) has the larger pd across it. I the thermistor temperature now increases, then the thermistor resistance will all. Now the fxed- value resistor will have the larger resistance and the pds will be reversed with the thermistor having the small voltage drop across it. A voltage sensor connected to a computer can be set to detect this voltage change and can activate an alarm i the thermistor has too high a temperature. You may be asking what the resistance o the fxed resistor should be. The answer is that it is normally set equal to that o the thermistor when it is at its optimum ( average) temperature. Then any deviation rom the average will change the potential dierence and trigger the appropriate change in the sensing circuit. The same principle can be applied to another sensor device, a lightdependent resistor (LD R) . This, like the thermistor, is made o semiconducting material but this time it is sensitive to photons incident on it. When the light intensity is large, charge carriers are released in the LD R and thus the resistance alls. When the intensity is low, the resistance is high as the charge carriers now re-combine with their atoms. Look at fgure 1 3 ( c) . You should be able to explain that when the LD R is in bright conditions then the pd across the fxed resistor is high.
Using a potential divider to give a variable pd A variable resistor circuit is shown in fgure 1 4( a) . It consists o a power supply, an ammeter, a variable resistor and a resistor. The value o each component is given on the diagram. We can predict the way this circuit will behave. When the variable resistor is set to its minimum value, 0 , then there will be a pd o 2 V across the resistor and a current o 0.2 A in the circuit. When the variable resistor is set to its maximum value, 1 0 , then the total resistance in the circuit is 2 0 , and the current is 0.1 A. This means that with 0. 1 A in the 1 0 fxed resistor, only 1 V is dropped across it. Thereore the range o pd across the fxed resistor can only vary
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E L E C T R I C I T Y AN D M AG N E T I S M rom 1 V to 2 V hal o the available pd that the power supply can in principle provide. You should now be able to predict the range across the variable resistor. A 2V fxed resistor 10
2V
slider
2V 1V
variable resistor 010 0V
(a) variable resistor
V fxed resistor 10
(b) potential divider
Figure 14
The limited range is a signifcant limitation in the use o the variable resistor. To achieve a better range, we could use a variable resistor with a much higher range o resistance. To get a pd o 0 . 1 V across the fxed resistor the resistance o the variable resistor has to be about 2 00 . I the fxed resistor had a much greater resistance, then the variable resistor would need an even higher value too and this would limit the current. The potential divider arrangement ( fgure 1 4( b) ) allows a much greater range o pd to the component under test than does a variable resistor in series with the component. In a potential divider, the same variable resistor can be used but the set up is dierent and involves the use o the three terminals on the variable resistor. ( A variable resistor is also sometimes called a rheostat.) O ne terminal is connected to one side o the cell, and the other end o the rheostat resistor is connected to the other terminal o the cell. The potential at any point along the resistance winding depends on the position o the slider ( or wiper) that can be swept across the windings rom one end to the other. Typical values or the potentials at various points on the windings are shown or the three blue slider positions on fgure 1 4( b) . The component that is under test ( again, a resistor in this case) is connected in a secondary circuit between one terminal o the resistance winding and the slider on the rheostat. When the slider is positioned at one end, the ull 2 V rom the cell is available to the resistor under test. When at the other end, the pd between the ends o the resistor is 0 V ( the two leads to the resistor are eectively connected directly to each other at the variable resistor) . You should know how to set this arrangement up and also how to draw the circuit and explain its use.
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Worked examples 1
A light sensor consists o a 6. 0 V battery, a 1 800 resistor and a light- dependent resistor in series. When the LD R is in darkness the pd across the resistor is 1 . 2 V.
2
a) C alculate the resistance o the LD R when it is in darkness.
Solution
b) When the sensor is in the light, its resistance alls to 2 400 . C alculate the pd across the LD R.
Solution a) As the pd across the resistor is 1 .2 V, the pd across the LD R must be 6 1 .2 = 4. 8 V. The current in the circuit is V 1 .2 = ____ = 0.67 mA. I = __ R 1 800 The resistance o the LD R is V 4. 8 __ = ________ = 72 00 . I 0.67 1 0 3
resistan ce across L DR 2 40 0 b) The ratio o ________________ = ____ = 1 .3 3 . 1 800 resistan ce across 1 8 0 0 pd across L DR This is the same value as ___________ . For the pd across 1 8 0 0
A thermistor is connected in series with a fxed resistor and a battery. D escribe and explain how the pd across the thermistor varies with temperature.
As the temperature o the thermistor rises, its resistance alls. The ratio o the pd across the fxed resistor to the pd across the thermistor rises too because the thermistor resistance is dropping. As the pd across the fxed resistor and thermistor is constant, the pd across the thermistor must all. The change in resistance in the thermistor occurs because more charge carriers are released as the temperature rises. Even though the movement o the charge carriers is impeded at higher temperatures, the release o extra carriers means that the resistance o the material decreases.
ratio o pds to be 1 .3 3 , the pds must be 2 .6 V and 3 .4 V with the 3 .4 V across the LD R.
Investigate! Variable resistor or potential divider?
Set up the two circuits shown in fgure 1 4. Match the value o the fxed resistor to the variable resistor they do not need to be exactly the same but should be reasonably close. Add a voltmeter connected across the fxed resistor to check the pd that is available across it.
Make sure that the maximum current rating or the fxed resistor and the variable resistor cannot be exceeded.
C heck the pd available in the two cases and convince yoursel that the potential divider gives a wider range o voltages.
Heating efect equations We saw earlier that the power P dissipated in a component is related to the pd V across the component and the current I in it: P = IV The energy E converted in time t is E = IV t When either V or I are unknown, then two more equations become available: V2 P = IV = I2 R = _ R
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E L E C T R I C I T Y AN D M AG N E T I S M and V2 E = IVt = I2 Rt = _ t R These equations will allow you to calculate the energy converted in electrical heaters and lamps and so on. Applications that you may come across include heating calculations, and determining the consumption o energy in domestic and industrial situations.
Worked examples 1
C alculate the power dissipated in a 2 5 0 resistor when the pd across it is 1 0 V.
Solution 2
V2 1 02 P = _ = _ = 0.40 W R 250 A 9. 0 kW electrical heater or a shower is designed or use on a 2 5 0 V mains supply. C alculate the current in the heater.
Solution 3
9000 P = _ P = IV so I = _ = 36 A V 250 C alculate the resistance o the heating element in a 2 .0 kW electric heater that is designed or a 1 1 0 V mains supply.
Solution V2 V2 1 1 02 P = _ ; R = _ = _ = 6.1 A. R P 2 000
Kirchhos frst and second laws This section contains no new physics, but it will consolidate your knowledge and put the electrical theory you have learnt so ar into a wider physical context.
I1 I5
I4
I3
I2
Figure 15 Kirchhos frst law.
We saw that the charge carriers in a conductor move into and out o the conductor at equal rates. I 1 0 6 fow into a conductor in one second, then 1 0 6 must fow out during the same time to avoid the buildup o a static charge. We also considered what happens when current splits into two or more parts at the j unction where a parallel circuit begins. This can be taken one step urther to a situation where there is more than one incoming current at the j unction too. Figure 1 5 shows a j unction with three incoming currents and two outgoing ones. O ur rule about the incoming charge equating to the outgoing charge must apply here too, so algebraically: I1 + I2 + I3 = I4 + I5 A general way to write this is to use the sign ( meaning add up everything) , so that for any j unction
I = 0
When using the notation, remember to get the signs o the currents correct. C all any current in positive, and any current out negative.
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
In words this can be written as the sum o the currents into a junction equals the sum o the currents away rom a junction or total charge fowing into a junction equals the total charge fowing away rom the junction. This important rule was frst quoted by the Prussian physicist Gustav Kirchho in 1 845 . It is known as Kirchhos frst law. It is equivalent to a statement o conservation o charge. Kirchho devised a second law which is also a conservation rule this time o energy. In any electric circuit there are sources o em ( oten a cell or a battery o cells or dc) and sinks o pd ( typically, lamps, heating coils, resistors, and thermistors) . A general rule in physics is that energy is conserved. Electrical components have to obey this too. So, in any electrical circuit, the energy being converted into electrical energy (in the sources o em) must be equal to the energy being transerred rom electrical to internal, by the sinks o pd. This is Kirchhos second law equivalent to conservation o energy. This second law applies to all closed circuits both simple and complex. In words, Kirchhos second law can be written as in a complete circuit loop, the sum o the ems in the loop is equal to the sum o the potential dierences in the loop or the sum o all variations o potential in a closed loop equals zero In symbols or any closed loop in a circuit =
IR R2
I2
E
G D
A
F R3
I1
I3
B
C R1
Figure 16 Kirchhofs second law.
Look at the circuit in fgure 1 6, it has parallel and series elements in it. A number o possible loops are drawn and analysed: Loop GABCDEG travelling anticlockwise round the loop This loop begins at the cell and goes around the circuit, through resistor R 1 and resistor R 2 fnally ending at the cell again. In this loop there is one source o em and two sinks o pd ( ignoring the leads, which we assume have zero resistance) .
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5
E L E C T R I C I T Y AN D M AG N E T I S M So ( the emf of the cell) = I1 R 1 + I2 R 2 ( the total pds aross the resistors) The direction o loop travel and the current direction are in all cases the same. We give a positive sign to the currents when this is the case. The em o the cell is driving in the same direction as the loop travel direction; it gets a positive sign as well. I the loop direction and the current or em were to be opposed then they would be given a negative sign. loop EFGE travelling clockwise round the loop This loop goes frst through resistor R 3 and the loop direction is in the same direction as the conventional current. Next the loop goes through resistor R1 but this time the current direction and the loop are dierent so there has to be a negative sign. There is no source o em in the loop so the Kirchho equation becomes 0 = I3 R 3 I2 R 2 Kirchhos frst law can be applied at point G. The total current into point G is I2 + I3 ; the total out is I1 . The application o the law is I1 = I2 + I3 . There are now three separate equations with three unknowns and these equations can be solved to work out the currents in each part o the circuit assuming that we know the value or the em o the cell and the values o the resistances in the circuit. B y setting up a series o loops it is possible to work out the currents and pds or complicated resistor networks, more complicated than could be done using the resistor series and parallel rules alone.
Ideal and non-ideal meters S o ar in this topic we have assumed ( witho ut mentioning it! ) that the me ters used in the circuits were ideal. Ideal meters have no eect on the circuits that they are me asuring. We would always want this to be true b ut, unortunate ly, the me ters we use in real circuits in the laboratory are not so ob liging and we nee d to know how to allow or this. Ammeters are placed in series with components so the ammeter has the same current as the components. It is undesirable or the ammeter to change the current in a circuit but, i the ammeter has a resistance o its own, then this is what will happen. An ideal ammeter has zero resistance clearly not attainable in practice as the coils or circuits inside the ammeter have resistance. Voltmeters are placed in parallel with the device or parts o a circuit they are measuring. In an ideal world, the voltmeter will not require any energy or its coil to move ( i it is a moving- coil type) or or its analogue to digital conversion ( i it is a digital meter) . The way to avoid current in the voltmeter is or the meter to have an infnite resistance again, not an attainable situation in practice. S ome modern digital meters can get very close to these ideals o zero ohm or ammeters and infnite ohms or voltmeters. D igital meters are used more and more in modern science.
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Worked examples 1
C alculate the unknown branch current in the ollowing j unctions.
to show the technique and to convince you that it works.)
a) 3.0 V 9A
5A I1 I
3A
9.0 3.0
I2 loop 2
junction A 6.0
I3
b) loop 1 11A I
C urrent directions have been assigned and two loops 1 and 2 and j unction A defned in the diagram.
5A
3A
For loop 1
7A
3 = 3 I1 + 9I2
Solution
( the em in the loop is 3 V)
a) C urrent into the j unction = 9 + 5 A
or loop 2
C urrent out o the j unction = 3 + IA I = 9 + 5 3 = 1 1 A out o the j unction. b) Current into the junction = I + 1 1 + 3 = I + 1 4A C urrent out o the j unction = 5 + 7 = 1 2 A I = 1 2 1 4 = 2 A, so the current I is directed away rom the j unction. 2
0 = 6I3 9I2
[equation 1 ]
[equation 2]
( there is no source o em in this loop, current I2 is in the opposite direction to the loop direction 0. For j unction A I1 = I2 + I3 so
C alculate the currents in the circuit shown.
0 = 9I2 6I1 + 6I2 0 = 1 5 I2 6I1
3.0 V
And rom equation 1 6 = 6I1 + 1 8I2 I1
9.0
I2
Adding the equations
3.0
6 = 3 3 I2 I2 = 0.1 8 A
6.0 I3
Solution (This circuit can be analysed using the resistors in series and parallel equations. It is given here
S ubstituting gives: I1 = 0.45 A and I3 = 0.2 7 A
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E L E C T R I C I T Y AN D M AG N E T I S M
Nature of science How times change! D igital meters are a good way to get very close to the ideal meter. The resistance between the input terminals can be made to be very large ( 1 0 1 2 or more) or a voltmeter and to have a very small value or an ammeter. The meters themselves are
0
scale
meter terminals thread spring resistance wire
path
pivot
Figure 17
easy to read without any j udgements required about what the pointer indicates. It was not always like this. An early orm o meter was the hot-wire ammeter. This uses the heating eect o a current directly to increase the temperature o a metal wire (the current fows through the wire) . There are a number o orms o the ammeter, but in the type shown here, a spring keeps the wire under tension. As the wire expands with the increase in temperature, any point on the wire moves to the right. A pivoted pointer is attached to the insulated thread. As the wire expands, the spring pulls the thread causing the pointer to rotate about the pivot; a reading o current can now be made. The scale is usually extremely non-linear.
Worked examples 1
A 2 5 0 resistor is connected in series with a 5 00 resistor and a 6.0 V battery. a) C alculate the pd across the 2 5 0 resistor. b) Calculate the pd that will be measured across the 250 resistor i a voltmeter o resistance 1 000 is connected in parallel with it.
Solution a) The pd across the 2 5 0 resistor Vin R 1 6 250 = __ = 2 .0 V. ( R 1 + R 2 ) (250 + 500) b) When the voltmeter is connected, the resistance o the parallel combination is R1 R 2 2 5 0 1 000 R = _ = __ = 2 00 1 250 ( R1 + R2 )
216
2 00 6 V = _ = 1 .7 V 700 2
An ammeter with a resistance o 5 .0 is connected in series with a 3 .0 V cell and a lamp rated at 3 00 mA, 3 V. C alculate the current that the ammeter will measure.
Solution V 3 Resistance o lamp = __ = ___ = 1 0 . Total I 0.3 resistance in circuit = 1 0 + 5 = 1 5 . S o current V 3 in circuit = __ = __ = 2 00 mA. This assumes that R 15 the resistance o the lamp does not vary between 0.2 A and 0.3 A
5 . 3 E LECTRI C CE LLS
5.3 Electric cells Understanding Cells Primary and secondary cells Terminal potential diference
Applications and skills Investigating practical electric cells (primary
Electromotive orce em Internal resistance
and secondary) Describing the discharge characteristic o a simple cell (variation o terminal potential diference with time) Identiying the direction current ow required to recharge a cell Determining internal resistance experimentally Solving problems involving em, internal resistance, and other electrical quantities
Equation em o a cell: = I(R + r)
Nature of science Scientists need to balance their research into more and more ecient electric cells with the long-term risks associated with the disposal o the cells. Some modern cells have extremely poisonous contents. There will be serious consequences i these
chemicals are allowed to enter the water supply or the ood chain. Can we aford the risk o using these toxic substances, and what steps should the scientists take when carrying orward the research?
Introduction E lectric currents can produce a chemical effect. This has great importance in chemical industries as it can be a method for extracting ores or purifying materials. However, in this course we do not investigate this aspect of the chemical effect. O ur emphasis is on the use of an electric cell to store energy in a chemical form and then release it as electrical energy to perform work elsewhere.
Cells C ells operate as direct- current ( dc) devices meaning that the cell drives charge in one direction. The electron charge carriers leave the negative terminal of the cell. After passing around the circuit, the electrons reenter the cell at the positive terminal. The positive terminal has a higher
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E L E C T R I C I T Y AN D M AG N E T I S M potential than the negative terminal so electrons appear to gain energy ( whereas positive charge carriers would appear to lose it) . The chemicals in the cell are reacting while current ows and as a result o this reaction the electrons gain energy and continue their j ourney around the circuit.
Nature of science Anodes and cathodes You will meet a naming system or anode and cathode which seems dierent or other cases in physics. In act it is consistent, but you need to think careully about it. For example, in a cathode- ray tube where the electrons in the tube are emitted rom a hot metal flament, the flament is called the cathode and is at a negative voltage. This appears to be the opposite notation rom that used or the electric cell. The reason is that the notation reers to what is happening inside the cell, not to the external circuit. The chemical reaction in the cell leads to positive
ions being generated at one o the electrodes and then owing away into the bulk o the liquid. S o as ar as the interior o the cell is concerned this is an anode, because it is an ( internal) source o positive ions. O course, as ar as the external circuit is concerned, the movement o positive charges away rom this anode leaves it negative and the electrode will repel electrons in the external circuit. As external observers, we call this the negative terminal even though ( as ar as the interior o the cell is concerned) it is an anode.
Primary and secondary cells Many o the portable devices we use today: torches, music players, computers, can operate with internal cells, either singly or in batteries. In some cases, the cells are used until they are exhausted and then thrown away. These are called p rimary cells. The original chemicals have completely reacted and been used up, and they cannot be recharged. E xamples include AA cells ( properly called dry cells) and button mercury cells as used in clocks and other small lowcurrent devices. O n the o the r hand, so me de vice s u se re charge ab le ce lls so that whe n the che mical re actio ns have inishe d, the ce lls can b e co nne cte d to a charge r. The n the che mical re actio n is re ve rse d and the o riginal che micals o rm again. Whe n as mu ch o the re - co nve rsio n as is p o ssib le has b e e n achie ve d, the ce ll is again availab le as a che mical e ne rgy sto re . Re charge ab le ce lls are kno wn as s eco nd ary cells . There many varieties o cell, here are two examples o the chemistry o a primary cell and that o a secondary cell. The Leclanch cell was invented by Georges Leclanch in 1 886. It is a primary cell and is the basis o many domestic torch and radio batteries.
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5 . 3 E LECTRI C CE LLS
-
+
zinc rod
carbon rod
+ gas vents -
porous pot (P) ammonium chloride solution mixture of carbon and manganese dioxide
positive terminal
negative terminal
lead oxide lead insulating case
sulfuric acid
glass vessel (a)
Leclanch cell
(b)
lead-acid accumulator
Figure 1 Leclanch cell and lead-acid accumulator.
Figure 1 ( a) shows one practical arrangement o the cell. It consists o a zinc rod that orms the negative terminal. Inside the zinc is a paste o ammonium chloride separated rom manganese dioxide ( the cathode) by a porous membrane. In the centre o the manganese dioxide is a carbon rod that acts as the positive terminal or the cell. Zinc atoms on the inside surace o the case oxidize to become positive ions. They then begin to move away rom the inside o the case through the chloride paste leaving the case negatively charged. When the cell is connected to an external circuit, these electrons move around the circuit eventually reaching the carbon rod. A reaction inside the cell uses these electrons together with the components o the cell eventually orming the waste products o the cell, which include ammonia, manganese oxide, and manganese hydroxide. S econdary cells are very important today. They include the leadacid accumulator, together with more modern developments such as the nickelcadmium ( NiC d) , nickelmetal- hydride ( NiMH) , and lithium ion cells. All these types can be recharged many times and even though they may have a high initial cost ( compared to primary cells o an equivalent em and capacity) , the recharge is cheap so that, during the proj ected lietime o the cell, the overall cost is lower than that o primary cells. Although the lead- acid accumulator is one o the earliest examples o a secondary cell, it remains important ( accumulator is an old word that implies the accumulation or collection o energy into a store) . It is capable o delivering the high currents needed to start internal combustion and diesel engines, and it is reliable in the long term to maintain the current to important computer servers i the mains supply ails. The leadacid cell ( fgure 1 ( b) ) is slightly older than the Leclanch cell, and was invented in 1 85 9 by Gaston Plant. In its charged state the cell consists o two plates, one o metallic lead, the other o lead( IV ) oxide ( PbO 2 ) immersed in a bath o dilute suluric
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5
E L E C T R I C I T Y AN D M AG N E T I S M acid. D uring discharge when the cell is supplying current to an external circuit, the lead plate reacts with the acid to form lead( II) sulfate and the production of two free electrons. The liquid surrounding the plates loses acid and becomes more dilute. At the oxide plate, electrons are gained ( from the external circuit) and lead( II) sulfate is formed, also with the removal of acid from the liquid. S o the net result of discharging is that the plates convert to identical lead sulfates and the liquid surrounding the plates becomes more dilute.
Worked examples 1
Explain the difference between a primary and a secondary cell.
Solution A primary cell is one that can convert chemical energy to electrical energy until the chemicals in the cell are exhausted. Recharging the cell is not possible. A secondary cell can be recharged and the chemicals in it are converted back to their original form so that electrical energy can be supplied by the cell again. 2
A cell has a capacity of 1 400 mA h. C alculate the number of hours for which it can supply a current of 1 .8 mA.
Solution C ell capacity = current time and therefore 1 400 = 1 .8t. In this case t = 780 hours.
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Recharging reverses these changes: electrons are forced from the positive plate by an external charging circuit and forced onto the negative plate. At the negative plate, atomic lead forms, at the positive, lead oxide is created, which restores the cell to its original state. The chargerecharge cycle can be carried out many times providing that the cell is treated carefully. However, if lead or lead oxide fall off the plates, they can no longer be used in the process and will collect at the bottom of the cell container. In the worst case the lead will short out the plates and the cell will stop operating. If the cell is overcharged ( when no more sulfate can be converted into lead and lead oxide) the charging current will begin to split the water in the cell into hydrogen and oxygen gas which are given off from the cell. The total amount of liquid in the cell will decrease, meaning that the plates may not be fully covered by the acid. These parts of the plates will then not take part in the reaction and the ability of the cell to store energy will decrease. Much industrial research effort is concentrated on the development of rechargeable cells. An important consideration for many manufacturers of electronic devices is the energy density for the battery ( the energy stored per unit volume) as this often determines the overall design and mass of an electronic device. Also a larger energy density may well provide a longer lifetime for the device.
Capacity of a cell Two cells with the same chemistry will generate the same electromotive force ( emf) as each other. However if one of the cells has larger plates than the other and contains larger volumes of chemicals, then it will be able to supply energy for longer when both cells carry the same current. The cap acity of a cell is the quantity used to measure the ability of a cell to release charge. If a cell is discharged at a high rate then it will not be long before the cell is exhausted or needs recharging, if the discharge current is low then the cell will supply energy for longer times. The capacity of a cell or battery is the constant current that it can supply for a given discharge. S o, if a cell can supply a constant current of 2 A for 2 0 hours then it said to have a capacity of 40 amp- hours ( abbreviated as 40 A h) . The implication is that this cell could supply 1 A for 40 hours, or 0.1 A for 400 hours, or 1 0 A for 4 hours. However, practical cells do not necessarily discharge in such a linear way and this cell may be able to provide a small discharge current of a few milliamps for much
5 . 3 E LECTRI C CE LLS
longer than expected rom the value o its capacity. C onversely, the cell may be able to discharge at 1 0 A or much less than 4 hours. An extreme example o this is the leadacid batteries used to start cars and commercial vehicles. A typical car battery may have a capacity o 1 00 A h. The current demand or starting the car on a cold winters day can easily approach 1 5 0 A. B ut the battery will only be able to turn the engine or a ew minutes rather than the two- thirds o an hour we might expect.
Discharging cells Investigate! Discharge of a cell
S ome cells have high capacity, and studying the discharge o such a cell can take a long time. This is a good example o how electronic data collection ( data- logging) can help an experimenter.
load resistor
Many data loggers need to be told how oten to take measurements (the sampling rate) . You will need to make some judgements about the overall length o time or which the experiment is likely to operate. Suppose you have a 1 .5 V cell rated at 1 5 00 mA h ( you can check the fgure by checking the cell specifcation on the manuacturers website) . It would seem reasonable that i the cell is going to supply a current o 2 5 0 mA, then it would discharge in a time o between 41 0 hours. You will need to set up the data logger accordingly with a suitable time between readings (the sampling interval) . D o not, however, exceed the maximum discharge current o the cell.
When the computer is set up, turn on the current in the discharge circuit and start the logging. Eventually the cell will have discharged and you can display the data as a graph. What are the important eatures o your graph?
Test other types o cell too, at least one primary and one secondary cell. Are there dierences in the way in which they discharge?
For at least one cell, towards the end o the discharge, switch o the discharge current while still continuing to monitor the terminal pd. D oes the value o the pd stay the same or does it recover? When the discharge is resumed what happens to the pd?
voltage sensor and data logger
Figure 2
The aim is to collect data to fnd how the terminal pd across the cell varies with time rom the start o the discharge. The basic circuit is shown in fgure 2 . Use a new cell to provide the current so that the initial behaviour o the cell can be seen too.
The results you obtain will depend on the types and qualities o the cells you test. A typical discharge curve looks like that in fgure 3 on p2 2 2 in which the terminal potential dierence o the cell is plotted against time since the discharge current began.
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E L E C T R I C I T Y AN D M AG N E T I S M 1.3
cell emf / V
1.2 midpoint voltage (1.18 volts)
1.1
1.0 end of discharge (127 minutes)
0.9 0
20
40
60 80 time (minutes)
100
120
140
Figure 3 Typical dischargetime graph for a cell.
Important features of this graph are that:
The initial terminal pd is higher than the quoted emf ( the value the manufacturer prints on the case) , but the initial value quickly drops to the rated emf ( approximately) .
For most of the discharge time the terminal pd remains more or less constant at or around the quoted emf. There is however sometimes a slow decline in the value of the terminal pd.
As the cell approaches exhaustion, the terminal pd drops very rapidly to a low level.
If the current is switched off, the terminal pd rises and can eventually reach the rated value again. However, when discharge is resumed, the terminal pd falls very quickly to the low value that it had before.
O ther experiments are possible, particularly for rechargeable cells to see the effects of repeated dischargecharge cycles on the cell. Figure 4 shows how the capacity of a NiC d cell changes as cell is taken through more and more cycles. cycle life at room temperature 5
capacity A h
4 3 2 1 0 0
222
500
1000 1500 cycle number
Figure 4 Cycle life of a rechargeable cell.
2000
2500
5 . 3 E LECTRI C CE LLS
Recharging secondary cells The clue to recharging a secondary cell is in the earlier descriptions o cell chemistry. The chemicals produce an excess o electrons at the negative terminal. D uring discharge these electrons then move through the external circuit transerring their energy as they go. When the electron arrives at the positive terminal, all o its energy will have been transerred to other orms and it will need to gain more rom the chemical store in the cell. To reverse the chemical processes we need to return energy to the cell using electrons as the agents, so that the chemical action can be reversed. When charging, the electrons need to travel in the reverse direction to that o the discharge current and you can imagine that the charger has to orce the electrons the wrong way through the cell. Part o a possible circuit to charge a 6- cell leadacid battery is shown in fgure 5 . +14 V direction o charge fow A
020 A
+ 1214 V dc discharged battery 0V
Figure 5 Charging circuit or a cell.
The direction o the charging current is shown in the diagram. It is in the opposite direction to that o the cell when it is supplying energy. An input pd o 1 4 V is needed, notice the polarity o this input. The lightemitting diode ( LED ) and its series resistor indicate that the circuit is switched on. The ammeter shows the progress o charging. When the battery is completely discharged, the charging current will be high, but as the charging level ( and thereore the em) o the battery rises, this current gradually alls. D uring the charging process the terminal voltage will be greater than the em o the cell. At ull charge the em o the battery is 1 3 .8 V. When the current in the meter is zero, the battery is ully charged.
Internal resistance and emf of a cell The materials rom which the cells are made have electrical resistance in j ust the same way as the metals in the external circuit. This internal resistance has an important eect on the total resistance and current in the circuit.
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r
E r
A
B
R (a)
TOK Simple assumptions The model o a cell with a fxed internal resistance and a constant em is an example o modelling. In this case the model is a simple one that cannot be realized in practice. Another model we used earlier was to suggest that there are ideal ammeters with zero resistance and ideal voltmeters with infnite resistance. Scientists requently begin with a very simple model o a system and then explore the possibilities that this model can oer in terms o analysis and behaviour. The next step is to make the model more complicated (but without being too intricate!) and to see how much complexity is needed beore the model resembles the real system that is being modelled. Do the simplifcations and assumptions o ideal behaviour orm a suitable basis or modelling?
224
(b)
Figure 6
Figure 6( a) shows a simple model or a cell. Inside the dotted box is an ideal cell that has no resistance o its own. Also inside the box is a resistance symbol that represents the internal resistance o the cell. The two together make up our model or a real cell. The model assumes that the internal resistance is constant ( or a practical cell it varies with the state o discharge) and that the em is constant ( which also varies with discharge current) . The model cell has an em and an internal resistance r in series with an external resistance R ( fgure 6( b) ) . The current in the circuit is I. We can apply Kirchhos second law to this circuit: the em o the cell supplying energy to the circuit
=
the sum o the pds
= IR + Ir
So = IR + Ir I the pd across the external resistor is V, then = V + Ir or V = Ir It is important to realize that V, which is the pd across the external resistance, is equal to the terminal pd across the ideal cell and its internal resistance ( in other words between A and B ) . The em f is the o p en circuit p d across the term inals of a p o wer so urce in other words, the term inal p d when no current is sup p lied. The pd between A and B is less than the em unless the current in the circuit is zero. The dierence between the em and the term inal p d ( the measured pd across the terminal o the cell) is sometimes reerred to as the lost pd or the lost volts . These lost volts represent the energy required to drive the electron charge carriers through the cell itsel. O nce the energy has been used in the cell in this way, it cannot be available or conversion in the external circuit. You may have noticed that when a secondary cell is being charged, or any cell is discharging at a high current, the cell becomes warm. The energy required to raise the temperature o the cell has been converted in the internal resistance.
5 . 3 E LECTRI C CE LLS
Investigate! Measuring the internal resistance of a fruit cell The method given here works or any type o electric cell, but here we use a citrus ruit cell ( orange, lemon, lime, etc.) or even a potato or the measurement. The ions in the esh o the ruit or the potato react with two dierent metals to produce an em. With an external circuit that only requires a small current, the ruit cell can discharge over surprisingly long times.
with the equation or a straight line y = c + mx A plot o V on the y-axis against I on the x-axis should give a straight line with a gradient o r and an intercept on the V-axis o .
V
Cu
A set o results or a cell ( not a ruit cell in this case) is shown together with the corresponding graph o V against I. The intercept or this graph is 1 .2 5 V and the gradient is 2 .4 V A 1 giving an internal resistance value o 2 .4 . 1.4
Zn
lemon
terminal pd /V
1.2 1 0.8 0.6 0.4 0.2 0 0
0.1
A
Figure 7
0.2
0.3 0.4 current/A
Terminal pd / V
Current / A
1.13
0.05
1.01
0.10
0.89
0.15
0.77
0.20
0.65
0.25
0.53
0.30
0.41
0.35
To make the cell, take a strip o copper oil and a strip o zinc oil, both about 1 cm by 5 cm and insert these, about 5 cm apart deep into the ruit. You may need to use a knie to make an incision unless the oil is sti enough. C onnect up the circuit shown in fgure 7 using a suitable variable resistor. Measure the terminal pd across the ruit cell and the current in it or the largest range o pd you can manage.
0.29
0.40
0.17
0.45
C ompare the equation
0.05
0.50
V = Ir
0.5
0.6
Figure 8
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Worked example 1
A cell o em 6.0 V and internal resistance 2 .5 is connected to a 7.5 resistor. C alculate: a) the current in the cell b) the terminal pd across the cell c) the energy lost in the cell when charge fows or 1 0 s.
2
A battery is connected in series with an ammeter and a variable resistor R. When R = 6.0 , the current in the ammeter is 1 .0 A. When R = 3 .0 , the current is 1 .5 A. C alculate the em and the internal resistance o the battery.
Solution Using
Solution
V = Ir
a) Total resistance in the circuit is 1 0 , so 6 current in circuit = __ = 0. 60 A 10
and knowing that V = IR gives two equations: 6 1 =1 r
b) The terminal pd is the pd across cell = IR = and
0.6 7.5 = 4.5 V c) In 1 0 s, 6 C fows through the cell and the energy lost in the cell is 1 . 5 J C 1 . The energy lost is 9.0 J.
3 1 .5 = 1 .5 r These can be solved simultaneously to give ( 6 4.5 ) = 0.5 r or r = 3 . 0 and = 9.0 V.
Power supplied by a cell
Worked example
The total power supplied by a non-ideal cell is equal to the power delivered to the external circuit plus the power wasted in the cell. Algebraically,
A battery o em 9.0 V and internal resistance 3 .0 is connected to a load resistor o resistance 6.0 . C alculate the power delivered to the external load.
P = I2 R + I2 r using the notation used earlier. The power delivered to the external resistance is 2 _ R ( R + r) 2
Solution Using the equation ______ R ( R + r) 2
2
9 leads to _____ 6 = 6.0 W (6+ 3) 2
power delivered to load resistor
2
0
r 0
load resistance R
Figure 9 Power delivered to a resistor.
Figure 9 shows how the power that is delivered to the external circuit varies with R. The peak o this curve is when r = R, in other words, when the internal resistance o the power supply is equal to the resistance o the external circuit. The load and the supply are matched when the resistances are equal in this way. This matching o supply and circuit is important in a number o areas o electronics.
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5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
5.4 Magnetic efects o electric currents Understanding Magnetic felds Magnetic orce
Applications and skills Sketching and interpreting magnetic feld patterns Determining the direction o the magnetic feld
based on current direction Determining the direction o orce on a charge moving in a magnetic feld Determining the direction o orce on a currentcarrying conductor in a magnetic feld Solving problems involving magnetic orces, felds, current and charges
Equations orce on a charge moving in a magnetic ield:
F = qvB sin orce on a current-carrying conductor in a magnetic ield: F = ILB sin
Nature of science Sometimes visualization aids our understanding. Magnetic ield lines are one o the best examples o this. Scientists began by visualizing the shape and strength o a magnetic ield through the
position and direction o the ictitious lines o orce (or ield lines) . The image proved to be so powerul that the technique was subsequently used with electric and gravitational ields too.
Introduction Electromagnetism is the third eect observed when charge moves in a circuit the electric current gives rise to a magnetic feld. B ut it was not the observation o a magnetic eect arising rom a current in a wire that began the ancient study o magnetism. Early navigators knew that some rocks are magnetic. As with the nature o electric charge, the true origins o magnetic eects remained obscure or many centuries. Only comparatively recently has an understanding o the microscopic aspects o materials allowed a ull understanding o the origins o magnetism. As with electrostatics, scientists began by using the concept o the feld to describe the behaviour o interacting magnets. However, magnetic felds dier undamentally in character rom electrostatic felds and are in some ways more complex.
Magnetic feld patterns The repulsion between the like poles o two bar magnets is amiliar to us. The orces between magnets o even quite modest strength are impressive. Modern magnetic alloys can be used to produce tiny magnets
227
5
E L E C T R I C I T Y AN D M AG N E T I S M ( less than 1 cm in diameter and a ew millimetres thick) that can easily attract another erromagnetic material through signifcant thicknesses o a non-magnetic substance. At the beginning o a study o magnetism, it is usual to describe the orces in terms o felds and feld lines. You may have met this concept beore. There is said to be a magnetic feld at a point i a orce acts on a magnetic pole ( in practice, a pair o poles) at that point. Magnetic felds are visualized through the construction o feld lines. Magnetic feld lines have very similar ( but not identical) properties to those o the electric feld lines in S ub- topic 5 .1 . In summary these are:
Magnetic feld lines are conventionally drawn rom the northseeking pole to the south-seeking pole, they represent the direction in which a north- seeking pole at that point would move.
The strength o the feld is shown by the density o the feld lines, closer lines mean a stronger feld.
The feld lines never cross.
The feld lines act as though made rom an elastic thread, they tend to be as short as possible.
Nature of science Talking about poles S
A word about notation: there is a real possibility o conusion when talking about magnetic poles. This partly arises rom the origin o our observations o magnetism. When we write magnetic north pole what we really mean is the magnetic pole that seeks the geographic north pole ( fgure 1 ( a) ) . We oten talk loosely about a magnetic north pole pointing to the north pole. Misunderstandings can occur here because we also know that like poles repel and unlike poles attract. S o we end up with the situation that a magnetic north pole is attracted to the geographic north pole which seems wrong in the context o two poles repelling. In this book we will talk about north- seeking poles meaning geographic north- seeking and south- seeking poles meaning geographic south seeking . O n the diagrams, N will mean geographic north seeking, S will mean geographic south seeking.
N
(a)
N
S
(b)
S
N
(c)
Figure 1 ( b) shows the patterns or a single bar magnet and ( c) and ( d) two arrangements o a pair o bar magnets o equal strength.
N
N
(d)
228
Figure 1 Magnetic feld patterns.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
In the pairs, notice the characteristic feld pattern when the two opposite poles are close ( fgure 1 ( b) ) and when the two north- seeking poles are close ( fgure 1 ( c) ) . When two northseeking poles are close ( or two south- seeking poles) , there is a position where the feld
is zero between the magnets ( called a null point) . When two opposite poles are close, the feld lines connect the two magnets. In this situation the magnets will be attracted, so this particular feld pattern implies an attraction between poles.
Investigate! Observing feld patterns o permanent magnets and electric currents
There are a number o ways to carry out this experiment. They can involve the scattering o small iron flings, observation o suspensions o magnetized particles in a special liquid, or other techniques. This Investigate! is based on the iron-fling experiment but the details will be similar i you have access to other methods.
Take a bar magnet and place a piece o rigid white card on top o it. You may need to support the card along its sides. C hoose a nonmagnetic material or the support.
Take some iron flings in a shaker ( a pepper pot is ideal) and, rom a height above the card o about 2 0 cm sprinkle flings onto the card. It is helpul to tap the card gently as the flings all onto it.
You should see the feld pattern orming as the magnetic flings all through the air and come under the inuence o the magnetic feld. S ketch or photograph the arrangement.
The iron flings give no indication o feld direction. The way to observe this is to use a plotting compass a small magnetic compass a ew centimetres in diameter that indicates the direction to which a north-seeking pole sets itsel. Place one or more o these compasses on the card and note the direction in which the north- seeking pole points.
Repeat with two magnets in a number o confgurations; try at least the two in fgure 1 .
Electric currents also give rise to a magnetic feld. However, currents small enough to be sae will only give weak felds, not strong enough to aect the flings as they all.
Figure 2 To improve the eect: cut a small hole in the centre o the card and run a long lead through it ( the lead will need to be a ew metres long) . Loop the lead in the same direction through a number o times ( about ten turns i possible) . This trick enables one current to contribute many times to the same feld pattern. You may need at least 2 5 A in total ( 2 . 5 A in the lead) to see an eect.
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Magnetic feld due to a current in conductors The magnetic feld pattern due to a current in a long straight wire is a circular pattern centred on the wire. This seems odd to anyone only used to the bar magnet pattern. Measurements show that the feld is strong close to the wire but becomes weaker urther away rom it. This should be clear in your drawings o the long wire feld pattern when the lines o orce are drawn at increased spacing as you move urther rom the wire. point o arrow indicates current leaving page towards you
tail o arrow indicates current entering page away rom you
lines o magnetic feld
south pole
north pole
Figure 3 Magnetic feld patterns around current-carrying conductors.
fngers curl around the conductor (indicating the direction o magnetic feld)
Your observations using the plotting compasses should have shown that the direction o the feld depends on the direction o the current. current ow
thumb points in direction o conventional current
Figure 4 Right-hand corkscrew rule.
230
Using the conventional current ( i.e. the direction that positive charges are moving in the wire) the relationship between the current and the magnetic feld direction obeys a right-hand corkscrew rule relationship . To remember this, hold your right hand with the fngers curled into the palm and the thumb extended away rom the fngers ( see fgure 4) . The thumb represents the direction o the conventional current and the fngers represent the direction o the feld. Another way to think o the currentfeld relationship is in terms o a
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
screwdriver being used to insert a screw. The screwdriver has to turn a right-handed screw clockwise to insert it and drive the screw orwards. The direction in which the imaginary screw moves is that o the conventional current, and the direction in which the screwdriver turns is that o the feld. Use whichever direction rule you preer, but use it consistently and remember that the rule works or conventional current. The strength o the magnetic feld can be increased by increasing the current in the conductor. The magnetic feld due to the solenoid is amiliar to you already. To understand how it arises, you need to imagine the long straight wire being coiled up into the solenoid shape taking its circular feld with it as the coiling takes place. With current in the wire, the circular feld is set up in each wire. This circular feld adds together with felds rom neighbouring turns in the solenoid. Figure 5 shows this; look closely at what happens close to the individual wires. The black lines show the feld near the wires, the blue lines show how the felds begin to combine, the red line shows the combined feld in the centre o the solenoid.
felds add together circular feld near turn consecutive turns on solenoid
Figure 5 Building up the feld pattern.
A feld runs along the hollow centre o the solenoid and then outside around the solenoid ( fgure 3 ) . O utside it is identical to the bar magnet pattern to the extent that we can assign north- and south- seeking poles to the solenoid. Again there is an easy way to remember this using an N and S to show the north- seeking and south-seeking poles. The arrows on the N and S show the current direction when looking into the solenoid rom outside. I you look into the solenoid and the conventional current is anticlockwise at the end o the solenoid closer to you then that is the end that is north- seeking. I the current is clockwise then that is the south- seeking end. The strength o the magnetic feld in a solenoid can be increased by:
increasing the current in the wire
increasing the number o turns per unit length o the solenoid
adding an iron core inside the solenoid.
anticlockwise current gives a north seeking pole
clockwise current gives a south seeking pole
Figure 6 Right hand corkscrew rule and pole direction.
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Worked example Four long straight wires are placed perpendicular to the plane o the paper at the edges o a square. The same current is in each wire in the direction shown in the diagram. D educe the direction o the magnetic feld at point P in the centre o the square.
Solution The our feld directions are shown in the diagram. The sum o these our vectors is another vector directed rom point P to the let.
A
B D B P
P
A C D
C
Nature of science But why permanent magnets? We have said nothing about the puzzle o permanent magnets or the magnetism o the Earth ( Figure 1 ( a) ) . The suggestion is that magnetism arises between charges when the charges move relative to each other. C an this idea be extended to help explain the reasons or permanent magnetism ( known as erromagnetism) ? In act, permanent magnetism is comparatively rare in the periodic table, only iron, nickel, and cobalt and alloys o these metals show it. The reason is due to the arrangement o the electrons in the atoms o these substances. Electrons are now known to have the property o spin which can be imagined as an orbiting motion around the atom. In iron, cobalt and nickel, there is a particular arrangement that involves an unpaired electron. This is the atomic origin o the moving charge that is needed or a magnetic feld to appear.
232
The second reason why iron, nickel, and cobalt are strong permanent magnets is that neighbouring atoms can co-operate and line up the spins o their unpaired electrons in the same direction. So, many electrons are all spinning in the same direction and giving rise to a strong magnetic feld. Deep in the centre o the Earth it is thought that a liquid-like metallic core containing ree electrons is rotating relative to the rest o the planet. Again, these are conditions that can lead to a magnetic feld. In which direction do you predict that the electrons are moving? However, this phenomenon is not well understood and is still the subject o research interest. Why, or example, does the magnetic feld o the Earth ip every ew thousand years? There is much evidence or this including the magnetic striping in the undersea rocks o the mid-Atlantic ridge and in the anomalous magnetism ound in some ancient cooking hearths o the aboriginal peoples o Australia.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
Forces on moving charges Force between two current-carrying wires We have used feld ideas to begin our study o magnetic eects, but these conceal rom us the underlying physics o magnetism. To begin this study we look at the interactions that arise between conductors when they are carrying electric current. +
foil strips
+
-
(a)
currents in same directions (b)
current into page
current out of page (c)
Figure 7
Figure 7( a) shows two oil strips hanging vertically. The current directions in the oils can be the same or opposite directions. When the currents are in the same direction, the strips move together due to the orce on one oil strip as it sits in the magnetic feld o the other strip. When the currents are in opposing directions, the strips are seen to move apart. You can set this experiment up or yoursel by using two pieces o aluminium oil about 3 cm wide and about 70 cm long or each conductor. The power supply should be capable o providing up to 2 5 A so take care! C onnections are made to the oils using crocodile (alligator) clips.
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5
E L E C T R I C I T Y AN D M AG N E T I S M The orces on the oil can be explained in terms o the interactions between the felds as shown in the fgure 7( b) . When the currents are in the same direction, the feld lines rom the oils combine to give a pattern in which feld lines loop around both oils. The notation used to show the direction o conventional current in the oil is explained on the diagram. Look back at fgure 1 ( d) which shows the feld pattern or two bar magnets with the opposite poles close. You know that the bar magnets are attracted to each other in this situation. The feld pattern or the oils is similar and also leads to attraction. Think o the feld lines as trying to be as short as possible. They become shorter i the oils are able to move closer together. When the currents are in opposite directions, the feld pattern changes ( fgure 7( c) ) . Now the feld lines between the oils are close together and in the same direction thus representing a strong feld. It seems reasonable that a strong magnetic feld ( like a strong electrostatic feld) represents a large amount o stored energy. This energy can be reduced i the oils move apart allowing the feld lines to separate too. Again, this has similarities to the bar magnet case but this time with like poles close together.
Force between a bar magnet feld and a current-carrying wire O ne extremely important case is the interaction between a uniorm magnetic feld and that produced by a current in a wire. Again we can start with the feld between two bar magnets with unlike poles close. In the centre o the region between the magnets, the feld is uniorm because the feld lines are parallel and equally spaced.
N
S
+
=
N
S orce on wire
(a)
(b)
Figure 8 The catapult feld.
S uppose a wire carrying a current sits in this feld. Figure 8( a) shows the arrangement and the directions o the uniorm magnetic feld and the feld due to the current. A orce acts downwards on the wire. The eect can again be explained in terms o the interaction between the two magnetic felds. The circular feld due to the wire adds to the uniorm feld due to the magnets to produce a more complicated feld. This is shown in fgure 8( b) . O verall, the feld is weaker below the wire than above it. Using our ideas o the feld lines, it is clear that the system ( the wire and the uniorm feld) can overcome this dierence in feld strength, either by attempting to move the wire downwards or by moving the magnets that cause the uniorm feld upwards. This eect
234
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
is sometimes called the catap ult feld because the feld lines above the wire resemble the stretched elastic cord o a catapult j ust beore the obj ect in the catapult is fred. This eect is o great importance to us. It is the basis or the conversion o electrical energy into kinetic energy. It is used in electric motors, loudspeakers, and other devices where we need to produce movement rom an electrical power source. It is called the motor eect. It is possible to predict the direction o motion o the wire by drawing the feld lines on each occasion when required but this is tedious. There are a number o direction rules that are used to remember the direction o the orce easily. O ne o the best known o these is due to the English physicist Fleming and is known as Flemings let-hand rule. orce (motion) along thuMb
feld along F irst fnger
F B I let hand
current along seCond fnger
Figure 9 Flemings let-hand rule.
To use the rule, extend your let hand as shown in fgure 9. Your frst ( index) fnger points in the direction o the uniorm magnetic feld and your second fnger points in the direction o the conventional current in the wire, then your thumb gives the direction o the orce acting on the wire.
The motor efect The explanation or the motor eect has been given so ar in terms o feld lines. This is, o course, not the complete story. We should be looking or explanations that involve interactions between individual charges, both those that produce the uniorm magnetic feld and those that arise rom the current in the wire. E lectrostatic eects ( as their name implies) arise between charges that are not moving. Magnetic eects arise because the sets o charges that produce the felds are moving relative to each other and are said to be in dierent rames o reerence. There have been no electrostatic eects because we have dealt with conductors in which there is an exact balance o positive and negative charges. Magnetism can be thought o as the residual eect that arises when charges are moving with respect to each other.
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Worked example Wires P and R are equidistant rom wire Q.
D escribe the direction o the orce acting on wire Q due to wires P and R.
Solution I
I
I
wire P
wire Q
wire R
Using the right-hand corkscrew rule, the feld due to wire P at wire Q is out o the plane o the paper, and the feld due to wire R at wire Q is also out o the plane o the paper.
Each wire carries a current o the same magnitude and the currents are in the directions shown.
Using Flemings let- hand rule, the orce on wire Q is in the plane o the paper and to the let.
The magnitude of the magnetic force
Investigate! Force on a current-carrying conductor current carrying lead
Arrange the apparatus as shown in the diagram.
Zero the balance so that the weight o the magnets is removed rom the balance reading.
The experiment is in two parts:
l
First, vary the current in the wire and collect data or the orce acting on the magnets, and thereore the balance, as a result. A trick to improve precision is to reverse the current in the wire and take balance readings or both directions. Then add the two together ( ignoring the negative sign o one reading) to give double the answer. D raw a graph to display your data.
Second, use two pairs o magnets side-by-side to double the length o the feld. Take care that the poles match, otherwise the orces will cancel out. This is not likely to work so well as you need to assume that the magnet pairs have the same strength. This will probably not be true. But, roughly speaking, does doubling the length o wire in the feld double the orce?
magnets on steel yoke
C
236
0
balance
Figure 10
This is an experiment that will give you an idea o the size o the magnetic orce that acts on typical laboratory currents. I carried out careully it will also allow you to see how the orce varies with the length o the conductor and the size o the current.
You will need some pairs o at magnets ( known as magnadur magnets, a sensitive top- pan balance, a power supply and a suitable long straight lead to carry the current.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
The result o experiments like the one above is that the orce acting on the wire due to the feld is proportional to:
the length ( l) o the wire
the current ( I) in the wire.
This leads us to a defnition o magnetic feld strength rather dierent rom that o electric feld strength and gravitational feld strength. We cannot defne the magnetic feld strength in terms o orce __ a single quantity because the orce depends on two quantities: current and length. Instead we defne magnetic feld strength orce acting on a current element = _____ current in the element length o the element B y element we mean a short section o a wire that carries a current. I a orce F acts on the element o length l when the current in it is I, then the magnetic feld strength B is defned by F B= _ Il
B
l
B
l
F = BIl sin
F = BIl (a)
(b)
Figure 11
The unit o magnetic feld strength is the tesla, abbreviated T and the tesla is equivalent to the undamental units kg s 2 A 1 . When a 1 metre long current element is in a magnetic feld and has a current o 1 A in it, i a magnetic orce o 1 N acts on it, then the magnetic feld strength is defned to be 1 T. The tesla can also be thought o as a shortened orm o N A- 1 m -1 . The tesla turns out to be a very large unit indeed. The largest magnetic feld strengths created in a laboratory are a ew kT and the magnetic feld o the Earth is roughly 1 0 4 T. The very largest felds are associated with some neutron stars. The feld strength can be order o 1 00 GT in such stars. Having defned B we can go on to rearrange the equation to give the orce that acts on a wire: F = BIl This applies when the feld lines, the current and the wire are all at 90 to each other as they are when you use the Fleming let-hand rule. I this is not the case (see fgure 1 1 (b) ) and the wire is at an angle to the lines then we need to take the appropriate component o I or l that is at 90 to the feld.
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5
E L E C T R I C I T Y AN D M AG N E T I S M In terms o the way the angle is defned in fgure 1 1 ( b) , the equation becomes F = BIl sin This equation is written in terms o the current in the wire. O course, the current is, as usual, the result o moving charge carriers. The equation can be changed to reect this. From S ub-topic 5 .1 we know that the current I is given by Q I= _ t where Q is the charge that ows through the current element taking a time t to do it. S ubstituting
( ) l = BQ ( _ ) sin t
Q F = B _ l sin t
The term ( _lt ) is the drift sp eed v o the charge carriers and making this substitution gives the expression F = BQv sin or the orce acting on a charge Q moving at speed v at an angle to a magnetic feld o strength B.
TOK Why direction rules? During this sub-topic we have introduced two direction rules: the corkscrew rule or magnetic felds around wires and Flemings rule or the motor eect. What is the status o these rules? Are they implicit in the way the universe operates, or are they simply indications that depend on the way we defne current direction and other undamental quantities in physics? To begin to answer this, consider what would have happened i Franklin had known that charge carriers in metals were electrons with a negative charge. What would this have changed (i anything)?
238
Notice the way that the angle is defned in the diagrams. It is the angle between the direction in which the charge is moving ( or the current direction the same thing) and the feld lines. D ont get this wrong and use cosine instead o sine in your calculations.
Nature of science The defnition o the ampere In Topic 1 you learnt that the ampere was defned in terms o the orce between two current- carrying wires. You may have thought this odd both then and when the relationship between current and charge was developed earlier in this topic. Perhaps it is clearer now. A precise measurement o charge is quite difcult. At one time it could only be achieved through chemical measurements o electrolysis and the levels o precision were not great enough to give a good value at that time. It is much easier to set up an experiment that measures the orce between two wires. The experiment can be done to a high precision using a device called a current balance ( as in the orce Investigate! above) in which the magnetic orce can be measured in terms o the gravitational orce needed to cancel it out. This is rather like oldashioned kitchen scales where the quantity being measured is j udged against a standard mass acted on by gravity.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
Worked example 1
When a charged particle o mass m, charge q moves at speed v in a uniorm magnetic feld then a magnetic orce F acts on it. D educe the orce acting on a particle mass m o charge 2 q and speed 2 v travelling in the same direction in the same magnetic feld.
Solution The equation or the orce is F = BQv sin In the equation, sin and B do not change but every other quantity does, so the orce is 4F. 2
Electric currents o magnitude I and 3 I are in wires 1 and 2 as shown. wire 1
wire 2
A orce F acts on wire 2 due to the current in wire 1 . D educe the magnitude o the orce on wire 1 due to the current in wire 2 .
Solution This problem can be solved by reerence to Newtons laws o motion, but an alternative is to consider the changes in magnetic feld. I the magnetic feld at wire 2 due to wire 1 is B then the magnetic feld at wire 1 due to wire 2 is 3 B. However, the current in wire 1 is one-third o that in wire 2 . S o orce at wire 1 due to wire 2 is I F = 3 B __ l = orce in wire 2 due to wire 1 . 3
The orces are the same.
I
3I
Nature of science Vectors and their products You may have ormed the view that vectors are j ust used or scale diagrams in physics. This is not the case, vectors come into their own in mathematical descriptions o magnetic orce. There is only one way to multiply scalars together: run a relay with our stages each o distance 1 00 m and the total distance travelled by the athletes is 400 m. Multiplying two scalars only gives another scalar. Vectors, because o the added direction, can be multiplied together in two ways:
to orm a scalar product ( sometimes called dot product) where, or example, orce and displacement are multiplied together to give work done ( a scalar) that has no direction. In vector notation this is written as F s = W. The multiplication sign is the dot ( hence the name) the vectors are written with a bold ont, the scalar in ordinary ont. to orm a vector product ( sometimes called cross product) where two vectors are
multiplied together to give a third vector, thus a b = c The multiplication o qv and B to orm the vector orce F is a vector product. The charge q is a scalar but everything else in the equation is a vector. A mathematician would write F = q ( v B) to show that the vector velocity and the vector magnetic feld strength are multiplied together. The order o v and B is important. There is a vector rule or the direction o F that is consistent with our observations earlier and the sin appears when the vector multiplication is worked out in terms o the separate components o the vector. Vector notation turns out to be an essential language o physics, because it allows a concise notation and because it contains all the direction inormation within the equations rather than orcing us to use direction rules. However we do not pursue the ull theory o vectors in IB Physics.
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Questions 1
( IB) Four point charges o equal magnitude, are held at the corners o a square as shown below. 2a
The electric orce between them is + F (i.e. attractive) . The spheres are touched together and are then returned to their original separation. a) Calculate the charge on X and the charge on Y.
+Q
+Q
b) C alculate the value o the electric orce between them ater being returned to their original separation. ( 7 marks) 2a
P
4
(IB) Two charged plastic spheres are separated by a distance d in a vertical insulating tube, as shown.
Q Q
tube spheres
The length o each side o the square is 2 a and the sign o the charges is as shown. The point P is at the centre o the square. a) ( i)
d
D etermine the magnitude o the electric feld strength at point P due to one o the point charges.
( ii) O n a copy o the diagram above, draw an arrow to represent the direction o the resultant electric feld at point P. The charge on each sphere is doubled. Friction with the walls o the tube is negligible.
( iii) D etermine, in terms o Q, a and k, the magnitude o the electric feld strength at point P. ( 7 marks)
D educe the new separation o the spheres. ( 5 marks)
2
(IB) Two point charges o magnitude +2Q and Q are fxed at the positions shown below. D iscuss the direction o the electric feld due to the two charges between A and B . S uggest at which point the electric feld is most likely to be zero. ( 3 marks) Q
A
3
+2Q +
+
(IB) a)
Electric felds may be represented by lines o orce. The diagram below shows some lines o orce.
B
(IB) Two identical spherical conductors X and Y are mounted on insulated stands. X carries a charge o +6.0 nC and Y carries a charge o 2.0 nC.
+6.0 nC
conductor X
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5
A
B
2.0 nC
insulated stands
conductor Y
( i)
State whether the feld strength at A and at B is constant, increasing or decreasing, when measured in the direction rom A towards B .
QUESTION S d) An alternative circuit or measuring the VI characteristic uses a potential divider.
( ii) E xplain why feld lines can never touch or cross. b) The diagram below shows two insulated metal spheres. Each sphere has the same positive charge.
( i)
D raw a circuit that uses a potential divider to enable the VI characteristics o the flament to be ound.
( ii) Explain why this circuit enables the potential dierence across the lamp to be reduced to 0 V. ( 1 3 marks)
+
+
7
(IB) The graph below shows the VI characteristic or two 1 2 V flament lamps A and B . lamp B
12
6
(IB) A lamp is at normal brightness when there is a potential dierence o 1 2 V across its flament and a current in the flament o 0.5 0 A.
potential diference/V
C opy the diagram and in the shaded area between the spheres, draw the electric feld pattern due to the two spheres. ( 8 marks)
lamp A
0 0
0.5 current/A
a) For the lamp at normal brightness, calculate: ( i) the power dissipated in the flament ( ii) the resistance o the flament. b) In order to measure the voltagecurrent ( VI) characteristics o the lamp, a student sets up the ollowing electrical circuit. 12 V battery
a) ( i)
1.0
E xplain why the graphs indicate that these lamps do not obey O hms law.
( ii) S tate and explain which lamp has the greater power dissipation or a potential dierence o 1 2 V. The two lamps are now connected in series with a 1 2 V battery as shown below. 12 V battery
S tate the correct positions o an ideal ammeter and an ideal voltmeter or the characteristics o the lamp to be measured. c) The voltmeter and the ammeter are connected correctly in the previous circuit. Explain why the potential dierence across the lamp ( i)
cannot be increased to 1 2 V
( ii) cannot be reduced to zero.
lamp A
lamp B
b) ( i) State how the current in lamp A compares with that in lamp B . ( ii) Use the VI characteristics o the lamps to deduce the total current rom the battery. ( iii) C ompare the power dissipated by the two lamps. ( 1 1 marks)
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( IB) a)
X
E xplain how the resistance o the flament in a flament lamp can be determined rom the VI characteristic o the lamp.
2.0
b) A flament lamp operates at maximum brightness when connected to a 6.0 V supply. At maximum brightness, the current in the flament is 1 2 0 mA. ( i)
E
( i) C opy the graph in ( a) , and draw the IV characteristics or the resistor R.
C alculate the resistance o the flament when it is operating at maximum brightness.
( ii) You have available a 2 4 V supply and a collection o resistors o a suitable power rating and with dierent values o resistance. C alculate the resistance o the resistor that is required to be connected in series with the supply such that the voltage across the flament lamp will be 6.0 V. ( 4 marks) 9
R
(IB) The graph below shows the IV characteristics or component X. I/A6
( ii) D etermine the total potential dierence E that must be applied across component X and across resistor R such that the current through X and R is 3 .0 A. ( 7 marks)
1 0 (IB) A student is to measure the currentvoltage ( IV) characteristics o a flament lamp. The ollowing equipment and inormation are available.
Battery Filament lamp
4
Voltmeter
2
Ammeter 0 -8
-6
-4
-2
0
2
4
6
-2 -4
8 V/V
Potentiometer
Information emf = 3.0 V, negligible internal resistance marked 3 V, 0.2 A resistance = 30 k, reads values between 0.0 and 3.0 V resistance = 0.1 , reads values between 0.0 and 0.5 A resistance = 100
a) For the flament lamp operating at normal brightness, calculate: ( i) its resistance
-6
( ii) its power dissipation. The student sets up the ollowing incorrect circuit.
The component X is now connected across the terminals o a battery o em 6. 0 V and negligible internal resistance. a) Use the graph to determine:
V
( i) the current in component X ( ii) the resistance o component X. b) A resistor R o constant resistance 2 .0 is now connected in series with component X as shown below.
A
b) ( i) E xplain why the lamp will not light. ( ii) S tate the approximate reading on the voltmeter. E xplain your answer. ( 6 marks)
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QUESTION S 1 1 ( IB) A particular flament lamp is rated at 1 2 V, 6.0 mA. It j ust lights when the potential dierence across the flament is 6.0 V.
a) ( i) S tate the value o the current or which the resistance o X is the same as the resistance o Y and determine the value o this resistance.
A student sets up an electric circuit to measure the IV characteristics o the flament lamp.
( ii) D escribe and suggest an explanation or the IV characteristic o conductor Y.
A
b) The two conductors X and Y are connected in series with a cell o negligible internal resistance. The current in the conductors is 0. 2 0 A.
100 k V
12 V
Use the graph to determine:
S
( i) the resistance o Y or this value o current ( ii) the em o the cell.
In the circuit, shown below, the student has connected the voltmeter and the ammeter into the circuit incorrectly.
( 8 marks)
1 3 ( IB) A cell o electromotive orce ( em) E and internal resistance r is connected in series with a resistor R, as shown below.
The battery has em 1 2 V and negligible internal resistance. The ammeter has negligible resistance and the resistance o the voltmeter is 1 00 k. The maximum resistance o the variable resistor is 1 5 . a) Explain, without doing any calculations, whether there is a position o the slider S at which the lamp will be lit. b) Estimate the maximum reading o the ammeter. ( 5 marks)
r
E
R
The cell supplies 8.1 1 0 3 J o energy when 5 .8 1 0 3 C o charge moves completely round the circuit. The current in the circuit is constant. a) C alculate the em E o the cell.
1 2 ( IB) The graph below shows the currentvoltage ( IV) characteristics o two dierent conductors X and Y. 0.50 0.45 0.40 0.35 0.30 Y X I/A 0.25 0.20 0.15 14 0.10 0.05 0.00 0.0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 V/V
b) The resistor R has resistance 6.0 . The potential dierence between its terminals is 1 .2 V. D etermine the internal resistance r o the cell. c) C alculate the total energy transer in R. d) D escribe, in terms o a simple model o electrical conduction, the mechanism by which the energy transer in R takes place. ( 1 2 marks)
( IB) A battery is connected in series with a resistor R. The battery transers 2 000 C o charge completely round the circuit. D uring this process, 2 5 00 J o energy is dissipated R and 1 5 00 J is expended in the battery. C alculate the em o the battery. ( 3 marks)
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5
E L E C T R I C I T Y AN D M AG N E T I S M 1 5 ( IB) A student connects a cell in series with a variable resistor and measures the terminal pd V o the cell or a series o currents I in the circuit. The data are shown in the table.
V/V 1.50 1.10 0.85 0.75 0.60 0.50
I/mA 120 280 380 420 480 520
The electric feld strength is 3 .8 1 0 5 V m 1 and the magnetic feld strength is 2 .5 1 0 2 T. C alculate the speed o the electron i the net orce acting on it due to the felds is zero. ( 3 marks)
1 8 ( IB) A straight wire lies in a uniorm magnetic feld as shown. current I magnetic feld
Use the data to determine the em and internal resistance o the cell. ( 5 marks)
1 6 ( IB) A battery is connected to a resistor as shown. The current in the wire is I and the wire is at an angle o to the magnetic feld. The orce per unit length on the conductor is F. D etermine the magnetic feld strength. ( 2 marks) 6.0 V
10
V
1 9 ( IB) A straight wire o length 0. 75 m carries a current o 3 5 A. The wire is at right angles to a magnetic feld o strength 0.05 8 T. C alculate the orce on the wire. ( 2 marks) When the switch is open the voltmeter reads 1 2 V, when the switch is closed it reads 1 1 . 6 V. a) E xplain why the readings dier. b)
( i) S tate the em o the battery. ( ii) C alculate the internal resistance o the battery.
c) C alculate the power dissipated in the battery. ( 6 marks)
1 7 ( IB) An electron enters a pair o electric and magnetic felds in a vacuum as shown in the diagram. region o magnetic feld
electron
244
E
+ B
2 0 ( IB) An ion with a charge o + 3 . 2 1 0 1 9 C and a mass o 2 . 7 1 0 26 kg is moving due south at a speed o 4.8 1 0 3 m s 1 . It enters a uniorm magnetic feld o strength 4. 6 1 0 4 T directed downwards towards the ground. D etermine the orce acting on the ion. ( 4 marks)
6 CI RCU LAR M O TI O N AN D G RAVI TATI O N Introduction Two apparently distinct areas of physics are linked in this topic: motion in a circle and the basic ideas of gravitation. B ut of course they are not distinct at all. The motion of a satellite about its planet involves both a consideration of the
gravitational force and the mechanics of motion in a circle. Man cannot travel beyond the E arth without a knowledge of both these aspects of Physics.
6.1 Circular motion Understanding Period, frequency, angular displacement, and
angular velocity Centripetal force Centripetal acceleration
Nature of science The drive to develop ideas about circular motion came from observations of the universe. How was it that astronomical objects could move in circular or elliptical orbits? What kept them in place in their motion? Scientists were able to deduce that there must be a force acting radially inwards for every case of circular motion that is observed. Whether it is a bicycle going around a corner or a planet orbiting its star, the physics is the same.
Applications and skills Identifying the forces providing the centripetal
forces such as tension, friction, gravitational, O B J TE XT_UND electrical, or magnetic Solving problems involving centripetal force, centripetal acceleration, period, frequency, angular displacement, linear speed, and angular velocity Qualitatively and quantitatively describing examples of circular motion including cases of vertical and horizontal circular motion
Equations speedangular speed relationship: v = r 4 2 r v2 centripetal acceleration: a = ____ = ________ r
mv 2 2 centripetal force: F = _______ r = m r
T2
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Moving in a circle Most children take great delight in an obj ect on a string whirling in a circle though they may be less happy with the consequences when the string breaks and the obj ect hits a window! Rides at a theme park and trains on a railway are yet more examples o movement in a circle. What is needed to keep something rotating at constant speed? The choice o term (as usual in physics! ) is very deliberate. In circular motion we say that the speed is constant but not the velocity is constant.
Figure 1
A airground carousel.
Velocity, as a vector quantity, has both magnitude and direction. The obj ect on the string has a constant speed but the direction in which the obj ect is moving is changing all the time. The velocity has a constant magnitude but a changing direction. I either o the two parts that make up a vector change, then the vector is no longer constant. Whenever velocity changes (even i it is only the direction) then the object is accelerated. Understanding the physics o this acceleration is the key to understanding circular motion. B ut beore looking at how the acceleration arises we need a language to describe the motion.
Angular displacement The angle moved around the circle by an obj ect rom where its circular motion starts is known as the angular disp lacement. Unlike the linear displacement used in Topic 2 , angular displacement will not be considered to be a vector in IB Physics. Angular displacement is the angle through which the obj ect moves and it can be measured in degrees ( ) or in radians ( rad) . Radians are more commonly used than degrees in this branch o physics. I you have not met radians beore, read about the dierences between radians and degrees.
Nature of science
s
Radians or degrees C alculations o circular motion involve the use o angles. In any science you studied beore starting this course you will almost certainly have measured all angles in degrees.
r s = r
1 1 ( degree) is defned to be ___ th o the way 360 around a circle.
In some other areas o physics (including circular motion) there is an alternative measure o angle that is much more convenient, the radian. Radians are based on the geometry o the arc o a circle. 1 radian ( abbreviated as rad) is defned as the angle equal to the circumerence o an arc o a circle divided by the radius o the circle. In symbols s =_ r
246
Figure 2
Defnition o radian.
Going around the circle once means travelling around the circumerence; this is a distance o 2 r. The angle in radians subtended by the 2 r whole circle is ___ r = 2 rad. S o 3 60 = 2 = 6.2 8 rad and 1 rad = 5 7.3
6 .1 CI R CU L AR M O TI O N
S ometimes, the radian numbers are let as ractions, so 1 90 = _ _ round the circle , 2 4
( ) 1 3 0 = _ ( _ round the circle ) 6 12 and so on.
To convert other values or yoursel, use the angle in de gre e an gle in radians equation ___________ = ___________ 2 360
There are some similarities between the sine o an angle and the angle in radians. The two quantities are compared in this Nature o S cience box which shows sin and in radians. Notice that, as becomes smaller, sin ( ) and become closer together. From angles o 1 0 down to 0, the dierences between sin and are very small and in some calculations and proos we treat sin and as being equal ( this is known as the small angle approximation) . For small angles cos approximates to zero radians.
To illustrate this, here are the values o sin and in radians or our angles: 90, 45 , 1 0, and 5 . Notice how similar the sine values and the radians are or 1 0 and 5 . _ sin ( 90) = 1 .000; rad = 1 .5 71 rad 2 _ sin ( 45 ) = 0.707; rad = 0.785 rad 4 _ sin ( 1 0) = 0.1 75 ; rad = 0.1 74 rad 18 _ sin ( 5 ) = 0.087; rad = 0.087 rad 36 Finally, a practical point: S cientifc and graphic calculators work happily in either degrees or radians ( and sometimes in another type o angular measure known as grad too) . B ut the calculator has to be told what to expect! Always check that your calculator is set to work in radians i that is what you want, or in degrees i those are the units you are using. You will lose calculation marks in an examination i you conuse the calculator!
, angular speed = t t = t 2 1
Angular speed In Topic 2 we used the term speed to mean linear speed. When the motion is in a circle there is an alternative: angular sp eed, this is given the symbol ( the lower-case Greek letter, omega) .
time t2
time t 1
angular displacement average angular speed = ____ time for the angular displacement to take place
Figure 3 shows how things are defned and you will see that in symbols the defnition becomes = _ t where is the angular displacement and t is the time taken or the angular displacement.
Figure 3
Angular speed.
Nature of science Angular speed or angular velocity? You may be wondering about the distinction between angular speed and angular velocity, and whether angular velocity is a vector similar to linear velocity. The answer is that angular velocity is a vector but an unusual one. It has a magnitude equal to the angular speed, but its direction is surprising! The direction is along the axis o rotation, in other
direction of direction rotation of angular velocity vector Figure 4 Angular velocity
direction.
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words, through the centre of the circle around which the obj ect is moving and perpendicular to the plane of the rotation. The direction follows a clockwise corkscrew rule so that in this example the direction of the
angular velocity vector is into the plane of the paper. In the IB course, only the angular speed the scalar quantity is used.
Period and frequency The time taken for the obj ect to go round the circle once is known as the p eriodic time or simply the p eriod of the motion, it has the symbol T. In one period, the angular distance travelled is 2 rad. S o, 2 T= _ When T is in seconds the units of are radians per second, abbreviated to rad s 1 . If you have already studied waves in this course, you might have met the idea of time period the time for one cycle. Another quantity that is associated with T is frequency. Frequency is the number of times an object goes round a circle in unit time (usually taken to be 1 second) , so one way to express the unit of frequency would be in per second or s 1 . However, the unit of frequency is re-named after the 1 9th century physicist Heinrich Hertz and is abbreviated to Hz. There is a link between T and f so that: 1 T= _ f This leads to a link between and f = 2 f
Worked example
Solution
A large clock on a building has a minute hand that is 4.2 m long.
a) The minute hand goes round once ( 2 rad) every hour.
C alculate: a) the angular speed of the minute hand b) the angular displacement, in radians, in the time periods
O ne hour is 3 600 s angular displacement angular speed = __ time taken 2 _ = = 0.001 75 rad s 1 3 600 2 1 b) ( i) 2 0 minutes is __ of 2 , so ___ rad 3 3
( i) 1 2 noon to 1 2 .2 0
( ii) 2 .5 h is 2 2 .5 = 5 rad
( ii) 1 2 noon to 1 4.3 0. c) the linear speed of the tip of the minute hand.
c) v = r = 4.2 0.001 75 = 0.007 3 3 m s 1 = 7. 3 mm s 1
Linking angular and linear speeds S ometimes we know the linear speed and need the angular speed or vice versa. The link is straightforward: When the circle has a radius r the circumference is 2 r, and T, is the time taken to go around once. S o the linear speed of the obj ect along the edge of the circle v is 2 r v= _ T
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6 .1 CI R CU L AR M O TI O N
Rearranging the equation gives 2 r T= _ v We have j ust seen that 2 T= _ so equating the two equations or T gives 2 r _ 2 _ v = C ancelling the 2 and rearranging gives v = r Notice that, in both this equation and in the earlier equation s = r, the radius r multiplies the angular term to obtain the linear term. This is a consequence o the defnition o the angular measure.
Centripetal acceleration E arlier we showed that an obj ect moving at a constant angular speed in a circle is being accelerated. Newtons frst law tells us that, or any obj ect in which the direction o motion or the speed is changing, there must be an external orce acting. In circular motion the direction is constantly changing and so the obj ect accelerates and there must be a orce acting on it to cause this to happen. In which direction do the orce and the acceleration act, and what are their sizes? The diagram shows two points P 1 and P 2 on the circle together with the velocity vectors vold and vnew at these points. The vectors are the same length as each other because the speed is constant. However, vold and vnew point in dierent directions because the obj ect has moved round the circle by an angular distance between P 1 and P 2 . Acceleration is, as usual, change o velocity ___ time taken or the change The change in velocity is the change-o- velocity vector v that has to be added to vold in order to make it become the same length and direction as vnew. Identiy these vectors on the diagram.
v vnew
vold
P2 vnew
v
2
vold v P1
vold
2
vnew Figure 5 Proof of centripetal
acceleration direction.
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N Notice that vold and vnew slide round the circle to meet. Where does the new vector v point? The answer is: to the centre o the circle. This is an averaging process to fnd out what the dierence is between vold and vnew hal- way between the two points. This averaging can be taken urther. The time, t, to go between P 1 and P 2 , and the linear distance around the circle between P 1 and P 2 ( which is r) are related by r t = _ v Using some trigonometry on the diagram shows that v _ = v sin _ 2 2 The size o the average acceleration a that is directed towards the centre o the circle is
( )
2 v sin ( ___ 2 ) v a = _= _ 2 r __ ___ t v
2
This can be written as sin ( ___ 2 ) v _ a= _ r ___ 2
2 sin ( 2 ) is almost exactly equal to 1 and When is very small, the ratio ______ ___ ___
2
so the instantaneous acceleration a when P 1 and P 2 are very close together is
Nature of science Linking it together Notice that some o these equations have interesting links elsewhere: mv is, or example, the magnitude o the linear momentum multiplied by . Try to be alert or these links as they will help you to piece your physics together.
v2 2 a=_ r = r = v directed to the centre o the circle. This acceleration is at 90 to the velocity vector and it points inwards to the centre o the circle.
The orce that acts to keep the obj ect moving in a circle is called the centrip etal force and this orce leads to a centrip etal acceleration. ( The origin o the word centripetal comes rom two Latin words centrum and petere literally to lead to the centre.)
Centripetal force Newtons second law o motion in its simpler orm tells us that F = ma using the usual symbols. The second law applies to the orce that provides the centripetal v 2 acceleration, so the magnitude o the orce = m __ r = m r = mv. The question we need to ask or each situation is: what orce provides the centripetal orce or that situation? The direction o this orce must be along the radial line between the obj ect and the centre o the circle. 2
Investigate! Investigating how F varies with m, v and r This experiment tests the relationship 2
v m_ r = Mg
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To do this a bung is whirled in a horizontal circle with a weight hanging rom one end o a string and mass (rubber bung) on the other end.
6 .1 CI R CU L AR M O TI O N
A paper clip is attached to the string below a glass tube. The clip is used to ensure that the radius o rotation o the bung is constant the bung should be rotated at a speed so that the paper clip j ust stays below the glass tube.
To veriy the equation you need to test each variable against the others. There are a number o possible experiments in each o which one variable is held constant (a control variable) , one is varied (the independent variable) , and the third (the dependent variable) is measured. One example is:
centripetal force apparatus r mass, m
Variation of v with r
In this experiment, m and M must be unchanged. Move the clip to change r, and or each value o r, measure v using the method given above.
Analysis: v2 _ r = constant A graph o v2 against r ought to be a straight line passing through the origin. Alternatively you could, or each experimental run, simply divide v2 by r and look critically at the answer (which should be the same each time) to see i the value is really constant. I going down this route, you ought to assess the errors in the experiment and v put error limits on your __ r value.
glass tube paper clip
string
weight (Mg) Figure 6 Centripetal
force, mass, and speed.
The tension in the string is the same everywhere ( whether below the glass tube or above in the horizontal part) . This tension is F in the equation and is equal to Mg where M is the mass o the weight ( hanging vertically) . Use a speed at which you can count the number o rotations o m in a particular time and rom this work out the linear speed v o mass m.
2
What are the other possible experimental tests?
In practice the string cannot rotate in the horizontal plane because o its own weight. How can you improve the experiment or the analysis to allow or this?
Nature of science Centripetal or centrifugal?
direction of car
When discussing circular motion, you will almost certainly have heard the term centriugal orce probably everywhere except in a physics laboratory! In this course we have spoken exclusively about centripetal orce. Why are there two terms in use? It should now be clear to you how circular motion arises: a orce acts to the centre o the circle around which the obj ect is moving. The alternative idea o centriugal orce comes rom common experience. Imagine you are in a car going round a circle at high speed. You will undoubtedly eel as i you are being fung outwards. O ne way to explain this is to imagine the situation rom the vantage point o a helicopter hovering
straight on direction at this instant
real centripetal force supplied by friction at tyres car Figure 7
Centripetal forces in a car seen from above.
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stationary above the circle around which the car is moving. From the helicopter you will see the passenger attempting to go in a straight line ( Newtons frst law) , but the passenger is orced to move in a circle through riction orces between passenger and seat. I the passenger were sitting on a riction- less seat and not wearing a seat belt, then he or she will not get the message that the car is turning. The passenger continues to move in a straight line eventually meeting the door that is turning with the car. I there were no door, what direction will the passenger take? Another way to explain this is to imagine yoursel in the car as it rotates. This is a rotating rame o reerence that is accelerating and as such cannot obey Newtons laws o motion. You instinctively think that the rotating rame is actually stationary. Thereore your tendency to go in what you believe to be a straight line actually eels like an outward orce away rom the centre o the circle ( remember the rest o the world now rotates round you, and your straight line is actually part o a circle) . Think about a cup o coee sitting on the oor o the car. I there is insufcient riction at the base o the cup, the cup will slide to the side o the car. In the inertial rame o reerence ( the Earth) the cup is trying to go in a straight line. In your rotating rame o reerence you have to invent a orce acting outwards rom the centre o the circle to explain the motion o the cup.
direction of movement observed by passenger
cup
force imagined to act on cup
centre of circle
Figure 8 Rotation
forces.
There are many examples o changing a reerence rame in physics: research the Foucault pendulum and perhaps go to see one o these ascinating pendulums in action. Look up what is meant by the C oriolis orce and fnd out how it aects the motion o weather systems in the northern and southern hemispheres. One o the tricks that physicists oten use is to change reerence rames its all part o the nature o science to adopt alternative rames o reerence to make explanations and theories more accessible. O ne last tip: D ont use explanations based on centriugal orce in an IB examination. The real orce is centripetal; centriugal orce was invented to satisy Newtons second law in an accelerated rame o reerence.
Centripetal accelerations and forces in action Satellites in orbit Figure 9 shows satellites in a circular orbit around the Earth. Why do they ollow these paths? Gravitational orces act between the centre o mass o the Earth and the centre o mass o the satellite. The direction o the orce acting on the satellite is always towards the centre o the planet and it is the gravity that supplies the centripetal orce. Figure 9
Satellites in orbit.
Amusement park rides Many amusement park rides take their passengers in curved paths that are all or part o a circle. How does circular motion provide a thrill? In the type o ride shown in fgure 1 0, the people are inside a drum that rotates about a vertical axis. When the rotation speed is large enough the people are orced to the sides o the drum and the oor drops away. The people are quite sae however because they are held against the inside o the drum as the reaction at the wall provides the centripetal orce to keep them moving in the circle. The people in the ride eel the reaction between their spine and the wall. Friction between the rider and the wall prevents the rider rom slipping down the wall.
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6 .1 CI R CU L AR M O TI O N
Reaction of wall on rider N
Figure 10
friction force weight
The rotor in action.
Turning and banking When a driver wants to make a car turn a corner, a resultant orce must act towards the centre o the circle to provide a centripetal orce. The car is in vertical equilibrium ( the driving surace is horizontal) but not in horizontal equilibrium.
Turning on a horizontal road For a horizontal road surace, the riction acting between the tyres and the road becomes the centripetal orce. The riction orce is related to the coefcient o riction and the normal reaction at the surace where riction occurs.
direction reaction, R
friction friction
centre of circle
elevation Figure 11
mg, W
plan
Car moving in a circle.
I the car is not to skid, the centripetal orce required has to be less than the rictional orce v m_ r < smg where s is the static coefcient o riction. Note that when the vehicle is already skidding the less than sign becomes an equality and the dynamic coefcient o riction should be used. 2
____
This rearranges to give a maximum speed o vmax = drg or a circle o radius r.
Banking Tracks or motor or cycle racing, and even ordinary roads or cars are sometimes banked (fgures 1 2 and 1 3) . The curve o the banked road surace is inclined at an angle so that the normal reaction orce contributes to the centripetal orce that is needed or the vehicle to go round the track
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N cos
C I R C U L AR M O T I O N AN D G R AVI TAT I O N at a particular speed. B icycles and motorcycles can achieve the same eect on a level road surace by leaning in to the curve. Tyres do not need to provide so much riction on a banked track compared to a horizontal road; this reduces the risk o skidding and increases saety.
N
Although you will not be asked to solve mathematical problems on this topic in your IB Physics examination, you do need to understand the principles that underpin banking.
N sin
mg Figure 12
Forces in banking.
Figure 1 2 shows orces acting on a small sphere rolling round a track. This is simplifed to a point obj ect moving in a circle to remove the complications o two or our wheels. A horizontal centripetal orce directed towards the centre o the circle is needed or the rotation. The other orces that act on the ball are the orce normal to the surace ( which is at the banking angle ) and its weight acting vertically down. The vector sum o the horizontal components o the weight and the normal orce must equal the centripetal orce. Looking at this another way, i N is the normal orce then the centripetal orce is equal to N sin
normal reaction
The normal orce resolved vertically is N cos and is, o course, equal mg sin = mg tan and opposite to mg. S o Fcentripetal = ( ____ co s ) 2
centripetal force
friction
2
v mv __ Fcentripetal = ___ r and thereore tan = gr
The banking angle is correct at a particular speed and radius. Notice that it does not depend on the mass o the vehicle so a banked road works or a cyclist and a car, provided that they are going at the same speed. weight
S ome more examples o banking:
C ommercial airline pilots y around a banked curve to change the direction o a passenger j et. I the angle is correct, the passengers will not eel the turn, simply a marginal increase in weight pressing down on their seat) .
S ome high- speed trains tilt as they go around curves so that the passengers eel more comortable.
Moving in a vertical circle
Figure 13
Cycle velodrome.
S o ar the examples have been o motion around a horizontal circle. People will queue or a long time to experience moderate ear on a airground attraction like the rollercoaster in fgure 1 4. The amount o thrill rom the ride depends on its height, speed, and also the orces that act on the riders. How is the horizontal situation modifed when the circular motion o the mass is in a vertical plane?
1 What are the forces acting when the motion is in a vertical circle? Imagine a mass on the end o a string that is moving in a vertical circle at constant speed. Look careully at fgure 1 5 and notice the way the tension in the string changes as the mass goes around.
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6 .1 CI R CU L AR M O TI O N
Figure 14 Theme park ride.
B egin with the case when the string is horizontal, at point A. The weight acts downwards and the tension in the string is the horizontal centripetal force towards the centre of the circle. The mass continues to move upwards and reaches the top of the circle at B . At this point the tension in the string and the weight both act downwards. Thus: v2 Tdown + mg = m _ r and therefore v2 Tdown = m _ r mg The weight of the mass combines with the tension to provide the centripetal force and so the tension required is less than the tension T when the string is horizontal. At C , the bottom of the circle, the tension and the weight both act vertically but in opposite directions and so v2 Tup = m _ r + mg At the bottom, the string tension must overcome weight and also provide the required centripetal force.
B Tdown
D
mg
T
T A
A mg
mg
Tup mg C Figure 15 Forces in
circular motion
in a vertical plane.
As the mass moves around the circle, the tension in the string varies continuously. It has a minimum value at the top of the circle and a maximum at the bottom. The bottom of the circle is the point where the string is most likely to break. If the maximum breaking tension of the string is Tbreak, then, for the string to remain intact, v2 Tbreak > m _ r + mg and the linear speed at the bottom of the circle must be less than ____________ r _ m ( Tbreak mg) If this seems to you to be a very theoretical idea without much practical value, think about a car going over a bridge. If you assume that the shape of the bridge is part of a circle, then there is a radius of curvature r. What is the speed at which the car will lose contact with the bridge?
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v
radius of curvature
Figure 16 Car going over a
bridge.
This is the case considered above, where the object, in this case the car, is at the top o the circle. What is the tension (in this case the orce between car and road) i the car wheels are to lose contact with the bridge? To answer this question, you might begin with a ree-body diagram. You __ should be able to show that the car loses contact at a speed equal to gr .
2 How does speed change when motion is in a vertical circle? Not all circular motion in a vertical circle is at a constant speed. As a mass moves upwards it slows as kinetic energy is transerred to gravitational potential energy ( i there is nothing to keep it moving at constant speed) . At the top o the motion the mass must not stop moving or even go too slowly, because i it did then the string would lose its tension. The motion would no longer be in a circle. The centripetal orce Fc needed to maintain the motion is v Fc = m __ r as usual, at the top o the circle, i Fc is supplied entirely by gravity then v2 Fc = mg = m _ r Just or an instant, the obj ect is in ree- all. 2
The equation can be rearranged to give __
vtop = gr
and this is the minimum speed at the top o the circle or which the motion will still be circular. The minimum speed does not depend on mass. Energy is conserved assuming that there are no losses ( or example, to internal energy as a result o air resistance as the mass goes round) . Equating the energies: kinetic energy at top + gravitational potential energy dierence between top and bottom = kinetic energy at the bottom and 1 1 mv 2 _ m v top 2 + mg( 2 r) = _ bottom 2 2 B y substituting or both tensions, Tbottom and Ttop , it is possible to show that Tbottom = Ttop + 6m You can fnd this proo on the website.
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Worked examples
Solution
1
A hammer thrower in an athletics competition swings the hammer on its chain round 7.5 times in 5.2 s before releasing it. The hammer describes a circle of radius 4.2 m and has a mass of 4.0 kg. Assume that the hammer is swung in a horizontal circle and that the chain is horizontal.
a) ( i) 7.5 revolutions = 1 5 rad
a) C alculate, for the rotation:
b) The thrower usually inclines the plane of the circle at about 45 to the horizontal in order to achieve maximum range. E ve n if the plane were horizontal, the n the we ight of the hammer would contribute to the system so that a component of the tension in the chain must allow for this. B oth assumptions are unlikely.
( i) the average angular speed of the hammer ( ii) the average tension in the chain. b) C omment on the assumptions made in this question.
1 5 angular speed = _ = 9.1 rad s 1 5.2 ( ii) Tension in the chain = centripetal force required for rotation centripetal force = mr 2 = 4.0 4.2 9.1 2 = 1 400 N
6.2 Newtons law of gravitation Understanding
Applications and skills
Newtons law o gravitation
Describing the relationship between
Gravitational feld strength
gravitational orce and centripetal orce Applying Newtons law o gravitation to the motion o an object in circular orbit around a point mass Solving problems involving gravitational orce, gravitational feld strength, orbital speed, and orbital period Determining the resultant gravitational feld strength due to two bodies
Nature of science Newtons insights into mechanics and gravitation led him to develop laws o motion and a law o gravitation. One o his motion laws and the law o gravitation are mathematical in nature, two o the motion laws are descriptive. None o these laws can be proved and there is no attempt in them to explain why the masses are accelerated under the inuence o a orce, or why two masses are attracted by the orce o gravity. Newtons ideas about motion have been subsequently modifed by the work o Einstein. The questioning and insight that leads to the development o laws are undamental to the nature o science.
Equations Mm Newtons law o gravitation: F = G ______ 2 r
F gravitational feld strength: g = ____ m
gravitational feld strength and the gravitational M constant: g = G ____ r2
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Nature of science Scientists from the past Three o the great names rom the history o astronomy and physics were C opernicus, Tycho B rahe, and Kepler. Their contributions were linked and ultimately led to the important gravitational work o Newton. Try to fnd out something about these astronomers. D uring the lietimes o these scientists, science was carried out in a very dierent way rom today. The realization that the Earth orbits the Sun rather than the Sun orbiting the Earth was one o the great developments in scientifc understanding.
Copernicus (Mikolaj Kopernik)
Isaac Newton Figure 1
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An astronomers portrait gallery.
Galileo Galilei and other scientists o the 1 6th century overcame cultural, philosophical and religious prejudices, and some even suered persecution or the scientifc truths they had discovered. As we study the work o these pioneers we should remember that scientists in past times were not always as ree as those today. Research the lie o Galileo ( we oten drop the second name) and explore why he and others came into conict with the Roman C atholic C hurch over their scientifc belies.
Johannes Kepler
Tycho Brahe
Galileo Galilei
6 . 2 N E W T O N S L AW O F G R AV I T AT I O N
Gravitational feld strength Like electrostatics, gravity acts at a distance and is an example o a orce that has an associated orce feld. Imagine two masses in deep space with no other masses close enough to inuence them. O ne mass ( call it A) is in the orce feld due to the second mass ( B ) and a orce acts on A. B is in the gravitational feld o A and also experiences a orce. These two orces have an equal magnitude ( even though the masses may be dierent) but act in opposite directions.
orce on A due to B
orce on B due to A
A B Figure 2
Gravitational orces between two masses.
I both masses are small, the size o a human, say, the orce o gravity is extremely weak. O nly when one o the masses is as large as a planet does the orce become noticeable to us. However, whatever the size o the mass we need a way to measure the strength o the feld to which it gives rise. Gravitational orces are the weakest o all the orces in nature and so require large amounts o mass or the orce to be elt.
large mass M
gravitational feld strength =
The strength o a gravitational feld is defned using the idea o a small test mass. This test mass has to be so small that it does not disturb the feld being measured. I the test mass is large then it will exert a orce o its own on the mass that produces the feld being measured. The test mass would then accelerate the other mass and alter the arrangement that is being measured. The gravitational orce that acts on the small test mass has both magnitude and direction. These are shown in the diagram. The test mass will accelerate in this direction i it is ree to move.
F m
orce, F small test mass, m
Figure 3
Defnition o gravitational feld strength.
Defning gravitational feld strength The concept o the small test mass leads to a defnition o the strength o the gravitational feld. I the mass o the test mass is m, and the feld is producing a gravitational orce o F on the test mass, then the gravitational feld strength ( given the symbol g) is defned as F g= _ m The units o gravitational feld strength are N kg 1 . S ince F is a vector and m a scalar, it ollows that g is a vector quantity and that its direction is that o F. A ormal defnition in words is that gravitational feld strength at a p oint is the orce p er unit mass exp erienced by a small p oint mass p laced at that p oint. This defnition requires that the test mass is not j ust small but is also an ( infnitesimally) small point in space. You might want to consider what the eect might be i the test mass has a shape and extends in space.
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N S o ar the discussion has been limited to the gravitational feld produced by one point mass. How does the situation change i there is more than one mass, excluding the test mass itsel?
resultant orce 43 = 1 unit feld strength B 3 units
feld strength A 4 units
Field strength is independent o the magnitude o the point test mass ( we divided F by m to achieve this) . S o the vector feld strengths can be added together ( fgure 4) .
g1
g2 resultant g
In IB Physics examinations you will only be asked to add the feld strengths o masses that all lie on the same straight line. B ut even in two dimensions, the addition is straightorward using the ideas o vector addition by drawing or by calculation.
g and the acceleration due to gravity Figure 4 Adding feld
vectorially.
strengths
Sometimes students are surprised that the symbol g is used or the gravitational feld strength, they think there might be a risk o conusion with g the acceleration due to gravity! However this does not happen. At the E arths surace ( using Newtons second law) acceleration due to gravity at the surace orce on a mass at the surace due to gravity = ____ size o the mass F F __ S o the acceleration = __ m but m is also the defnition o gravitational feld strength so the acceleration due to gravity = gravitational feld strength = g.
The magnitude o the gravitational feld strength ( measured in N kg 1 ) is equal to the value o the acceleration due to gravity ( measured in m s 2 ) . You should be able to show that N kg 1 m s 2 . ( The symbol means is equivalent to.)
Newtons law of gravitation an inverse-square law Isaac Newton ( 1 642 1 72 7) was a B ritish scientist who consolidated the work o others and who added important insights o his own. D uring his lie he contributed to the study o optics, mechanics and gravitation. O ne o his greatest triumphs was work on gravity that he developed using the data and ideas o the German astronomer Johannes Kepler and others about the motion o the planets. Newton realized that the gravitational orce F between two obj ects with masses M and m whose centres are separated by distance r is:
always attractive
1 proportional to __
proportional to M and m.
r2
This can be summed up in the equation Mm F _ r2 1 Laws that depend on __ are known as inverse- square. I the distance r2
between the two masses is doubled without changing mass, then the orce between the masses goes down to one-quarter o its original value. To use this equation numerically a constant o proportionality is needed and is given the symbol G,
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6 . 2 N E W T O N S L AW O F G R AV I T AT I O N GMm F= _ r2 G is known as the universal gravitational constant and it has an accepted value o 6.67 1 0 1 1 N m 2 kg2 . Gravity is always attractive so i the distance is measured from the centre o mass M to mass m then the orce on m due to M is towards M. In other words, the orce is in the opposite direction to the direction in which the distance is measured. You may see some books where a negative sign is introduced to predict this direction. The IB Diploma Programme physics course does not attribute negative signs to attractive orces, the responsibility o keeping track o the orce direction is with you.
Nature of science A universal constant? What does it mean to say that G is a universal gravitational constant? Newton did not know what the size o the constant was ( as the value was determined over a hundred years later by C avendish) . In a sense, it did not matter. Newton realized that all obj ects are attracted to the Earth the apocryphal story o him seeing a alling apple reminds us that he knew this.
The insight that Newton had was to realize that the orce o gravity went on beyond the apple tree and stretched up into the sky, to the Moon, and beyond. He realized that the Moon was alling continuously towards the E arth under the inuence o gravity and because it was also moving horizontally it was in continuous orbit.
Worked examples 1
Solution
C alculate the orce o attraction between an apple o mass 1 00 g and the E arth. Mass o Earth = 6. 0 1 0 24 kg. Radius o Earth = 6.4 Mm
GMm F= _ r 6. 7 1 0 1 1 1 . 7 1 0 27 9.1 1 0 31 = ____ ( 1 . 5 1 0 1 0 ) 2 2
Solution GMm 6.7 1 0 1 1 0.1 6.0 1 0 24 F = _ = ___ r ( 6. 4 1 0 6 ) 2
= 4. 6 1 0 48 N
2
( C ompare this with the electrostatic attraction o the electron and proton at this separation o about 1 0 8 N.)
= 1 .1 N 2
C alculate the orce o attraction between a proton o mass 1 .7 1 0 27 kg and an electron o mass 9.1 1 0 3 1 kg a when they are at a distance o 1 . 5 1 0 1 0 m apart.
Gravitational feld strengths re-visited Knowledge o Newtons law o gravitation means that the defnition o gravitational feld strength can be taken urther to help our study o real situations.
The feld strength at a distance r rom a point mass, M The simplest case is that o a single point mass M placed, as usual, a long way rom any other mass. In practice we say that the point mass is at an infnite 1 distance rom any other mass because i r is very large then __ is extremely r 1 __ small (mathematicians say that as r tends to infnity, tends to zero) . 2
r
2
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N As usual, the mass o our small test obj ect is m. This means that the magnitude o the orce F between the two masses M and m is GMm F=_ r2 so that the gravitational feld strength g is F GM _ g=_ m = r2 As beore, the direction is measured outwards from M but the orce on the point mass is in the opposite direction towards M.
The feld strength at a distance r rom the centre outside a sphere o mass M It turns out that the answer or g outside a spherical planet is exactly the same as g or the point mass j ust quoted: GM g= _ 2 r
I we are outside the sphere, all the mass acts as though it is a point mass o size M positioned at the centre o the sphere (the centre o mass) . We only need one equation or both point masses and spheres.
Nature of science And inside the Earth...? Science is all about asking questions. Outside the 1 Earth, the gravitational feld strength varies as __ . But r what happens to gravity inside the Earth? Is it zero? Is it a constant? Does it become larger and larger, reaching infnity, as we get closer to the centre? You might, at frst sight, expect this rom Newtons law. 2
this is the part that accounts or gravity
out. The second thing that happens is that only the mass inside the smaller Earth contributes to the gravitational pull. The mass o this smaller Earth varies with r3 assuming a constant density or the Earth. B ecause the gravitational orce varies 1 with __ , together these give an overall dependence r o r. The whole graph or the variation o g or a planet (inside and out) is given in fgure 6, gs is the gravitational feld strength at the surace 2
tunnel
this part does not contribute Figure 5 Journey
to the centre o the Earth.
In Journey to the centre of the Earth the novelist Jules Verne imagined going through a volcanic tunnel to the centre o the planet. Visualize his travellers halway down the tunnel as they stand on the surace o a smaller Earth defned by their present distance rom the centre. Two remarkable things happen: The frst is that the contribution rom the shell o Earth above the travellers makes no contribution to the gravitational feld; all the dierent parts o the outer shell cancel
262
gravitational feld strength, g
gs
1g 4 s 1g 9 s
0
R 2R 3R distance rom centre o planet o radius R
4R
Figure 6 Gravitational feld strength inside and outside the Earth.
6 . 2 N E W T O N S L AW O F G R AV I T AT I O N
Linking orbits and gravity The gravitational orce o a planet provides the centripetal orce to keep a satellite in orbit. cannon hill
impact i Earth is fat impact i Earth is a sphere in orbit Earth
constant distance rom surace Figure 7
Newtons cannon how he thought about orbits.
Newton had an insight into this too. He used the example o a cannon on a high mountain (fgure 7) . The cannon fres its cannonball horizontally and it accelerates vertically downwards. On a at Earth, it will eventually hit the ground. Newton knew that the Earth was a sphere and that the curvature o the earth allowed the ball to travel urther beore hitting the ground. He then imagined the ball being fred at larger and larger initial speeds. E ventually the shell will travel horizontally at such a high speed that the curvature o the E arth and the curve o the traj ectory will be exactly the same. When this happens the distance between shell and surace is constant and the shell is in orbit around the E arth. What do you expect to happen to the trajectory o the cannon ball i it is fred at even greater speeds? To check your conclusions, fnd an applet on the Internet that will allow you to vary the fring speed o Newtons cannon. A good starting point or the search is applet Newton gun.
Worked examples 1
( Mass o Moon = 7.3 1 0 22 kg; radius o the Moon = 1 .7 1 0 6 m)
Solution GM g=_ r2 6.7 1 0 1 1 7.3 1 0 22 = ___ ( 1 .7 1 0 6 ) 2 = 1 .7 N kg 1
This motion o a satellite around the Earth can be analysed by combining the ideas o centripetal orce and gravitational attraction. The gravitational attraction FG provides the centripetal orce Fc , so ( ignoring signs) GM E m mv 2 _ Fc = FG = _ r = r2 where ME is the mass o the Earth, r is the distance rom the satellite to the centre o the Earth, v is the linear speed o the satellite, and m is its mass. These equations can be simplifed to ____
v=
GM _ r E
This equation predicts the speed o the satellite at a particular radius. Notice that the speed does not depend on the mass o the satellite. All satellites travelling round the Earth at the same distance above the surace have the same speed. As an exercise, use the equations on page 2 5 0 to show that GM E 2 = _ r3
C alculate the gravitational feld strength at the surace o the Moon.
2
C alculate the gravitational feld strength o the S un at the position o the Earth. (Mass o Sun = 2.0 1 0 30 kg; EarthS un distance = 1 .5 1 0 1 1 m.)
Solution GM g= _ r2 1 1 6. 7 1 0 2 .0 1 0 30 = ___ ( 1 .5 1 0 1 1 ) 2 = 6.0 mN kg 1
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N where is the angular speed, and that 4 2 r 3 T2 = _ GM E where T is the orbital period ( time for one orbit) of the satellite. This result: ( orbital period of a satellite) 2 ( orbital radius) 3 is known as Kep lers third law. It is one of the three laws that he discovered when he analysed Tycho B rahes data.
Worked examples 1
C alculate the orbital period of Jupiter about the Sun. Mass of Sun = 2.0 1 0 30 kg; radius of Jupiters orbit = 7.8 1 0 1 1 m)
Solution
2
The orbital time period of the Earth about the S un is 3 .2 1 0 7 s. C alculate the orbital period of Mars. ( radius of Earth orbit = 1 . 5 1 0 1 1 m; radius of Mars orbit = 2 .3 1 0 1 1 m)
2 3
4 r T2 = _ GM S
Solution
T = 3 .7 1 0 8 s ( about 1 2 years)
r 3M T2M _ _ = r 3E T2E ___ r 3M TM = TE _ = 6. 1 1 0 7 s ( about 1 .9 years) r 3E
Nature of science What Newton knew S everal decades before Newton began working on his ideas of gravitation, Kepler had used his own and others astronomical data to show that, for the planets, ( radius of the orbit) 3 ( time period of orbit) 2 This was an empirical result ( meaning that it came from experimental data) . Newton was able to show that this proportionality arose from his own theory; this was a theoretical result ( meaning that it came from a theory, not data) . Here are some data for the satellites of Jupiter.
Moon Io Europa Ganymede Callisto
Distance from centre of Jupiter/10 3 km 420 670 1070 1890
Orbital period/ days 1.8 3.6 7.2 16.7
Use these data to show that ( radius of the orbit) 3 ( time period of orbit) 2 3
( radiu s o f the o rb it) O r, put another way, that _______________ is a constant. ( tim e p e rio d o f o rb it) 2
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QUESTION S
Questions 1
A particle P is moving in a circle with uniorm speed. D raw a diagram to show the direction o the acceleration a and velocity v o the particle at one instant o time.
5
The S ingapore Flyer is a large Ferris wheel o radius 8 5 m that rotates once every 3 0 minutes.
P
2
State what provides the centripetal orce that causes a car to go round a bend.
3
State the centripetal orce that acts on a particle o mass m when it is travelling with linear speed v along the arc o a circle o radius r.
4
( IB) At time t = 0 a car moves o rom rest in a straight line. O il drips rom the engine o the car with one drop every 0.80 s. The position o the oil drops on the road are drawn to scale on the grid below such that 1 .0 cm represents 4.0 m. The grid starts at time t = 0.
a) C alculate the linear speed o a point on the rim o the wheel o the Flyer. b) ( i) D etermine the ractional change in the weight o a passenger on the Flyer at the top o the ride. ( ii) Explain whether the passenger has a larger or smaller apparent weight at the top o the ride. c) The capsules need to rotate to keep the oor o the cabin in the correct place. C alculate the angular speed o the capsule about its central axis.
Direction of motion 1.0 cm
6 a) ( i) S tate the eature o the diagram that indicates that the car accelerates at the start o the motion.
a) Quito in E cuador ( 1 4 minutes o arc south o the Equator)
( ii) D etermine the distance moved by the car during the frst 5 .6 s o its motion. b) The car then turns a corner at constant speed. Passengers in the car who were sitting upright eel as i their upper bodies are being thrown outwards. ( i) Identiy the orce acting on the car, and its line o action, that enables the car to turn the corner. ( ii) Explain why the passengers eel as i they are being thrown outwards.
The radius o the E arth is 6400 km. D etermine the linear speed o a point on the ground at the ollowing places on E arth:
b) Geneva in S witzerland ( 46 north o the Equator) c) the South Pole. 7
A school bus o total mass 65 00 kg is carrying some children to school. a) D uring the j ourney the bus needs to travel round in a horizontal curve o radius 1 5 0 m. The dynamic coefcient o riction between the tyres and the road surace is 0.7. Estimate the maximum speed at which the driver should attempt the turn.
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N b) Later in the j ourney the driver needs to drive across a curved bridge with a radius o curvature o 75 m. Estimate the maximum speed i the bus is to remain in contact with the road. 8
A velodrome used or bicycle races has a banking angle that varies continuously rom 0 to 60. Explain how the racing cyclists use this variation in angle to their advantage in a race.
D ata needed or these questions: Radius o Earth = 6.4 Mm; Mass o E arth = 6.0 1 0 24 kg; Mass o Moon = 7.3 1 0 22 kg; Mass o S un = 2 .0 1 0 30 kg; E arthMoon distance = 3.8 1 0 8 m; S unEarth distance = 1 . 5 1 0 1 1 m; G = 6.67 1 0 1 1 N m 2 kg 2 9
D educe how the radius R o the circular orbit o a planet around a star o mass m s relates to the period T o the orbit.
1 0 A satellite orbits the Earth at constant speed as shown below.
satellite
1 1 Determine the distance rom the centre o the Earth to the point at which the gravitational feld strength o the Earth equals that o the Moon. 1 2 The ocean tides on the Earth are caused by the tidal attraction o the Moon and the S un on the water in the oceans. a) C alculate the orce that acts on 1 kg o water at the surace o the sea due its attraction by the ( i) Moon ( ii) S un. b) Optional difcult. E xplain why there are two tides every day at many coastal points on the Earth. [Hint: there are two parts to the answer, why a tide at all, and why two every day.] 1 3 There are two types o communication satellite. O ne type o communication satellite orbits over the poles at a distance rom the centre o the Earth o 7400 km; the other type is geostationary with an orbital radius o 3 6 000 km. Geostationary satellites stay above one point on the equator whereas polar- orbit satellites have an orbital time o 1 00 minutes. C alculate: a) the gravitational feld strength at the position o the polar- orbit satellite
Earth
b) the angular speed o a satellite in geostationary orbit c) the centripetal orce acting on a geostationary satellite o mass 1 .8 1 0 3 kg.
a) Explain why, although the speed o the satellite is constant, the satellite is accelerating. b) D iscuss whether or not the gravitational orce does work on the satellite.
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7 ATOMIC, NUCLEAR, AND PARTICLE PHYSICS Introduction In this topic we consider the composition o atoms. We look at extra- nuclear electrons and the nucleus and the particles o which the nucleus is composed. We see the vast array o particles that are now known to exist and how
these particles can be classifed and grouped. As is oten the case, energy plays in important role in the atom and undamental to this is the tendency or particles to be most stable when their energy is minimized.
7.1 Discrete energy and radioactivity Understanding Discrete energy and discrete energy levels Transitions between energy levels Radioactive decay
Applications and skills Describing the emission and absorption spectra
Alpha particles, beta particles, and gamma rays Hal-lie Constant decay probability
Absorption characteristics o decay particles Background radiation
o common gases Solving problems involving atomic spectra, including calculating the wavelength o photons emitted during atomic transitions Completing decay equations or alpha and beta decay Determining the hal-lie o a nuclide rom a decay curve Investigating hal-lie experimentally (or by simulation)
Equations Photon energyrequency relationship: E = hf hc Planck relationship or wavelength: = _____ E
Nature of science Unintentional discoveries Physics is scattered with examples o experimenters making signifcant discoveries unintentionally. In the case o radioactivity the French physicist Henri Becquerel, in 1896, was investigating a possible link between X-rays and phosphorescence. He had stored a sample o uranium salt (which releases light over a period o time having being exposed to light) in a drawer with a photographic plate (the orerunner to flm) wrapped in an opaque covering. Becquerel discovered that the plate had become exposed even though it had been kept in the dark. Rather
than ignore this unpredicted outcome, Becquerel sought an explanation and ound other uranium salts produced the same result suggesting that it was the uranium atom that was responsible. He went on to show that the emissions rom uranium ionize gases but dier rom X-rays in that they are deected by electric and magnetic felds. For his work, which developed as the result o a lucky accident, Becquerel was awarded a share o the Nobel Prize or Physics in 1903 (with the Curies or their work also on radioactivity) .
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS
TOK Assumptions in physics Topic 5 showed that protons and electrons are charged. A bigger question is why do charges behave in this way? The simple answer to this is that we dont know we do know that some objects have a property that we call charge and that charges interact at a distance through an electromagnetic feld. Yet we dont know what charges are and what the dierence between positive and negative actually is. This may seem a huge gap in our understanding. However, we use the same concept o the implied labelling o objects or properties or example, when we ask the question what is a chair?, our answers will inevitably give us a description o a chair along the lines o something we sit on, having our legs, it has a seat and back etc. Thereore, what we have done is described the properties and characteristics o chairs it remains the case that a chair is a chair. So, although we only have a uzzy idea o what electrons and charges actually are, does this prevent us rom developing eective theories that are based on poorly understood concepts?
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Introduction Although many texts take a historical tour o the development o atomic and nuclear physics, the IB Physics syllabus does not lend itsel to such a treatment. Instead, there is a ocus on the common aspects o the constituents o the atom. It is likely that you are already amiliar with the structure o an atom as consisting o a positively charged nucleus surrounded by negatively charged electrons in fxed orbits. You are probably also be aware that the nucleus consists o positively- charged protons and uncharged neutrons. We will take these aspects as read, but we will also revisit them throughout the sections o Topic 7 ( and, or those students studying HL Physics, Topic 1 2 too) .
Energy levels Returning to the nuclear model o an atom, the orbiting electrons cannot occupy any possible orbit around the nucleus. D ierent orbits correspond to dierent amounts o energy, or energy levels, and the electrons are restricted to orbits with specifc energies. Electrons change energy so that they can j ump rom one energy level to another, but they can only occupy allowed energy levels. Although you may ask what makes an allowed orbit or allowed energy level? it takes an understanding o the wave nature o electrons and quantum mechanics to attempt to explain this consistently. We return to this or HL students in Topic 1 2 , but or now it is maybe better to accept that this is how nature operates. Hydrogen is the simplest o all elements, with the hydrogen atom normally consisting o a proton bound to a single electron by the electromagnetic orce ( an attractive orce between oppositely charged particles) . There are other isotop es o hydrogen, with a nucleus including one or more neutrons. I a hydrogen atom completely loses the electron it becomes a positively- charged hydrogen ion. It is the single proton that defnes this nucleus to be hydrogen any atom with a single proton must be hydrogen. In the same way any atom having two protons is an isotope o helium, and any with three protons will be lithium, etc. A common visualization o the atom is the planetary model in which the electron orbits the proton mimicking the E arth orbiting the S un. D espite weaknesses in this model, it still gives a good introduction to energy levels in the atom. The energy levels or the hydrogen atom are shown in the table below:
Energy level ( n )
Energy/eV
1
13.58
2
3.39
3
1.51
4
0.85
5
0.54
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
The energy levels are usually shown diagrammatically as in fgure 1 . This diagram shows the energy level and the energy o an electron that occupies this level. An electron in the ground state o the hydrogen atom must have exactly 1 3 . 5 8 eV o energy; one in the second energy level must have 3 . 3 9 eV, etc. An atom with an electron occupying an energy level higher than the ground state is said to be excited; so an electron in level 2 is in the frst excited state, one in level 3 is in the second excited state, etc. The energy levels in atoms are said to be quantized which means they must have discrete fnite values. An electron in a hydrogen atom cannot have 5 . 92 eV or 1 0. 2 1 eV or any other value between 1 3 . 5 8 and 3 . 3 9 eV; it m ust have one o the values in the table.
Transitions between energy levels
Note n=5 n=4 second excited state n = 3
-3.39 eV
frst excited state n = 2
-13.58 eV
ground state n = 1
When an electron in the hydrogen atom jumps rom the ground state to the frst excited state it must gain some energy; this cannot be any randomly chosen amount o energy it must be exactly the right amount. The electron needs to gain ( 3.39) ( 1 3.5 8) = 1 0.1 9 eV o energy. The electron does not physically pass through the space between the two energy levels. B eore gaining the energy it must have 1 3.58 eV and immediately ater gaining the energy it must have 3.39 eV. The energy has to be transerred to the electron in one discrete amount; it cannot be gradually built up over time. The energy needed to excite an atom can come rom absorption o light by the atom. In order to understand this we must consider light to be a packet or quantum called a photon the rationale or this is discussed in Topic 1 2 . The energy, E, carried by a photon is related to the requency o the radiation by the equation E = hf where h is the Planck constant ( = 6.63 1 0 34 J s) and f is the requency in Hz. The wave equation c = f ( where c is the speed o electromagnetic waves in a vacuum) also applies to the photons and, by combining the two equations, we can c c relate the energy to the wavelength E = hf = h __ or = h __ . We have seen E that an electron needs to absorb 1 0.1 9 eV in order to j ump rom the ground state to the frst excited state. What, then, will be the wavelength o the electromagnetic radiation needed to do this? First we need to convert 1 0.1 9 eV into j oules. This is simply a matter o multiplying the energy in eV by the charge on an electron, so 1 0.1 9 eV = 1 0.1 9 1 .6 1 0 1 9 J ~ 1 .6 1 0 1 8 J.
When an electron has been excited in this way it is likely to be quite unstable and will very quickly (in about a nanosecond) return to a lower energy level. In order to do this it must lose this same amount o energy. This means that a photon o energy 1 0.1 9 eV and wavelength 1 20 nm must be emitted by the atom. The electron will then jump to the lower energy level.
The energies are given in units o
The energies are all negative
values this is because the potential energy o two objects is zero only when they are at infnite separation. Because there is an attractive orce between a proton and an electron the electron has been moved rom infnity until it is in its orbit. The protonelectron system has lost some energy and, since it was zero to start with, it is now negative.
Energy level 1 is the lowest energy
level having the most negative and, thereore, smallest amount o energy. It is the most stable state and is called the ground state.
-18
This radiation in the ultraviolet part o the spectrum.
Figure 1 Energy level diagram or hydrogen.
electron volt (1 eV = 1.6 10 19 J) . This is a very common unit (along with the multipliers keV and MeV) or atomic physics and it avoids having to use very small numbers. However, you might fnd that a question is set in joule.
8
c 3 .00 1 0 Now using = h __ we get = 6.63 1 0 - 34 J ________ = 1 .2 1 0 - 7 m E 1 .6 1 0 or 1 2 0 nm.
ionization
0 -0.54 eV -0.85 eV -1.51 eV
n is known as the principal
quantum number and we will return to its values in Topic 12.
For the transition rom the second energy level to the third to take place only 1 .88 eV needs to be absorbed by the atom and 0.66 eV to jump rom the third level to the ourth. When the electron is given the ull 1 3.5 8 eV it
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS is completely removed rom the nucleus (or taken to infnity) and the atom has become ionized. The energy supplied to the electron in the ground state to take it to infnity is called the (frst) ionization energy.
Worked example Singly- ionized helium ( He + ) is said to be hydrogen- like in that it only has one electron ( although it has two protons and two neutrons) . The energy levels dier rom those o hydrogen as shown in the ollowing table:
Energy level ( n )
Energy/eV
1
54.4
2
13.6
3
6.0
4
3.4
5
2.2
a) Explain why the ground state has a much higher energy level than that o hydrogen ( 1 3 .6 eV) . b) (i) D etermine the requency o the photon emitted by an electron transition rom energy level 4 to energy level 2 . (ii) State which region o the electromagnetic spectrum the emitted photon belongs to.
Solution a) The hydrogen atom has a single proton in the nucleus but the helium ion has two. This means the attractive orce between the nucleus and an electron is greater or the helium nucleus. The helium electron is more tightly bound to the nucleus and, thereore, requires more energy to remove it to infnity. b) (i) E = ( 1 3 . 6) ( 3 .4) = 1 0.2 eV ( the minus sign tells us that there is a loss o energy during the emission o photons) . 1 0.2 eV = 1 0.2 1 .6 1 0 1 9 J = 1 .6 1 0 1 8 J 1 . 6 1 0 -18 E E = h = _ = __ h 6.63 1 0 - 34 = 2 .4 1 0 1 5 Hz (ii) We have seen this energy or the hydrogen atom previously and so know that this requency corresponds to ultraviolet. You can calculate the wavelength by dividing the speed o electromagnetic waves by your known requency.
Nature of science Patterns in physics You may have noticed that the calculation or the energy o photon emitted by the helium ion in the worked example above is identical to an energy value or the hydrogen atom. This sort o coincidence is not unusual in physics and suggests that there may be a link between the energy dierences. In act in 1 88 8 the S wedish physicist Johannes Rydberg proposed a ormula or hydrogen- like atoms: singly
ionized helium, doubly ionized lithium, triply ionized beryllium, etc. based on observation o spectral line patterns emitted by these elements. Rydbergs theory, based on his practical work, utilized the idea o wave number ( the reciprocal o the wavelength) this was o undamental importance to the D anish physicist Niels B ohr in explaining quantization o energy levels in 1 9 1 3 .
Emission spectra When energy is supplied to a gas o atoms at low pressure the atoms emit electromagnetic radiation. In the laboratory the energy is usually supplied by an electrical discharge ( an electrical current passing through the gas when a high voltage is set up between two electrodes
270
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
across the gas) . I the radiation emitted by the gas is incident on the collimating slit o a spectrometer, it can then be dispersed by passing it though a diraction grating or a glass prism ( see fgure 5 ) . O bserving the spectrum through a telescope, or proj ecting it onto a screen, will give a series o discrete lines similar to those shown in fgure 2 .
Figure 2 Line emission spectra.
electron transitions for the hydrogen atom n=7 n=6 n=5 n=4
Brackett series
n=3
Paschen series
n=2
Balmer series
Figure 2 Line emission spectra.
Spectral lines appear in series in the dierent regions o the electromagnetic spectrum (inra-red, visible, and ultraviolet) . Each series o lines is dependent on the energy level that the electrons all to. In the case o hydrogen the series are named ater the frst person to discover them: in the Lyman series the electrons all to the ground state (n = 1 ) this series is in the ultraviolet region o the electromagnetic spectrum; in the Balmer series the electrons all to the frst excited level (n = 2) this is in the visible region; the Paschen series relates to n = 3 and is in the inrared region.
n=1
Lyman series
Figure 3 Electron transitions for hydrogen.
Absorption spectra The electrons in solids, liquids, and dense gases can also be excited they tend to glow when heated to a high temperature. When the emitted light is observed it is seen to consist o a spectrum o bands o colours rather than lines. In the case o solids this will give a continuous spectrum in which the colours are merged into each other and so are not discrete. This is typical o matter in which the atoms are closely packed. The neighbouring atoms are very close and this causes the energy levels in the atoms to change value. When there are many atoms, the overall energy levels combine to orm a series o very similar, but dierent, energies which makes up an energy band. When the object emitting a continuous spectrum is surrounded by a cool gas, then the continuous spectrum is modifed by the surrounding gas. The continuous spectrum is streaked by a number o dark lines. When a heated tungsten flament is viewed through hydrogen gas the absorption spectrum
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS shown in the lower part o fgure 4 can be seen. Further inspection shows that the black lines occur at exactly the same positions as the lines o the hydrogen emission spectrum. This pattern cannot simply be a coincidence.
Figure 4 Emission and absorption spectra or hydrogen.
Absorption occurs when an electron in an atom o the absorbing material absorbs a photon. The energy o this photon must be identical to the dierence between the energy levels. The material removes photons o this requency rom the continuous range o energies emitted by the light source. Naturally, this will make the absorbers atoms become unstable and they will revert to a lower energy level by emitting photons these will include the absorbed requencies but they will be emitted in random directions and not necessarily in the original direction. This will reduce the intensity o those specifc requencies in the original direction giving the black lines seen crossing the continuous spectrum in fgure 4. Absorption spectra or sodium can be demonstrated with the apparatus shown in fgure 5 . A white light source emits light which is incident on a diraction grating on the turntable o the spectrometer. A continuous spectrum is seen through the telescope or displayed on a computer monitor ( using an appropriate sensor and sotware) . A burner is used to heat sodium chloride ( table salt) . The black absorption lines o sodium should be detected in the yellow region o the continuous spectrum. ame with table salt added
white light source
telescope or sensor linked to computer
collimator burner
difraction grating
spectrometer
Figure 5 Demonstration o sodium absorption spectra.
Nature of science Fraunhofer lines
272
Figure 6 Stamp commemorating the lie and work o Fraunhoer.
The English scientist William Woolaston originally observed absorption lines in the Suns spectrum in 1 802 . However it was the German physicist, Joseph von Fraunhoer, who, in 1 81 4, built a spectrometer and invented the diraction grating with which he was able to observe and analyse these absorption lines now known as Fraunhofer lines. These lines were the frst lines in a spectrum to be observed. Fraunhoer labelled the most prominent o the lines as AK. These lines provide astronomers with a means to study the composition o a star.
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
Worked example a) The element helium was frst identifed rom the absorption spectrum o the S un.
diraction grating or a prism in order to disperse it into its component wavelengths. This gives evidence about the elements in the gases in the outer part o the Sun.
(i) Explain what is meant by the term absorption spectrum. (ii) O utline how this spectrum may be experimentally observed. b) O ne o the wavelengths in the absorption spectrum o helium occurs at 5 88 nm. (i) S how that the energy o a photon o wavelength 5 88 nm is 3 .3 8 1 0 - 1 9 J. (ii) The diagram below represents some o the energy levels o the helium atom. Use the inormation in the diagram to explain how absorption at 5 88 nm arises. (iii) Mark this transition on a copy o the energy level diagram below.
b)
hc 6.63 1 0 - 34 3 .00 1 0 8 (i) E = _ = ___ 5 88 1 0 - 9 = 3 .3 8 1 0 - 1 9 J (ii) As this is absorption, the electron is being raised to a higher energy level. The dierence between the levels must be equal to 3 .3 8 1 0 - 1 9 J and so this is between ( 5 . 80 1 0 - 1 9 J) and (2.42 1 0 -19 J) levels, i.e., (2.42 1 0 -19 J) (5 .80 1 0 - 1 9 J) = (+ ) 3.38 1 0 - 1 9 J
Note that the energy levels are given in joules and so there is no need to do a conversion rom electronvolts in this question.
0
(iii) The transition is marked in red.
0
-2.42 -3.00
-1.59 -5.80
-7.64
energy/10 -19 J
energy/10 -19 J
-1.59
-2.42 -3.00 -5.80
Solution a)
(i) An absorption spectrum consists o a continuous spectrum that has a number o absorption lines crossing it. These lines correspond to the requencies o the light in the emission spectrum o the elements within the substance that is absorbing the light. (ii) The light rom the Sun can be projected onto a screen ater passing through a
-7.64
Note that light rom the Sun needs to be fltered and should not be observed directly through a telescope.
Radioactive decay Radioactive decay is a naturally occurring process in which the nucleus o an unstable atom will spontaneously change into a dierent nuclear confguration by the emission o combinations o alpha particles, beta
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS particles, and gamma radiation. There are ewer than our hundred naturally occurring nuclides ( nuclei with a particular number o protons and neutrons) but only about 60 o these are radioactive. The trend is or these nuclei to become more stable although this may take a very long time. In radioactive decay the nuclide decaying is reerred to as the p arent and the nuclide( s) ormed as the daughter(s) . As we have discussed, the nucleus contains protons and neutrons these are jointly called nucleons. These are bound together by the strong nuclear orce, which must overcome the electrostatic repulsion between the positively charged protons. The presence o the neutrons moderates this repulsion the strong nuclear orce (which has a very short range 1 0 1 5 m) acts equally on both the protons and neutrons. For the nuclei with ew nucleons, having approximately equal numbers o protons and neutrons corresponds to nuclides being stable and not radioactive. As we will soon see, heavier nuclei need a greater proportion o neutrons in order to be stable.
Nature of science Nuclear radiation and safety Many people are apprehensive about any exposure to radiation rom radioactive sources. The danger rom alpha particles is small unless the source is ingested into the body. B eta particles and gamma rays are much more penetrating and can cause radiation burns and long term damage to D NA. Any sources used in schools are very weak but should still be treated with respect. They should always be lited with long tongs, never held near the eyes and should be kept in lead-lined boxes and stored in a locked container. D uring the 1 00 plus years since its discovery nuclear radiation has been studied extensively
and can be saely controlled. Those working with radioactive sources use radiation monitoring devices to record exposure levels. Whenever there may be concerns about nuclear radiation, exposure is limited by:
distance keeping as ar away rom a radiation source as possible
shielding placing absorbers between people and sources
time restricting the amount o time or which people are exposed.
Nuclide nomenclature This is a shorthand way o describing the composition o a nuclide. The element is described by its chemical symbol H, He, Li, B e, B , etc. The structure o the nucleus is denoted by showing both the proton number Z ( otherwise called the atomic number) and the nucleon number A ( or mass number) . This always takes the orm: AZ X so, or example, the isotopes o carbon, carbon-1 2 and carbon-1 4, are written as 1 62 C and 14 6 C . C arbon-1 2 has 6 protons and 1 2 nucleons and so it has 6 neutrons; carbon-1 4 has 6 protons and 1 4 nucleons making 8 neutrons. Isotopes are nuclides o the same element ( and so have the same number o protons) having dierent numbers o neutrons.
Alpha () decay
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In alpha-particle decay, an unstable nuclide emits a particle o the same confguration as a helium nucleus 42 He ( having two protons and two neutrons) . Many nuclides o heavy elements decay primarily by alphaparticle emission.
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
E xamples: 238 92
U 42 He +
234 90
234 90
Th 42 He +
Th
230 88
Ra
As a consequence of the conservation of charge and massenergy, the equation must balance so that there are equal numbers of protons and nucleons on either side of it. The alpha particle is sometimes written as 42 instead of 42 He.
Negative beta ( ) decay In negative beta- particle emission, an unstable nuclide emits an electron. The emission of a beta particle does not change the nucleon number of the parent nuclide. A neutron is converted to a proton and an electron is ej ected. This decay occurs for those nuclides with too high a neutron proton ratio. The decay is accompanied by an electron antineutrino which we will discuss in Sub- topic 7. 3 . E xamples: 1 31 53
I
0 -1
234 90
Th
e+ 0 -1
1 31 54
_
Xe +
234 91
e+
_
Pa +
The negative beta particle is sometimes written as
0 -1
instead of
0 -1
e.
Again, the equation must balance. The daughter nuclide has the same nucleon number as the parent but has one extra proton meaning that its proton number increases by 1 . The antineutrino has no proton or _ nucleon number and is often written as 00 .
Positron ( + ) decay In positron or positive beta- particle emission, an unstable nuclide emits a positron. This is the antiparticle of the electron, having the same characteristics but a positive charge instead of a negative one. The emission of the positron does not change the nucleon number of the parent nuclide. A proton is converted to a neutron and a positron is ej ected. This decay occurs for those nuclides with too high a proton neutron ratio. The decay is accompanied by an electron neutrino which, again, we will discuss in S ub- topic 7.3 . E xamples: 11 6 21 11
C Na
0 +1
e+
0 +1
11 5
e+
B +
21 10
Ne +
The positron is sometimes written as
0 +1
instead of
0 +1
e.
Again, the equation balances. The daughter nuclide has the same nucleon number as the parent but has one less proton, meaning that its proton number decreases by 1 . The neutrino has no proton or nucleon number and is alternatively written as 00 .
Gamma ray emission Gamma rays are high-energy photons often accompanying other decay mechanism. Having emitted an alpha or beta particle the daughter nucleus
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS is oten let in an excited state. It stabilizes by emitting gamma photon(s) thus losing its excess energy. Examples: 60 27
Co
0 -1
e+
60 28
60 28
Ni *
60 28
Ni +
_
Ni * + +
Here the cob alt- 6 0 decays b y b e ta e mission into an excite d nicke l- 6 0 * nuclide 60 28 Ni . This is accomp anied b y a gamma photo n. The nickel- 6 0 de - excite s by e mitting a se cond gamma photo n ( o die rent e nergy rom the original pho ton) . Gamma photons have no proton or nucleo n numb er and are sometimes writte n as 00 . I we had wished to summarize the cobalt decay in one equation we could write it as: 60 27
Co
0 -1
e+
60 28
_
Ni + + 2
This tells us that overall there are two gamma photons emitted.
Worked examples 1
A nucleus o strontium-90 ( 90 38 S r) decays into an isotope o yttrium ( Y) by negative beta emission. Write down the nuclear equation or this decay.
When the iron- 5 5 ( 55 26 Fe) nucleus captures an electron in this way the nucleus changes into a manganese ( Mn) nucleus. Write a nuclear equation to summarise this.
Solution
Solution
This is a normal beta negative decay so the equation will be:
This is a case o balancing a nuclear equation with the electron being identical to a negative beta particle. The electron is present beore the interaction and so appears on the let-hand side o the equation giving:
90 38
2
Sr
0 -1
e+
90 39
_
Y + 00
Under certain circumstances a nucleus can capture an electron rom the innermost shell o electrons surrounding the nucleus.
55 26
Fe +
0 -1
e
55 25
Mn
Half-life Radioactive decay is a continuous but random process there is no way o predicting which particular nucleus in a radioactive sample will decay next. However, statistically, with a large sample o nuclei it is highly probable that in a given time interval a predictable number o nuclei will decay, even i we do not know exactly which particular ones. We say that the nuclide has a constant probability of decay and this does not depend upon the size o the sample o a substance we have. Another way to look at this is in terms o half-life. The hal-lie is the time taken or hal the total number o nuclei initially in a sample to decay or or the initial activity o a sample to all by hal. This varies signifcantly rom nuclide to nuclide: the hal-lie o uranium-238 is 4.5 1 0 9 years, while that o phosphorus-30 is 2.5 minutes, and that o the artifcially produced nuclide ununoctium-294 is 5 milliseconds.
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7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y The nucleus o an atom has a diameter o the order o 1 0 1 5 m and is essentially isolated rom its surroundings , as atoms themselves are separated by distances o about 1 0 - 1 0 m. This means that the decay o a nucleus is independent o the physical state o the nuclide ( whether it is solid, liquid, or gas or in a chemical compound) and the physical conditions such as temperature and pressure. O nly nuclear interactions such as a collision with a particle in a particle accelerator can infuence the hal-lie o a nuclide. Figure 7 shows how the parent nuclei decay during a time equal to our hal- lives. D uring this time the percentage o the parent nuclide present has allen to 6.2 5 % . This graph shows an exponential decay the shape o this decay curve is common to all radioactive isotopes. The curve approaches the time axis but never intercepts it ( it is said to be asymptotic) .
percentage of parent isotope remaining
parent nuclide (red)
100
50
25 13 6 0
daughter nuclide (grey)
1
2 number of half-lives
3
4
Figure 7 Radioactive half-life.
C onsider a sample o radioactive material with N 0 undecayed nuclei at time t = 0. N In one hal- lie this would become __ 2 0
N In two hal- lives it would become __ 4 0
N In three hal- lives it would be __ 8 0
N And in our hal-lives it would become __ ( = 6.2 5 % o N 0 ) 16 0
1 4 1 Alternatively you could take hal to the power o our to give ( __ = __ 2 ) 16 or 6.2 5 %
At S L you will only be given calculations which have integral numbers o hal- lives, those o you studying HL will see how we deal with nonintegral hal- lives in Sub- topic 1 2 .2 .
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Investigate! Modelling radioactive decay D ifferent countries have varying regulations regarding the use of radioactive sources. It may be that you have the opportunity to investigate halflife experimentally, for example, using a thorium generator or by the decay of protactinium. If this is possible there is merit, as always, in performing a real experiment. If this is not possible then you should use software or dice to simulate radioactive decay.
Repeat this until you have just a few dice left.
278
100 100 100 100 400 92 88 79 88 347 84 79 71 80 314 67 59 61 70 257 52 51 56 57 216 48 41 36 50 175 40 36 33 42 151 36 33 28 31 128 32 28 24 21 105 28 20 16 20 84 23 18 15 19 75 17 15 11 17 60 13 15 10 12 50 11 37 6 10 10 33 9 6 9 9 28 7 3 9 9 26 7 3 8 8 21 4 2 7 8 19 4 1 7 7 18 4 1 7 6 14 4 0 5 5 13 4 0 4 5 10 3 0 2 5 6 0 0 2 4 6 0 0 2 4 6 0 0 2 4
Figure 8 Dice can be used to model radioactive decay.
0 7 7 6 3 7 5 4 6 6 4 3 3 3 2 3 3 3 3 3 3 3 3 2 2 2
number of dice remaining against number of throws 450 400 350 300 250
half-life ( T 1 ) is consistently 4 throws
N
N (student A)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
uncertainty
throw (t)
N (all students)
N (student D)
Throw the dice and remove all those which show one.
Make a note of the number of dice that you have removed.
N (student C)
N (student B)
It is possible to carry out a very simple modelling experiment using 1 00 dice ( but the more the merrier, if you have lots of time) .
2
200 150 T1
100
2
50
T1
T1
2
2
0 0
Figure 9 Modelling radioactive decay using a spreadsheet.
5
10
15 t
20
25
30
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
Record the remaining number o dice on a spreadsheet.
Use the spreadsheet sotware to draw a line graph ( or scatter chart) showing the number o dice remaining against the number o throws.
Draw a trend line or the points.
You will see that, with a small sample o just 1 00 dice, there will be quite a large variation rom point to point.
I several groups o students repeat this experiment you will be able to include error bars on your graph.
From the trend line you should calculate an average value or the hal-lie (see fgure 9) .
Measuring radioactive decay
Note
For beta and gamma radiation the count rate near to a source is measured using a Geiger counter. Strictly, this should be called a GeigerMller (GM) tube and counter. The GM tube is a metal cylinder flled with a low-pressure gas. At one end o the tube is a thin mica window (that allows radiation to enter the tube) . A high voltage is connected across the casing o the tube and the central electrode, as shown in fgure 1 0. The beta or gamma radiation entering the tube ionizes the gas. The ions and electrons released are drawn to the electrodes, thus producing a pulse o current that can be measured by a counting circuit. Alpha particles will be absorbed by the window o a GM tube but can be detected using a spark counter such as that shown in fgure 1 1 . A very high voltage is connected across the gauze on the top and a flament positioned a ew millimetres under the gauze. When the alpha particles ionize the air a spark jumps between the gauze and the flament.
to calculate the hal-lie, you should make at least three calculations and take the average o these. Students oten halve a value and halve that again and again; as a consequence o doing this the changes are getting very small and will not be reliable. It is good practice to calculate three hal-lives, starting rom dierent values, beore averaging as shown in fgure 9.
-
When you use a graph, in order
+
In the example shown the hal-lie
consistently takes our throws the number o throws would be equivalent to a measurement o time in a real experiment with a decaying source.
supply casing insulator central electrode thin mica window
For clarity, the error bars have
been omitted rom this graph. gas at low pressure
Figure 10 GeigerMller tube.
to counting circuit -source leads to high voltage supply gauze
Background count When a GM tube is connected to its counter and switched on it will give a reading even when a source o radioactivity is not present. The device will be measuring the background count. I this is carried out over a a series o equal, short time intervals, the count or each interval will vary. However, when measurements are taken over periods o one minute several times, you should achieve a airly constant value or your location. Radioactive material is ound everywhere. Detectable amounts o radiation occur naturally in the air, rocks and soil, water, and vegetation. The count rate varies rom place to place but the largest single source is rom the radioactive gas radon that can accumulate in homes and in the
spark
flament
Figure 11 Spark counter with alpha source.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS background radiation
radon gas rom the ground ood and drink cosmic rays
building and the artifcial ground sources
workplace. Figure 1 2 is a pie chart that shows the contribution o dierent sources o background radiation to the total. The sievert ( S v) is a radiation unit that takes the ionizing eects o dierent radiations into account. Most people absorb between 1 . 5 and 3 . 5 millisievert per year largely rom background radiation. There are places where the background dose o radiation is in excess o 3 0 mS v yr 1 . However, there is no evidence o increased cancers or other health problems arising rom these high natural levels. The sievert will not be examined on the IB D iploma Programme physics course.
Absorption of radiation medical
other sources
nuclear power and weapons test
Figure 12 Sources o background radiation.
The dierent radioactive emissions interact with materials according to their ionizing ability. Alpha particles ionize gases very strongly, have a very short range in air and are absorbed by thin paper; this is because they are relatively massive and have a charge o + 2 e. B eta particles are poorer ionizers but have a range o several centimetres in air and require a ew millimetres o aluminium to absorb them. They are much lighter than alpha particles and have a charge o 1 e. Gamma rays, being electromagnetic waves, barely interact with matter and it takes many metres o air or several centimetres o lead to be able to absorb them. The apparatus shown in fgure 1 3 is a simple arrangement or measuring the thickness o materials needed to absorb dierent types o radiation. The source is positioned opposite the window o the GM tube so that as high a reading as possible is achieved. D ierent thicknesses o materials are then inserted between the source and the GM tube until the count rate is brought back in line with the background count. When this happens, the absorption thickness o the named material is ound. The ollowing table summarizes the properties o , , and sources. The inormation represents rule o thumb values and there are exceptions to the suggested ranges.
counter
GM tube lead absorber source
280
Figure 13 Apparatus to measure absorption o nuclear radiation.
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
Emission
Composition
Range
Ionizing ability
a helium nucleus (2 protons and 2 neutrons)
low penetration, biggest mass and charge, absorbed by a few centimetres of air, skin or thin sheet of paper
very highly ionizing
high energy electrons
moderate penetration, most are absorbed by 25 cm of air, a few centimetres of body tissue or a few millimetres of metals such as aluminium
moderately highly ionizing
very high frequency electromagnetic radiation
highly penetrating, most photons are absorbed by a few cm of lead or several metres of concrete
poorly ionizing usually secondary ionization by electrons that the photons can eject from metals
few photons will be absorbed by human bodies
Worked example
a) D efne the terms (i) nuclide and (ii) hal- lie. b) C opy and complete the ollowing reaction 227 equation. 227 89 Ac 90 Th + + c)
(i) D raw a graph to show the variation with time t ( or t = 0 to t = 72 days) o the number o nuclei in a sample o thorium-2 2 7 (ii) D etermine, rom your graph, the count rate o thorium ater 3 0 days. (iii) O utline the experimental procedure to measure the count rate o thorium- 2 2 7.
Solution a)
c)
ecay
(i) count rate/counts per second
Actinium-2 2 7 ( 227 89 Ac) , decays into thorium- 2 2 7 ( 227 90 Th) . Thorium- 2 2 7 has a hal- lie o 1 8 days and undergoes - decay to the nuclide radium- 2 2 3 . O n a particular detector a sample o thorium- 2 2 7 has an initial count rate o 3 2 counts per second.
b) The proton number has increased by one so this must be negative beta decay. 227 89
Ac
227 90
Th +
0 -1
0_ 0
e+
30 25 20 15 10 5 0 0
20
40 time/days
60
80
(ii) Marked on the graph in green the activity is 1 0 units. There is likely to be a little tolerance on this type of question in an examination. (iii) You would not be expected to give great detail in this sort o question. You should include the ollowing points:
Use o a GM tube as detector.
Measuring the average background count rate in counts per second (this can be done by timing or 1 00 seconds with no source nearby) . This should be done three times, averaged and divided by 1 00 to give the counts per second.
Measuring the count rate three times with the source close to the detector.
C orrecting the count rate by subtracting the background count rate rom the count rate with the source in position.
(i) A nuclide is a nucleus with a particular number o protons and neutrons. (ii) Hal-lie is the time or the count rate to halve in value O R the time or hal the number o nuclei to decay into nuclei o another element.
35
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS
7.2 Nuclear reactions Understanding
Applications and skills
The unifed atomic mass unit Mass deect and nuclear binding energy Nuclear fssion and nuclear usion
Solving problems involving mass deect and
binding energy Solving problems involving the energy released in radioactive decay, nuclear fssion, and nuclear usion Sketching and interpreting the general shape o the curve o average binding energy per nucleon against nucleon number
Equations The massenergy relationship: E = m c 2
Nature of science Predictions about nuclides In 1869, the Russian chemist Dmitri Mendeleev took the atomic masses o chemical elements and arranged them into a periodic table. By studying his patterns, Mendeleev was able to predict missing elements; he let gaps in the table to be completed when the elements were discovered. Since the advent o the periodic table, patterns o atoms have allowed scientists to make predictions about the nuclides o which the elements are composed.
For nuclear physicists the graph o proton number against neutron number tells them whether a nuclide is likely to be unstable and, i it is, by which mechanism it will decay. The graph o nuclear binding energy per nucleon against nucleon number tells them how much energy can be generated by a particular fssion or usion reaction. These patterns have taken the guesswork out o physics and have proved to be an invaluable tool.
Patterns or stability in nuclides When a graph of the variation of the neutron number with proton number is plotted for the stable nuclei, a clear pattern is formed. This is known as the zone o f stab ility. Nuclides lying within the zone are stable, while those outside it are unstable and will spontaneously decay into a nuclide tending towards the stability zone. In this way it is possible to predict the mechanism for the decay: , - , + (or electron capture) . Nuclides having low proton numbers are most stable when the neutron proton ratio is approximately one. In moving to heavier stable nuclides the neutronproton ratio gradually increases with the heaviest stable nuclide, bismuth 2 09, having a ratio of 1 .5 2 .
282
7. 2 N U C L E A R R E A C T I O N S
Figure 1 shows this plot o neutron number against proton number. neutron number against proton number for stable nuclides 140
-emitters
neutron number (N)
120
zone of stability
100 - emitters
80 60
+ emitters or electron capture
40 20 0 0
20
40 60 80 proton number (Z)
100
Figure 1 Graph of neutron number against proton number for some stable nuclides.
Unstable nuclides lying to the let o the zone o stability are neutron rich and decay by - emission. Those nuclides to the right o the zone o stability are proton rich and decay by + emission or else by electron capture ( this is, as the name suggests, when a nucleus captures an electron and changes a proton into a neutron as a result thus increasing the neutronproton ratio) . The heaviest nuclides are alpha emitters since emission o both two protons and two neutrons reduces the neutronproton ratio and brings the overall mass down. A second pattern that is seen to aect the stability o a nucleus is whether the number o protons and neutrons are even. Almost hal the known stable nuclides have both even numbers o protons and neutrons, while only fve o the stable nuclides have odd numbers o both protons and neutrons. The elements with even numbers o protons tend to be the most abundant in the universe. The third stability pattern is when either the number o protons or the number o neutrons is equal to one o the even numbers 2 , 8, 2 0, 2 8, 5 0, 82 , or 1 2 6. Nuclides with proton numbers or neutron numbers equal to one o these magic numbers are usually stable. The nuclides, where both the proton number and the neutron number are magic numbers, are highly stable and highly abundant in the universe.
The unifed atomic mass unit The unifed atomic mass unit ( u) is a convenient unit or masses measured on an atomic scale. It is defned as one-twelth o the rest mass o an unbound atom o carbon-1 2 in its nuclear and electronic ground state, having a value o 1 .661 1 0 27 kg. C arbon was chosen because it is abundant and present in many dierent compounds hence making it useul or precise measurements. With carbon- 1 2 having six protons and six neutrons, this unit is the average
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS mass o nucleons and is, thereore, approximately equal to the mass o either a proton or a neutron. In some areas o science ( most notably chemistry) the term unifed atomic mass unit has been replaced by the term dalton ( D a) . This is an alternative name or the same unit and is gaining in popularity among scientists. For the current IB D iploma Programme Physics syllabus the term unifed atomic mass unit will be used.
Binding energy As discussed in Sub- topic 7.1 the strong nuclear orce acts between neighbouring nucleons within the nucleus. It has a very short range 1 0 1 5 m or 1 m. The stability o many nuclei provides evidence or the strong nuclear orce. In order to completely dismantle a nucleus into all o its constituent nucleons work must be done to separate the nucleons and overcome the strong nuclear orce acting between them. This work is known as the nuclear binding energy. 1 a ree proton and a ree neutron collide proton
neutron
2 the proton and neutron combine to orm a deuteron with the binding energy being carried away bey a photon deuteron photon 3 a photon o energy greater than the binding energy o the deuteron is incident on the deuteron photon
deuteron
4 the proton and neutron separate with their total kinetic energy being the diference between the photon energy and the binding energy needed to separate the proton and neutron
5 the ree proton and neutron have a greater total rest mass than the deuteron
deuteron ree proton and neutron
Figure 2 Binding energy o a deuteron.
284
S uppose we could reverse the process and construct a nucleus rom a group o individual nucleons. We would expect there to be energy released as the strong nuclear orce pulls them together. This would be equal to the nuclear binding energy needed to separate them. This implies that energy is needed to deconstruct a nucleus rom nucleons and is given out when we construct a nucleus.
Mass defect and nuclear binding energy Energy and mass are dierent aspects o the same quantity and are shown to be interchangeable through perhaps the most amous equation in physics ( Einsteins massenergy relationship) : E = m c2 where E is the energy, m the equivalent mass and c the speed o electromagnetic waves in a vacuum. This equation has huge implications or physics on an atomic scale. When work is done on a system so that its energy increases by an amount + E then its mass will increase by an amount + m given by: E m = _ c2 Alternatively when work is done by a system resulting in its energy decreasing by an amount - E then its mass will decrease by an amount - m given by: -E -m = _ c2 These relationships are universal but are only signifcant on an atomic scale. When energy is supplied to accelerate a rocket, there will be an increase in the mass o the rocket. In an exothermic chemical reaction there will be a decrease in the mass o the reactants. However, these are insignifcant amounts and can be ignored without j eopardizing the calculations. It is only with atomic and nuclear changes that the percentage mass change becomes signifcant.
7. 2 N U C L E A R R E A C T I O N S
It ollows rom Einsteins relationship that the total mass of the individual nucleons making up a nucleus must be greater than the mass of that nucleus since work needs to be done in order to break the nucleus into its component parts. This dierence is known as the mass defect which is the mass equivalent o the nuclear binding energy.
Mass and energy units for nuclear changes We have seen in the last sub- topic that the electronvolt ( eV) is commonly used as the unit o energy on the atomic and nuclear scale. Nuclear energy changes usually involve much more energy than that needed or electron energy changes and so MeV ( = eV 1 0 6 ) is the usual multiplier. In the massenergy relation E = m c 2 , when the energy is measured in MeV, the mass is oten quoted in the unit MeV c 2 . This is not an SI unit but is very convenient in that it avoids having to convert to kg when the energy is in electronvolts. O n this basis, the unifed atomic mass unit can also be written as 93 1 .5 MeV c 2 . To fnd the equivalent energy ( in MeV) we simply have to multiply by c2 !
Worked example The nuclide a)
24 11
Na decays into the stable nuclide
24 12
Mg.
into a proton and an electron with the electron being emitted as a negative beta particle) .
(i) Identiy this type o radioactive decay. (ii) Use the data below to determine the rest mass in unifed atomic mass units o the particle emitted in the decay o a sodium- 2 4 nucleus 24 1 1 Na. rest mass o
24 11
Na = 2 3 . 990 96u
rest mass o
24 12
Mg = 2 3 .985 04u
(ii) The energy released is equivalent to a mass o 5 . 002 1 60 MeV c 2 . 1 u is equivalent to 93 1 . 5 MeV c 2 S o the energy is equivalent to a mass 5 .002 1 60 o _______ = 0.005 3 7u 931 .5 The energy mass o the electron must thereore be
energy released in decay = 5 . 002 1 60 MeV
23.990 96u - (23.985 04u + 0.005 37u) = 0.000 5 5 u
b) The isotope sodium- 2 4 is radioactive but the isotope sodium- 2 3 is stable. Suggest which o these isotopes has the greater nuclear binding energy.
Solution a)
(i) This is an example o negative beta decay since the daughter product has an extra proton ( a neutron has decayed
This is consistent with the value or the mass o an electron ( given as 0. 000 5 49u in the IB Physics data booklet) . b) As sodium- 2 4 has 2 4 nucleons and sodium- 2 3 has 2 3 nucleons, the total binding energy or sodium 2 4 is going to be greater than that o sodium 2 3 .
Variation of nuclear binding energy per nucleon The nuclear binding energy o large nuclei tends to be larger than that or smaller nuclei. This is because, with a greater number o nucleons, there are more opportunities or the strong orce to act between nucleons. This means more energy is needed to dismantle a nucleus into its component nucleons. In order to compare nuclei it is usual to plot the average binding energy per nucleon o nuclides against the nucleon number as shown in fgure 3 .
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS The average binding energy per nucleon is found by dividing the total binding energy for a nucleus by the number of nucleons in the nucleus. Although there is a general pattern, with most nuclides having a binding energy of around 8 MeV per nucleon, there are wide differences from this. most stable nuclides (e.g. Fe, Ni)
10
average binding energy per nucleon (MeV)
9 8 heavier nuclides (e.g. U, Pu)
7 6 5 4 3 2
lighter nuclides (e.g. H, He)
1 0 0
50
100
150 200 nucleon number
250
300
Figure 3 Plot of average binding energy per nucleon against the nucleon number.
O n the left of the plot, those nuclides of low nucleon number, such as hydrogen-2 and helium, are less tightly bound than the more massive nuclides ( hydrogen-1 or normal hydrogen consists of a single proton and has no nuclear binding energy) . As we move to the central region of the plot we reach the maximum binding energy per nucleon with nuclides such as iron-5 8 and nickel-62 . These nuclides, having the highest nuclear binding energies per nucleon, are the most stable nuclei and therefore are abundant in the universe. Further to the right than these nuclei, the pattern reverses and the heavier nuclei are less tightly bound than lighter ones. Towards the extreme right we reach the heavy elements of uranium and plutonium in these the binding energy per nucleon is about 1 MeV less than those in the central region.
Nuclear fusion The j oining together ( or fusing) of small nuclei to give larger ones releases energy. This is because the total nuclear binding energy of the fused nuclei is larger than the sum of total nuclear binding energies of the component nuclei. The difference in the binding energies between the fusing nuclei and the nucleus produced is emitted as the kinetic energy of the fusion products. Although it doesnt actually happen in this way, it is useful to think of the energy being released as the difference between the energy emitted in constructing the fused nucleus and the energy required in deconstructing the two nuclei.
286
7. 2 N U C L E A R R E A C T I O N S
When two nuclei o masses m 1 and m 2 use to orm a nucleus o mass m 3 the masses do not add up as we might expect and m 1 + m 2 > m 3 S ince the total number o nucleons is conserved this means that m 3 has a smaller mass than the total or m 1 + m 2 . This loss o mass is emitted as the kinetic energy o the usion products in other words: E = ( ( m 1 + m 2 ) - m 3 ) c 2 Lets look at an example: A helium- 4 nucleus is composed o two protons and two neutrons. Let us consider the mass deect o helium-4 compared with its constituent particles. The mass o helium- 4 nucleus = 4. 002 602 u The mass o 2 ( individual) protons = 2 1 . 007 2 76u = 2 .01 4 5 5 2 u The mass o 2 ( individual) neutrons = 2 1 . 008 665 u = 2 .01 7 3 3 u The total mass o all the individual nucleons = 4.03 1 882 u S o the mass deect is 4.031 882 u 4. 002 602 u = 0.02 9 2 8u This is equivalent to an increased binding energy o 0.02 9 2 8 93 1 .5 MeV or 2 7.3 MeV this energy is given out when the nucleus is ormed rom the individual nucleons. You can probably see that the potential or generating energy is immense, but being able to produce the conditions or usion anywhere except in a star is a signifcant problem. Initially, the repulsion between the protons means that energy must be supplied to the system in order to allow the strong nuclear orce to do its work. In reality, j oining together two protons and two neutrons is not a simple task to achieve on Earth. It seems that Earth- based usion reactions are more likely to occur between the nuclei o deuterium ( hydrogen- 2 ) and tritium ( hydrogen-3 ) .
Nuclear fssion The concept o nuclear fssion is not quite as straight-orward as usion although fssion has been used practically or over 70 years. I we are able to take a large nucleus and split it into two smaller ones, the binding energy per nucleon will increase as we move rom the right-hand side to the centre o fgure 3 . This means that energy must be given out in the orm o the kinetic energy o the fssion products. S o, in nuclear fssion, the energy released is equivalent to the dierence between the energy needed to deconstruct a large nucleus and that emitted when two smaller nuclei are constructed rom its components. In terms o masses, the total mass o the two smaller nuclei will be less than that o the parent nucleus and the dierence is emitted as the kinetic energy o the fssion products. S o we see that as m 1 + m 2 < m 3 then E = ( m 3 - ( m 1 + m 2 ) ) c 2 There are some nuclei that undergo fssion spontaneously but these are largely nuclides o very high nucleon number made synthetically. Uranium- 2 3 5 and uranium- 2 38 do undergo spontaneous fssion but this has a low probability o occurring. Uranium- 2 3 6 however is not a common naturally occurring isotope o uranium but it does undergo
TOK The use o fssion and usion In the last twenty years there has been a development in techniques to extract ossil uels rom locations that were not either physically or fnancially viable previously. Given the debate relating to nuclear energy, which is the more moral stance to take: to use nuclear uel with its inherent dangers or to risk damage to landscapes and habitats when extracting ossil uels?
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS spontaneous fssion. Uranium-2 3 6 can be produced rom uranium-2 3 5 when it absorbs a low energy neutron. Uranium-2 3 6 then undergoes spontaneous fssion to split into two lighter nuclides and at the same time emits two or three urther neutrons: 235 92
U + 10 n
236 92
U
1 41 56
Ba +
92 36
Kr + 3 10 n
These isotopes o barium and krypton are j ust one o several possible pairs o fssion products o uranium- 2 3 6.
Nature of science 92 36
Fission chain reaction
Kr n n
n 235 92 92 36
Kr
141 56
U
92 36
n
Ba
Kr n
neutron
n
n
n 235 92
U
n 141 56
235 92
141 56
U
Ba
Ba 92 36
Kr n
n n 235 92
141 56
U
Ba
Figure 4 Uranium chain reaction.
The use o nuclear fssion in nuclear power stations is discussed in Topic 8. O ne o the key aspects o a continuous power production plant is that the nuclear uel is able to fssion in a controlled chain reaction. The neutrons produced in the reaction above must be capable
o producing more nuclear fssion events by encountering urther uranium nuclei. When there is sufcient mass o uranium- 2 3 5 the reaction becomes sel- sustaining, producing a great deal o kinetic energy that can be transormed into electricity in the power station.
Worked example a) C ompare the process o nuclear fssion with nuclear usion. b) Helium-4 ( 42 He) and a neutron are the products o a nuclear usion reaction between deuterium ( 21 H) and tritium ( 31 H) . 2 1
288
3 1
4 2
1 0
The masses o these nuclides are as ollows: 2 1
H
2.014 102u
3 1
H
3.016 050u
He
4.002 603u
4 2
H + H He + n + energy Show that the energy liberated in each reaction is approximately 2 .8 1 0 - 1 2 J.
7. 2 N U C L E A R R E A C T I O N S
Solution a) Nuclear usion involves the joining together o light nuclei while nuclear fssion involves the splitting up o a heavy nucleus. In each case the total nuclear binding energy o the product(s) is greater than that o the initial nuclei or nucleus. The dierence in binding energy is emitted as the kinetic energy o the product(s) . In relation to the plot o nuclear binding energy per nucleon against nucleon number, fssion moves nuclei rom the ar right towards the centre whereas usion moves nuclei rom the ar let towards the centre both processes involve a move up the slopes towards higher values. b) To simpliy the calculation you should break it down into several steps. Remember to include all o the steps with so many signifcant fgures, short cuts could cost you marks in an exam. Total mass on let-hand side o equation = 2 .01 4 1 02 u + 3 .01 6 05 0u = 5 .03 0 1 5 2 u
Looking up the mass o the neutron in the data booklet ( = 1 .008 665 u) Total mass on right- hand side o equation = 4.002 603 u + 1 .008 665 u = 5 .01 1 2 68u Thus there is a loss o mass on the right-hand side ( as the binding energy o the helium nucleus is higher than that o the deuterium and tritium nuclei) Mass dierence ( m) = 5 .03 0 1 5 2 u 5 .01 1 2 68u = 0.01 8 884u As u = 93 1 .5 MeV c 2 , m = 0.01 8 884 93 1 .5 MeV c 2 = 1 7.5 9 MeV c 2 m E = ___ = 1 7.5 9 MeV c2
= 1 7.5 9 1 0 6 1 .60 1 0 - 1 9 J E = 2 .81 1 0 - 1 2 J
Nature of science Alternative nuclear binding energy plot 0
average binding energy per nucleon (MeV)
-1
50
100
nucleon number 150 200
250
300
light nuclides: H, He, etc.
-2 -3 -4 -5 -6 -7 -8 -9
- 10
0
heavy nuclides: U, Pu, etc. most stable nuclides: Fe, Ni, etc.
It is quite usual to see the binding energy plot, shown in fgure 3 , drawn upside down as shown in fgure 5 . The reason or plotting this inverted graph is that when the nucleons are at infnite separation they have no mutual potential energy but, because the strong nuclear orce attracts them, they lose energy as they become closer meaning that their potential energy becomes negative. This implies that the more stable nuclei are in a potential valley. Fission and usion changes that bring about stability, will always move to lower ( more negative) energies. This convention is consistent with gravitational and electrical potential energies which are discussed in Topic 1 0. In line with most text books at IB D iploma level we will continue to use the plot shown in fgure 3 .
Figure 5 Alternative plot of binding energy per nucleon against the nucleon number.
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7.3 The structure of matter Understanding Quarks, leptons, and their antiparticles
Exchange particles
Hadrons, baryons, and mesons
Feynman diagrams
The conservation laws o charge, baryon number,
Confnement
lepton number, and strangeness The nature and range o the strong nuclear orce, weak nuclear orce, and electromagnetic orce
The Higgs boson
Applications and skills Describing the RutherordGeigerMarsden
Describing the mediation o the undamental
experiment that led to the discovery o the nucleus Applying conservation laws in particle reactions Describing protons and neutrons in terms o quarks Comparing the interaction strengths o the undamental orces, including gravity
orces through exchange particles Sketching and interpreting simple Feynman diagrams Describing why ree quarks are not observed
Particle properties Charge
Quarks
___ 2 e
u
c
t
- ___13 e
d
s
b
3
Baryon number
Charge
___ 1
-1
e
0
e
3 ___ 1 3
All quarks have a strangeness number o 0 except the strange quark that has a strangeness number o 1 Gravitational Weak Particles All Quarks, leptons experiencing Particles Graviton W+ , W - , Z 0 mediating
Leptons
All leptons have a lepton number o 1 and antileptons have a lepton number o 1 Electromagnetic Strong Charged
Quarks, gluons
Gluons
Nature of science Symmetry and physics Symmetry has played a major part in the development o particle physics. Mathematical symmetry has been responsible or the prediction o particles. By searching the bubble chamber tracks produced by cosmic rays or generated by particle accelerators many o the predicted 290
particles have been ound. Theoretical patterns have been very useul in developing our current understanding o particle physics and, increasingly, experiments are conirming that these patterns are valid even though their proound signiication is yet to be determined.
7. 3 T H E S T R U C T U R E O F M AT T E R
Introduction At the end o the nineteenth century, physicists experimented with electrical discharges through gases at low pressure (see fgure 1 ) . to vacuum pump fuorescence
cathode rays cathode + anode high voltage
Figure 1 Discharge tube.
In 1 86 9 , the German physicist, Johann Hittor, observed a glow coming rom the end o a discharge tube, opposite the cathode. He suggested that radiation was being emitted rom the cathode and this caused the tube to uoresce. The radiation causing the eect was later called a beam o cathode rays . E ight years later the B ritish physicist, S ir Joseph Thomson, discovered that cathode rays could be deected by both electric and magnetic felds. Thomsons experiment showed that cathode rays were charged. Ater urther experiments with hydrogen gas in the tube, Thomson concluded that cathode rays were beams o particles coming rom atoms. With atoms being neutral, Thompson deduced that the atom must also carry positive charge. Figure 2 shows Thomsons plum pudding or current bun model o the atom, consisting o a number o electrons buried in a cloud o positive charge.
+ -
Although Thomsons model was short-lived, it was the frst direct evidence that atoms have structure and are not the most elementary building blocks o matter as had been previously thought.
The scattering of alpha particles In 1 909, the German, Johannes Geiger, and the E nglishNew Zealander, E rnest Marsden, were studying at Manchester University in E ngland. E rnest Rutherord ( another New Zealander) was supervising their research. The students were investigating the scattering o alpha particles by a thin gold oil. They used a microscope in a darkened room to detect ashes o light emitted when a alpha particles collided with a zinc sulfde screen surrounding the apparatus. Geiger and Marsden were expecting the alpha particles to be deected by a very small amount as a result o the electrostatic eects o the charges in the atom. At Rutherords suggestion they moved their detector to the same side o the oil as the alpha source and were astonished to fnd that approximately one in every eight thousand alpha particles appeared to be reected ( or back- scattered ) by the thin oil. With the alpha particles travelling at about 3 % o the speed o light, reection by electrons surrounded by a cloud o positive charge was unthinkable and Rutherord described the eect as:
+ -
+
+
- +
-
-
+
+
-
+
+ -
spherical cloud o positive charge
+
-
-
+
Figure 2 Thomsons plum pudding model o the atom. scattered particles
beam o particles
source o particles
electron
most particles are undefected
thin gold oil
circular fuorescent screen
Figure 3 The RutherordGeigerMarsden apparatus.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS ... quite the most incredible event that has ever happened to me in my lie. It was almost as incredible as i you fred a 1 5-inch shell at a piece o tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result o a single collision, and when I made calculations I saw that it was impossible to get anything o that order o magnitude unless you took a system in which the greater part o the mass o the atom was concentrated in a minute nucleus. It was then that I had the idea o an atom with a minute massive centre, carrying a charge. Rutherford proposed that, as the alpha particles carry a positive charge, back-scattering could only occur if the massive part of the atom was also positively charged ( for the mechanics of this see S ub-topic 2 .4) . Thus the atom would consist of a small, dense positive nucleus with electrons well outside the nucleus. Rutherfords calculations showed that the diameter of the nucleus would be of the order of 1 0 1 5 m while that of the atom as 10 a whole would be of the order of 1 0 m; this meant that the atom was almost entirely empty space.
TOK Physicists and knowing
As will be discussed in Topic 1 2, for electrons to be able to occupy the orbits required by later models of the atom, a theory that contradicts classical physics is required. The electrons that are accelerating because of their circular motion should emit electromagnetic radiation and spiral into the nucleus. The consequence of this would be that all atoms should collapse to the size of the nucleus and matter could not exist. Since it is obvious that matter does exist, it is clear that Rutherfords model is not the whole story.
Now I know what the atom looks like. a quote rom Sir Ernest Rutherord, ater publishing the results o his alpha scattering experiment. When scientists devote time to investigating a specifc area o research, they are oten keen to publish their fndings. Although peer appraisal is an important aspect o research there have been many instances in which a scientist adheres to one chosen model without being objective about the work o others. Many scientists pride themselves on their objectivity, but is it possible to be truly objective when those who are deemed to be successul are rewarded with prestigious awards and unding?
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electron cloud
alpha particles
nucleus
Figure 4 Paths of several alpha particles in the scattering experiment.
The particle explosion In 1 928, the B ritish physicist, Paul Dirac, predicted an antiparticle of the electron (the positron) . Particles and their associated antiparticles have
7. 3 T H E S T R U C T U R E O F M AT T E R
identical rest mass but have reversed charges, spins, baryon numbers, lepton numbers, and strangeness (see later or these) . The positron was discovered in cosmic rays by the American, C arl Anderson, in 1 932. C osmic rays consist mainly o high-energy protons and atomic nuclei ejected rom the supernovae o massive stars. Anderson used a strong magnetic feld to deect the particles and ound that their tracks were identical to those produced by electrons, but they curved in the opposite direction.
non-ionizing photon (leaving no track)
electron track positron track
When an electron collides with a positron the two particles annihilate and their total mass is converted into a pair o photons o identical energy emitted at right angles to each other. The inverse o this process is called p air p roduction. This is when a photon interacts with a nucleus and produces a particle and its antiparticle; or this event to occur the photon must have a minimum energy equal to the total rest mass o the particle and the antiparticle. 1 93 2 was a signifcant year in the development o our understanding o the atom with other developments in addition to the positron discovery. In 1 93 0, B othe and B ecker, working in Germany, had ound that an unknown particle was ej ected when beryllium is bombarded by alpha particles. James C hadwick at C ambridge University proved experimentally that this particle was what we now know to be the neutron.
scattered atomic electron track more energetic electron-positron pair track
Figure 5 Electron and positron tracks in a cloud chamber.
In 1 93 6, Anderson and his Ph. D . student S eth Neddermeyer at the C aliornia Institute o Technology went on to discover the muon, using cloud chamber measurements o cosmic rays. Although at the time they believed this particle to be the theoretically predicted pion, this was the start o the recognition that there were more particles than protons, neutrons, electrons and positrons. The understanding that electrons had antiparticles suggested to D irac that the same should be true or protons and, in 1 95 5 , the antiproton was discovered at the University o C aliornia, B erkeley by Emilio S egr and O wen C hamberlain. In their experiment protons were accelerated to an energy o approximately 6 MeV beore colliding with urther protons in a stationary target. The reaction is summarized as: _
p+ p p+ p+p+p
_
Here p represents a proton and p an antiproton. The kinetic energy o the colliding protons ( let- hand side o the equation) is sufcient to produce a urther proton and an antiproton using E = m c 2 . We have encountered the neutrino and antineutrino in beta decay. According to the standard B ig B ang model, these particles are thought to be the most numerous in the universe. The electron neutrino was frst proposed in 1 93 0 by the German physicist, Wolgang Pauli, as an explanation o why electrons emitted in beta decay did not have quantized energies as is the case with alpha particles. Pauli suggested that a urther particle additional to the electron should be emitted in beta decay so that energy and momentum are conserved. It was the Italian, Enrico Fermi, who named the particle little neutral one or neutrino and urther developed Paulis theory. As the neutrino has no charge and almost no mass it interacts only minimally with matter and was not discovered experimentally until 1 95 6.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS B y the end o the 1 960s over 3 00 particles had been discovered and physicists were starting to classiy them. A ull theory explaining the structure o matter and the nature o the orces that hold particles together has become a major goal o scientifc research. The current theory is that the universe is made up o a relatively small number o undamental particles.
Classifcation o particles the standard model The S tandard Model came about through a combination o experimental discoveries and theoretical developments. Although ormulated towards the end o the twentieth century, it is supported by the experimental discoveries o the bottom quark in 1 977, the top quark in 1 995 , the tau particle in 2 001 , and the Higgs boson in 2 01 2 . The S tandard Model has been described as being the theory o ( almost) everything. O n the whole the model has been very successul, but it ails to ully incorporate relativistic gravitation or to predict the accelerating expansion o the universe. The model suggests that the only undamental particles are leptons, quarks, and gauge bosons. All other particles are believed to consist o combinations o quarks and antiquarks.
Leptons The leptons are members o the electron amily and consist o the electron ( e - ) , the muon ( ) , the tau ( ) , their antiparticles plus three neutrinos associated with each o the particles and three neutrinos associated with the antiparticles. Electrons, muons, and taus are all negatively charged and their antiparticles are positively charged. Neutrinos and antineutrinos are electrically uncharged. 1 The electron is known to have a mass o about ____ th o the mass o a 1 800 proton, making it a very light particle. The muon is also light, having a mass o about 2 00 times that o the electron. The tau is heavier and has a mass similar to that o a proton. For reasons that will be shown later, each lepton is given a lepton number. Leptons have lepton number + 1 and antileptons 1 .
Particle
Leptons e _
_
_
Antiparticles Neutrinos
e e
Antineutrinos
_
_
_
e
Charge/e 1
Lepton number (L) +1
+1 0
1 +1
0
1
Quarks S peculation about the existence o quarks began ater scattering experiments were perormed with accelerated electrons at the S tanord Linear Accelerator C enter ( S LAC ) at S tanord University in the US A between 1 967 and 1 973 . The electrons were accelerated up to energies o 6 GeV beore colliding with nuclei. In a similar way to alpha scattering, some o the electrons were scattered through large angles by the nucleons which suggested that the nucleons are not o uniorm density but have discrete charges within them. These experiments
294
7. 3 T H E S T R U C T U R E O F M AT T E R supported the theories put orward independently by the Russian American physicist George Zweig and the American Murray Gell-Mann in 1 964. Gell-Mann proposed that the charges within nucleons were grouped in threes and coined the term quarks ( which he pronounced qworks) in deerence to a quote rom Finnegans Wake by the Irish novelist James Joyce: Three quarks for Muster Mark! Sure he hasn't got much of a bark. And sure any he has it's all beside the mark. Zweig reerred to quarks as aces and believed ( incorrectly, as it turned out) there were our o them as the aces in a pack o cards. As with leptons there are six quarks and six antiquarks. The quarks are labelled by their avour which has no physical signifcance apart rom identiying the quark. These avours are called up (u) , down (d) , strange (s) , charm (c) , bottom (b) , and top (t) . Quarks each carry a charge o 2 1 2 either + __ e or - __ e, and antiquarks each carry a charge o either - __ e 3 3 3 1 __ or + 3 e. O ten the e is omitted rom charges and they are written as relative 1 2 charge + __ , - __ , etc. 3 3 These quarks are split into three generations o increasing mass. The frst generation contains the up and down quarks, which are the lightest quarks. The second contains the strange and charm quarks, and the third the bottom and top quarks the heaviest quarks. The up, down, and strange quarks were the frst to be discovered, with the up and down quarks combining to orm nucleons.
Quarks with charge - ___1 e
u c t
d s b
3
3
__ _ _
Antiquarks carry the opposite charge and are denoted by u , d , c , etc.
Quark confnement It is thought that quarks never exist on their own but exist in groups within hadrons. Hadrons are ormed rom a combination o two or three quarks ( called mesons and baryons) this is known as quark confnement. The theory that explains quark confnement is known as quantum chromodynamics ( QC D ) something that is not included in the IB D iploma Programme Physics syllabus. To hold the quarks in place they exchange gluons ( the exchange particle or the strong nuclear orce see later in this sub- topic) . Moving a quark away rom its neighbours in the baryon or meson stores more energy in the interaction between the quarks and thereore requires increasing amounts o energy to increase their separation. I more and more energy is ed into the system, instead o breaking the orce between quarks and separating them ( which is what classical physics might suggest) , more quarks are produced this leaves the original quarks unchanged but creates a new meson or baryon by the massenergy relationship E = m c 2 .
u
u
d proton u d u anti-proton
d
u
d neutron u
s
u d
u
+
0 c
d
c
s
d K0
lambda
J/
Figure 6 Three quark and two quark hadrons.
Baryons qqq and antibaryons qqq These are a few of the many types of baryons. Symbol
Name
p p n
proton antiproton neutron lambda omega
V
Quarks with charge + ___2 e
u
-
Quark content uud uud udd uds sss
Electric charge +1 -1 0 0 -1
Mesons qq These are a few of the many types of mesons. Symbol
Name
+ K+ B0 c
pion kaon rho B-zero eta-c
Quark content ud su ud db cc
Electric charge +1 -1 +1 0 0
Figure 7 Baryons and mesons.
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Hadrons
strangeness n
0
-
-1
strangeness
0 0
-1
K0
+1 -
+1
+
K-
K0 -1
In order to explain which particles can exist and to explain the outcome o observed interactions between particles, the quarks are assigned properties described by a numerical value. The quark is given a baryon 1 . number ( B ) o 13 and or an antiquark the baryon number is __ 3
__
K+
0
charge
+
charge
-1
Hadrons are particles composed o quarks and include baryons (which are made up o three quarks) or mesons (which comprise o quark-antiquark pairs) . The strong interaction (see later) acts on all hadrons but not on leptons while the weak interaction acts on both leptons and hadrons. Some particle physicists have hypothesized pentaquarks consisting o our quarks and one antiquarks although experiments at the start o the century looked promising, these have yet to be confrmed experimentally.
0
-2
0
p
0
+1
Figure 8 Example patterns for some baryons and mesons (The Eightfold Way) .
S trangeness ( S) is a property that was initially defned to explain the behaviour o massive particles such as kaons and hyperons. These particles are created in pairs in collisions and were thought to be strange because they have a surprisingly long lietime o 1 0 1 0 s instead o the expected 1 0 23 s. A strange quark has a strangeness o 1 , and a strange antiquark has a strangeness o + 1 . This property o strangeness is conserved when strange particles are created but it is not conserved when they subsequently decay. Figure 8 shows eight baryons each with a baryon number +1 and eight mesons each with baryon number 0. Gell-Mann used these diagrams as a means o organizing baryons and mesons. Particles on the same horizontal level have the same strangeness, while those on the same diagonal have the same charge. Using this system, Gell-Mann predicted the eta () particle in 1 961 , which was discovered experimentally a ew months later. The name Eightold Way reers to Noble Eightold Path a way towards enlightenment in B uddhism.
Nature of science Symmetry and physics In the same way that the periodic table gives the pattern of the elements based on their electron structure, the Standard Model gives the pattern for the fundamental particles in nature. It is believed that there are six leptons and six quarks and these occur in pairs. A further symmetry is seen by each particle having its own antiparticle (which will have the same rest mass as the particle but other properties such as charge and spin are reversed) . When an electron collides with its antiparticle the positron, the two particles annihilate producing pairs of photons that travel in opposite directions (thereby conserving momentum and energy) . It is not possible for annihilating particles to produce a single photon.
296
Figure 9 Particle collisions shown in a bubble chamber.
7. 3 T H E S T R U C T U R E O F M AT T E R
Examples of baryons Protons and neutrons are important in atoms and you should know their quark composition and be able to work out that of their antiparticles. S ince they are both baryons, they each consist of three quarks and have a baryon number + 1 . The proton consists of two up quarks and one down quark ( uud) , which means it has charge
( + _23 + _23 - _13 ) = + 1 The neutron consists of two down quarks and one up quark ( ddu) , which gives it a charge of 1 - _ 1 + _ 2 = 0 -_ 3 3 3
(
)
An antiproton has a baryon number of 1 and consists of two __ ___ antiup quarks and one antidown quark ( u u d ) . This gives a charge 2 1 2 of ( - __ - __ + __ = -1 3 3 ) 3 You may wish to prove to yourself that the antineutron has a charge of zero ( and baryon number of 1 ) . Since none of these particles has any strange quarks their strangeness is 0.
Examples of mesons In questions you will always be given the quark composition of mesons, so there is no need to try to remember these or the composition of any baryons apart from the proton and neutron. A + meson is also called _ a positive pion and consists of an up quark and 2 1 an antidown quark ( u d ) . The positive pion has charge ( + __ + __ = +1 . 3 ) 3 As it is not a baryon, its baryon number is 0 and, as it has no strange quarks, its strangeness is 0 too. A K + meson or positive kaon is the lightest strange meson and consists of _ an up quark and an antistrange quark ( u s ) . The positive kaon has charge ( + __23 + __13 ) = + 1 . Again, a meson is not a baryon and so it has baryon number 0. However, this particle does contain an antistrange quark and so it has strangeness of + 1 .
Conservation rules When considering interactions between particles or their decay, the equivalence of mass and energy must be taken into account. Mass may become some form of energy and vice versa using the equation E = m c 2 . In addition to the expected conservation of massenergy and momentum, no interaction that disobeys the conservation of charge has ever been observed; the same is true for baryon number ( B ) and lepton number ( L) . All lep tons have a lep ton number of + 1 and antilep tons have a lep ton number of 1 . As mentioned before, the conservation rules for strangeness are different and we will consider them later. You may be asked to decide whether an interaction or decay is feasible on the basis of the conservation rules.
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Worked examples 1
a) Show that, when a proton collides with a __ negative pion ( u d) , the collision products can be a neutron and an uncharged pion.
This suggests that the neutral pion is very __ short lived since the combination u u would mutually annihilate. In act this particle has a lietime o about 8 1 0 - 1 7 s and annihilates to orm two gamma ray photons or, very occasionally, a gamma ray photon, an electron and a positron.
b) D educe the quark composition o the uncharged pion.
Solution a) The equation or the interaction is p + - n + 0 Q: + 1 1 0 + 0
2
Explain whether a collision between two protons could produce two protons and a neutron.
Solution
B: +1 + 0 +1 + 0
Writing the equation or the baryons:
L: 0 + 0 0 + 0 This interaction is possible on the basis o conservation o charge, baryon number and lepton number. b) Writing the equation in terms o quarks: __
uud + u d ddu + ?? __
?? = u u in order to balance this equation.
p+ p p+ p+ n Q: + 1 + 1 + 1 + 1 + 0 L: 0 + 0 0 + 0 + 0 B: +1 + 1 +1 + 1 + 1 S o this interaction ails on the basis o baryon number.
Fundamental forces C urrent theories suggest that there are j ust our undamental orces in nature. In the very early universe, when the temperatures were very high, it is possible that at least three o these our orces originated as a single unifed orce. The our undamental orces are:
The gravitational force is weak, has an infnite range and acts on all particles. It is always attractive and over astronomic distances it is the dominant orce on an atomic or sub- atomic scale it is negligible.
The electromagnetic force causes electric and magnetic eects such as the orces between electrical charges or bar magnets. Like gravity, the electromagnetic orce has an infnite range but it is much stronger at short distances, holding atoms and molecules together. It can be attractive or repulsive and acts between all charged particles.
The strong nuclear force or strong interaction is very strong, but has very short- range. It acts only over ranges o 1 0 1 5 m and acts between hadrons but not leptons. At this range the orce is attractive but it becomes strongly repulsive at distances any smaller than this.
The weak nuclear force or weak interaction is responsible or radioactive decay and neutrino interactions. Without the weak interaction stars could not undergo usion and heavy nuclei could not be built up. It acts only over very short ranges o 1 0 1 8 m and acts between all particles.
As these our undamental orces have dierent ranges it is impossible to generalize their relative strengths or all situations. The table below shows a
298
7. 3 T H E S T R U C T U R E O F M AT T E R comparison o the eects that the our orces have on a pair o protons in a nucleus (since all our orces will act on protons at that range) .
Force
Range
Relative strength
Gravitational
1
binding planets, solar system, sun, stars, galaxies, clusters o galaxies
Weak nuclear
10 1 8 m
10 24
(W+ , W- ) : transmutation o elements (Wo ) : breaking up o stars (supernovae)
10 3 5
binding atoms, creation o magnetic felds
10 1 5 m
10 3 7
binding atomic nuclei, usion processes in stars
Electromagnetic Strong nuclear
Roles played by these forces in the universe
Exchange particles A very successul model that has been used to explain the mechanism o the undamental orces was suggested by the Japanese physicist, Hideki Yukawa, in 1 93 5 . Yukawa proposed that the orce between a pair o particles is mediated ( or transmitted) by particles called gauge bosons. The our undamental orces have dierent ranges and a dierent boson is responsible or each orce. The mass o the boson establishes the range o the orce. The bosons carry the orce between particles. Figure 1 0 shows how an electron can exchange a photon with a neighbouring electron, leading to electromagnetic repulsion. The exchange particle is said to be a virtual particle because it is not detected during the exchange. The exchange particle cannot be detected during its transer between the particles because detection would mean that it would no longer be acting as the mediator o the orce between the particles. The larger the rest mass o the exchange particle is, the lower the time it can be in fight without it being detected and, thereore, the lower the range o the orce. A
(a)
(b)
A
B
B
-
-
A
B
-
-
A
B
-
-
A
B
virtual photon exchanged
electrons separate
Figure 10 Exchange of virtual photon between two electrons.
A
A
electrons A and B approach
B
B
Figure 11 Analogy of exchange particles.
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Nature of science Feynman diagrams and cloud chamber tracks Dont be confused by thinking that Feynman diagrams are like the pictures of cloud or bubble chamber interactions they are not and they really should not be thought of as spacetime diagrams! The diagrams help theoretical physicists work out the probability of interactions occurring, but for the IB course we treat them as illustrations of the interactions between complex particle events.
electrons repel
time
e
-
e-
e-
efrst incident electron
virtual photon
position
second incident electron
Figure 12 Feynman diagram o the electromagnetic orce between two electrons.
time
p
Exchange particle
Acts on
Gravitational
gravitons (undiscovered)
all particles
Weak nuclear
W + , W and Z 0 bosons
quarks and leptons
Electromagnetic
photons
electrically charged particles
Strong nuclear
gluons (and mesons)
quarks and gluons (and hadrons)
Feynman diagrams These are graphical visualizations, developed by the American physicist Richard Feynman, that represent interactions between particles. These diagrams are sometimes known as spacetime diagrams. They have the time axis going upwards and the space or position axis to the right ( although many particle physicists draw these axes with space going upwards and time to the right so be careful when you are researching interactions) . S traight lines represent particles and upwards arrows show particles moving orwards in time ( downward arrows indicate an antiparticle also moving orwards in time) . Wavy or broken lines that have no arrows represent exchange particles. Points at which lines come together are called vertices ( plural o vertex) and, at each vertex, conservation o charge, lepton number and baryon number must be applied.
The electromagnetic force Figure 1 2 shows a Feynman diagram or the electromagnetic orce between two electrons. The exchange particle that gives rise to the orce is the photon. Photons have no mass and this equates to the orce having an infnite range.
n
The strong force
neutral pion ( 0 ) is exchanged between a proton (p) and a neutron (n) mediating the strong nuclear orce between these particles in the nucleus
Figure 13 Feynman diagram o the strong orce between a proton and a neutron.
300
Force
This diagram shows two electrons moving closer and interacting by the exchange o a virtual photon beore moving apart.
position
The table shows the exchange particles or the our undamental orces.
n 0
p
The simplest analogy to explain how a repulsive orce can be produced by transer o a particle is to picture what happens when a heavy ball is thrown backwards and orwards between people in two boats (see fgure 1 1 ) . The momentum changes as the ball is thrown and caught and to someone who cannot seen the ball travelling, a repulsive orce seems to exist as they move apart. In order to explain the attractive orce we need to imagine that a boomerang is being thrown between the people in the boats in this case the change o momentum brings the boats together. To understand this more ully requires the use o the uncertainty principle discussed in Topic 1 2.
This is the strongest o the orces and acts between quarks and, thereore, between nucleons. The exchange particles responsible are pions ( + , or 0 ) . Figure 1 3 shows the Feynman diagram or the strong orce between a proton and a neutron. In this case a neutral pion is exchanged between the proton and the neutron that ties them together. In hadrons, the pion carries gluons between the quarks the gluons are the exchange particles or the colour orce acting between quarks ( colour is not included on the IB D ip loma Programme Physics syllabus and so questions will be limited to the exchange of mesons between hadrons) .
7. 3 T H E S T R U C T U R E O F M AT T E R
Weak nuclear orce
ep time
This is responsible for radioactive decay by beta emission. In negative beta decay a neutron decays to a proton. In this process a W boson is exchanged as a quark changes from down to up. The W boson then immediately decays into an electron and an electron antineutrino.
e
n position
The W and Z particles responsible for weak interactions are massive. The weak interaction is the only mechanism by which a quark can change into another quark, or a lepton into another lepton.
A neutron decays into a proton by the emission of an electron and an electron antineutrino (shown with the arrow moving downwards because it is an antiparticle) . The decay is mediated by the negative W boson.
Conservation o strangeness We are now in a position to discuss conservation of strangeness. The strange quark has a strangeness of 1 and particles containing a single strange quark will also have a strangeness of 1 . The antistrange quark has a strangeness of + 1 and particles containing a single antistrange quark will also have a strangeness of + 1 . One type of K 0 (neutral kaon) has a strangeness of + 1 and so it contains an antistrange quark. A particle containing two strange quarks would have a strangeness of 2 and one containing two antistrange quarks would have a strangeness of + 2 , etc.
W-
Figure 14 Feynman diagrams for negative beta decay.
Note The shorter the range o the exchange orce the more massive the exchange particle so the exchange particles or gravitation and the electromagnetic interaction (both o infnite range) must have zero rest masses. The weak interaction will have the heaviest boson because it range is the shortest; the strong interaction has an exchange particle o intermediate mass.
S trangeness is not conserved when strange p articles decay through the weak interaction. For example, strangeness is not conserved when a strange quark decays into an up quark. S trangeness is conserved when there is a strong interaction. This is why strange particles are always produced in pairs. If two particles interact to produce a strange particle then a strange antiparticle must also appear. Figure 1 5 shows a stationary proton interacting through the strong interaction with a negative pion ( short green line) at the bottom right0 hand side of the image. These particles create a neutral kaon ( K ) and a second neutral particle called a lambda particle ( 0 ) . As these particles are neutral they produce no tracks in a bubble chamber ( a device that forms trails of bubbles along the path of a charged particle) . The K 0 track is shown as a purple broken line and the 0 as a blue broken line. The reaction is given by: p + - K0 + 0 There are no strange particles on the left-hand side of this equation. The K 0 has a strangeness of + 1 so the 0 must have a strangeness of 1 for the reaction to be viable by the strong interaction. In terms of quarks the equation for the reaction is _
__
uud + u d d s + uds Lets check this to see that charge, lepton number, baryon number, and strangeness are conserved: 2 1 1 2 2 Q: left- hand side ( + __ + __ - __ + ( - __ - __ = 0 and 3 3 ) 3 ) 3 3
right hand side
(
1 - __ 3
+
1 __ 3
)+(
2 + __ 3
-
1 __ 3
-
1 __ 3
Figure 15 Proton-pion interaction.
)=0
As 0 = 0, charge is conserved. L: as none of the particles are leptons, the lepton numbers on both sides are zero meaning lepton number is conserved.
301
7
ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS 5 1 1 1 1 1 B : left-hand side ( + __ + __ + __ + ( + __ + __ = __ and 3 3 ) 3 3 3 3 ) 5 1 1 1 1 1 right hand side ( + __ + __ + ( + __ + __ + __ = __ 3 ) 3 3 ) 3 3 3 5 5 As __ = __ , baryon number is conserved ( as it will always be if there are 3 3 the same number of quarks on both sides of the equation.
S: there are no strange quarks on the left-hand side so the strangeness = 0; on the right- hand side there is one strange and one antistrange quark so the total strangeness = - 1 + 1 = 0. S o strangeness is also conserved. With all four quantities conserved the interaction is viable.
You should check each of these equations to show that charge, baryon number, and lepton number are conserved. Strangeness should not be conserved because each of the two interactions involves the weak nuclear force.
The kaon subsequently decays into positive and negative pions and the lambda particle decays into a negative pion and a proton. As each of these decays is a weak interaction strangeness is not conserved. These reactions are given by _K 0 + + _ __ or in terms of quarks d s u d + u d and 0 - + p __ or in terms of quarks uds u d + uud
Worked example D raw Feynman diagrams to show the following interaction:
In this case the proton and the electron collide and produce a neutron and an electron neutrino. This interaction is mediated by the W boson.
a) positive beta ( positron) decay: p n + e + + e b) protonelectron collision: p + e - n + e
c) n
c) the two types of neutronelectron neutrino collision: n + e e + n and n + e p + e -
time
e
Solution a) time
W
+
position e p
p time
eposition
b)
time
n
e We-
p position
302
W
+
n
e
Figure 16 Feynman diagram for positive beta (positron) decay. In this decay the proton decays into a neutron and emits a positron and electron neutrino. The decay is mediated by the positive W boson ( W + ) .
n
e
e+ n
Z0
Figure 17 Feynman diagram for an protonelectron collision.
position
Figure 18 Feynman diagrams for the two types of neutronelectron neutrino collision. The most likely collision between a neutron and an electron neutrino is one in which the Z 0 boson mediates the collision and the neutrino effectively bounces off the electron this is known as a neutral current interaction. The electron neutrino can occasionally also interact through the W boson by changing a neutron into a proton. These are the charged current interactions.
7. 3 T H E S T R U C T U R E O F M AT T E R
The Higgs boson The Standard Model has been very successul at tying together theory and experimental results. The original theory predicted that leptons and quarks should have zero mass this clearly did not agree with the experimental results or leptons and bosons in which fnite masses o the particles had been already measured. To solve this problem the B ritish physicist Peter Higgs introduced a theory that explained the mass o particles including the W and Z bosons. B y introducing the Higgs mechanism, the equations o the Standard Model were changed in such a way as to allow these particles to have mass. According to this theory, particles gain mass by interacting with the Higgs feld that permeates all space. The theory about the Higgs mechanism could be tested experimentally because it predicted the existence o a new particle not previously seen. This particle, called the Higgs particle, is a boson-like orce mediator; however, it does not, in this case, mediate any orce. The mass o this particle is very large and, thereore, requires a great deal o energy to be produced. With the opening o the large hadron collider (LHC ) at the European C entre or Nuclear Research (C ERN) in Geneva, a particle accelerator became available that was capable o providing sufcient energy to produce the Higgs boson. In July 2 01 2 , scientists both at C E RN and Fermilab announced that they had established the existence o a Higgs- like boson and, by March 1 4 2 01 3 , this particle was tentatively confrmed to be positively charged and to have zero spin; these are two o the properties o a Higgs boson. The discovery o the Higgs boson leads to the adaptation o the diagram o the S tandard Model as shown in fgure 1 9.
Figure 19 The Standard Model including the Higgs boson.
Nature of science International collaboration The history o particle physics is one o international collaboration and the particle research acility at C ERN is an excellent example o how the international science community can co-operate eectively. C ERN is run by 2 0 European member states with our nations waiting to j oin; in addition there are more than 5 0 countries with agreements with C ERN. The
C ERN website claims that hal the worlds particle physicists and over ten thousand scientists rom more than a hundred countries do research there. With unds at a premium it makes fnancial sense to pool resources and work internationally. In terms o international cooperation, the work undertaken by particle physicists has much in common with the ideals o the IB .
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7
ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS
Questions 1
D iagram 2 shows some o the principal energy levels o atomic hydrogen.
(IB) a) Light is emitted rom a gas discharge tube. O utline briey how the visible line spectrum o this light can be obtained.
a) S how, by calculation, that the energy o a photon o red light o wavelength 65 6 nm is 1 . 9 eV.
The table below gives inormation relating to three o the wavelengths in the line spectrum o atomic hydrogen.
Wavelength / 10 9 m
Photon energy / 10 19 J
1880
1.06
656
3.03
486
4.09
b) O n a copy o diagram 2 , draw arrows to represent: ( i)
the electron transition that gives rise to the red line ( label this arrow R)
( ii) a possible electron transition that gives rise to the blue line (label this arrow B ) . ( 4 marks)
b) Deduce that the photon energy or the wavelength o 486 1 0 9 m is 4.09 1 0 1 9 J. The diagram below shows two o the energy levels o the hydrogen atom, using data rom the table above. An electron transition between these levels is also shown. -2.41 10 -19 J photon emitted, wavelength = 656 nm
3
(IB) A nucleus o the isotope xenon, Xe- 1 3 1 , is produced when a nucleus o the radioactive isotope iodine I- 1 3 1 decays. a) Explain the term isotopes. b) Fill in the boxes on a copy o the equation below in order to complete the nuclear reaction equation or this decay. 1 31
-5.44 10 -19 J
c) ( i)
O n a copy o the diagram above, construct the other energy level needed to produce the energy changes shown in the table above.
( ii) D raw arrows to represent the energy changes or the two other wavelengths shown in the table above. ( 9 marks)
I 1 53 41 Xe + - +
c) The activity A o a reshly prepared sample o the iodine isotope is 3.2 1 0 5 B q. The variation o the activity A with time t is shown below. 3 .5 3 .0
2
(IB) D iagram 1 below shows part o the emission line spectrum o atomic hydrogen. The wavelengths o the principal lines in the visible region o the spectrum are shown. diagram 1 red (R)
656 nm
blue (B)
diagram 2 violet (V)
486 nm 434 nm
0 -0.54 -0.85 -1.5 -3.4
wavelength
energy/eV
A/10 5 Bq
2.5 2.0 1.5 1.0 0.5 0
0
5
10
15
20 25 t/days
30
35
40
45
On a copy o this graph, draw a best-ft line or the data points. d) Use the graph to estimate the hal-lie o I- 1 3 1 . ( 8 marks)
-13.6
304
QUESTION S (IB) O ne isotope o potassium is potassium-42 ( 42 1 9 K) . Nuclei o this isotope undergo radioactive decay with a hal- lie o 1 2 .5 hours to orm nuclei o calcium. a)
C omplete a copy o the nuclear reaction equation or this decay process. 42 19
K
20
Ca +
b) The graph below shows the variation with time o the number N o potassium- 42 nuclei in a particular sample.
10 average binding energy per nucleon/MeV
4
56 Fe
9 16 O
8
138 Ba
208 Pb 235 U
7
9 Be
6
6 Li
5 4 3H
3 2
2H
1 0 0
50
N0
0
10
20
30 40 t/hours
50
60
(IB) a) Use the ollowing data to deduce that the binding energy per nucleon o the isotope 3 2 He is 2 . 2 MeV.
0 70
nuclear mass o 32 He = 3 .01 6 03 u mass o proton = 1 . 007 2 8u
The isotope o calcium ormed in this decay is stable.
mass o neutron = 1 .008 67u
O n a copy o the graph above, draw a line to show the variation with time t o the number o calcium nuclei in the sample.
b) In the nuclear reaction 2 2 3 1 1 H + 1 H 2 He + 0 n energy is released. ( i)
Use the graph in ( c) , or otherwise, to determine the time at which the ratio
is equal to 7.0. ( 7 marks)
( iii) With reerence to your graph, explain why energy is released in the nuclear reaction above.
(IB)
( 9 marks)
a) Explain what is meant by a nucleon. b) D efne what is meant by the binding energy o a nucleus. The plot below shows the variation with nucleon number o the binding energy per nucleon.
S tate the name o this type o reaction.
( ii) S ketch the general orm o the relationship between the binding energy per nucleon and the nucleon number.
number o calcium nuclei in sample ____ number o potassium-42 nuclei in sample
5
250
( 5 marks)
N0
6
c)
200
c) With reerence to the graph, explain why energy can be released in both the fssion and the usion processes.
N 1 2
100 150 nucleon number
7
a) D istinguish between nuclear fssion and radioactive decay. b) A nucleus o uranium- 2 3 5 ( 235 92 U) may absorb a neutron and then undergo fssion to produce nuclei o strontium-90 ( 90 38 Sr) 42 and xenon-1 42 ( 1 54 Xe) and some neutrons.
305
7
ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS The strontium- 90 and the xenon- 1 42 nuclei both undergo radioactive decay with the emission o particles. ( i)
Write down the nuclear equation or this fssion reaction.
( ii) S tate the eect, i any, on the nucleon number and on the proton number o a nucleus when the nucleus undergoes decay. ( 6 marks) 8
a) A neutron collides with a nucleus o uranium- 2 3 5 and the ollowing reaction takes place. 2 35 92
U + 10 n
96 37
Rb +
1 38 55
C s + 2 10 n
S tate the name o this type o reaction. b) Using the data below, calculate the energy, in MeV, that is released in the reaction. mass o
235 92
mass o
96 37
mass o
1 38 55
U nucleus = 2 3 5 . 043 9u
Rb nucleus = 95 .93 42 u C s nucleus = 1 3 7. 91 1 2 u
mass o 10 n nucleus = 1 . 0087u c) S uggest the importance o the two neutrons released in the reaction. d) The rest mass o each neutron accounts or about 2 MeV o the energy released in the reaction. E xplain what accounts or the remainder o the energy released. ( 9 marks) 9
The diagram below illustrates a proton decaying into a neutron by beta positive ( + ) decay.
1 0 a) Possible particle reactions are given below. In each case apply the conservation laws to determine whether or not the reactions violate any o them. ( i)
e +
( ii) p + n p + 0 ( iii) p + + b) S tate the name o an exchange particle involved in the weak interaction. (1 0 marks) 1 1 (IB) When a negative kaon ( K ) collides with a proton, a neutral kaon ( K 0 ) , a positive kaon ( K + ) and a urther particle ( X) are produced. 0
K + p K + K+ + X The quark structure o kaons is shown in the table.
Particle K K+ 0 K
Quark structure _ su _ us _ ds
a) S tate the amily o particles to which kaons belong. b) S tate the quark structure o the proton. c) The quark structure o particle X is sss. Show that the reaction is consistent with the theory that hadrons are composed o quarks. (4 marks) 1 2 The Feynman diagram below represents a decay via the weak interaction process. Time is shown as upwards. The wiggly line represents a virtual exchange particle.
particle Y neutron
eparticle X
+
e
u proton d
S tate the name o: a) S tate what is meant by virtual exchange particle.
a) the orce involved in this decay b) the particle X c) the particle Y involved in the decay. ( 3 marks)
b) Determine whether the virtual particle in the process represented by the Feynman diagram is a W + , a W , or a Z 0 boson. (4 marks)
306
8 E N E RGY PRO D U CTI O N Introduction In Topic 2 we looked at the principles behind the transer o energy rom one orm to another. We now look in detail at sources that provide the energy we use every day. The provision o energy is a global issue. O n the one hand, ossil uel reserves are limited and these uels can be a source o pollution and greenhouse gases, yet they are a convenient and energy- rich resource. The
development o renewable energy sources continues but they are not yet at a point where they can provide all that we require. Political rhetoric and emotion oten obscure scientifc assessments about energy resources. E veryone not j ust scientists needs a clear understanding o the issues involved in order to make sound j udgements about the uture o our energy provision.
8.1 Energy sources O B J TE XT_UND
Understanding Primary energy sources
Applications and skills Describing the basic eatures o ossil uel
Renewable and non-renewable energy sources Electricity as a secondary and versatile orm
o energy Sankey diagrams Specifc energy and energy density o uel sources
Nature of science We rely on our ability to harness energy. Our largescale production o electricity has revolutionized society. However, we increasingly recognize that such production comes at a price and that alternative sources are now required. There are elements o risk in our continued widespread use o ossil uels: risk to the planet and risk to the supplies themselves. Society has to make important decisions about the uture o energy supply on the planet.
power stations, nuclear power stations, wind generators, pumped storage hydroelectric systems, and solar power cells Describing the dierences between photovoltaic cells and solar heating panels Solving problems relevant to energy transormations in the context o these generating systems Discussing saety issues and risks associated with the production o nuclear power Sketching and interpreting Sankey diagrams Solving specifc energy and energy density problems
Equations energy power = __
time 1 Av 3 wind power equation = ___ 2
307
8
EN ERGY PRO D U CTI O N
Primary and secondary energy We use many dierent types o energy and energy source or our heating and cooking, transport, and or the myriad other tasks we undertake in our daily lives. There is a distinction between two basic types o energy source we use: primary sources and secondary sources. A p rimary source is one that has not been transormed or converted beore use by the consumer, so a ossil uel coal, or example burnt directly in a urnace to convert chemical potential energy into the internal energy o the water and surroundings is an example o a primary source. Another example is the kinetic energy in the wind that can be used to generate electricity ( a secondary source) or to do mechanical work such as in a windmill ( a device used, or example, to grind corn or to pump up water rom underground) . The defnition o a secondary source o energy is one that results rom the transormation o a primary source. The electrical energy we use is generated rom the conversion o a primary source o energy. This makes electrical energy our most important secondary source. Another developing secondary source is hydrogen, although this is, at the moment, much less important than electricity. Hydrogen makes a useul uel because it burns with oxygen releasing relatively large amounts o energy ( you will know this i you have ever observed hydrogen exploding with oxygen in the lab) . The product o this reaction ( water) has the advantage that it is not a pollutant. However, hydrogen does not exist in large quantities in the atmosphere. So energy rom a primary source would have to be used to orm hydrogen rom hydrocarbons, or by separating water into hydrogen and oxygen. The hydrogen could then be transported to wherever it is to be used as a source o energy.
Renewable and non-renewable energy sources The primary sources can themselves be urther divided into two groups: renewable and non-renewable. Renewable sources, such as biomass, can be replenished in relatively short times ( on the scale o a human lietime) , whereas others such as wind and water sources are continually generated rom the S uns energy. Non- renewable sources, on the other hand, can be replaced but only over very long geological times. A good way to classiy renewable and non-renewable resources is by the rates at which they are being consumed and replaced. Coal and oil, both non-renewable, are produced when vegetable matter buried deep below ground is converted through the eects o pressure and high temperature. The time scale or production is hundreds o millions o years (the deposits o coal on Earth were ormed rom vegetation that lived and died during the Carbonierous geological period, roughly 300 million years ago) . There are mechanisms active today that are beginning the process o creating ossil uels in suitable wetland areas o the planet, but our present rate o usage o these ossil uels is ar greater than the rate at which they are being ormed. O n the other hand, renewable uels such as biomass use biological materials such as trees that were only recently alive. S uch sources can be grown to maturity relatively quickly and then used or energy conversion. The rate o usage o the uels can be similar to the rate at which they are being grown.
308
8.1 EN ERGY SOURCES There are urther advantages in the use o biomass and other renewable sources, where the material has been produced recently. When these renewable resources are converted, they will release carbon dioxide ( one o the greenhouse gases) back into the atmosphere. B ut this is new carbon that was taken rom the atmosphere and trapped in the biomass material relatively recently. The conversion o ossil uels releases carbon dioxide into the atmosphere that was fxed in the ossil uels hundreds o millions o years ago. The carbon dioxide content o the atmosphere was more than 1 5 0% greater in the C arbonierous period than it is today, and thus the burning o ossil uels increases the overall amount o this greenhouse gas in the present- day atmosphere.
Types of energy sources In Topic 2 there was a list o some o the important energies available to us. S ome were mechanical in origin. O ther energies were related to the properties o bulk materials and atomic nuclei. O particular importance are the nuclear reactions, both fssion and usion, that you met in Topic 7 .
Primary sources The table below gives an indication o many o the primary sources that are used in the world today although not all sources can be ound in all locations. The use o geothermal energy, or example, requires that the geology o the location has hot rocks suitably placed below the surace.
Energy sources source Nonrenewable sources
Nuclear fuels Fossil fuels
Renewable sources
Energy form
uranium-235 crude oil coal natural gas
nuclear
Sun water wind biomass geothermal
radiant (solar) kinetic kinetic chemical potential internal
chemical potential
Not all the primary sources in the table are necessarily used to provide electricity as a secondary source, it depends on local circumstances. A water wheel using owing water in a river can be used to grind corn in a arming community rather than be harnessed to an electrical generator. A solar urnace may be used in an Arican village to boil water or to cook, while photovoltaic cells may produce the electrical energy required by the community. In some situations, this is oten a better solution than that o changing all the solar energy to an electrical orm that has to be reconverted subsequently. S ome energy is always degraded into an internal orm in a conversion. Nevertheless, many o the worlds primary sources are used to provide electricity using a power station where the secondary source output is in the orm o electrical energy.
309
8
EN ERGY PRO D U CTI O N
Primary energy use D ata or the present usage o various energy sources are readily available rom various sources on the Internet. Figure 1 shows two examples o data released by the US D epartment o Energy. primary energy use World energy consumption by fuel, 19902035 (quadrillion BTU)
2010 World marketed energy use 500 quadrillion Btu (12,603 Mtoe) renewables 10% nuclear 6%
liquids 35%
coal 26%
(a) Figure 1
natural gas 23%
energy/quadrillion BTU
250
projections
200 liquids 150
coal natural gas
100
renewables 50 0 1990
(b)
history
nuclear 2000
2008
2015 year
2025
2035
Total world energy usage (US Energy Information Administration, report #DOE/EIA-0484 (2010) ) . The frst example ( fgure 1 (a) ) is a chart that shows the various energy sectors that account or the total world usage o energy sold on the open market in 2 01 0. There are two units here that are common in energy data. When you carry out your own investigations o global energy usage, you will almost certainly come across these non-SI units.
The B ritish Thermal Unit ( B TU) is still used in energy resource work; it was historically very important. The B TU is defned as the energy required to raise one B ritish pound o water ( about 0.5 kg) through 1 F ( a change o roughly 0.6 K) and is equivalent to about 1 000 J.
Mtoe stands or million tonnes o oil equivalent. O ne tonne o oil equivalent is the energy released when one tonne ( 1 000 kg) o crude oil is burnt; this is roughly 42 GJ, leading to a value o 5 1 0 20 J or 2 01 0 usage total.
The US chart uses an unusual multiplier the energy total or 2 01 0 is given as 5 00 quadrillion. The quadrillion is either 1 0 1 5 or 1 0 24 depending on the defnition used. In this particular case it is 1 0 1 5 and the total world energy usage in 2 01 0 was about 5 1 0 1 7 B TU, which also leads to about 5 1 0 20 J. The second chart ( fgure 1 ( b) ) gives a proj ection rom 1 990 up to the year 2 03 5 or world energy usage. Adding the various contributions to the total indicates that the US D epartment o Energy predicts a world total energy usage in 2 03 5 o about 7.5 1 0 20 J; a 5 0% increase on the 2 01 0 fgure. O ne eature o this graph ( in the light o the enhanced greenhouse eect that we shall discuss later) is that the relative proportions o the various energy sources do not appear to be changing greatly over the timescale o the prediction.
310
8.1 EN ERGY SOURCES
Investigate! Where are we today?
Textbooks are o necessity out-o-date! They represent the position on the day that part o the book was completed. So the examples here are to give you some insight into the type and quantity o data that are available to you. But they are not to be regarded either as the last word or inormation that you must memorize or the examination. All the resources shown here were accessed on the Internet without difculty. You should search or the latest tables and graphs using the source inormation printed with the graphs. ( This is known as a bibliographic reerence and it is essential to quote this when you use other peoples data.)
S earch or the latest data and discuss it in class. D ivide up the j obs so that students bring dierent pieces o data to the discussion. Ask yourselves: What are the trends now? Has the position changed signifcantly since this book was written?
Science does not stand still, and this area o environmental science is moving as quickly as any other discipline. You need to have accurate data i you are to make inormed judgments.
Predictions are based on assumptions. They cannot predict critical events that might alter the situation e.g. a disaster triggered by a tsunami.
Specifc energy and energy density Much o the extraction o ossil uels involves hard and dangerous work in mines or on oilrigs whether at sea or on land. O n the ace o it, the eort and risk o mining ossil uels does not seem to be j ustifed when there are other sources o energy available. So why are ossil uels still extracted? The answer becomes more obvious when we look at the energy available rom the ossil uel itsel. There are two ways to measure this: specifc energy and energy density. The word specifc has the clear scientifc meaning o per unit mass, or ( in S I) per kg. S o, sp ecifc energy indicates the number o j oules that can be released by each kilogram o the uel. Typical values or a particular uel can vary widely, coal, or example, has a dierent composition and density depending on where it comes rom, and even depending on the location o the sample in the mining seam. D ensity is a amiliar concept; it is the amount o quantity possessed by one cubic metre o a substance. E nergy density is the number o j oules that can be released rom 1 m 3 o a uel.
Fuel Wood Coal Gasoline (petrol) Natural gas at atmospheric pressure Uranium (nuclear fssion) Deuterium/tritium (nuclear usion) Water alling through 100 m in a hydroelectric plant
Specifc energy/ Energy density/ MJ kg1 MJ m 3 16 1 10 4 2060 (20-60) 10 6 45 35 10 6 55 3.5 10 4 8 10 7 1.5 10 1 5 3 10 8 6 10 1 5 10 3
10 3
311
8
EN ERGY PRO D U CTI O N The table shows comparisons between some common uels ( and in the case o usion, the possible energy yields i usion should become commercially viable) . You should look at these and other values in detail or yoursel. Notice the wide range o values that appear in this table. Explore data or other common uels. As you learn about dierent uels, fnd out data or their specifc energy and energy density and add these values to your own table.
Worked examples 1
A ossil-uel power station has an efciency o 25 % and generates 1 2 00 MW o useul electrical power. The specifc energy o the ossil uel is 5 2 MJ kg 1 . C alculate the mass o uel consumed each second.
Solution I 1 2 00 MW o power is developed then, 1 200 1 00 including the efciency fgure, _________ = 4800 25 MW o energy needs to be supplied by the ossil uel. The specifc energy is 5 2 MJ kg 1 , so the mass o 48 00 uel required is ____ = 92 kg s 1 . ( That is roughly 52 1 tonne every 1 0 s, or one railcar ull o coal every 2 minutes. ) 2
When a camping stove that burns gasoline ( petrol) is used, 70% o the energy rom the uel reaches the cooking pot. The energy density o the gasoline is 3 5 GJ m 3 .
a) C alculate the volume o gasoline needed to raise the temperature o 1 litre o water rom 1 0 C to 1 00 C . Assume that the heat capacity o the pot is negligible. The specifc heat capacity o the water is 4.2 kJ kg 1 K 1 . b) Estimate the volume o uel that a student should purchase or a weekend camping expedition.
Solution a) 1 litre o water has a mass o 1 kg so the energy required to heat the water is 42 00 1 90 which is 0. 3 8 MJ. Allowing or the inefciency, 0.5 4 MJ o energy is required and this is a volume o uel 0.5 4 1 0 o ________ = 1 .6 1 0 4 m 3 or about 2 00 ml. 6
35 1 09
b) Assume that 2 litres o water are required or each meal, and that there will be 5 cooked meals during the weekend. S o 2 litres o uel should be more than enough.
Thermal power stations A thermal powe r station is one in which a primary source o energy is converted frst into internal ene rgy and then to electrical energy. The primary uel can include nuclear uel, ossil uel, biomass or other uel that can produce inte rnal e nergy. ( In principle , a se condary source could be used to provide the initial inte rnal e nergy, but this would not be sensible as it incurs substantial extra losses. ) We will discuss the die re nt types o primary conversion later. Howe ver, once the primary energy has been converte d to internal energy, all thermal power stations use a common approach to the conversion o internal into ele ctrical energy: the energy is used to heat water producing steam at high temperatures and high pressures. Figure 2 shows what happe ns. Energy rom the primary uel heats water in a pressure vessel to create steam. This steam is superheated. This means that its temperature is well above the amiliar 1 00 C boiling point that we are used to at atmospheric pressure. To attain such high temperatures the steam has to be at high pressure, hundreds o times more than atmospheric pressure. At these high pressures the water in the vessel does not boil in the
312
8.1 EN ERGY SOURCES electricity generator
turbines
kinetic energy (rotation)
superheated steam (internal energy)
electrical energy
boiler condenser water
internal energy rom primary source transerred because o the temperature diference between the source and the boiler Figure 2
pylon
Energy conversions inside a power station.
way that is amiliar to us, it goes straight into the steam phase without orming the bubbles in the liquid that you see in a cooking pot on a stove. However, we will continue to use the term boiler or the vessel where the water is converted to steam even though the idea o boiling is technically incorrect. The high- pressure steam is then directed to a turbine. Turbines can be thought o as reverse ans where steam blows the blades around ( whereas in a an the blades turn to move the gas) . There is usually more than one set o blades and each set is mounted on a common axle connected to an alternating current ( ac) electricity generator. In the generator, the electrical energy is produced when coils o wire, turned by the turbine, rotate in a magnetic feld. The energy that is generated is sent, via a network o electrical cables, to the consumers. You can fnd the details o the physics o electrical energy generation and its subsequent transmission in Topic 1 1 . There are really three energy transers going on in this process: primary energy to the internal energy o water, this internal energy to the kinetic energy o the turbine, and kinetic energy o the turbine to electrical energy in the generator. It is easy to orget the kinetic energy phase and to say that the internal energy goes straight to electrical. O course, the original internal energy is produced in dierent ways in dierent types o thermal power station. In ossil and biomass stations, there is a straightorward combustion process where material is set alight and burnt. In nuclear stations, the process has to be more complicated. We shall look at the dierences between stations in the initial conversion o the energy later.
Tip Do not conuse the roles o turbine and dynamo in a power station. The energy conversions they carry out are quite diferent.
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EN ERGY PRO D U CTI O N
Sankey diagrams D ierent types o thermal power station have dierent energy losses in their processes and dierent overall efciencies. The S ankey diagram is a visual representation o the ow o the energy in a device or in a process ( although in other subj ects outside the sciences, the S ankey diagram is also used to show the ow o material) . There are some rules to remember about the S ankey diagram:
Each energy source and loss in the process is represented by an arrow.
The diagram is drawn to scale with the width o an arrow being proportional to the amount o energy it represents.
The energy ow is drawn rom let to right.
When energy is lost rom the system it moves to the top or bottom o the diagram.
Power transers as well as energy ows can be represented.
Here is an example to demonstrate the use o a S ankey diagram. coal (primary electricity energyn source) (secondary energy)
electricity (fnal energy)
light radiation (useul energy) 1.0
2
65 Figure 3
2
2
1
28
lampshade
lamp internal energy
domestic wiring
distribution
100
power generation losses
30
31
33 transmission
35
1
Sankey diagram or a lamp.
It shows the ows o energy that begin with the conversion o chemical energy rom ossil uels and end with light energy emitted rom a flament lamp. The red arrows represent energy that is transerred rom the system in the orm o energy as a result o temperature dierences. It is important to recognize that in any process where there is an energy transormation, this energy is lost and is no longer available to perorm a useul j ob. This is degraded energy and there is always a loss o energy like this in all energy transers. O the original primary energy ( 1 00% shown in blue at the bottom o the diagram) , only 3 5 % appears as useul secondary energy. The remaining 65 % ( shown in red) is lost to the surroundings in the process, shown by an arrow that points downwards o the chart. The secondary energy is then shown with the losses involved in the transmission and distribution o the electricity, and to losses in the house wiring. In the lamp itsel, most o the energy ( 2 8% o the original) is transerred to the internal energy o the surroundings. Finally only 1 % o the original primary energy is let as light energy or illumination.
314
8.1 EN ERGY SOURCES A S ankey diagram is a useul way to visualize the energy consumption o nations. There are many examples o this available on the Internet. Figure 4 shows the energy ows associated with the C anadian economy. Look on the Internet and you will probably fnd the S ankey diagram or your own national energy demand. Energy equivalents In terms o energy equivalencies, 1 exajoule ( EJ) is equal to: e xp o rt 7 .6 1 5 u ra n iu m 8 .4 2 3 im p o rt 0 .0 9 4 0 .8 1 3 h y d ro
e l e ctric p o we r 3 .7 4 1
1 .2 0 1 0 .3 4 1
160 million barrels o oil
energy consumed annually by 15 million average Canadian single detached homes
energy produced annually by 14 Pickering-sized nuclear stations operating at nominal capacity.
e xp o rt 0 .1 2 4
d is trib u te d e l e ct ricit y 2 .0 1 3 tra n s m is s io n l o s s e s 0 .1 7 3
1 .0 2 1 e l e ct rica l s y s t e m e n e rgy l o s s e s 1 .8 2 3 0 .0 0 3 1
re s id e n t ia l/ co m m e rcia l 2 .6 4 1
1 .2 5 1 3 .9 2 3
0 .0 8 1 ,2
e xp o rt 4.1 5 4 1 .1 6 1
im p o rt s 0 .3 8 4
0 .1 4 1
0 .43 4 0 .0 1 1
- 0 .7 0 3
1 .4 7 3 ,4
co a l
0 .5 0 1
useul energy 4.87 3
in d u s t ria l 4.1 4 1 ,4
2 .5 2 3
0 .1 8 1
1 .3 5 3
im p o rts 0 .6 1 4
lost energy 4.27 3
1 .8 7 3
1 .6 2 3
0 .8 6 1
0 .1 1 1 0 .1 8 4
b io m a s s & o th e r
0 .7 7 3
0 .2 6 1
1 .7 2 4
n a tu ra l ga s 7.69 3 ,4
the energy produced annually by over 1400 square kilometres o state-o-the-art solar cells operating under nominal conditions, enough to cover the entire Toronto urbanized area.
0 .0 1 4
e xp o rt 0 .7 4 4
p ip e l in e s
0 .0 0 1 4 0 .1 9 4
1 .8 9 3 0 .4 7 3
0 .8 8 1 4.1 0 3
p e tro l e u m 5 .9 6 3 ,4
2 .3 4 1
Version 1August 2006
0 .0 1 1 im p o rts 2 .5 9 4
Figure 4 The energy
tra n s p o rt a tio n 2.36 1
e xp o rt 4.4 5 4
0 .4 7 4 n o n -u e l 0 .9 0 3
Sources: 1) NRCan Energy Handbook2005 2) Canadian wind energy association 3) Calculated value 4) Statistics Canada 5) Natural Resources Canada
fow in Canada in 2003. The values are in exajoule. (http://ww2.nrcan.ge.ca/es/oerd) .
Nature of science Sankey diagram in another context Here is a completely dierent use o the S ankey diagram constructed in a historical context. It shows the progress o Napoleons army to and rom his Russian campaign 1 81 2 1 81 3 . The width o the strips gives the number o men in the army: brown in the j ourney to Moscow, black on the return. The graph below the diagram gives the
average temperature experienced by the army on its return j ourney in Raumur degrees. The Raumur temperature scale is named ater the French scientist who suggested a scale that had fxed points o 0 at the reezing point o water and 80 at the boiling point o water.
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EN ERGY PRO D U CTI O N
Figure 5 A Sankey
diagram showing the change in size of Napoleons army during his Russian campaign.
Worked examples 1
An electric kettle of rating 2 .0 kW is switched on for 90 s. D uring this time 2 0 kJ of energy is lost to the surroundings from the kettle. D raw a Sankey diagram for this energy transfer.
remainder of the energy is wasted in the engine, the gearbox and the wheels. Use these data to sketch a S ankey diagram for the car.
Solution Solution The energy supplied in 90 s is 2 1 000 90 = 1 80 kJ. 2 0 000 J is lost to the surroundings; this is 1 1 % of the total.
180 kJ supplied to kettle
160 kJ to heating water
Of the 1 00% of the original fuel energy, 1 2% is lost in the exhaust, and 34% is useful energy. This leaves 54% energy lost in the engine and the transmission. A convenient way to draw the diagram is on squared paper. Use a convenient scale: 1 0% 1 large square is a reasonable scale here.
1000 J as chemical energy 880 J as kinetic energy in the engine
20 kJ to surroundings
2
316
In a petrol- powered car 3 4% of the energy in the fuel is converted into the kinetic energy of the car. Heating the exhaust gases accounts for 1 2 % of the energy lost from the fuel. The
120 J as exhaust
340 J useful energy to move the car 540 J wasted in moving all the engine parts
8.1 EN ERGY SOURCES
Primary sources used in power stations This section compares the dierent ways in which the initial energy required by a thermal power station can be generated.
Fossil fuels Modern ossil-uel power stations can be very large and can convert signifcant amounts o power. The largest in the world at the time o writing has a maximum power output o about 6 GW. The exact process required beore the uel is burnt diers slightly depending on the uel used, whether coal, oil, or natural gas. The gas and oil can be burnt readily in a combustion chamber that is thermally connected to the boiler. In the case o coal, some pre-treatment is normally required. O ten the coal is crushed into a fne powder beore being blown into the urnace where it is burnt. There are obvious disadvantages to the burning o ossil uels. S ome o these are environmental, but other disadvantages can be seen as a misuse o these special materials:
The materials have taken a very large time to accumulate and will not be replaced or equally long times.
The burning o the uels releases into the atmosphere large quantities o carbon dioxide that have been locked in the coal, oil, and gas or millions o years. This has a maj or impact on the response o the atmospheric system to the radiation incident on it rom the S un ( the greenhouse and enhanced greenhouse eects) .
Fossil uels have signifcant uses in the chemical industry or the production o plastics, medicines, and other important products.
It makes sense to locate power stations as close as possible to places where ossil uels are recovered; however, this is not always possible and, in some locations, large- scale transportation o the uels is still required. A need or transport leads to an overall reduction in the efciency o the process because energy has to be expended in moving the uels to the power stations.
Nuclear fuel Sub-topic 7.2 dealt with the principles that lie behind nuclear fssion. It explained the origin o the energy released rom the nucleus when fssion occurs and showed you how to calculate the energy available per fssion. In this course we will only consider so- called thermal fssion reactors, but there are other types in requent use. A particularly common variety o the thermal reactors is the pressurized water reactor ( PWR) . Uranium- 2 3 5 is the nuclide used in these reactors. As with all our other examples o power stations, the aim is to take the energy released in the nuclear fssion and use this to create high-pressure steam to turn turbines connected to an electrical generator. However, the energy is not gained quite so easily as in the case o the ossil uels. Figure 6 shows a schematic o a PWR with the fnal output o steam to, and the return pipe rom, the turbines on the right-hand side o the diagram. The remainder o the power station is as in fgure 2 on p3 1 3 .
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EN ERGY PRO D U CTI O N
control rods
steam to drive turbines
containment building graphite moderator
uranium fuel rods steel reactor vessel
cold water from turbines
Figure 6 Basic features of a pressurised water reactor ( PWR) . Uranium is mined as an ore in various parts o the world, including Australia, C anada, and Kazakhstan ( which together produce about 6 0 % o the worlds ore every year) . The US , Russia, and parts o Arica also produce smaller amounts o uranium ore. About 9 9 % o the ore as it comes directly rom the ground is made up o uranium- 2 3 8 , with the remainder being U- 2 3 5 ; it is the U- 2 3 5 not U- 2 3 8 that is required or the fssion process. This means that an initial extraction process is required to boost the ratio o U- 2 3 5 : U- 2 3 8 . The uel needs to contain about 3 % U- 2 3 5 beore it can be used in a reactor. This is because U- 2 3 8 is a good absorber o neutrons and too much U- 2 3 8 in the uel will prevent the fssion reaction becoming sel- sustaining. The uel with its boosted proportions o U- 2 3 5 is said to be enriched.
Fission product fssion ragments decay o fssion ragments emitted gamma rays emitted neutrons
energy/MeV 160 21 7 5
The enriched material is then ormed into uel rods long cylinders o uranium that are inserted into the core o the reactor. Most o the energy ( about 2 00 MeV or 3 1 0 1 1 J per fssion) is released in the orm o kinetic energy o the fssion ragments and neutrons emitted during the fssion. Immediately ater emission, the neutrons are moving at very high speeds o the order o 1 0 4 km s 1 . However, in order or them to be as eective as possible in causing urther fssions to sustain the reaction they need to be moving with kinetic energies much lower than this, o the order o 0. 02 5 eV ( with a speed o about 2 km s 1 ) . This slower speed is typical o the speeds that neutrons have when they are in equilibrium with matter at about room temperatures. Neutrons with these typical speeds are known as thermal neutrons. The requirement o removing the kinetic energy rom the neutrons is not only that the neutrons can stimulate urther fssions eectively, but that their energy can be efciently transerred to the later stages o the power station. The removal o energy is achieved through the use o a moderator, so-called because it moderates (slows down) the speeds o the neutrons. Typical moderators or the PWR type include water and carbon in the orm o graphite. The transer o energy is achieved when a ast- moving neutron strikes a moderator atom inelastically,
318
8.1 EN ERGY SOURCES
transerring energy to the atom and losing energy itsel. Ater a series o such collisions the neutron will have lost enough kinetic energy or it to be moving at thermal speeds and to have a high chance o causing urther fssion. A urther problem is that U-2 3 8 is very eective at absorbing high- speed neutrons, so i the slowing down is carried out in the presence o the U- 2 3 8 then ew ree neutrons will remain at the end o it. S o reactor designers have moderators close to, but not part o, the uel rods; the neutrons slow down in the presence o moderator but away rom the U- 2 3 8. The uel rods and the moderating material are kept separate and neutrons move rom one to the other at random. The reactor vessel and its contents are designed to acilitate this. The criteria or a material to be a good moderator include not being a good absorber o neutrons ( absorption would lower the reaction rate and possibly stop the reaction altogether) and being inert in the extreme conditions o the reactor. S ome reactor types, or example, use deuterium ( 21 H) in the orm o deuterium oxide ( D 2 O , called heavy water) rather than graphite. You should be able to recognize that the best moderator o all ought to be a hydrogen atom ( a single proton in the nucleus) because the maximum energy can be transerred when a neutron strikes a proton. However, hydrogen is a very good absorber o neutrons and it cannot be used as a moderator in this way. There is a need to regulate the power output rom the reactor and to shut down its operation i necessary. This is achieved through the use o control rods. These are rods, oten made o boron or some other element that absorbs neutrons very well, that can be lowered into the reactor. When the control rods are inserted a long way into the reactor, many neutrons are absorbed in the rods and ewer neutrons will be available or subsequent fssions; the rate o the reaction will drop. B y raising and lowering the rods, the reactor operators can keep the energy output o the reactor ( and thereore the power station) under control. The last part o the nuclear power station that needs consideration is the mechanism or conveying the internal energy rom inside the reactor to the turbines. This is known as the heat exchanger. The energy exchange cannot be carried out directly as in the ossil- uel stations. There needs to be a closed- system heat exchanger that collects energy rom the moderator and other hot regions o the reactor, and delivers it to the water. The turbine steam cannot be piped directly through the reactor vessel because there is a chance that radioactive material could be transerred outside the reactor vessel. Using a closed system prevents this. The pressurized water reactor is given its name because it transers the energy rom moderator and uel rods to the boiler using a closed water system under pressure. Water is not the only substance available or this. In the Advanced Gas-cooled Reactors ( AGR) used in the UK, carbon dioxide gas is used rather than water, but the principle o transerring energy saely through a closed system is the same.
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EN ERGY PRO D U CTI O N
Nature of science Fast breeder reactors A remarkable design or nuclear reactors comes with the development o the ast breeder reactor. Plutonium-239 (Pu-239) is the fssionable material in this case just as uranium-235 is used in PWRs. However, a blanket o uranium-238 surrounds the plutonium core. Uranium-238 does not easily fssion and is a good absorber o neutrons (remember that its presence is undesirable in a PWR) . This U-238 absorbs neutrons lost rom the reactor core and is transmuted into Pu-239 the uel o the ast breeder reactor! The reactor is making its own uel
and generating energy at the same time. It has been reported that, under the right conditions, a ast breeder reactor can produce 5 kg o fssionable plutonium or every 4 kg used in fssion. This is a good way to convert the large stockpiles o virtually useless uranium-238 into something o value. There are drawbacks o course: large amounts o high- level radioactive waste rom the ast breeder reactor need to be dealt with and, in the wrong hands, the plutonium can be used or nuclear weapons.
Investigate! Running your own reactor
No schools are likely to have their own nuclear reactor or students to use! However you can still investigate the operation o a nuclear power station.
There are a number o simulations available on the Internet that will give you virtual control o the station. A suitable starting point is to enter nuclear reactor applet into a search engine.
Worked examples
Solution
1
The role o the moderator is to remove kinetic energy rom neutrons so that there is a high probability that urther fssions will occur. When neutrons are moving at high speeds, there is a very high probability that uranium-2 3 8 nuclei will absorb them without fssion occurring. So the removal o the moderator will mean that neutrons are no longer slowed down, and will be absorbed by U- 2 3 8. The fssion reaction will either stop or its rate will be reduced.
When one uranium-2 3 5 nucleus undergoes fssion, 3 .2 1 0 1 1 J o energy is released. The density o uranium-2 3 5 is 1 . 9 1 0 4 kg m 3 . C alculate, or uranium- 2 3 5 : a) the specifc energy b) the energy density.
Solution a) The mass o the atom is 2 3 5 u ( ignoring the mass o the electrons in the atom) .
3
This is equivalent to a mass o 1 .7 1 0 - 27 2 3 5 = 4.0 1 0 - 25 kg. So the specifc energy o uranium- 2 3 5 3.2 1 0 is ________ = 8.0 1 0 1 3 J kg - 1 -11
When a moving neutron strikes a stationary carbon-1 2 atom head-on, the neutron loses about 3 0% o its kinetic energy. Estimate the number o collisions that would be required or a 1 MeV neutron to be slowed down to 0.1 eV.
4. 0 1 0 - 2 5
1 b) 1 kg o uranium-235 has a volume o _______ 1 .9 1 0 4
5 10
-5
3
m . Thereore the energy density
o pure uranium-2 3 5 = 8. 0 1 0 = 1 .6 1 0 1 8 J m 3 2
320
13
5 10
-5
Explain what will happen in a pressured water reactor i the moderator is removed.
Solution Ater one collision the remaining kinetic energy o the neutron will be 0.7 1 MeV. Ater two collisions the energy o the neutron will be 0.7 0.7 1 MeV = 0. 7 2 1 MeV
8.1 EN ERGY SOURCES
Ater n collisions the energy o the neutron will be 0.7 n 1 MeV.
S o 46 collisions are required to reduce the neutron speed by this actor o 1 0 7 .
S o 0.7 n = 0.1 1 0 6 ,
( In practice, about 1 00 are required; you might want to consider why the actual number is larger than the estimate.)
-7 or n = ______ = 45 . 2 lo g 0 . 7 10
Saety issues in nuclear power
TOK
There needs to be a range o saety measures provided at the site o a nuclear reactor to protect the work orce, the community beyond the power station, and the environment.
The reactor vessel is made o thick steel to withstand the high temperatures and pressures present in the reactor. This has the beneft o also absorbing alpha and beta radiations together with some o the gamma rays and stray neutrons.
The vessel itsel is encased in layers o very thick reinorced concrete that also absorb neutrons and gamma rays.
There are emergency saety mechanisms that operate quickly to shut the reactor down in the event o an accident.
The uel rods are inserted into and removed rom the core using robots so that human operators do not come into contact with the spent uel rods, which become highly radioactive during their time in the reactor.
The issue o the disposal o the waste produced in the nuclear industry is much debated in many parts o the world. S ome o the waste will remain radioactive or a very long time but, as you know rom Topic 7, this implies that its activity will be quite low during this long period. There are however other problems involving the chemical toxicity o this waste material which mean that it is vital to keep it separate rom biological material and thus the ood chain. The technology required to achieve this is still developing.
O all the scientifc issues o our time, perhaps nuclear energy invokes the greatest emotional response in both scientists and nonscientists alike. It is vital to carry out accurate risk assessments or all energy sources, not just nuclear. Is it possible we could orget the location o the waste sites in 50, 100, or 1000 years? Are human errors part o the equation? How can such assessments be carried out in an emotionally charged debate?
At the end o the lie o a reactor ( o the order o 2 5 5 0 years at the moment) , the reactor plant has to be decommissioned. This involves removing all the uel rods and other high- activity waste products and enclosing the reactor vessel and its concrete shield in a larger shell o concrete. It is then necessary to leave the structure alone or up to a century to allow the activity o the structure to drop to a level similar to that o the local background. S uch long-term treatment is expensive and it is important to actor these maj or costs into the price o the electricity as it is being produced during the lietime o the power station.
Nature of science Society and nuclear power The use o nuclear power has been growing throughout the world since the late 1 940s. However, society has never been truly comortable
with its presence and use. There are many reasons or this, including a lack o public understanding o the fssion process ( both advantages and
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EN ERGY PRO D U CTI O N
disadvantages) , public reaction to accidents that have occurred periodically over the years, and a ear o radioactive materials. Maj or accidents have included the C hernobyl incident, the event at Three Mile Island, and the Fukushima accident o 2 01 1 that was triggered by a tsunami and possibly also an associated earthquake.
understand the scientifc acts and statistics that surround the issue o nuclear power. It is important also to understand the meaning o risk, not j ust in the context o the nuclear power industry, but also in relation to things we do every day. D o some research or yoursel. Find out the risks to your health o:
At the time o writing there has been a withdrawal o approval or nuclear plans by some governments as a result o public opinion changes ater the Fukushima accident. C ontinuing decisions about how we generate energy will be required by society as time goes on, and as our energy- generation technology improves.
living within 2 0 km o a nuclear power station
ying once rom E urope to a country on the Pacifc Rim
smoking 2 0 cigarettes every day
driving 1 000 km in a car.
Try to fnd numerical estimates o the risk, not j ust written statements.
When you debate these issues, ensure that you
Wind generators Wind generators can be used successully in many parts o the world. Even though the wind blows erratically at most locations, this can be countered by the provision o separate wind arms, each with large numbers o individual turbines all connected to the electrical power grid. There are two principal designs o generator: horizontal- axis and vertical- axis. In both cases a rotor is mounted on an axle that is either horizontal or vertical, hence the names. The rotor is rotated by the wind and, through a gearbox, this turns an electrical generator. The electrical energy is ed either to a storage system ( but this increases the expense) or to the electrical grid.
radius o rotor rotor blade radius o rotor
gearbox generator nacelle rotor height
fxed pitch rotor blade
tower
gearbox horizontal axis
322
Figure 7 Wind turbine confgurations.
generator vertical axis
8.1 EN ERGY SOURCES
The horizontal-axis machine can be steered into the wind. The verticalaxis type does not have to be steered into the wind and thereore its generator can be placed o-axis.
blade radius r
air density,
It is possible to estimate the maximum power available rom a horizontal- axis wind turbine o blade area A. In one second, the volume o air moving through the turbine is vA when the speed o the wind is v ( fgure 8) . The mass o air moving through the turbine every second is vA where is the density o the air, and the kinetic energy o the air arriving at the turbine in one second is
wind speed v v
Figure 8 Volume of air entering a wind turbine in one second.
1 1 1 __ (mass) (speed) 2 = __ ( vA) v 2 = __ Av 3 2 2 2
I a wind turbine has a blade radius r then the area A swept out by the blades is r 2 and the maximum theoretical kinetic energy arriving at the turbine every second ( and hence the maximum theoretical power) is 1 __ r2 v3 2
This is a maximum theoretical value of the available power as there are a number o assumptions in the proo. In particular, it is assumed that all the kinetic energy o the wind can be used. O course, the wind has to move out o the back o the turbine and so must have some kinetic energy remaining ater being slowed down. Also, i the turbine is part o a wind arm then the presence o other nearby turbines disturbs the ow o the air and leads to a reduction in the energy rom turbines at the rear o the array. The turbine power equation suggests that a high wind speed and a long blade ( large A or r) will give the best energy yields. However, increasing the radius o the blade also increases its mass and this means that the rotors will not rotate at low wind speeds. The blade radius has to be a compromise that depends on the exact location o the wind arm. Many wind arms are placed o- shore; the wind speeds are higher than over land. This is because the sea is relatively smooth compared with the buildings, hills and so on that are ound on land. However, installation and subsequent maintenance or o- shore arrays are more expensive because o the access issues. Another ideal place or wind arms is at the top o a hill. The eect o the land shape is to constrict the ow o the wind into a more confned volume so that in consequence the wind speed rises as the air moves up the hill. Wind speeds thereore tend to be greater at the top o hills and, because the power output o the turbine is proportional to v3 , even a small increase in average wind speed is advantageous. S ome people obj ect to both on- and o- shore arrays on the grounds o visual pollution. There are also suggestions that wind arms compromise animal habitats in some places and that the turbines are noisy or those who live close by.
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8
EN ERGY PRO D U CTI O N The actors or and against wind being used as an energy source are summarized in this table:
Advantages
Disadvantages
No energy costs
Variable output on a daily or seasonal basis
No chemical pollution
Site availability can be limited in some countries
Capital costs can be high but reduce with economies o scale Easy to maintain on land; not so easy of-shore
Noise pollution Visual pollution Ecological impact
Worked examples 1
a) C alculate the maximum possible output o the wind turbine i the density o air is 1 .3 kg m - 3 .
A wind turbine produces a power P at a particular wind speed. I the efciency o the wind turbine remains constant, estimate the power produced by the turbine:
b) O utline why your estimate will be the maximum possible output o the turbine.
a) when the wind speed doubles b) when the radius o the blade length halves.
Solution a) Using the wind turbine equation:
Solution The equation or the kinetic energy arriving at the 1 wind turbine every second is __ r2 v3 . 2
the kinetic energy arriving at the wind turbine 1 every second is __ r2 v3 , 2
a) When the wind speed v doubles, v3 increases by a actor o 8, so the power output will be 8P.
this will be the maximum power output and 1 is __ 1 .3 2 5 2 1 1 3 = 1 .7 MW. 2
2
b) When the radius o the blade halves, r will go down by a actor o 4 and ( i nothing else P changes) the output will be __ . 4 2
A wind turbine with blades o length 2 5 m is situated in a region where the average wind speed is 1 1 m s 1 .
b) Mechanical and electrical inefciencies in the wind turbine have not been considered. The calculation assumes that all the kinetic energy o the wind can be utilized; this is not possible as some kinetic energy o the air will remain as it leaves the wind turbine.
Pumped storage There are a number o ways in which water can be used as a primary energy resource. These include:
pumped storage plants
hydroelectric plants
tidal barrage
tidal ow systems
wave energy.
All these sources use one o two methods:
324
The gravitational potential energy o water held at a level above a reservoir is converted to electrical energy as the water is allowed to all to the lower level ( used in hydroelectric ( fgure 9( a) ) , pumped storage, and tidal barrage) .
8.1 EN ERGY SOURCES
The kinetic energy o moving water is transerred to electrical energy as the water ows or as waves move ( river or tidal ow or wave systems) . Figure 9( b) shows a picture o the C anadian B eauharnois run-o- the- river power station on the S t-Laurent river that can generate 1 .9 GW o power.
(a)
In this topic we ocus on the p ump ed storage system. The wind arms and nuclear power stations we have discussed so ar are known as baseload stations. They run 2 4 hours a day, 7 days a week converting energy all the time. However, the demand that consumers make or energy is variable and cannot always be predicted. From time to time the demand exceeds the output o the baseload stations. Pumped storage is one way to make up or this defcit.
intake
(b)
reservoir
power cable access main access tunnel
surge chamber
discharge
Figure 9 Water as a primary energy resource. ( a) A hydroelectric plant in Thailand. (b) A run-of-the-river plant in Canada.
powerplant chamber transformer
Figure 10 A pumped-storage generating station.
A pumped storage system ( Figure 1 0) involves the use o two water reservoirs sometimes a natural eature such as a lake, sometimes a man-made lake or an excavated cavern inside a mountain. These reservoirs are connected by pipes. When demand is high, water is allowed to run through the pipes rom the upper reservoir to the lower via water turbines. When demand is low, and electrical energy is cheap, the turbines operate in reverse to pump water back rom the lower to the upper reservoir. S ome pumped storage systems can go rom zero to ull output in tens o seconds. The larger systems take longer to come up to ull power, however, substantial outputs are usually achieved in only a ew minutes rom switch on. For a pumped storage system that operates through a height dierence o h, the gravitational potential energy available = mg h where m is the mass o water that moves through the generator and g is the gravitational feld strength. S o the maximum power P available rom the water is equal to the rate at which energy is converted in the machine and is m V P = __ g h = ( __ ) g h t t
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EN ERGY PRO D U CTI O N where t is the time or mass m to move through the generator, V is the volume o water moving through the generator in time t and is the density o the water.
Worked examples 1
Solution
Water rom a pumped storage system alls through a vertical distance o 2 60 m to a turbine at a rate o 600 kg s 1 . The density o water is 1 000 kg m 3 . The overall efciency o the system is 65 %. C alculate the power output o the system.
Solution In one second the gravitational potential energy lost by the system is mg h = 600 g 2 60 = 1 .5 MJ. The efciency is 65 % . 65 O utput power = 1 .5 1 0 6 _ = 0.99 MW 1 00 2 In a tidal barrage system water is retained behind a dam o height h. Show that the gravitational potential energy available rom the water stored behind the dam is proportional to h 2 .
Assume that the cross- sectional area o the dam is A and that the cross- section is rectangular. The volume o water held by the dam is Ah. The mass o the water held by the dam is Ah where is the density o water. When the dam empties completely the centre o mass o the water alls through a distance h __ ( because the centre o mass is hal way up the 2 height o the dam) . The gravitational potential energy o the water is h 1 mgh = ( Ah) g __ = __ Ah 2 . 2 2 Thus, gpe h 2
Solar energy Although most energy is ultimately derived rom the Sun, two systems use photons emitted by the Sun directly in order to provide energy in both large- and small-scale installations.
Solar heating panels S olar heating panels is a technique or heating water using the S uns energy. Sun
hot water, underfoor heating, and central heating
solar panel
pump and solar controller
bo ile r
mains cold water eed hot water storage cylinder
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Figure 11 Solar thermal-domestic hot water system.
8.1 EN ERGY SOURCES
A solar heating p anel contains a pipe, embedded in a black plate, through which a glycolwater mixture is circulated by a pump ( glycol has a low reezing point, necessary in cold countries) . The liquid heats up as inra-red radiation alls on the panel. The pump circulates the liquid to the hot- water storage cylinder in the building. A heatexchanger system transers the energy to the water in the storage cylinder. A pump is needed because the glycolwater mixture becomes less dense as it heats up and would thereore move to the top o the panel and not heat the water in the cylinder. A controller unit is required to prevent the system pumping hot water rom the cylinder to the panel during the winter when the panel is cold.
Solar photovoltaic panels The frst solar photocells were developed around the middle o the nineteenth century by Alexandre- E dmond B ecquerel ( the ather o Henri the discoverer o radioactivity) . For a long time, the use o solar cells, based on the element selenium, was restricted to photography. With the advent o semiconductor technology, it was possible to produce p hotovoltaic cells ( as they are properly called, abbreviated to PV) to power everything rom calculators to satellites. In many parts o the world, solar panels are mounted on the roos o houses. These panels not only supply energy to the house, but excess energy converted during sunny days is oten sold to the local electricity supply company.
electrons to external circuit
light
p-type layer
The photovoltaic materials in the panel convert electromagnetic radiation rom the S un into electrical energy. A ull explanation o the way in which this happens goes beyond the IB syllabus, but a simplifed explanation is as ollows. The photovoltaic cell consists o a single crystal o semiconductor that has been doped so that one ace is p- type semiconductor and the opposite ace is n- type. These terms n- type and p-type indicate the most signifcant charge carriers in the substance ( electrons in n- type, positive holes an absence o electrons in p-type) . Normally there is equilibrium between the charge carriers in both halves o the cell. However, when energy in the orm o photons alls on the photovoltaic cell, then the equilibrium is disturbed, electrons are released and gain energy to move rom the n- region to the p- region and hence around the external circuit. The electrons transer this energy to the external circuit in the usual way and do useul work.
Figure 12 A domestic photovoltaic panel system.
n-type layer
electron
Figure 13 Cross-section of a photovoltaic cell.
Figure 14 Connecting photovoltaic cells together.
O ne single cell has a small em o about 1 V ( this is determined by the nature o the semiconductor) and so banks o cells are manuactured in order to produce usable currents on both a domestic and commercial scale. Many cells connected in series would give large ems but also large internal resistances; the compromise usually adopted is to connect the cells in a combination o both series and parallel. The efciencies o present-day solar cells are about 2 0% or a little higher. However, extensive research and development is being carried out in many countries and it is likely that these efciencies will rise signifcantly over the next ew years.
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EN ERGY PRO D U CTI O N The advantages o both solar heating panels and photovoltaic cells are that maintenance costs are low, and that there are no uel costs. Individual households can use them. D isadvantages include high initial cost, and the relatively large inefciencies ( so large areas o cells are needed) . It goes without saying that the outputs o both types o cell are variable, and depend on both season and weather. We will look at the reasons or these variations in S ub- topic 8.2 . The mathematics o both photovoltaic cells and solar heating panels is straightorward. At a particular location and time o day, the S un radiates a certain power per square metre I ( also known as the intensity o the radiation) to the panels. Panels have an area A, so the power arriving at the surace o the panel will be IA. Panels have an efciency which is the raction o the energy arriving that is converted into internal energy ( o the liquid in the heating panels) or electrical energy ( photovoltaics) . S o the total power converted by the panel is IA.
Worked examples 1
A house requires an average power o 4.0 kW in order to heat water. The average solar intensity at the Earths surace at the house is 65 0 W m 2 . C alculate the minimum surace area o solar heating panels required to heat the water i the efciency o conversion o the panel is 2 2 % .
Solution 4000 W are required, each 1 m 2 o panel can produce 22 65 0 ___ = 1 40 W 1 00 40 0 0 Area required = ____ 1 40
= 2 8 m2 2
Identiy the energy changes in photovoltaic cells and in solar heating panels.
Solution A solar heating cell absorbs radiant energy and converts it to the internal energy o the working uid. A photovoltaic cell absorbs photons and converts their energy to electrical energy.
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R
8.2 Thermal energy transfer Understanding Conduction, convection, and thermal radiation Black-body radiation Albedo and emissivity Solar constant Greenhouse efect Energy balance in the suraceatmosphere
system
Nature of science The study o the Earths climate illustrates the importance o modelling in science. The kinetic theory or an ideal gas is a good model or the way that real gases actually behave. Scientists model the Earths climate in an attempt to understand the implications o the release o greenhouse gases or global warming. The climate is a much more complex system than a simple gas. Issues or scientists include: the availability o data or the planet as a whole, and greater computing power means that more sophisticated models can be tested. Collaboration between research groups means that debate about the accuracy o the models can take place.
Applications and skills Sketching and interpreting graphs showing
the variation o intensity with wavelength or bodies emitting thermal radiation at diferent temperatures Solving problems involving SteanBoltzmann and Wiens laws Describing the efects o the Earths atmosphere on the mean surace temperature Solving albedo, emissivity, solar constant, and the average temperature o the Earth problems
Equations Stean-Boltzmann equation: P = eAT4
2.90 10 - 3 Wiens Law: m a x (metres) = ___ T(kelvin) power intensity equation: I = __ A total scattered power albedo = _____ total incident power
Introduction We considered some of the energy resources in use today in Sub-topic 8.1 . In the course of that sub-topic some of the pollution and atmospheric effects of the resources were mentioned. In this sub-topic we look in more detail at the physics of the Earths atmosphere and the demands that our present need for energy are making on it. After a review of the ways in which energy is transferred due to differences in temperature, we discuss the Suns radiation and its effect on the atmosphere. Finally, we look at how changes in the atmosphere modify the climate.
Thermal energy transfer Any object with a temperature above absolute zero possesses internal energy due to the motion of its atoms and molecules. The higher the temperature of the object, the greater the internal energy associated with
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EN ERGY PRO D U CTI O N the molecules. Topic 3 showed that in a gas the macroscopic quantity that we call absolute temperature is equivalent to the average o the kinetic energy o the molecules. Given the opportunity, energy will spontaneously transer rom a region at a high temperature to a region at a low temperature. Heat, we say rather loosely, fows rom hot to cold. In this sub-topic we look at the ways in which energy can fow due to dierences in temperature. There are three principal methods: conduction, convection, and thermal radiation. All are important to us both on an individual level and in global terms.
Thermal conduction There are similarities between electrical conduction and thermal conduction to the extent that this section is labelled thermal conduction or correctness. However, or the rest o our discussion we will use the term conduction, taking the word to mean thermal conduction. We all know something o practical conduction rom everyday experience. B urning a hand on a camping stove, plunging a very hot metal into cold water which then boils, or melting ice in the hand all give experience o energy moving by conduction rom a hot source to a cold sink. cool water ice good conductor warm water
Figure 1 Laboratory conduction demonstrations.
Metals are good thermal conductors, j ust as they are also good electrical conductors. Poor thermal conductors such as glass or some plastics also conduct electricity poorly. This suggests that there are similarities between the mechanisms at work in both types o conduction. However, it should be noted that there are still considerable dierences in scale between the very best metal conductors ( copper, gold) and the worst metals ( brass, aluminium) . There are many lab experiments that you may have seen designed to help students recognize the dierent thermal properties o good and poor conductors. faster vibrating ions collide with slower vibrating ions
free electrons transfer energy through the metal
hot
cold
atom
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electron
Figure 2 How conduction occurs at an atomic level.
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
In conduction processes, energy ows through the bulk o the material without any large- scale relative movement o the atoms that make up the solid. C onduction ( electrical and thermal) is known as a transport phenomenon. Two mechanisms contribute to thermal conduction:
Atomic vibration occurs in all solids, metal and non-metal alike. At all temperatures above 0 K, the ions in the solid have an internal energy. So they are vibrating about their average fxed position in the solid. The higher the temperature, the greater is the average energy, and thereore the higher their mean speed. Imagine a bar heated at one end and cooled at the other (see fgure 2 ) . At the hot end, the ions vibrate at a large amplitude and with a large average speed. At the cold end the amplitude is lower and the speed is smaller. At the position where the bar is heated the ions vibrate with increasing amplitude and collide with their nearest neighbours. This transers internal energy and the amplitude o the neighbours will increase; this process continues until the bar reaches thermal equilibrium. In this case the energy supplied to each ion is equal to that transerred by the ion to its neighbours in the bar or the surroundings. Each region o the bar will now be at the same uniorm temperature.
C onduction can occur in gases and liquids as well as solids, but, because the inter- atomic connections are weaker and the atoms ( particularly in the gases) are arther apart in uids, conduction is much less important in many gases and liquids than is convection.
Although thermal conduction by atomic vibration is universal in solids, there are other conduction processes that vary in importance depending on the type o solid under discussion. As we saw in Topic 5 , electrical conductors have a covalent (or metallic) bonding that releases free electrons into what is essentially an electron gas flling the whole o the interior o the solid. These ree electrons are in thermal equilibrium with the positive ions that make up the atomic lattice o the solid. The electrons can interact with each other and the energy rom the hightemperature end o the solid diuses along the solid by interactions between these electrons. When an electron interacts with an atom, then energy is transerred back into the atomic lattice to change the vibrational state o the atom. This ree-electron mechanism or conduction depends critically on the numbers o ree electrons available to the solid. Good electrical conductors, where there are many charge carriers (ree electrons) available per unit volume, are likely also to be good thermal conductors. For example, in copper there is one ree electron per atom. You should be able to use Avogadros number, the density o copper and its relative atomic mass, to show that there are 8.4 1 0 28 electrons in every cubic metre o copper.
Nature of science Thermal and electrical resistivity Analogies are oten used in science to aid our understanding o phenomena. Electrical and thermal conduction are closely linked and so it ought to be possible to transer some ideas rom
one to the other. There is an analogy between thermal and electrical eects when thermal conduction is compared with electrical conduction or a wire o cross- section area A and length l:
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EN ERGY PRO D U CTI O N
Thermal
where is the electrical resistivity o the material.
The rate o transer o internal energy Q is proportional to temperature gradient:
The thermal resistivity k o a material corresponds to the electrical resistivity and the temperature gradient corresponds to the electric potential gradient. The rate at which both internal energy and charge are carried through the wire is related to the presence o ree electrons in the metal. It is more than a coincidence o equations, physics at the atomic scale is involved. Are good electrical conductors also good thermal conductors? C ompare the values o and k or dierent metals.
( )
Q A T _ = -_ _ k l t
where k is the thermal resistivity o the material.
Electrical The rate o transer o charge q ( the current) is proportional to potential gradient:
( )
Q A _ V _ = -_ l t
Convection Convection is the movement o groups o atoms or molecules within uids (liquids and gases) that arises through variations in density. Unlike conduction, which involves the microscopic transer o energy, convection is a bulk property. Convection cannot take place in solids. An understanding o convection is important in many areas o physics, astrophysics and geology. In some hot countries, houses are designed to take advantage o natural convection to cool down the house in hot weather.
(a)
chimney
(b)
potassium permanganate crystal
(c)
potassium permanganate water
candle
glass fronted box Figure 3
Convection currents.
Figure 3 shows three lab experiments that involve convection. In all three cases, energy is supplied to a uid. Look at the glass-ronted box (a) frst. A candle heats the air underneath a tube (a chimney) that leads out o the box. The air molecules immediately above the ame move urther apart decreasing the air density in this region. With a smaller density these molecules experience an upthrust and move up through the chimney. This movement o air reduces the pressure slightly which pulls cooler air down the other chimney. Further heating o the air above the ame leads to a continuous current o cold air down the right- hand chimney and hot air up the let-hand tube. This is a convection current.
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R Similar currents can be demonstrated in liquids. Figure 3(b) shows a small crystal o a soluble dye (potassium permanganate) placed at the bottom o a beaker o water. When the base o the beaker is heated gently near to the crystal, water at the base heats, expands, becomes less dense, and rises. This also leads to a convection current as in fgure 3(c) where a glass tube, in the shape o a rectangle, again with a small soluble coloured crystal, can sustain a convection current that moves all around the tube.
day time warm air
cool air
This is the mechanism by which all the water heated in a saucepan on a stove eventually reaches a uniorm temperature.
night time
Examples of convection
warm air
cool air
There are many examples o convection in action. Figures 4 and 5 show examples rom the natural world; there are many others.
Sea breezes I you live by an ocean you will have noticed that the direction o the breeze changes during a 2 4- hour period. D uring the day, breezes blow on- shore rom the ocean, at night the direction is reversed and the breeze blows o-shore rom the land to the sea.
Figure 4 Sea
breezes.
C onvection eects explain this. D uring the day the land is warmer than the sea and warm air rises over the land mass, pulling in cooler air rom above the ocean. At night the land cools down much more quickly than the sea ( which has a temperature that varies much less) and now the warmer air rises rom the sea so the wind blows o- shore. ( You might like to use your knowledge o specifc heat capacity to explain why the sea temperature varies much less than that o the land. ) A similar eect occurs in the ront range o the Rocky Mountains in the USA. The east-acing hills warm up frst and the high-pressure region on the plains means that the wind blows towards the mountains. Later in the day, the east-acing slopes cool down frst and the eect is reversed.
Convection in the Earth
subduction
subduction mid-ocean ridge
convection
Figure 5 Convection
upwelling
At the bottom o the Atlantic O cean, and elsewhere on the planet, new crust is being created. This is due to convection eects that are occurring below the surace. The Earths core is at a high temperature and this drives convection eects in the part o the planet known as the upper mantle. Two convection currents operate and drive material in the same direction. Material is upwelling at the top o these currents to reach the surace o the Earth at the bottom o the ocean. This creates new land that is orcing the Americas, Europe, and Arica apart at the rate o a ew centimetres every year. In other parts o the world convection currents are pulling material back down below the surace ( subduction) . These convection currents have, over time, given rise to the continental drit that has shaped the continents that we know today.
convection
currents in the Earths
mantle.
Why the winds blow The complete theory o why and how the winds blow would occupy a large part o this book, but essentially the winds are driven by uneven heating o the Earths surace by the Sun. This dierential heating can be due to many causes including geographical actors and the presence o cloud. However,
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EN ERGY PRO D U CTI O N where the land or the sea heat up, the air just above them rises and creates an area o low pressure. Conversely, where the air is alling a high-pressure zone is set up. The air moves rom the high- to the low-pressure area and this is what we call a wind. There is a urther interaction o the wind velocity with the rotation o the Earth (through an eect known as the Coriolis orce) . This leads to rotation o the air masses such that air circulates clockwise around a high-pressure region in the northern hemisphere but anti-clockwise around a high-pressure area in the southern hemisphere.
Nature of science Modelling convection using a scientifc analogy Faced with a hot cup o morning coee and little time to drink it, most o us blow across the liquid surace to cool it more quickly. This causes a more rapid loss o energy than doing nothing and thereore the temperature o the liquid drops more quickly too. This is an example o forced convection when the convection cooling is aided by a draught o air. D oing nothing and allowing the convection currents to set up by themselves is natural convection. Newton stated an empirical law or cooling under conditions o orced convection. He suggested that the rate o change o the temperature o d the cooling body __ was proportional to the dt temperature dierence between the temperature o the cooling body and the temperature o the surroundings s .
The key to understanding this equation is to realize that it is about the temperature difference between the hot obj ect and its surroundings; we call this the temperature excess. Newtons law o cooling leads to a hal- lie behaviour in j ust the same way that radioactive hal- lie ollows rom the radioactive equation dN _ N dt where N is the number o radioactive atoms in a sample. Using radioactivity as an analogy, there is a cooling hal- lie so that the time or the temperature excess over the surroundings to halve is always the same or a particular situation o hot obj ect and surroundings. This is another analogy that helps us to understand science by linking apparently dierent phenomena.
In symbols, d _ ( - s) dt
Worked examples 1
E xplain the role played by convection in the fight o a hot- air balloon.
Solution The air in the gas canopy is heate d rom below and as a result its te mperature increases. The hot air in the balloon expands and its de nsity decreases below that o the cold air outside the gas e nvelope. There is thereore an upward orce on the balloon. I this e xceeds the weight o the balloon ( plus basket and occupants) then the balloon will accelerate upwards. 2
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S uggest two reasons why covering the liquid surace o a cup o hot chocolate with marshmallows will slow down the loss o energy rom the chocolate.
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
Solution The marshmallows, having air trapped in them, are poor conductors so they allow only a small fow o energy through them. The upper surace o marshmallow will be at a lower temperature than the lower surace. This reduces the amount o convection occurring at the surace as the convection currents that are set up will not be so strongly driven as the dierential densities will not be so great.
Thermal radiation Basics Thermal radiation is the transer o energy by means o electromagnetic radiation. E lectromagnetic radiation is unique as a wave in that it does not need a medium in order to move ( propagate) . We receive energy rom the S un even though it has passed through about 1 5 0 million km o vacuum in order to reach us. Radiation thereore diers rom conduction and convection, both o which require a bulk material to carry the energy rom place to place. Thermal radiation has its origins in the thermal motion o particles o matter. All atoms and molecules at a temperature greater than absolute zero are in motion. Atoms contain charged particles and when these charges are accelerated they emit photons. It is these photons that are the thermal radiation.
Investigate! Black and white surfaces
Take two identical tin cans and cut out a lid or each one rom polystyrene. Have a hole in each lid or a thermometer. Paint one can completely with matt black paint, paint the other shiny white.
thermometer
black ca n
white ca n
wood block
wood block
Figure 6 Comparing emission of radiation from two surfaces.
Fill both cans with the same volume o hot water at the same temperature, replace the lids and place the thermometers in the water. Keep the cans apart so that radiation rom one is not incident on the other.
C ollect data to enable you to plot a graph to show how the temperature o the water in each can varies with time. This is called a cooling curve.
You could also consider doing the experiment in reverse, beginning with cold water and using a radiant heater to provide energy or the cans. In this case, you must make sure that the heater is the same distance rom the surace o each can and that the shiny can is unable to refect radiation to the black one.
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EN ERGY PRO D U CTI O N Experiments such as the one in Investigate! suggest that black suraces are very good at radiating and absorbing energy. The opposite is true or white or shiny suraces; they reect energy rather than absorb it and are poor at radiating energy. This is why dispensers o hot drinks are oten shiny it helps them to retain the energy.
Nature of science Radiator or not?
Making a saucepan
In many parts o the world, houses need to be heated during part or all o the year. O ne way to achieve this is to circulate hot water rom a boiler through a thin hollow panel oten known as a radiator. B ut is this the appropriate term?
We all need pans to cook our ood. What is the best strategy or designing a saucepan?
The outside metal surace o the panel becomes hot because energy is conducted rom the hot water through the metal. The air near the surace o the panel becomes hotter and less dense; it rises, setting up a convection current in the room. There is some thermal radiation rom the surace but as its temperature is not very dierent rom that o the room, the net radiation is quite low certainly lower than the contributions rom convection. Should the radiator be called a radiator?
The pan will be placed on a at hot surace heated either by ame or radiant energy rom an electrically heated flament or plate. The energy conducts through the base and heats the contents o the saucepan. The base o the pan needs to be a good conductor to allow a large energy ux into the pan. The walls o the saucepan need to withstand the maximum temperature at which the pan will be used but should not lose energy i possible. Giving them a shiny silver fnish helps this. The handle o the pan needs to be a poor conductor so that the pan can be lited easily and harmlessly. D ont make it solid, make it strong and easy to hold but as thin as possible ( think how electrical resistance varies with the shape and thickness o a conductor) . C onclusion: a good pan will have a thick copper base ( a good conductor) , sides, and handle o stainless steel ( a poor conductor) and the overall fnish will be polished and silvery.
Black-body radiation
dull black cylinder incident ray
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Figure 7 A black-body enclosure.
The simple experiments showed that black suraces are good radiators and absorbers but poor reectors o thermal energy. These poor reectors lead to a concept that is important in the theory o thermal radiation: the black-body radiator. A black body is one that absorbs all the wavelengths o electromagnetic radiation that all on it. Like the ideal gas that we use in gas theory, the black body is an idealization that cannot be realized in practice although there are obj ects that are very close approximations to it. O ne way to produce a very good approximation to a black body is to make a small hole in the wall o an enclosed container ( a cavity) and to paint the interior o the container matt black. The container viewed through the hole will look very black inside.
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
S ome o the frst experiments into the physics o the black body were made by Lummer and Pringsheim in 1 8 9 9 using a porcelain enclosure made rom fred clay. When such enclosures are heated to high temperatures, radiation emerges rom the cavity. The radiation appears coloured depending on the temperature o the enclosure. At low temperatures the radiation is in the inra- red region, but as the temperature rises, the colour emitted is frst red, then yellow, eventually becoming white i the temperature is high enough. The intensity o the radiation coming rom the hole or cavity is higher when the cavity is at a higher temperature. The emission rom the hole is not dependent on the material rom which the cavity is made unlike the emission rom the surace o a container. This can be seen in the picture o the interior o a steel urnace ( see fgure 8) . In the centre o the urnace at its very hottest point, the colour appears white, at the edges the colour is yellow. At the entrance to the urnace where the temperature is very much lower, the colour is a dull red.
colour and temperature/K 1000 2000 2500 3200 3300 3400 3500 4500 4000 5000
Figure 8 Interior of furnace.
The emission spectrum from a black body Although there is a predominant colour to the radiation emitted rom a black-body radiator, this does not mean that only one wavelength emerges. To study the whole o the radiation that the black body emits, an instrument called a spectrometer is used. It measures the intensity o the radiation at a particular wavelength. Intensity is the power emitted per square metre. As an equation this is written: P I= _ A where I is the intensity, P is the power emitted, and A is the area on which the power is incident. The units o intensity are W m 2 or J s 1 m 2 . A typical intensitywavelength graph is shown in fgure 9 or a black body at the temperature o the visible surace o the S un, about 6000 K. The S un can be considered as a near- perect black- body radiator. The
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EN ERGY PRO D U CTI O N
A potter needs to know the temperature o the inside o a kiln while the clay is being fred to transorm it into porcelain. S ome potters simply look into the kiln through a small hole. They can tell by experience what the temperature is by seeing the radiating colour o the pots inside. O ther potters use an instrument called a pyrometer. A tungsten flament is placed at the entrance to the kiln between the kiln interior and the potters eye. An electric current is supplied to the flament and this is increased until the flament disappears by merging into the background. At this point it is at the same temperature as the interior o the kiln. The flament system will have previously been calibrated so that the current required or the flament to disappear can be equated to the flament temperature.
relative intensity
The potters kiln
483 nm
max T = 2.898 10 -3 m K The wavelength of the peak of the black-body radiation curve gives a measure of temperature
6000 K
visible
0
wavelength
Figure 9 Intensity against wavelength for a black body at the temperature of the Sun.
graph shows how the relative intensity o the radiation varies with the wavelength o the radiation at which the intensity is measured. There are a number o important eatures to this graph:
There is a peak value at about 5 00 nm ( somewhere between green and blue light to our eyes) . ( Is it a coincidence that the human eye has a maximum sensitivity in this region?)
There are signifcant radiations at all visible wavelengths.
There is a steep rise rom zero intensitynotice that the line does not go through the origin.
At large wavelengths, beyond the peak o the curve, the intensity alls to low levels and approaches zero asymptotically.
Figure 1 0 shows the graph when curves at other temperatures are added and this gives urther perspectives on the emission curves. 483 nm
relative intensity (arbitrary units)
Nature of science
0
visible 580 nm
6000 K 5000 K 724 nm
0
966 nm (IR)
4000 K 3000 K
wavelength
Figure 10 Intensity against wavelength for black bodies at four temperatures.
This diagram shows our curves all at dierent temperatures. As beore the units are arbitrary, meaning that the graph shows relative and not absolute changes between the curves at the our temperatures.
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R This amily o curves tells us that, as temperature increases:
the overall intensity at each wavelength increases ( because the curve is higher)
the total power emitted per square metre increases ( because the total area under the curve is greater)
the curves skew towards shorter wavelengths ( higher requencies)
the peak o the curve moves to shorter wavelengths.
The next step is to ocus on the exact changes between these curves.
Wiens displacement law In 1 893 Wilhelm Wien was able to deduce the way in which the shape o the graph depends on temperature. He showed that the height o the curve and the overall width depends on temperature alone. His ull law allows predictions about the height o any point on the curve but we will only use it to predict the peak o the intensity curve. Wiens disp lacement law states that the wavelength at which the intensity is a maximum max in metres is related to the absolute temperature o the black body T by
Tip Notice that the unit or b is metre kelvin and must be written with a space between the symbols, take care not to write it as mK which means millikelvin and is something quite diferent!
max = bT where b is known as Wiens displacement constant. It has the value 2 .9 1 0 3 m K.
SteanBoltzmann law The scientists S tean and B oltzmann independently derived an equation that predicts the total power radiated rom a black body at a particular temperature. The law applies across all the wavelengths that are radiated by the body. S tean derived the law empirically in 1 879 and B oltzmann produced the same law theoretically fve years later. The S tefanB oltzmann law states that the total power P radiated by a black body is given by P = AT 4 where A is the total surace area o the black body and T is its absolute temperature. The constant, , is known as the SteanB oltzmann constant and has the value 5 .7 1 0 8 W m 2 K 4. The law reers to the power radiated by the obj ect, but this is the same as the energy radiated per second. It is easy to show that the energy radiated each second by one square metre o a black body is T 4. This variant o the ull law is known as S teans law.
Grey bodies and emissivity In practice obj ects can be very close to a black body in behaviour but not quite 1 00% perect in the way they behave. They are oten called grey obj ects to account or this. A grey obj ect at a particular temperature will emit less energy per second than a perect black body o the
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EN ERGY PRO D U CTI O N same dimensions at the same temperature. The quantity known as emissivity, e, is the measure o the ratio between these two powers: power emitted by a radiating object e = ________ power emitted by a black body o the same dimensions at the same temperature B eing a ratio, emissivity has no units. For a real material, the power emitted can be written as P = eAT 4
Material water snow ice soil coal
Emissivity 0.60.7 0.9 0.98 0.40.95 0.95
using the same symbols as beore. A perect black body has an emissivity value o 1 . An object that completely reects radiation without any absorption at all has an emissivity o 0. All real objects have an emissivity somewhere these two values. Typical values o emissivity or some substances are shown in the table. Emissivity values are a unction o the wavelength o the radiation. It is surprising that snow and ice, although apparently white and reective, are such eective emitters ( and absorbers) in the inra- red.
Nature of science Building a theory B y the end o the 1 9 th century, the graph o radiation intensity emitted by a black body as a unction o wavelength was well known. Wiens equation ftted the experiments but only at short wavelengths. Rayleigh attempted to develop a new theory on the basis o classical physics. He suggested that charges oscillating inside the cavity produce standing electromagnetic waves as they bounce backwards and orwards between the cavity walls. S tanding waves that escape rom the cavity produce the observed black- body spectrum. Rayleighs model fts the observations at long wavelengths but predicts an ultraviolet catastrophe o an infnitely large intensity at short wavelengths. Max Planck varied Rayleighs theory slightly. He proposed that the standing waves could
not carry all possible energies but only certain quantities o energy E given by nhf where n is an integer, h is a constant ( Plancks constant) and f is the requency o the allowed energy. Plancks model ftted the experimental results at all wavelengths and thus, in 1 9 00, a new branch o physics was born: quantum physics. Planck limited his theory to the space inside the cavity, he believed that the radiation was continuous outside. Later, E instein proposed that the photons outside the cavity also had discrete amounts o energy. Planck was the scientifc reeree or E insteins paper and it is to Plancks credit that he recognised the value o Einsteins work and accepted the paper or publication even though it overturned some o this own ideas.
Worked examples 1
The S un has a surace temperature o 5 800 K and a radius o 7. 0 1 0 8 m. C alculate the total energy radiated rom the S un in one hour.
Solution P = AT 4 S urace area o Sun = 4r2 = 4 3 .1 4 ( 7 1 0 8 ) 2 = 6.2 1 0 1 8
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R ___________________
S o power = 5 .7 1 0
-8
6.2 1 0
18
5 800
= 4.0 1 0 26 W In one hour there are 3600 s, so the energy radiated in one hour is 1 .4 1 0 30 J. 2
A metal flament used as a pyrometer in a kiln has a length o 0. 05 0 m and a radius o 1 . 2 1 0 3 m. D etermine the temperature o the flament at which it radiates a power o 48 W.
Solution The surace area o the flament is 2 rh = ( 2 1 .2 1 0 - 3 ) 0.05 0 = 3 . 8 1 0 4 m 2 S o the power determines the temperature as 48 = 5 .7 1 0 - 8 3 .8 1 0 4 T 4
T=
4
3
48 = 1 2 00 K. ___ 5 .7 1 0 3 .8 1 0 4
-8
-4
A spherical black body has an absolute temperature T1 and surace area A. Its surroundings are kept at a lower temperature T2 . D etermine the net power lost by the body.
Solution The power emitted by the body is; AT1 4 the power absorbed rom the surroundings is AT 2 4. S o the net power lost is A( T1 4 - T2 4) . Note that this is not the same as A( T1 - T2 ) 4.
Sun and the solar constant The Sun emits very large amounts o energy as a result o its nuclear usion reactions. B ecause the Earth is small and a long way rom the Sun, only a small raction o this arrives at the top o the Earths atmosphere. A black body at the temperature o the Sun has just under hal o its radiation in our visible region, roughly the same amount in the inrared, and 1 0% in the ultraviolet. It is the overall dierence between this incoming radiation and the radiation that is subsequently emitted rom the Earth that determines the energy gained by the Earth rom the Sun. This energy is used by plants in photosynthesis and it drives the changes in the worlds oceans and atmospheres; it is crucial to lie on this planet. The amount o energy that arrives at the top o the atmosphere is known as the solar constant. A precise defnition is that the solar constant is the amount of solar radiation across all wavelengths that is incident in one second on one square metre at the mean distance of the Earth from the Sun on a plane perpendicular to the line joining the centre of the Sun and the centre of the Earth. The energy rom the Sun is spread over an imaginary sphere that has a radius equal to the EarthSun distance. The Earth is roughly 1 .5 1 0 1 1 m rom the Sun and so the surace area o this sphere is 2 .8 1 0 23 m 2 . The S un emits 4 1 0 26 J in one second. The energy incident in one second on one square metre at the distance o the E arth rom the 4. 0 1 0 S un is ________ = 1 400 J. The answer is quoted to 2 s.., a reasonable 2 .8 1 0 precision or this estimate and represents about 5 1 0 - 1 0 o the entire output o the Sun.
radiation rom Sun
26 23
The value o the solar constant varies periodically or a number o reasons:
The output o the Sun varies by about 0.1 % during its principal 1 1 -year sunspot cycle. The Earths orbit is elliptical with the E arth slightly closer to the S un in January compared to July; this accounts or a dierence o about
one square metre at top o atmosphere
Figure 11 Defning the solar constant.
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EN ERGY PRO D U CTI O N 7% in the solar constant. ( Note that this dierence is not the reason or summer in the Southern Hemisphere the seasons occur because the axis o rotation o the Earth is not perpendicular to the plane o its orbit around the Sun.)
O ther longer- period cycles are believed to occur in the S un with periods ranging rom roughly hundreds to thousands o years.
Energy balance in the Earth surface atmosphere system The solar constant is the power incident on the top o the atmosphere. It is not the power that arrives at ground level. As the radiation rom the Sun enters and travels through the atmosphere, it is subj ect to losses that reduce the energy arriving at the Earths surace. Radiation is absorbed and also scattered by the atmosphere. The degree to which this absorption and scattering occurs depends on the position o the S un in the sky at a particular place. When the S un is lower in the sky ( as at dawn and sunset) , its radiation has to pass through a greater thickness o atmosphere and thus more scattering and absorption takes place. This gives rise to the colours in the sky at dawn and dusk. Even when the energy arrives at ground level, it is not necessarily going to remain there. The surace o the Earth is not a black body and thereore it will reect some o the energy back up towards the atmosphere. The extent to which a particular surace can reect energy is known as its albedo (rom the Latin word or whiteness) . It is given the symbol a: energy reected by a given surace in a given time a = _____ total energy incident on the surace in the same time Like emissivity, albedo has no units, it varies rom 0 or a surace that does not reect any energy ( a black body) to 1 or a surace that absorbs no radiation at all. Unless stated otherwise, the albedo in the Earth system is normally quoted or visible light ( which as we saw earlier accounts or nearly a hal o all radiation at the surace) . The average annual albedo or the whole Earth is about 0.3 5 , meaning that on average about 3 5 % o the Suns rays that reach the ground are reected back into the atmosphere. This fgure o 0. 3 5 is, however, very much an average because albedo varies depending on a number o actors:
Surface Ocean Fresh snow Sea ice Ice Urban areas Desert soils Pine forest Deciduous forest
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Albedo 0.06 0.85 0.60 0.90 0.15 0.40 0.15 0.25
It varies daily and with the seasons, depending on the amount and type o cloud cover ( thin clouds have albedo values o 0. 3 0.4, thick cumulo- nimbus cloud can approach values o 0.9) .
It depends on the terrain and the material o the surace. The table gives typical albedo values or some common land and water suraces.
The importance o albedo will be amiliar to anyone who lives where snow is common in winter. Fresh snow has a high albedo and reects most o the radiation that is incident on it the snow will stay in place or a long time without melting i the temperature remains cold. However, sprinkle some earth or soot on the snow and, when the sun shines, the snow will soon disappear because the dark material on its
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
surace absorbs energy. The radiation provides the latent heat needed to melt the snow. Albedo eects help to explain why satellites ( including the International S pace S tation) in orbit around the Earth can take pictures o the Earths cloud cover and surace in the visible spectrum.
Worked examples 1
Four habitats on the E arth are: orest, grassland ( savannah) , the sea, an ice cap. D iscuss which o these have the greatest and least albedo.
D ata Incident intensity rom the S un = 3 40 W m 2 Reected intensity at surace
= 1 00 W m 2
Radiated intensity rom surace = 2 40 W m 2
Solution A material with a high albedo reects the incident visible radiation. Ice is a good reector and consequently has a high albedo. On the other hand, the sea is a good absorber and has a low albedo. 2
The data give details o a model o the energy balance o the Earth. Use the data to calculate the albedo o the Earth that is predicted by this model.
Re- radiated intensity rom atmosphere back to surace
= 2 W m 2
Solution The defnition o albedo is clear. p o w e r re e cte d b y a give n su race It is ________________________ to tal p o w e r incide nt o n the su race 1 00 S o in this case the value is ___ = 0.2 9 3 40
The greenhouse efect and temperature balance The Earth and the Moon are the same average distance rom the S un, yet the average temperature o the Moon is 2 5 5 K, while that o the E arth is about 2 90 K. The discrepancy is due to the Earth having an atmosphere while the Moon has none. The dierence is due to a phenomenon known as the greenhouse effect in which certain gases in the E arths atmosphere trap energy within the Earth system and produce a consequent rise in the average temperature o the E arth. The most important gases that cause the eect include carbon dioxide ( C O 2 ) , water vapour ( H 2 O ) , methane ( C H 4) , and nitrous oxide ( dinitrogen monoxide; N 2 O ) , all o which occur naturally in the atmosphere. O zone ( O 3 ) , which has natural and man- made sources, makes a contribution to the greenhouse eect. It is important to distinguish between:
the natural greenhouse eect that is due to the naturally occurring levels o the gases responsible, and
the enhanced greenhouse eect in which increased concentrations o the gases, possibly occurring as a result o human- derived processes, lead to urther increases in the E arths average temperature and thereore to climate change.
The principal gases in the atmosphere are nitrogen, N 2 , and oxygen, O 2 , ( respectively, 70% and 2 0% by weight) . B oth o these gases are made up o tightly bound molecules and, because o this, do not absorb energy rom sunlight. They make little contribution to the natural greenhouse eect. The 1 % o the atmosphere that is made up o the C O 2 , H 2 O , C H 4 and N 2 O has a much greater eect.
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EN ERGY PRO D U CTI O N The molecular structure o greenhouse- gas molecules means that they absorb ultraviolet and inra- red radiation rom the S un as it travels through the atmosphere. Visible light on the other hand is not so readily absorbed by these gases and passes through the atmosphere to be absorbed by the land and water at the surace. As a result the temperature o the surace rises. The E arth then re- radiates j ust like any other hot obj ect. The temperature o the E arths surace is ar lower than that o the S un, so the wavelengths radiated rom the E arth will peak in the long- wavelength inra- red. The absorbed radiation had, o course, mostly been in the visible region o the electromagnetic spectrum. S o, j ust as gases in the atmosphere absorbed the S uns inrared on the way in, now they absorb energy in the inra- red rom the E arth on the way out. The atmosphere then re- radiates the energy yet again, this time in all directions meaning that some returns to E arth. This energy has been trapped in the system that consists o the surace o the E arth and the atmosphere.
Nature of science Other worlds, other atmospheres The dynamic equilibrium in our climate has been very important or the evolution o lie on Earth. Venus and Mars evolved very dierently rom Earth. Venus has similar dimensions to the Earth but is closer to the Sun with a very high albedo at about 0.76. Its atmosphere is almost entirely carbon dioxide. In consequence, the surace temperature reaches a 730 K and a runaway greenhouse eect acts. Venus and Mars are clear reminders to us o the ragility o a planetary climate.
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The whole system is a dynamic equilibrium reaching a state where the total energy incident on the system rom the S un equals the energy total being radiated away by the Earth. In order to reach this state, the temperature o the Earth has to rise and, as it does so, the amount o energy it radiates must also rise by the SteanB oltzmann law. Eventually, the Earths temperature will be such that the balance o incoming and outgoing energies is attained. O course, this balance was established over billions o years and was steadily changing as the composition o the atmosphere and the albedo changed with changes in vegetation, continental drit, and geological processes.
Why greenhouse gases absorb energy Ultraviolet and long- wavelength inra- red radiations are absorbed by the atmosphere. Photons in the ultraviolet region o the electromagnetic spectrum are energetic and have enough energy to break the bonds within the gas molecules. This leads to the production o ionic materials in the atmosphere. A good example is the reaction that leads to the production o ozone rom the oxygen atoms ormed when oxygen molecules are split apart by ultraviolet photons. The energies o inra-red photons are much smaller than those o ultraviolet and are not sufcient to break molecules apart. When the requency o a photon matches a vibrational state in a greenhouse gas molecule then an eect called resonance occurs. We will look in detail at the vibrational states and resonance in carbon dioxide, but similar eects occur in all the greenhouse gas molecules. In a carbon dioxide molecule, the oxygen atoms at each end are attached by double bonds to the carbon in a linear arrangement. The bonds resemble springs in their behaviour. The molecule has our vibrational modes as shown in fgure 1 2. The frst o these modes a linear symmetric stretching does not cause inrared absorption, but the remaining three motions do. Each one has
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
O
C
O
equilibrium position
symmetric stretching bending modes anti-symmetric stretching
Figure 12 Vibrational states in the carbon dioxide molecule.
a characteristic requency. I the requency o the radiation matches this, then the molecule will be stimulated into vibrating with the appropriate mode and the energy o the vibration will come rom the incident radiation. This leads to vibrational absorption at wavelengths o 2.7 m, 4.3 m and 1 5 m. These eects o these absorptions can be clearly seen in fgure 1 3 which shows part o the absorption spectrum o carbon dioxide. In this diagram a peak indicates a wavelength at which signifcant absorption occurs.
Modelling the climate balance
relative absorption
2
3 4 5 wavelength/m
6
Figure 13 Part of the absorption spectrum for carbon dioxide.
We said earlier that about 1 400 J alls on each square metre o the upper atmosphere each second: the solar constant. We use the physics introduced in this topic to see what the consequences o this are or the E arths suraceatmospheric system. The ull 1 400 J does not o course reach the surace. O the total, about 2 5 % o the incident energy is reected by the clouds and by particles in the atmosphere, about 2 5 % is absorbed by the atmosphere, and about 6% is reected at the surace. The incoming radiation alls on the portion o the Earths surace which is normal to the Suns radiation i. e. a circle o area equal to ( radius of Earth) 2 , as only one side o the E arth aces the S un at any one time. However this radiation has to be averaged over the whole o the surace which is 4 ( radius of Earth) 2 . S o the mean power arriving at each 1 40 0 square metre is ____ = 3 5 0 W. 4 The albedo now has to be taken into account to give an eective mean power at one square metre o the surace o ( 1 a) 3 5 0 For the average E arth value or a o about 0. 3 , the mean power absorbed by the surace per square metre is 2 45 W.
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energy intensity (arbitrary units)
EN ERGY PRO D U CTI O N
2 45 = T4
0
transmittance
Figure 14 Intensity and transmittance for a completely transparent atmosphere.
energy transmitted wavelength energy not transmitted
transmittance energy intensity (arbitrary units)
0
transmittance 0
energy not transmitted
wavelength
290 K
256 K
0
wavelength
100%
wavelength
Figure 15 Intensity and transmittance for an atmosphere opaque to infra-red and ultraviolet radiation.
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We made the assumption that the Earth emits 2 45 W m - 2 and that this energy leaves the surace and the atmosphere completely. This would be true or an atmosphere that is completely transparent at all wavelengths, but Earths atmosphere is not transparent in this way. Figure 1 4 shows the relative intensitywavelength graph or a black body at 2 5 6 K. As expected, the area under this curve is 2 45 W m 2 and represents the predicted emission rom the E arth. It shows the response o an atmosphere modelled as perectly transparent at all wavelengths. ( Technically, this graph shows the transmittance o the atmosphere as a unction o wavelength, a value o 1 00% means that the particular wavelength is completely transmitted, 0 means that no energy is transmitted at this wavelength.) Not surprisingly all the blackbody radiation leaves the Earth because the transmittance is 1 or all wavelengths in this model.
256 K
100%
2 45 = 256 K _ 5 .67 1 0 4
This is very close to the value or the Moon, which has no atmosphere. We need to investigate why the mean temperature o the E arth is about 3 5 K higher than this.
wavelength
0
(b)
So T =
100%
energy intensity (arbitrary units)
(a)
_________
wavelength
0
The knowledge o this emitted power allows a prediction o the temperature o a black body T that will emit 2 45 W m 2 . Using the S teanB oltzmann law:
256 K
In act the atmosphere absorbs energy in the inra- red and ultraviolet regions. A simple, but slightly more realistic model or this absorption will leave the transmittance at 1 00% or the visible wavelengths and change the transmittance to 0 or the absorbed wavelengths. Figure 1 5 shows how this leads to an increased surace temperature. When the transmittance graph is merged with the graph or blackbody radiation to give the overall emission rom the E arth into space, the area under the overall emission curve will be less than 2 45 W m 2 because the inra- red and ultraviolet wavelengths are now absorbed in the atmosphere and these energies are not lost ( Figure 1 5 ( a) ) . This defcit will be re- radiated in all directions; so some returns to the surace. In order to get the energy balance correct again, the temperature o the emission curve must be raised so that the area under the curve returns to 245 W m 2 . As the curve changes with the increase in temperature, the area under the curve increases too. The calculation o the temperature change required is difcult and not given here. However, or the emission rom the surace to equal the incoming energy rom the Sun, allowing or the absorption, the surace temperature must rise to about 2 90 K. The net eect is shown in Figure 1 5 (b) with a shited and raised emission curve compensating or the energy that cannot be transmitted through the atmosphere. The suggestion that the atmosphere completely removes wavelengths above and below certain wavelengths is an over- simplifcation.
8 . 2 TH E RM AL E N E R G Y TRAN S FE R Figure 1 6 shows the complicated transmittance pattern in the inrared and indicates which absorbing molecules are responsible or which regions o absorption.
transmittance (percent)
100 80 60 40 20 0 0
1
O2
2
H 2O
3
4
5
CO 2 H 2 O CO 2 O 3
6
7 8 9 wavelength/m
10
11
H 2O CO 2 O 3 absorbing molecule
12
13
14
H 2 O CO 2
15
CO 2
Figure 16 Transmittance o the atmosphere in the inra-red.
The energy balance of the Earth The suraceatmosphere energy balance system is very complex; fgure 1 7 is a recent diagram showing the basic interactions and you should study it careully.
global energy fows W m -2 102
refected solar radiation 101.9 W m -2 refected by clouds and atmosphere
341
79 emitted by 169 atmosphere
79
outgoing longwave radiation 238.5 W m -2
239
incoming solar radiation 341.3 W m -2
40 30
atmospheric window greenhouse gases
absorbed by 78 atmosphere latent 80 heat 17 refected by surace 23
161 absorbed by surace
356 396 80 17 surace thermals evaporadiation transpiration net absorbed 0.9 W m -2
40
333 back radiation 333
absorbed by surace
Figure 17 Factors that make up the energy balance o the Earth (ater Stephens and others. 2012. An update on earths energy balance in light of the latest global observations. Nature Geoscience.) .
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EN ERGY PRO D U CTI O N
Global warming There is little doubt that climate change is occurring on the planet. We are seeing a signifcant warming that may lead to many changes to the sea level and in the climate in many parts o the world. The act that there is change should not surprise us. We have recently ( in geological terms) been through several Ice Ages and we are thought to be in an interstadial phase at the moment ( interstadial means between Ice Ages ) . In the seventeenth century a Little Ice Age covered much o northern E urope and North America. The River Thames regularly roze and the citizens held airs on the ice. In 1 6 08, the D utch painter Hendrick Avercamp painted a winter landscape showing the typical extent and thickness o the ice in Holland ( fgure 1 8) .
Figure 18 Winter landscape with skaters (1608) , Hendrick Avercamp.
Many models have been suggested to explain global warming, they include:
changes in the composition o the atmosphere ( and specifcally the greenhouse gases) leading to an enhanced greenhouse eect
increased solar are activity
cyclic changes in the Earths orbit
volcanic activity.
Most scientists now accept that this warming is due to the burning o ossil uels, which has gone on at increasing levels since the industrial revolution in the eighteenth century. There is evidence or this. The table below shows some o the changes in the principal greenhouse gases over the past 2 5 0 years.
Gas Carbon dioxide Methane Nitrous oxide Ozone
Pre-1750 concentration 280 ppm 700 ppb 270 ppb 25 ppb
Recent concentration 390 ppm 1800 ppb 320 ppb 34 ppb
ppm = parts per million; ppb = parts per billion
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% increase since 1750 40 160 20 40
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
The recent values in this table have been collected directly in a number o parts o the world ( there is a well- respected long- term study o the variation o carbon dioxide in Hawaii where recently the carbon dioxide levels exceeded 400 ppm or the frst time or many thousands o years) . The values quoted or pre- 1 75 0 are determined rom a number o sources:
Analysis o Antarctic ice cores. C ores are extracted rom the ice in the Antarctic and these yield data or the composition o the atmosphere during the era when the snow originally ell on the continent. C ores can give data or times up to 400 000 years ago.
Analysis o tree rings. Tree rings yield data or the temperature and length o the seasons and the rainall going back sometimes hundreds o years.
Analysis o water levels in sedimentary records rom lake beds can be used to identiy historical changes in water levels.
An enhanced greenhouse eect results rom changes to the concentration o the greenhouse gases: as the amounts o these gases increase, more absorption occurs both when energy enters the system and also when the surace re- radiates. For example, in the transmittancewavelength graph or a particular gas, when the concentration o the gas rises, the absorption peaks will increase too. The surace will need to increase its temperature in order to emit sufcient energy at sea level so that emission o energy by Earth rom the top o the atmosphere will equal the incoming energy rom the Sun. Global warming is likely to lead to other mechanisms that will themselves make global warming increase at a greater rate:
the ice and snow cover at the poles will melt, this will decrease the average albedo o Earth and increase the rate at which heat is absorbed by the surace. a higher water temperature in the oceans will reduce the extent to which C O 2 is dissolved in seawater leading to a urther increase in atmospheric concentration o the gas.
Nature of science An international perspective There have been a number o international attempts to reach agreements over the ways orward or the planet. These have included:
The Kyoto Protocol was originally adopted by many ( but not all) countries in 1 997 and later extended in 2 01 2 .
The Intergovernmental Panel on Climate Change.
AsiaPacifc Partnership on C lean D evelopment and C limate.
The various other United Nations C onventions on C limate C hange, e.g. C ancn, 2 01 0.
Do some research on the Internet to fnd what is presently agreed between governments.
O ther human- related mechanisms such as deorestation can also drive global warming as the amount o carbon fxed in the plants is reduced. This is a problem that has to be addressed at both an international and an individual level. The world needs greater efciency in power production and a maj or review o uel usage. We should encourage the use o non- ossil-uel methods. As individuals we need to be aware o our personal impact on the planet, we should be conscious o our carbon ootprint. Nations can capture and store carbon dioxide, and agree to increase the use o combined heating and power systems. What everyone agrees is that doing nothing is not an option.
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EN ERGY PRO D U CTI O N
Questions 1
b) The ow rate o water in the pipe is 400 m 3 s 1 . C alculate the power delivered by the alling water.
(IB) a) A reactor produces 2 4 MW o power. The efciency o the reactor is 3 2 % . In the fssion o one uranium- 2 3 5 nucleus 3 .2 1 0 1 1 J o energy is released.
3
D etermine the mass o uranium- 2 3 5 that undergoes fssion in one year in this reactor.
The energy losses in a pumped storage power station are shown in the ollowing table.
b) D uring its normal operation, the ollowing set o reactions takes place in the reactor. 1 0 239 92 239 93
n+
238 92
U
U
Np
239 93 239 94
239 92
Source of energy loss
U
Np + Pu +
0 -1 0 -1
_
e+ v _
e+ v
C omment on the international implications o the product o these reactions. 2
(IB)
Percentage loss of energy
friction and turbulence of water in pipe
27
friction in turbine and ac generator
15
electrical heating losses
5
a) C alculate the overall efciency o the conversion o the gravitational potential energy o water in the tank into electrical energy.
(IB) The diagram shows a pumped storage power station used or the generation o electrical energy.
b) S ketch a Sankey diagram to represent the energy conversion in the power station. tank
4
A nuclear power station uses uranium-2 3 5 ( U- 2 3 5 ) as uel.
pipe
310 m
(IB)
a) O utline: turbine
( i)
( ii) the role o the heat exchanger o the reactor and the turbine in the generation o electrical energy.
lake
Water stored in the tank alls through a pipe to a lake through a turbine that is connected to an electricity generator. The tank is 5 0 m deep and has a uniorm area o 5 .0 1 0 4 m 2 . The height rom the bottom o the tank to the turbine is 3 1 0 m. 3
3
The density o water is 1 .0 1 0 kg m . a) S how that the maximum energy that can be delivered to the turbine by the alling water is about 8 1 0 1 2 J.
350
the processes and energy changes that occur through which the internal energy o the working uid is increased
b) Identiy one process in the power station where energy is degraded. 5
(IB) The intensity o the Suns radiation at the position o the Earth is approximately 1 400 W m 2 . Suggest why the average power received per unit area o the Earth is 3 5 0 W m 2 .
QUESTION S 6
a) State the region o the electromagnetic spectrum to which the resonant requency o nitrogen oxide belongs.
(IB) The diagram shows a radiation entering or leaving the Earths surace or a simplifed model o the energy balance at the Earths surace.
b) Using your answer to ( a) , explain why nitrogen oxide is classifed as a greenhouse gas.
atmosphere TA = 242 K
9 transmitted through radiated by atmosphere 245 W m -2 atmosphere radiated by Earths 0.700 T4A surface = T4E Earths surface TE
a) State the emissivity o the atmosphere. b) D etermine the intensity o the radiation radiated by the atmosphere towards the Earths surace.
( IB) The diagram shows a simple energy balance climate model in which the atmosphere and the surace o Earth are treated as two bodies each at constant temperature. The surace o the E arth receives both solar radiation and radiation emitted rom the atmosphere. Assume that the Earths surace and the atmosphere behave as black bodies. 344 W m -2
c) C alculate TE . e = 0.720
8
a)
O utline a mechanism by which part o the radiation radiated by the Earths surace is absorbed by greenhouse gases in the atmosphere. Go on to suggest why the incoming solar radiation is not aected by the mechanism you outlined.
a = 0.280
atmospheric radiation
242 K
solar radiation
Earths surface
288 K
b) C arbon dioxide ( C O 2 ) is a greenhouse gas. S tate one source and one sink ( that removes C O 2 ) o this gas.
D ata or this model:
(IB)
average temperature o the atmosphere o E arth = 2 42 K
The graph shows part o the absorption spectrum o nitrogen oxide ( N 2 O ) in which the intensity o absorbed radiation A is plotted against requency f.
emissivity e o the atmosphere o Earth = 0.72 0 average albedo a o the atmosphere o E arth = 0.2 80 solar intensity at top o atmosphere = 344 W m 2 average temperature o the surace o E arth = 2 88 K a) Use the data to determine:
A/arbitrary units
7
atmosphere
( i) the power radiated per unit area o the atmosphere ( ii) the solar power absorbed per unit area at the surace o the Earth.
2
3
4
5
f/ 10 13 Hz
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EN ERGY PRO D U CTI O N b) It is suggested that, i the production o greenhouse gases were to stay at its present level, then the temperature o the Earths atmosphere would eventually rise by 6 K. C alculate the power per unit area that would then be ( i) radiated by the atmosphere ( ii) absorbed by the Earths surace. c) Estimate the increase in temperature o the Earths surace.
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1 0 ( IB) It has been estimated that doubling the amount o carbon dioxide in the Earths atmosphere changes the albedo o the Earth by 0.01 . Estimate the change in the intensity being refected by the Earth into space that will result rom this doubling. S tate why your answer is an estimate. Average intensity received at E arth rom the Sun = 3 40 W m 2 Average albedo = 0.3 0
9 WAVE PH E N O M E N A ( AH L) Introduction In this topic we develop many of the concepts introduced in Topic 4. In general, a more mathematical approach is taken and we consider
the usefulness of modelling using a spreadsheet both for graphing and developing relationships through iteration.
9.1 Simple harmonic motion Understanding The defning equation o SHM Energy changes
Nature of science The importance o SHM The equation or simple harmonic motion (SHM) can be solved analytically and numerically. Physicists use such solutions to help them to visualize the behaviour o the oscillator. The use o the equations is very powerul as any oscillation can be described in terms o a combination o harmonic oscillators using Fourier synthesis. The modelling o oscillators has applications in virtually all areas o physics including mechanics, electricity, waves and quantum physics. In this sub-topic we will model SHM using a simple spreadsheet and see how powerul this interpretation can be.
Applications and skills Solving problems involving acceleration,
velocity and displacement during simple harmonic motion, both graphically and algebraicallyO B J TE XT_UND Describing the interchange o kinetic and potential energy during simple harmonic motion Solving problems involving energy transer during simple harmonic motion, both graphically and algebraically
Equations
2 angular velocityperiod equation: = _ 2
defning equation or shm: a = - x
T
displacementtime equations: x = x 0 sin t;
x = x 0 cos t velocitytime equations: v = x 0 cos t; v = - x 0 sin t velocity-displacement _______ equation:
v = ( x 0 2 - x 2 )
1 m 2 ( x0 2 - x2 ) kinetic energy equation: E K = ___ 2 1 total energy equation: E = ___ m 2 x 2 T
2
0
__
period o simple pendulum: T = 2 ___gl __
m period o massspring: T = 2 ____ k
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Introduction In this sub-topic we treat S HM more mathematically but restrict ourselves to two systems the simple pendulum and o a mass oscillating on a spring. Each o these systems is isochronous and is usually lightly damped; this means that a large number o oscillations occur beore the energy in the system is transerred to the internal energy o the system and the surrounding air.
Angular speed or frequency ()
screen metal sphere turntable light source drive belt
Figure 1 Comparison of SHM and circular motion.
In Sub-topic 6.1 we considered the angular speed in relation to circular motion; it is the rate o change o angle with time and is also called angular frequency. It is measured in radians per second (rad s - 1 ) . This quantity is important when we deal with simple harmonic motion because there is a very close relationship between circular motion and SHM. This relationship can be demonstrated using the apparatus shown in fgure 1 . A metal sphere is mounted on a turntable that rotates at a constant angular speed. A simple pendulum is arranged so that it is in line with the sphere and oscillates with the same periodic time as that o the turntable a little trial and error should give a good result here. The pendulum and the turntable are illuminated by a light that is projected onto a screen. The shadows projected, onto the screen, o the circular motion o the sphere and oscillatory motion o the pendulum show these motions to be identical.
Circular motion and SHM In mathematical terms the demonstration in fgure 1 is equivalent to proj ecting the two- dimensional motion o a point onto the single dimension o a line. Imagine a point P rotating around the perimeter o a circle with a constant angular speed . The radius o the circle r j oins P with the centre o the circle O . At time t = 0 the radius is horizontal and at time t it has moved through an angle radians. For constant angular speed = _t , rearranging this gives = t. Proj ecting P onto the y- axis gives the vertical component o r as r sin . Proj ecting P onto the x- axis gives the horizontal component o r as r cos . The variation o the vertical component with time or angle is shown on the right o fgure 2 and takes the orm o a sine curve. B ecause the rate o rotation is constant, the angle or t is proportional to time. I we drew a graph o y against t the quantities T 2 and would be replaced by T and __ respectively. 2 y
y
y = r sin
r P r = t
0 x = r cos
x
0
2
t
-r
Figure 2 Projection of circular motion on to a vertical line. The equations o the proj ections are y = y 0 sin t and x = x 0 cos t.
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9 .1 S I M PLE H ARM O N I C M O TI O N
Here y 0 and x 0 are the maximum values o y and x, which in this case are identical to r. These are the amplitudes o the motion. You may remember rom S ub- topic 4.1 that in S HM the displacement, velocity and acceleration all vary sinusoidally with time. Thus, the proj ection o circular motion on the vertical or horizontal takes the same shape as SHM. This is very useul in analysing S HM.
The relationship between displacement, velocity, and acceleration In S ub-topic 4.1 we saw that, starting rom the displacementtime curve, we could derive the velocitytime and accelerationtime curves rom the gradients. This can be done graphically, but another technique is to dierentiate the equations with respect to time ( dierentiation is equivalent to fnding the gradient) . We have seen rom the comparison with circular motion that the displacement takes the orm: x = x0 sin t or x = x0 cos t depending on when we start timing. It really doesnt matter whether the projection is onto the x-axis or the y-axis thereore we can use x and y interchangeably. Lets start with x = x 0 sin t
Note
You will meet diferential
calculus in your Mathematics or Mathematics Studies course. This is not the place to teach you to diferentiate you will not be expected use calculus on your IB Physics course. However, or students studying physics, engineering and allied subjects at a higher level, calculus will orm a major aspect o your course. In this case, we will diferentiate the equations, but it is the results that are important not the method o obtaining them.
dx = x 0 cos t, where x 0 This means that the velocity is given by v ( = __ dt ) is the amplitude and the angular requency.
The maximum value that cosine can take is 1 so the maximum velocity is v 0 = x 0 thus making the equation or the velocity at time t become v = v 0 cos t We know that the acceleration will be the gradient o a velocitytime graph so we have dv a = __ = - v 0 sin t = - x 0 2 sin t dt
As or cosine, the maximum value that sine can take is 1 so a 0 = v 0 = x 0 2 giving a = - a 0 sin t C omparing the equations x = x 0 sin t and a = - x 0 2 sin t we can see a common actor o x 0 sin t meaning that a = - 2 x This may remind you, in S ub- topic 4.1 , we saw that a = - kx or S HM. S o the constant k must actually be 2 ( the angular requency squared) . We will look at the signifcance o 2 very soon. In examinations, you can be asked to fnd maximum values or velocity and acceleration this makes the calculations easier because the maximum o sine and cosine are each 1 and, thereore, we dont need to include the sine or cosine term in our calculations. Thus the maximum velocity will be v0 = x0 and the magnitude o the maximum acceleration will be a 0 = x0 2 (the direction o a 0 will be opposite that o x0) .
Modelling SHM with a spreadsheet Much can be learned about SHM by using a spreadsheet to graph it. Figure 3 is a screen shot o part o a spreadsheet set up or this purpose
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WAVE P H E N O M E N A ( AH L ) B C D A time/s displacement/m velocity/m s1 acceleration/m s2
6.30 6.10 5.52 4.58 3.36 1.93 0.37 1.21 2.71 4.05 5.12 5.87 6.26 6.24 5.84 5.06 3.96 2.62 1.11 0.48 2.03 3.45 4.66 5.57 6.13 6.30 6.07 5.46 4.51 3.27 1.83 0.26 1.31 2.81 4.13 5.18 5.91 6.27 6.23 5.79 5.00 3.88 2.52 1.00 0.58 2.13 3.54 4.73 5.62
0.00 1.98 3.83 5.45 6.71 7.56 7.92 7.79 7.16 6.09 4.62 2.87 0.93 1.06 2.99 4.73 6.17 7.22 7.81 7.92 7.52 6.64 5.35 3.72 1.85 0.13 2.11 3.95 5.54 6.78 7.60 7.93 7.76 7.11 6.00 4.51 2.74 0.84 1.20 3.11 4.84 6.25 7.28 7.84 7.90 7.47 6.57 5.25 3.60
G
H
I
J
K
L
M
N
O
P
Q
x0 = 5 0 0.2 = 1.26
SHM displacement-time graph 6.00 4.00 2.00 x/m
0.00 1.25 2.41 3.43 4.23 4.76 4.99 4.91 4.51 3.83 2.91 1.81 0.59 0.67 1.88 2.98 3.89 4.55 4.92 4.99 4.73 4.18 3.37 2.34 1.17 0.08 1.33 2.49 3.49 4.27 4.79 5.0 4.89 4.48 3.78 2.24 1.73 0.50 0.75 1.96 3.05 3.94 4.58 4.94 4.98 4.71 4.14 3.31 2.27
F
t1 = t =
t/s 0.00 0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
2.00 4.00 6.00
SHM velocity-time graph 8.00 6.00 4.00 v/m s 1
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 20. 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0 6.2 6.4 6.6 6.8 7.0 7.2 7.4 7.6 7.8 8.0 8.2 8.4 8.6 8.8 9.0 9.2 9.4 9.6
E
2.00 t/s
0.00 2.00
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
4.00 6.00 8.00
SHM acceleration-time graph 10.00 8.00 6.00 4.00 2.00 0.00 2.00 4.00 6.00 8.00 10.00
a/m s 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
t/s 0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
Figure 3 Spreadsheet for SHM. ( as before this uses Microsoft E xcel but other spreadsheets will have similar functions) . C olumn A contains incremental times starting from zero ( cell H1 is copied into A2 this means that the starting time can be changed) . C ell H2 determines the time increments ( in this case 0.2 s) by adding the contents of H2 to each previous cell; the times can be increased down the column. The formula in cell A3 , in this case, is = A2 + $ H$ 2 and the formula in cell A4 is = A3 + $ H$ 2 , etc.
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9 .1 S I M PLE H ARM O N I C M O TI O N
The times generated, together with chosen values o x0 and , are used to generate the displacement, velocity and time curves. The values or x0 and are inserted into cells J1 and J2 respectively these can then be changed at will. The equation or the displacement, x = x0 sin t is converted into the ormula =$J$1 *SIN(A2*$J$2) , which is copied into column B. The equation or the velocity, v = x0 cos t is converted into the ormula =$J$1 *$J$2*COS(A2*$J$2) which is copied into column C. Finally, the equation or the acceleration, a = - x0 2 sin t, is converted into the ormula =(- 1 ) *$J$1 *($J$2) ^ 2*SIN(A2*$J$2) and is copied into column D.
Note We have used a spreadsheet to show the solutions to the SHM equation. Later we will discuss how we can actually solve the SHM equation using iteration.
To produce the curves shown in fgure 3 , you now need to:
insert a scatter chart
choose the data series
do a little ormatting to improve the size and position o the chart and label the axes, etc.
The SHM equation and 2 Re- visiting the equation a = - 2 x discussed earlier in this sub- topic, we see that it fts the defnition o S HM ( motion in which the acceleration is p rop ortional to the disp lacement rom a fxed p oint and is always directed towards that fxed p oint) . The equation is the defning equation or SHM. The sinusoidal graphs provide the solutions to this equation with respect to time. This may seem strange because there is no apparent time actor in a = - 2 x. This is where 2 comes in. We saw with circular motion that represents the angular speed. Thereore 2 is simply the square o 2 1 this measured in rad 2 s 2 . 2 has dimensions equivalent to ( ____ . tim e ) In circular motion we defned as being __ or simply __t when it is t constant. For a complete revolution the angle will be 2 radians and the 2 time will be the periodic time T. This means that = ___ or, alternatively, T 1 __ = 2 f; comparing this equation with f = T explains why we can call the angular requency.
Worked example An obj ect perorms S HM with a period o 0.40 s and has amplitude o 0.2 0 m. The displacement is zero at time zero. C alculate:
We are only asked to fnd the magnitude o the velocity so it doesnt matter which o the cosine curves the motion really takes.
a) the maximum velocity
Using v = x 0 cos t ( 2 0.1 0) 2 t rad t = _ = __ _ T 0.40 2 2 _ _ S o v = 0. 2 0 = 0 cos 0.40 2 This could have been done without the ull calculation once we had decided that it was a cosine. 0. 1 0 s represents the time or a quarter o a period ( 0. 40 s) ; the value o any cosine at a quarter o a period ( or __ radian) is zero. 2
b) the magnitude o the velocity ater 0.1 0 s c) the maximum acceleration o the obj ect.
Solution 2 a) v 0 = x 0 = x 0 _ T 2 _ v 0 = 0.2 0 = 3 .1 m s 1 0.40 b) As the displacement is zero at time zero this must be a sine or negative sine wave. Thus the velocity will be a cosine or negative cosine.
(
2 c) a 0 = x 0 2 = 0.2 0 _ 0.40
)
2
49 m s 1
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The velocity equation The derivation o the SHM equation is or reerence and you dont need to reproduce it ollowing it through, however, will help you to understand what is going on. We have already seen that v = x 0 cos t and x = x 0 sin t 2 and you may know that sin + cos 2 = 1 . ________
This means that cos = 1 - sin ( dont orget that squaring either the positive or negative cosine will give ( + ) cos 2 . This means that must be included in the square root o this. 2
________
S o v = x 0 1 - sin 2 t
x x 2 2 __ B ut sin t = __ x so sin t = ( x ) 0
v = x0
0
_______
( ) x 1- _ x0
2
=
___________
x
2 0
( )
x - x0 2 _ x0
2
______
= x 0 2 - x 2
______
The equation v = x 0 2 - x 2 is useul or fnding the velocity at a particular position when you know the amplitude and period (or requency or angular requency) you dont need to know the time being considered.
Worked example
Note
An obj ect oscillates simple harmonically with requency 60 Hz and amplitude 2 5 mm. C alculate the velocity at a displacement o 8 mm.
Solution = 2 f = 1 2 0 ( there is no need to calculate the value here but do not leave in the answer in an examination.
tells us that the obj ect could be going in either o the two opposite directions. D ont orget x 0 2 - x 2 ( x 0 - x ) 2 ; it is a common mistake or students to equate these two expressions.
______
v = x 0 2 - x 2
____________________
= 1 2 0 ( 2 5 1 0 - 3 ) 2 - ( 8 1 0 - 3 ) 2 = 8.9 m s 1
Simple harmonic systems 1. The simple pendulum
The simple pendulum represents a straightorward system that oscillates with SHM when its amplitude is small. When the pendulum bob (the mass suspended on the string) is displaced rom the rest position there is a component o the bobs weight that tends to restore the bob to its normal rest or equilibrium position. A condition o a system oscillating simple harmonically is that there is a restoring orce that is proportional to the displacement rom the equilibrium position this is, in eect, tying in with the equation defning SHM because F = ma and a = - 2 x this means that F = - m 2 x.
l Ft
m
mg sin mg cos
Figure 4 Restoring force for simple pendulum.
358
Figure 4 shows the orces acting on a pendulum bob. The bob is in equilibrium along the radius when the tension in the string F t equals the component o the weight in line with the string ( = mgcos). The component o the weight perpendicular to this is not in equilibrium and provides the restoring orce.
9 .1 S I M PLE H ARM O N I C M O TI O N
S o the restoring orce must be equal to the mass multiplied by the acceleration according to Newtons second law o motion:
Note
mg sin = ma
The period of a simple
pendulum is independent of the mass of the pendulum.
x sin _ l g _ x = ma rearranging gives - m l
For a small angle
( )
pendulum is independent of the amplitude of the pendulum.
( The minus sign is because the displacement ( to the right) is in the opposite direction to the acceleration ( to the let) in fgure 4. C ancelling m
( )
g a= - _ x l
g 2 = _ l
l is the length of the
pendulum from the point of suspension to the centre of mass of the bob.
this compares with the defning equation or S HM a = - 2 x leading to
The equation only applies to
small oscillations (swings making angle of less than 10 with the rest position) .
__
2 As the periodic time or SHM is given by T = _ , this shows that T = 2
The period of a simple
_gl .
Investigate! 1
Experimenting with the simple pendulum:
attach a piece o thread o length about 2 metres to a pendulum bob ( a lump o modelling clay would do or this)
split cork
clamp
suspend the thread through a split cork to provide a stable point o suspension align a pin mounted in another piece o modelling clay with the rest position o the bob to act as a reerence point ( sometimes called a fducial marker)
ruler l timer
set the bob oscillating through a small angle and start timing as the bob passes the fducial marker (you should consider why your measurements are likely to be more reliable when the bob is moving at its astest)
G-clamp bob
time thirty oscillations (remember that each oscillation is rom A to C and back to A)
measure the length o the thread rom the point o suspension to the centre o mass o the bob ( take this to be the centre o the bob)
repeat the procedure and fnd an average period or the pendulum
< 10
A
C B
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2
repeat your measurements or a range o lengths
frst plot a graph o periodic time T against length l
reerring to the simple pendulum equation you may see that this graph should not be linear what should you plot in order to linearize your results?
how can you use the graph to calculate a value or the acceleration o reeall g?
dont orget to do an error analysis and compare your result with the accepted value o g.
increment the change in velocity will be the acceleration multiplied by the time increment when you add this to your previous velocity you will get the new velocity ( actually its the average velocity through the time increment)
because velocity is the change in displacement divided by the time increment, you can fnd the change in displacement by multiplying your current velocity value by your chosen time increment
the new value o the displacement will be the previous value added to the change in displacement
This now eeds back into fnding the next value o the acceleration and the cycle repeats and you generate your data or which you can plot displacement- time, velocity- time and acceleration-time graphs ( a example o the spreadsheet is included on the website) .
Use o iteration Making use o the powerul iterative unctionality o a spreadsheet can be a valuable learning aid. With iteration a graphical investigation is possible rom basic principles and basic equations without knowing the solution and without advanced mathematics.
you will need to set up a worksheet with the headings time, displacement, acceleration, change in velocity, velocity and change in displacement you now need to set up the initial conditions: 1 . choose a time increment or the iteration ( make it airly large until your spreadsheet is working well) 2 . start your time column at zero and decide how long you want to run the iteration or ( make this, say, 1 0 time increments initially) 3 . choose an amplitude value or your oscillation 4. set the initial displacement value equal to the amplitude
The ollowing ow chart illustrates the iteration:
n =0 choose t, tn , xn , k n = 0 a n+1 = -k xn = a n+1 t n+1 = n +
n
n+1
x = n+1 t xn+1 = xn + x
enough increments ?
no
yes
2
5 . choose a constant ( k or ) 6. set the initial velocity to be zero
360
the acceleration will always be - k multiplied by the displacement
because acceleration is the change in velocity divided by your time
fnish
S ubscripts n represent the current value o a variable and n + 1 the next value; at each loop the next value always replaces the previous one.
9 .1 S I M PLE H ARM O N I C M O TI O N
2. Massspring system We have ocused on a simple pendulum as being a very good approximation to S HM. A second system which also behaves well and gives largely undamped oscillations is a massspring system. We will consider a mass being oscillated horizontally by a spring ( see fgure 5 ) ; this is more straightorward than taking account o including the eects o gravity experienced in vertical motion. We will assume that the riction between the mass and the base is negligible. The mass, thereore, exchanges elastic potential energy ( when it is ully extended and compressed) with kinetic energy ( as it passes through the equilibrium position) .
Note
The period of the
mass-spring system is independent of amplitude (for small oscillations) .
The period of the
mass-spring system is independent of the acceleration of gravity.
restoring force relaxed spring mass
extended spring
mass
base
base initial position of left edge
initial position of left edge
x
position of left edge when spring extended
Figure 5 Restoring force for mass-spring system. When a spring ( having spring constant k) is extended by x rom its equilibrium position there will be a restoring orce acting on the mass given by F = - kx ( the orce is in the opposite direction to the extension) . Using Newtons second law ma = - kx which can be written as
( )
k a= - _ m x this compares with the defning equation or S HM a = - 2x
( )
k 2 = _ m
As the periodic time or S HM is given by 2 T= _ this shows that __
T = 2
m
_k
When the spring is compressed the quantity x represents the compression o the spring. When the mass is to the let o the equilibrium position the compression is positive but the restoring orce will be negative. This, thereore, leads to the same outcomes as or extensions.
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Energy in SHM systems
Worked example The massspring system is used in many common accelerometer designs. A mass is suspended by a pair of springs which displaces when acceleration occurs. An accelerometer contains a mass of 0.080 kg coupled to a spring with spring constant of 4.0 kN m -1 . The amplitude of the mass is 20 mm. Calculate:
We have seen that, in a simple pendulum, there is energy interchange between gravitational potential and kinetic; in a horizontal massspring system the interchange is between elastic potential and kinetic energy. B ecause each system involves kinetic energy we will focus on this form of energy and bear in mind that the potential energy will always be the difference between the total energy and the kinetic energy at a particular time. The total energy will be equal to the maximum kinetic energy. From Topic 2 we know that the kinetic energy of an obj ect of mass m, moving at velocity v, is given by 1 m v2 EK = _ 2
a) the maximum acceleration b) the natural frequency of the mass.
Solution
( )
We also know that the equation for the velocity at a particular position is
k a) a = - _ m x
(
______
v = x 0 2 - x 2
)
4.0 1 0 3 a = - _ 2 0 1 0-3 0.08 = 1 000 m s 2 __
m 1 b) T = 2 _ so f = _ T k
__
1 = _ 2
k _ m
________
1 f= _ 2
C ombining these equations gives the kinetic energy at displacement x: 1 E K = _ m 2 ( x0 2 - x2 ) 2 This tells us that the maximum kinetic energy will be given by 1 m 2 x 2 E Kmax = _ 0 2 and this must be the total energy ( when the potential energy is zero) so we can say
4.0 1 0 = 3 6 Hz _ 0.08 3
1 m 2 x 2 ET = _ 0 2 The potential energy at any position will be the difference between the total energy and the kinetic energy so 1 m 2 x 2 - _ 1 m 2 x 2 - x2 = _ EP = ET - EK = _ ( 0 ) 1 m 2 x2 0 2 2 2 Figure 6 illustrates the variation with displacement of energy: the green line shows the total energy, the red the potential energy and the blue the kinetic energy. At any position the total energy is the sum of the kinetic energy and the potential energy as we would expect. 1.200 1.000
energy/J
0.800 0.600 0.400 0.200 0.000 0.30
0.20
0.10
0.00 0.10 displacement/m
Figure 6 Variation of energy with displacement.
362
0.20
0.30
9 .1 S I M PLE H ARM O N I C M O TI O N
The variation o the potential energy and kinetic energy with displacement are both parabolas. With all the quantities in the total energy equation being constant, the total energy is, o course, constant or an undamped system. In addition to looking at the variation o energy with displacement we should consider the variation o energy with time. Again, let us start with the kinetic energy. The velocity varies with time according to the equation v = x 0 cos t so the kinetic energy will be 1 m v2 = _ 1 m( x cos t) 2 EK = _ 0 2 2 or 1 m x 2 2 cos 2 t EK = _ 0 2 When the cosine term equals 1 , this gives the maximum kinetic energy. The maximum kinetic energy occurs when the potential energy is zero and so is numerically equal to the total energy at that instant. 1 E T = _ m x0 2 2 2 The potential energy will be the dierence between the total energy and the kinetic energy so EP = ET - EK 1 1 m x 2 2 co s 2 t = _ m x0 2 2 - _ 0 2 2 1 = _ m x 0 2 2 si n 2 t 2 This relationships are shown on fgure 7. The green line represents the total energy, the red curve the potential energy and the blue curve the kinetic energy.
Note
1.20
the sum of the kinetic and potential energies.
1.00
0.80
energy/J
The total energy is always
The graphs (unlike sine
and cosine) never become negative.
0.60
The period of the energy
change is half that of the variation with time of displacement, velocity, or acceleration.
0.40
0.20
0.00 0.0
2.0
4.0
6.0
8.0 time/s
10.0
12.0
The frequency of the energy
change is twice that of the variation with time of displacement, velocity, or acceleration.
Figure 7 Graph showing variation of energy with time.
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9.2 Single-slit difraction Understanding
Applications and skills
The nature o single-slit diraction
Describing the eect o slit width on the
diraction pattern Determining the position o frst intererence minimum Qualitatively describing single-slit diraction patterns produced rom white light and rom a range o monochromatic light requencies
Nature of science Development o theories That rays travel in straight lines is one o the frst theories o optics that students encounter. It comes as a surprise when this theory cannot explain the diraction seen at the edges o shadows cast by small objects illuminated using point sources. Although partial shadows can be explained by considering light sources to be extended, this cannot account or the diraction rom a point source. The wave theory o diraction and how diraction can be explained in terms o waveront propagation rom secondary sources is a good example o how theories have been developed in order to explain a wider variety o phenomena.
angle between frst minimum and central maximum = ____ a
Introduction
intensity
Io
Equations
In S ub- topic 4. 4 we introduced diffraction and saw that when a wave passes through an aperture it spreads into the geometric shadow region. We also saw that the diffraction pattern consists of a series of bright and dark fringes. We will now consider the diffraction pattern in more detail.
5% Io angle 3
2
1
0
1
2
3
central maximum
Figure 1 Variation o intensity with angle or a difraction pattern.
Graph o intensity against angle Figure 1 shows a single-slit diffraction pattern together with a graph of the variation of the intensity of the diffraction pattern with the angle measured from the straight- through position. 1 , 2 , and 3 are the angles with the straight- through position made by the minima.
Note
The central maximum has twice the angular width o
the secondary maxima (each o these have the same angular width) .
No light reaches the centre o the minima but, in going
towards the maxima, the intensity gradually increases it is very dicult to decide the exact positions o the maxima and minima.
The intensity alls o quite signifcantly rom the
principal maximum to the secondary maxima the intensity o the frst secondary maximum is approximately 5% o that o the principal maximum,
364
the second is about 2% and the third is about 1% (fgure 1 is not drawn to scale the secondary maxima are all larger than a scale-diagram would show) .
The intensity is proportional to the square o the amplitude.
9 . 2 S I N G LE- S LI T D I FFR ACTI O N
The single-slit equation
TOK
Returning to the work o S ub- topic 4. 4 we saw how waves can superpose to give constructive intererence ( when they meet in phase) or destructive intererence ( when they meet anti- phase) .
Small angle approximation
(a)
incident plane waves a
(b)
single slit
top o slit
towards frst minimum a
waveront
bottom o slit
path dierence = a sin
Figure 2 Deriving the single slit equation. Figure 2 (a) shows a single-slit o width a. Each point on the waveront within the slit behaves as a source o waves. These waves interere when they meet beyond the slit. For the two waves coming rom the edges o the slit making an angle with the straight through, there is a path dierence o a sin . Waves rom a point halway along the slit will have a a path dierence o __ sin rom the waves coming rom each o the 2 edges. When this path dierence equals hal a wavelength, the waves rom halway along the slit will interere destructively with the waves coming rom the bottom edge o the slit. It ollows that or each point in the bottom hal o the slit there will be a point in the top hal o the slit a a at a distance __ sin rom it. This means that when __ sin = __ , there is 2 2 2 destructive intererence between a wave coming rom a point in the upper hal o the slit and an equivalent wave rom the lower hal o the slit.
We say that diraction is most eective when the aperature is o the same order o magnitude as the width o the slit; we can demonstrate this eectively using a ripple tank. In the equation = ____ a i we make a, then = 1 radian (or 57.3) or, i you avoid the small angle approximation, sin = 1 and = 90. Examine how closely diraction in a ripple tanks agrees with small angle approximation prediction. With poor agreement (as this should show) why do we continue to use the equation = ____ a?
B ecause we are dealing with small angles we can approximate sin to and, by cancelling the actor o two, we arrive at the equation = _ a n The position o other minima ( shown in fgure 1 ) will be given by = ___ a ( where n = 2 , 3 , 4, ...) but you will not be examined on this relationship.
Note
This is the equation or the angle o the frst minimum. n that the minima are not actually separated It ollows rom the equation = _____ a
by equal distances. However, or values o that are less than about 10, it is a good approximation to consider the minima to be equally spaced.
The principal or central maximum occurs because the pairs o waves rom the
top and bottom halves o the slit will travel the same distance and have no path dierence.
The angular width o the principal maximum (rom frst minimum on one side
to frst minimum on the other) is 2.
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Worked example
Solution
a) Explain, by reerence to waves, the diraction o light at a single slit.
a) The waveront within the slit behaves as a series o point sources which spread circular wave ronts rom them. These secondary waves superpose in ront o the slit and the superposition o the waves produces the diraction pattern.
b) Light rom a heliumneon laser passes through a narrow slit and is incident on a screen 3 .5 m rom the slit. The graph below shows the variation with distance x along the screen o intensity I o the light on the screen. I
b) (i) In this case the graph is drawn or distance not angle, so we need to calculate the angle in order to be able to use = __ a. -3
s 3.0 1 0 = __ = ________ D 3.5
= 0.86 1 0 3 rad with s being the distance rom the centre o the principal maximum to the frst minimum. It is more reliable to measure 2 rom the frst minimum on one side o the principal maximum to the frst minimum on the other side o the principal maximum. D is the distance rom the slit to the screen. -9
630 1 0 a = __ = _________ 0.86 1 0 -3
10
5
0 x/mm
5
(i) The wavelength o the light emitted by the laser is 63 0 nm. D etermine the width o the slit. (ii) S tate two changes to the intensity distribution o the central maximum when the single slit is replaced by one o greater width.
10
= 0.73 1 0 3 m or 0. 73 mm (ii) With a wider slit more light is able to pass through. This will result in an increase in the intensity o the beam and so the peaks will all be higher. a in the equation = __ a increases but remains constant; the angle will decrease which means the principal maximum will become narrower and the minima will move closer together.
Single slit with monochromatic and white light
Figure 3 Single slit with monochromatic and white light.
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Figure 3 shows two images o the light emerging rom a singleslit. The upper image is obtaine d using monochromatic ( one requency) green light and the lower image is obtained using white light. B oth the angular width o the ce ntral maximum and the angular se paration o successive secondary maxima depend on the wavelength o the light this is the re ason why the secondary maxima produced by the diraction o white light are coloured. You will se e that, or the secondary maxima, the blue light is le ss deviated than the other colours as it has the shortest wavele ngth. The e dges o the principal maximum are coloured rather than pure white . This is because the principal maxima or the colours at the blue end o the visible spectrum are le ss spread than the colours at the red end; the e dges are thereore a combination o red, orange, and ye llow and have an orange hue.
9.3 IN TERFEREN CE
Investigate! Diraction at a single slit
Using a laser pointer, shine the light on the gap between the j aws o a pair o digital or vernier callipers ( orming a slit) and proj ect it onto a white screen. The screen needs to be at least 3 m rom the callipers.
Measure the distance between the callipers and the screen, D, using a tape- measure.
C alculate the wavelength o the light using s a as we know that 2 _ D B oth the dependence o the width o the central maximum and the separation o the maxima on the wavelength o light can be investigated by using laser pointers with dierent colours.
Adj ust the callipers so that the image is clear.
Measure the separation, s, o the frst minima on the screen using a metre ruler. D
Research the wavelength ranges o the laser light and then suggest how you could modiy the experiment to check the calibration o the callipers at small j aw separation.
s
2 laser callipers
We will return to diraction when we look at resolution in S ub-topic 9.4.
screen
Figure 4 Investigating laser light passing through slit.
9.3 Interference Understanding Youngs double-slit experiment
Applications and skills Qualitatively describing two-slit intererence
Modulation o two-slit intererence pattern by
one-slit diraction eect Multiple slit and diraction grating intererence patterns Thin flm intererence
Nature of science Thin flm intererence The observation o colour is not simply a question o colour pigmentation. Certain mollusc and beetle shells, buttery wings, and the eathers o hummingbirds and kingfshers can produce beautiul colours as a result o thin flm intererence known as iridescence. The observed colour changes depend on the angle o illumination or viewing, and are caused by multiple reections rom the suraces o flms with thicknesses similar to the wavelength o light. These natural aesthetics require the analysis o the physics o intererence.
patterns, including modulation by one-slit diraction eect Investigating Youngs double-slit experimentally Sketching and interpreting intensity graphs o double-slit intererence patterns Solving problems involving the diraction grating equation Describing conditions necessary or constructive and destructive intererence rom thin flms, including phase change at interace and eect o reractive index Solving problems involving intererence rom thin flms
Equations D ringe separation or double slit: s = ______ d diraction grating equation: n = d sin light reection by parallel-sided thin flm: constructive intererence 2dn = (m + ___12 ) destructive intererence 2dn = m
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Introduction In Sub-topic 4.4 we considered intererence o the waves emitted by two coherent sources. We saw that intererence is a property o all waves, and we considered the equal ringe spacing in the Young double-slit experiment. In this sub-topic we ocus on the intensity variation in a double-slit experiment and see how intererence is achieved using multiple slits. We then look at a second way o achieving intererence, using division o amplitude instead o division o waveront as with double-slit intererence.
Intensity variation with the double-slit Figure 1 shows the image o the light rom a heliumneon laser that has passed through a double slit. The alternate red and dark ringes are equally spaced as we saw in Sub-topic 4. 4. However, looking at the image closely we see that there are extra dark regions.
Figure 1 Double-slit difraction pattern or light rom HeNe laser. We know rom Sub-topic 9.2 that a single slit produces a diraction pattern with a very intense principal maximum and much less intense secondary maxima. A double slit is, o course, two single slits so each o the slits produces a diraction pattern and the waves rom the two slits interere. The two eects mean that the intensity o the intererence pattern is not constant, but is modifed by the diraction pattern to produce the intensity. Figure 2(a) below shows how the relative intensity would vary or a doubleslit intererence pattern without any modifcation due to diraction. By using relative intensity we avoid the need to think about the actual intensity values and units. Figure 2(b) shows the variation o relative intensity with angle or a single slit. Figure 2(c) shows the superposition o the two eects so that the single-slit diraction pattern behaves as the envelope o the intererence pattern. Shaping a pattern in this way is called modulation and is important in the theory o AM (amplitude modulation) radio. You will note that the ringe spacing does not change between fgures 2(a) and (c) but that the bright ringes occurring between 1 1 and 1 4 in fgure 2(a) are reduced to a much lower intensity. An intererence maximum coinciding with a diraction minimum is suppressed and does not appear in the overall pattern. We saw in S ub- topic 4.4 that the ringe separation s or light o wavelength is given by D s=_ d where D is the distance rom the double slit to the screen and d is the separation o the slits. The value o s used in this equation is or the intererence pattern not the modulated pattern caused by diraction.
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9.3 IN TERFEREN CE
relative intensity (a)
20
15
10
5 0 10 5 angle rom straight through position/
15
20
15
20
15
20
relative intensity (b)
20
15
10
5 0 5 10 angle rom straight through position/
relative intensity (c)
20
15
10
5 0 10 5 angle rom straight through position/
Figure 2 Combination o difraction and intererence.
Multiple-slit interference We have now seen the interference patterns produced both by a single slit and a double slit. What happens if there are more than two slits? The answer is that the bright fringes, which come from constructive interference of the light waves from different slits, remain in the same positions as for a double slit but the pattern becomes sharper. The bright fringes are narrower and their intensity is proportional to the square of the number of slits. Why is this? At the centre of the principal maximum the waves reaching the screen from all of the slits are in phase and so it is very bright here. B y moving to a position close to the centre of this maximum the path difference between the light from two adj acent slits has changed by such a small amount that it hardly affects the interference and so it is still bright. Slits further away from each other will have more likelihood of a greater path difference and therefore meeting out of phase.
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WAVE P H E N O M E N A ( AH L ) Imagine having three slits: at the principal maximum the path dierence between the waves will be small as their paths are nearly parallel. Each o the three wave trains will meet nearly in phase at positions around the principal maximum peak, which makes it airly wide. Now imagine having 1 0 0 slits: or every slit there will a second slit somewhere that transmits a wave with a hal wavelength path dierence rom the frst. When the waves interere at a position close to the centre o the maximum, destructive intererence occurs and reduces the overall intensity. This will be true or the waves coming rom the two slits adj acent to this pair, the two next to those and so on. Increasing the number o slits will give destructive intererence close to the centre o the maxima this reduces the width o the central maximum ( and the other maxima too) . The extra energy that is needed to increase the intensity o the maxima must come rom the regions that are now darker, so the maxima are more intense. The mathematics o modulation o waves is quite complex and or the purpose o the IB Physics course you will simply need to recognize that the modulation is happening and remember the eect o increasing the number o slits. Figure 3 shows the intererence patterns obtained when red laser light passes through various numbers o slits o identical width. Figures 3 and 4 show that the single-slit diraction pattern always acts as an envelope or the multiple-slit intererence patterns.
Figure 3 The efect o increasing the number o slits on an intererence pattern. In fgure 4 the relative intensities have been drawn the same size in order to show the increased sharpness; however, or two, three, and fve slits the actual intensities are in the ratio 1 : 9 : 2 5 .
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9.3 IN TERFEREN CE
single slit
double slit
relative intensity
20
15 10 5 0 5 10 15 angle rom straight through position/
relative intensity
20
20
three slits
15 10 5 0 5 10 15 angle rom straight through position/ fve slits
relative intensity
20
15 10 5 0 5 10 15 angle rom straight through position/
20
relative intensity
20
20
15 10 5 0 5 10 15 angle rom straight through position/
20
Figure 4 The eect o increasing the number o slits on variation o intensity.
white light source
The difraction grating A diraction grating is a natural consequence o the eect on the intererence pattern when the number o slits is increased. D iraction gratings are used to produce optical spectra. A grating contains a large number o parallel, equally spaced slits or lines ( normally etched in glass or plastic) typically there are 600 lines per millimetre ( see fgure 5 ) . When light is incident on a grating it produces intererence maxima at angles given by
diraction grating
Figure 5 Diraction grating.
n = d sin The spacing between the slits is small, which makes the angle large or a fxed wavelength o light and n. This means that we cannot use the small angle approximation or relating the wavelength to the position o the maxima as we did or a double-slit. Figure 6 shows a section o a diraction grating in which three consecutive slits deviate the incident waves towards a maximum. The slits are so narrow and the screen is so ar away rom the grating that the angles made by the diracted waves are virtually identical. Providing the path dierence between waves coming rom the same part o successive slits is an integral number o wavelengths, the waves will reach the screen in phase and give a maximum. In the case o a diraction grating the distance d is taken as the length o both the transparent and opaque sections o the slit (they are taken to be o equal width) . From fgure 6 we see that the path dierence between those waves coming rom A and B and the path between waves coming B and C will each be d sin . This means
A d
B C
portion o grating
light diracted
Figure 6 Deriving the diraction grating equation.
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WAVE P H E N O M E N A ( AH L ) that n = d sin (where n is the order o the maximum and is zero or the central maximum, 1 or the frst maximum on each side o the centre, etc.) . We can see that the angular positions o the intererence maxima depend on the grating spacing, d. The shape o the diraction envelope, however, is determined by the width o the clear spaces ( as or a single slit) . The maxima must lie within the envelope o the single slit diraction pattern i they are to be relatively intense; this sets an upper limit on how wide the transparent portions o the grating can be.
Figure 7 Dispersing white light with a difraction grating.
O ne o the uses o a diraction grating is to disperse white light into its component colours: this is because dierent wavelengths produce maxima at dierent angles. Figure 7 shows that light o greater wavelength ( or any given order) is deviated by a larger angle. This is in line with what we would predict rom using n = d sin . Each successive visible spectrum repeats the order o the colours o the previous one but becomes less intense and more spread out.
Grating spacing and number o lines per mm It is usual or the number o lines per mm or ( N) to be quoted or a diraction grating, rather than the spacing. N must be converted into d in order to use the diraction grating equation n = d sin . 1 This is straightorward because d = __ but N must frst be converted to N the number o lines per metre by multiplying by 1 000 beore taking the reciprocal.
Nature of science
For a diraction grating to produce an observable pattern, the grating spacing must be comparable to the wavelength o the waves. The wavelength o visible light is between approximately 400700 nm. A grating with 600 lines per mm has a spacing o approximately 2 000 nm or our times the wavelength o green light. This spacing produces very clear images.
The wavelengths o low- energy X-rays is around 1 0 - 1 0 m or 0.1 nm. The 600 lines per millimetre optical grating will not produce any observable maximum . .. thereore, what will diract X- rays? The spacing o ions in crystals is o the same order o magnitude as X- rays and the regular lattice shape o a crystal can perorm the same task with X- rays as a diraction grating does with visible light. The diracted patterns are now commonly detected with charge-coupled device ( C C D ) detectors.
Worked example
Solution
A diraction grating having 6 0 0 lines per millimetre is illuminated with a parallel beam o monochromatic light, which is normal to the grating. This produces a se cond- order maximum which is observed at 42 . 5 to the straightthrough direction. C alculate the wave length o the light.
N = 6.00 1 0 5 lines per metre.
Appropriate wavelengths or efective dispersion
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1 1 d = __ = ________ = 1 .67 1 0 - 6 m. N 6.00 1 0 5
-6
dsin 1 . 6 7 1 0 sin 42 . 5 n = d sin so = ____ = ________________ ( dont n 2
orget to have your calculator in degrees) = 5 .64 1 0 - 7 = 5 64 nm ( all data is given to three signifcant fgures so the answer should also be to this precision) .
9.3 IN TERFEREN CE
Nature o science Using a spectrometer A spectrometer is a useul, i expensive, piece o laboratory equipment that allows the wavelengths o light emitted by sources to be analysed. The instrument consists o a collimator, turntable and telescope. A diraction grating or other light disperser is placed on the turntable, which is careully levelled. The collimator uses lenses to produce a parallel beam o light rom a source this light beam is then incident on the grating that disperses it. The end o the collimator urthest rom the grating has a vertical adj ustable slit that serves as the source when we talk about spectral lines we iner that a spectrometer is being
used because the lines are the dispersed images o the collimator slit. B eore measurements are made the telescope is ocused on the collimator slit this means that the dispersed light will also be in ocus. It is usual or sources that are said to be monochromatic to actually emit a variety o dierent wavelengths. B y using the spectrometer, the angle in the diraction grating equation can be measured. This is usually done by measuring the angle between the two frst-order images on either side o the straight- through position i.e. 2 . With a knowledge o the grating spacing d, the wavelength can be determined.
adjustable slit collimator difraction grating turntable
light source angular scale
te l e
sco
pe
Figure 8 Key eatures o a spectrometer.
Interference by division of amplitude It was mentioned in the introduction to this sub-topic that there are two ways o providing coherent sources that are able to interere. Youngs double slit and multiple slits all derive their interering waves by taking waves rom dierent parts o the same waveront. Because the interering waves have all come rom the same waveront they will be in phase with each other. Wherever the waves meet they will interere and a ringe pattern can be obtained anywhere in ront o the sources (the slits) . Since this intererence can be ound anywhere the ringes are said to be non-localized. D ivision o amplitude is a method o achieving intererence using two waves that have come rom the same point on a waveront. Each wave has a portion o the amplitude o the original wave. In order to achieve intererence by division o amplitude, the source o light must come rom a much bigger source than the slit used or division o waveront intererence. The image produced will, however, be localized to one place instead o being ound anywhere in ront o the sources.
Thin flm intererence Figure 9 shows a wave incident at an angle to the normal to the surace o a flm o transparent material ( such as low- density oil or detergent) having reractive index n. This diagram is not drawn to scale
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WAVE P H E N O M E N A ( AH L ) and t are both very small so that incident wave is eectively normal to the surace. The incident wave partially reects at the top surace o the flm and partially reracts into the flm. This reracted wave, on reaching the lower surace o the flm, again partially reects ( remaining in the flm) and partially reracts into the air below the flm. This process can occur several times or the same incident wave. A
incident wave
B
C
Waves reected by the flm (you may be tested on this in IB Physics examinations) In this case A has been reected rom the top surace o the flm and, because the reection is at an optically denser medium, there is a phase change o radian (equivalent to hal a wavelength) . The wave B travels an optical distance o 2tn beore it reracts back into the air. Thus the optical path dierence between A and B will be 2tn. I there had been no phase change then this optical distance would equal m or constructive intererence. However, because o As phase change at the top surace the overall eect will be destructive intererence. Thus or the light reecting rom the flm when 2tn = m there will be destructive intererence and when 2tn = (m + ___12 ) there will be constructive intererence.
air transparent medium o reractive index n
t
air A
B
C
Figure 9 Intererence at parallel-sided thin flm.
Waves transmitted through the flm (you will not be tested on this in IB Physics examinations) I the flm is thin and is small the (geometrical) path dierence between waves that have passed through the flm (i.e. between A' and B' or between B' and C') will be very nearly 2t, in other words B' travels an extra 2t compared with A'. Because the waves are travelling in a material o reractive index n, the waves will slow down and the wavelength becomes shorter. This means that, compared with travelling in air, they will take longer to pass through the flm. This is equivalent to them travelling through a thicker flm at their normal speed. The optical path dierence will, thereore, be 2tn. I this distance is equal to m where m = 0, 1, 2 etc. (we are using m to avoid conusion with the reractive index n) then there will be constructive intererence. I the optical path dierence is equal to an odd number o hal wavelengths, (m + ___12 ), then there will be destructive intererence.
Nature of science Coating o lenses When light is incident on a lens some o it will be reected and some transmitted. The reected light is eectively wasted, reducing the intensity o any image ormed ( by the eye using corrective lenses, in a camera or in a telescope) . B y coating the lens with a transparent material o quarter o a wavelength thickness, the light reected by the coating and the light reected by the lens can be made to interere destructively and thus eliminate the reection altogether. Magnesium uoride is oten used or the coating and is optically denser than air but less dense than
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glass ( having a reractive index o 1 . 3 8) . When it is used to coat a lens the waves reecting rom both the magnesium uoride and the glass will undergo a phase change o radian . . . so the phase changes eectively cancel each other out. The optical path dierence between the waves will be equal to the reractive index o magnesium uoride multiplied by twice the thickness o the coating. For destructive intererence the optical path dierence is 2 nt = __ 2 so the thickness o the coating is given by t = __ . 4n With white light there will be a range o
9.3 IN TERFEREN CE
wavelengths and so it is impossible to match the thickness o the coating to all o these wavelengths. I the thickness is matched to green light, then red and blue will still be reected giving the lens the appearance o being magenta
( = red + blue) . B y using multiple layers o a material o low reractive index and one o higher reractive index, it is possible to reduce the amount o light reected to as little as 0. 1 % or a chosen wavelength.
Worked example
Solution
a) Name the wave phenomenon that is responsible or the ormation o regions o dierent colour when white light is reected rom a thin flm o oil oating on water.
a) Although there is reection involved, the colours come about because o intererence.
b) A flm o oil o reractive index 1 .45 oats on a layer o water o reractive index 1 .3 3 and is illuminated by white light at normal incidence. illumination air oil water
When viewed at near normal incidence a particular region o the flm looks red, with an average wavelength o about 65 0 nm. (i) Explain the signifcance o the reractive indices o oil and water with regard to observing the red colour. (ii) C alculate the minimum flm thickness.
b) (i) n is the reractive index o the oil. Because the waves are travelling in oil, they move more slowly than they would do in air and so the eective path dierence between the waves reected at the airoil interace and the waves at the oilwater interace is longer by a actor o 1 .45. The wave reected at the airoil interace undergoes a phase change equivalent to __ (oil is denser 2 than air) . The waves reected at the oil water interace undergo no phase change on reection at the less dense medium. So or the bright constructive red intererence 1 2tn = ( m + __ ) 2 i m = 0, 2 tn = _ or = 4tn 2 65 0 (ii) t = _ = 1 1 0 nm ( 4 1 .45 )
Nature of science Vertical soap flms Figure 1 0 shows the colours o light transmitted by a vertical soap flm. Over a ew seconds the flm drains and becomes thinner at the top and thicker at the bottom. When the flm is illuminated with white light, the reected light appears as a series o horizontal coloured bands. The bands move downwards as the flm drains and the top becomes thinner. The top o the flm appears black just beore the flm breaks. It has now become too thin or there to be a path dierence between the waves coming rom the two suraces o the flm. The phase change that occurs or the light reected by the surace o the flm closest to the source means that, or all colours, there is cancellation and so no light can be seen.
Figure 10 Thin flm intererence in a vertical soap flm.
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TOK The aesthetics of physics The colour in the soap flm makes a beautiul image. Does physics need to rely upon the arts to be aesthetic? Herman Bondi the Anglo-Austrian mathematician and cosmologist implied that Einstein believed otherwise when he wrote: What I remember most clearly was that when I put down a suggestion that seemed to me cogent and reasonable, Einstein did not in the least contest this, but he only said, Oh, how ugly. As soon as an equation seemed to him to be ugly, he really rather lost interest in it and could not understand why somebody else was willing to spend much time on it. He was quite convinced that beauty was a guiding principle in the search for important results in theoretical physics. H. Bondi How does a mathematical equation convey beauty? Is the beauty in the mathematics itsel or what the mathematics represents?
9.4 Resolution Understanding The size o a diracting aperture The resolution o simple monochromatic
two-source systems
Applications and skills Solving problems involving the Rayleigh
criterion or light emitted by two sources diracted at a single slit Describing diraction grating resolution
Equations Rayleighs criterion: = 1.22 ____ b resolvance o a diraction grating: R = _______ = mN
Nature of science How ar apart are atoms? When we view objects, we are limited by the wavelength o the light used to make an observation shorter wavelengths such as X-rays, gamma rays and ast-moving electrons will improve the resolution that is achievable. However, even using the shortest wavelengths obtainable does not allow us to locate the exact position o objects on the
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atomic and sub-atomic scale. The process o making an observation using these waves disturbs the system and increases the uncertainty with which we can locate an object. Heisenbergs uncertainty principle places a limit on how close we can be to fnding the exact position o objects on the quantum scale.
9.4 RESOLU TI ON
Introduction Resolution is the ability o an imaging system to be able to p roduce two sep arate distinguishable images o two sep arate obj ects. The imaging system could be an observers eye, a camera, a radio telescope, etc. Whether obj ects can be resolved will depend on the wavelength coming rom the obj ects, how close they are to each other and how ar away they are rom the observer.
Difraction and resolution
images ully resolved
We have seen that when light passes through an aperture a diraction pattern is ormed. For an optical system the aperture could be the pupil o the observers eye or the obj ective lens o a telescope. When there are two sources o light two diraction patterns will be ormed by the system. How close can these patterns be or us to still recognize that there are two sources? Two obj ects observed through an aperture will produce two diracted images which may or may not overlap. In the late nineteenth century the E nglish physicist, John William Strutt, 3 rd B aron Rayleigh, proposed what is now known as the Rayleigh criterion or resolution o images. This states that two sources are resolved i the p rincip al maximum rom one diraction p attern is no closer than the frst minimum o the other p attern.
images just resolved
The limit to resolution is when the principal maximum o the diraction pattern rom one source lies on the frst minimum diraction pattern rom the second source ( and vice versa) . D iracted images urther apart than this limit will be resolved and those closer will be unresolved. Figure 1 shows the diraction intensity patterns produced by two obj ects the same distance apart but viewed through a circular aperture rom dierent distances. The variation o intensity with angle or each o the diraction patterns is shown below the image. According to the Rayleigh criterion, the uppermost pair o images are ully resolved because the principal maximum o each diraction pattern lies urther rom the other than the frst minimum. The central images are j ust resolved since the principal maximum o one diraction pattern is at the same position as the frst minimum o the second diraction pattern. The bottom images are unresolved as the principal maximum o one diraction pattern lies closer to the second pattern than its frst minimum.
images unresolved
Figure 1 Difraction intensity patterns o two objects viewed through a circular aperture.
Nature of science Other criteria or resolution? Rayleighs criterion is not the only one used in optics. Many astronomers believe that they can resolve better than Rayleigh predicts. C M S parrow developed another criterion or telescopes that leads to an angular separation
at resolution about hal that o Rayleigh. His criterion is that the two diraction patterns when added together give a constant amplitude in the regions o the two central maxima.
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Resolution equation For single slits we saw in S ub-topic 9.2 that the frst minimum occurs when the angle with the straight- through position is given by = __ a where is the wavelength o the waves and b is the slit width. With a circular aperture the equation is modifed by a actor o 1 .22, but derivation o this actor is beyond the scope o the IB Physics course. So we have = 1 .2 2 __ a
in this case a is the diameter o the circular aperture (or very commonly the diameter o the lens or mirror orming the image) . The pupil o the eye has a diameter o about 3 mm and, taking visible light to have a wavelength in the order o 6 1 0 7 m, the minimum 1 .22 6 1 0 2 1 0 - 4 rad. angle o resolution or the eye is = ___________ 3 10 O ptical deects in the eye mean that this limit is probably a little small. -7
-3
The primary mirror o the Hubble Space Telescope has a diameter o 2.4 m. 1 .2 2 6 1 0 For this telescope, the minimum angle o resolution is = ___________ 2.4 3 1 0 - 7 rad. This is a actor o about a thousand smaller than that achieved by the unaided eye which means that the Hubble Space Telescope is much better at resolving images. As this telescope is in orbit above the atmosphere, it avoids the atmospheric distortion which degrades images achieved by Earth- based telescopes. These concepts are covered in more detail in O ption C ( Imaging) . -7
Worked example A student observes two distant point sources o light. The wavelength o each source is 5 5 0 nm. The angular separation between these two sources is 2 . 5 1 0 4 radians subtended at the pupil o a students eye.
Note
You may wish to support
your answer by drawing the two intensityangle curves i you think your answer may not be clear.
It is common or
students to miss out the word difraction in their answers this is crucial to score all marks.
a) S tate the Rayleigh criterion or the two images on the retina to be j ust resolved. b) E stimate the diameter o the circular aperture o the eye i the two images are j ust to be resolved.
Solution a) The images will be j ust resolved when the diraction pattern rom one o the point sources has its central maximum at the same position as the frst minimum o the diraction pattern o the other point source. 1 .2 2 5 5 0 1 0 b) = 1 .2 2 __ = > b = 1 .2 2 __ = _____________ b 2 .5 1 0 -9
-4
= 2 .7 1 0 3 m 3 mm
Nature of science Difraction and the satellite dish When radiation is emitte d rom a transmitting satellite dish the waves diract rom the dish. The diameter o the dish behaves as an aperture. The angle made by the frst
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minimum with the straight- through is related to diameter b by the equation = 1 .22 _ b
9.4 RESOLU TI ON
When the wavelength increases so does the angle through which the waves become diracted. This means that the diracted beam now covers a larger area. However, increasing the diameter o a dish narrows the beam and it covers a smaller area. In satellite communications the footp rint is the portion o the E arths surace over which the satellite dish delivers a specifed amount o signal power. The ootprint will be less than the region covered by the principal maximum because the signal will be too weak to be useul at the edges o the principal maximum. Figure 2 shows the ootprint o a communications satellite.
Small values o the dish diameter will give a large ootprint but the intensity may be quite low since the energy is spread over a large area.
The ootprint o a satellite has social and political implications ranging rom unwarranted observation to sharing television programmes. satellite
ootprint
Figure 2 Satellite ootprint.
Resolvance o difraction gratings We have seen that diraction gratings are used to disperse light o dierent colours. Such gratings are usually used with spectrometers to allow the angular dispersion or each o the colours to be measured. When using a spectrometer, knowing the number o lines per millimetre on the diraction grating and measuring the angles that the wavelengths o light are deviated through allows the wavelengths o the colours to be determined. We have seen that there is a limit to how close two objects can be beore their diraction images are indistinguishable the same is true o the dierent wavelengths that can be resolved using a particular diraction grating.
double-slit angular dispersion o principal maximum
When we previously looked at intererence patterns or multiple slits in S ub- topic 9.3 , we saw that increasing the number o slits improves the sharpness o the maxima ormed. Figure 3 shows that, when light o the same wavelength is viewed rom the same distance x, the angular dispersion D or the principal maximum with the double slit is larger than angular dispersion G or the diraction grating. A sharper principal maximum is one with less angular dispersion. With wider maxima there is more overlap o images rom dierent sources and lower resolution.
D X
Using this argument we see that, when beams o light are incident on a diraction grating, a wider beam covers more lines ( a greater number o slits) and will produce sharper images and better resolution.
difraction grating angular dispersion o principal maximum
The resolvance R or a diraction grating ( or other device used to separate the wavelengths o light such a multiple slits) is defned as the ratio o the wavelength o the light to the smallest dierence in wavelength that can be resolved by the grating .
G X
The resolvance is also equal to Nm where N is the total number o slits illuminated by the incident beam and m is the order o the diraction. R = _ = Nm The larger the resolvance, the better a device can resolve.
Figure 3 Angular dispersion o principal maximum or difraction grating compared with a double-slit.
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Nature of science Resolution in a CCD C harge- coupled devices ( C C D s) were originally developed or use in computer memory devices but, today, appear in all digital cameras and smartphones. When you buy a camera or phone you will no doubt be interested in the number o pixels that it has. The pixel is a picture element and, or example, a 2 0 megapixel camera will have 2 1 0 7 pixels on its C C D . The resolution o a C C D depends on both the number o pixels and their size when compared to the proj ected image. The smaller the camera, the more convenient it is to carry, and ( at present) cameras with pixels o dimensions as small as 2 .7 m 2 . 7 m are mass- produced. In general C C D images are resolved better with larger numbers o smaller pixels. Figure 4 shows two images o the words
IB Physics the top uses a small number o large pixels, while the lower one is ar better resolved by using a large number o small pixels. The upper image is said to be pixellated.
B Ph sics
Figure 4 Images with a small number of large pixels and a large number of small pixels.
Worked example Two lines in the emission spectrum o sodium have wavelengths o 5 89. 0 nm and 5 89. 6 nm respectively. C alculate the number o lines per millimetre needed in a diraction grating i the lines are to be resolved in the second- order spectrum with a beam o width 0. 1 0 mm.
Solution 5 89.0 R = ___ = _____ ( both values are in nanometres so 0.6
this actor cancels) You would be equally j ustifed in using 5 89.6 ( or 5 89.3 the mean value) in the numerator here.
380
R = 981 .7 ( no units) Thus 981 .7 = Nm = N 2 N = 498.8 lines This is in a beam o width 0.1 0 1 0 3 m so, in 1 mm, there needs to be 4988 5 000 lines. S ince diraction gratings are not normally made with 4988 lines mm 1 , the sensible choice is to use one with 5 000 lines mm 1 !
9 . 5 TH E D O PPLE R E FFE CT
9.5 The Doppler efect Understanding The Doppler eect or sound waves and
light waves
Nature o science From water waves to the expansion o the universe In his 1842 paper, ber das farbige Licht der Doppelsterne (Concerning the coloured light o the double stars) , Doppler used the analogy o the measurement o the requency o water waves to reason that the eect that bears his name should apply to all waves. Three years later, Buys Ballot verifed Dopplers hypothesis or sound using stationary and moving groups o trumpeters. During his lietime Dopplers hypothesis had no practical application; one hundred years on it has arreaching implications or cosmology, meteorology and medicine.
Applications and skills Sketching and interpreting the Doppler eect
when there is relative motion between source and observer Describing situations where the Doppler eect can be utilized Solving problems involving the change in requency or wavelength observed due to the Doppler eect to determine the velocity o the source/observer
Equations Doppler equation or a moving source: v f = f __ v us or a moving observer: v uo f = f __ v
(
)
(
)
or electromagnetic radiation:
_ = _ v _ c f
Introduction When there is relative motion between a source of waves and an observer, the observed frequency of the waves is different to the frequency of the source of waves. The apparent change in pitch of an approaching vehicle engine and a sounding siren are common examples of this effect. The D oppler effect has wide- ranging implications in both atomic physics and astronomy.
The Doppler efect with sound waves Although we use general equations for this effect, we build up the equations under different conditions before combining them. In the following derivations ( that need not be learned but will help you to understand the equations and how to answer questions on this topic) the letter s refers to the source of the waves ( the obj ect giving out the sound) and the letter o refers to the observer; f will always be the frequency of the source and f the apparent frequency as measured by the observer.
Note It is only the component o the wave in the source observer direction that is used; perpendicular components o the motion do not alter the apparent requency o the wave.
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1. Moving source and stationary observer v f
v f us
A
B
S
S' us f
Figure 1 The Doppler efect or a moving source and stationary observer.
At time t = 0 the source is at position S and it emits a wave that travels outwards in all directions, with a velocity v as shown in fgure 1 . At time t = T ( i.e. one period later) the wave will have moved a distance equivalent to one wavelength or ( using v = f) a distance = _vf to reach positions A and B . When the source is moving to the right with a u velocity u s , in time T it will have travelled to S , a distance u s T( = __ ) . To f a stationary observer positioned at B it will appear that the previously emitted crest has reached B but the source that emitted it has moved orwards and now is at S . Thereore, to the observer at B , the apparent u wavelength ( ) is the distance SB = _vf - __ f s
s
v - us = _ f
Thus, the wavelength appears to be squashed to a smaller value. This means that the observer at B will hear a sound o a higher requency than would be heard rom a stationary source the sound waves travel at speed v ( which is unchanged by the motion o the source) so v v f = _ = f _ v - us
(
)
To an observer at A the wavelength would have appeared to be stretched to a longer value given by S A meaning that v + us = _ f and the observed requency would be lower than that o a stationary source, meaning that v v f = _ = f _ v + us
(
)
The two equations or f can be combined into a single equation v f = f _ v us
(
)
in which the sign is changed to - or a source moving towards a stationary observer and to + or a source moving away rom a stationary observer.
2. Moving observer and stationary source In this case the source remains stationary but the observer at B moves towards the source with a velocity u o . The source emits crests at a
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9 . 5 TH E D O PPLE R E FFE CT
frequency f but the observer, moving towards the source, encounters the crests more often, in other words at a higher frequency f. Relative to the observer, the waves are travelling with a velocity v + u o and the frequency is f. The wavelength of the crests does not appear to have changed and will be = _vf . The wave equation ( v = f) applied by the observer becomes v + u o = f _vf meaning that the frequency measured by the observer will be v + uo f = f _ v
(
)
When the observer moves away from the source the wave speed appears to be v - u o and so f = f
v- u (_ v ) o
Again the two equations for f can be combined into a single equation
(
v uo f = f _ v
)
in which the sign is changed to - for an observer moving away from a stationary source and to + for an observer moving towards a stationary source.
Worked example A stationary loudspeaker emits sound of frequency of 2 .00 kHz. A student attaches the loudspeaker to a string and swings the loudspeaker in a horizontal circle at a speed of 1 5 m s 1 . The speed of sound in air is 3 3 0 m s 1 . An observer listens to the sound at a close, but safe, distance from the student. a) Explain why the sound heard by the observer changes regularly. b) D etermine the maximum frequency of the sound heard by the observer.
Solution a) As the loudspeaker approaches the observer, the frequency appears higher than the stationary frequency because the apparent wavelength is shorter meaning that the wavefronts are compressed. As it moves away from the observer, the frequency appears lower than the stationary frequency because the apparent wavelength is stretched. The overall effect is, therefore, a continuous rise and fall of pitch heard by the observer. b) The maximum observed frequency occurs when the speaker is approaching the observer so: v 330 _______ f = f( _____ v - u ) = 2 000 ( 3 3 0 - 1 5 ) = 2 1 00 Hz ( this is 2 s.f. precision in s
line with 1 5 m s 1 )
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The Doppler efect with light The D oppler eect occurs not only with sound but also with light ( and other electromagnetic waves) ; in which case the requency and colour o the light diers rom that emitted by the source. There is a signifcant dierence in the application o D oppler eect or sound and light waves. S ound is a mechanical wave and requires a medium through which to travel; electromagnetic waves need no medium. Additionally, one o the assumptions or postulates o special relativity is that the velocity o light waves is constant in all inertial reerence rames this means that, when measured by an observer who is not accelerating, the observer will measure the speed o light to be 3 . 00 1 0 8 m s - 1 irrespective o whether the observer moves towards the source or away rom it and, thereore, it is impossible to distinguish between the motion o a source and an observer. This is not true or sound waves, as we have seen. Although the D opple r eect e quations or light and sound are derived on complete ly dierent principles, providing the speed o the source ( or observer) is much less than the speed o light, the equations give approximately the corre ct results or light or sound. The quantities u s and u o in the D oppler e quations or sound have no signifcance or light, and so we need to de al with the re lative velocity v between the source and observer. Thus, in either o the equations v f = f _ v us
(
)
(
)
or v uo f = f _ v
the wave speed is that o electromagnetic waves ( c) and one o the velocities u s or u o is made zero while the other is replaced by the relative velocity v.
Note
Substituting values into
the second equation will give the same resulting relationship you may like to try this but remember you do not need to know any of these derivations.
The equation is only
valid when c v and so cannot usually be used with sound (where the wave speed is 300 m s 1 unless the source or observer is moving much more slowly than this speed) .
384
The frst o these equations becomes
(
)
c v -1 1 _ f = f _ = f _ v = f(1 + c ) _ c+ v 1 + c
(
)
This can be expanded using the binomial theorem to approximate to v v _ f = f ( 1 - _ c ) f- f c ( ignoring all the terms ater the second in the expansion) . This can be written as v f- f f _ c or v f f _ c This equation is equivalent to v _ c
9 . 5 TH E D O PPLE R E FFE CT
Worked example As the S un rotates, light waves received on E arth rom opposite ends o a diameter show equal but opposite D oppler shits. The speed o the edge o the S un relative to the E arth is 1 . 9 0 km s - 1 . What wavelength shit should be expected in the helium line having wavelength 5 87 . 5 6 1 8 nm?
3
1 .90 1 0 = 5 87.5 61 8 ________ = 0.003 7 nm 3.00 1 0 8
S ince one edge will approach the E arth the shit rom this edge will be a decrease in wavelength ( blue shit) and that o the other edge ( receding) will be an increase in wavelength ( red shit) .
Solution v Using f = f _vc this is equivalent to _ c. 1 .90 km s 1 = 1 .90 1 0 3 m s 1
Nature of science Applications o the Doppler efect 1. Astronomy The D oppler eect is o particular interest in astronomy it has been used to provide evidence about the motion o the obj ects throughout the universe. The Doppler eect was originally studied in the visible part o the electromagnetic spectrum. Today, it is applied to the entire electromagnetic spectrum. Astronomers use Doppler shits to calculate the speeds o stars and galaxies with respect to the Earth. When an astronomical body emits light there is a characteristic spectrum that corresponds to emissions rom the elements in the body. B y comparing the position o the spectral lines or these elements with those emitted by the same elements on the Earth, it can be seen that the lines remain in the same position relative to each other but shited either to longer or shorter wavelengths (corresponding to lower or higher requencies) . Figure 2 shows an unshited absorption spectrum imaged rom an Earth-bound source together with the same spectral lines rom distant astronomical objects. The middle image is red-shited indicating that the source is moving away rom the Earth. The lower image is blue-shited showing the source to be local to the Earth and moving towards us. Frequency or wavelength shits can occur or reasons other than relative motion. Electromagnetic waves moving close to an obj ect with a very strong gravitational feld can be red- shited this, unsurprisingly, is known as
unshifted
redshifted
blueshifted
Figure 2 The Doppler shifted absorption spectra.
gravitational red-shift and is discussed urther in O ption A. The cosmological red-shit arises rom the expansion o space ollowing the B ig B ang and is what we currently detect as the cosmic microwave background radiation ( this is urther discussed in O ption D ) .
2. Radar Radar is an acronym or radio detection and ranging. Although it was developed or tracking aircrat during the S econd World War, the technique has wide- ranging uses today, including:
weather orecasting
ground- penetrating radar or locating geological and archaeological arteacts
providing bearings
radar astronomy
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use in salvaging
collision avoidance at sea and in the air.
transmitterreceiver ultrasound beam blood cells v
direction o blood ow
blood vessel
Figure 3 Radar screen used in weather orecasting.
Radar astronomy diers rom radio astronomy as it can only be used or Moon and planets close to the Earth. This depends on microwaves being transmitted to the object which then reects them back to the Earth or detection. In order to use the Doppler equations, we must recognize that the moving object frst o all behaves as a moving observer and then, when it reects the microwaves, behaves as a moving source. This means, or microwave sensing, the equation f f _vc is adapted to become f 2 f _vc .
3. Measuring the rate o blood fow The D oppler eect can be used to measure the speed o blood ow in blood vessels in the body. In this case ultrasound ( i.e. longitudinal mechanical waves o requency above
Figure 4 The Doppler efect used to measure the speed o blood cells.
approximately 2 0 kHz) is transmitted towards a blood vessel. The change in requency o the beam reected by a blood cell is detected by the receiver. The speed o sound and ultrasound is around 1 5 00 m s 1 in body tissue so the equation f f _vc is appropriate or blood cells moving at speeds o little more than 1 m s 1 . As the blood does not ow in the direction o the transmitter receiver, there needs to be a actor that will give the necessary component o the blood velocity in a direction parallel to the transmitter receiver. As with radar, there will need to be a actor o two included in the equation the shit is being caused by the echo rom a moving reector. As can be seen rom fgure 4, the equation or the v co s rate o blood ow will be f 2 f _____ c . The D oppler eect owmeter may be used in preerence to in- line owmeters because it is non- invasive and its presence does not aect the rate o ow o uids. The act that it will remain outside the vessel that is carrying the uid means it will not suer corrosion rom contact with the uid.
Worked example Microwaves o wavelength 1 5 0 mm are transmitted rom a source to an aircrat approaching the source. The shit in requency o the reected microwaves is 5 .00 kHz. C alculate the speed o the aircrat relative to the source.
Using 8
c 3.00 1 0 = ________ = 2 .00 1 0 9 Hz. c = f gives f = __ 150 10 -3
Using the D oppler shit equation or radar fc 5.00 1 0 3.00 1 0 f 2 f _vc = > v ___ = _________________ 2f 2 2.00 1 0 1 = 3 75 m s 3
8
9
Solution The data gives a wavelength and a change o requency so we need to fnd the requency o the microwaves.
386
QUESTION S
Questions 1
b) Explain why the magnitude of the tension in the string at the midpoint of the oscillation is greater than the weight of the pendulum bob.
(IB) The variation with displacement x of the acceleration a of a vibrating obj ect is shown below. 3000
c) The pendulum bob is moved to one side until its centre is 2 5 mm above its rest position and then released.
a/m s 2
point of suspension rigid support
2000 1000 x/mm 0.6 0.4 0.2
0
0.2
0.4
0.6
0.80 m
1000 2000 25 mm 3000 pendulum bob
a) State and explain two reasons why the graph indicates that the obj ect is executing simple harmonic motion.
(i) S how that the speed of the pendulum bob at the midpoint of the oscillation is 0.70 m s 1 .
b) Use data from the graph to show that the frequency of oscillation is 3 5 0 Hz.
(ii) The mass of the pendulum bob is 0.057 kg. The centre of the pendulum bob is 0.80 m below the support. Calculate the magnitude of the tension in the string when the pendulum bob is vertically below the point of suspension.
c) State the amplitude of the vibrations. ( 9 marks) (IB) a) A p e ndu lum co nsists o f a b o b that is susp e nde d fro m a rigid sup p o rt b y a light ine xte nsib le string. The p e ndulu m b o b is mo ve d to o ne side and the n re le ase d. The ske tch grap h sho ws ho w the disp lace me nt o f the p e ndulu m b o b unde rgo ing simp le harmo nic mo tio n varie s with time o ve r o ne time p e rio d. displacement
0
3
( 1 0 m arks ) (IB) a) A particle of mass m attached to a light spring is executing simple harmonic motion in a horizontal direction. State the condition (relating to the net force acting on the particle) that is necessary for it to execute simple harmonic motion. b) The graph shows how the kinetic energy E K of the particle in (a) varies with the displacement x of the particle from equilibrium.
0
0.07 EK /J
time
0.06 0.05 0.04 0.03
C opy the sketch graph and on it clearly label
0.02
(i) a point at which the acceleration of the pendulum bob is a maximum.
x/m
05 0.
03
02
01
04 0.
0.
0.
0.
0.
0
0.
01
02
0.
03
04 0.
05 0.
(ii) a point at which the speed of the pendulum bob is a maximum.
0.01
2
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WAVE P H E N O M E N A ( AH L ) (i) On a copy o the axes above, sketch a graph to show how the potential energy o the particle varies with the displacement x.
Q
Z W
(ii) The mass o the particle is 0.30 kg. Use data rom the graph to show that the requency f o oscillation o the particle is 2.0 Hz.
b
Y
X
P
( 8 m arks ) 4
(IB) a) D escribe what is meant by the diffraction of light.
The angle is small.
b) A parallel beam o monochromatic light rom a laser is incident on a narrow slit. The diracted light emerging rom the slit is incident on a screen.
a) O n a copy o the diagram, label the hal angular width o the central maximum o the diraction pattern. b) S tate and explain an expression, in terms o , or the path dierence ZW between the rays ZP and XP.
screen slit parallel light wavelength 620 nm
0.40 mm
c) D educe that the hal angular width is given by the expression = _ b d) In a certain demonstration o single slit diraction, = 45 0 nm, b = 0.1 5 mm and the screen is a long way rom the slits.
C
1.9 m
The centre o the diraction pattern produced on the screen is at C . Sketch a graph to show how the intensity I o the light on the screen varies with the distance d rom C . c) The slit width is 0.40 mm and it is 1 .9 m rom the screen. The wavelength o the light is 62 0 nm. D etermine the width o the central maximum on the screen. ( 8 marks) 5
(IB) Plane waveronts o monochromatic light are incident on a narrow, rectangular slit whose width b is comparable to the wavelength o the light. Ater passing through the slit, the light is brought to a ocus on a screen. The line XY, normal to the plane o the slit, is drawn rom the centre o the slit to the screen. The points P and Q are the frst points o minimum intensity as measured rom point Y. The diagram also shows two rays o light incident on the screen at point P. Ray ZP leaves one edge o the slit and ray XP leaves the centre o the slit.
388
screen
slit
C alculate the angular width o the central maximum o the diraction pattern on the screen. ( 8 marks) 6
(IB) Monochromatic parallel light is incident on two slits o equal width and close together. Ater passing through the slits, the light is brought to a ocus on a screen. The diagram below shows the intensity distribution o the light on the screen. I
A
B distance along the screen
a) Light rom the same source is incident on many slits o the same width as the widths o the slits above. O n a copy o the diagram, draw a possible new intensity distribution o the light between the points A and B on the screen.
QUESTION S a) S tate the phase change that occurs when light is reected rom
A parallel beam o light o wavelength 45 0 nm is incident at right angles on a diraction grating. The slit spacing o the diraction grating is 1 .2 5 1 0 6 m.
( i) surace A ( ii) surace B .
b) D etermine the angle between the central maximum and frst order principal maximum ormed by the grating.
The light incident on the plastic has a wavelength o 62 0 nm. The reractive index o the plastic is 1 .4.
( 4 marks) 7
b) C alculate the minimum thickness o the flm needed or the light reected rom surace A and surace B to undergo destructive intererence.
(IB) Light o wavelength 5 90 nm is incident normally on a diraction grating, as shown below.
( 5 marks)
grating 6.0 10 5 lines per metre
9 frst order
light wavelength 590 nm
(IB) The two point sources A and B emit light o the same requency. The light is incident on a rectangular narrow slit and, ater passing through the slit, is brought to a ocus on the screen.
zero order frst order
The grating has 6.0 1 0 5 lines per metre. a) D etermine the total number o orders o diracted light, including the zero order, that can be observed.
A
b) The incident light is replaced by a beam o light consisting o two wavelengths, 5 90 nm and 5 89 nm. State two observable dierences between a frst-order spectrum and a second-order spectrum o the diracted light. ( 6 marks) 8
(IB) Monochromatic light is incident on a thin flm o transparent plastic as shown below.
C
monochromatic light
A
B flm
The plastic flm is in air. Light is partially reected at both surace A and surace B o the flm.
B point sources slit screen
a) B is covered. S ketch a graph to show how the intensity I o the light rom A varies with distance along the screen. Label the curve you have drawn A. b) B is now uncovered. The images o A and B on the screen are j ust resolved. Using your axes, sketch a graph to show how the intensity I o the light rom B varies with distance along the screen. Label this curve B . c) The bright star Sirius A is accompanied by a much ainter star, Sirius B. The mean distance o the stars rom Earth is 8.1 1 0 1 6 m. Under ideal atmospheric conditions, a telescope with an objective lens o diameter 25 cm can just resolve the stars as two separate images. Assuming that the average wavelength emitted by the stars is 5 00 nm, estimate the apparent, linear separation o the two stars. ( 6 marks)
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WAVE P H E N O M E N A ( AH L ) 1 0 (IB) a) Explain what is meant by the term resolvance with regards to a diraction grating. b) A grating with a resolvance o 2 000 is used in an attempt to separate the red lines in the spectra o hydrogen and deuterium. ( i)
The incident beam has a width o 0.2 mm. For the frst order spectrum, how many lines per mm must the grating have?
( ii) E xplain whether or not the grating is capable o resolving the hydrogen lines which have wavelengths 65 6.3 nm and 65 6. 1 nm.
a) Explain, using a diagram, any dierence between f and f. b) The requency f is 3 .00 1 0 2 Hz. An observer moves towards the stationary car at a constant speed o 1 5 .0 m s 1 . C alculate the observed requency f o the sound. The speed o sound in air is 3.30 1 0 2 m s -1 . (5 marks) 1 3 (IB) The wavelength diagram shown below represents three lines in the emission spectrum sample o calcium in a laboratory. A
B
C
( 6 marks) 1 1 A source o sound approaches a stationary observer. The speed o the emitted sound and its wavelength, measured at the source, are v and respectively. C ompare the wave speed and wavelength, as measured by the observer, with v and . Explain your answers. ( 4 marks) 1 2 The sound emitted by a cars horn has requency f, as measured by the driver. An observer moves towards the stationary car at constant speed and measures the requency o the sound to be f.
390
wavelength
A distant star is known to be moving directly away rom the Earth at a speed o 0.1 c. The light emitted rom the star contains the emission spectra o calcium. C opy the diagram and sketch the emission spectrum o the star as observed in the laboratory. Label the lines that correspond to A, B , and C with the letters A*, B *, and C *. Numerical values o the wavelengths are not required. ( 3 marks)
10 F I E L D S ( A H L ) Introduction E arlier in the book you were introduced to gravitation and electrostatics. At that stage, these were treated as separate sets of ideas, but there are strong similarities between the
concepts in both topics. In this topic we are going to discuss these similarities and develop the concepts of gravitation and electrostatics in parallel.
10.1 Describing felds Understanding Gravitational felds Electrostatic felds Electric potential and gravitational potential Field lines Equipotential suraces
Applications and skills Representing sources o mass and charge,
lines o electric and gravitational orce, and feld patterns using an appropriate symbolism Mapping felds using potential Describing the connection between equipotential suraces and feld lines
Equations workelectric potential equation: W = QVe workgravitational potential equation: W = mVg
Nature of science Our everyday experience o orces is that they are direct and observable. They act directly on objects and the response o the object is calculable. Acceptance o the feld concept means acceptance o action at a distance. One object can inuence
another without the need or contact between them. The paradigm shit rom one world view to another is dicult. It took a signifcant eort in the history o science and demands a leap in conceptual understanding rom physicists.
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Nature of science Scalar and vector felds S trictly, the felds here are vector felds as the orces have magnitude and direction. O ther felds are dierent in that the quantities they represent only have magnitude they are scalar felds. The temperature associated with each point around a camp fre is an example o a scalar feld.
The parts o this chapter devoted to electric felds have a blue background or a blue line in the margin; the sections dealing with gravity have a green background or a green line in the margin. You can concentrate on one type o feld by sticking to one colour. B eore we introduce new ideas, here is a review o those ideas that we considered in the earlier topics.
Fields A feld is said to exist when one obj ect can exert a orce on another obj ect at a distance.
Electric felds An electrostatic orce exists between two charged objects.
A gravitational orce exists between two objects that both have mass.
There are two types o charge:
A gravitational feld is associated with each mass. Any other mass in this feld has a gravitational feld acting on it.
negative, corresponding to a surplus o electrons in the object
positive, corresponding to a defcit o electrons in the object.
Nature of science Action-at-a-distance The action-at-a-distance envisaged here is imagined to act simultaneously even though it does not. Topic 1 2 teaches you that contemporary models o physics explain interactions using the concept o exchange particles that have a fnite ( and calculable) lietime.
Gravity felds
When the two objects have the same sign o charge, then the orce between them is repulsive.
Gravity is always an attractive orce. Repulsion between masses is never observed.
When the objects have opposite signs, then the orce between them is attractive. All electric charges give rise to an electric feld. An electrostatic orce acts on a charge that is in the feld o another charge.
Field strength Electric felds
Gravity felds
The defnition o feld strength is similar in both types o feld. It arises rom a thought experiment involving the measurement o orce acting on a test object. Near the surace o the Earth, the gravitational feld strength is 9.8 N kg- 1 and, on a clear day, the electric feld strength is about 100 N C- 1 . Both felds point downwards. In both gravity and electrostatics there is a problem with carrying out the measurement o feld strength practically. The presence o a test object will distort and alter the feld in which it is placed since the test object carries its own feld.
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orce acting on positive test charge electric feld strength = ___________________________________________________ magnitude o test charge
orce acting on test mass gravitational feld strength = _____________________________________ magnitude o test mass
The direction o the feld is the same as the direction o the orce acting on a positive charge. We have to pay particular attention to direction in electrostatics because o the presence o two signs o charge.
Gravitational orce is always attractive. This means that the direction o feld strength is always towards the mass that gives rise to the feld.
The unit o electric feld strength is N C1 .
The unit o gravitational feld strength is N kg1 .
10.1 DE SCRI BI N G FI ELD S
Energy ideas so ar Electric felds
Potential diference
Gravity felds
Potential energy
In Topic 5 electric potential dierence In Topic 2 gravitational potential energy was used in electrostatics and current was used as the measure o energy electricity as a measure o energy transer. transer when a mass m is moved in a gravitational feld. W V= _ Q EP = mgh V is the electric potential dierence and W is the work done on a positive test charge o size Q.
Where EP is the change in gravitational potential energy, m is the mass, g is the gravitational feld strength, and h is the change in vertical height. EP is measured in joules.
V is measured in volts, which is equivalent to joule coulomb 1 .
This equation applies when g is eectively constant over the height change being considered. Remember that g has to be the local value o the feld strength near the Earths surace 9.8 N kg1 is the value to use, but urther out a smaller value would be required.
The electric potential dierence reers to work done per unit charge, gravitational potential energy reers to work done = ( force (weight) distance moved) .
Now read on ... The recap o earlier work at the start o this topic should have reminded you o the similarities and dierences between gravitational and electric felds. The dierences include:
Gravity is most noticeable when masses are very large the size o planets, stars, and galaxies.
Electric orces largely control the behaviour o atoms and molecules but, because there is usually a complete balance between numbers o positive and negative charges, the eects are not easily noticed over large distances.
Gravitational orces on the other hand are perceived to act over astronomical distances.
D espite these dierences, the two felds share an approach that allows the study o one to inorm the other. S tudying the two types o feld in parallel will help you understand the links. I you need to concentrate on either gravity or electric felds, look at the relevant background colour or line in the margin.
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Field lines Field lines help us to visualize the shapes o electric felds that arise rom static charges. They also aid the study o the magnetic felds associated with moving charges. Magnetic felds and their consequences are discussed urther in Topic 1 1 . I an obj ect in a feld has the relevant feld property ( mass or gravity, charge or electricity, etc.) , then it will be inuenced by the feld. In a Topic 5 Investigate!, small particles ( semolina grains) were suspended in a uid. This experiment gives an impression o the feld shape between charged parallel plates and other arrangements o charge. However the experiments are tricky to carry out and give no quantitative data. Here is another qualitative experiment that gives more insight into the behaviour o electrostatic felds.
Investigate! Field between parallel plates
A small piece o oil that has been charged can be used to detect the presence o an electric feld.
The detector is made rom a rod o insulator a plastic ruler or strip o polythene are ideal. Attached to the rod is a small strip o oil: thin aluminium or gold oil or D utch metal are suitable. The dimensions o the oil need to be about 4 cm 1 cm and the oil can be attached to the rod using adhesive tape.
Touch the oil briey to one o the plates. This will charge the oil. You should now see the oil bend away rom the plate it touched.
The angle o bend in the oil indicates the strength o the electric feld. E xplore the space between the plates and outside them too. Notice where the feld starts to become weaker as the detector moves outside the plate region. D oes the orce indicated by the detector vary inside the plate region or is it constant?
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Set up two vertical parallel metal plates connected to the terminals o a power supply that can deliver about 1 kV to the plates. Take care when carrying out this experiment ( use the protective resistor in series with the supply i necessary) . B egin with the plates separated by a distance about one- third o the length o their smaller side.
Turn o the supply and change the spacing between the plates. D oes having a larger
separation produce a larger or a smaller feld?
C hange the pd between the plates. D oes this aect the strength o the feld?
An alternative way to carry out the experiment is to use a candle ame in the space between the plates. What do you notice about the shape o the ame when the feld is turned on? C an you explain this in terms o the charged ions in the ame?
The oil detector itsel can also be used to explore the feld around a charged metal sphere such as the dome o a van der Graa generator.
oil detector -
+ + + + charged plate ++ + + + + + +
- + high voltage power supply
Figure 1 Electric feld detector.
10.1 DE SCRI BI N G FI ELD S
E xperiments such as this suggest that the feld between two parallel plates:
is uniorm in the region between the plates
becomes weaker at the edges ( these are known as edge effects) as the feld changes rom the between- the- plates value to the outside- the- plates value ( oten zero) . You should be able to use the properties o the feld lines rom Topic 5 to explain why there can be no abrupt change in feld strength.
+
+
+
+
+
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+
-
-
-
-
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-
-
Figure 2 Electric feld line pattern or parallel plates.
A close study o the feld shows feld lines like those in fgure 2 . At the edge, the feld lines curve outwards as the feld gradually weakens rom the large value between the plates to the much weaker feld well away rom them. For the purposes o this course, you should assume that this curving begins at the end o the plate ( although in reality it begins a little way in as you may have seen in your experiment with the feld detector) . Try to predict the way in which the shape o the feld lines might change i a small conductive sphere is introduced in the middle o the space between the two plates. Look back at some o the feld pattern results rom Topic 5 and decide how the feld detector used here would respond to the felds there. S ometimes you will see or hear the feld lines called lines o orce. It is easy to see why. The feld detector shows that orces act on charges in a feld. The lines o orce originate at positive charge ( by defnition) and end at negative charge as indicated by the arrow attached to the line. I the line is curved, the tangent at a given point gives the direction o the electric orce on a positive charge. The density o the lines ( how close they are) shows the strength o the orce.
Linking feld lines with electric potential The Investigate! indicates that the strength o the electric feld depends on the potential dierence between them. Another experiment will allow you to investigate this in more detail.
Investigate! Measuring potentials in two dimensions
This experiment gives a valuable insight into the way potential varies between two charged parallel plates.
You need a sheet o a paper that has a uniorm graphite coating on one side, one type is called Teledeltos paper. In addition, you will need a 6 V power supply ( domestic batteries are suitable) , two strips o copper oil, and two bulldog clips. Finally, you will need a high- resistance voltmeter ( a digital meter or an oscilloscope will be suitable) and connecting leads.
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C onnect the circuit shown in the diagram. Take particular care with the connections between the copper strips and the paper. O ne way to improve these is to paint the connection between copper and paper using a liquid consisting o a colloidal suspension o graphite in water.
Press the voltage probe onto the paper. There should be a potential dierence between the lead and the 0 V strip.
C hoose a suitable value or the voltage ( say 3 V) and explore the region between the copper strips. Mark a number o points where the voltage is 3 V and draw on the paper j oining the points up. Repeat or other values o voltage. Is there a consistent pattern to a line that represents a particular voltage?
Try other confgurations o plates. O ne important arrangement is a point charge, which can be simulated with a single point at 6 V and a circle o copper oil outside it at 0 V. O ne way to create the point is to use a sewing needle or drawing pin ( thumb tack) . You will need the colloidal graphite to make a good connection between the point and the paper.
An alternative way to carry out the experiment, this time in threedimensions, is to use the same circuit with a tank o copper( II) sulate solution and copper plates. voltage probe at end o fying lead
0V 0V +
-1.5 V
+
+
+
+6.0 V +4.5 V + +
- 3.0 V -4.5 V
6V cells
voltmeter
+3.0 V - -6.0 V
-
-
-
0V
+1.5 V
Figure 3 Equipotentials in two-dimensions.
Figure 4 Lines o equal potential.
Tip Think about why the feld line and the equipotential must be at 90 to each other. As a hint, remember the defnition o work done in terms o orce and distance moved rom Topic 2.
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The lines o equal potential dierence between two parallel plates drawn in this experiment resemble those shown in fgure 4. These lines are called equip otentials because they represent points on the two-dimensional paper where the voltage is always the same. When a charge moves rom one point to another along an equipotential line, no work is done. Figure 4 also shows the electric feld lines between the plates. You will see a simple relationship between the feld lines and the equipotentials. The angle between them is always 90. I you know either the shape o the equipotential lines or the shape o the feld lines then you can deduce the shape o the other. The same applies to the relationship between rays and waveronts in optics.
10.1 DE SCRI BI N G FI ELD S
You will have noticed that the word dierence has disappeared and that we are using the plain word potential. As long as we always reer to some agreed reerence point ( in this case the 0 V copper strip on the paper) , the absolute values o potential measured on the voltmeter are identical to the values o potential dierence measured across the terminals o the meter. To illustrate this, the parallel- plate equipotentials in fgure 4 are re- labelled on dierent sides o the diagram taking a dierent point in the circuit to be zero. You will see that the potential dierences are unchanged, but the change o reerence position shits all the potential values by a fxed amount. This defnition o a reerence position or potential will be considered in the next sub- topic.
Tip Calculate the change in potential going rom -4.5 V to -3.0 V as well as rom +1.5 V to +3.0 V. In both cases, we go rom a low potential point to a higher potential point so the potential dierence should be positive and the same in both cases (+1.5 V) . Remember dierences are calculated rom fnal state minus initial state.
+ -
+ -
+
+
-
+
+
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-
+ +
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+ -
Figure 5 Equipotentials in three dimensions.
The conducting-paper experiment is two-dimensional. The oil detector allowed you to explore a three-dimensional feld. The equipotentials between the charged plates in three dimensions are in the orm o sheets parallel to the plates themselves always at 90 to the feld lines. So another consequence o extending to three dimensions is that it is possible to have equipotential suraces and even volumes. An example o this is a solid conductor: i there is a potential dierence between any part o the conductor and another, then charge will ow until the potential dierence has become zero. All parts o the conductor must thereore be at the same potential as each other the whole conductor is an equipotential volume. C onsequently the feld lines must emerge rom the volume at 90 whatever shape it is. In summary, equipotential suraces or volumes:
link points having the same potential
are regions where charges can move without work being done on or by the charge
are cut by feld lines at 90
should be reerred to a zero o potential
do not have direction (potential, like any energy, is a scalar quantity)
can never cross or meet another equipotential that has a dierent value.
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Equipotentials and feld lines in other situations Figure 6 shows the equipotentials and electric feld lines or a number o common situations.
+
+
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-
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(b)
(a)
-80 V outer conductor -90 V -100 V equipotential
equipotentials
+ + + + + + +
inner conductor (c)
-
(d)
(e)
Figure 6 Equipotentials around charge arrangements.
(a) Field due to a single point charge Earlier we looked at the feld due to a single point charge. In the case o a positive charge, the feld lines radiate out rom the point and the feld in this case is called a radial feld. For a negative charge the only change is that the arrows now point inwards to the charge. What are the equipotentials or this case? The key lies in the 90 relationship between feld line and equipotential. Figure 6( a) shows the arrangement o equipotentials around a single point charge. They are a series o concentric shells centred on the point charge. This arrangement is shown as a two-dimensional arrangement, but it is important always to think in three dimensions. Two- dimensional diagrams are however easier to draw and perectly acceptable in an examination.
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(b) Field due to two point charges of the same and opposite sign These arrangements bear some similarities to the magnetic feld patterns in the space between two bar magnets; they demonstrate the use o the rules or the electric feld lines.
(c) Field due to a charged sphere A hollow or solid conducting sphere is at a single potential.The feld outside the sphere is identical to that o a point charge o the same magnitude placed at the centre o the sphere. This time however, there are no feld lines inside the sphere.
(d) Field due to a co-axial conductor C o- axial conductors are commonly used in the leads that connect a domestic satellite dish to a TV. There is a central conductor with an earthed cylinder outside it, separated and spaced by an insulator. The symmetry o the arrangement is dierent rom that o a sphere and so the symmetry o feld pattern changes too.
(e) Field between a point charge and a charged plate The key to this diagram is to remember that the feld lines are radial to the point charge when very close to it and the lines must be at 90 to the surace o the plate.
Field and equipotential in gravitation We did not use the concept o gravitational feld lines in Topic 6. However, feld lines can be used to describe gravitational felds. The concepts o feld lines and equipotentials transer over to gravity without difculty. Although there is no easy experiment that can make the gravitational feld lines visible ( unlike magnetic or electric feld lines) , the direction can be determined simply by hanging a weight on a piece o string! The direction o the string gives the local direction o the feld pointing towards the centre o the planet. The arrangement o gravitational feld lines that surround a point mass or a spherical planet is similar to the pattern or electric feld lines around a negative point charge. Again, the feld is radial or both cases with the feld lines directed towards the mass because gravitational orce can only be attractive. All that has been written about equipotentials in electric felds also applies to gravitational equipotentials. Imagine a mass positioned at a vertical height above the E arths surace, say 1 0 m. I the mass moves to another location which is also 1 0 m above the surace, then no work has to be done ( other than against the non- conservative orces o riction) to move it. The mass begins and ends its journey on the same equipotential and no work has been transerred. These two points lie on the surace o a sphere whose centre is at the centre o the planet.
Figure 7 Field lines and equipotentials around a planet.
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F I E L D S ( AH L ) There is another way to interpret gravitational equipotentials and feld lines. It is possible that you regularly use equipotentials without even realizing it!
line of steepest descent
Figure 8 Contour map (contour lines have been removed for clarity) .
The contour lines on this map are equipotentials. I you walk along a contour line then your vertical height does not change, and you will neither gain nor lose gravitational potential energy. Mountain walkers have to be experts in using contour maps, not j ust to know where they are, but to choose where it is sae to go. Looking across the contours where they are closest together gives the line o steepest descent. This gives the direction o the gravitational feld down the slope at that point. Just as or electric felds, a gravitational feld is uniorm and directed downwards at 90 normal to the surace when close to the surace o the obj ect.
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Worked examples 1
An electrostatic device has electric feld lines as shown in the diagram.
2
The diagram shows electric equipotential lines or an electric feld. The values o the equipotential are shown. Explain where the electric feld strength has its greatest magnitude.
10 V
20 V
D raw the diagram and add to it fve possible equipotential lines or the arrangement.
30 V
Solution
40 V
The equipotential lines must always meet the feld lines at 90. O ne possible line is a vertical line in the centre o the diagram. O n each side o this the lines must bend to meet the lines appropriately.
50 V
Solution The work done in moving between equipotentials is the same between each equipotential. The work done is equal to force distance. So the orce on a test charge is greatest where the distance between lines is least. This is in the region around the base o the 2 0 V equipotential. Providing that potential change is the same between neighbouring equipotential lines, then the closer the equipotential lines are to each other, the stronger the electric feld strength will be.
Potential at a point Gravitational potential S o ar the terms potential and potential difference have been used almost interchangeably by saying that potential needs a reerence point. B ut this needs clarifcation. We need to defne potential careully or both gravitation and electrostatics and, in particular, we need to decide on the reerence point we will use or our zero. It is easier to begin with gravity: Gravitational p otential dierence is the work done in moving a unit mass (1 kg in our unit system) between two p oints. To calculate the change in gravitational potential o an obj ect when it leaves one point and arrives at another, we need to know both the
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F I E L D S ( AH L ) gravitational energy needed to achieve the move and also the mass o the obj ect. Then the change in the gravitational potential o the w o rk do ne to m o ve th e o b j e ct arrival point relative to the departure point is _____________________ . m ass o the o b j e ct
In symbols W Vg = _ m where Vg is the change in gravitational potential, W is the work done, and m is the mass o the obj ect. The unit o gravitational potential is j oule kilogram 1 ( J kg 1 ) . You can expect to use MJ kg 1 or changes in potentials on a planetary scale. What makes a good zero reerence or potential? O ne obvious reerence point could be the surace o the oceans. When travel is based only on E arth this is a reasonable reerence to choose. We reer to heights as being above or below sea level so that Mount E verest is 8848 m asl ( above sea level) . D espite being an appropriate local zero o potential or E arth- based activities, sea level is not good enough as soon as astronauts leave the E arth. It makes no sense to reer planet Mars to the E arth sea level we need another measure. The reerence point that makes most sense although it might seem implausible on frst meeting is infnity. Infnity is not a real place, it is in our imagination, but that does not prevent it rom having some interesting and useul properties! Gravitational p otential is defned to be zero at infnity. Newtons law o gravitation tells us that the orce o attraction F between two point masses is inversely proportional to the square o their separation r 1 F _2 r The defnition o gravitational potential reects the act that there is no interaction between two masses when they are separated by this imaginary infnite distance. Newtons law predicts that i the separation between two masses increases by one thousand times, then the attractive orce goes down by a actor o one million. S o i r is very large, then the orce becomes very small indeed. At infnity we know that the orce is zero even though we cannot go there to check. Now we can study what happens as the two masses begin to approach each other rom this infnite separation. Imagine that you are one o the masses and that the second mass ( originally at infnite distance) is moved a little closer to you. This mass is now gravitationally attracted to you ( and vice versa) . Work would now need to be done on the system to push the mass away rom you against the orce o attraction back to its original infnitely distant starting point. Energy has to be added to the system to return the mass to infnity. O n arrival at infnity the mass will have returned to zero potential
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10.1 DE SCRI BI N G FI ELD S
( because it is at infnity) and thereore whenever the masses are closer than infnity, the system o the two masses has a negative potential. The more closely they approach each other, the more negative the stored energy becomes because we have to put increasing amounts o energy back into the system to return the masses to an infnite separation ( fgure 9) . Gravitational p otential at a p oint is defned to be equal to the work done p er unit mass (kilogram) in moving a test mass rom infnity to the p oint in question. 3m
+30J kg-1
2m
+20J kg-1
1m
+10J kg-1
Tip At infnity, the gravitational potential is maximum and equal to zero, by the convention. At separations less than this the masses are bound to each other gravitationally and so have negative gravitational potential.
0J kg-1
-6.3 10 7 J kg-1 -4 -1 10 7 J kg-1 0
10 7 J
-5 kg-1
10 7 J
kg-1 8.0 Mm
10 Mm
40 Mm
Figure 9 School lab and the Earth.
O n the Earth, sea level corresponds to a gravitational potential o 6.2 5 1 0 7 J kg 1 . ( We will calculate this in the next section.) Near the surace, raising one kilogram through one metre rom sea level requires 9.8 ( N) 1 ( m) o energy, in other words, about 1 0 J o energy or each kilogram raised. S o the potential becomes more positive by 1 0 J kg 1 or each metre raised. You could label the wall o your school laboratory in potential values! C omparing the gravitational potential at sea level with the gravitational potential at the top o Mount Everest shows that the potential at the top o the mountain is 86 71 0 J kg 1 greater than at sea level. This makes the gravitational potential at the top o Everest equal to 6.2 4 1 0 7 J kg 1 . Not apparently a large change, but then infnity is a good deal urther away! I we know the change in potential, then we can use it to calculate the total gravitational potential energy required or a change in height. For a mountaineer and kit o total mass 1 2 0 kg, the gravitational potential energy required to scale Mount Everest is about 1 0 7 J ( that is 86 71 0 J kg 1 1 2 0 kg) starting at sea level.
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F I E L D S ( AH L ) Returning to the idea o our contour maps, we could re-label all the contours changing the values rom heights in metres to potentials in joule kg 1 . The shape would be the same. Instead o knowing the dierence in height, we now know the amount o energy that will be used or released when each kilogram o the mass moves rom one place on the map to another.
Electrical potential The situation needs to be modifed or electric potential. Again, the zero o potential is defned to be at infnity and the change in electric potential Ve is defned to be
Tip These charges are not bound to each other. They have a high electric potential energy. When returned to infnity they will have zero electric potential energy and this value is a minimum.
work done in moving charge between two points W Ve = ____ = _ Q magnitude of charge The energy stored in the system per unit charge is what we call the potential o the system. I we have two charges o opposite sign then they attract each other, and the situation is exactly the same as in gravitation. In other words, when the charges are some fnite distance apart, energy is required to move them to infnite separation. As the charges separate, the electric potential will gradually climb to zero. The situation changes when the two charges have the same sign: now both charges repel each other.
Tip Dont orget there are three separate points in this defnition:
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work done in moving unit charge
the test charge is positive
the test charge moves rom infnity to the point
Imagine two negative charges held at some fnite distance apart. Energy was stored in the system when the charges were brought together rom infnity. I they are released, the two charges will y apart ( to infnity where the repulsive orce becomes zero) . The system does work on the charges. There must have been a positive amount o energy stored in the system beore the charges were allowed to separate. The potential is still defned to be zero at infnity, but this time as one charge is moved rom infnity towards the other charge, the overall value o the potential rises above the zero level to become positive. Work has been done on the charge. The electric p otential at a p oint is the work done in bringing a unit p ositive test charge rom infnity to the point.
1 0 . 2 F I E L D S AT W O R K
10.2 Fields at work Understanding Potential and potential energy
Applications and skills Determining the potential energy o a point
Potential gradient Potential dierence
Escape speed
Orbital motion, orbital speed, and orbital energy
Forces and inverse-square law behaviour
Nature of science Electric charge is invisible to our senses, yet we accept its existence. We are aware o it through the orce laws that it obeys and via the ield properties that we assign to it. In the same way, the motion o planets and satellites on a larger scale also requires that scientists develop ways to visualize and then report their theories to non-scientists.
mass and the potential energy o a point charge Solving problems involving potential energy Determining the potential inside a charged sphere Solving problems involving the speed required or an object to go into orbit around a planet and or an object to escape the gravitational feld o a planet Solving problems involving orbital energy o charged particles in circular orbital motion and masses in circular orbital motion Solving problems involving orces on charges and masses in radial and uniorm felds
Equations potential energy
equations:
kQ Ve = _____ r V
feld strength
equations: relation between
feld strength and potential:
GM Vg = ______ r Vg
e E = _______ r
g = _______ r
Ep = qVe
Ep = mVg
kQq = _______ r q q
GMm = ________ r m m
1 2 1 2 FE = k ________ FG = G __________ r2 r2
orce laws:
___
2GM escape speed: ve s c = ________ r __ ______ GM orbital speed: v = o rb it
r
Introduction In the second part o this topic we look at the underlying mathematics o felds and how it applies to gravitational and electric felds. Parts o this chapter devoted to electric felds have a blue background or margin line; the sections dealing with gravity have a green background or margin line. You can concentrate on one type o feld by sticking to one colour.
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Forces and inverse-square law behaviour Electric felds
Gravity felds
Both felds obey an inverse-square law in which the orce between two objects is inversely proportional to the distance between them squared. In a vacuum: kq 1 q 2 FE = + __ r2 where FE is the orce between two point charges q 1 and q 2 and r is the distance between them. This is known as Coulombs law. The constant k in the equation is _1_ 4 0 where 0 is known as the permittivity o a vacuum or the permittivity o ree space.
Gm 1 m 2 FG = __ r2 where FG is the orce between two point masses m 1 and m 2 and r is the distance between them. This is known as Newtons law o gravitation. The constant in the equation is G known as the universal gravitational constant. The value o G is a universal constant.
I the feld is in a medium other than a vacuum then the 0 in the equation is replaced by , the permittivity o the medium. The sign in the equation is positive. The sign o the overall result o a calculation indicates the direction in which the orce acts. Negative indicates attraction between the charges; positive indicates repulsion.
Nature of science Inverse-square laws B oth felds obey the inverse-square law in which 1 the orce depends on _______ . Sub-topic 4.4 distance showed how the inverse-square law also arises in the context o radiation rom the geometry o space. 2
1 The value o n in __ has been tested a number o r times since the inverse- square law behaviour o electric and gravitational felds was suggested, and n is known to be very close to 2 ( or C oulombs law, within 1 0 - 1 6 o 2 ) n
Tip Charges q1 and q2 can be positive or negative. A positive orce FE indicates a repulsive orce. A negative orce FE indicates an attractive orce. The masses m 1 and m 2 are always positive, and so in this notation a positive orce FG always indicates an attractive orce.
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One consequence o a orce law being inversesquare is that the orce between the charges or masses becomes weaker as the distance increases, but never becomes zero at any real distance. We said earlier that the orce was zero when the objects were separated by an infnite distance. There is no actual infnity point in space, but there is one in our imagination. We use infnity as a useul concept or our energy ideas. In one sense, infnity can be pushed urther away as the instruments that measure feld strength become more precise!
We saw in Topic 5 that, close to any surace, the feld is uniorm. The argument was that, locally, the surace appears at and that there is cancellation o the components o feld parallel to the surace due to each charge. This leaves only components perpendicular to the surace to contribute to the feld. The electric felds between parallel plates or the gravitational or electric felds close to a curved surace are uniorm and are useul in developing ideas about felds.
1 0 . 2 F I E L D S AT W O R K
Investigate! Charged parallel plates
This experiment illustrates how the feld strength and other actors are connected or parallel plates and will help you to understand these actors.
Repeat the measurements o distance apart and charge stored. D ont orget to switch o the power supply beore you measure the separation o the plates.
S et up a pair o parallel plates, a 5 kV power supply, a well-insulated ying lead, and a coulombmeter ( a meter that will measure charge directly in coulombs) as shown in the circuit. Later you will need to replace the parallel plates with ones that have dierent areas. You will also need to record the distance between the plates.
C arry out another experiment where the power supply em and the plate separation are unchanged but plates o dierent areas are used. Record the plate areas.
C arry out another experiment where the em V o the power supply is changed but the plate area and distance are not changed.
Plot your results as Q versus V, Q versus A, and Q versus d. where Q is the charge on the plates, V is the potential o the plate, A is the area o the plates, and d is the separation o the plates.
S et and record a suitable voltage V on the power supply. Your teacher will tell you the value to use. Take great care at all times with the high voltages used in this experiment.
Measure and record the distance d between the parallel plates.
Touch the ying lead to the right- hand plate briey and then remove it rom the plate. This charges the plates.
d a rea A cha rge Q
Zero the coulombmeter and then touch its probe to the right-hand plate that you j ust charged. Record the reading Q on the meter. You may wish to repeat this measurement as a check. Change the distance between the plates without changing the setting on the power supply.
-
+ + + + + + + + +
0V
pla te
cou lom b m e ter
V f y in g lea d
0V
-
+
power su pply
Figure 1 How the charge stored on parallel plates varies. -
The results o the experiment show that
+ V
QV
QA 1 Q_ . d These results can be expressed as
-
+
-
+ F
+Q
VA Q_ d The constant in the equation turns out to be 0 ; in principle careul measurements in this experiment could provide this answer too.
-
+
-
+
-
+
-
+
So VA Q = 0 _ d where 0 = 8.85 4 1 0 - 1 2 m - 3 kg - 1 s 4 A 2 .
d
Figure 2 Field strength between parallel plates.
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F I E L D S ( AH L )
Electric feld strength and potential gradient The uniorm electric feld between two charged plates provides us with another way to think about electric feld strength.
In Topic 5 (page 1 77) we mentioned that electric feld strength could be measured as the change in potential divided by the change in distance (we call this the potential gradient) . We can now show that this result is correct. In fgure 2, a positive charge that has a size Q is in a feld between two charged plates. This feld has a strength E and the orce F that acts on the charge is thereore (rom the defnition o electric feld strength given in Sub-topic 5.1 ) F = EQ It is easy to determine the work done by the feld on the charge when the charge is moved at constant speed because the feld is uniorm and the orce is constant. As usual, the work done is force distance moved and the lines o orce are in the same direction as the distance moved so there is no component o distance to worry about Work done = EQ x where x is the distance moved. When the charge goes rom one plate to the other the distance moved is d. The work done on the charge is EQd. The potential dierence V between the plates is (see Topic 5.1 ) work done in moving a charge V = ___ magnitude of charge So EQd work done V= _ =_ Q Q leading to V E= _ d This shows that electric feld strength can be written and calculated in two ways:
force actin g on ch arge ______________ , this gives the unit N C 1 m agn itu de of ch arge
poten tial ch an ge ___________ , this gives the unit V m1 . distan ce m oved
So as well as two equations, there are two possible units or electric feld strength and both are correct. Writing the equations in ull or the uniorm feld between parallel plates V _ E=_ = F Q d
In practice, it is easier to measure a potential dierence with a voltmeter in the laboratory than to measure a orce. This is why an electric feld strength is commonly expressed in volt metre -1 .
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1 0 . 2 F I E L D S AT W O R K
Worked examples 1
A pair o parallel plates with a potential dierence between them o 5 .0 kV are separated by 1 2 0 mm. C alculate:
a) C alculate the potential dierence required to produce a orce o 3 .6 1 0 1 4 N on the droplet.
a) the electric feld strength between the plates
b) The plates are now moved closer to each other with no change to the potential dierence. The orce on the droplet changes to 1 .4 1 0 1 3 N. C alculate the new separation o the plates.
b) the electric orce acting on a doubly ionized oxygen ion between the plates.
Solution V 5 000 a) E = __ = ____ = 4. 2 1 0 4 V m 1 0.1 2 d
Solution
F = QE = 3.2 1 0 1 9 4.2 1 0 4 = 1 .34 1 0 1 4 N
3 .6 1 0 1 4 F a) E = _ = 3 .2 1 0 4 N C 1 = __ Q 1 1 .2 1 0 1 9 V = Ed = 3 .2 1 0 4 0.080 = 2 600 V
The orce is directed towards the negative plate.
b) The orce changes by a actor o
b) The charge o the ion is 2 e = + 3 .2 1 0 1 9 C .
2
A pair o parallel plates are separated by 80 mm. A droplet with a charge o 1 1 .2 1 0 1 9 C is in the feld.
1 . 4 1 0 1 3 ________ = 3 .9 3 . 6 1 0 1 4
The separation decreases by this actor, to 80 ___ = 2 1 mm. 3.9
Electric feld strength and surace charge density The expression VA Q = 0 _ d rearranges to Q V _ = 0 _ A d Q __ is , the surace charge density on the plate. We say that or the A
uniorm feld between the plates, the electric feld strength E can be V written as __ . d S o between the plates E=_ 0 Remember, this equation is or the feld between two parallel plates. Each plate contributes hal the feld, so the electric feld E close to the surace o any conductor is equal to E=_ 2 0
Tip
where is the charge per unit area on the surace and 0 has its usual meaning. This is dimensionally correct because the units o are C m 2 and the C m _______ 1 units o 0 are C 2 N 1 m 2 , __ is C N m which simplifes to N C , the 2
0
2
1
2
same units as or E.
Generally, the term gradient refers to a change per unit of distance. The term rate refers to a change per unit to time. (Sometimes gradient is used a word for slope)
You will be given this equation in an examination i you need to use it.
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10
F I E L D S ( AH L ) 0V
(a)
2V
4V
6V
+
-
+ +
-
Figure 3(a) is a reminder o the feld pattern between charged parallel plates.
+
(b)
Graphical interpretations o electric feld strength and potential
8 V 10 V 12 V
The equipotentials are equally spaced. Figure 3 ( b) shows the data rom measurements o the potential at a series o points between the plates plotted as a graph o potential against distance rom the zero potential plate. It is a straight line because the electric feld is uniorm. A graph o electric feld strength against distance rom the 0 V plate is a straight line parallel to the distance axis ( Figure 3 ( c) ) .
+ 12
potential/V
10 8 6
ch an ge in V The gradient o the potentialdistance graph is ________ and because it is ch an ge in x V __ a straight line, this is equal to d , in other words, E.
4 2
This leads to:
0 0
1 2 3 4 5 distance rom 0 V plate/cm
E, electric feld strength = p otential gradient =
6
V _ e
x
( , as usual, stands or change in) .
electric feld strength/NC - 1
(c)
distance rom 0 V plate/cm 0 1 2 3 4 5
6
0 -100 -200 -300
Figure 3 Electric feld strength and potential.
(a) orce
F work done = area under graph = F r r
distance
electric orce
(b) work done = area under graph
V The equation E = ___ can be re- written as E x = V, and in this x orm has a graphical interpretation.
electric feld strength
work done per unit charge = potential change between r1 & r2 potential at r3 A r1
B
to
r2 r3 distance
Figure 4 Work done and work done per unit charge.
410
You may be wondering whether the argument changes i the moving charge is an electron. I an electron moves rom 0 V towards the 1 2 V plate (a position o higher potential) its potential energy is reduced because the potential energy change Ep is given by eV as usual but here e is negative and V is positive. The electron will accelerate in this direction i ree to do so gaining kinetic energy rom the feld at the expense o the electrical potential energy. V In numerical terms (fgures 3( b) and ( c) ) , the gradient ___ is + 2 00 V m - 1 x -1 and so the electric feld strength is - 2 00 N C .
distance (c)
The minus sign that appears in the equation needs an explanation. It means that the direction o the vector electric feld is always opposite to the variation o the potential ( Ve ) o a positive charge. S o, travel in the opposite direction to that o the feld means moving to a position o higher potential and thus a positive gain in potential. In Figure 3 ( a) the motion o the positive charge is to the let, Ve is negative ( fnal state minus initial state: 0 V - 1 2 V) ; so, according to the equation, E will be positive and this tells us that the motion is in the direction o the feld. In other words, going upstream in the feld ( against the feld) means going to higher potential so a gain in potential ( positive change in Ve ) . Going downstream, Ve is negative and E is positive ( product o two minus signs) indicating the motion is in the direction o feld.
A graph o orce against distance or a constant orce is shown in fgure 4( a) . The work done when this orce moves an obj ect through a distance r is equal to the area under the graph. In symbols F r = W When an electric orce varies with distance then the work done is still the area under the graph ( Figure 4( b) ) . Electric feld strength ( Figure 4( c) ) is the orce per unit charge, and so the area under a graph o electric feld strength against distance is equal to the work done per unit charge in other words the change in potential.
1 0 . 2 F I E L D S AT W O R K
kQ E=_ r2 Q here is the charge creating the feld ( not a charge that would react to the feld) and show that Ve is given by Q Ve = _ or 4 0 r
kQ _ r 1 using the constant k = _. 4 0 This is the expression or the potential V at a distance r rom a point charge. It predicts that the closer we are to a (positive) charge, the greater (more positive) is the potential. C onversely, the closer we are to a negative charge, the more negative is the potential. These predictions correspond to the conclusions we reached when we considered fgure 3.
1 Notice the relative shapes of the curves: ____ r ___ 1 is more sharply curving than the r of the potential. Notice also the links between the two graphs. 2
(a) electric feld
The equation Ve E = _ r can be written in calculus orm as dVe E = _ dr This new orm can be used to take the equation or the feld strength
Tip
V = area = E r (r small) E r
distance
(b) electric potential
Figure 4( c) shows the variation o electric feld strength against distance or a single point positive charge this is an inverse- square variation. Two areas are shown on this graph, one area ( A) that shows the potential change between r1 and r2 , the other ( B ) shows the potential at r3 in this case the area included goes all the way to infnity.
feld E = V
r
V (r small) r
distance
Figure 5 Electric feld strength and potential against distance or a positive point charge.
Potential is the work done in moving a positive unit charge ( rom infnity) to a particular point, so the work required to move a charge o size q rom infnity to the point will be: Ve q. This is the potential energy that charge q possesses due to its position in the feld that is giving rise to the potential. I a charge q is in a feld that arises rom another single point charge Q then its potential energy Ep = qVe , or kqQ qQ _ Ep = _ r = 4 r 0 Graphs o feld strength and potential against distance or the feld due to a single positive point charge are shown in fgure 5 .
Worked examples 1
Two parallel metal plates have a potential dierence between them o 1 5 00 V and are separated by 0.2 5 m. C alculate the electric feld strength between the plates.
Solution
potential dierence Electric feld strength = __ separation 1 500 = _ = 6 kV m1 (or kN C 1 ) 0.25
2
A point charge has a magnitude o 0.48 nC . C alculate the potential 1 .5 m rom this charge.
Solution Q Ve = _ 4 0 r S ubstituting 0. 48 1 0 9 Ve = __ = 2 .9 V 4 0 1 .5
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F I E L D S ( AH L )
Nature of science Linking electrostatics and gravity The conclusion reached here arises because o the inversesquare law. It also applies to a hollow planet. We can show, in a similar way, that the gravitational feld strength is zero. This helps us later to show how the gravitational feld strength varies inside a solid planet o uniorm density.
Potential inside a hollow conducting charged sphere S o ar we have considered point charges and the parallelplate arrangement in detail. Now we turn to another important charge confguration: the hollow conducting charged sphere. O utside a charged conducting sphere, the feld is indistinguishable rom that o the point charge. The argument is that because the feld lines leave the surace o the sphere at 90 these lines must be radial, so an observer outside the sphere cannot distinguish between a charged sphere and a point charge with the same magnitude o charge placed at the centre o the sphere.
The feld inside the sphere B ecause the sphere is a conductor, all the surplus charge must reside on the outside o the sphere. This ollows because:
the charges will move until they are as ar apart as possible
the charges will move until they are all in equilibrium ( which in practice means that they have to be equidistant on the surace) .
To fnd the feld strength inside the conductor we need to compute the orce that acts on a positive test charge Q placed anywhere inside the conductor. We choose a point inside the conductor at random and place our test charge there. This is shown in fgure 6. The next step is to fnd the orce acting on this test charge. To make this easier we consider two cones that meet at the test charge. These cones have the same solid angle at the apex. O ne o these cones is large and the other is small. The test charge is close to the conductor surace on the side o the small cone but not so close on the side o the large cone.
surface of charge conducting sphere 2xl rl
+Q rs
cm -2 , surface charge density
Figure 6 Field inside a charged conductor.
2xs
C all the distance rom the test charge to the sphere rs or the small cone and rl or the large cone. The cone radii are xs and xl or the small and large cones respectively. The surace charge density on the cone is and is the same over the whole sphere. Thereore
For the small cone
For the large cone
Area of the end of the cone
xs 2
xl 2
Charge on the end of the cone
xs 2
xl 2
rs
rl
Distance of area from test charge Force on test charge due to area
kQxs 2 __ rs 2
2 kQx l _ _ rl 2
Directed away from the surface
Directed away from the surface
The geometry o the cones is such that xs xl = _ _ tan _ = rs rl 2
( )
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1 0 . 2 F I E L D S AT W O R K
and thereore kQxs2 kQxl2 _ _ is equal to rs2 rl 2
E 1 r2
The orces acting on the test charge due to the two charge areas are equal in size and opposite in direction. Remarkably, they completely cancel out leaving no net orce due to these two areas o charge. We chose the angle o the cones arbitrarily so this result applies or any angle we could have chosen. S imilarly, we chose the position or the test charge at random. Thereore this proo must apply or any pair o cones and any test point inside the sphere. There is thereore no net orce on the test charge anywhere inside the sphere and consequently there is no electric feld either. This proo relies on the inverse- square law and the act that the area o the ends o the cones also depends on the ( distance of the test charge from the area) 2 . B ecause the electric feld is zero the potential inside the sphere must be a V is zero inside the sphere. V being constant constant since the gradient ___ r must mean that V is zero so no work is required to move a charge at constant speed inside the charged sphere. The potential inside is equal to the potential at the surace o the sphere.
r
0
V 1 r
0
r
Figure 7 Field and potential inside and outside the conductor.
S o we can now plot both the electric feld strength and the electric potential or a charged conducting sphere and these graphs are shown in fgure 7.
Nature of science Potential and potential energy The connection between feld strength and potential gradient is universal and applies to all felds based on an inverse-square law. It tells us about the undamental relationship between orce and the work the orce does and it tells us this in a way that is independent o the mass (or the charge) o the test object that is moved into the feld. Thus, feld strength is a concept that represents orce but with the mass or charge term o the test object
removed. O course, the mass or charge element that gives rise to the feld in the frst place is still important and remains in E and g. So Mass/charge independent Field strength Potential
Mass/charge dependent force p otential energy
Gravitational potential Gravitation As with electric felds Vg g = _ r where g is the gravitational feld strength, Vg is the gravitational potential, and r is distance. For the case o the feld due to a single point mass M with a point mass m a distance r rom it, the potential Vg at r is GM Vg = _ r
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10
F I E L D S ( AH L ) The potential energy o m at r rom mass M is
distance
(a)
GMm Ep = mVg = _ r
1 r2 gravitational feld strength distance 1 r gravitational potential (b)
These equations lead to graphs similar to those or electric felds, which are illustrated in fgure 8( a) . The orces are always attractive or gravity and we saw that this has the consequence that the potential is always negative with a maximum o zero at infnity. The graphs in fgure 8 are plotted to reect this. The graphs should be shown in three-dimensions, but are plotted in two dimensions or clarity. When thinking o the plotted surace o this 1 negative __ r potential in 3 -D , treat it as a potential well ( fgure 8( b) ) . This well will trap any particle that has insufcient kinetic energy to escape the mass. Later when we discuss how spacecrat leave the Earth you can imagine that a spaceship without enough energy can only climb up so ar. From this point the spaceship can circle around the mass ( that is orbit the mass) , but climb no higher.
Figure 8 Gravitational feld strength and potential variation with distance.
Worked examples 1
At a point X the gravitational potential is 8 J kg 1 . At a point Y the gravitational potential is 3 J kg 1 . C alculate the change in gravitational potential energy when a mass o size 4 kg moves rom X to Y.
Solution A change in potential is calculated rom fnal state minus initial state so change here is ( 3 ) ( 8) = + 5 J kg 1 . S o energy required is + 2 0 J, moving rom a lower point in the potential well to a higher point ( climbing toward infnity where the value is maximum and equal to zero) . 2
Which diagram shows the gravitational equipotential suraces due to two spheres o equal mass? A.
B.
C.
D.
Solution The spheres attract and so the gravitational feld lines j oining them resemble the pattern between two unlike charges ( this pattern is similar to that in A) . The equipotential suraces must cut these feld lines at 90 so the best ft to the appropriate pattern is C . 3
414
C alculate the potential at the surace o the Earth. The radius o the Earth is 6.4 Mm and the mass o the Earth is 6.0 1 0 24 kg.
1 0 . 2 F I E L D S AT W O R K
Solution GM Vg = _ r 6.7 1 0 1 1 6.0 1 0 24 ___ Vg = 6.4 1 0 6 = 6.3 1 0 7 J kg 1
distance rom centre o Earth
rE
Potential inside a planet In a charged conducting sphere the mobile charges move to the outside o the sphere and this results in zero feld inside. This is not the same or the gravitational feld o a solid massive sphere (massive in this context means having mass not having a large mass) . This case was mentioned in Topic 6 in the context o a tunnel drilled rom the surace to the centre o the Earth. The essential physics is that i we are at some intermediate depth between surace and centre we can think o the E arth as having two parts: the solid sphere beneath our eet and the spherical shell above our head. The inner solid sphere, a miniature Earth with a smaller radius, behaves as a normal Earth but with a dierent gravitational feld strength given by G redu ced m ass of th e Earth g = ___________________ radiu s of sm all Earth
outside Earth
outside Earth
g inside Earth
Figure 9 Gravitational feld inside the Earth.
Nature of science Delivering the mail!
2
The outer spherical shell does not give rise to a gravitational feld. The argument is similar to that which we used to prove that the electric feld inside a charged conductor was zero. 4 The reduced mass of the Earth is equal to __ r 3 where is the density o 3 the Earth ( assumed constant) , so
4G r g = _ 3 where r is the reduced radius o the small E arth The gravitational feld is proportional to the distance rom the centre o the Earth until we reach the surace. This is an unexpected result. The inverse- square law actually gives rise to a linear relationship because 1 the __ behaviour o the orce cancels with the r3 variation o the mass o r the reduced Earth. 2
Leaving the Earth Picture a spacecrat stationary on its launch pad. A space launch rom Earth may put a satellite into Earth orbit. Or it may be a mission to explore a region ar away rom Earth requiring the crat to escape Earths gravity and eventually come under the inuence o other planets and stars. What minimum energy is needed to allow the crat to escape the Earth?
Orbiting In Topic 6 there are some basic ideas about circular orbits o a satellite about a planet. We showed that the linear orbital speed v o a satellite about a planet is ____
vorbit =
GM _ r
The acceleration g is always directed toward the centre o Earth and is proportional to r. So a vehicle inside a tunnel drilled right through the Earth will perorm simple harmonic motion i riction is neglected (see Topic 9) . The vehicle will stop momentarily when arriving at the antipode 84 minutes ater release. This is also the period o a satellite that goes round the Earth in low-earth orbit (technically, just skimming the surace) . Get the timing right and mail bags could be exchanged at the surace! A chordal tunnel dug between Vancouver and Montreal or between New York and Rio de Janeiro or between London and Johannesburg ( not passing through the center o the Earth) also allows the mail to be delivered. Use some o the ideas and equations in Topic 9 to assess the engineering issues with this plan!
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10
F I E L D S ( AH L ) and that the orbital time T or this orbit is _____
Torbit =
4 r _ GM 2 3
here M is the mass o the planet and r is the radius o the orbit.
(b)
(a)
rotation of Earth
geostationary orbit
polar orbit polar axis
Figure 10 Polar and geostationary orbits (not to scale) and the track of a geosynchronous orbit.
Worked example C alculate the orbital time or a satellite in polar orbit.
Solution The orbital time or these satellites is obtained by using the appropriate value or r in the T equation. The radius o the orbit is the orbital height (1 00 km) plus the radius o the Earth (6.4 Mm) , so r = 6.5 1 0 6 m. _______________
T=
4 (6.5 1 0 ) __________________ 6.67 1 0 6.0 1 0 2
The p olar orbit is used or satellites close to the E arths surace ( close in this context means about 1 00 km above the surace) . The orbit is called polar because the satellites in this orbit are oten put into orbit over the poles. Imagine you are viewing the E arth and the orbiting satellites rom some way away in space. You will see the satellite orbiting in one plane ( intersecting the centre o the E arth) with E arth rotating beneath it. In the course o 2 4 hours, the satellite will view every point on the E arth. For the orbital radius o a polar orbit, the linear speed o the satellite is 7800 m s 1 or 2 8 000 km per hour.
6 3
1 1
24
= 5 2 00 s ( about 87 min)
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Although satellites can be put into orbits o any radius ( provided that the radius is not so great that a nearby astronomical body can disturb it) , there are two types o orbit that are o particular importance.
Geosynchronous satellites orbit at much greater distances rom the E arth and have orbital times equal to one sidereal day which is roughly 2 4 hours. This means that the geosynchronous satellite can be made to stay in the same area o sky and typically ollows a fgureo- eight orbit above a region o the planet. A typical track in the southern hemisphere is shown in fgure 1 0( b) . This track shows where the satellite is overhead during the 2 4 hour cycle. B ecause the plane o the orbit coincides with the centre o the E arth but is not through
1 0 . 2 F I E L D S AT W O R K
the E quator, the overhead position wanders. To see this, use a globe imagining the satellite going around its orbit as the E arth rotates below it at the same rate. A geostationary orbit on the other hand is a special case o the geosynchronous orbit. In this case the satellite is placed in orbit above the plane o the Equator and will not appear to move i viewed rom the surace. This has the advantage that a receiving antenna ( aerial) or satellite dish on E arth also does not have to track the transmitting antenna on the satellite. A TV satellite dish pointing towards the south ( equator) in the northern hemisphere will be always aligned with the geostationary satellite. These satellites are generally used or communication purposes and or collecting whole- disk images o the E arth or weather orecasting purposes.
Worked examples
____________________________
r=
a) A geostationary satellite has a orbital time o 2 4 hours. C alculate the distance o the orbit rom the surace o the Earth.
86 400 6.67 1 0 6.0 1 0 ____ 4 1 1
2
24
3
2
= 42 1 0 6 m = 42 000 km
b) C alculate the gravitational feld strength at the orbital radius o a geostationary satellite.
This is ( 42 000 6400) = 3 6 000 km above the surace ( the subtraction is to fnd the distance rom the surace) .
Solution
b)
a) 2 4 hours is 86 400 s.
GME g = _2 ( rE + r) 6. 7 1 0 1 1 6.0 1 0 24 = ___ ( 4. 2 1 0 7 ) 2 = 0. 2 3 N kg 1
Rearranging the orbital time equation ______
r=
T GM _ 4 2
3
2
Escaping the Earth The total energy o a satellite is made up o the gravitational potential energy and the kinetic energy ( ignoring any energy transerred to the internal energy o the atmosphere by the satellite) . To escape rom the surace o the Earth, work must be done on the satellite to take it to infnity. For an unpowered proj ectile this is simply equal to the kinetic energy meaning that the total o the ( negative) gravitational potential energy and the ( positive) kinetic energy must add up to zero. Thus to reach infnity gravitational potential energy + kinetic energy = 0 and thereore GME m S 1 m v2 = 0 _ + _ RE 2 S E where RE is the distance o the satellite rom the centre o the Earth, vE is the escape speed, and ME and m s are the masses o the Earth and the satellite respectively. The gravitational potential energy is negative because this is a bound system. Kinetic energy is always positive. ( It is important to keep track o the signs in this proo.)
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F I E L D S ( AH L ) To escape the Earths gravitational feld completely the total energy o the satellite must be ( at least) zero. For the case where it is exactly zero ( or the satellite to j ust reach infnity) vE =
_____
2 GME _ RE ____
= 2 gR E
where g is the gravitational feld strength at the surace. This speed must be about 1 1 2 00 m s 1 ( 400 00 km hour 1 ) or something to escape. It is important to be clear about the true meaning o escape speed. This is the speed at which an unpowered obj ect, something like a bullet, would have to be travelling to leave the Earth rom the surace. In theory, a rocket with enough uel can leave the E arth at any speed. All that is required is to supply the 63 MJ or each kilogram o the mass o the rocket ( the gravitational potential o the Earths feld at the surace is 62 . 5 MJ kg 1 ) . However, in practice it is best to reach the escape speed as soon as possible. Earth
Moon
distance
0 L
gravitational potential
- 0.77 mJ kg-1
-63 mJ kg-1
Figure 11 Gravitational potential between the Earth and the Moon (not to scale) .
S imilarly, i a spacecrat begins its j ourney rom a parking orbit, then less uel will be required rom there because part o the energy has already been supplied to reach the orbit. The graph o potential against distance helps here. The graph in fgure 1 1 shows the variation in gravitational potential rom the Earth to the Moon. B eore takeo, a spacecrat sits in a potential well on the Earths surace. When the rockets are fred the spacecrat gains speed and moves away rom the Earth. It needs enough energy to reach the maximum o the potential at point L (this is known as the Lagrangian point and it is also the point where the gravitational feld strengths o the Earth and Moon are equal and opposite) . Once at L the spacecrat can all down the potential hill to arrive at the Moon.
Orbit shapes We have now identifed two speeds: orbital speed close to the surace ( 7. 8 km s 1 ) and the escape speed ( 1 1 . 2 km s 1 ) . At speeds between these values the spacecrat will achieve dierent orbits. These are shown in fgure 1 2 .
v > 11.2 km s -1 hyperbola v = 7.8 km s -1 circle
v = 11.2 km s -1 parabola 7.8 km s -1 < v < 11.2 km s -1 ellipse
418
Figure 12 Orbital shapes of satellites.
1 0 . 2 F I E L D S AT W O R K
At speeds less than 7.8 km s 1 the crat will return to Earth.
At a speed o 7.8 km s 1 the crat will have a circular path.
At speeds between 7.8 km s 1 and 1 1 . 2 km s 1 the orbit will be an ellipse. At a speed o 1 1 .2 km s 1 the crat will escape ollowing a parabolic path. At speeds greater than 1 1 .2 km s 1 the crat will escape the Earth along a hyperbolic path. ( It will not however necessarily escape the S un and leave the S olar S ystem. )
You will only have to consider circular orbits or the examination.
Nature of science Using the Earths rotation and the planets Rotation Many launch sites or rockets are close to the Equator. This is because the Equator is where the linear speed o the E arths surace is greatest. This equatorial velocity is about 5 00 m s 1 rom west to east meaning that a rocket taking o to the east has to attain a relative speed o 1 1 .2 0. 5 = 1 0.7 km s 1 , a signifcant saving in uel.
Slingshots Another technique much used in missions to distant planets is to use slingshot techniques in which a spacecrat is attracted towards a planet and then swings around it to shoot o in a dierent direction possibly at a greater speed. The interaction is eectively a collision between the crat and the planet in which momentum is conserved but the crat gains some energy rom the orbital energy o the planet. The spacecrat Voyager 2 was launched in 1 977 and in the course o its j ourney has passed Jupiter, S aturn, Uranus, and Neptune, using each planet to change its traj ectory achieving a Grand Tour o the planets. The spacecrat is still travelling and is now a considerable distance rom Earth ( it has its own Twittereed i you wish to know where
it is today) . It is expected to lose all its electrical energy sometime in 2 02 5 and will then continue, powered down, through space carrying messages rom the people o Earth to any other lie orm that may encounter it.
Gases in the atmosphere The Earths atmosphere only contains small amounts o the lighter gases such as hydrogen and helium. All gases in the atmosphere are at the same average temperature __ T and thus have the same mean kinetic energy EK or the molecules __ ( because, rom Topic 3 , EK T) . 1 Kinetic energy is __ mv2 , thereore the mean speed 2 o the molecules o a particular element with mass 1 __ . C alculations show that m is proportional to ___ m the lightest gases in the atmosphere have mean speeds that are close to ( but less than) escape speed. B ut even so, the very astest o these molecules can escape the atmosphere so that eventually almost all the light gases disappear. The Moon has no atmosphere, because its escape speed is lower than that o the E arth and ultimately almost all its gas molecules leave ater release. The low-density atmosphere o Mars is slowly leaking away into space.
Charges moving in magnetic and electric felds Charges moving in magnetic felds We saw in Topic 5 that a orce acts on a charge moving in a magnetic feld. In Figure 1 3 an electron moves to the right into a uniorm magnetic feld in which the feld lines are oriented at 90 to the direction in which
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10
F I E L D S ( AH L ) the electron moves ( the feld lines are out o the plane o the paper in the fgure) .
magnetic feld out o plane o peper current orce magnetic feld out orce
o plane o peper current
conventional
orce
current -
magnetic feld out o plane o peper
Figure 13 Force acting on an electron in a uniorm magnetic feld.
Flemings let-hand rule predicts the eect on the electron. The orce will be at right angles to both the velocity and the direction o the magnetic feld. In using Flemings rule, remember that it applies to conventional current, and that here the electron is moving to the right so the conventional current is initially to the let. The electron accelerates in response to the orce, and its direction o motion must change. The direction o this change is such that the electron will still travel at right angles to the feld and the magnetic orce will continue to be at right angles to the electrons new direction. This is exactly the condition required or the electron to move in a circle. The magnetic orce acting on the electron is providing the centripetal orce or the electron. As the electron continues in a circle so the magnetic orce direction alters as shown in fgure 1 3 . The orce acting on the electron depends on its charge q, its speed v, and the magnetic feld strength B. The centripetal orce will lead to a circular orbit o radius r and m ev2 _ r = Bqv S o the radius o the circle is ( written in a number o ways) 1 _
m ev ( 2 m e Ek ) 2 p r= _= _= _ Bq Bq Bq
Figure 14 Electrons moving in a circle demonstrated in a fne-beam tube.
420
where p is the momentum and m e is the mass o the electron. It is possible to use observations o this electron motion to measure the specifc charge on the electron. This is the charge per unit mass measured in C kg 1 . A fne- beam tube is used in which a beam o
1 0 . 2 F I E L D S AT W O R K
electrons is fred through a gas at very low pressure ( which means that the electrons do not collide with too many gas atoms) . When a uniorm magnetic feld is applied at right angles to the beam direction, the electrons move in a circle and their path is shown by the emission o visible light rom atoms excited by collisions with electrons along the path. The electrons o mass m e are accelerated using a potential dierence V beore entering the feld. Their energy is 1 _ m v2 = qV 2 e So
y helical path
____
v=
_ m 2 qV e
which with
B
Bqr v= _ m e gives q 2V _ _ m e = B 2 r2
+ +q z
In a case where the electron beam is moving into a magnetic feld that is not at 90 to the beam direction, then it is necessary to take components o the electron velocity both perpendicular to the feld and parallel to it. The perpendicular component leads to a circular motion exactly as beore. The only dierence will be that m e v sin r= _ Bq
x
Figure 15 A positively-charged particle moving in a magnetic feld not at 90 to its direction o motion.
where is the angle between the feld direction and the beam direction. S o the radius o the circle will be smaller than the perpendicular case. The parallel component o velocity does not lead to a circular motion. The electrons will continue to move at the component speed ( v cos ) in this direction. The overall result is that as the beam enters the feld it will begin to move in a helical path. Large particle accelerators can use magnetic felds to steer the charged particles as they are accelerated to higher and higher speeds. The acceleration is carried out using electric felds.
Charges moving in electric felds The situation is dierent or electrons moving into the uniorm electric feld produced by a pair o charged parallel plates. +V
parabolic path d
-
-
E=
V d
X 0V
Figure 16 Motion o an electron in a uniorm electric feld.
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10
F I E L D S ( AH L ) The orce on the electron now acts parallel to the feld lines. In the case shown in the fgure 1 6 the electron is accelerated vertically. B ecause the orce is constant in both magnitude and direction ( uniorm feld) the acceleration will also be constant. Mechanics equations rom Topic 2 are used to analyse the situation: V E = __ as usual or a uniorm feld, and d
qE qV F _ _ a vertical = _ me = me = m d e where m e is the mass o the electron. There is no horizontal acceleration and the horizontal component o the velocity does not change. The time t taken to travel between the plates o length x is thereore x t= _ vhorizontal S o the vertical component o the velocity as the electron leaves the plates is ( using v = u + at) qV x vvertical = a vertical t = _ _ vhorizontal m ed To obtain the fnal speed with which the electron leaves the feld, it is necessary to combine the two components in the usual way as ____________
v 2vertical + v 2horizonta l The deection s o the electron while in the feld is proportional to 1 t2 ( s = __ at2 ) . This orm is equivalent to the equation or a parabola, 2 so the traj ectory o the electron is parabolic not circular unlike the case with a magnetic feld. This derivation ignores the eects o gravity on the electron. Is this j ustifed?
Charge moving in magnetic and electric felds D espite the act that magnetic felds lead to circular orbits and electric felds lead to parabolic traj ectories, it is possible to use them in a particular confguration to do a useul j ob. The trick is to combine the magnetic and electric felds at right angles to each other.
+
E-
FE B +
FB -
422
Figure 17 Crossed electric and magnetic felds can cancel out.
1 0 . 2 F I E L D S AT W O R K For the arrangement shown in the diagram, the electric orce is vertically upwards and the magnetic orce is downwards. When these orces are equal then there will be no net orce acting on the charged particle. For this to be the case FE = FB ;
qE = Bqv
leading to E v= _ B This means that or a particular ratio o E : B there is one speed at which the orces will be balanced. C harged particles travelling more slowly than this speed will have a larger electric orce than magnetic orce, and the net orce will be upwards on the diagram. For aster electrons the magnetic orce will be larger and the electron will be deected downwards. This provides a way to flter the speeds o charged particles and the arrangement is known as a velocity selector.
Nature of science Bainbridge mass spectrometer The B ainbridge mass spectrometer is a good example o a number o aspects o electric and magnetic felds being brought together to perorm a useul j ob. There are two parts to the instrument: a velocity selector with crossed electric and magnetic felds and a deection chamber with j ust a magnetic feld ( in this diagram, into the paper) . It is let as an exercise or the reader to see that the motion o equally charged ions o the same speed but dierent mass travel in circles o dierent radii in the deection chamber. The ions arrive at the photographic plate or electronic sensor at dierent positions and, by measuring these positions, their mass ( and relative abundance) can be determined.
photographic plate or electronic sensor path o ion with smaller m q vacuum + cathode anode + ion source
gas
positive ions velocity selector
magnetic feld into paper defection chamber
Figure 18 Bainbridge mass spectrometer.
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F I E L D S ( AH L )
Concept map or feld theory Figure 1 9 gives a visual summary o all the relationships presented in this topic. It applies to both gravitational and electric felds. B oth sets o equations are represented on it. The diagram shows the cycle o feld strength potential potential energy orce and orce the relationships between them.
Comparison o gravitational and electric felds The table below sets out some o the important similarities between electric and gravitational orces and summarizes the contents o Topics 5, 6, and 1 0. E
orce = - gradient - dE = dr
F
r
r
F=
kQq r2
F=
g=
GM r2
V
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g =
Potential energy = potential quantity
FIELD STRENGTH
- dV dr
GM Eg = r Ve =
r1
g
kQ r
r
V = area under graph
Quantity = charge or mass g
Figure 19 Relationships between feld quantities.
kQq r
Potential = potential energy quantity
Potential = area under feld strength-distance graph
r
GMm r
Ee =
POTENTIAL
Field strength = - gradient o potential-distance graph
kQ E= 2 r
r1
Force = feld strength quantity
r
Eg =
POTENTIAL ENERGY
FORCE
Field strength = orce quantity
r
Potential energy = area under orce- distance graph
Force = - gradient o potential energy- distance graph
GMm r2
energy = area under graph
1 0 . 2 F I E L D S AT W O R K
Electric Qq F = __ (Coulombs law) 4 0 r2
GMm _(Newtons law) F= _ r2
Modifcation when not in a vacuum
Replace 0 with
No change
Defnition
F E= _ q
F g=_ m
Unit
N kg1 or m s 2
Distance r rom a point object
N C1 or V m 1 Q E = __ 4 0 r2
At r rom centre o sphere o radius R, r R
Q E = __ 4 0 r2
GM g = _ r2
At r rom centre o sphere o radius R, r < R
E= 0
4Gr g = __ 3
Defnition
Electric potential energy per unit charge
Gravitational potential energy per unit mass
Unit
V J C1
J kg1
For two point charges or masses
Q Vp = __ 4 0 r
GM Vg= _ r
Force law Field strength
Potential
Dierences
value at infnity: Attractive: zero and maximum Repulsive: zero and minimum
Gravitational
Constant in orce law
Between charges Attractive and repulsive depending on charge sign
_1_(= Coulomb constant k) 4 0
GM g=_ r2
Between masses Only attractive
G
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F I E L D S ( AH L )
Questions 1
b) The graph shows the variation o the gravitational potential V with distance r rom the centre o Mars. R is the radius o Mars which is 3 .3 Mm. ( Values o V inside the planet are not shown. )
( IB) The mass o Earth is ME and the radius o Earth is RE . At the surace o Earth the gravitational feld strength is g. A spherical planet o uniorm density has a mass o 3 ME and a radius 2 R E . C alculate the gravitational feld strength at the surace o the planet. ( 1 mark)
( IB) A spacecrat travels away rom a planet in a straight line with its rockets switched o. At one instant the speed o the spacecrat is 5 400 m s 1 when the time t = 0. When t = 600 s, the speed is 5 1 00 m s 1 . C alculate the average gravitational feld strength acting on the spacecrat during this time interval. ( 1 mark)
3
(IB) a)
0.0 - 1.0 - 2.0 - 3.0 - 4.0 - 5.0 - 6.0 - 7.0 - 8.0 - 9.0 - 10.0 - 11.0 - 12.0 - 13.0
1
2
3
4
5
6
7
V/MJ kg- 1
2
r/R 0
State, in terms o electrons, the dierence between a conductor and an insulator.
b) Suggest why there must be an electric feld inside a current- carrying conductor.
A rocket o mass 1 2 Mg lits o rom the surace o Mars.
c) The magnitude o the electric feld strength inside a conductor is 55 N C 1 . Calculate the orce on a ree electron in the conductor.
( i) C alculate the change in gravitational potential energy o the rocket at a distance 4R rom the centre o Mars.
d) The electric orce between two point charges is a undamental orce that applies to charges whereas gravity is the gravitational orce that applies to two masses. State one similarity between these two orces and two other dierences.
( ii) D etermine the magnitude o the gravitational feld strength at a distance 4R rom the centre o Mars. c) D etermine the magnitude o the gravitational feld strength at the surace o Mars.
e) The orce on a mass o 1 . 0 kg alling reely near the surace o Jupiter is 2 5 N. The radius o Jupiter is 7.0 1 0 7 m.
d) The gravitational potential at the surace o Earth is 63 MJ kg 1 . Without any urther calculation, compare the escape speed required to leave the surace o Earth with that o the escape speed required to leave the surace o Mars.
( i) State the value o the magnitude o the gravitational feld strength at the surace o Jupiter. ( ii) C alculate the mass o Jupiter. ( 1 3 marks)
( 1 0 marks) 4
5
An astronaut in orbit around Earth is said to be weightless. Explain why this is.
( IB) This question is about the gravitational feld o Mars. a) D efne the gravitational potential energy o a mass at a point.
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6
( IB) A small sphere X o mass M is placed a distance d rom a point mass. The gravitational orce on sphere X is 90 N. S phere X is removed and a second sphere Y o mass 4M is placed a distance 3 d rom the same point mass. C alculate the gravitational orce on sphere Y.
11
ELECTRO M AGN ETI C I N D U CTI O N ( AH L)
Introduction The physics of electromagnetic induction has profound implications for the way we generate electrical energy and therefore for the way we live. E very year throughout the world about 1 0 2 0 J of energy are converted into an electrical form using electromagnetism. This vast
co nversion of energy enables us to feed our ever- growing appetite for technologies that require electricity. It also raises considerab le issues about the sustainability of the energy sources used for the conversion.
11.1 Electromagnetic induction Understandings Electromotive orce (em) Magnetic ux and magnetic ux linkage Faradays law o induction Lenzs law
Nature of science Much o the physics in this sub-topic was discovered through painstaking experimentation by a handul o scientists. Pre-eminent among them was Michael Faraday who observed currents induced in a coil when magnetic felds were varied nearby. These observations led him to the laws o electromagnetic induction that we use in our largescale generation o electrical energy.
Applications and skills Describing the production o an induced em by
a changing magnetic ux and within a uniorm magnetic feld Solving problems involving magnetic ux, magnetic ux linkage, and Faradays law Explaining Lenzs law through the conservation o energy
Equations Flux: = BA cos
Faraday's / Neumann's equation: = - N _ t
em induced in moving rod: = Bvl in side o coil with N turns: = BvlN
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Inducing an emf In Topic 5 we saw that when an electric charge moves in a magnetic feld, then a orce acts on the charge ( and thereore on the conductor in which it is moving) . In a reverse sense, a movement or change in a magnetic feld relative to a stationary charge gives rise to an electric current. This phenomenon is called electromagnetic induction.
Investigate! Making a current B egin with a magnet and a coil or a solenoid o wire. Arrange the coil horizontally and connect it to a galvanometer ( a orm o sensitive ammeter, with the zero in the middle o the scale) . You can use a laboratory coil, or you can wind your own rom suitable metal wire using a cylinder as a ormer.
Move the bar magnet so that its north- seeking pole approaches one end o the coil and observe the eect on the meter. Record the direction o the current as indicated by the meter and the peak value shown.
Repeat the movement, moving the bar magnet with its south- seeking pole towards the coil.
Move the bar magnet away rom the coil.
C hange the speed with which you move the magnet.
C ompare the current directions or each case and also the size o the current produced.
Now try moving the coil and keeping the magnet still. Does this change your observation?
Now try moving the coil and the magnet at the same speed and in the same direction. What is the size o the current now? I your coil allows it, you might also try making the magnet enter the coil at an angle rather than along its axis. You could also try moving the magnet completely through the coil and out o the other side. Try to interpret this complex situation when you have understood the simpler cases.
Relate the direction o current ow in the coil to the magnetic poles produced at the ends o the coil using the ideas in Topic 5 . ( Figure 1 ( b) reminds you o the rule. )
(a)
(b)
(c) S
N
S
N
N
S
N
S
N
S
N
Figure 1 (a) the experiment, (b) the N and S rule, and (c) typical results.
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S
11 .1 E LE CTR O M AG N E TI C I N D U CTI O N
The results or this simple experiment are shown in fgure 1 ( c) . You should ocus both on the direction o the conventional current ow and on the magnitude o the current. You are observing results similar to those made by Faraday in the 1 9th century. A number o conclusions emerge rom these simple experiments:
The current only appears when there is relative motion between the coil and the magnet; either or both can move to produce the eect. However, i both coil and magnet move with no relative motion between them then no current occurs. O nly movement o the wire in the coil relative to the feld is important.
When the north- seeking pole o the magnet is inserted into the coil, the current in the coil tends to reduce the magnets motion by producing another north-seeking pole at the magnet end o the coil. Push a south- seeking pole in and another south pole appears at the magnet end o the coil. It is as though the system acts to repel the bar magnet and to reduce its movement. The system appears to oppose any change in the magnetic ux; the greater the rate o change, the greater is this opposition. We will look at this again later.
In the same way, when a magnetic pole is moved away rom the coil, the opposite pole is ormed by the current in the coil in an attempt to attract the magnet and reduce its speed o motion.
Moving the coil at greater speeds relative to the magnet increases the sizes o the currents. The eect is at a maximum when the axis o the magnet between its poles is perpendicular to the area o cross-section o the coil.
The keys to understanding these eects lie in what we said earlier in Topic 5 about the motion o charges in a magnetic feld and what we know about the internal structure o the conducting wire that makes up the coil. Figure 2 ( a) shows electrons in a metal rod that is moving through a uniorm, unchanging magnetic feld. Free electrons in the wire are moving upwards with the rod in an external magnetic feld that acts into the page. A orce acts on each electron to the right. This orce is equivalent in direction to that which would act on a positive charge moving down the page ( a conventional current downwards) . This direction is perpendicular to both the feld and the direction o motion o the rod and is determined using Fleming' s let- hand rule. magnetic feld into page
L
(a)
R
L
- - - R
+ + +
(b)
L
conventional current
R
(c)
electron ow
Figure 2 Electrons orced to move in a magnetic feld experience a orce. In fgure 2 ( b) , with no external connection between L and R the electrons accumulate at the right- hand end ( R) o the rod making it negatively charged, and a lack o electrons at the let-hand end ( L)
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) makes it positive. A potential dierence exists between L and R, L being at the higher potential. When there is no external connection between L and R, no current will circulate. Charges will accumulate at the end o the rod, that is electrons will move to one end leaving the other end positive. Without a current in a closed circuit no work is required, no transormation o energy takes place. I the circuit is closed externally between L and R ( fgure 2 ( c) ) , a ow o electrons will occur as shown. Inside the rod the conventional current ows rom R to L. The electrons ow out o the right-hand end o the rod and this is a conventional current in the external circuit rom L to R. A current has been generated, or induced, in the circuit. The system is acting to move the electrons through the resistor and, because this is a transormation o energy into an electric orm, the source o the energy is termed an electromotive orce (em) . As usual, we can identiy the amount o energy transerred or each coulomb o charge that moves around the circuit and we use the term induced emf to show that the em arises rom an induction eect. (The word induction is another term used in the early days o the study o electromagnetism.)
Nature of science Electromagnetic force Perhaps you can now see why the term electromagnetic orce arose in the early days o electromagnetic induction and why it still persists. S ome physicists obj ect that no orce acts when the term em is applied to electric cells, batteries, piezoelectric devices, and so on thereore, they say, em is not a good expression. It is true that it is difcult to see how the word orce can apply in the case o a cell. B ut in the case o electromagnetic induction, it is clear that a orce is acting on the electrons in the conductor that is being moved and so the term em continues to be used in physics. The act that we oten use the abbreviation em rather than the ull expression is a reminder that we should not ocus on the term orce but rather on the units o em: J C - 1 .
Lenzs law An important aspect o electromagnetic induction is that the induced em exists whether the charge fows in a complete circuit or not. In the case shown in fgure 2 ( b) the circuit is incomplete. E lectrons ow along the rod until an excess o them sets up an electric feld which repels urther electrons, stopping urther ow. It is only when the circuit is complete that charge ows. An induced em is always generated by the system and the induced current will exist only i there is a complete circuit. We can look at the system in terms o a possible direction rule. Flemings let-hand rule was able to predict the orce on, and thereore a ow o, the electrons. This ow o electrons is equivalent to a conventional current
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11 .1 E LE CTR O M AG N E TI C I N D U CTI O N
motion o the conductor
orce F
Nature of science Another rule
magnetic feld B
magnetic feld
right hand left hand Flemings let-hand rule or motor eects
current I
induced current Flemings right hand rule or induction eects
Figure 3 Flemings rules. acting in the opposite direction. You have two choices: either use Flemings let-hand rule to work out the orce direction rom frst principles and let this lead you to the conventional current direction (the argument is given above) , or you can use another rule (Flemings right-hand rule) , which uses the symmetry between our let and right hands to give the relationship between the motion o the conductor, the direction o the feld and the direction o the induced conventional current. It' s your choice!
Another possibility is to use the right hand with its thumb in the direction o the current and fngers in the direction o the magnetic feld. B y pushing you get the direction o the orce on a positive charge or a conventional current. I an electron is the moving charge, point the thumb in the opposite direction to get the orce in opposite direction This convention can also be used to get the direction o the magnetic feld due a conventional current e.g. by curling the fngers around a wire with current.
The observations you made in the Investigate! can be interpreted by looking closely at the directions o current in the coil relative to the movement o feld and coil. A rule that describes this was summed up by the German scientist Heinrich Lenz in 1 833. He stated that the direction of the induced current is such as to op p ose the change that created the current. C heck your experimental notes ( or the diagrams in fgure 1 that sum them up) and see i your results confrm this law. In act, Lenzs law is little more than the conservation o energy. Suppose that, rather than opposing the induced eect, the change were to enhance it. This would imply an attraction instead o a repulsion between magnet and coil; the magnet would be pulled into the coil, accelerating as it goes. This would increase the speed and lead to an even greater acceleration. The magnet would move aster and aster into the coil, gaining kinetic energy rom nowhere. C onservation o energy tells us this cannot happen. Another way to look at the consequences o Lenz' s law is to realize that you cannot do work without having some opposition. The induced current in the coil is such that the induced feld produced by this current opposes the motion o the magnet you are holding. I the circuit is open, there is no current, no opposition, and no electric energy produced. I you move the magnet very ast you will clearly eel the opposite orce acting on you! E lectrons had not been discovered in Lenzs time and we can see how his law arises rom frst principles. In Topic 5 we saw that when a current ows within a magnetic feld the current produces a magnetic feld which distorts the original feld pattern. The result was that a orce acted on ( and could accelerate) the current- carrying conductor. This was the eect that led us to the basis o an electric motor.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) B X W
B -
constant speed
motor eect
conventional charge ow
Z Y rod rolling to right
motor equal i v external eect and is to be orce orce opposite constant to
Figure 4 Conducting rod rolling in a magnetic feld. In electromagnetic induction, there is a current in the conductor as a result o the motion o the conductor. Figure 4 extends the example o the moving rod that is now rolling to the right at a constant speed through the magnetic feld on a pair o rails. The rails conduct and orm part o a complete electrical circuit WXYZ. ( Rolling means that we do not have to worry about riction between the rod and the rails. ) C harges, driven by the induced em, ow around the circuit giving rise to an induced current in the direction shown. This induced current interacts with the uniorm magnetic feld to give rise to a orce - the motor eect orce that we discussed in Topic 5 ( page 2 3 4) . I you now use Fleming' s let- hand rule you will see that the induced current leads to a motor eect ( a orce) acting to the let in fgure 4, that is, opposite to the direction in which the conductor is moving. I the rod is to move at a constant speed then ( by Newtons frst law o motion) an external orce must be exerted on it. This is where the energy conversion comes in. The work done by the external orce to keep the conductor moving at a constant speed appears as electrical energy in the conductor.
Magnetic fux and fux density We can use the physics rom Topic 5 to extend these qualitative ideas. The magnitude o this magnetic orce is BIl where B is the magnetic feld strength, I is the induced current in the rod, and l is the length o the rod. Flemings let- hand rule shows that the magnetic orce arising rom the induced current opposes the original orce. In other words, the opposing magnetic orce is to the let i the original applied orce is to the right. The net orce is zero and the rod moves at constant speed. It is possible or work to be done because o the opposition to the motion presented by the external magnetic feld that produces a magnetic orce on the induced current. No opposition, no work possible ( or needed! ) . From Newtons frst law, to keep the rod in fgure 4 moving at constant velocity, a constant orce equal to BIl must act on the rod to the right.
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11 .1 E LE CTR O M AG N E TI C I N D U CTI O N
The energy we have to supply thereore in a time t is orce distance moved which is BIl x where x is the distance moved to the right by the rod. The induced em is equal to the energy per coulomb supplied to the system. In other words energy supplied __ = charge moved
BIl x _ BIl x =_ = Q I t
and thereore
= Blv where v is the speed o the rod. C ancelling and rearranging gives B A =_ = B t
rate o change o area
This is because the area that is swept through by the rod in a time t is lx = A ( because it is the amount by which the area changes in time t) S o, in words, induced em = magnetic fux density rate o change o area This introduces you to an alternative term in magnetism or what was earlier called the magnetic feld strength magnetic fux density. The magnetic eld strength is numerically equivalent to magnetic fux density. In Topic 5 we used the term magnetic feld strength because in that topic we were concerned with basic ideas o feld. In both electrostatics orce orce _____ and gravity, the term feld strength has a meaning o ____ m ass or ch arge
depending on the context. We defned magnetic feld strength in Topic 5 as orce __ current length However, this defnition does not take account o the old but helpul view o magnetic feld lines as lines directed rom a north- seeking to a south- seeking pole with a line density that is a measure o the strength o the feld. It is this visualization o a feld in terms o lines ( close together when the feld is strong, and well- separated when the feld is weak) that links magnetic feld strength to magnetic ux density. When lines o orce (feld lines) are close together, then the magnetic feld strength is large. There will be many lines through a given area. We say also that the magnetic ux density is large. The total number o lines per square metre is a measure o the magnetic ux density and thereore the total number o lines in a given area is a measure o the magnetic ux. Flux is an old E nglish word that has the meaning o ow and one way to think about ux is to imagine a windsock used to show the direction and speed o wind at an airfeld. I the wind is strong then the ux
normal to area B, magnetic fux density
= BA cos area, A
Figure 5 Flux and fux density.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) density is high. The ux is the number o streamlines going through the sock. I the wind has the same speed or two windsocks o dierent sizes then the windsock with the larger opening will have a larger ux even though the ux density is the same. I magnetic ux density is defned as the number o ux lines per unit area, then ux itsel must be equivalent to fux density area so that, in symbols, =B A This equation assumes that B and A are at right angles. Figure 5 shows the relationship between area and ux density when this is not the case. In this case, we need to consider the component o the feld normal to the plane. A normal to the area is constructed and this normal makes an angle with the lines o ux. S o, = BA cos so that when = 0 ( area normal to feld lines) = BA, and when = 90, = 0. To sum up:
Magnetic ux density B is related to the number o feld lines per unit area. It is a vector quantity.
Magnetic ux is equal to BA ( also written as ) . It is a scalar quantity.
The equation = BA cos is used i the area is not normal to the lines.
The unit o ux is the weber ( Wb) and is defned in terms o the em A induced when a magnetic feld changes. The equation = B ___ can be t
ch an ge in fu x ( BA) or ______________ and so re-written as = _____ tim e taken or ch an ge t
a rate o change o fux o one weber p er second induces an em o one volt across a conductor For a particular conductor, knowledge o a magnetic feld in terms o the magnetic ux and the rate at which it changes allows a direct calculation o the magnitude o the induced em that will appear. We can now make a direct link between ux density and feld strength. The magnetic fux density
) o rc e _____________ strength ( ).
fux , measured in ( __________________ a re a o ve r w h ich it a c ts
weber metre 2 is numerically equal to the magnetic eld cu rre n t le n g th
O ne tesla (T) one weber p er square metre (Wb m 2 ) We thereore also have a link between changes in the magnetic feld strength and the induced em.
Magnetic fux linkage A Finally, one more quantity appears. O ur derivation o = B ___ above t used a single rod rolling along two rails. This is equivalent to a single rectangular coil o wire that is gradually increasing in area. Imagine this single turn o coil gradually increasing its area to include more and more feld lines. As beore the em across the ends o the coil will be
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equal to the rate o change o area multiplied by the magnetic ux density. I there are N turns o wire in the coil then the induced em ( N ) A will be N times greater so that = NB ___ = _____ . N is known as the t t magnetic fux linkage. The unit o ux linkage is oten written as weber turns, although this is entirely equivalent to weber because the number o turns is simply a number. The relationships between these interlinked quantities can be shown as: magnetic ux linkage, measured in Wb-turns
N
magnetic ux measured in Wb
= BA
magnetic ux density measured in Wb m -2
B
magnetic feld strength measured in T
Magnetic induction is summed up in a law devised by Faraday himsel that is known as Faradays law, it states that the induced em in a circuit is equal to the rate o change o magnetic fux linkage through the circuit. In our usual notation this is written algebraically as N =- _ t The negative sign is added to include Lenzs law. In its ull mathematical orm, the equation is also known as Neumanns equation. This equation combines Faradays and Lenzs ideas. It reminds us that, ( as we shall see in the next section) magnetic ux can be changed in three dierent ways: by changing the area o
[
]
( co s ) A cross- section with time ( ___ , by changing with time ______ or by t ) t B changing magnetic ux density with time ( ___ ). t
Nature of science Cutting lines of force Faraday frst introduced the fe ld- line model though his model is not quite the same as our modern inte rpre tation o a magnetic feld. He considered the lines o orce to be at the edges o tubes o orce , like elongated elastic bands. At the time, an invisible ae ther , thought to have elastic properties, was conside re d to fll space. Later on, Faraday and others took the concept o the feld line urther by suggesting that it was the action o the conductor
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cutting the tubes o orce that led to an induced em. This is a helpul way to think o the process, although it conceals the link between a charge being moved in a magnetic feld and the magnetic orce that acts on the charge as a result. B ut we need always to remember that Faraday and the others did not know o the existence o the electron, and that they were very amiliar with the ideas o feld lines. In the nineteenth century, Maxwell refned these models by including both electrostatic and electromagnetic orces in one set o equations. This illustrates two things about the nature o science: the way in which scientists allow a discovery to illuminate prior knowledge in a dierent way, and the power o the visual image to help us to comprehend a phenomenon.
Changing felds and moving coils An em can be electromagnetically induced in a number o ways that, on the ace o it, appear dierent rom each other:
A wire or coil can move in an unchanging magnetic feld ( the example o the rolling rod above) .
The magnetic feld can change in strength but the conductor does not move or change.
A coil can change its size or orientation in an unchanging magnetic feld.
C ombinations o these changes can occur.
We will look at these cases in the context o a rectangular coil interacting with a uniorm ( but possibly changing) magnetic feld.
X
Y
B, feld out o page
area, A
X
Y X zero feld
X
= 2BA
= BA (a)
(b)
G (c)
Figure 6 Moving coils and changing felds.
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ipped though 180
Y
Y
11 .1 E LE CTR O M AG N E TI C I N D U CTI O N
Case 1 : Straight wire moving in a uniorm feld This is the case o the rolling rod above. The change in area per second is lv, the length l o the rod multiplied by v, the speed o the rod. The induced em is thereore = Bvl when the wire moves at 90 to the feld lines. I the wire motion is not at 90 to the feld then, as usual, the component o feld at 90 to the direction should be used.
Case 2: Coil moving The coil can move as shown in fgure 6(a) rom one position in a magnetic feld to another position where the feld may be dierent. I the coil begins and ends in positions where the feld is identical, then there is no change in the ux linkage and there is no induced em. Although the coil is cutting lines, the same number is being cut on opposite sides o the coil. Two ems are induced but in opposite senses and they thereore cancel out. I the coil moves rom a position where the ux is to a position where the ux is zero, the change in ux linkage is N and the induced em is N = __ time taken for change
An interesting variant o this occurs where a coil in a feld is ipped through 1 80 ( fgure 6( b) ) . To visualize this, look at things rom the point o view o the coil. The feld lines appear to reverse their direction through the coil, and so the change in ux is ( ) , in other words, 2 N 2 . S o the em induced will be equal to _______________ . tim e taken to rotate coil When a coil rotates in a feld the em produced instantaneously depends on the rate o change o the ux linkage, and this in turn depends on the angle the coil makes instantaneously with the feld. I the coil rotates at a constant angular speed, then the em output varies in a sinusoidal way. This is the basis o an alternating current ( ac) generator as we shall see later.
Case 3: Magnetic feld changes S ometimes the feld changes rom one value to another it gets stronger or weaker but the coil does not move. The act o cutting feld lines is not so obvious here. S uppose the feld is being turned on rom zero. B eore the feld begins to change there are no feld lines inside the coil. You can think o the lines as moving rom the outside into the area bounded by the coil. The change stops when the ux density is at its fnal value. In so doing the lines must have cut through the stationary coil. N B Again, the solution is not difcult. Now = - ____ becomes = - NA ___ t t because only B is changing. You need to know the rate at which the feld is changing with time ( or the total change and the total time over which it happens) . Another example is the case o two coils ace- to-ace as in fgure 6( c) . O ne coil is connected to a galvanometer alone, the other coil is connected to a circuit with a cell, a variable resistor and a switch. In what way will you expect the galvanometer reading to change when the switch is closed and remains closed? When the switch is opened? When the switch is closed and the resistance in the circuit is varied?
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Worked examples 1
The graph shows the variation o magnetic ux with time through a coil o 5 00 turns. 2.5
the gradient o this graph ( as always) as the ux is proportional to the time we can use any corresponding values o and t. 2 mWb = _ _ = 0. 5 V 4 ms t thus the induced em = 5 00 0.5 V = 2 5 0 V
2.0
2
1.5
A small cylindrical magnet and an aluminum cylinder (which is non-magnetic) o similar shape and mass are dropped rom rest down a vertical copper tube o length 1 .5 m.
/10 -3 Wb
a) Show that the aluminum cylinder will take about 0.5 s to reach the bottom o the tube. b) The magnet takes 5 s to reach the bottom o the tube. E xplain why the obj ects take dierent times to reach the bottom.
1.0
Solution 1 a) Use a kinematic equation, e.g. s = ut + __ a t2 . 2 1 __ 2 Then 1 .5 = 2 9.8 t
0.5
t = 0. 5 5 s. 0.0 0
1
2 t/10
3 -3
4
5
s
C alculate the magnitude o the em induced in the coil.
Solution The change in ux is 2 1 0 3 Wb and this occurs in a time o 4.0 ms. the rate o change o ux is
b) As the magnet alls the copper tube experiences a changing magnetic ux, and as a result an em is induced in the walls o the tube. This em results in a current in the tube. The current leads to another magnetic feld that opposes the motion o the magnet by Lenzs law. There is an upward orce on the magnet so that its acceleration is less than the value or ree all. In the case o the aluminium cylinder no current arises and it alls with the usual acceleration.
Nature of science Applications of electromagnetic induction There are many applications o electromagnetic induction over and above the generation o electrical energy. They include electromagnetic braking, which is used in large commercial road vehicles; the use o an induction coil to generate the large pds required to provide the spark that ignites the petrolair mixture in a car engine; and the generation o the signal in geophones and metal detectors. In each o these examples, a changing magnetic feld leads to the generation o an em and demonstrates the physics developed in this sub-topic.
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11.2 Power generation and transmission Understanding Alternating current (ac) generators Average power and root mean square (rms)
Applications and skills Explaining the operation o a basic ac generator,
values o current and voltage Transormers Diode bridges Hal-wave and ull-wave rectifcation
Nature of science
The provision o abundant, cheap electrical energy has been one o the ways in which an abstract science and its technological development have directly aected the lives o many people on the planet. Who could have imagined the enormous impact that Faraday's discoveries would make? ...certainly not Faraday himsel! This is an example o research in pure science leading to great practical applications.
including the eect o changing the generator requency Solving problems involving the average power in an ac circuit Solving problems involving step-up and stepdown transormers Describing the use o transormers in ac electrical power distribution Investigating a diode bridge rectifcation circuit experimentally Qualitatively describing the eect o adding a capacitor to a diode bridge rectifcation circuit
Equations rms and peak values: I0 __ I = _______ r.m .s .
2
0 __ potential dierence: Vr.m .s . = _______
V
2
0 r.m.s. = ________ resistance: R = ____ I I
V
0
V
r.m.s.
maximum power: Pm a x = I 0 V0 1 I V average power: Pa ve ra ge = ___ 2 0 0
p _____ Np Is ___ transormer equation: _ s = Ns = Ip
Introduction We have seen that a rod rolling along two parallel rails generates induced em and induced current; this is hardly a sensible way to generate electrical energy on a large scale. The practicalities o generation were solved by scientists rom the 1 83 0s onwards, frst or direct current, and later or alternating current.
Alternating current (ac) generators In this IB course we ocus on the ac generator because it is commonly used or the generation o energy. S uch a generator consists o a coil with a large number o turns; the coil rotates relative to a magnetic feld.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
(a)
G
fxed coil N
magnet
S
turn table
(b) ux linkage
or time
rate o change o ux linkage and induced em (c)
time
Magnetic feld lines cut the coil as the turntable rotates and an induced em is generated. However the em varies as the turntable rotates. Figure 1 ( b) shows how the ux linkage varies with , the angle between the normal to the coil and feld lines. When is equal to 90, the feld lines lie in the plane o the coil and so the ux through the coil is zero ( cos 90 = 0 in equation = BA cos ) . When is equal to 0 the feld lines are at 90 to the plane o the coil and the ux through the coil is a maximum. When the turntable rotation speed is constant, this graph also shows how the ux varies with time. The graph is a sine curve.
ac generator
N
S
slip rings V brush (d) N
For the moment we will imagine a fxed coil placed between the poles o a U- shaped magnet that stands at the centre o a rotating turntable ( fgure 1 ( a) ) . The turntable can turn at dierent angular speeds and the coil can have dierent numbers o turns and cross-sectional areas. The coil is connected to a galvanometer or to a data logger that registers the em across the coil. When the magnetic ux in the coil is maximum, the em induced ( current) is minimum ( 0) and vice- versa. C hanging the speed o the turntable changes the requency o the em as well as its amplitude. Increasing the number o turns or increasing the area o the coil will increase the amplitude o the em but leave the requency unchanged i the turntable speed does not change.
S N
S N
Figure 1 Basic ac generator.
The em induced in the coil is equal to - ___ , in other words the negative t o the gradient o the ux-time graph. Figure 1 ( b) also shows how the induced em varies with time; this graph can be obtained either by dierentiation or by a consideration o the gradient o the ux-time graph. When the normal to the coil and the feld lines are parallel ( is 90) then the em is zero because the coil does not instantaneously cut the feld lines at all. However, while some ac generators have a rotating magnetic feld, others have fxed magnets and a rotating coil ( Figure 1 ( c) ) . The principle is however the same. The direction o charge ow in the coils varies with the position o the coil. The use o a direction rule shows this. When the let- hand wire is moving upwards as shown in Figure 1 ( d) , the conventional current direction in this wire is to the back o the S coil. The right- hand wire is moving downwards at the same instant and the current in this wire is towards the ront o the coil. C harge ows clockwise ( looking rom above) in the coil and out into the external circuit. Hal a cycle later the sides o the coil will have exchanged position. C harge is again owing clockwise, but because the coil has rotated, the current ( so ar as the meter is concerned) is in the opposite direction. Figure 1 ( d) shows this too. I there were wires permanently connecting the coil to the meter they would quickly become twisted. Energy needs to be extracted rom the generator without this happening. S lip rings are used or this. The ends o the coil terminate in two rings o metal that rotate with the coil about the same axis. Two stationary brushes, connected to the external part o
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the circuit, press onto the rotating rings and charge ows out into the circuit through these connections ( Figure 1 ( c) ) . The essential requirements or an ac generator are thereore:
a rotating coil
a magnetic feld
relative movement between the coil and the magnetic feld
a suitable connection to the outside world.
Real-lie generators are more sophisticated than our simple models and an Internet search will allow you to see many dierent types o ac generator. For real generators, there is another issue that arises because an induced current is generated in the coils. As we saw in S ub- topic 1 1 . 1 , any moving conductor carrying an induced current in a magnetic feld has two orces acting on it: the orce that moves it, and an opposite orce that arises because o the induced current. This also applies to the rotating current- carrying coil in the ac generator. Flemings lethand rule and Lenzs law show that this orce opposes whatever is turning the coil. I a generator coil is being rotated clockwise by an external agent, then the magnetic orces that arise rom the induced current interacting with the magnetic feld will exert a turning orce anticlockwise on the coil. S ome o the energy provided by the external agent turning the coil has to be used to overcome this anticlockwise magnetic eect. This reduces the induced current that can be made available to the external circuit. However, remember that with no opposition, no work is done and no energy will be transerred rom the agent ( doing the turning) to the coil ( and its associated electrical circuit) . This is easily demonstrated using a bicycle dynamo ( a device similar to our frst rotating-magnet generator) ( see fgure 2 ) . In this type o dynamo a permanent magnet is rotated in the gap inside a coil. When the lamp is switched o so that no induced current is produced, the dynamo is relatively easy to turn ( remember, there will still be an induced em across the terminals) . When the dynamo supplies current and lights the lamp, more eort is required to rotate the dynamo at the same speed since an opposing magnetic orce will act on the current ( coil) . This is Lenzs law in action.
S
N
m o t i o n o f wh e el
iron core S
N
rotating magnet coil to bicycle lamp
Figure 2 Bicycle dynamo.
Nature of science Modelling an ac generator Although the ollowing derivation will not be required in the examination, it shows you how Faradays law can be used to model the behaviour o a simple ac generator.
The coil has an average length l and an average width o w with N turns ( these dimensions are shown in fgure 3 ( a) ) . The area o the coil A is thereore l w.
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Thereore the ux linkage varies with t as NB cos t and a graph showing how ux linkage varies with time has a cosine shape with maximum and minimum values o NBA at t = 0 and NBA when the coil is halway through one cycle.
(a)
w B l
(b) fux linkage N
At the instant when the normal to the plane o the coil is at an angle to the magnetic feld, the ux linkage through the coil is N , which is N BAcos . The coil spins at a constant angular speed In time t the angle swept is ( = t) .
em
A supply in which the current and voltage vary as a sine wave is an alternating supply. Although is used or convenience in the mathematics o the em induced in the generator, or everyday purposes we use the requency. This has the same meaning as elsewhere in physics: the number o cycles that occur each second. A generator that rotates 50 times in one second has an alternating current output at a requency o 5 0 Hz.
em
time
time
Figure 3
The model above shows that the output em o a generator is sinusoidal and that one complete rotation o the coil through 3 60 gives one cycle o the alternating current. time
when the rotation speed is doubled:
time
Figure 4 Increasing the rotation speed o the coil.
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rate o change o fux linkage and induced em
The value o the induced em at any instant is equal to the ( - ) rate o change o the ux linkage ( N) and this is the the ( - ) gradient o the ux linkagetime graph. S o the graph o induced em is a ( negative) sine curve with an equation o which has maximum and minimum values or the em o 0 = BAN.
Figure 4 shows the eect on the em o changing the angular speed o the coil ( without changing any other eature o the coil or feld) . I the angular speed o the coil is increased then:
the coil will take a shorter time to complete one cycle and so there will be more cycles every second hence the requency increases.
the time between maximum and minimum ux linkage will decrease and thereore ( as the ux linkage is constant) the rate o change increases and hence the peak em increases.
O ther ways to increase the output o the em, but without changing requency, include increasing:
the magnetic feld strength ( B)
the number o turns on the rotating coil ( N)
the larger coil area ( A) .
1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
Worked example A coil rotates at a constant rate in a uniorm magnetic feld. The variation o the em E with angle between the coil and the feld direction is shown. C opy the graph below and, on the same axes, add the em that will be produced when the rate o rotation o the coil is doubled. Explain your answer.
The rate o change o the ux linkage will double, so the magnitude o the peak em will also double. This is because Faradays law states that the induced em is proportional to the rate o change o ux linkage. However, the coil now rotates in hal the time, so the time or one cycle will be halved. O n the same scales, the new graph is: E 0
0
180
360
/degree
0
180
Measuring alternating currents and voltages The current and voltage o an alternating supply change constantly throughout one cycle. Measuring these quantities is not straightorward because, with positive and negative hal- cycles, the average value or current or voltage over one cycle is zero.
360
/degree
current
E 0
Solution
0
time
O ne way around this problem is to consider the power supplied to a resistance R connected to the generator. The instantaneous power dissipated in the resistance is I2 R where I is the instantaneous current.
The powertime graph is always positive ( which we would expect because the power is I2 R and a number squared is always positive) .
The power graph cycles at twice the requency o the current.
To see this, suppose that the time period o the ac generator is very large taking 1 0 s to turn once through one cycle. I you watch a flament lamp supplied with an ac supply o such a low requency you will see the lamp ash on and o twice in each cycle. The lamp is on when the em is near its maximum (positive) and minimum (negative) values.When we look at a flament lamp powered by the AC mains, persistence o vision prevents us seeing its ashing like this because it is switched on and o at 1 00 or 1 20 times per second (twice the normal mains requency o 50 or 60 Hz) . Alternating values are measured using the equivalent direct current that delivers the same power as the ac over one cycle. A lamp supplied rom a dc supply would have the same brightness as the average brightness o our ac lamp ashing on and o twice a cycle. This equivalent dc current is that which gives the average value o the power in the powertime graph. B ecause the average value o a sin 2 graph is halway between the peak and zero values and the curve is symmetrical about this line (shown on fgure 5 (b) ) , the areas above and below the average line are the same.
(a)
power dissipated in resistor
Figure 5 shows both the currenttime graph and the power dissipatedtime graph with the same time axes. Notice the dierence between the two:
(b) 0
mean value of power
time
Figure 5 Currenttime and power time graphs for ac.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) 1 2 So the mean power that an ac circuit supplies is __ I R where Imax is the 2 max peak_____ value of the current. The dc current required to give this power I 1 I 2 which is ___ __ . is _ 2 2 max This value is known as the root mean square (rms) current I V ___ __ . In a similar way V __ Irms = ___ rms = . The power P dissipated in a
m ax
m ax
m ax
2
2
resistance = Irms Vrms
I m a x ___ Vm ax I m a x Vm a x __ __ = ______ = ___ 2 2
2
with the usual equivalents: V2rms 1 2 1 _ __ P = IrmsVrms = __ I R = rms 2 2 R You will not need to know how to prove the relationship between peak values and rms values for the examination. Many countries use alternating current for their electrical supply to homes and industry. We will look at some of the reasons for this later, but different countries have made differing decisions about the potential differences and frequencies at which they transmit and use electrical energy. Thus, in some parts of the world the supply voltage is about 1 00 V; in others it is roughly 2 5 0 V. Likewise, frequencies are usually either 5 0 Hz or 60 Hz.
Worked examples 1
Solution
The diagram below shows the variation with time t of the emf E generated in a rotating coil.
a) the peak value of the emf is 3 60 V, so the rms 360 __ = 2 5 5 V value is ___ 2
1 1 b) f = _ = ____ = 5 0 Hz T 0.02
E/V 360
2 0 0
10
30
20
t/ms
-360
C alculate: a) the rms value of the emf b) the frequency of rotation of the coil.
A resistor is connected in series with an alternating current supply of negligible internal resistance. The peak value of the supply voltage is 1 40 V and the peak value of the current in the resistor is 9.5 A. C alculate the average power dissipation in the resistor.
Solution
peak current peak pd The average power = __ 2 1 = _ 1 40 9.5 = 670 W 2
Transformers One of the reasons why ac supplies are so common is because a device called a transformer can be used to change alternating supplies from one pd to another. Transformers come in many shapes and sizes ranging from devices that convert voltages at powers of many megawatts down to the small devices used to power domestic devices that need a low-voltage supply. A transformer consists of three parts:
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an input ( or primary) coil
an output ( or secondary) coil
an iron core on which both coils are wound.
1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
(a)
primary winding Np turns primary current
laminations
m agn e ti c f u
secondary winding Ns turns
(b)
x,
Ip
secondary l current s
primary voltage Vp secondary voltage Vs
(c) transormer core
Figure 6 Transormers in theory and practice. Figure 6 shows a schematic diagram o a transormer ( a) , a real-lie transormer ( b) , and also the transormer symbol used in IB examinations ( c) . The operation o a transormer, like the ac generator, relies on electromagnetic induction.
Alternating current is supplied to the primary coil.
A magnetic feld is produced by the current in the primary coil and this feld links around a core made rom a magnetic material, usually sot iron. (The basic ideas behind the production o this feld were covered in Topic 5 .)
B ecause the primary current is alternating, the magnetic feld in the core alternates at the same requency also. The feld goes frst in one direction around the core and then reverses its direction.
The transormer is designed so that the changing ux also links the secondary coil.
The secondary coil has a changing feld inside it and an induced alternating em appears at its terminals. When the coil is connected to an external load, charge will ow in the circuit o the secondary coil and its load.
Energy has been transerred rom the primary to the secondary circuit through the core.
Suppose that an alternating pd with a peak value o Vp is applied to the primary coil and that this results in a ux o in the core. The ux linked to the secondary coil o Ns turns is thereore Ns and the induced em in the secondary coil is Vs = Ns ___ . There is also a ux linkage to the t primary coil even though the primary current was originally responsible or setting up the feld in the frst place. This gives rise to p = N p ___ t
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) where p and N p are the induced em and the number o turns in the primary coil. This induced primary em will oppose the applied pd. I we assume that the resistance o the primary coil is negligible, then p is the same or both will be equal in magnitude to Vp . Thus, because ___ t V V coils __ = __ = __ and N N N p
p
s
p
p
s
p Vp Np _ _ _ s = Vs = N s
This is known as the transormer rule. It relates the number o turns on the coils to the input and output voltages:
When N s > N p the output voltage is greater than the input voltage. This is known as a step -up transormer.
When N s < N p the output voltage is less than the input voltage. This is known as a step -down transormer.
The terms step- up and step-down reer to changes in the alternating voltages not to the currents.
You may wonder how zero current in the primary coil can lead to any energy transer to the secondary. The answer is: it cannot. The equation applies to the case where there is no current in the secondary ( in other words it has not yet been connected to a load) . O nce again Lenzs law has a part to play. When the secondary coil supplies current ( because a resistor is now connected across its terminals) , then because charge ows through this secondary coil, another magnetic feld is set up in the coil and through the core. This magnetic feld tries to oppose the changes occurring in the system and so tends to reduce the ux in the core. This in turn reduces the opposing em in the primary ( oten called a back em or this reason) and so now there will be an overall current in the primary that allows energy to be transerred. Electrical engineers use a more complex theory o the transormer than the one presented here to take account o this and other actors. B ut or our simple theory, which assumes that the transormer loses no energy, the energy entering the primary is equal to the energy leaving the secondary so Ip Vp = Is Vs where Ip and Is are the currents in the primary and secondary circuits respectively. This is a second transormer equation that you should know be able to use. In act, many transormers have an efciency that is close to 1 00% as energy losses can be reduced by good design. The efciency o a transormer is energy supplied by secondary coil ____ 1 00% energy supplied to primary coil Ways to improve the efciency include:
Laminating the core
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Iron is a good conductor and the changing ux in the transormer produces currents owing inside the iron core itsel. These are known as eddy currents. To prevent these, transormer designers use thin
1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N layers o insulating material that separate layers (sheets) o iron in the core (shown in Figure 6(a) ) . This has the result that although the magnetic properties o the iron are largely unaltered, the electrical resistance o the core is signifcantly increased. The currents are orced to travel along longer paths within the layers hence increasing electrical resistance and reducing the current. Laminations reduce the energy losses that result rom a reduction in the amount o ux and rom a rise in temperature o the iron that would occur i the eddy currents were large.
Note Hard and soft magnetic materials We sometimes talk about magnetic materials being "sot" or "hard". A sot material, such as iron, can be easily magnetised by another magnet or a current-carrying coil. When the magnetic inuence is removed, however, the iron loses all or most o its magnetism easily. Such materials are excellent or the cores o transormers because they respond well to the variations in magnetic feld. Hard materials such as some iron alloys (a steel) do not magnetise easily but are good at retaining the magnetism. These materials are best or the manuacture o permanent magnets such as the ones you use in the laboratory.
Choosing the core material The magnetic material o the core is a sot magnetic material. It can be magnetized and demagnetized very readily and also maintain high uxes too. These are all desirable properties or the core.
Choosing the wire in the coils Low- resistance wires are used in the primary and secondary coils as high resistances would lead to heating losses ( called j oule heating) in the coils.
Core design It is important that ux is not allowed to leak out o the core. As much ux as possible should link both coils so that the maximum rate o change o ux linkage occurs.
Worked example A transormer steps down a mains voltage o 1 2 0 V to 5 V or a tablet computer. The computer requires 1 .0 W o power to operate correctly. There are 2 300 turns on the primary coil o the transormer. a) C alculate the number o turns on the secondary coil. b) C alculate the current in the secondary coil when it is operating correctly.
c) The input current to the primary is 0.0090 A. C alculate the efciency o the transormer.
Solution V N V N 5 2300 a) __ = __ so N s = ______ = _______ = 96 turns V N V 120 p
p
s
s
s
p
p
tab le t p o w e r 1 .0 b) C urrent in the secondary Is = _________ = ___ 5 V = 0. 2 0A s
p o w e r su p p lie d b y se co ndary co il c) Efciency = _______________________ p o w e r su p p lie d to p rim ary co il 5 0.2 = __________ = 0.93 ( or 93 % ) 1 20 0.0090
Investigate! Transormer action
In this experiment you are asked to investigate the basic ideas behind the transormer. You will require an ac lowvoltage power supply to provide an input, an oscilloscope to view the output o your coils, and the coils themselves. O ne coil should have signifcantly less turns than the other,
typical values might be 2 40 turns on one coil and 1 1 00 turns on the other. You will also need additional apparatus or part o the experiment. ( S ome coil kits have special laminated iron cores that clip together to give a continuous magnetic circuit, )
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C onnect the coil with ewer turns to the supply (this is the primary circuit) and the other coil to the oscilloscope (this is the secondary) . Set the coils close so that the secondary coil is linked to the fux produced by the primary. Look closely at the output rom the secondary coil and compare the relative sizes o the input and output and also their phases. Try the ollowing changes to see how they aect the output:
Alter the separation and orientation o the coils ( reverse one coil relative to the other and observe the dierence) . Alter the requency o the primary current. Link the
coils by inserting a piece o iron or several iron nails through the centre o both coils.
With the coils linked with iron compare the output voltages when the 2 40 turn coil is the primary and, later, when the 1 1 0 turn coil is the primary. I you have access to other coils with dierent numbers o turns collect data to complete the table.
Number Primary Number of Secondary of primary pd, secondary pd, V / V s turms Ns turns Np Vp / V
Np Vp _ _ Vs Ns
oscilloscope low-voltage power supply
3 2 1 0
4
5 6 7 8
9 10 11 12
primary coil. 240 T
secondary coil, 1100 T
V
Figure 7
Transformers in action Many countries have an electrical grid system so that each separate community within the country does not have to provide its own energy. In the event o power ailures in one part o the grid system, energy can be diverted where needed. When energy has to be sent over long distances it is advantageous to send it at very high voltages. This is because transmitting at high voltages and low currents helps to reduce the energy lost in the power transmission lines that are used in the grid. A laboratory example will help here: Figure 8 shows two alternative ways to transmit electrical energy rom one point in a school laboratory to another. In both circuits a power supply acts as the generator providing energy at an alternating potential dierence o 1 2 V. This energy is transmitted through two transmission lines ( leads strung between two retort stands) . In one case ( Figure 8( a) ) , the energy is tramsitted at 1 2 V, but in the other ( Figure 8( b) ) a step- up transormer with a turns ratio o 1 : 1 0 is used to transmit the energy at 1 2 0 V. At the other end o this transmission line, a step-down transormer returns the pd supply to the voltage level required by the lamps in the circuit.
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1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
dimly lit lamps
(a)
(b)
1:10 step-up
12 V ac power supply
120 V pd used in transmission line
10:1 step-down
12 V ac power supply
brightly lit lamps
Figure 8 Transmission at high voltages. A numerical calculation will help you to compare the two cases: In each case, there are three lamps each rated at 0.3 A, 1 2V (a total power requirement o 3.6 W) and a transmission line o total resistance 1 .5 ohms. Without the transormers, the total current required by lamps will be 0.9 A and power loss in transmission line = I2 R = 0.81 1 .5 = 1 .2 W. With the voltage stepped up to 1 2 0 V along the transmission line, the total current will be stepped down to 0.09 A during transmission. This means the power loss in the transmission line = I2 R = 0.081 1 .5 = 0.01 2 W. S o, stepping up the voltage by a actor o 1 0 reduces the current by a actor o 1 0 and thereore reduces the power lost during transmission by a actor o 1 00 ( i.e. 1 0 2 ) . This is an important saving or electricity supply companies ( and their customers! ) . Transormers play a crucial role in increasing the efciency o transmission. In practice, grid systems are usually like the one shown in fgure 9. Numbers are not given on this diagram as they vary rom country to country. Find out the details o the grid voltages used where you live. high voltage transmission
power generation
generator transformer
transformer lower voltage distribution
transformer small commercial and residential
light industry medium factories
heavy industry large factories
Figure 9 A grid system.
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Nature of science HVDC and international collaboration The argument that high transmission voltages lead to better efciency does not apply simply to ac. The argument is still valid or dc but historically the transormation between voltages was easier using ac. However, things are now changing. As the physics and engineering o electrical transmission improve, so it is advantageous to use high-voltage direct- current transmission ( HVD C ) . Although the cost o the equipment to convert between two dc voltages is greater than the cost o a transormer, there are other actors in the equation. Countries use dierent supply requencies and this is a major problem when eeding electricity rom one country into the grid o another. Using undersea cables over long distances with ac also involves larger currents than might be expected as the cables have capacitance and induction eects. Additional currents are required to move charges every cycle.
S ome o the longest HVD C transmission paths in the world include the 2 400 km long connection between the Amazon B asin and south- eastern B razil that carries about 3 GW o electrical power, and the Xiangj iabaS hanghai system in C hina that carries 6.4 GW over a distance o 2 000 km. Governments work together to maintain electricity supplies. There are many examples o electrical links between countries provided so that one nation can supply energy to another nation during times o shortage. These are not necessarily times o crisis. There are short- term uctuations in the demand or electricity. S ometimes energy is ed rom country to country at one time o the day and then back again later. E xamples o such links include the electrical links rom the Netherlands to the UK and the HVD C link between Italy and Greece.
Rectifying ac The convenience o ac or domestic distribution is clear, but it raises the question o how devices that can only operate on dc, including computers and electronic equipment, can be made to unction when connected to an ac supply. The process o converting an ac supply into dc is called rectifcation. A device that carries this out is known as a rectifer. We consider two varieties o rectifer in this course: hal wave and ull wave. In hal wave, only hal o each cycle o the current is used whereas in ull wave a more complex circuit leads to the use o both halves o the ac cycle.
Hal-wave rectifcation Figure 1 0( a) shows the basic circuit. A single diode is connected in series with the secondary terminals o a transormer and the load resistor. The load resistor represents the part o the circuit that is being supplied with the rectifed current. D iodes are devices that only allow charge to ow through them in one direction ( the symbol or the diode has an arrowhead that points in the direction o conventional current allowed by the device) . We say that the diode is forward biased when it is conducting. When no charge can move through it ( because the pd across the device has the wrong polarity) we say that the diode is reverse biased. The load resistor will only have a pd across it or about hal a cycle ( fgure 1 0( b) ) . The waveorm is not quite hal a cycle wide because the diode does not conduct rom exactly 0 V; it requires a small orward pd or conduction to begin.
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1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
The way in which the circuit operates is straightorward. When terminal X on the transormer is positive with respect to terminal Y, then there will be a current through the diode in its orward conducting direction. When the ac switches so that X is negative with respect to Y, then the diode is reverse biased and there is no conduction or current. An obvious problem is that although the charge now ows in only one direction, the current is not constant. We need a way to smooth the waveorm so that it more closely resembles the constant value o a dc supply. O ne way to achieve this is to use a resistor and capacitor connected in parallel between the diode and the load ( fgure 1 0( c) ) . D uring the frst hal-cycle, the capacitor charges up and the potential dierence across it approaches the peak o the transormer output em. When the current is zero in the second ( negative) hal- cycle, the capacitor discharges through the resistor at a rate determined by the time constant o the circuit ( Figure 1 0( d) ) . I the time constant is much larger than the time or hal a cycle, the amount o charge released by the capacitor ( and thereore the pd across it) will be small and the pd will not change very much. When the diode conducts again in the next hal- cycle, the charge stored on the capacitor will be topped up. From now on there will be a dischargecharge cycle with the pd across the capacitor varying much less than in the basic circuit ( Figure 1 0( e) ) . When a capacitor is used in this way, it is oten reerred to as a smoothing or reservoir capacitor. The small variation in the output pd is known as the ripple voltage. The choice o capacitor and resistor values depends on the application in use. A large time constant provides good smoothing, but at the cost o a more expensive capacitor and a trade- o in the shape o the waveorm due to the large charging currents that are drawn rom the transormer.
X
Y (a)
output pd
time
(b)
input
+ ++ ++ -- -(c) charging
++ -+ (d) discharging
Full-wave rectifcation For some applications, the amount o ripple in a hal- wave rectifer cannot be tolerated. In such cases, ull-wave rectifers are used. Figure 1 1 ( a) below shows one way to achieve ull- wave rectifcation. This arrangement uses two diodes and requires a centre- tap on the transormer. This means that there is a connection hal way along the length o the wire that has been wound to make the secondary coil. This can be clearly seen at Y on the transormer symbol in the diagram. Also in the diagram is the resistorcapacitor pair that will smooth the rectifed wave.
load resistor
10 V ac
ripple
pd
time
(e)
Figure 10 Hal-wave rectifer.
The best way to understand how the circuit works is to imagine that terminal Y is always at zero potential. Then, or hal the time X will be positive and Z will be negative relative to point Y. For the other hal o the cycle, these polarities reverse. When X is positive relative to Y, diode D 1 will conduct. Notice that only hal o the secondary coil connected to D 1 takes part in conduction at any time. One hal cycle later, D 1 will not conduct because X will be negative and Z will be positive relative to Y. D 2 conducts during this hal cycle. Current is again supplied to a capacitorresistor combination during both halves o the cycle leading to ull-wave rectifcation.
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(a)
C R
pd
time
pd
time
Y
D2 Z X (b) Y
D4 D3
secondary o transormer
X
Y
Z
D1 D2
+
A B
Figure 11 Full-wave rectifcation. In order to achieve a particular value o peak pd, twice as many turns are required on the secondary compared to the hal- wave arrangement. This disadvantage can be overcome using a diode bridge ( fgure 1 1 ( b) ) . The ull secondary coil is now used and the diode arrangement allows the whole o the coil to supply current throughout the cycle. The disadvantage is the need or our diodes and a more complex circuit arrangement. When X is positive relative to Z then the j unction between D 1 and D 4 is positive relative to the j unction between D 2 and D 3 . O D 1 and D 4, D 1 is the one that conducts so that point A o the capacitor will be positive too. S imilarly, D 3 conducts and point B o the capacitor becomes negative. The capacitor charges and current is supplied to the rest o the circuit. When the polarity o the secondary coil switches, X becomes negative relative to Z and the conducting diodes are now D 2 and D 4. The connections are arranged so that the direction o the conventional current in D 4 is away rom B and the direction in D 2 is towards A. This is the same as in the previous hal cycle and the polarity o charge delivered to the capacitor is unchanged. The best way to learn this arrangement is not necessarily to memorize the orientation o the diodes but by understanding how they unction in order to provide a consistent pattern o polarity at the capacitor during the ull cycle. These examples o rectiying circuits are said to be passive, meaning that none o the devices ampliy or modiy the waveorms. Some modern power supplies or computers are active devices. Switched- mode power supplies change the requency o the supplied ac to a much higher value to allow the signal to be modifed by an electronic circuit. The advantage is that less energy is wasted in resistance but this is achieved only at the expense o much greater complexity in the electronic circuits.
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Nature of science The war of the currents In the 1 880s a commercial battle broke out between the direct-current distribution system developed by Edison and the alternating-current system o Westinghouse. Both companies were attempting to corner the US market. In those days, electricity was principally used to provide energy or flament lamps. Edisons system had the advantage that batteries could be used as a backup i the power ailed (which was a requent event when electrical distribution began) . On the other hand, Westinghouses ac system could be transmitted with lower energy losses using larger and ewer distribution stations. Westinghouse made ull use o scientists rom around the world in developing ac
distribution, including Nikola Tesla who had made much progress in transormer design. The ull story o the battle is complex, but more and more companies ( principally the newly ormed General E lectric C ompany) began to ollow Westinghouses lead and eventually dc supplies largely disappeared. However, as late as the 1 980s, direct current was generated in the UK to supply the dc powered printing presses in London. In the US , the hotel where Tesla lived used dc into the 1 960s. C onsolidated E dison ( as Edisons company became known) shut down its last dc supply on 1 4 November 2 007.
Wheatstone and Wien bridge circuits The our- diode rectifer arrangement is one o a class o circuits known as bridge circuits. The Wheatstone bridge ( fgure 1 2 ) was popularized by S ir C harles Wheatstone in 1 83 3 and is a completely resistive arrangement that can be used to estimate the resistance o an unknown resistor. It is normally used in a dc context. The working o the circuit is straightorward ( fgure 1 2 ) . Three known resistors R 1 , R 2 and R 3 , are connected in the circuit with a ourth unknown resistor RX. O ne o the known resistors is variable and is adj usted until the current in a galvanometer that " bridges" the pairs o resistors is zero. This galvanometer is usually a centre-zero meter ( that shows negative and positive values with the 0 marking in the centre o the scale) with high sensitivity. When the current in the galvanometer is 0, the bridge is said to be balanced. No charge is owing through the galvanometer at the balance point because the potentials at B and D are identical so there is no potential dierence across B D . This means that the potential dierences across R1 and R X are the same as each other, and the potential dierences across R2 and R 3 are also identical. V1 = I1 R 1 = I2 R X
A RX
R1 + B
D
G
R2
R3 C
Figure 12 The Wheatstone bridge.
V2 = I1 R 2 = I2 R 3
R I R S o __ = __ = __ I R R 1
X
3
2
1
2
thereore R1 R3 RX = _ R2 Knowledge o three o the resistors means that that R X can be calculated. For accurate work, the bridge requires that the known resistors have accurately measured values and that the ammeter is sufciently sensitive to give precise results.
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Investigate! Using the Wheatstone bridge
The details o this experiment will depend on the equipment that you have in your school. Your teacher will advise you on this.
2
RX
1 m wire
R The way to determine the ratio __ is to fnd the R 3 2
G
R3
O ne common way to carry out the experiment is to use a long ( 1 m) straight wire o uniorm cross- section perhaps taped along a metre ruler. The resistance per unit length o such a wire should be constant along its length. Figure 1 3 shows the way the wire is connected in the circuit.
point on the metre wire at which the current in the ammeter is zero. Then the ratio o the lengths o the two parts o the wire is equal to the ratio o their resistances.
tapping key
l2
3
ammeter ( the galvanometer) is zero.
Set up the circuit you are to use. C hoose R 1 so that it is similar in value to the unknown resistor.
l1
R S et the ratio __ so that the current in the R
The equation or the bridge becomes l R 3 = R 1 __ l 1
2
where l1 and l1 are the relevant lengths on the metre wire.
Figure 13
Nature of science Unbalanced Wheatstone bridges It is possible to make use o the Wheatstone bridge when it is out o balance. Suppose the unknown resistor RX is replaced by a thermistor. At a particular temperature the bridge can be made to balance with a particular combination o three fxed resistances and the thermistor itsel. When the temperature changes rom this starting point, the thermistor resistance will also change, increasing i the temperature drops, and becoming smaller i the temperature rises. These changes will take the bridge out o balance with a
CX R3 RX
C2 R4
Figure 14 Wien bridge.
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R2
potential dierence appearing across the meter that bridges the resistor pairs. The whole circuit can be calibrated so that a known temperature change at the thermistor gives rise to a known out-o-balance voltage across the meter. Such a circuit can be made to be extremely sensitive i the sensitivity o the meter and the resistor values are chosen careully. Other sensors can be chosen too: or example, a light-dependent resistor or changes in light intensity, and a length o resistance wire under tension (a strain gauge) or changes in length.
The Wien bridge circuit is a modifcation o the Wheatstone arrangement to allow the identifcation o resistance and capacitance values or an unknown component. This bridge operates with an alternating power supply. The bridge is modifed by the addition o a capacitor in series with R 2 . Again the current in the centre arm o the bridge is minimized. R2 , C2 and the requency o the supply need to be adj usted or the minimum current. Knowing the values o the components allows the operator to calculate the value o RX and CX. The theory or this bridge will not be tested in the examination.
1 1 . 3 C APACI TAN CE
11.3 Capacitance Understanding Capacitance Dielectric materials Capacitors in series and parallel
Applications and skills Describing the eect o dierent dielectric
Resistorcapacitor (RC) series circuits Time constant
Nature of science This sub-topic contains many important links and analogies. Some are straightorward, the link between the energy stored in a spring and the energy stored in a capacitor, or example. Others, however, reect the importance o the over-arching ideas o exponential growth and decay that underpin many areas o sciences and social sciences. Rates o reaction in chemistry, changes in living populations in biology, and radioactive decay in physics are all related by the important idea that the rate o change depends on the total instantaneous number and a constant probability o change. This is one o the many ways in which scientists use mathematics to model reality.
materials on capacitance Solving problems involving parallel-plate capacitors Investigating combinations o capacitors in series or parallel circuits Determining the energy stored in a charged capacitor Describing the nature o the exponential discharge o a capacitor Solving problems involving the discharge o a capacitor through a fxed resistor Solving problems involving the time constant o an RC circuit or charge, voltage, and current
Equations defnition o capacitance: C = ___V Q
combining capacitors in parallel:
C p a ra l l e l = C 1 + C 2 + ... 1 1 1 series: ________ = ____ + ____ + ... C series C1 C2 capacitance o a parallel-plate capacitor: C = ___A d
1 2 energy stored in a capacitor: E = ___ CV 2
time constant: = RC ___t
exponential discharge charge: Q = Q 0 e - ___t
current: I = I 0 e -
___t potential dierence: V = V0 e -
Introduction In Topic 1 0, a pair o charged parallel plates was used to produce a uniorm feld. A charge was transerred to the plates using a power supply. In this topic we look in detail at this transer o charge onto, and rom, the plates.
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Capacitors in theory
beore charging
-
+
-
+
-
electron + movement +
-
-
C harge is being separated by the system and stored. This requires energy and, not surprisingly, this is provided by the cell, the only source o em in the circuit. Energy is being stored on the plates as the electrons arrive there.
-
V< E
during charging
-
E
-
+ + + + + + + + + E
charging fnished
E
Figure 1 shows an arrangement o two parallel plates ( in a vacuum or simplicity) connected to a cell. The plates are initially uncharged. When the switch is closed, electrons begin to ow. There is no current between the plates because o the insulation between them. E lectrons are removed rom the plate connected to the positive terminal o the cell and are transerred to the plate connected to the negative.
E
+
The arrangement o parallel plates in which the two plates are separated by an insulator is called a capacitor. The insulator might be a vacuum, it could also be air or another gas providing that a spark (and thereore charge) cannot jump between the plates. It could also be a non-conducting material such as a plastic.
Figure 1 Charging a capacitor.
To understand this, imagine the frst electron that moves through the cell rom the positive plate in the diagram to the negative one. The cell will not need much energy to do this as the plates were initially uncharged. However, the second electron to move will fnd it slightly more difcult ( fgure 1 ) . This is because the frst electron is now on the right- hand plate and repels this second electron. The cell has to use more energy to move the second electron. The energy required rom the cell increases with each subsequent electron until fnally there is insufcient potential energy available in the cell to move any more electrons. At this point, i we were to disconnect the cell and measure the potential dierence across the capacitor then we would fnd that the capacitor pd is equal to the em o the cell. Had this been done earlier in the charging, then the unconnected capacitor pd would be less than the cell em. The capacitor pd cannot exceed the em o the power supply. The charge Q on the capacitor is stored at a potential dierence V. The ratio o the charge stored to the potential dierence between the plates is defned to be the cap acitance o the capacitor. charge stored on one plate Q capacitance, C = ___ = _ V potential difference between plates The charge stored on one plate is the same as the charge transerred through the cell and moved rom one plate to the other.
Tip Notice that, even though one plate has charge +Q and the other -Q, the charge stored is Q not 2Q. This is because it is only one group of charges (electrons) that are being moved between the plates.
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The unit of capacitance is the coulomb per volt, or farad (symbol: F) and is named ater Michael Faraday, the English physicist, who developed so much o our understanding about current electricity and magnetism. In undamental units the arad is equivalent to A2 S 4 kg 1 m 2 . O ne arad is a very large unit o capacitance since 1 C o charge is a very large amount o charge and you will usually use capacitances measured in milliarads ( mF) , microarads ( F) and picoarads ( pF) . Try not to conuse mF and F.
1 1 . 3 C APACI TAN CE
Worked examples 1
A pair o parallel plates store 2 .5 1 0 6 C o charge at a potential dierence o 1 5 V. C alculate the capacitance o this capacitor.
Solution
2
C alculate the potential dierence across a capacitor o capacitance 0.1 5 F when it stores a charge o 7.8 1 0 8 C .
Solution
Q 2 .5 1 0 C = _ = _ = 1 . 7 1 0 - 7 F = 1 70 nF. V 15 -6
Q 7.8 1 0 - 8 V= _ = _ = 0.5 2 V C 1 .5 1 0 - 7
Nature of science Changing names The name capacitance has not always been used. In the late nineteenth century and even later it was called capacity in English ( in French capacit dun condensateur and in Spanish la capacidad de un condensador) These were in many ways obvious names, and the idea o the capacity o a container can provide a model to help you understand capacitance ideas. Imagine two containers o water the same height, one narrow, the other wide. B oth containers, initially empty, are flled with water at the same rate ( the same volume o water every second) . O bviously, the narrower container overows frst. Thinking in terms o the top o both containers having the same gravitational potential, the liquid in the narrow container reaches this potential frst. Put another way, when both containers are ull, or the same potential ( height o the water surace
table
Figure 2 Water container analogy.
above the table is equivalent to the potential dierence o the capacitor) the wider container will hold more liquid ( equivalent to more charge) than the other. This is another link within physics that leads to analogous relationships.
Investigate! Charging a capacitor with a constant current
It is unusual or a capacitor to charge or discharge with a constant current. B ut this exercise will help you to understand the charging process or any capacitor.
S et up the circuit. A value o 470 F is suitable or this experiment.
You also require a clock with a second hand. You may wish to use a data logger with a
voltage sensor in place o the voltmeter and the clock ( the data logger can make both measurements or you) .
B egin by shorting out the capacitor with the ying lead so that it is completely discharged. The voltmeter should indicate zero.
In the experiment you need to alter the resistance o the variable resistor to keep the
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(a)
charging current at a constant value. I the resistance remains at a constant value, the current will all, reaching zero when the capacitor is ully charged so you will need to decrease the resistance as the experiment proceeds.
A
470 F
S et the variable resistor to its maximum value. C lose the switch with the ying lead connected across the capacitor. Remove the ying lead, start the clock, and record measurements o time elapsed t and potential dierence V across the capacitor. Record the value o the constant current I. Use your measurements to plot a graph to show how the charge Q stored on the capacitor varies with potential dierence V. To obtain the charge stored, remember that Q = It.
V flying lead
(b) 3.0 2.5 charge/mC
100 k
20 1.5 1.0 0.5 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 pd/V
Figure 3(a) and (b) Charging a capacitor with constant current.
The results o this experiment are shown in fgure 3(b) . The graph is a straight line with a gradient that is equal to C. The potential dierence across the plates o a capacitor is directly proportional to the charge it carries.
Energy stored in a capacitor The graph shown in fgure 3(b) can give more inormation than the gradient. The potential dierence V is the energy per unit o charge stored on the capacitor and Q is the charge, so the product o these two quantities is the energy stored and this quantity is equal to the area under the graph. Algebraically, this area is 1 1 Q V _ ( base height) = _ 2 2 There are two additional ways that this expression or the energy can be written: Q2 1 1_ 1 Energy stored on capacitor = _ QV = _ = _ CV2 2 2 C 2 This is an idea that is analogous to the potential energy stored in a stretched 1 string where there was also a actor o __ . (We looked at this in Topic 2.) 2
Worked examples 1
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C alculate the energy that can be stored on a 5 000 F capacitor that is charged to a potential dierence o 2 5 V.
2
A capacitor o value 1 00 mF stores an energy o 2 5 0 J. C alculate the pd across the capacitor.
Solution
Solution
1 CV2 = 0.5 5 1 0 - 3 2 5 2 Energy stored = _ 2 = 1 .6 J
Energy stored V =
_________
2 energy __ = C
____
5 00 = 71 V _ 0. 1
1 1 . 3 C APACI TAN CE
Capacitance of a parallel-plate capacitor In Topic 1 0 we saw that or a pair o parallel plates
Worked example 1
Q V _ = 0 _ A d where Q is the charge stored, A is the area o overlap o the plates, V is the potential dierence between the plates, and d is the separation o the plates. Rearranging this expression gives
Two parallel plates both have an area 0.01 5 m 2 and are placed 2 . 0 1 0 3 m apart. C alculate the capacitance o this arrangement.
Solution
Q A _ = 0 _ V d Q and __ is the capacitance C o the two- plate system. V
Finally,
A C = 0 _ d 0.01 5 = 8.9 1 0 - 1 2 _ 2 1 0 -3 = 6. 7 1 0 - 1 1 F
A C = 0 _ d This enables us to calculate the capacitance o a capacitor given the area o overlap o the plates and their separation. However, in real capacitors, edge eects will reduce this value. E arlier we mentioned the edge eects, the feld distortions at the edge o the plates that lead to dierences between the theoretical results here and what happens in practice with real capacitors. You saw some o the results o edge eects when you considered electric feld- line shapes at the edges o plates in Topic 1 0. I the gap between the plates is flled with an insulator o permittivity , this becomes A C= _ d
Capacitors in practice C apacitors are used extensively in electronic circuits. They store charge and can be used to reduce the eects o uctuations in a circuit. They are also the basis o timing circuits. This means that electric component designers need to be able to design capacitors o various sizes and capacitance values.
__
The equation C = 0 Ad gives the designers a basis or this. It is evident that there are three ways to increase the capacitance o the plates:
Make the plate overlap area larger because C A
1 Set the plates closer together because C __ d
C hange the value o the constant in the equation ( in other words change the permittivity rom that o a vacuum to some other value) .
Increasing the area o plate overlap increases capacitance because more charge can be stored or a given potential dierence between the plates Q and thereore __ increases. V The second way to increase C is to move the plates closer together. To understand the mechanism here, imagine the plates charged and
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plate
++++++-
++++++-
++++++-
++++++-
+ + + + + + + +
dielectric
Figure 4 How a dielectric increases capacitance.
Dielectric strength of dielectrics I the potential dierence across a capacitor is increased, a maximum feld strength will be reached ater which the dielectric becomes a conductor. A spark will then jump between the plates. This maximum feld is called the dielectric strength o the dielectric. For air this is about 1 kV mm - 1 (the exact value depends on temperature, pressure and how dry the air is) ; or teon the dielectric strength is 60 kV mm - 1 . I during thunderstorms the electric feld strength in air between charged clouds or between a charged cloud and the ground exceeds this critical value, then a lightning strike may occur.
isolated, i.e. not connected to a power supply. The charge on the plates cannot change as there is no route along which the charges can move between them. The plates attract because they are positive and negative. I they are allowed to move closer together at a constant speed ( without touching) then they will do work on whoever is moving them. This energy must come rom somewhere, the only possible source is the capacitor itsel and so the potential dierence must drop and once Q increases. again __ V The third way to change capacitance is to insert a material between the plates that replaces the air ( or vacuum) that we have so ar imagined to fll the space. This means that the new material can do two things or the capacitor designer: it can be used to separate the plates by a fxed amount, and also change the properties o the capacitor at the same time. In Topic 5 we mentioned that materials have their own permittivity which is greater than that o a vacuum. To treat this mathematically, when a material o permittivity is present we replace 0 in the equations where it occurs with alone. Some typical values or were also given in Topic 5 or a number o dierent materials. A particularly useul thing happens when a dielectric material is inserted between capacitor plates to fll the whole space between them. A dielectric is an electrical insulator that is polarized when placed in an electric feld. The origin o the polarization is in the molecules o the dielectric substance. Each molecule is slightly more positive at one end than the other. This means that when a molecule is in an electric feld, it responds either by moving slightly or by rotating so that the more positive end o the molecule moves in the direction o the electric feld. I the molecules are in a solid and the solid is packed between the plates o a capacitor then this reduces the electric feld strength between the plates. When the dielectric is in place ( fgure 4) , the molecules inside it respond to the feld produced by the capacitor. Notice the feld directions careully. In the diagram the original feld Ecap o the capacitor is rom right to let, but the dielectric feld Edielectric ( indicated by the charges at the surace o the dielectric) is rom let to right. Thereore the net feld between the plates is equal to the capacitor feld minus the dielectric feld. The overall feld Enet in the dielectric V ( V is the potential is reduced and because Enet = E cap - E dielectric = __ d dierence) then V decreases too as d is fxed. The insertion o the dielectric reduces the potential dierence between the plates because some o the stored energy o the capacitor has been used to align the dielectric molecules. The overall charge stored is unchanged Q so __ is increased and hence so is the capacitance o the capacitor. V D ielectrics increase the capacitance o a capacitor. Another way to describe the action o the dielectric is by saying that the presence o the dielectric raises the potential o the negative plate and lowers the potential o the positive plate; this reduces the potential difference between the plates.
This explanation o the dielectric eect is a simplifed one. There are other reasons or the increase in capacitance, but in many cases the explanation can be given in terms o potential change as here.
460
1 1 . 3 C APACI TAN CE Figure 5 ( a) shows some typical designs used or practical capacitors. A very common type o capacitor is the electrolytic capacitor shown as a cutaway in fgure 5 ( b) . The designer here uses the dielectric advantageously by using it to space the two metal oils apart. The capacitor layers are then rolled up together ( rather like rolling up three carpets initially on top o each other) . This type o capacitor must be connected into a circuit correctly. The dielectric material is a chemical ( the electrolyte in the diagram) that is a good dielectric when the electric feld direction is correct. In this design the layers are very thin ( good because d is small) and the plate area is large ( even better) giving some o the largest values o capacitance possible in a given volume.
(a)
Materials used or the dielectric include; paper, mica ( a mineral that can be cut into thin layers) , Teon, plastics, ceramics, and the oxides o various metals such as aluminium.
(b)
negative charge connection positive charge connection
The table shows how good some o these materials can be at improving the ability o a capacitor to store charge at a given voltage. For each material the number given is the ratio o the permittivity o the material to that o a vacuum, in symbols __ . ( This ratio is called the relative permittivity or sometimes the dielectric constant, but you will not be asked about this in the examination.)
dielectric
0
Material vacuum air paper mica polystyrene ceramic aluminium oxide tefon paran water (pure)
metal plate aluminum
_
plastic insulation
0 1 1.000 54 1 4 5 3 10015 000 911 2.1 2.3 80
paper separator
Tip I you are using electrolytic capacitors in the laboratory, take care that the polarity o the capacitor in the circuit is correct. I the polarity is reversed, the dielectric can become hot and cause the capacitor to explode.
+ oil oxide layer oil
tissue soaked in electrolyte
Figure 5 Practical capacitors.
Worked example 1
In a laboratory experiment, two parallel plates, each o area 1 00 cm 2 , are separated by 1 .5 cm. C alculate the capacitance o the arrangement i the gap between the plates is flled with:
b) Polystyrene has a dielectric constant o 3 , which means that = 3 8. 9 1 0 - 1 2 . C apacitance is now 1 .8 1 0 1 1 F.
a) air b) polystyrene.
Solution a) C onvert the centimetre units to metres: area = 1 00 1 0 4 = 1 0 2 m 2 ; separation = 0.01 5 m 1 0 -2 A = 8.9 1 0 - 1 2 _ C = 0 _ = 5.9 1 0 - 1 2 F 0.01 5 d
2
C alculate the area o overlap o two capacitor plates separated by a thickness o 0.01 0 m o air. The capacitance is 1 nF.
Solution Cd 1 1 0 - 9 0.01 __ A= _ = 8.9 1 0 - 1 2 = 1 .1 m
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Combining capacitors in parallel and series A urther way to modiy capacitance values is to combine two or more capacitors together in much the same way that resistors were combined in Topic 5 . Like resistors, capacitors can be connected in parallel and in series. C1 (a)
+Q 1
V
-Q 1 V +Q
+Q 2
-Q
-Q 2 C C2
V1
C = C1 + C2
V2
(b) +Q
-Q +Q
-Q
+Q
-Q
V C1
C2
C 1
C
=
1
C1
+
1
C2
Figure 6 Capacitors in parallel and series.
Parallel Parallel capacitors have the same potential dierence across them when connected as in fgure 6(a) . The total charge stored is Q 1 + Q 2 (as shown on the diagram) and these charges are Q 1 = VC1 and Q 2 = VC2 . The single capacitor that is equivalent to the two parallel ones has a charge o Q = VC.
Worked example C alculate the capacitance o the network below.
So Q = Q 1 + Q 2 ( conservation o charge) and thereore
C
VC = VC1 + VC 2 3C
C ancelling the V terms gives C = C1 + C2
2C
Solution
When two cap acitors are connected in p arallel, the total cap acitance is equal to the sum of the cap acitances.
Series
C ombining the parallel capacitors and 3 C gives:
C apacitors in series ( fgure 6( b) ) store the same amount o charge Q as each other because the same current charges both capacitors ( an example o Kirchhos frst law in action) . Kirchhos second law tells us that the potential dierences across the capacitors V1 and V2 must add up to give the em o the cell V.
1 1 1 2 ___ = __ + __ = __ so Ctotal= 1 . 5 C. C 3C 3C 3C
Thus
The capacitors in parallel have capacitance o C + 2 C = 3 C
to ta l
V = V1 + V2 ( conservation o energy)
462
1 1 . 3 C APACI TAN CE
Using the defnition o capacitance Q Q Q _ = _+ _ C C1 C2 Q cancels to give
_1 = _1 + _1 C
C1
C2
S o this time, the recip rocal of the total cap acitance is equal to the sum of the reciprocals of each cap acitance. This is a reversal o the equations or combining resistors and can be a convenient way to remember the equations.
Discharging and charging a capacitor In an earlier Investigate! a capacitor was charged with a constant current. This is an unusual situation that required continuous changes to the total resistance in the circuit to achieve a constant ow o charge. When the total resistance is constant then the charging current varies with time. This section examines the nature o this variation.
Investigate! Discharging a capacitor
S et up the circuit shown in fgure 7( a) . This circuit has two unctions: to charge the capacitor and then to discharge it with the power supply disconnected rom the circuit. S uitable values or the components are: capacitance, 1 00 F; resistance 470 k. C harge the capacitor by connecting the ying lead to point X. Then begin the discharge by disconnecting the ying lead.
R
X fying lead (b)
Record the variation o the potential dierence across the capacitor with time. The capacitance and resistance values are designed to allow you to carry out the experiment by yoursel. However, you may fnd it even easier with two people, one recording the data. Alternatively use a voltage sensor together with a data logger to collect and display the data.
7 6 5 4 3 2 1
Plot the graph o potential dierence against time or the discharge. Your results will probably resemble those shown in fgure 7( b) .
V
C
pd/V
(a)
0 0
20
40
60 80 time/s
100 120 140
Figure 7 Circuit or capacitor discharge and results.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Discharging You can think o the capacitor as taking the place o a power source and, at any instant, the current I and the pd VC across the capacitor are related by
R
VC = IR C Vc I
Here both VC and I change with time. This equation is obtained by applying Kirchhos second law to the circuit loop in Figure 8. Thereore
Figure 8 Discharging circuit.
VC Q _ = _ R t where Q is the charge on the capacitor and t is the time that has elapsed since discharging began. S o, because Q VC = _ C Q Q _ = -_ RC t A negative sign has appeared in the right- hand side o the expression. As the capacitor discharges, the charge on the capacitor alls and in each t the change in the charge is a negative value. Rearranging gives Qt Q = -_ RC and we replace RC by the constant where = R C so that Q t Q = -_
7
This new equation allows us to analyse how the charge on the capacitor varies with time during the discharge. The frst step is to recognize what the equation says. It predicts that the loss of charge 1 rom the capacitor in a time interval t will be equal to __ o the total charge that was stored on the capacitor at the beginning o the time interval.
6
charge/mC
5 4 3
Figure 9 shows what this means in terms o a graph o Q against t. It has Q the same shape as fgure 7( b) because C = __ and so Q V. V
2
Q t
1 0 0
20
40
60 80 time/s
100 120 140
Figure 9 Charge versus time for a discharging capacitor.
The next step is to model the discharge using Q = ____ . We will use a spreadsheet to do this and begin by frst concentrating on the part o the Qt graph j ust ater the discharge begins ( at time t = 0) at which time the original charge on the capacitor is Q 0 . I the charge on the capacitor did not change, then ater a time interval t the charge would still be Q 0 and the graph would be parallel to the time axis. B ut this is not what happens, the equation tells us that the Q t charge goes down (remember the minus sign! ) by ____ . This change is shown on the graph. 0
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Q t The charge remaining on the capacitor ater t will be Q 0 - ____ .
The graph line has a gradient o - __ during the irst time interval t.
0
Q0
1 1 . 3 C APACI TAN CE
What happens next? A second time interval begins and charge continues to move off the capacitor plates. B ut at the start of this interval the charge and, therefore, the pd across the capacitor are less than when t was 0. C all the new charge stored Q 1 . The change in charge this time Q t because the charge is no longer Q 0 . B ecause Q 1 is less is Q = - ____ than at the start, Q will also be less over this second time interval. The Q gradient of the graph becomes less and has a new value of __ . 1
1
B
C
D
t/s
delta Q/C
Q/C
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160
2.13E- 07 1.67E-07 1.32E-07 1.04E-07 8.17E-08 6.43E-08 5.06E-08 3.99E-08 3.14E-08 2.47E-08 1.95E-08 1.53E-07 1.21E-08 9.49E-09 7.47E-09 5.88E-09 4.63E-09
1.00E-06 7.87E-07 6.20E-07 4.88E-07 3.84E-07 3.02E-07 2.38E-07 1.87E-07 1.48E-07 1.16E-07 9.14E-08 7.20E-08 5.67E-08 4.46E-08 3.51E-08 2.76E-08 2.18E-08
E
F
G
H
I
4.70E+05 1.00E-04 10 1.00E-06
R C delta t Q0
J
K
L
M
N
O
how charge stored on a capacitor varies with time during discharge of the capacitor 1.20E-06 1.00E-06 8.00E-07 Q/C
A
6.00E-07 4.00E-07 2.00E-07 0.00E+00 0
10
20
-13.6
30
40 t/s
50
60
70
80
40 discharge time/s
60
-13.8 - 14 -14.2 -14.4 -14.6 -14.8
0.25
-15
0.20 current/A
In (Q/C)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
-15.2 -15.4
0.15 0.10 0.05
-15.6 -15.8
0.00 0
20
40 60 discharge time/s
80
100
0
20
80
Figure 10 Spreadsheet model for capacitor discharge.
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11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) In each successive t the change in the charge will be less than in the previous time interval and as time goes on the graph line will curve.
C olumn A shows the time increasing in steps o 1 0 s ( this is a deliberately large time increment, 1 s or 0.1 s would be better) . Q t
C olumn B shows the calculation o Q using Q = - ___ . RC
C olumn C shows the new value o Q at the end o the time interval.
The value o Q in column C is then applied as the initial charge in the next time interval on the next row o the sheet ollowing a loop with initial conditions changing at each cycle.
The program calculates successive values or the charge that remains on the capacitor or every 1 0 s o the discharge. The program then plots a graph o charge o the capacitor against time. ( I you want to produce your own version o this spreadsheet, the ormulae are:
in cell B 3 and below = C 3 * $G$3 /($G$1 * $G$2 )
in cell C 9 and below = C 3 -B 3
the symbol $ is required to ensure that the spreadsheet uses the same row and column or C, R and t when the columns are copied downwards rom row to row i these were not there then G3 would become G4 then G5 then G6 and so on as the rows are copied down the sheet.
At the top let-hand corner o the spreadsheet are the constants required in the solution: R, C, t and Q 0 . R and C have the same value as in the Investigate! so this solution should match up with the results you obtained then. Look closely at this graph ( which only has the frst 80 s o the discharge plotted) and compare it to your own results or the discharging experiment. The graphs have interesting properties. Examine the time it takes the charge on the capacitor to halve. The initial charge at t = 0 is 1 1 0 6 C (written as 1 .00E-6 in Excel notation) and hal this value (5 .00E-7) occurs at about t = 2 9 s. C ompare this with the time taken or the charge to halve rom 5 .00E-7 to 2 .5 0E-7. This is rom t = 2 9 s to t = 5 8 s another time o 2 9 s. The next halving rom 2 .5 0E-7 to 1 .2 5 E-7 takes another 29 s. For this combination o R and C it always takes 2 9 s or the charge to halve. I you used values or 1 00 F and 470 k in your experiment check that your results give a similar result. This behaviour is similar to that shown by decaying radioactive nuclei. The activity o the radioactive substance halves in one hal-lie. For a capacitor, hal the charge leaves the capacitor during a hal- lie. S uch behaviour is characteristic o exp onential decay. The equation Q 1 t _ = -_ Q has a similar orm to the radioactivity decay equation
466
1 1 . 3 C APACI TAN CE N _ = - t N (N is the number o nuclei present at a given instant and is the probability o decay) and thereore we expect them to share the same solution N = N 0 e - t and _t
Q = Q0 e -
The capacitor discharge equation can also be written ( by taking natural logs o both sides o the expression) t log e Q = log e Q 0 _ 1. A graph o log e Q against t should be a straight line with gradient - _ The log e Q t graph or the spreadsheet results is shown in fgure 1 0. It has an intercept on the y-axis o 1 3 .8 and a gradient o 0.02 1 s 1 . This tells us that log e ( Q 0 /C ) = 1 3 .8 ( remember that the units are written inside the brackets to make the whole log number unitless) . Thereore Q 0 = e - 1 3.8 = 1 .0 C which was the original charge the model gave the capacitor. R was 470 000 and C was 1 00 F in the model, 1 so RC = ___ = 47 s and the model predicts Gk to be 0.02 1 s 1 . This is G k also confrmed by the gradient. The computer spreadsheet allows us to model one other aspect o capacitor discharge. As the charge ows, discharging the capacitor, there is a current in the circuit. The current is equal to the change in charge per second over each 1 0 s time interval. This can be modelled by assigning the average current during a time interval to the value o the current at a time halway through the time interval ( i.e. at 5 s or the frst 01 0 s interval, at 1 5 s or the second 1 02 0 s interval and so on) . Another graph in fgure 1 0 shows that the current j ust like charge and potential dierence decays exponentially with a hal- lie behaviour. The product R C = is known as the time constant or the circuit. The name is ully j ustifed because the units o RC are ampere second volt coulomb coulomb _ __ ohm arad = _ = _ ampere ampere = ampere volt = second. The capacitor time constant is not the same as a hal-lie. Ater a time o RC since the start, Q will be Q = Q 0 e -1 because t = RC = So Q 1 = 0.3 7 _ = _ e Q0 D uring each time interval the charge drops to 3 7% o its value at the start o the interval. Ater 2 this will be ( 0.3 7) 2 or 1 3 .7% . Ater 5 the charge remaining on the capacitor will be less than 1 % o its initial value.
Tip To obtain the instantaneous current in the resistor rom the graph o charge on the capacitor against time, remember that the rate at which charge leaves the capacitor is also the rate at which charge ows through the resistor. In other words, the gradient o the tangent o the Q-t graph will give you the value o the current in the resistor.
Tip There are three equations that give the variation o charge, current and potential diference with time or a discharge circuit: t _
Q = Q0 e- t _
I = I0 e -
t _
V = V0 e -
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11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Worked examples 1
A 2 2 0 F capacitor charged to 3 0 V discharges through a 3 3 0 k resistor. a) Calculate the time taken or the capacitor to discharge to 1 0 V. b) C alculate the charge moved rom the capacitor in this time.
2
C alculate the number o multiples o the time constant that are required or a capacitor to lose 90% o its charge.
Solution Q = Q 0e
t -_ RC t ___
t For this case 0.1 = e - , thereore - ___ = ln 0.1 RC = 2 . 3 , making t = 2 .3 RC. RC
Solution t -_
t - ___ 3 -6
a) V = V0 e RC so 1 0 = 3 0 e 330 1 0 220 1 0 . t and t = 80 s. Thereore ln 0.3 3 = - ____ 72.6
2 .3 time constants are required to lose 90% o charge.
b) Q = CV. So Q = 220 1 0 - 6 20 = 4.4 mC.
Charging Again we approach the solution o these equations through an experiment ollowed by making a spreadsheet model.
Investigate! Charging a capacitor
This is very similar to the experiment during which you varied the total resistance in a circuit to achieve a constant current. This time, however, the resistance does not change. As the potential dierence at which the capacitor is storing charge increases, the current in the circuit will decrease.
the capacitor every 5 s until the pd becomes constant.
Plot a graph o pd against time and examine it closely. To what extent does the behaviour resemble the discharge case? R
Suitable values or the fxed resistor and capacitor are, as beore, 1 00 F and 470 k.
Make sure the capacitor is discharged, by short- circuiting it with the ying lead, and then disconnect the ying lead to begin the experiment. Wait or a data logger to do its j ob or take manual readings o the voltage across
Vem
fying lead
C
V
Figure 11
A model for charging The capacitor is in series with a resistor o resistance R and a cell o em Vem. When the circuit is switched on with the capacitor uncharged, charge begins to ow. Using Kirchhos second law Vem = VC + VR and thereore VR = Vem - VC = IR
468
1 1 . 3 C APACI TAN CE
so ( Vemf - VC ) I = __ R In the time interval t, the change in the charge on the capacitor Q is related to the change in the potential difference between the plates V by Q CV I= _= _ t t therefore ( rearranging) (Vemf - VC ) t V = __ RC This expression can be integrated mathematically, but a numerical solution illustrates more of the physics of the charging.
B
C
delta V/C
Vc/C
0 1.28E+00 1.00E+00 7.91E-01 6.23E-01 4.90E-01 3.86E-01 3.04E-01 2.39E-01 1.88E-01 1.48E-01 1.17E-01 9.19E-02 7.23E-02 5.69E-02 4.48E-02 3.53E-02 2.78E-02 2.19E-02 1.72E-02 1.36E-02 1.07E-02 8.40E-03 6.61E-03 5.21E-03 4.10E-03 3.23E-03 2.54E-03 2.00E-03 1.57E-03 1.24E-03
0 1.28E+00 2.28E+00 3.07E+00 3.70E+00 4.19E+00 4.57E+00 4.88E+00 5.11E+00 5.30E+00 5.45E+00 5.57E+00 5.66E+00 5.73E+00 5.79E+00 5.83E+00 5.87E+00 5.90E+00 5.92E+00 5.94E+00 5.95E+00 5.96E+00 5.97E+00 5.98E+00 5.98E+00 5.98E+00 5.99E+00 5.99E+00 5.99E+00 5.99E+00 6.00E+00
D
E
F
G
H
I
J
K
L
R 4.70E+05 C 1.00E-04 delta t 10 Vemf 6
7 6 5 Vc/V
A 1 2 time/s 3 0 4 10 5 20 6 30 7 40 8 50 9 60 10 70 11 80 12 90 13 100 14 110 15 120 16 130 17 140 18 150 19 160 20 170 21 180 22 190 23 200 24 210 25 220 26 230 27 240 28 250 29 260 30 270 31 280 32 290 33 300 34
4 3 2 1 0 0
50
100 150 200 250 300 350 400 t/s
Figure 12 Spreadsheet model for capacitor charging.
469
11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) The three columns of the spreadsheet model are:
time is increased row by row by an amount t; the formula for A4 is =A3 + $F$3 ( as before the $ signs force the spreadsheet to use the same row and column each time) .
deltaVc is the change in Vc since the last time; it is given by the equation above and B4 translates into = ($F$4-B 3 ) * $F$3 / ($F$1 * $F$2 ) in the spreadsheet language. Look carefully at the contents of each cell and you should see why this is correct.
Vc is the new value of Vc that incorporates the deltaVc from the previous line; it is a simple addition = C 3 + B 4
The spreadsheet is copied for about 5 0 rows ( not all shown here) and then the values of t and Vc are used to plot a graph. D oes this graph have similarities to the one you plotted from experimental data? What are the similarities between charge and discharge?
TOK Exponential decay The similarity between radioactivity and capacitor discharge is shared with many other phenomena throughout the whole o science and extends into economic theory. In radioactivity the reason or the behaviour is that each nucleus o a particular isotope has an identical probability o decay per second. Thereore the number decaying in one second is directly proportional to the number o nuclei that remain undecayed. In capacitor discharge, the potential diference (which is proportional to the charge) between the plates is related to the discharge current Q through V = IR so ___QC = R _______ . We can use our water analogy again, this time or t water owing rom the bottom o a large tank. When the tank is ull the pressure (potential) rom the weight o water at the bottom is large and the ow rate is large too. When the tank is nearly empty the pressure is lower and the ow rate smaller. There are other examples o exponential changes in the natural world. Sea animals with shells grow at a rate that depends on the mass o ood they can eat and this mass depends on their mouth size. So the shell grows exponentially and the shape o the shell can be in the orm o an exponential curve. Population growth is an exponential change i there are no predators to remove a species. Why do so many academic areas have this linking relationship?
470
QUESTION S
Questions 1
S tate the magnitude o the magnetic ux linking the coil.
( IB) A bar magnet is suspended above a coil o wire by means o a spring, as shown below.
B
spring
area S
magnet
3
( IB)
coil electromagnet
The ends o the coil are connected to a sensitive high-resistance voltmeter. The bar magnet is pulled down so that its north pole is level with the top o the coil. The magnet is released and the variation with time t o the velocity v o the magnet is shown below.
The current in the circuit is switched on. a) S tate Faradays law o electromagnetic induction and use the law to explain why an em is induced in the coil o the electromagnet. b) S tate Lenzs law and use the law to predict the direction o the induced em in part a) .
v
0
c) Magnetic energy is stored in the electromagnet. State and explain, with reerence to the induced em, the origin o this energy. ( 8 marks)
t
0
a) C opy the diagram and on it: (i) mark with the letter M, one point in the motion where the reading o the voltmeter is a maximum (ii) mark with the letter Z, one point where the reading on the voltmeter is zero.
4
( IB) A small coil is placed with its plane parallel to a long straight current- carrying wire, as shown below.
current-carrying wire
b) Explain, in terms o changes in ux linkage, why the reading on the voltmeter is alternating. ( 4 marks) 2
A uniorm magnetic feld o strength B completely links a coil o area S. The feld makes an angle to the plane o the coil.
small coil
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11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) a) Use Faradays law o electromagnetic induction to explain why, when the current in the wire changes, an em is induced in the coil.
b) O utline why the core is laminated. c) The primary coil o an ideal transormer is connected to an alternating supply rated at 2 3 0 V. The transormer is designed to provide power or a lamp rated as 1 2 V, 42 W and has 45 0 turns o wire on its secondary coil. D etermine the number o turns o wire on the primary coil and the current rom the supply or the lamp to operate at normal brightness. ( 7 marks)
current
0
magnetic fux
0
em
The diagram below shows the variation with time t o the current in the wire.
0
0
t
0
t
0
t
6
(IB) The graph shows the variation o potential dierence V with time t across a 2 2 0 F capacitor discharging through a resistor. C alculate the resistance o the resistor. ( 3 marks) 14 12 10
(i) C opy the diagrams and sketch, on the axes, graphs to show the variation with time t o the magnetic fux in the coil.
V/V
b)
c) Such a coil may be used to measure large alternating currents in a high-voltage cable. Identiy one advantage and one disadvantage o this method. ( 8 marks) 5
6 4
(ii) S ketch, on your axes, a graph to show the variation with time t o the em induced in the coil. (iii) S tate and explain the eect on the maximum em induced in the coil when the coil is urther away rom the wire.
8
2 0
7
a)
0
10
20
30 t/s
40
50
60
A circuit is used to charge a previously uncharged capacitor. The supply has an em o 1 2 . 0 V and negligible internal resistance and is in series with resistance R. The graph shows how the potential dierence across the capacitor varies with time ater the switch is closed. S
(IB) The diagram below shows an ideal transormer. laminated core 12.0 V secondary coil primary coil
a) Use Faradays law to explain why, or normal operation o the transormer, the current in the primary coil must vary continuously.
472
V 200 F
QUESTION S
C alculate:
15
(i) the initial energy stored by the capacitor (ii) the efciency o the system.
10 pd/V
( 1 0 marks) 5
9
0 0
50
100 time/s
150
200
(i) D etermine the time taken or the potential dierence across the capacitor to reach hal the maximum value. (ii) C alculate R. (iii) C alculate the initial charging current. b) A 1 00 F capacitor is added in series with the 2 00 F capacitor. (i)
( 1 0 marks) a)
1 0 In a timer, an alarm sounds ater a time controlled by a discharge circuit powered by a 9.0 V cell o negligible internal resistance. The alarm time is varied using resistor R. The capacitor is charged by moving the twoway switch to position S 1 . The timing starts when the switch is moved to S 2 . S1
X
S2
C alculate the eective capacitance o the combination.
(ii) D raw a graph showing how the potential dierence across the combination varies with time when the combination is charged.
8
A capacitor stores a charge o 3 0 C when the pd across it is 1 5 V. C alculate the energy stored by the capacitor when the pd across it is 1 0 V. ( 2 marks)
(i) A capacitor has a capacitance o 1 F. O utline what this tells you about the capacitor. (ii) S ketch a graph to show how the charge on the 1 F capacitor varies with the potential dierence across it over a range o 6 V. (iii) E xplain how you could determine the energy stored by the capacitor or a potential dierence o 6 V.
b) A 0.047 F capacitor is charged to a pd o 2 0 V, disconnected rom the supply and connected to a small motor without discharging. The motor can then lit an obj ect o mass 0.2 5 kg through a height o 0.90 m beore the capacitor is ully discharged.
R
6.0 V
to alarm
2200 F
Y
An alarm rings when the potential dierence across R reaches 3 .0 V. a) In one setting the time constant o the circuit when the capacitor is discharging is 3 . 0 minutes. Sketch a graph to show how the potential dierence across R varies with time or two time constants o the discharge. b) Use your graph to state the time at which the alarm will sound. c) C alculate the resistance o the variable resistor when the time constant is 3 .0 minutes. d) D etermine the maximum value o the resistance R that is needed or the timer to operate or up to 5 minutes. e) S tate how a capacitor could be connected to the circuit to increase the range o the timer. ( 1 1 marks)
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) 11
1 2 Three capacitors, one of unknown value, are connected in a circuit. The supply has a negligible internal resistance. The total charge stored on the capacitors is 400 C when the potential difference between A and B is 1 2 .0 V.
switch A
R
V
200 F
8.0 V
S 20 F
A
12 V
X
switch
60 F
150 k
B
B X
a) R
V
8.0 V
(i) C alculate the total capacitance of the circuit. (ii) C alculate X. (iii) C alculate the potential difference across the 2 0 F capacitor.
200 F
The diagram shows two graphs of the variation of p.d. with time for the discharge through resistor of value R of ( i) a 2 2 0 F capacitor and ( ii) the same 2 2 0 F capacitor in series with a capacitor of unknown capacitance. 12
b) The supply is disconnected and the capacitors discharge through the 1 5 0 resistor. C alculate the: (i) time taken for the potential difference between A and B to fall to 6. 0 V (ii) potential difference between A and B 8.0 s after opening the switch.
10
( 1 0 marks) V/V
8 6
1 3 A parallel plate capacitor is made from two circular metal plates with an air gap of 1 .8 mm between them. The capacitance was found to be 2 .3 1 0 1 1 F.
A 4 B 2
C alculate: 0 0
10
20 time/s
30
40
a) O utline why the p.d. in experiment B falls more rapidly than in experiment A. b)
(i) D etermine R. (ii) D etermine the capacitance of the unknown capacitor. ( 8 marks)
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a) the diameter of the plates b) the energy stored when the potential difference between the plates is 6.0 V. ( 5 marks)
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Q U AN TU M AN D N U CLE AR P H YS I C S ( A H L )
Introduction This topic develops material we frst met in Topic 7. The physics included here was ground-breaking when it was frst proposed (mostly in the early to mid 20th century) ; there were many highly respected physicists who totally rejected much o
the theory. Today these ideas are seen as being airly uncontroversial but there are still those who believe that reality should be modelled by theories that are simpler or more universal. Time will, no doubt, tell whether we have currently got it right!
12.1 The interaction of matter with radiation Understandings Photons The photoelectric efect Matter waves Pair production and pair annihilation Quantization o angular momentum in the Bohr
model or hydrogen The wave unction The uncertainty principle or energy and time, and position and momentum Tunnelling, potential barrier, and actors afecting tunnelling probability
Applications and skills Discussing the photoelectric efect
experiment and explaining which eatures o the experiment cannot be explained by the classical wave theory o light Solving photoelectric problems both graphically and algebraically Discussing experimental evidence or matter waves, including an experiment in which the wave nature o electrons is evident Stating order o magnitude estimates rom the uncertainty principle
Equations
Nature of science Scientists increasing dependence on quantum phenomena Much o what is described by quantum mechanics depends upon probability and seems in many ways counter-intuitive. Einstein, who remained unconvinced about quantum mechanical interpretations o matter, said that he did not believe that God plays dice.
Planck relationship: E = hf Einstein photoelectric equation: Em a x = hf -
13.6 n nh quantization o angular momentum: mvr = _ 2 Probability density: P(r) = || 2 V Heisenberg relationships: positionmomentum h xp _ 4 h energytime: Et _ 4
Bohr orbit energies: E = - _ eV 2
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Introduction In Topic 4 we treated light as a wave, but in Topic 7 we needed to modiy our views o this in order to describe energy changes that happen in atoms and to explain atomic spectra. In S ub- topic 7.3 we looked at the Rutherord model o the atom with a nucleus surrounded by orbiting electrons. We said that this model did not obey the laws o classical physics but had much to commend it in terms o agreement with experiment. We now consider the urther leap o aith that needed to be made in order to reconcile the Rutherords experimental results with theory; this leap o aith meant abandoning aspects o classical physics and our everyday experiences and to accept many outlooks which go against our intuition. We will now consider the photoelectric eect and go on to look at B ohrs old quantum theory and its development into the modern interpretation o the quantum theory; acets o physics that would mean that physicists could never be truly certain o anything again!
Nature o science In the later part o the nineteenth century physicists were attempting to explain the radiation emitted by a black body a perect emitter and absorber o radiation. The radiation emitted by a black body was modelled using classical physics by the RayleighJeans law, whereby the intensity is proportional to the square o the requency. This theory worked well or the visible and inrared parts o the spectrum, where it matched the practical curve, but it ailed in the ultraviolet region by implying that ultraviolet radiation would be emitted with an infnite intensity something that clearly was not possible ( see fgure 1 ) . In 1 900, the German physicist Max Planck suggested that the ultraviolet catastrophe would be rectifed i electrons oscillating in the atoms o hot bodies were to have energies that were quantized in integral values o hf where f is the requency o the electrons and h is Plancks constant. The theory that he developed worked well at interpreting black- body radiation, but it
intensity of radiation/arbitrary units
The ultraviolet catastrophe the ultraviolet catastrophe 5000K
Rayleigh-jeans law
Planck radiation formula 1000 2000 wavelength of radiation/nm
3000
Figure 1 The ultraviolet catastrophe the experimental curve and one based on classical physics. was not well- received by the physics community because he was unable to j ustiy why the energies o electrons should be quantized. It was not until Einstein used similar assumptions to those o Planck in order to explain the photoelectric eect that physicists began to take Plancks ideas seriously.
The photoelectric efect Demonstration o the photoelectric efect The photoelectric eect can be demonstrated using a gold-lea electroscope (or a coulombmeter) . A reshly cleaned sheet o zinc should be mounted on the electroscope plate and the sheet charged negatively by connecting it to a high negative potential (o around 3 kV) . When a range o
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1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
electromagnetic radiation rom inra-red to ultraviolet is incident on the sheet the divergence o the lea only alls when ultraviolet radiation is used; the lea then immediately collapses. This is because the zinc sheet and electroscope lea are discharging by the emission o electrons.
negatively charged zinc plate
ultraviolet radiation
Explanation o the photoelectric efect E instein explained the photoelectric eect in the ollowing way:
Light can be considered to consist o photons, each o energy = hf
Each photon can only interact with a single electron.
There is a minimum photon requency called the threshold frequency ( f0 ) below which no electron can be emitted.
Energy is needed to do the work to overcome the attractive orces that act on the electron within the metal this energy is called the work function ( ) .
Any urther energy supplied by a photon becomes the kinetic energy o the emitted electron ( oten called a photoelectron) .
Increasing the intensity o light simply increases the number o photons incident per second.
gold lea alls immediately that the zinc plate is illuminated with ultraviolet radiation
Figure 2 Gold-lea electroscope demonstration o the photoelectric efect.
Explaining observations from the gold leaf experiment The zinc sheet has a certain work unction and photons must have a greater energy than this to be able to emit electrons. The ultraviolet radiation has the highest requency o the radiation used and so it is these photons that are able to ree the electrons rom the metal. a) With very intense visible or infra-red light incident on the sheet the leaf remains diverged. Increasing the intensity only increases the number o photons incident per second; with long wavelength ( or low requency) none o the photons have enough energy to liberate electrons so the lea remains diverged. b) If low intensity ultraviolet is used, the leaf still falls immediately. O ne photon interacts with one electron so no time is needed to build up the energy to release an electron; as soon as the electron absorbs a sufciently energetic photon it will be ej ected rom the metal. c) Placing a sheet of glass between the ultraviolet source and the zinc p revents the leaf from falling. Glass will only transmit low energy visible photons; it absorbs the higher energy ultraviolet ones so none reach the zinc.
hf < no electrons emitted
hf = electrons brought to surace but stay there
d) If the zinc sheet is charged p ositively the leaf remains diverged for all wavelengths of radiation. Raising the potential o the metal presents a much deeper potential well or the electrons to escape rom, meaning that the photons no longer have sufcient energy to liberate the electrons. This is similar
hf > electrons ejected
metal o work unction
Figure 3 Photoelectric emission and the work unction.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) to replacing the zinc with a metal o greater work unction. This is analogous to someone being trapped in a well with vertical sides and only being able to j ump out in one leap.
Einsteins photoelectric equation We are now in a position to look at how E instein expressed the photoelectric eect in a single equation. His equation takes the orm: Emax = hf - In this equation Emax represents the maximum kinetic energy o the emitted electron, hf is the energy o the incident photon, and is the work unction o the metal. Each o the terms in the equation represents a quantity o energy that could be measured in j oule. However, it is quite common to express the energies in electronvolts simply because this avoids using very small powers o ten. The kinetic energy is expressed as a maximum value because the work unction is defned as the minimum energy required to liberate an electron those embedded urther inside the metal will take more energy than this to be liberated.
Worked examples 1
The work unction or aluminium is 6.6 1 0 1 9 J. a) C alculate the photoelectric threshold requency or aluminium. b) C alculate the maximum kinetic energy o the electrons emitted when photons o requency 1 . 2 1 0 1 5 Hz are incident on the aluminium surace.
Solution
b) C alculate the maximum speed with which a photoelectron may be emitted rom a potassium surace by photons o this energy. The work unction or potassium is 1 .5 eV.
Solution a) First the energy needs to be converted into joules: 2 .2 eV = 2 . 2 1 .6 1 0 - 1 9 = 3 .5 1 0 - 1 9 J
The work unction or aluminium is 6.6 1 0
1 9
J.
3 .5 1 0 E Then E = hf f = __ = _________ h 6.63 1 0 -19
-34
a) Using Einsteins equation Emax = hf - at the threshold value Emax = 0 so = hf0 where f0 is the threshold requency. 6. 6 1 0 - 1 9 = __ f0 = _ h 6.63 1 0 - 34 = 1 .0 1 0 1 5 Hz This is to two signifcant fgures in line with the work unction value. b) Using Emax = hf - Emax = 6.63 1 0 - 34 1 .2 1 0 1 5 - 6.6 1 0 - 1 9 = 1 .4 1 0 - 1 9 J 2
Photons incident on a metal surace have energy 2 .2 eV. a) C alculate the requency o this radiation.
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= 5.3 1 0
14
Hz
b) Again using Emax = hf - but sticking to electronvolts to start with: Emax = 2 . 2 - 1 .5 = 0. 7 eV S o the maximum kinetic energy = 0.7 eV = 0.7 1 .6 1 0 - 1 9 J = 1 .1 2 1 0 - 1 9 J 1 This means that __ m v2 = 1 .1 2 1 0 - 1 9 J 2 e
__________
So v =
2 1 .1 2 1 0 ____________ m -19
e
Looking up the value or the mass o electron in the data booklet gives m e = 9.1______________ 1 1 0 - 31 kg 2 1 .1 2 1 0 - 1 9 v = __ 9.1 1 1 0 - 31 = 5 .0 1 0 5 m s - 1
1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
Investigate! Millikans photoelectric experiment In 1 91 6, the American physicist Robert Millikan designed an elegant experiment with which to test Einsteins photoelectric equation. We can use a photocell or essentially the same experiment with an arrangement as shown in fgure 4.
potential and e is the elementary charge on an e lectron) .
Einsteins equation Emax = hf - can now be rearranged to give eVs = hf - hf0 where f0 is the threshold requency.
The flters are usually provided with a range o transmitted wavelengths which gives a value or their uncertainties. This means E insteins equation needs urther rearrangement into
anode Vs +
- electrons -
pA
vacuum
hc hc eVs = _ - _ o
-
where o is the threshold wavelength and c is the speed o electromagnetic waves in a vacuum.
pA cathode
D ividing throughout by e gives
potential of anode is made negative so electrons cannot quite reach it and the picoammeter reading becomes zero
(
hc hc hc _ 1 1 _ Vs = _ - _ = _ e - e e o o This is o the orm:
Figure 4 Photocell for measuring the Planck constant. A variety o coloured flters are used with a white light source to allow incident photons o dierent requencies to all on the cathode o the photocell. The electrons are emitted rom the photocell and travel across the vacuum towards the anode.
y = mx + c
1 A graph o Vs against __ :
hc will be o gradient __ e and
1 1 have an intercept on the __ axis equal to __ o
have an intercept on the Vs axis equal to hc - ___ ( as shown in fgure 5 ) e o
In this way the electrons complete the circuit and a small current o a ew picoamps registers on the picoammeter.
Vs gradient =
The maximum kinetic energy is obtained by adj usting the voltage o the anode with respect to the cathode using a potential divider ( not shown on the diagram) the anode is actually made negative! When the pote ntial die rence across the tube is j ust sufcie nt to preve nt electrons rom crossing the tube , the maximum kinetic energy o the emitte d electrons is equal to eVs ( whe re Vs is the stopping
)
0
0
1 0
hc e
1
-
hc e 0
Figure 5 Plancks constant graph.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L )
Note Dierent metal cathodes will have dierent work unctions so they will always give graphs o the same gradient as fgure 5 but they will produce parallel lines with dierent intercepts.
The wave theory and the photoelectric eect The reason that the wave theory ails to account or the photoelectric eect is explained by the instantaneous nature o the emission that occurs when light alls onto a metal surace. Waves provide a continuous supply o energy, the intensity o which is proportional to the square o the wave amplitude. According to classical wave theory, when low intensity electromagnetic radiation o any requency is incident on a metal surace, given sufcient time, enough energy should eventually accumulate to allow an electron to escape rom its potential well. However we fnd that actually a small number o photons with requencies above the threshold will always ej ect electrons rom a metal surace immediately; below this requency, no electrons are ej ected. This is completely contradictory to wave theory. The outcome o the photoelectric eect leaves us in a slightly uncomortable position o needing waves to describe some properties o light, such as intererence and diraction, but needing particle theory to explain the photoelectric eect. This does not mean that either theory is wrong rather that both are incomplete. Light appears to have characteristics that can be attributed to either a wave or a particle we call this waveparticle duality and is another example o the outcomes o experiments ailing to be in accord with our everyday perception.
Worked example
Solution 15
Photons o requency o 1 . 2 1 0 Hz are incident on a metal o work unction o 1 . 8 eV. a) C alculate the energy transerred to the metal by each photon. b) C alculate the maximum kinetic energy o the emitted electrons in electronvolts. c) D etermine the stopping potential or the electrons.
a) Using E = hf to calculate the photon energy in j oules ( we are not asked or this to be in eV so it is fne to leave it in j oules) . E = 6.63 1 0 - 34 1 . 2 1 0 1 5 = 8.0 1 0 - 1 9 J b) We now need to work in electronvolts so the 8.0 1 0 photon energy = ________ = 5 .0 eV 1 .6 1 0 -19 -19
With = 1 .8 eV this means the maximum kinetic energy is ( 5 . 0 - 1 .8) = 3 .2 eV c) Working in electronvolts now really comes into its own since the stopping potential will be 3.2 V.
Matter waves In his 1 924 PhD thesis, the French physicist, Louis de Broglie (pronounced de broy) , used the ideas o symmetry to suggest that i something classically considered to be a wave had particle-like properties, the opposite would also be true. Matter could, thereore, also have wave-like properties. He suggested that the wavelength associated with a particle is given by h = _ p Here h is the Planck constant and p is the momentum o the particle (= mv) . This wavelength is known as the de Broglie wavelength. D e B roglie used ideas rom the special theory o relativity and the photoelectric eect in order to derive this equation or light ( you dont need to learn this derivation) .
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1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
The total energy o an obj ect ( rom the special theory o relativity) is the _________ total o the rest energy and kinetic energy, given by E = p 2 c 2 + m 0 2 c 4 ( the frst term in the equation is a kinetic energy term and the second term is the rest mass energy term) . For___ a photon the second term is zero 2 2 ( photons have no rest mass) so E = p c or E = pc. As we have seen rom the photoelectric eect E = hc h pc = __ or = __ p
__h c equating these
With this equation derived or light, de B roglie simply speculated that the same thing would be true or matter, and very soon he was shown to be correct!
Electron difraction In 1 92 5 , two American physicists, C linton D avisson and Lester Germer, demonstrated de B roglies hypothesis experimentally by observing intererence maxima when a beam o electrons was reected by a nickel crystal. In 1 92 8, the B ritish physicist George Thomson independently repeated D avisson and Germers work at the University o Aberdeen. Figure 6( a) and ( b) shows a laboratory arrangement or demonstrating the eect using the transmission o electrons through a thin slice o crystal. E lectrons rom a heated cathode pass through a thin flm o carbon atoms ( the crystal) . I the electrons behaved like particles they would be only slightly deviated by collisions with the carbon atoms and would orm a bright region in the centre o the screen. evacuated tube crystal
power supply and connections not shown
series o bright and dark rings heated cathode difracted electron beams
(a)
(b)
Figure 6 Electron difraction tube.
The bright rings indicate where the electrons land on the screen. Where there is a bright glow there is a high probability o electrons reaching that point where there is darkness there is a low probability o the electrons reaching that point. The same pattern builds up slowly, even i there are only a ew electrons travelling in the tube at any one time. This pattern is very similar to the intererence pattern that is obtained with light using a diraction grating and can be explained by assuming that electrons behave in a similar way to waves: When electrons are accelerated through a potential dierence they gain kinetic energy 1 mv2 eV = _ 2
(
)
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) Assuming that the accelerated electrons do not travel close to the speed of light then the momentum of the electrons is given by p = mv, meaning that p 2 = ( mv) 2 2
p 1 so __ mv 2 = ___ = eV 2 2m
_____
and p = 2 meV using the de B roglie relationship h = _ p h _____ this gives = _ 2 meV
Worked example C alculate the de B roglie wavelength of electrons accelerated through a potential difference of 3 . 00 kV.
Solution In an IB question on this you would be asked to develop this equation step by step and then do this calculation. h 6.63 1 0 - 34 _____ = ____ ________________________________ = _ 2 meV 2 9.1 1 0 1 0 - 31 1 .60 1 0 - 1 9 3 000 = 2 .2 4 1 0 1 1 m
This is similar to the wavelength of the X- rays used to form diffraction patterns when they are incident on crystals. Increasing the accelerating voltage increases the energy and momentum of the electrons. The wavelength, therefore, decreases and so produces smaller diameter rings with smaller spacing between them. This is analogous to light passing through a diffraction grating: the diffraction angle ( ) in the equation n = d sin is reduced when light of a shorter wavelength is used.
Nature of science Waveparticle duality From our discussion of the photoelectric effect and electron diffraction we have seen that both electrons and photons sometimes behave as waves and sometimes as particles. This gives rise to questions such as is matter a wave or a particle? and is light a wave or a particle?. The answer to each of these questions is neither: matter is matter and light is light, therefore they are neither waves nor particles. However, in each case we need the wave model to explain some properties of each and we need the particle model to explain other properties. Nobody knows the mechanisms
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by which electrons and photons behave, because they do not enter into the realms of everyday experience. We interpret their behaviour by using mathematics ( of increasing complexity) , but it is impossible to unravel the properties of waves or electrons by relating these to everyday occurrences. That individual photons can pass through individual slits and produce a pattern consisting of regions of high photon densities and regions of low photon densities makes no sense when we think of the interference of particles yet this is what happens in the quantum world.
1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
The Bohr model In order to interpret the scattering o alpha particles as discussed in S ubtopic 7.3 , Niels B ohr proposed a model o an atom in which electrons could only occupy orbits o certain radii. His model was based on the ollowing three assumptions: 1
E lectrons in an atom exist in stationary states.
2
3
C ontradicting classical physics, electrons could remain in these orbits without emitting any electromagnetic radiation.
Electrons may move from one stationary state to another by absorbing or emitting a quantum of electromagnetic radiation
I an electron absorbs a quantum o radiation it can move rom one stationary state to another o greater energy; when an electron moves rom a stationary state o higher energy to one o lower energy it emits a quantum o radiation.
The dierence in energy between the stationary states is given by E = hf.
The angular momentum of an electron in a stationary state is quantized in integral values of 2h
___
This can be represented mathematically by: nh mvr = _ 2 Angular momentum is the ( vector) product o the momentum o a particle and the radius o its orbit so, or a particle in a circular orbit, the angular momentum will be constant. This assumption is equivalent to suggesting that an integral number o de B roglie-type wavelengths fts the electrons orbital: h h _ = _ p p( = mv) = The circumerence o an orbit ( o radius r) = 2 r When the number o complete waves ftting this orbit is n then each
visualization o electron waves or frst three Bohr orbits electron wave resonance n = 1, 1 = 2r1
2 r = _ n
n = 2, 2 2 = 2r2 n = 3, 3 3 = 2r3
Equating the two values or gives h 2 r _ _ p = n
n=3
or, rearranging and substituting mv or p, gives nh mvr = _ 2
n=2 n=1
This pattern is shown in fgure 7 and corresponds to standing waves.
Energies in the Bohr orbits O ne o the triumphs o the B ohr atom was that it produced an equation that agreed with the experimental equation or the spectrum o the hydrogen atom. B y measuring the total kinetic and potential energy o
Figure 7 Standing waves in an atom or n = 2 and n = 3.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) the hydrogen atom we fnd that, or an electron in the nth energy level ( where n = 1 represents the ground state, n = 2 the frst excited state, etc. and is called the p rincip al quantum number) , the total energy E in electronvolts at each level is given by:
n= n=4 n=3
E =0 E = -0.85 eV E = -1.51 eV
n=2
E = -3.40 eV
n=1
E = -13.6 eV
Figure 8 Energy levels in a hydrogen atom.
1 3.6 E = -_ n2 S ee fgure 8. This total energy is negative because the electron is bound to the nucleus and energy must be supplied to the system in order to completely separate the electron rom the proton. B ohr went on to modiy this equation so that it could accommodate other hydrogen- like ( i.e. one electron) systems such as singly ionized helium and doubly ionized lithium, etc. The model ailed, however, to be extended to more complicated systems o atoms and it could not explain why certain allowed transitions were more likely to occur than others. D espite its aws, the B ohr model proved to instigate a more undamental approach to the atom; this is now called quantum mechanics.
Worked example In his theory o the hydrogen atom, B ohr reers to stable electron orbits. a) S tate the B ohr postulate that determines which stable orbits are allowed.
When in a stable orbit an electron does not emit radiation.
b) D escribe how the existence o such orbits accounts or the emission line spectrum o atomic hydrogen.
When an electron drops to a lower energy level or orbit it emits a photon.
The requency o the emitted photon is proportional to the dierence in energy between the two levels.
A transition between two levels results in a spectral line o a single wavelength being emitted.
The B ohr model o the hydrogen atom can be extended to singly ionized helium atoms. The model leads to the ollowing expression or the energy En o the electron in an orbit specifed by the integer n. k En = _2 n where k is a constant. In the spectrum o singly ionized helium, the line corresponding to a wavelength o 3 6 2 nm arises rom ele ctron transitions betwe en the orbit n = 3 to the orbit n = 2 . c) D educe the value o k.
Solution a) The angular momentum ( mvr) o an electron in a stationary state is quantized in integral h values o ___ 2
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b) There are many things that can be said here but some of the key points are:
hc k c) E = __ = __ so the dierence in energy n between the two levels will be 2
(
)
1 1 = k _ - _ = 0.1 3 9 k 4 9 hc 6.63 1 0 - 34 3 1 0 8 k = _ = ___ 0.1 3 9 3 62 1 0 - 9 0.1 3 9 = 3 . 95 1 0 - 1 8 J
1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
Schrdingers equation
intensity/arbitrary units
22
Waveparticle duality explains a bright intererence ringe as being the place where there is a high probability o fnding a particle. The position o particles is described mathematically by probability waves. As with classical waves, probability waves superpose with one another to produce the expected intererence pattern. A low-intensity beam o photons incident on a single slit one at a time will build up a distribution that is identical to the expected diraction pattern, provided that we wait a sufciently long time. A similar pattern is obtained by fring electrons at a slit o suitably small width (see fgure 1 0) . From this distribution we can predict the places where electrons will not reach but we are unable to predict where an individual electron will be detected although, statistically, there will be approximately 2 2 times the number o electrons arriving at the area around the principal diraction maximum compared to the number around the secondary maximum.
1 (a)
(c)
0
(b)
1
2 3 4 5 6 difraction angle/degrees
7
Figure 9 Difraction pattern intensity distribution.
(d)
Figure 10 Difraction pattern being built up by individual electrons.
The concept o probability was developed into the quantum mechanical model o the atom in 1 92 6 by the Austrian ( and later AustrianIrish) physicist, E rwin Schrdinger. S chrdingers wave function describes the quantum state o particles. His wave equation has many similarities to a classical wave equation but it is not a derived equation. It has, however, been thoroughly verifed experimentally and its solution or the single electron hydrogen atom agrees with the B ohr relationship 1 3.6 E = - ____ . n 2
The wave unction is not a directly observable quantity but its amplitude is very signifcant. With light waves we observe the intensity, not the amplitude, and we have seen that the intensity is proportional to the square o the amplitude. For the wave unction, where the square o the amplitude is a maximum there is the greatest probability o fnding a photon. When the square o the amplitude is zero there is zero probability o fnding the photon. The quantity may be thought o as the amplitude o the de B roglie wave corresponding to a particle ( although it does not have any physical signifcance) ; however, the square o the amplitude o the wave unction 2 is proportional to the
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) probability per unit volume o fnding the particle this is known as the probability density. Mathematically we write this as: P( r) = 2 V Here P( r) is the probability o fnding a particle a distance r rom a chosen origin and V is the volume being considered. For double slit intererence, in terms o the probability wave, the wave unction is considered to be such that a single photon or electron passes through both slits and be everywhere on the screen until it is observed or measured. When this happens, the wave unction collapses to the classical case and the particle is detected. This is known as the Copenhagen interpretation, so named by the German physicist, Werner Heisenberg. It relates to the interpretation o quantum mechanics used by Heisenberg, B ohr and their co-workers between 1 92 4 and 1 92 7. It can be summarized as nothing is real unless it is observed. So matter or light can be considered to be a wave or a particle. I it behaves like a particle then it is a particle. I it behaves like a wave, then it is a wave.
energy/eV 0 n=3
n=2 n=1 -13.6
Figure 12 Electron standing waves in a potential well.
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Figure 11 The Copenhagen interpretation.
In the simplifed ( one-dimensional) version o the hydrogen atom, as shown in fgure 1 2 , an electron would be detected somewhere between the nucleus and the outside edge o the atom these are shown by the edges o a potential well. A more realistic model would show the potential 1 varying as the inverse o the distance rom the nucleus ( V - __ r ). Within the well, the electron energy must be such that the wave unction has nodes at the sides. In the electron wave model the probabilities o fnding an electron within the nucleus or outside
1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
the atom are both zero so the wave amplitude is zero at these points. The electron is most likely to be ound ( highest probability) where the amplitude is maximum; this is midway between the nodes.
Worked example The graph below shows the variation with distance r rom the nucleus o the square o the wave unction, 2 , o an electron in the hydrogen atom according to the S chrdinger theory. The nucleus is assumed to be a single point.
S tate where the electron is most likely to be ound. Explain whether the electron could be ound either in the nucleus or at positions corresponding to the largest value o r shown on the graph.
Solution
2
0 0
a
r
The position around a has the highest value o the square o the wave unction. This is the position where there is the highest probability o fnding the electron. B ecause the square o the wave unction is zero at the position o the nucleus, the electron defnitely cannot be ound there but there is a fnite chance o fnding the electron at large values o r since the graph has not allen to zero. The electrons probability cloud would be densest at a, but it will still have a little density at large values o r.
The Heisenberg uncertainty principle We have seen that when a quantum is diracted it is only possible to predict its subsequent path in terms o the probability o the wave unction. The outcome o this experiment is in line with Heisenbergs uncertainty principle. h This is usually written as x p ___ and places a limit on how precisely 4 we are able to know the position and momentum o something in the quantum mechanical realm.
I we wish to know where an electron is positioned at a given time, the principle tells us that there is an uncertainty ( given by x) with which we can know that position. I x is very small, then the uncertainty ( p) in knowing the momentum o the electron is very large. Thereore, it is not possible to precisely determine the position o the electron and its momentum at the same time; the product o the two uncertainties h will always be greater than or equal to ___ . 4 I we imagine the wave unction o a ree electron (one that is not in any feld that would change its motion) to be a sine wave then we can measure its wavelength perectly. Since we know its wavelength perectly, we also h know its momentum perectly (p = __ ) . This implies that the position o the electron has an infnite uncertainty and is spread out over all o space. In order to detect a quantum particle it would be necessary to use something that has a comparable size to that particle. Using radiation with which to detect a nucleus would need a wavelength o 1 0 1 5 m. h We know rom the de B roglie relationship ( = __ p ) that the shorter the
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) wavelength is, the greater the momentum so, with a wavelength o 6.63 1 0 1 0 1 5 m, the momentum will be _________ 1 0 - 1 8 N s. 10 - 34
Note
-15
In the limit when one o the quantities is known perectly the other quantity has an infnite uncertainty.
O n a nuclear level this is a very large momentum and would mean that any radiation o this wavelength would impart energy to the nucleus which would then make its position eectively immeasurable. In dealing with electrons diracting through a narrow gap, perhaps o size 1 0 1 8 m, the uncertainty principle also applies. In passing through the gap the uncertainty o the electrons position in the gap will be hal the gap width. This then puts a limit on the precision with which we can know the component o the momentum o the electron parallel to the gap (and h 1 1 0 - 1 6 N s and thereore its wavelength) . Thus p ____________ 4 0 . 5 1 0 h ___ -17 p 1 0 m ( or approximately ten times the gap size. That is a large uncertainty in relation to the gap size! ) -18
Worked example The diagrams show the variation, with distance x, o the wave unction o our dierent electrons. The scale on the horizontal axis in all our A
0
C
B
0
0
D
0
x
0
diagrams is the same. For which electron is the uncertainty in the momentum the largest?
x
0
x
0 x
0
Solution The square o the wave unction is proportional to the probability o fnding the electron. Each o B , C and D have positions where the square o the wave unction is greatest ( close to zero in B and centrally in C and D ) . This means that there is a high probability o fnding the electron at one position and, thereore, there is low uncertainty in ( the Heisenberg) position but, as a consequence,
there will be large uncertainty in momentum. For A there are many positions where the electron could be ound with equal probability at each o the maxima or minima there is large uncertainty in position but low uncertainty in momentum. B ecause C has the position where the electron is most well- defned it must have the largest uncertainty in its momentum.
Pair production and annihilation C lose to an atomic nucleus, where the electric feld is very strong, a photon o the right energy can turn into a particle along with its antiparticle. This could be an electron and a positron or a proton and
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an antiproton. The outcome will always be a particle and an antiparticle in order to conserve charge, lepton number, baryon number and strangeness. The particle and antiparticle are said to be a pair and the eect is known as p air p roduction. The antiparticle will have a mass equal to that o the particle meaning that the photon must have enough energy to create the masses o the two particles. The minimum energy needed to do this is given by:
time e e
-
+
E = 2 mc2 where m is the ( rest) mass o the particle/antiparticle and c is the speed o electromagnetic waves in a vacuum. Figure 1 3 shows a Feynman diagram or the production o an electronpositron pair. The gamma ray photon ( ) must have an energy o at least 1 .02 MeV ( which is twice the rest energy o an electron) . Any photon energy in excess o this amount is converted into the kinetic energy o the electronpositron pair and the original electron. Pair production can also occur in the vicinity o an orbital electron but in this case more energy will be needed as the orbital electron itsel gains considerable momentum and kinetic energy. Figure 1 4 shows pair production taking place near an atomic electron. The photon is non- ionizing and leaves no track but the newly ormed electron and positron can be seen spiralling in opposite directions in the applied magnetic feld. The recoiling electron gains a great deal o kinetic energy and this is why it hardly bends in the magnetic feld although it can be seen to bend in the same direction as the other electron. Theory shows that the threshold energy needed or this type o pair production is 4mc2 ( = 2 .04 MeV) .
Figure 13 Feynman diagram o electron positron pair production.
The equation or this interaction is: + e- e - + e - + e + When a particle meets its antiparticle they annihilate, orming two photons. The total energy o the photons is equal to the total mass-energy o the annihilating particles. S ometimes a pair o particles annihilate but then one o the photons produces another pair o particles. The positron that is ormed in the interaction, as shown in fgure 1 4, quickly disappears as it is re- converted into photons in the process o annihilation with another electron in matter.
position o original electron
newly ormed positron original electron gains signifcant kinetic energy
path o incident gamma ray photon (no track actually seen)
newly ormed electron
Figure 14 Bubble chamber tracks o electronpositron pair production.
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Pair production and the Heisenberg uncertainty principle There are pairs o variables, other than x and p, to which the uncertainty principle also applies o these energy and time are two important conj ugate variables. This version o the uncertainty principle is written as: h E t _ 4 S ome interesting things can happen in the quantum world. It is ound experimentally that the threshold energy required or the production o an electronpositron pair can be much less than the expected 1 .02 MeV in the presence o a heavy nucleus. Imagine a 1 0 eV photon in the vicinity o a heavy nucleus: the low energy photon produces an electronpositron pair. A very short while later the electron and the positron collide to produce two 5 eV photons. This looks like a huge violation o the conservation o mass- energy but it is allowable under the uncertainty principle. D uring the lietime o the composite electronpositron pair there is uncertainty regarding the total energy. I the uncertainty was equal to 1 .02 MeV, what would be the limit on the uncertainty o the lietime o the pair? The time t can be calculated by the energytime ormulation o the uncertainty principle: h 6.63 1 0 - 34 t = _ = ___ = 3 .2 1 0 - 2 2 s 1 .02 1 0 6 1 .6 1 0 - 1 9 4 E 4 Here we have converted the energy in MeV into J by multiplying by the electronic charge. This lietime is so short that a measurement o the energy o the pair would have an uncertainty o at least 1 . 02 MeV and the experiment would not be able to detect the violation o the law o conservation o energy. I we cannot perorm an experiment to detect a violation o the conservation law, then quantum mechanics says there is some probability o the process occurring. Another way o explaining this example is to say that nature will cheat i it can get away with it! This example has been verifed experimentally but the theory behind it is beyond that covered in the IB D iploma physics course.
Quantum Tunnelling According to quantum mechanics, a particles wave unction has a fnite probability o being everywhere in the universe at the same time. The probability may be infnitesimally small away rom the eect that we would expect rom classical physics but it is, nevertheless, fnite. This means that, or example, an electron in the ground state o a hydrogen atom could escape the attraction o the nucleus with less than the expected 1 3 . 6 eV. In agreement with the uncertainty principle a particle can eectively borrow energy rom its surroundings, pass through a barrier and then pay the energy back, providing it does not take too long. Figure 1 5 shows a situation in which the wave unction o the electron extends across a physical barrier giving the electron a fnite possibility o
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it existing there ( ignoring any modifcation o the probability curve by the presence o the barrier) .
probability
Quantum tunnelling is responsible or the relatively low temperature usion that occurs in main sequence stars such as the S un. The repulsive orces between a pair o protons that are to use means that they require kinetic energies o j ust over 1 MeV this requires them to be at a temperature o around 1 0 1 0 K. This is much higher than the temperature o the core o the S un ( which is around 3 1 0 7 K) . As the result o the high pressures and quantum tunnelling there is a small chance that hydrogen atoms can use at a temperature below that expected. B ecause o the immense numbers o atoms in the S un, even with a very low probability o quantum tunnelling usion occurring, there is still a great deal going on. In the case o the S un there is estimated to be in excess o our million tonnes o hydrogen using in this way in every second.
physical barrier
fnite probability displacement
Figure 15 Quantum tunnelling to pass through a physical barrier.
The scanning tunnelling microscope ( STM) , invented in 1 981 by IB M Zurich, has revolutionized the study o material suraces and has been used to manipulate individual strands o D NA thus oering the potential to repair genetic damage. STMs use the currents generated when electrons tunnel into a surace in order to map out the structure o the surace. Quantum tunnelling is currently used in quantum tunnelling composites ( QTC s) . QTC s are the basis o touch screen technology. This technology has applications in smartphones, computer tablets, cameras and monitors. The entanglement o quantum particles is starting to bear ruit in photon teleportation, quantum cryptography and computing.
Worked example The graph shows the variation with distance x o the wave unction o an electron at a particular instant o time. The electron is confned within a region o length 2 .0 1 0 1 0 m.
Solution a) The wave unction is a property o the electron everywhere in space. The square o the wave unction is proportional to the probability o fnding the electron somewhere. b) ( i) the ( de B roglie) wavelength o the electron is
0
0.0
0.5
1.0
1.5
x/10 - 10 m 2.0
2 1 0-10 _______ = 3 .3 1 0 1 1 m 6
and so h 6 .6 3 1 0 p = ( __ = _________ = ) 2.0 1 0 23 N s 3 .3 1 0 - 34
1 1
( ii) The electron is confned to 2 .0 1 0 1 0 m, this means the uncertainty in its position ( x) = 2 .0 1 0 1 0 m. a) S tate what is meant by the wave function o an electron. b) Using data rom the graph estimate, or this electron: ( i) its momentum ( ii) the uncertainty in its momentum.
Using the uncertainty principle this means that the uncertainty in momentum will be at least: 6.63 1 0 - 34 h p = _ = __ = 4 2 .0 1 0 - 1 0 4 x
(
)
2 .6 1 0 25 N s
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12.2 Nuclear physics Understanding Rutherord scattering and nuclear radius Nuclear energy levels
Applications and skills Describing a scattering experiment including
The neutrino The law o radioactive decay and the decay
constant
Nature of science Why we need particle accelerators When Rutherord, Geiger and Marsden perormed their scattering experiments they needed to use naturally occurring sources o alpha particles with relatively low energies o between 3 and 7 MeV. This is not enough energy or them to penetrate the electrostatic potential energy barrier around the nucleus. Although there are cosmic rays reaching the Earth with energies a million times greater than can be produced in a particle accelerator (10 2 1 eV compared with 10 1 5 eV on Earth) , these are unpredictable and cannot be used. In CERNs large hadron collider (LHC) the energy available has been raised to 7 TeV. By using particle accelerators, detectors and sophisticated computers the number o known particles has increased rom the proton, neutron and electron to the enormous number possible in the standard model. I our knowledge o the atom is to progress still urther, particle accelerators are the most likely way orward. Plans are already in place to construct a very large hadron collider (VLHC) with a 240 km circumerence chamber and beam energy o at least 50 TeV.
location o minimum intensity or the diracted particles based on their de Broglie wavelength Explaining deviations rom Rutherord scattering in high energy experiments Describing experimental evidence or nuclear energy levels Solving problems involving the radioactive decay law or arbitrary time intervals Explaining the methods or measuring short and long hal-lives
Equations relationship between radius o nucleus and ___ 1
nucleon number: R = R0 A 3 decay equation or number o nuclei at time t: N = N0 e - t decay equation or activity at time t: A = N0 e - t angle o electron diraction frst minimum: sin ____ D
Introduction In S ub- topic 7.3 we looked at the Rutherford model of the atom with a nucleus surrounded by orbiting electrons. We will now look more closely at the implications of the alpha scattering experiment and see how similar experiments have provided a much better understanding of the nucleus. We will then consider a more mathematical approach to the radioactive decay of nuclei.
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Rutherford scattering and the nuclear radius In S ub- topic 7.3 we saw that the main results o the alpha scattering experiment were that:
most o the alpha particles passed through the gold lea undefected
some alpha particles were defected through very wide angles
some alpha particles rebounded in the opposite direction.
The interpretations o these results were:
most o the atom is empty space
the atom contains small dense regions o electric charge
these small dense regions are positively charged.
We will now look at the analysis o this ground- breaking experiment in some detail.
The method of closest approach Alpha particles that backscatter are those colliding head- on with a gold nucleus. As only about 1 in 8000 alpha particles are scattered through large angles it must mean that the probability o a head- on collision is very small and that the nucleus occupies a very small portion o the total atomic volume ... but how much? maximum electrical potential energy
-particle
Au (79e)
path of alpha particle
rc position of closest approach
Figure 1 Method of closest approach.
Figure 1 shows an alpha particle that is incident head- on with a gold nucleus. As the alpha particle becomes closer to the nucleus its kinetic energy alls and its electrical potential energy increases. When the alpha particle is at its closest to the nucleus, its kinetic energy has allen to zero and it has momentarily stopped moving. Taking an alpha particle o kinetic energy E at the position o closest approach ( = distance rc rom the nucleus) we can equate its kinetic energy to the electrical potential energy giving: kZe 2 e E = _ rc where k is the C oulomb constant, Z is the proton number o gold ( making Ze the charge on the gold nucleus) and 2 e represents the alpha particle charge.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) S o for an alpha particle k2 Ze2 rc = _ E In Rutherfords experiment E = 7.68 MeV and Z = 79, this gives a value for rc of 8.99 1 0 9 2 79 ( 1 .60 1 0 - 1 9 ) 2 rc = ____ = 2 .96 1 0 1 4 m ( 7.68 1 0 6 1 .60 1 0 - 1 9
Note You should not conuse the Fermi radius (which is related to a nucleus) with the Bohr radius (which is related to an atom) : there is a actor o 1 0 5 diference in these values.
This derivation is an approximation because we have treated the gold nucleus as being a point mass. If the alpha particle were to have penetrated the nucleus, then the C oulomb force would not have been applicable as we now know that the strong nuclear force is dominant within the nucleus. Rutherford went on to obtain an expression for the number of alpha particles scattered through a variety of angles and he also arrived at values that were in agreement with the upper limit of the gold nucleus being approximately 3 1 0 1 4 m. However, further experiments using more energetic alpha particles have approached the nucleus closer than this value. At this separation the strong nuclear force becomes dominant and the alpha particle must have penetrated the nucleus. As the volume V of a nucleus must be proportional to the number of its nucleons, we would expect V A ( where A is the nucleon number) and __ so the nuclear radius R would be expected to be A . This gives 1 3
1 _
R = R0 A 3 Here R0 is called the Fermi radius and has a value of 1 .2 1 0 1 5 m and is measured experimentally.
Nuclear density If we imagine the nucleus to be spherical, it follows that its volume can be calculated using the following equation: 4 4 AR 3 V = _ R 3 = _ 0 3 3 The density of nuclear material will be given by M Au 3u = _= _ = _3 4 __ 3 V 4R 0 AR 0 3 where u is the uniform atomic mass and Au is the total mass of a nucleus of nucleon number A. As each of the quantities in the equation is a constant, it implies that the density of any nucleus is independent of the number of nucleons in the nucleus. S ubstituting values into this equation gives: 3 1 .66 1 0 - 27 = __ = 2 . 3 1 0 1 7 kg m 3 4 ( 1 .2 1 0 - 1 5 ) 3 This is an incredibly dense material a volume of 1 cm 3 would have a mass of 2 00 000 tonne! In nature the only obj ect that has a nuclear density of this value is a neutron star a type of stellar remnant resulting from the gravitational collapse of a massive star, which is composed almost entirely of neutrons.
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Deviations rom Rutherord scattering
The method o closest approach gives an approximation o the size o a nucleus. More reliable values or the size o a nucleus can be ound using electron diraction.
repulsive 0 attractive
The scattering experiments perormed by Rutherord, Geiger, and Marsden were limited by the energies o the alpha parties emitted by the radioactive sources available to them. When their experiments are repeated using more energetic, accelerated alpha particles it is ound that, at these higher energies, the Rutherord scattering relationship does not agree with experimental results. At higher energies the alpha particles were able to approach the target nucleus so closely that the strong nuclear attractive orce overcomes the electrostatic repulsion. Figure 2 shows how the strong nuclear orce and the repulsive coulomb orce ( between protons) vary with distance.
orce between nucleons/10 - 4 N
+3.0
coulomb repulsion between two protons 1.0
2.0
3.0
nucleon separation/m
strong nuclear orce between two nucleons
-3.0 equilibrium position or protons sum o two orces = zero
Figure 2 Variation o the strong nuclear orce and coulomb orce with distance.
Electron difraction As electrons are leptons ( and not hadrons) they are not aected by the strong nuclear orce but are aected by the charge distribution o the nucleus. High-energy electrons have a short de B roglie wavelength o the order o 1 0 1 5 m. As this is also the order o magnitude o the size o a nucleus, it means that diraction analogous to that observed with light incident on a narrow slit or small obj ect can be observed. For light incident on a small circular obj ect o diameter D, the angle that the frst diraction minimum makes with the straight- through position ( = 0) is given by sin _ D
where is the wavelength o the light
The elastic scattering o high energy electrons by a nucleus produces a similar eect. With the arrangement shown in fgure 3 ( a) , the intensity o the diracted beam is seen to be a maximum in the straight-through position, alling to a minimum beore slightly increasing again. The minimum diers rom light because it never reaches zero or scattered electrons. Again, the relationship can be approximated by
Note With angles greater than 10 the small angle approximation that sin cannot be applied to the electron scattering.
sin _ D Here D is the nuclear diameter and is the de B roglie wavelength o the electrons. electron intensity
electron beam detector thin metal sample in a vacuum amplifer and meter (a) outline o experiment
min angle o diraction (b) typical results
Figure 3 Experimental arrangement and results.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) To achieve an appropriate de B roglie wavelength the electrons used in this scattering experiment need energies in the region o 400 MeV. At these energies, electrons are travelling close enough to the speed o light to mean that relativistic corrections should be applied to both the electron momentum and the associated wavelengths. With the electron rest energy o approximately 0.5 MeV, the total energy can be taken as 400 MeV without any serious error consideration. The wavelength o an electron is given by hc = _ E For an electron with energy 400 MeV 6.63 1 0 - 34 3 .00 1 0 8 = ___ = 3 .1 1 0 - 1 5 m 400 1 0 6 1 .60 1 0 - 1 9 This is the same order o magnitude as the size o a nucleus.
Worked example The nuclear radius o calcium- 40 has an accepted value o 4.5 4 m. It is being investigated using a beam o electrons o energies 42 0 MeV. a) How closely does the value o the radius o calcium-40, obtained using the relationship __ = R 0 A , agree with the accepted value? 1 3
b) C alculate the de B roglie wavelength o an electron having energy 42 0 MeV. c) Determine the angle that the frst minimum in the diraction pattern makes with the straightthrough direction. d) Explain why 5 0 MeV electrons would be unlikely to provide reliable results in this experiment.
Solution a) A is the nucleon number that, or calcium- 40, ___ __ 3 is 40. Thus A = 40 = 3 . 42 which gives a value or R o 3 . 42 R 0 = 3 .42 1 .2 m = 4.1 0 m. 1 3
There is an error o 0. 44 m between the accepted value and that obtained using
the relationship given ( 4.5 4 4. 1 0) . As a 0 . 44 = 9.7% 1 0% . percentage this is = ____ 4. 5 4 hc 6. 63 1 0 - 34 3 . 00 1 0 8 b) = _ = ___ E 42 0 1 0 6 1 .60 1 0 - 1 9 = 2 . 9 1 0-15 m c) Using sin __ sin - 1 ( __ ) D D
Thus, as the radius is 4.5 4 m, the diameter is 9.08 m giving 2 . 9 1 0-15 sin - 1 __ = 1 8. 6 9. 08 1 0 - 1 5 d) As the de B roglie wavelength is given by hc = __ , 5 0 MeV electrons would have a de E 0 B roglie wavelength o 42 = 8.4 times greater 50 than that o 420 MeV electrons. This means that
___
(
2 . 4 1 0-14 sin - 1 __ 9.08 1 0 - 1 5
)
This is not calculable as sin cannot be greater than one. The de B roglie wavelength is too long to be diracted by a calcium nucleus.
Using electrons of higher energies When electrons o much greater energies than 42 0 MeV are used in scattering experiments, something very dierent happens. The collisions are no longer elastic ( the bombarding electrons lose kinetic energy) . This energy is converted into mass as several mesons are emitted rom the nucleus. At still higher energies, the electrons penetrate deeper into the nucleus and scatter o the quarks within protons and neutrons this is
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known as deep inelastic scattering and provides direct evidence for the quark model of nucleons.
Energy levels in the nucleus Much of our evidence for the nucleus having energy levels comes from the radioactive decay of nuclides. The emission of gamma radiation is analogous to the emission of photons by electrons undergoing energy level transitions. The emission of alpha or beta particles by radioactive parent nuclei often leaves the daughter nucleus in an excited state. The daughter nucleus then emits one or more gamma ray photons as it reaches the ground state. Figure 4 shows some of the decay routes of americium-2 41 . E ach nucleus emits an alpha particle having one of a number of possible energies ( three of which are shown) to become a nucleus of neptunium- 2 3 7. D epending on the energy of the emitted alpha particle, the neptunium nucleus can be in the ground state or an excited state. From this it will decay into the ground state by emitting a single gamma photon or, when it decays in two steps, two photons. From the differences in the energies of the alpha particles we can see that the energy level E1 will be (5 .5 45 5 .486) = 0.05 9 MeV above the ground state (E0 ) and energy level E2 will be (5 .5 45 - 5 .443) = 0.1 02 MeV above the ground state. From the differences in the energy levels we can calculate the energies of each of the three gamma ray photons that could be emitted they will be 0.1 02 MeV, 0.05 9 MeV, and 0.043 MeV. 241 Am 95
5.443 MeV 5.486 MeV
E2
5.545 MeV
E1
237 Np 93
E0 ground state
Figure 4 Decay of americium-241.
As we see from the decay of americium, the energies of the alpha particles are also quantized and provide evidence for the nucleus having energy levels. The mechanism by which the alpha particle leaves the nucleus is more complex than that producing the emission of the gamma ray photons. Alpha particles form as clusters of two protons and two neutrons inside the nucleus well before they are emitted as alpha particles. The nucleons are in random motion within the nucleus but their kinetic energies are much smaller than those needed to escape from the nucleus. This is because the strong nuclear force provides a potential energy barrier which the alpha particle needs to overcome before it can escape from the nucleus ( when the electrostatic repulsion will ensure that it accelerates away from the nucleus) . From a classical
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) mechanics point o view the alpha particle simply should not leave the nucleus ( see fgure 5 ) . From the quantum mechanical standpoint, the wave unction or the alpha particle is not localized to the nucleus and allows an overlap with the potential energy barrier provided by the strong nuclear orce. This means that there is a fnite but very small probability o observing the alpha particle outside the nucleus. Although the probability is extremely small, some alpha particles will tunnel out o the nucleus. Experimentally it is ound that, with a higher potential barrier and greater thickness to cross, a nucleus will have a longer lietime. This explains the very long hal- lives o uranium and polonium. The Russian- born American physicist, George Gamow, was the frst person to describe alpha decay in terms o quantum tunnelling ( discussed in S ub- topic 1 2 . 1 ) . When the wave unction is at its maximum the probability o tunnelling is greatest, meaning that alpha particles with specifc energies are most likely to be emitted. energy energy
kinetic energy o uppermost cluster
wave unction outside nucleus has less momentum and longer wavelength
alpha particle cluster with insufcient kinetic energy to penetrate the potential energy barrier
distance
distance
wave unction in potential energy well has large kinetic energy (and momentum) and short wavelength
edge o nucleus where nucleons are tightly bound by the strong nuclear orce - inside the nucleus there are nucleons with a variety o kinetic energy
Figure 5 Classical mechanics view o alpha decay.
Figure 6 Quantum tunnelling view o alpha decay.
Negative beta decay We looked at beta decay in S ub- topic 7.1 and returned to it in S ubtopic 7.3 when it was explained that an anti- neutrino accompanies the electron emitted in negative beta decay. The frst theory o beta decay was proposed in 1 93 4 by Fermi; at this time the existence o quarks was unknown and neutrinos were hypothetical. Experiments show that beta particles emitted by a source have a continuous energy spectrum and are not o discrete single energy as are alpha particles and gamma photons. Figure 7 shows a typical negative beta- energy spectrum.
intensity
energy spectrum o beta decay electrons rom 210 Bi
0
0.2
0.4 0.6 0.8 1.0 kinetic energy, MeV
Figure 7 Negative beta-energy spectrum.
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1.2
Possible explanations or this spectrum were that massenergy and momentum were not conserved in beta decay. These were very unlikely solutions since both o these principles are considered to be undamental to physics. Pauli suggested that, i a third particle was to be emitted in the decay, not only would this solve the massenergy and momentum problems but it would also allow spin angular momentum to be conserved in the emission. The emission (o what has proved to be) an electron antineutrino meant that or a particular nucleus the energy would be shared between the electron (the beta particle) and the antineutrino.
1 2 . 2 N U C L E A R P H YS I C S
TOK Awe and wonder or turn-of? Rumour has it that Murray Gell-Mann, when searching or a name to call what is now known as the quark, had a epiphany upon seeing the word quark in the novel Finnegans Wake by the Irish novelist James Joyce. The poet, John Updike, on reading about neutrinos chose to write the ollowing:
Cosmic Gall Neutrinos, they are very small. They have no charge and have no mass And do not interact at all. The earth is just a silly ball To them, through which they simply pass, Like dustmaids down a draughty hall Or photons through a sheet o glass. They snub the most exquisite gas, Ignore the most substantial wall, Cold shoulder steel and sounding brass, Insult the stallion in his stall, And scorning barriers o class, Infltrate you and me! Like tall And painless guillotines, they all Down through our heads into the grass. At night they enter at Nepal And pierce the lover and his lass From underneath the bed you call It wonderul; I call it crass. Telephone Poles and Other Poems, John Updike, (Knop, 1960)
Figure 8 Super-Kamioka Neutrino Detection Experiment Mount Kamioka near Hida, Japan.
Updike clearly was not a believer in the neutrino. Does he have a strong case or his disbelie? Is the concept o something as challenging as the neutrino inspirational or is it too anciul to be credible?
The law of radioactive decay As we saw in S ub-topic 7.1 , radioactive decay is a random and unp redictable process. There is no way of telling which nucleus in a sample of material will decay next. What we do know is that the more radioactive nuclei present, the greater the probability of some decaying. With a sample of many millions of nuclei the rules of statistics can be applied with a virtual certainty. The probability that an individual nucleus will decay in a given time interval (of one second, one minute, one hour, etc.) is known as the decay constant, . The units for are time 1 ( s 1 , minute 1 , h 1 , etc. ) . The activity of a sample A is the number of nuclei decaying in a second it is measured in becquerel (B q) . In a sample of N undecayed
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) nuclei, the activity will be equal to the number of nuclei present multiplied by the probability that one will decay in a second. In equation form: A = N N will decrease with time; this relationship is often written with a minus sign ( although this is not the case on the IB D iploma Programme physics syllabus)
Note Again, we can make use iteration to avoid using calculus when solving the decay equation. The ollowing fow chart shows an algorithm or doing this:
Using calculus notation, this is can be written as: dN _ = - N dt This is the relationship that gives an exponential decay and is analogous to capacitor discharge as seen in S ub- topic 1 1 .3 with the constant being analogous to - 1 .
n=0 choose t, tn , Nn ,
In general, when the rate of change of a quantity is proportional to the amount of the quantity left to change, an exponential relationship will always be obtained. We will now show this for radioactive decay the process will involve integral calculus and so you will not be tested on this in examinations.
Nn = - Nn t N n+1 = N n + Nn
A = - N
n
n+ 1
Rearranging the equation dN _ = - N dt
t n+1 = t n + t
gives enough increments? yes fnish
no
dN _ = - dt N this can be integrated from time t = 0 to time t = t when the number of undecayed nuclei will fall from N = N0 to N = N N
N0
t
dN _ = N
dt 0
Integrating this gives [ln N] NN = - t 0
So ln N - ln N0 = - t or
( )
N ln _ = - t N0 Raising both sides to the power of e gives N _ = e - t or N = N0 e - t N0 As A is proportional to N, this equation can be written in terms of the activity to give A = A 0 e - t Here A 0 is the activity of sample of radioactive material at time t = 0 and this can also be written as A = N0 e - t
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1 2 . 2 N U C L E A R P H YS I C S
Worked example 1 mol of a substance contains 6.02 1 0 23 particles so this quantity amounts to
Radium-2 2 6 emits alpha particles. The decay constant is 1 .3 5 1 0 1 1 s 1 What mass of radium 2 2 6 is needed to give an activity of 2 2 00 B q?
1 .63 1 0 1 4 __ = 2 . 7 1 0 - 1 0 mol 6.02 1 0 23 As a mol of radium 2 2 6 has an approximate mass of 2 2 6 g the sample has a mass of
Solution 2 2 00 A A = N so N = _ = 1 . 63 1 0 1 4 = __ 1 .3 5 1 0 - 1 1
2 .7 1 0 - 1 0 2 2 6 1 0 - 3 kg = 6.1 1 0 - 1 1 kg
Decay constant and half-life We dealt with integral numbers of half-lives in S ub- topic 7.1 . Now we consider cases where the number of half-lives is not a whole number. The half-life is the time that it takes for the number of radioactive nuclei N to halve. So, in this time, N falls from N0 to __ . 2 0
- t N S ubstituting these values into N = N0 e - t gives __ = N0 e where t1 /2 is 2 the half-life. 0
1 __ 2
When we rearrange this and take logs to base e we get
( ) N0 __
( )
2 1 ln _ = ln _ = - t1 /2 N0 2 1 C alculating log to base e of __ gives 2
- 0.693 0.693 t1 /2 = _ = _ -
Worked example A laboratory prepares a 1 0 g sample of caesium- 1 3 4. The half- life of caesium- 1 3 4 is approximately 2 . 1 years. a) D etermine, in s 1 , the decay constant for this isotope of caesium. b) C alculate the initial activity of the sample. c) C alculate the activity of the sample after 1 0.0 years.
Solution 0.693 0.693 a) = _ = ___ t 1 /2 2 .1 3 65 2 4 3 600 = 1 .05 1 0 - 8 s - 1 ( = 0.3 3 y- 1 ) 1 34
b) 1 mol of C s has a mass of approximately 1 3 4 g so 1 0 g 10 10 comprises of _______ mol or = 7. 5 1 0 - 8 mol. 1 34 -6
This means that there are 7.5 1 0 - 8 6. 02 1 0 23 = 4. 49 1 0 1 6 atoms of C s A = N = 1 .05 1 0 - 8 4. 49 1 0 1 6 = 4. 7 1 0 8 B q c) A = A 0 e - t = 4.7 1 0 8 e - 0.33 1 0 ( working in years) A = 1 .7 1 0 7 B q
Note A second way to tackle this is to calculate the number of half10 lives that have elapsed = ______ 2.1 = 4.76 The amount that will be left is then 0.5 4.7 6 = 0.037, the activity will then be 0.037 4.7 10 8 = 1.7 1 0 7 Bq
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L )
Investigate! Measuring short half-lives The GM tube was discussed in Sub-topic 7.1 . This is a very important instrument used to measure hal-lives o radioactive materials. For those with airly short hal-lives (rom a ew seconds to a ew hours) it is a straightorward process to measure the count rate using a GM tube and counter or a data logger.
Measure the background count rate with your apparatus with no radioactive sources in the vicinity.
Position the source close to the window o the G- M tube so that almost none o the radiation is absorbed by the air.
Take readings o the count rate at appropriate time intervals until the count rate is the same as the background count.
S ubtract the background count rate rom your readings to give the corrected count rate ( R) ; assuming that the radiation is emitted equally in all directions, the count rate shown on the counter will be proportional to the activity o the source.
Plot a graph o the natural log o the corrected count rate, ln( R) , against time.
As R A we can write R = R 0 e - t Taking natural logs o this gives ln R = ln R0 - t So a graph o ln R ( on the y- axis) against t will be o gradient - and intercept on the ln R axis o ln R0 as shown in fgure 9.
Note the notation or the unit o a log quantity logs have no units but R does so the unit is bracketed to R. The corrected count rate is in counts per second, as count has no unit this is equivalent to s 1 .
What is the advantage o plotting this log graph when compared with plotting count rate against time?
How do you use the gradient to calculate the hal- lie o the radioactive nuclide?
C ompare your value with the accepted value to give you an idea o the uncertainty.
What is the dierence between the activity o source and the count rate?
In (R/s -1 ) ln(R0 )
t/s
Figure 9 Graph of natural log of the corrected count rate against time.
Measuring long half-lives S ome nuclides have very long hal- lives, or example uranium- 2 3 8 has one o j ust under 4.5 billion years. When a radioactive nuclide has a hal-lie that is long compared to the time interval over which radioactive decay observations are possible, there is no apparent rate o decay and it is not possible to measure the hal- lie in the manner suggested using a GM tube.
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In these cases a pure sample o the nuclide in a known chemical orm needs to be separated, its mass measured and then a count rate taken. From this reading the activity can be calculated by multiplying the count rate by the ratio: area o sphere o radius equal to the position o the G- M tube window _______ area o G- M tube window The decay constant is then determined rom the mass o the specimen using the method shown in the frst worked example in this section.
Investigate! Decay in height of water column setting so, i the clip is not ully undone, you will need to count how many turns you make rom the clip being ully tight.
Perspex tube with open top
As the water level in the long tube reaches a previously chosen point near the top o the tube, start measuring the height at regular time intervals you will need to perorm a trial experiment to allow you to j udge what a sensible time interval should be one that will give you a minimum o six points on your graph.
Record your results in a table.
Repeat the readings twice more always starting the clock when the water level is at the same mark on the scale.
Find the average o the readings or each time interval.
Plot graphs o height against time and the natural log o height against time.
From each o these graphs calculate the hallie and decay constant o the water.
D etermine which o the two values is the most reliable.
Consider how this experiment models radioactive decay and in which ways it is dierent.
height metre ruler
adjustable clip
capillary tube rubber tube
Figure 10 Decay in height of a water column.
This is an analogue o radioactive decay using a long vertical tube o water, which is connected to a capillary tube.
Make sure the clip is closed, then fll the long tube almost to the brim with water. Undo the clip completely to allow the water to ow through the capillary tube you will need to repeat the experiment with the same
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L )
Questions 1
When light is incident on a metal surace, electrons may be ej ected. The ollowing graph shows the variation with requency f o the maximum kinetic energy Emax o the ej ected electrons.
( i) S tate the connection between photon energy and the energy o the emitted electron. ( ii) Using your answer to ( i) calculate the work unction o the surace o the metal plate.
Emax
0
( 8 marks)
3 0
f
a) Sketch the graph obtained or the variation with requency o the maximum kinetic energy o the emitted electrons when a dierent metal with a lower threshold requency is used. b) Explain why you have drawn the graph in this way. ( 4 marks)
2
(IB) In order to demonstrate the photoelectric eect, the apparatus shown below is used.
a) S tate one aspect o the spectrum o atomic hydrogen that B ohrs model did not explain. B ohr proposed that the electron could only have certain stable orbits. These orbits are specifed by the relation nh mvr = _ 2
with n = 1 , 2 , 3
where m is the mass o the electron, v its speed, r the radius o the orbit and h the Planck constant. This is sometimes known as B ohrs frst assumption.
B y using Newtons second law and C oulombs law in combination with the frst assumption, it can be shown that
glass tube vacuum metal plate V
Monochromatic light is incident on the metal plate. The potentiometer is adj usted to give the minimum voltage at which there is zero reading on the microammeter. a) State and explain what change, i any, will occur in the reading o the microammeter when: ( i) the intensity o the incident light is increased but the requency remains unchanged ( ii) the requency o the light is increased at constant intensity. b) For light o wavelength 5 40 nm, the minimum reading on the voltmeter or zero current is 1 .9 V.
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In 1 91 3 Niels Bohr developed a model o the hydrogen atom which successully explained many aspects o the spectrum o atomic hydrogen.
b) S tate a second assumption proposed by B ohr.
A
monochromatic light
(IB)
n2h2 r= _ 4 2 mke2
1 . where k = _ 4 0
It can also be shown that the total energy En o the electron in a stable orbit is given by ke2 En = _. 2r c) Using these two expressions, deduce that the total energy En may be given as K where K is a constant. En = _2 n d) S tate and explain what physical quantity is represented by the constant K. e) O utline how the S chrdinger model o the hydrogen atom leads to the concept o energy levels. ( 1 0 marks)
QUESTION S 4
d) The radius ( in metre) R o a nucleus with nucleon number A is given by
(IB) A beam o electrons is incident normally to the plane o a narrow slit as shown below.
1 _
R = 1 .2 1 0 - 1 5 A 3 . ( i)
slit
beam of electrons
x
State in terms o the unifed atomic mass unit u, the approximate mass o a nucleus o mass number A.
( ii) The volume o a sphere o radius R 4 R . D educe that the is given by v = ____ 3 density o all nuclei is approximately 2 1 0 1 7 kg m 3 . 3
( 8 marks)
The slit has width x equal to 0. 01 mm. As an electron passes through the slit, there is an uncertainty x in its position. a) C alculate the minimum uncertainty p in the momentum o the electron. b) Suggest, by reerence to the original direction o the electron beam, the direction o the component o the momentum that has the uncertainty p. ( 3 marks)
6
(IB) A nucleus o the nuclide xenon, Xe- 1 3 1 , is produced when a nucleus o the radioactive nuclide iodine, I-1 3 1 decays. a) Explain the term nuclide. b) C omplete the nuclear reaction equation or this decay. 1 3 1 I 1 3 1 Xe + - + 54
5
(IB) An - particle approaches a nucleus o palladium. The initial kinetic energy o the - particle is 3 .8 MeV. The particle is brought to rest at point P, a distance d rom the centre o the palladium nucleus. It then reverses its incident path. palladium nucleus
-particle P d
a) C alculate the value, in j oules, o the electric potential energy o the - particle at point P. Explain your working. b) The proton number o palladium is 46. C alculate the distance d.
c) The activity A o a reshly prepared sample o I- 1 3 1 is 6.4 1 0 5 B q and its hal-lie is 8.0 days. ( i)
Sketch a graph to show the variation o the activity o this sample over a time o 25 days.
( ii) D etermine the decay constant o the isotope I- 1 3 1 ( in day - 1 ) . The sample is to be used to treat a growth in the thyroid o a patient. The isotope should not be used until its activity is equal to 0.5 1 0 5 B q. ( iii) C alculate the time it takes or the activity o a reshly prepared sample to be reduced to an activity o 0.5 1 0 5 B q ( 1 1 marks)
c) Gold has a proton number o 79. Explain whether the distance o closest approach o this - particle to a gold nucleus would be greater or smaller than your answer in ( b) .
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) 7
(IB)
( iii) Use the following data to determine the maximum energy, in J, of the
a) A stable isotope of argon has a nucleon number of 3 6 and a radioactive isotope of argon has a nucleon number of 3 9.
particle in the decay of a sample of argon- 3 9.
( i)
S tate what is meant by a nucleon.
Mass of argon-39 nucleus = 38.96431 u
( ii)
O utline the quark structure of nucleons.
Mass of K nucleus = 3 8. 963 70 u
( iii)
c) The half- life of argon- 3 9 is 2 70 years. ( i)
S uggest, in terms of the number of nucleons and the forces between them, why argon- 3 6 is stable and argon- 3 9 is radioactive.
( ii) Explain how you would calculate the half-life using the quantities you have stated in ( i) .
b) Argon- 3 9 undergoes decay to an isotope of potassium ( K) . The nuclear reaction equation for this decay is
39 18
State what quantities you would measure to determine the half-life of argon- 3 9.
( 2 1 marks)
Ar K + - + x
( i)
S tate the proton number and the nucleon number of the potassium nucleus and identify the particle x.
( ii)
The existence of the particle x was postulated some years before it was actually detected. Explain the reason, based on the nature of energy spectra, for postulating its existence.
8
(IB) a) O utline a method for the measurement of the half-life of a radioactive isotope having a half- life of approximately 1 0 9 years. b) A radioactive isotope has a half-life T1 /2 . D etermine the fraction of this isotope that remains in a particular sample of the isotope after a time of 1 .6 T1 /2 . ( 5 marks)
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A RELATI VI TY Introduction Earlier we described the rules of classical mechanics as developed by Newton and others. These rules provided the basis for physics for 3 00 years. Ultimately, however, Newtonian and Galilean mechanics struggles: it cannot deal with things that move very fast or with obj ects that are
very small. The insights of Einstein and others around the beginning of the twentieth century enabled physicists to change their understanding of time and space this type of change is called a paradigm shift. This topic describes our present view of what we now called spacetime.
A.1 The beginnings of relativity Understandings Reerence rames Galilean relativity and Newtons postulates
concerning time and space Maxwell and the constancy o the speed o light Forces on a charge or current
Nature of science Einsteins great insight was to realize that the speed o light is constant or all inertial observers. This has enormous consequences or our understanding o space and time. A paradigm shit occurred and the Newtonian view o time was overturned. There are many other examples o paradigm shit in science but perhaps none quite as proound as this.
Applications and skills Using the Galilean transormation equations Determining whether a orce on a charge or
current is electric or magnetic in a given rame o reerence Determining the nature o the felds observed by dierent observers
Equations Galilean transormation equations:
x' = x - vt u' = u - v
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Reference frames A reerence rame allows us to reer to the position o a particle. Reerence rames consist o an origin together with a set o axes. In this course we generally use the Cartesian reerence rame, in which position is defned using three distances measured along axes that are at 90 to each other. The axes o a three-dimensional graph make up a Cartesian reerence rame. O ther rames are, o course, available. Sailors use latitude and longitude; these together with the distance o an obj ect rom the centre o the Earth constitute a dierent reerence rame. Astronomers usually use angles when defning the position o a star that is being observed; they only need two angles because the distance to the star ( or observational purposes) is irrelevant. For some o the rames o reerence in use, one or more o Newtons laws o motion do not hold. Our own rame on the surace o a rotating planet shows this straight away. Everything o-planet appears to be spinning around us. An object that is at rest relative to us on planet Earth is not moving at a constant velocity at all. This has consequences that we have already seen in earlier topics; the fctitious centriugal orce and the Coriolis orce used to explain the movement o weather systems are two cases in point. For Newtons frst law to be valid we need to be careul about the nature o the reerence rame in which we use the law. We defne an inertial frame of reference as a rame in which an object obeys Newtons frst law: so that it travels at a constant velocity because no external orce acts on it. Do inertial rames exist? The best way to fnd one is to take a spaceship out into deep space, well away rom the gravitational eects o planets and stars, and then turn o the engines. No orces act rom outside or inside the spacecrat and this will be a true inertial rame o reerence.
Galilean relativity and Newtons postulates E ven though we have to take some trouble to reach one, there are an infnite number o inertial rames o reerence in the universe and there are a number o ways in which we can move between them. y
y'
y
The frst obvious way is to step sideways rom one rame to another. This is known as a translation ( fgure 1 ( a) ) .
The set o axes can be rotated to orm another set. This is known as a rotation (fgure 1 (b) ) . We shall not consider these in detail in this course.
O ne rame can move relative to another rame with a constant relative velocity. This is known as a boost. ( fgure 1 ( c) )
y'
x'
X
x'
x'
x (a) y
x (b)
x
later time t
y'
t=0 vt
v x'
x x x = vt + x'
x'
(c)
Figure 1 Translations, rotations, and boosts.
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It is easy to see that i an object moves with constant velocity in one reerence rame then under any o these three conditions it will be measured as having a constant (but dierent) velocity in the other reerence rame. The principle o relativity is oten associated with Albert Einstein. In act, Galileo was probably the frst person to discuss the principle. He describes how, in a large sailing ship, butteries in a cabin with no windows would be observed to y at random whether the ship were moving at constant velocity or not. An observer in the cabin could not deduce by observing
A . 1 T H E B E G I N N I N G S O F R E L AT I V I T Y
the butteries whether the ship was moving. And the butteries would certainly not be pinned against the back wall o the cabin. This principle o relativity as described by Galileo tells us about the nature o our universe.
The translation rule is equivalent to saying that there is no special place in the universe; position is relative.
The rotation rule says that there is no special direction; direction is relative.
The boost rule says that there is no special velocity. Stationary is not an absolute condition and any obj ect can be described as stationary with respect to another obj ect. ( As we shall see, the disagreement between Galileos ideas and those o Einstein is crucial here.)
Thinking, or a translation, in terms only o the x-direction (that is, in one dimension) , i the distance between the origins o two inertial rames S' and S is X then a position x in S is related to position x' in rame S' by x' = x - X For a boost, i one inertial rame moves relative to the other by a constant relative velocity v along the x-axis, then the distance between the origins o the reerence rames must be changing by v every second and this distance is vt where t is the time since the rames coincided. Imagine that the origins o the two reerence rames S and S' ( Figure 1 (c) ) were at the same position ( that is, they were coincident) at time t = 0. Technically, we say that clocks in the rames were adjusted so that x = x = 0 when t = t = 0. At a later time t, the origins o S and S' will be separated by vt where v is the velocity o rame S' relative to rame S. Thereore a position x in rame S will be related to position x in S' by x = x' + vt and x' = x - vt The velocities also transorm in an obvious way. I the velocity in S is u and the velocity in S' is u' , then u' = u - v This set o equations that link two reerence rames by their relative velocity are known as the Galilean transformations. Newton developed Galileos ideas urther in his Principa Mathematica by suggesting two important postulates ( a postulate is an assertion or assumption that is not proved and acts as the starting point or a proo) :
Newton treated space and time as fxed and absolute. This is implied in our use o t in both equations above (t' does not appear, only t) . A time interval between two events described in rame S is identical to the time interval between the same two events as described in rame S' . The evidence o our senses seems to confrm this (but remember that we do not travel close to the speed o light in everyday lie) . Newton recognized that two observers in separate inertial rames must make the same observations o the world. In other words, they will both arrive at the same physical laws that describe the universe.
Nature of science The way Newton put it. These postulates were expressed somewhat dierently by Newton in the Principa Mathematica the book ( in Latin) that he wrote to publish some o his discoveries. In translation his postulates were: I. Absolute, true, and mathematical time, o itsel, and rom its own nature, ows equably without relation to anything external . II. Absolute space, in its own nature, without any relation to anything external, remains always similar and immovable. ( Translation by Mottes ( 1 971 ) , revised by C aj orio, University o C aliornia Press.)
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Worked example 1
Solution
In a laboratory an electron travels at a speed o 2 1 0 4 m s 1 relative to the laboratory and another electron travels at 4 1 0 4 m s 1 relative to the laboratory. Use a Galilean transormation to calculate the speed o one electron in the rame o the other when they are travelling a) in opposite directions b) in the same direction.
a) The closing speed o the electrons is 2 1 0 4 ( 4 1 0 4) = 6 1 0 4 m s 1 b) The relative speed o one electron relative to the other is 2 1 0 4 ( 4 1 0 4) = 2 1 0 4 m s 1
TOK
Charges and currents a puzzle
Changing perspectives
The choice o inertial rame can make a radical dierence to the perception o a situation.
In Topic 2 we began by describing Newtons second law o motion in a simple way as: orce = mass acceleration Later we showed that this was better expressed as
C onsider a positively- charged particle moving initially at velocity v some distance away rom and parallel to a wire carrying a current. In the wire the electrons are also moving with speed v. The positive charges in the wire are stationary. We will use two inertial reerence rames in this example. O ne rame is at rest relative to the positive charges in the wire. The other rame is at rest relative to the moving charge q. (a)
(b)
orce = rate o change o momentum A similar change o expression is possible here: Newtons frst law is usually given as a variant o Every object continues in its state o rest or uniorm motion unless net external orces act on it (and you should continue to use this or a similar wording in your own work) . But this is not the only possibility. A succinct and interesting way to express Newtons frst law is as Inertial rames exist. To what extent do concise orms o scientifc laws help or hinder our understanding?
v
v
+q +
v
+q
-
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
v
v
v
+
-
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-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
+
v
Figure 2 Moving charges in reference frames. To an observer at rest ( fgure 2 ( a) ) with respect to the stationary positive charges the situation in the wire seems clear. The numbers o negative and positive charges in the wire are equal and thereore the lone moving charge + q does not experience an electrostatic ( electric) orce. The movement o the electrons in the wire to the right ( a conventional current to the let) gives rise to circular magnetic feld lines centred on the wire and going into the page above the wire. The single moving charge is thereore moving perpendicularly with respect to this feld. Flemings let-hand rule indicates that there is a magnetic orce acting perpendicularly outwards on the charge and thereore the charge will be accelerated in this direction. The stationary observer detects a magnetic repulsive orce acting between the charge and the current-carrying wire. What is the situation rom the point o view o the moving charge q? An observer moving with the charge ( fgure 2 ( b) ) sees the single charge + q as stationary; electrons in the wire that also appear to be stationary, but the positive charges in the wire appear to be moving to the let at speed v. We could imagine that the moving positive charges lead to a
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magnetic feld around the wire j ust as beore but this does not help! To the moving observer the charge + q is stationary relative to the resulting magnetic feld and thereore no magnetic orce should arise. What happens is that ( as we shall see later) the relative movement o the positive charges in the wire leads to a contraction in the spacing o these charges as perceived by the observer moving with + q. There are more positive charges than negative charges per unit length as detected by the moving observer and so there is eectively a net repulsive orce acting on + q ( we shall see how this change arises rom length contraction in a later sub- topic) . Now the observer moving with q explains the orce acting on q as electrostatic in origin, not magnetic as beore. However, both observers report a orce and in both cases the orce acts outwards rom the wire. The physical result is the same even though the explanations dier. A mathematical analysis rom the standpoint o the two inertial rames also confrms that the magnitude o the orce is identical in both cases. Another situation is that o two point electric charges moving in parallel directions at the same speed as each other. This case (fgure 3) is not identical to that o the charge moving near a current-carrying wire. There is no balance o positive and negative charges to complicate matters this time. Again, we consider the situation rom the standpoint o two separate inertial rames: that o an observer stationary relative to the point charges, and that o an observer moving at a dierent velocity relative to the charges. Observations made in the rame o the point charges are easier to understand (fgure 3(a) ) : the observer moving with the charges sees two positive charges that repel and no magnetic attraction between the two. This observer describes the repulsion as purely electrostatic.
v
+q
+q
(a)
stationary, observes a magnetic feld due to both charges moving
+q
+q
(b)
Figure 3 Two point charges moving parallel.
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TOK Validating a paradigm shift The development of Einsteins relativity ideas required a shift in the view that scientists took of the physical rules that govern the universe. How do scientists ensure that the need to shift perspectives is valid?
TOK Maxwells advances Einstein wrote: The precise ormulation o the timespace laws was the work o Maxwell. Imagine his eelings when ...equations he had ormulated proved to him that electromagnetic felds spread in the orm o polarised waves, and at the speed o light!... it took physicists some decades to grasp the ull signifcance o Maxwell's discovery, so bold was the leap that his genius orced upon the conceptions o his ellow workers. (Albert Einstein, Science, May 24, 1940) To what extent do you think this is true?
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For an observer no longer stationary relative to the point charges ( fgure 3 ( b) ) , the situation is changed. From this point o view, the repulsive electric feld is increased ( through relativistic length contraction) . There is also a magnetic feld that was not apparent when the observer was stationary relative to the charges. This is because a moving charge gives rise to a magnetic feld. E ach charge now appears to be moving within the magnetic feld due to the other charge and consequently there is an attraction that the observer describes as magnetic in origin. There is both an increased ( electrostatic) repulsion and a new ( magnetic) attraction compared with the stationary observer rame. Again, a mathematical analysis shows that the orce between the charges is identical or all observer inertial rames o reerence. This is what we expect given that all physical systems observed in inertial rames must obey the same laws. As Einstein himsel said: What led me ... to the special theory o relativity was the conviction that the electromagnetic orce acting on a body in motion in a magnetic feld was nothing else but an electric feld.
Maxwell and electromagnetism In 1 861 James Maxwell established the connection between electrostatics, electromagnetic induction, and the speed o light. He developed our equations that between them describe the whole o electrical and magnetic theory and lead to the recognition that light is a orm o electromagnetic radiation. The our equations incorporate the value o the speed o light travelling in a vacuum ( ree space) in a undamental way. The conclusion that must be drawn rom Maxwells equations is that, i observers in dierent inertial rames make observations o the speed o light then, i they are to agree about physical laws, they must observe identical values or the speed o light. B ut the Galilean transormations predict a dierent result rom this. The Galileo predicts that the speed o light diers in dierent rames by the magnitude o the relative velocity between the rames. S o the inescapable conclusion that ollows rom Maxwell ( same speed o light or all observers) is directly contrary to the assumption o absolute time and absolute space as postulated by Newton, and as embodied in the Galilean transormations. Physics had reached an impasse; it required the genius o Maxwell to recognize the problem. It required another genius, Einstein, to move the subj ect orward again hal a century later.
A . 2 L O R E N T Z T R A N S F O R M AT I O N S
A.2 Lorentz transformations Understanding The two postulates of special relativity Clock synchronization The Lorentz transformations Velocity addition Invariant quantities (spacetime interval, proper
time, proper length, and rest mass) Time dilation Length contraction The muon decay experiment
Nature of science Einsteins theory of relativity stems from two postulates. He deduced the rest of the theory mathematically. This is an example of pure deductive science at work.
Applications and skills Solving problems involving velocity addition Solving problems involving time dilation and
length contraction Solving problems involving the muon decay experiment
Equations Lorentz transformation equations:
1 _____ = ___ v2 1- _ c2 x' = (x - vt) ; x' = (x - vt)
(
)
(
vx ; t' = t - _ vx t' = t - _ 2 2 c uv __ u' = uv 1- _ c2 t = t 0
c
)
The two postulates of special relativity Newtons two postulates o space and time rom Sub- topic A.1 require an absolute time that is the same or all observers in inertial rames. This implies, through the Galilean transormation, that a moving observer will observe a dierent value or the speed o light in ree space rom that o a stationary observer. Maxwells discoveries began to prompt serious questions about the peculiar behaviour o light and the validity o Galilean transormations. In 1 887, two US scientists, Michelson and Morley, using very precise apparatus showed experimentally that any such dierence between moving and stationary observers was below the measurement limits o their experiment. Scientists such as Lorentz, Fitzgerald, and Poincar attempted to explain both Maxwells conclusions and the results o the Michelson-Morley experiment. Einsteins great leap orward was to recognize that i Maxwells our electromagnetic equations were to be true in all inertial rames (which had to be the case) then some modifcations o Newtons postulates were required. He was able to show that some equations that had already been developed by Lorentz as a way o avoiding the problems o the null result o Michelson and Morley could, alternatively and more properly, be derived assuming only Einsteins own modifcations o Newtons postulates.
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R E L AT I VI T Y Einsteins two postulates are:
The laws o physics are the same in all inertial rames o reerence ( Newtons frst postulate too) .
The speed o light in ree space ( a vacuum) is the same in all inertial rames o reerence ( replacing the concept o absolute time and space, Newtons second postulate) .
The Lorentz transformation Earlier we saw that, or boosts, the Galilean transormations lead to the equations x = x X, x= x vt, and u= u v. There is an additional new transormation that arises rom Newtons postulate that time is absolute. This is represented as t'= t. To make the Maxwells equations consistent or all inertial reerence rames, Lorentz introduced a actor given by 1 _____ = _ v2 1- _ c2
v is ( as usual) the speed o one inertial rame relative to the other and c is the speed o light in ree space, is called the Lorentz factor. As v tends to c, tends to infnity. Lorentz then used this actor to modiy the Galilean expressions so that x' = ( x - vt) and t' = t 8 7 6 5 4 3 2 1 0 0
0.2
0.4
0.6 v c
0.8
1
1.2
Figure 1 How the Lorentz factor varies with speed. Figure 1 shows how the Lorentz actor changes with speed. The scale on the x- axis is in the ratio o _vc and shows us that, or speeds up to 2 0% o that o light, the Lorentz actor remains close to 1 . I the Galilean transormations were true or all speeds then the graph would show a horizontal line at value 1 or all values o _vc . S trictly, the equation involving x and x is used to translate rom a position in one rame to the position in the other rame at the same instant in time. In the Galilean transormation, lengths measured within one rame transorm without change into the same length in any other rame because length x is the dierence between two positions, and x = x1 - x2 = ( x1' - vt) - ( x2' - vt) = x1' - x2' = x' .
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In the Lorentz transformation this equality of x and x' is no longer true and x' = ( x - v t) This is a crucial result because it tells us that if one inertial reference frame is moving at a constant velocity relative to another, then an observer in one frame making a length measurement of an obj ect in the other frame will not agree with the measurement made by an observer in the other frame. S pace is no longer absolute. The expression x' = ( x - vt') gives the position x of the obj ect as observed in the moving reference frame. Sometimes we observe the position in the moving frame x' and we need the value of x in the stationary frame. This is given by x = ( x' + vt) and is known as the inverse Lorentz transformation. The Lorentz transformation for time is v t' = t - _2 x c -v with an inverse transformation of t = ( t + ___ x ) c
(
)
2
Like space, time has also lost the property of being absolute. Time measured in different frames differs when there is relative velocity between the frames. Also, terms in x now appear in the time equations and terms in t in the expressions for x' and x. We have assume d so far that there is no relative motion betwee n the frames in the y or z directions. If this is true the n the re will be no relativistic changes in these directions e ither ( this can be proved formally) . The assumption of no motion in directions y and z and the pre vious expressions lead to the complete set of Lore ntz transformations which are here compared with their Galilean e quivale nts.
Galilean
Lorentz
Inverse Lorentz
x' = x - vt
v ct x' = ( x - _ c )
v ct' x = ( x' + _ c )
y' = y
y' = y
y = y'
z' = z
z' = z
z = z'
t' = t
-v x t= t- _ c2 vx ct' = ( ct - _ c )
(
)
-v x t= t+ _ c2 v x' ct = ( ct' + _ c )
(
)
In each case, if v c then 1 and the Lorentz equations reduce to the Galilean equations with which we are already familiar. An additional change in the Lorentz equations in this table is the expression of time using ct rather than t alone ( this gives the time equation the dimensions of distance) . A second change is to include the speed of light, c, twice in the distance equations. These changes make the equations appear more symmetric and help to explain why later in this topic we use axes of ct against x to draw spacetime diagrams.
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Worked examples 1
a) The distance between x1 and x2
C alculate the Lorentz factor for an obj ect travelling at 2 .7 1 0 8 m s 1 .
b) The time interval between t1 and t2 .
Solution
Solution 5 a) = _ = 1 .67 3 From the Lorentz transformation
This speed is 0.9c. 1 1 _____ = __ ________ = 2 .3 = _ 2 ( 1 0.9 2 ) v 1- _ 2 c
2
x2 - x1 = ( ( x2 - x1 ) - v( t2 - t1 ) ) S ubstituting gives
C locks in two frames S and S are adj usted so that when x = x = 0, t = t = 0. Frame S as a speed of 0. 8c relative to S. Event 1 occurs at x1 = 5 0 m, y1 = 0, z1 = 0, and t1 = 0.3 s. Event 1 occurs at x2 = 80 m, y2 = 0, z2 = 0, and t2 = 0.4 s. C alculate, as measured in S,
x2 - x1 = 1 .67 ( ( 80 - 5 0) - 0.8 3 1 0 8 (4 1 0 -7) - (3 1 0 -7) ) x2 - x1 = 1 0 m
(
)
( x2 - x1 ) v_ b) t2 - t1 = ( t2 - t1 ) - _ c c v ( t2 - t1 ) = 1 . 67 ( t2 - t1 ) - _c _ c
(
)
t2 - t1 = 3 0 ns ( 2 s.f. )
Velocity addition An obj ect is moving in frame A with a constant velocity u A. Frame A itself is moving with a constant velocity v with respect to frame B . What do the Lorentz equations have to say about the velocity u B of the obj ect when it is viewed by an observer in frame B ?
v uA
frame B
frame A
Figure 2 Relativistic relative velocities. Galilean relativity has no problem with this, the answer is simple: u B = u A + v. B ut this cannot be correct from a relativistic view. S uppose the rocket in frame A is moving at the speed of light. Then if v is positive u B will exceed c; this is not allowed by Einsteins second postulate. x We need to use the Lorentz equations. The speed u A is equal to __ when t viewed in frame A ( here xA and tA correspond to x' and t' in the earlier transformation equations) . A
A
x S imilarly u B = __ ( x and t from before) t B
B
so ( x A + vt A ) xB u B = _ = __ vx A tB t A + _ c2
(
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)
A . 2 L O R E N T Z T R A N S F O R M AT I O N S
using the Lorentz transormations. The positive signs are there because we are expressing xB and tB in terms o xA and tA rather than the other way round as we did beore.
Worked example 1
You should satisy yoursel that substituting xA = u AtA into this expression gives uA + v uB = _ u Av 1 + _ c2 This is the case where an observer at rest in frame B is measuring the speed o the obj ect. The inverse case o an observer in rame A ( sitting in the rame A rocket in fgure 2 ) measuring the speed o the obj ect in rame B gives a similar expression with a change in sign. uB - v uA = _ u Bv 1 - _ c2 A way to get the signs correct or a particular combination o rame velocities is to begin with the Galilean transormation ( that is, when uv c) and see what signs you would expect in this case. Then remember that the sign matches top and bottom o the equation.
Jean and Phillipe are in separate rames o reerence, neither o which is accelerating. Jean observes a spacecrat moving to his right at 0.8c. Phillippe observes a spacecrat moving to his let at 0.9c. C alculate the velocity o Phllippes rame o reerence relative to Jeans.
Solution S uppose Jean is the observer at rest and Phillippe is moving relative to Jean with speed v. u - v u A = ______ u v can be re___ B
Invariant quantities
1 -
Spacetime interval
B
c2
u - u arranged to give v = ______ u u ____ B
In developing his theory o special relativity, Einstein realized that absolute time and absolute space are not invariant (unchanging) properties when moving rom one inertial reerence rame to another. However, not all quantities change when moving between inertial rames. In the previous section we chose to express time in the Lorentz transormations not as plain t but as the product o the speed o light and time, ct. This quantity ct has the dimensions o length and this leads us to an invariant quantity known as the spacetime interval s that is defned or motion in the x direction as s 2 = ( c t) 2 - x2 ( sometimes called the invariant interval) . ( You may see this defned in some books as x2 ( c t) 2 , in other words, as the negative o our defnition.)
1 -
A
B
A
c2
This gives the relative velocity o the reerence rame in terms o the known individual speeds in the reerence rames. So
( 0 . 8 c ( 0 . 9 c) ) v = ___________ 0.7 2 ____ 1 +
c2
1 .7c = 0.989c. = ____ 1 .72
The spacetime interval is invariant because in another rame o reerence
(
v v v x' _ _ ct = c( t2 - t1 ) = ( ct2 ' + _ c x2 ' ) - ( ct1 ' + c x1 ' ) = c t' + c
)
and v v _ x = x2 - x1 = ( x2 ' + _ c ct2 ' ) - ( x1 ' + c ct1 ' ) = ( x' + v t' ) thereore
(
)
2
v x' s2 = 2 c t' + _ - 2 ( x' + v t' ) 2 c
(
)
v2 = 2 ( c 2 - v 2 ) t' 2 - 2 1 - _ x' 2 c2 = ( c t' ) 2 - x' 2 This is obviously identical to the original defnition using the same quantities ( and no others) measured in the new rame.
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R E L AT I VI T Y In three dimensions the spacetime interval becomes s2 = ( ct) 2 - x2 - y2 - z2 .
TOK So what about time travel? The spacetime interval has a bearing on the cause and eect relationship between two objects or events. Intervals can be classifed as space-like, time-like, or light-like depending on the value o s 2 or the two events. Space-like intervals I s2 is (ct) 2 . This means that the distance between the events is too great or light (or anything travelling slower) rom one event to have any eect on the other. They do not occur in each others past or uture and although there is a reerence rame in which they occur at the same time, there is no reerence rame in which the events can occur at the same place. Time-like intervals In this case, s2 is > 0 (ie positive) and (ct) 2 > (x2 + y 2 + z2 ) . Now there is sucient time or there to be a cause and eect relationship between the events because the time part o the spacetime interval is greater than the spatial separation. Light-like intervals The intermediate case is where (ct) 2 = (x2 + y 2 + z2 ) and thereore s2 = 0. The spatial distance and the time interval are exactly the same. Such events are linked by, or example, a photon travelling at the speed o light. Time travel has always been a ascination o science-fction authors. The spacetime interval can tell us the extent to which two events in space and time can aect each other. To what extent do fctional works that you know mirror scientifc truth?
Rest mass A second invariant quantity is the rest mass m 0 o a particle. This is an important quantity and is defned as the mass o a particle in the rame in which the particle is at rest. This will be discussed at more length in a later sub- topic, as the concept o mass is bound up with what we mean by energy.
Proper time t = 2L c
L
D
L
mirror position 1
D
mirror position 2
S ome o the most dramatic dierences between our everyday perceptions o space and time and the predictions made by special relativity concern the time and length dierences that arise between rames o reerence moving relative to each other.
vt'
Figure 3 A light clock from stationary and moving frames of reference.
518
Figure 3 shows a simple light clock that consists o two mirrors acing each other across a room. An observer sits at rest in the room and watches light reect between the mirrors. The distance across the room is L and so the time taken or the light to return to a mirror is t = 2cL .
___
A . 2 L O R E N T Z T R A N S F O R M AT I O N S
Anothe r observer moves to the le t parallel to the mirrors at constant ve locity v relative to the mirrors, watching the reections. The righthand diagram shows the bottom mirror at two positions as se en by the moving observer: when the light leaves the bottom mirror and when it returns. In the rame o this observer, the light appears to travel to the right at an angle to the direction o motion ( but, o course, at the same spee d o light) . The distance travelle d by the light is now 2 D and the time observe d between reections at the same 2D mirror is now t' = ___ c . The distance travelled horizontally by the moving observer in time t is vt and by an application o Pythagoras theorem: v 2 t 2 D2 = L 2 + _ 42 ct' D=_ 2 Rearranging or t gives 2L _ c _ t' = ______ v2 1 - _2 c and so t ______ t' = _ v2 1 - _ c2
which reduces to t' = t. This result also ollows directly rom the Lorentz transormations: Two events ( the light leaving rom, and returning to, the bottom mirror) occur at t1 and t2 . These events occur at the same place ( the mirror) so we do not need to include the x terms in our proo ( because x1 = x2 = x) . The time interval between these events is t = t2 t1 . In the observer rame thereore vx vx t' = t2 - t1 = t2 - _ - t1 - _ c2 c2
(
(
) ) = t
The same result as beore and time interval in the observer rame = time in the mirror rame t = t0 The quantity t0 , the time interval in the stationary rame, is known as the p rop er time. Its defnition is the time interval between two events measured in the reerence rame where both events occur at the same position. Proper time is our third invariant quantity. Alternatively, it is the shortest possible measured time interval between two events. The result shows that time measured in a moving rame is always longer than the time measured in a rame that is stationary relative to a clock. The eect is known as time dilation ( dilated means expanded) . I the moving observer in our example here also has a clock then an observer stationary with respect to the mirror rame observes the moving clock running slower than the mirror clock. The situation is symmetric. We will discuss this point urther in the next section.
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R E L AT I VI T Y
Worked example 1
Tom is ying a plane at 0.9c. The landing lights o the plane ash every 2 s as measured in the reerence rame o the plane. S am watches the plane go by. C alculate the time between ashes as observed by Sam.
Proper length The length o an obj ect also changes when observed in rames moving relative to the obj ect. When discussing proper time we had to be careul to speciy that the positions at which time was measured were the same or both measurements. This time, the length L o an obj ect ( where L = x2 x1 ) must have x1 and x2 measured at the same time so that t1 and t2 in the Lorentz equations are both equal to t. In reerence rame S, x1 and x2 represent the ends o an obj ect o length L 0 . This is the frame in which the object is at rest. In S' these ends become x1 ' and x2 ' with a length L' . The reerence rame o S moves at speed v relative to S' . In S
Solution When v = 0.5 c, = 2 .3 ( this was calculated in an earlier example) . The proper time is 2 s. The time between ashes or Sam = t = 2 .3 2 = 4.6 s. The time is dilated in S ams rame o reerence.
L 0 = x2 - x1 which in S' using the Lorentz transormations x2 + vt2 x1 + vt1 _____ - _ _____ = _ 2 v v2 1- _ 1- _ 2 c c2
The ends o the rod are measured at the same time and so t1 and t2 are equal. So L' ______ L0 = _ v2 1 - _ c2
and L0 L = _ This leads to the ourth invariant quantity ( L 0 ) called the p rop er length defned as the length o an obj ect measured by an observer at rest relative to the obj ect. B y implication the two measurement events have to be made at the same time. The proper length can also be regarded as the longest measured length that can be determined or an obj ect. There is direct experimental evidence or time dilation and length contraction ( which are two sides o the same coin) . Muons are particles that can be created either in high- energy accelerators or ( more cheaply! ) in the upper atmosphere when cosmic rays strike air molecules. These muons have very short mean lietimes o about 2 .2 s. When travelling at 0.98c, the distance the muon will travel in one mean lietime is roughly 660 m ar less than the height o 1 0 km above the Earths surace where they are created. O n a Newtonian basis very ew muons would be expected to reach the surace as the time to reach it is about 1 5 mean lietimes. Yet a considerable number o muons are detected at the surace. This is due to time dilation ( or length contraction, whichever 1 _______ = 5 . S o, in the viewpoint you choose) . At a speed o 0.98c, = ________ 1 - 0 . 9 8 2
reerence rame o the Earth, the mean lietime becomes 1 1 s. The time to travel 1 0 km at 0.98c is 3 3 s so that a signifcant number o muons will remain undecayed at the surace.
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A . 2 L O R E N T Z T R A N S F O R M AT I O N S
In the rame o reerence o the muon, the 1 0 km ( as measured by an 1 0 km observer on the E arth) rom atmosphere to Earths surace is only _____ ( to the muon) . This is 2 km in the muons rest rame corresponding to a travel time o about 3 mean lietimes allowing many more muons to reach the surace than Galilean relativity would suggest. D epending on the observers viewpoint, either time dilation or length contraction can be used to explain the observed large number o muons at the surace.
Clock synchronization The problem o synchronizing the clocks we have used in our discussions is an important one. S uppose an observer is standing close to a clock and simultaneously viewing another clock 1 km away. B ecause the speed o light is invariant, the distant clock even i originally synchronized 10 will appear to register a time _____ = 3 0 s later than the nearby clock. 3 10 However, i the two clocks are in the same inertial rame, they will be synchronized ( tick at the same rate) .
Worked example 1
C lare and Phillippe y identical spacecrat that are 1 6 m long in their own rame o reerence. C lares spacecrat is travelling at a speed o 0.5 c relative to Phillippes. Calculate the length o a) C lares aircrat according to Phillippe b) Phillippes aircrat according to C lare.
4
8
O ne way to achieve this synchronization is to synchronize both clocks when they are close together and then move one to its fnal position, but to do it very slowly so that the two clocks continue ( approximately) to share a reerence rame with each other. Another method would be to have both clocks at their fnal positions and to use a third clock moving slowly between them to transer the times rom one to another.
Solution = 1 .1 5 or this relative speed. a) The length o C lares 16 aircrat is ____ m according 1 .1 5 to Phillippe, this is 1 3 .9 m. b) B ecause the situation is symmetrical C lare will also think that Phillippes spacecrat is 1 3.9 m long.
Nature of science GPS a study in relativity S atellite navigation units ( satnavs) in cars and other domestic devices that use the global positioning system are now very common. They can pinpoint their position to within a ew metres. At the time o writing there is a network o 2 4 satellites orbiting at about 2 1 0 7 m above the Earth with orbital periods o about 1 2 hours. The orbits are such that at least our satellites are above the horizon at any point on the surace at all times. Inside the satellite is an atomic clock that is accurate to about 1 0 9 s and transmits a signal to the receivers on or above the Earth.
The receivers triangulate the signals rom satellites above the horizon to arrive at a positional fx to within metres within a ew seconds. Wait a little longer with some special receivers and this precision can rise to orders o millimetres. It is now routine or a satnav in a moving vehicle to show its speed and heading in real time. The design o both the satellite transmitters and the GPS receivers need to take account o relativity. The atomic clocks are adj usted so that once in orbit they run at the same rate as Earth- bound clocks. The GPS receivers have microcomputers that carry out the required calculations to make the relativistic corrections.
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A.3 Spacetime diagrams Understanding
Applications and skills
Spacetime diagrams
Representing events on a spacetime diagram
Worldlines The twin paradox
as points Representing the positions o a moving particle on a spacetime diagram by a curve (the worldline) Representing more than one inertial reerence rame on the same spacetime diagram Determining the angle between a worldline and the time axis or specifc speed on a spacetime diagram Solving problems on simultaneity and kinematics using spacetime diagrams Representing time dilation and length contraction on spacetime diagrams Describing the twin paradox Resolving o the twin paradox through spacetime diagrams
Nature of science The spacetime (or Minkowski) diagram is a perect example o a picture adding value to conceptual understanding. At frst, Einstein did not realize the ull potential o these diagrams but came to see that they had considerable value in making some o the dicult ideas o relativity including simultaneity more obvious.
Equations v angle o worldline to ct axis = tan - 1 ( _ ) c
Spacetime diagrams and worldlines In 1 908, Minkowski introduced a way to visualize the concept of spacetime. This will help you to understand many of the ideas and concepts that arise in special relativity. Physicists are well used to graphs as ways to visualise data. Minkowski attempted to represent the four- dimensional nature of spacetime using a graphical picture known as a sp acetime diagram or sometimes as a Minkowski diagram. t/s t t
P'
worldline of Q
3
R'
speed = c
2 1 x P (a)
Figure 1 Spacetime diagrams.
522
x/m 0 (b)
4
8
R
12 (c)
x
A. 3 S PACE TI M E D I AG R AM S
S pacetime diagrams show the position o an obj ect in one dimension ( x) at a time ( t) in an inertial rame. The axes themselves constitute the inertial rame. The diagram resembles ( but should not be conused with) the ordinary distancetime graphs with which you are amiliar in mechanics, except that time is plotted on the y- axis and position on the x- axis. Figure 1 ( a) shows the spacetime diagram or a particle that is stationary with respect to the inertial rame that is represented by the diagram. At t = 0 the particle P is on the x-axis. As time goes on, because the object is stationary, it does not change its position (x) in the reerence rame. Time is, o course, increasing. Line PP shows the traj ectory o the particle through spacetime and is known as the worldline o the particle. Figure 1 ( b) shows a dierent particle Q moving at a constant velocity in the reerence rame o this spacetime diagram. At t = 0, Q is at the origin o the diagram ( x = 0) and it is moving at 4 m s - 1 to the right. Each second ater the origin time, Q is 4 m urther to the right and so its worldline in the spacetime diagram is a line at an angle to the axes. It should be easy or you to see that i a urther particle R were to be accelerating relative to the reerence rame o the diagram, then the worldline o R would be a curve ( fgure 1 ( c) ) . There must be a limit to the R worldline because nothing can exceed the speed o light in ree space. So the gradient o the dotted line on fgure 1 ( c) shows the maximum limiting speed o R. This dotted line also represents the world line o a photon in the diagram ( the minimum gradient o RR) . Returning to particle Q which is moving with constant velocity in the reerence rame o P, we could think o Q as being stationary in its own reerence rame and in this event the spacetime diagram or Q in its own rame would be identical in shape to fgure 1 ( a) . We can combine these two separate spacetime diagrams or dierent inertial rames moving at constant speed relative to each other and this combination o axes will be o particular help later in this sub- topic when we resolve some o the paradoxes that special relativity appears to create.
t
t' worldline o Q x'
Figure 2 shows the combined spacetime diagram or Q and P drawn or the reerence rame o P. Notice what has happened to the Q axes (t' x' ) , they have swung away rom t and x and become closed up together. The Q x'-axis is now at the same angle as the previous Q worldline in the P reerence rame. Q is, o course, moving along a worldline parallel to the time axis in the Q reerence rame. We have so ar used t and x or our axes, but this is not the only convention used. S ometimes you will see time plotted in units o ct. This is convenient because it means that the limiting line that represents the speed o light will have a gradient o 1 on the spacetime diagram. You can expect to see both conventions used in IB examinations. An additional convention is that sometimes physicists defne c to be equal to 1 so that, in calculations, large values or the answers do not trouble them. Equally, expect to see speeds quoted as, or example, 0.95 c meaning 95 % o the speed o light in ree space ( 2 .85 1 0 8 m s 1 ) .
x
Figure 2 Two inertial rames one spacetime diagram.
ct
particle X worldline c
cT x
Figure 3 Defnition o .
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R E L AT I VI T Y S ome simple geometry ( Figure 3 ) shows that when we are using ct-x axes, then the angle between the worldline o a particle and the ct axis is given by opposite v X 1 _ X=_ tan = _ = _ =_ c c cT T adjacent or v = tan - 1 ( _ c) When v = c, then = 45 ( tan 45 = 1 ) and the worldline or a photon starting at the origin o the spacetime diagram is a line at 45 to both ct and x axes.
Simultaneity There are signifcant changes to our ideas about the order in which things happen or whether two events happen simultaneously under special relativity. This is because the speed o light is always observed to have the same value by observers in dierent rames. The classic thought experiment to illustrate this is the example o a train carriage moving at constant velocity past an observer standing on a station platorm ( Figure 4) . A person in the carriage ( Jack) switches on a lamp that hangs rom the centre o the ceiling. Jack observes that the light rom the lamp reaches the two end walls o the carriage ( R and L) at the same moment. railway carriage lamp R
L Jack
Jill
platform
simultaneous
ct'
not simultaneous
ct L
R
L
R
x' x (a)
Jacks frame
x (b)
Jills frame (ct' - x')
Figure 4 Simultaneity at work in a spacetime diagram. Jill who is on the platorm, does not agree with this observation. The light rom the lamp moves to both ends o the carriage at the same speed c. However, in the time it takes the light to get to the ends, the let-hand end o the carriage has moved towards the light and the right-hand end has
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A. 3 S PACE TI M E D I AG R AM S
moved away. The consequence is that the light (according to Jill) hits the let-hand end frst (event L) beore the right (event R) . This result becomes clear in a spacetime diagram. In the reerence rame o Jack (ct-x) the events R and L occur at the same instant because they are on a line parallel to the x axis and are at the same ct coordinate (fgure 4(a) ) . In Jills rame plotted on fgure 4(b) (ct-x) it is clear that L occurs beore R when you look these events in terms o the ct axis. There is a danger o conusion here because it is possible to misunderstand and think that the loss o simultaneity is to do with the transmission o the inormation. In other words, that this dierence o opinion between Jack and Jill arises because the light travels through dierent distances rom the ends o the carriage to their eyes. That is not the explanation o what is happening. The lack o simultaneity arises because the speed o light is always constant even i a particular observer is moving relative to the light source. As ar as Jack is concerned, he is always midway between the carriage endwalls. As ar as Jill is concerned, once the photons have let the lamp, then they travel at c and the carriage will continue to move while the photons themselves are in transit. O ne o the reasons or this conusion is the use o the term observer, which is a very common one in books and articles about relativity. We oten think o an observer as being located at one point in the inertial reerence rame. This is not the true meaning. It is better to think o the observer as being in overall charge o an ( infnitely) large number o clocks and rulers that are located throughout the observers rame. Jack ( the stationary observer in this case) can take a reading at the instant when the light hits the end wall o the carriage without having to worry about the time taken or this inormation to travel rom the carriage to his position. Another way is to think about the observer as being a whole team o observers with each one able to make measurements o his or her immediate region o the reerence rame. This can be explained in terms o the Lorentz transormations. Imagine that two events are simultaneous in one rame o reerence. This will mean that or both events the time coordinate will be ct or any value o x. However, in a rame moving at v relative to the frst rame, the Lorentz transormation shows that: vx t- _ c2 _ t' = ______ v2 1 - _2 c
Thereore unless x is the same or both events then t cannot be the same or both events they will not be simultaneous. Conversely, this also tells us that i two events are simultaneous in the second, moving rame then they must be occurring at the same position (x) .
Time dilation and length contraction re-visited Figure 5 ( a) shows the spacetime diagram or an observer in rame B who is viewing an automobile moving at a constant velocity v relative to B . The diagram shows the stationary rame A or the automobile with the B spacetime axes included. We need to be quite clear about what we mean by the term time interval here. It is the time between two events
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R E L AT I VI T Y measured at the same place in the reerence rame ( in other words, the proper time) . The spacetime diagram should show that a measurement in any other rame leads to a time interval greater than the proper time. ct
proper time time in A
ct ct'
ct'
time in B
time that A thinks B measures
proper time
x' frame B x
x' x
frame A (a)
(b)
Figure 5 Measuring times in two reference frames. At t = 0 the origins o both reerence rames coincide and the automobile is at the origin in both rames. The automobile is stationary in rame A and its worldline lies along the time axis as usual. Frame B is moving at constant velocity and its axes are modifed as usual in rame A. The event that marks the end o the proper time interval is transormed along the x axis to meet the ct axis in order to obtain the time that B will measure at what B thinks is the same instant in the moving rame ( fgure 5 ( a) ) . The question o what A (in the stationary rame) thinks is the same instant is dierent (fgure 5 (b) ) . However, whichever view we take o the measurement in rame B (whether rom the A or B standpoint) , the time measured in B is always greater than the proper time because the length o the line in the spacetime diagram is always along the hypotenuse o the triangle, whereas the proper time is one o the other sides o the triangle. A spacetime diagram also helps understanding o length contraction. To see why distance measurements change in a moving rame, again we must understand what is being measured. We need to consider the distance between two points at the same instant in time as j udged by the two observers in dierent rames. ct frame B
ct' frame A
worldline of 0 cm in B
worldline of 30 cm in B worldline of 30 cm in A
worldline of 0 cm in A
p ro p e r le
n gth
29 30 26 27 28 23 24 25 20 21 22 17 18 19 14 15 16 13 12 9 10 11 7 8 4 5 6 1 2 3
x2 '
CM
simultaneous in A simultaneous in B
x1 ' x' frame A
x1
x2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 CM
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Figure 6 Measuring lengths in two reference planes.
x frame B
A. 3 S PACE TI M E D I AG R AM S
Figure 6 shows a ruler stationary in rame A. The ruler lies along the x- axis because this is where we will determine the proper length. The rules or measuring proper length indicate that the two position determinations must be made simultaneously. The other rame B is considered to be the stationary rame or the purposes o the spacetime diagram. Thus, A must be moving relative to B and as a result has the axes ct- x. The diagram shows what happens. The proper length o the ruler is the distance measured between the worldlines and parallel to the x axis. The equivalent simultaneous measurement in B will always be shorter than that in A the length is contracted.
The twin paradox Many o the ideas we have discussed so ar in the study o special relativity come together in a series o paradoxes. The twin paradox is the most amous o these and can be very simply stated: ct
Mark and Maria are twins. Mark decides to go on a journey to a distant star at a high speed with Lorentz actor equal to . Ater reaching the star taking a time T in Marias rame o reerence. Mark returns with his journey also taking a time T (to Maria) to do so. At the end o his journey 2T Maria has aged 2T but she is amazed to fnd that Mark has only aged by __ . This is nothing more than time dilation so where does the paradox arise? Think about Marks experience. He sits in his spacecrat his rame and watches Maria and the E arth move away at high speed ( with the same as beore) so why is Maria not younger than him on his return? We would expect some symmetry between the two rames. The answer to the paradox is that there is no symmetry at all between the two cases. Maria has remained in an inertial rame throughout Marks j ourney. Mark has not. He needs to accelerate our times during the journey: at the start o the trip, when he slows down at the star, when he accelerates back up to top speed and fnally when he decelerates to arrive at the Earth. Moving out o an inertial rame o reerence even once breaks the symmetry and as a consequence Mark and Maria age at dierent rates relative to each other over the whole journey. The spacetime diagram (fgure 7(a) ) shows what happens. The rame or Maria is ct- x, the rame or Mark is ct- x. Mark leaves Maria at the origin o her reerence rame and she remains here throughout. O course, she moves along the ct axis as time increases. Meanwhile, Mark moves along his worldline which is at his origin x = 0 or (in Marias rame) at x = vt where v is Marks speed relative to Maria. Mark reaches the star at event P. We can draw two lines o simultaneity; one or Mark and one or Maria. In Marias rame she thinks Mark arrives at the star at time Q. R is the time that Mark thinks Maria observes when he arrives at the star. They disagree about the simultaneity o events Q and R as we should expect. At this stage, both Mark and Maria think that the other is younger by a actor o as predicted by the time dilation result rom earlier. There is a problem in thinking about the return j ourney because i it is to happen at all, Mark has to change speed. It is not necessary or him to do this, though. We can imagine that as soon as he reaches the star, he synchronizes his clock with another clock on a spacecrat that belongs
Mark worldline
P ct'
Q
Mark arrives at star
R
x' x
Maria worldline
(a)
ct Jay worldline Mark worldline
S Q
Mark arrives at star
ct'
R
x' x
Maria worldline
(b)
Figure 7
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R E L AT I VI T Y to Jay. Jay is already on his way to Earth ( and thereore Maria) at the same speed that Mark had when approaching the star. Figure 7( b) shows the added worldline or Jay. As Jay leaves the star, he thinks that Maria is at S. I Mark were to slow down at the star, turn round, and go back to the Earth, the acceleration procedure would make Maria appear to rapidly age rom Q to S.
Worked example In the distant uture a network o our warning beacons W, X, Y and Z is set up to warn spaceship commanders o the approach lanes or planet Earth. The beacons ash in sequence. The spacetime diagram shows the reerence rame in which the beacons are at rest and one cycle o the sequence. The worldline or a spaceship is also shown.
ct ct'
Y
X
ct ct' worldline of spaceship
W
Z x
( ii) The order o the ashes in the spaceship rame has to be obtained rom constructing lines parallel to the x axis. In this rame, Z is observed to ash frst, W and X then ash simultaneously, fnally Y ashes. x W
X
beacon frame
Y
Z
a) D etermine the order in which the our beacons ash according to:
b) To decide on the arrival o the light rom the beacons it is necessary to add the photon worldlines to the diagram. These are lines that begin at the beacon ash and travel at 45 to the axes. ct
( i) an observer stationary in the rame o the beacons ( ii) an observer on the spaceship b) D etermine the order in which the observer on the spaceship sees the beacons ash. c) C alculate the speed o the spaceship.
light from Y
Solution a) ( i) T h e s p a ce tim e dia g ra m in th e ra m e o th e b e a co n s in dica te s th e ch ro n o lo g ica l o rde r in w h ich the b e a co n s la s h : W a n d Z s im u lta n e o u s ly, th e n X, a n d th e n Y.
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light from W
light from X
light from Z x
The intersection o the photon worldline with the ct axis gives the arrival time at the spacecrat. The order is Z, Y, X, W. v c) tan = _c = 0.5 8 v = 0.5 8c
A . 4 R E L AT I V I S T I C M E C H A N I C S ( A H L )
A.4 Relativistic mechanics (AHL) Understanding Total energy and rest energy Relativistic momentum Particle acceleration Electric charge as an invariant quantity Photons MeV c 2 as the unit o mass and MeV c 1 as the
unit o momentum
Applications and skills Describing the laws o conservation o
momentum and conservation o energy within special relativity Determining the potential dierence necessary to accelerate a particle to a given speed or energy Solving problems involving relativistic energy and momentum conservation in collisions and particle decays
Nature of science A urther paradigm shit occurred in physics when Einstein realized that some conservation laws (momentum and energy) broke down as inviolate laws o physics unless modifcations were made to them. This led him (amongst other things) to ormulate one o the most amous equations in the whole o physics.
Equations total relativistic energy E = m 0 c 2 rest-mass energy E0 = m 0 c 2 relativistic kinetic energy EK = ( - 1) m 0 c 2 relativistic momentum p = m 0 v energy-momentum relation E2 = p 2 c 2 + m 0 2 c 4 acceleration through pd V qV = EK
So far we have dealt almost exclusively with the basic concepts of space and time and we have shown that they change their nature when we have regard to the relative motion of inertial frames. Now we have to identify and explain the changes that need to take place in the other laws of physics for them to be invariant for all inertial observers.
Total energy and rest energy In a previous sub- topic we mentioned that the rest mass m 0 of a particle was an invariant quantity. E instein, in one of his famous 1 905 papers, proved that when an obj ect loses energy its mass changes and by so doing he suggested that energy and mass are related. This led him to possibly the most well- known physics equation in the world and possibly the most misunderstood, E = mc2 . When a particle is viewed from its rest frame then its mass will be observed as the rest mass m 0 . This means that the rest energy of the particle is E0 = m 0 c2 . Like the rest mass, this is an invariant quantity for a particular particle. Some physicists also use the concept of relativistic mass, meaning the mass observed when the particle is in a reference frame moving relative
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R E L AT I VI T Y to the observer. However, the consequences of energy and mass having an equivalence means that it is not necessary to use both relativistic mass and energy; we will only use the term total energy to signify both these ideas. The total energy can easily be converted into an equivalent mass if required using Einsteins equation ( but we will not do so in this course) . The total energy E of a particle is equal to the sum of the rest energy E0 and the kinetic energy EK: E = EK + E 0 ignoring potential energy and its changes, and energy dissipation. This leads to a set of equations relating the energies and masses: E = m 0 c2 and therefore EK = mc2 - m 0 c2 = m 0 ( - 1 ) c2 10 special relativity newton
kinetic energy EK
8 6 4 2 0 0
0.5
1 speed/v/c
1.5
2
Figure 1 Relativistic kinetic energy against speed. Figure 1 shows the shape of the graph of EK against speed ( expressed as a fraction of c) . When the speed is zero, the energy is the rest energy. At high speeds approaching the speed of light, the curve becomes asymptotic to the line _vc = 1 .
Relativistic momentum Momentum must still be conserved within the special theory. However in order to achieve this, the expression for momentum must incorporate to change from the Newtonian mu to m 0 u. In fact, both momentum and energy are j ointly conserved within the theory with the proviso that energy has to include the whole bundle of energies associated with a particle including its rest energy. The momentum p of the particle is expressed as p = mu and ( as before) the total energy is E = m 0 c2 . C ombining these equations and eliminating u leads to an expression for the total energy: E2 = p 2 c2 + m 0 2 c4 = ( pc) 2 + ( m 0 c2 ) 2
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A . 4 R E L AT I V I S T I C M E C H A N I C S ( A H L ) This is known as the energymomentum relation. It has an important property. I the equation is rearranged as
Worked example 1
( m 0 c2 ) 2 = E2 - ( pc) 2 then the quantity on the let- hand side is the invariant mass it does not change between inertial rames. Thereore the right- hand side must also be invariant, so we can write ( m 0 c2 ) 2 = E2 - ( pc) 2 = E' 2 - ( p' c) 2 The knowledge that ( E2 p 2 c2 ) in the rame is invariant allows us to, or example, move easily between the particle rame in a particle accelerator and the lab rame. Many tests o these equations were made rom the time o E insteins frst suggestions o the relationships. The equations were always verifed but some o the determinations were indirect and had many sources o error. Perhaps the most direct and convincing experiment was that developed by B ertozzi in 1 964. He accelerated electrons to a high speed and measured their speed as they travelled through a vacuum. Immediately ater passing through the speed- measuring apparatus, the electrons were absorbed by an aluminium disc, thus transerring their energy to the internal energy o the disc. The temperature o the disc increased as a result and the energy o the incident electrons could be measured directly. B ertozzis results are shown in fgure 2 . This was a direct and convincing verifcation o Einsteins theory.
C alculate the speed at which a particle must travel or its total energy to equal fve times its rest mass energy.
Solution E = mc2 = 5 mc2 1 _____ As = 5 = _ v2 1- _ c2 2 v v2 1 24 _ _ 1- _ = and = __ 25 25 c2 c2
Thereore, v = 0.98c.
1.1 1.0
v2 /c 2
0.9 0.8 special relativity
0.7
bertozzi 0.6 0
5
10
15
20
25
30
kinetic energy/mc 2
Figure 2 Results of Bertozzis experiment.
Nature of science Operating a particle accelerator Although one o the most direct verifcations o special theory was only carried out in the 1 960s, it was clear that the theory was the appropriate one to use. Cyclotrons had already been designed and built, and in these particle accelerators the circulating particles gain considerable energy. The energy that they gain (in our rame o reerence) is ound not to
1 mv2 and a relativistic correction or this vary with __ 2 is required. Essentially, the particles become more massive than would be expected rom a Newtonian consideration. As cyclotrons were developed that could supply more and more energy there came a point where a synchronization mechanism was required to allow or the relativistic changes. This led to the development o synchrocyclotrons and other high-energy accelerators.
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Particle acceleration The obvious way to accelerate a charged particle is to place it in an electric feld. As we saw in Topic 5 this leads to an electric orce on the particle and to a gain in energy that can be expressed in terms o the charge on the particle and the potential dierence through which it moves. The key here is that charge q is our fth invariant quantity so the term does not enter into our specifcation o charge in a moving rame. Thus the charge o the particle during the acceleration does not change and we can see directly ( as in earlier parts o this book) that qV = EK where EK is the change in kinetic energy and V is the potential dierence. This leads to a new set o units that are extensively used in particle and relativistic physics. Rather than use kg or mass and kg m s 1 or momentum it is much easier to think and work in terms o the energy equivalent o these units. Thus, or mass we use eV c 2 ( or, more commonly, multiples o this, MeV c2 and GeV c 2 ) and or momentum MeV c 1 and GeV c 1 . What has eectively happened is that c in the energymomentum relation has been made equal to 1 . It is as though the equation has been written: E2 = p 2 + m 0 2 One o the worked examples below is a repeat o an earlier example to show you how this change in units works.
Photons Photons have zero mass. What does this mean or their properties within the special theory? The starting point is the energymomentum relation once again, but this time m 0 is zero and so the right-most term disappears leaving E2 = p 2 c2 hf _ h E= _ S o the momentum o a photon is p = _ c c =
TOK
Worked examples 1
Conservation laws This discussion o conservation rules that have needed to be modifed in the light o the special theory calls into question the nature o all conservation rules. Do the laws o natural science dier rom the laws that exist in other branches o knowledge, or example, in economic theory?
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A particle o charge + e has a rest mass o 9. 1 1 0 31 kg. It is accelerated rom rest to a speed o 2 . 4 1 0 8 m s 1 . C alculate, or this particle: a) the rest energy in MeV b) the total energy ater acceleration c) the kinetic energy ater acceleration d) the potential dierence through which it must be accelerated to reach this speed.
Solution 8.2 1 0 - 1 4 a) The rest energy = m 0c2 = 8.2 1 0 1 4 J; this is __ = 0.5 1 MeV. 1 .6 1 0 - 1 9
A . 4 R E L AT I V I S T I C M E C H A N I C S ( A H L )
b) The speed is equivalent to = 1 .7. The total energy = m0 c2 = 1 .7 8.2 1 0 14 = 1 .4 1 0 13 J = 0.87 MeV c) The kinetic energy = ( 1 .7 1 ) 8.2 1 0 1 4 = 5 .5 1 0 1 4 J = ( 0.87 0.5 1 ) MeV = 0.3 6 MeV d) To attain a kinetic energy o 3 60 keV must require a pd o 3 60 kV. 2
C alculate the momentum o a photon o visible light o wavelength 5 60 nm.
Solution h p= _ 6.6 1 0 - 34 SO p = _ = 1 . 2 1 0 - 27 kg m s 1 . 5 .6 1 0 - 7 3
A small insect with a mass o 1 .5 1 0 3 kg fies at 0.48 m s 1 . C alculate the momentum and kinetic energy o the insect.
Solution This is a non- relativistic solution as the speed is much less than that o light. 1 eV 1 .6 1 0 - 1 9 J and 1 kg 9 1 0 1 6 J = 9 1 0 4 TJ. 1 Kinetic energy = _ 1 .5 1 0 3 0. 48 2 = 1 .7 1 0 4 J 2 -4 1 .7 1 0 = __ = 1 .1 1 0 1 5 eV = 1 . 1 1 0 3 TeV 1 .6 1 0 - 1 9 Momentum = 1 .5 1 0 3 0.48 = 7.2 1 0 4 kg m s - 1 = 7.2 1 0 4 3 1 0 8 1 .6 1 0 1 9 = 1 .3 5 1 0 24 eV c 1 = 1 . 4 1 0 1 2 TeV c 1 .
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A.5 General relativity (AHL) Understanding
Applications and skills
The equivalence principle
Using the equivalence principle to deduce and
The bending o light Gravitational redshit and the Pound-Rebka
Snider experiment Schwarzschild black holes Event horizons Time dilation near a black hole Applications o general relativity to the universe as a whole
Nature of science Ater publishing his special theory, which applied to non-accelerated reerence rames, Einstein tackled the general theory incorporating the efects o acceleration and gravity. This required intuition, imagination and creative thinking on his part. He needed to modiy his ideas o spacetime in the light o the efects o mass and the curvature that it produces. In this way he was responsible or yet another paradigm shit in physics.
explain light bending near massive objects Using the equivalence principle to deduce and explain gravitational time dilation Calculating gravitational requency shits Describing an experiment in which gravitational redshit is observed and measured Calculating the Schwarzschild radius o a black hole Applying the ormula or gravitational time dilation near the event horizon o a black hole
Equations
f gh Gravitational requency shit: _ = __ 2 f c 2GM _ Schwarzchild radius: Rs = c2 t0 _____ gravitational time dilation t = __
1 -
Rs ____ r
The equivalence principle In Topic 2 we introduced the idea that there are two types o mass: inertial and gravitational and we suggested that these are equivalent. Mass that is gravitationally attracted is taken to be the same as the mass that responds to a orce by accelerating. This equivalence had been recognized since the time o Galileo but was frst discussed in detail by the German physicist Mach at the end o the nineteenth century ( having been touched on by various philosophers beore him) . E instein and others named the central idea: Machs principle. Mach had rej ected Newtons view o absolute time and space, taking a relational view o the universe in which any motion can only be seen with respect to other obj ects in the universe. Thus we cannot say merely that an obj ect is rotating but must reer to the axis about which it rotates. Machs writings, although controversial amongst scientists, partly led to E instein developing the general theory o relativity that he published in 1 91 6. In this he proposed a principle o equivalence: E insteins p rincip le of equivalence states that gravitational effects cannot be distinguished from inertial effects.
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A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) A thought experiment helps to explain the equivalence principle; the experiment involves two observers and an elevator ( lit) . The elevator is a long way rom any other mass so that it is essentially a gravity- ree zone. O ne o the observers ( X) is in the elevator and carries an obj ect that has mass; X cannot see out o the elevator. Another observer ( Y) is outside the elevator and not connected to it in any way but can view what happens inside.
X
Y
Y
X
(a) b
Figure 1 The elevator "thought" experiment.
Initially the elevator is moving at constant velocity. X releases the obj ect and it stays exactly where it is placed. ( Remember that there are no nearby masses or planets to attract the mass.) X repeats the experiment ( Figure 1 ( a) ) but this time, as X releases the obj ect, the elevator begins to accelerate at 9.8 m s 2 in the direction o the elevator roo ( we will call this " upwards" or brevity) . X will think that the obj ect accelerates downwards at 9.8 m s 2 . Y observes rom outside that the obj ect stays where it is in space and that the lit is accelerating upwards around it. C onsider the same experiment repeated ( Figure 1 (b) ) with the elevator stationary relative to and close to the surace o the Earth. When X releases the obj ect it will be accelerated downwards at 9.8 m s 2 . Y will explain this acceleration as due to the attraction o the Earth. The important point here is that in both experiments the obj ect accelerated downwards, in one case due to an inertial ( acceleration) eect and in the other due to the eects o gravity. B ut to observer X in the elevator these two cases are identical and cannot be distinguished. It is this inability to decide what causes the acceleration that lies at the heart o the principle o equivalence. There is in principle no experiment that the observer in the lit can carry out to decide which eect is which. In fgure 1 ( b) we can either regard the lit as accelerating within a universe ( i. e. everything else) that is stationary or we can view the lit as stationary in the rame o observer X with the rest o the universe accelerating around it. According to E instein there is no absolute motion, only relative motion exists. This means that the special status held by observers in inertial rames o reerence in the special theory no longer exists in the general theory and accelerated or not all observers have the same status and obey the same laws. However, it may be that some rames allow the laws to be stated in a more straightorward way.
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Gravitational redshift The general theory predicts that gravity can aect the motion o light itsel. The equivalence principle helps us understand why.
h
F F
R
(a)
R
(b)
Figure 2 Equivalence in action.
Imagine a spaceship that is stationary ( or moving at constant velocity) relative to the rest o the universe. Inside the ship are an observer and two identical light sources, one light source at the rear o the ship ( lamp R) and one close to the observer at the ront ( lamp F) . The spaceship begins to accelerate Figure 2(a) ) . At the instant that the acceleration begins, both light sources start to emit light o the same requency. (It helps to imagine that the light emission begins as the spaceship starts to accelerate, but this is not an essential part o the argument.) h The light rom R take s a inite time t = ___ c to re ach the obse rve r ( whe re h is the distance b e twe e n the light source s) . D uring this time the ve locity o the space ship and ob se rve r will have change d by an amount v = g t, whe re g is the accele ration. Howe ve r, the light in transit rom R to the ob se rve r will not change its spe e d be cause ( as usual) the spe e d o light is inde pe nde nt o the ob se rve r. The obse rve r will obse rve a longe r wave le ngth or the light rom R compare d to the light rom F as lamp R will appe ar to have b e e n D opple r re dshite d.
This can be made quantitative. Merging the two equations above gives g h v = g t = _ c v is the change in speed o the observer over the time period, t is relative to the speed o light source R when it emitted the observed light, and so the requency shit f observed is f _ g h v _ = c = _ f c2 where f is the requency o the emitted light. The equivalence principle predicts that the eects in a gravitational feld should be indistinguishable rom an inertial system, so we do not need necessarily to say more. However, it is instructive to consider the problem rom another dierent ( but ultimately identical) standpoint. Figure 2(b) shows the spaceship at rest relative to, and sitting on, the surace o the Earth. Again lamp R emits light to the observer. Each emitted light
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A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) photon has energy hf when it leaves the source. To reach the observer some o this energy has to be traded o against the gravitational feld as gravitational potential energy. This loss o photon energy (reduction in f) is equivalent to an increase in the wavelength o the light (f = c) and so again, the light rom R is redshited compared to the light rom lamp F. These discussions o requency and wavelength shits lead to the idea o gravitational time dilation. We can regard the arrival o the wave as a series o ticks o a clock. A redshit means that the clock is observed to tick more slowly than the original. S o the observer at the ront o the rocket thinks that lamp R is ticking more slowly than lamp F and the observer at the top o a mountain on Earth thinks that time is running more slowly or an observer at sea level. This eect is gravitational time dilation. It is dierent rom the time dilation eect observed when inertial rames are moving relative to each other. We mentioned the GPS system earlier ( p 5 2 1 ) in the context o special relativity, which predicts that the clocks on the satellites all behind ground clocks by about 7 s every 2 4 hours due to their relative motion with respect to the surace. General relativity, on the other hand, predicts that the satellite clocks should advance compared with the surace clocks by about 45 s in the same time period. The net result is that the clocks in the GPS satellite gain on clocks back on E arth by about 3 8 s every day. This actor swamps the 2 0 ns accuracy required o the E arth-bound GPS receivers. I relativistic eects are not taken into account, then the errors in a position become serious ater about 1 00 s and accumulate at a rate o tens o kilometres every day. This would be completely unacceptable or any navigational systems. The GPS receivers in our cars and on our mobile phones are constantly carrying out relativistic corrections to adj ust or the unavoidable time changes due to relativistic eects.
Worked examples 1
The requency o a line in the emission spectrum o sodium is measured in a rame o reerence in which the sodium source is stationary and well away rom gravitational inuences. C alculate the ractional requency shit that will be measured by an observer also stationary with respect to the sodium source but placed 1 km above the source close to the surace o Earth.
This is a redshit so the wavelength is increased by this ractional amount. 2
C alculate the dierence in time per day due to gravitational redshit o a clock at the top o E verest compared to a clock at sea level. Mount Everest is 8800 m above sea level. Assume that the acceleration due to gravity is constant over this height dierence.
Solution
Solution
f T ___ = __ where T is the time or one day. T f g h gT h T This means that ___ = ____ and T = ____ = T c2 c2 9 . 8 8 6 40 0 8 8 0 0 _______________ this is 80 ns every day. The
The ractional requency shit is given by f _ g h _ = f c2
( 3 1 0 8) 2
g h 9.8 1 000 so the wavelength shit is ___ = ____ = ________ c 9 10 -13 1 .1 1 0 2
16
observer on the mountain thinks that the sea-level clock is running more slowly so the mountain clock appears to gain time.
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Nature of science Tidal orces, and a more precise defnition The earlier explanation o the elevator thought experiment was too simplistic. The equivalence o gravity and acceleration is true only or uniorm gravitational felds. Real gravitational felds will almost always be non-uniorm and inhomogeneous. At the surace o the E arth with human-sized experiments, the feld is very close to uniorm and nearby obj ects appear to all in parallel directions with the same acceleration. B ut two obj ects in a very large elevator will converge towards the centre o the Earth as they all. An observer in the elevator sees the two obj ects apparently moving closer together and will wonder why. This is known as a tidal eect ( fgure 3 ) and means that a precise statement o the principle o equivalence should include a clause stating that it applies when tidal eects can be
Figure 3 Tidal efects.
neglected. Alternatively, we can say that over an infnitesimally small spacetime region the laws o physics in general relativity are equivalent to those under special relativity.
Although the need or relativistic corrections in the GPS satellite navigation system indicates the truth o Einsteins general theory, there have been ormal scientifc tests made over the past century to veriy that it is correct. They include experiments some o which were suggested by Einstein himsel:
gamma detector
gamma source
gamma source
gamma detector
Figure 4 PoundRebkaSnider experiment.
The precession o the perihelion o Mercury ( the point in its orbit where Mercury is closest to the S un)
D eection o light by the S un
Gravitational redshit o light
O ne o the most recent o these is the experiment devised by Pound, Rebka and S nider in 1 95 9 to measure gravitational redshit. This experiment was the last o Einsteins suggestions to be attempted using the ( then) recently discovered technique o Mossbauer spectroscopy. This technique allows small shits in the requencies o gamma rays to be measured. In a very precise experime nt, Pound and his co- workers fred a beam o gamma rays ve rtically upwards towards a detector place d about 2 2 m above the gamma source. The y repeated the experiment with the gamma source fring downwards to the de tector. I the upward and downward ractional changes in energy o the gammas are compared then E (_ E )
upwards
( )
E - _ E
downwards
2 g h = _ c2
For the PoundRebkaSnider experiment, h = 2 2 .6 m and g = 9.81 m s 2 which leads to a theoretical dierence in the ractional energies o 4.9 1 0 1 5 . They measured the dierence to be ( 5 .1 0.5 ) 1 0 1 5 which compares well with the theoretical value. This was a convincing verifcation o the general theory.
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A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) I a single planet orbits a star, then Newtonian gravitation and mechanics predict that the orbit o the planet will ollow the same elliptical path orever. The presence o other nearby planets and moons, however, disturbs this motion. In practice, planets in the solar system orbit the S un in an ellipse that rotates gradually around the S un; this is known as precession. The rate at which this rotation occurs can be predicted very accurately on the basis o Newtonian mechanics. Observations made in 1 85 9 showed that the precession o Mercury is aster than predicted by Newtonian mechanics. The change to the precession rate is small but much larger than the error in its measurement; there is no doubt that the precession prediction is incorrect. One o the frst successes o Einsteins general theory was that it correctly predicted the value o the observed precession rate o Mercury. E insteins third suggestion was that a massive star would deect light rom its straight- line path. S hortly ater Einstein published his theory, the English physicist E ddington travelled to the west coast o Arica with colleagues to observe stars during the 1 91 9 total eclipse o the S un. This enabled them to confrm that the mass o the S un deects light according to the general theory. This will be discussed in more detail in the next section. A ourth eect was teste d by Irwin S hapiro in 1 9 6 4. E instein had p redicted that the time or a radar ( microwave) pulse o radiation to go past the S un and return to E arth ater relection rom Venus and Mercury would take longe r than expected b ecause o the eect o the S uns gravity. S hapiro measured this time and made a similar measure ment when the signal did not travel close to the S un ( because the planets had moved on in their orbits) . He ound that the delay existed as E instein had predicted and that its magnitude was as expected. S imilar experiments have been repeated in various ways since the 1 9 6 0 s, and always indicate a time delay as predicted b y E instein.
The bending of light S o ar we have discussed the general theory in terms o the equivalence it suggests between gravitational and inertial eects and the changes it predicts or astronomical observations. The general theory, o course, oers much more than this. For Newtonian gravitation, the gravity feld is a model o reality that can be analysed using the law o gravitation. In the general theory, Einstein constructed a set o ten equations known as the Einstein feld equations. These equations indicate that gravity is the eect observed when spacetime is curved ( distorted) by the presence o mass and energy. An analogy or spacetime curvature is that o a rubber sheet. The sheet represents a two-dimensional space in a three-dimensional spacetime continuum. ( This means that it is only an analogy to the real world because one dimension has been suppressed.) In the absence o energy or mass, the sheet is at, horizontal and undistorted. However, i a mass is placed on the sheet, then the sheet deorms under the inuence o the mass as shown in fgure 5 .
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R E L AT I VI T Y light path
star apparent position o star apparent light path sun
earth
Figure 6 A primary star and lensing efect.
Figure 5 Rubber sheet analogy or gravity.
The eect o this deormation on the passage o light rom a distant star can be seen in fgure 5 . Light comes rom the real location o the star but, as it passes near the Sun, the direction o the light is altered by the warping o spacetime. Ater leaving the vicinity o the S un the light appears to have come rom a dierent direction. In 1 91 9 E ddington was able to measure this apparent shit in the star position when the S un moves close to the line between the Earth and the star. Measurements were made in B razil and Arica ( both places where the eclipse was total) and they confrmed the predictions made by the general theory. However the data were hard to collect and it is only recently that reworkings o the data have confrmed that Eddingtons conclusion was not aected by observational errors and confrmation bias. O course, the defection eect is happening in the our dimensions o spacetime, not the three o our analogy. This means that a threedimensional ( to us) gravitational lensing can be observed. Figure 6 is a striking image taken by the Faint O bj ect C amera o a European S pace Agency satellite showing the primary image o a star and our additional images o it ormed by this lensing eect.
event horizon
rs
R photon emitted
The Einstein feld equations do not always have exact solutions (in the sense that two simultaneous equations with two unknowns have an exact solution) . For example, at present, the equations cannot provide an exact solution or the spacetime o two binary stars one o the commonest star arrangements. Physicists usually make simpliying approximations when studying the implications o the equations. I, or example, the assumptions o low speeds and very weak gravity are made, then the feld equations can be manipulated to give Newtons law o gravitation which, thereore, proves to be a special case o general relativity.
Schwarzschild black holes mass M
Figure 7 The Schwarzschild radius.
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Within a ew weeks o the publication o the general theory, Karl S chwarzchild was able to produce one o the frst exact solutions o the Einstein feld equations. He assumed the presence in spacetime
A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) o a spherical, non-rotating, uncharged mass and was able to derive equations or the gravitational feld that surrounds such an obj ect. Using S chwarzchilds equations, the photon redshit that results rom the eect o the gravitational feld o the mass can be derived and is given ( approximately) by Rs gM _ = _= _ 2r rc2
Worked example C alculate the S chwarzchild radius or: a) the Earth ( mass = 6 1 0 24 kg) b) the Sun ( mass = 2 .0 1 0 30 kg) .
where is the emitted wavelength shited by and r is the distance rom the centre o the mass to the point at which the photon is emitted 2 GM . This approximate equation assumes that r is and the constant Rs = ____ c much greater than Rs . Notice the similarity between this equation and the earlier expression or the ractional change in requency 2
f _ g h v _ = c = _ f c2 The constant in the expression, R s , has the dimensions o length and is 2 GM . known as the S chwarzschild radius; it is equal to ____ c
Solution 2 GM a) The S chwarzchild radius = ____ c 2
2 6 .6 7 1 0- 11 6 1 0 24 = ___________________ 9 1 016
= 8.9 mm b) A similar substitution or the mass o the S un gives: 3. 0 km.
2
O utside the S chwarzchild radius ( r > R s) gravity obeys the usual rules, but inside the spherical region defned Rs where r < R s the normal structure o spacetime does not apply. The change between the two regimes where r = R s is known as an event horizon. Events happening inside the event horizon cannot aect an observer outside the horizon. Although the event horizon is not a true boundary ( observers are able to pass through it rom outside to inside) , the event horizon represents the surace where the gravitational pull is so large that nothing can escape not even light. A dierent interpretation o the event horizon is that it is the surace at which the speed needed to escape rom the mass becomes equal to the speed o light. Light emitted rom inside the event horizon cannot escape. The strong gravitational feld ( extreme warping o spacetime) associated with the black hole, attracts nearby mass to it. Mass will appear to collapse towards the centre o the black hole. C locks in a strong gravitational feld run more slowly than in the absence o gravity. The same phenomenon is observed near an event horizon. As a clock moves towards the event horizon rom the outside, an external observer will see the clock slow down with the clock never quite passing through the horizon itsel. The light emitted rom the clock will also be increasingly gravitationally redshited as the clock approaches the S chwarzschild radius ( we already know that this is equivalent to time dilation) . We might expect the situation in the rame o the clock to be dierent and, indeed, the clock ( in its rame) will pass through the horizon in a fnite amount o proper time. The proper time interval to or an observer at distance R o within the gravitational feld o the mass that has a Schwarzchild radius o R s is related to the time interval td measured by a distant observer by ______
to = td
Rs 1 - _ R0
This equation assumes that the obj ect giving rise to the event horizon does not rotate.
Worked example Ticks separated by intervals o 1 .0 s are emitted by a clock that is 2 1 0 5 m rom the event horizon o a black hole o mass 3 1 0 31 kg. C alculate the number o ticks detected by an observer in a distant rocket in a period o 1 0 minutes in the rame o reerence o the rocket.
Solution The S chwarzchild radius 2 6 .6 7 1 0 3 10 is ___________________ = 9 10 -11
31
16
4.5 1 0 4 m_____
Rs
tp = to 1 - __ and when the R measured time interval on the rocket ship is 600 s the proper time at the horizon is 0
__________
600
1 -
4. 5 1 0 4 ________ 2 .0 1 0 31
= 5 2 9 s. There will be 5 92 ticks.
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R E L AT I VI T Y This applies to all bodies that have mass and the worked examples show typical S chwarzchild radii or planetary and solar bodies. In theory, any object that can be compressed sufciently or all its mass to be inside the event horizon will demonstrate these eects. The crucial point is whether this compression is possible or not. For almost all obj ects it is not, because the gravitational sel-attraction o a planets mass is insufcient to overcome the repulsion o the electrons in the atomic shells o atoms. However, or very dense objects, the mass can ft inside the Schwarzchild radius and this object then becomes a black hole. A star needs, typically, to be about three times the mass o the Sun or this to occur. B lack holes are massive, extremely dense obj ects rom which matter and radiation cannot escape. S pacetime inside the event horizon is so warped that any path taken by light inside the S chwarzchild radius will curve arther into the black hole.
Nature of science Links between classical and quantum physics For many years ater Einsteins discovery it was thought that nothing could escape a black hole. It is now recognized that this is only true or black holes under a classical theory. Under quantum theory a black hole can be shown to radiate in a similar way to a black body. This result was unexpected and led to other connections between black holes and the classical study o thermodynamics. The work o the physicist S tephen Hawking and others has led to an understanding o the thermodynamics and mechanics o black holes,
so that it is now recognized that black holes ( or rather the strong gravity feld near them) can lead to the emission o Hawking radiation. This would in principle allow a black hole to be observed and studied. In practice, the radiation emitted by mass being accelerated towards the event horizon will swamp the Hawking radiation. This emission o this radiation implies that black holes can evaporate over time leading to a dynamic process o hole creation and disappearance,
Applications of general relativity to the universe as a whole There are many topics in cosmology that rely on general relativity. S ome o these have been solved; some remain the subj ect o active research. O thers remain as tantalizing theoretical predictions.
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S tudies o the E instein feld equations lead to a number o dierent solutions ( o which S chwarzchilds was one o the frst) in particular the solutions o Friedmann, Lemaitre, Robertson and Walker ( FLRW) . These allow the behaviour o the universe over its lietime to be modelled and the equations have been highly successul in their application. Many aspects o the early universe are illustrated in the FLRW solutions including the large- scale structure o the universe, the way in which chemical elements were created, and the presence o the cosmic background radiation. Additionally, knowledge o the rate at which the universe is expanding allows the total mass o the universe to be estimated.
A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) The answer the equations produce does not agree with the amount o matter that we can actually see around us in the universe. Physicists now suggest the presence o an ( at present) unobservable dark matter and dark energy.
There is a suggestion that single super- massive black holes can be ound at the centre o galaxies. The mass o these obj ects can range up to a mass o several billion times that o our S un. Such black holes will have been inuential in the ormation o galaxies and other larger structures in the universe. As interstellar dust and other materials all into the galactic black holes, a number o arteacts appear, many o which are predicted by the general theory. These include the emission o j ets o very energetic particles at speeds close to that o light that can in principle be observed. Astronomers look careully or evidence o black holes at the centre o galaxies. There is a strong candidate in the Milky Way with an obj ect ( Sagittarius A*) that has a diameter similar to the radius o Uranus but a mass about 4 million times greater than that o the S un.
B inary black holes, normally in orbit around each other, can merge. The general theory suggests that the resulting event should lead to the emission o gravitational waves. O ther events in the universe may also cause gravitational-wave emission. There are experiments in operation and others being devised that will, it is hoped, detect the eects o these gravitational waves and provide urther verifcation o some o E insteins predictions.
TOK The unnecessary constant One orm o the Einstein feld equations is 8G T 1 g R+ g = _ R - _ 2 c4 (this is not going to be tested in the examination!) . R and g tell us about the curvature o spacetime, and T is concerned with the matter and energy in the universe. The constants G and c have their usual meaning. The reason or including this equation is to draw your attention to the third term on the let-hand side o the equation and the constant (capital lambda) that it contains. Einstein called the cosmological constant. In the early part o the twentieth century when Einstein was working on the general theory, the scientifc view was that the universe was stationary and unchanging. Einstein realized that his theory predicted a universe that was neither static nor unchanging according to the equations it should be growing larger. Einstein added the term to adjust the theory to predict a stationary state. A ew years later the astronomer Edwin Hubble ound strong evidence that the universe is expanding. At this point, Einstein realized that his inclusion o had been unnecessary. He later described putting the constant into the equation as my greatest blunder. What other examples are there o scientists whose work was initially rejected (either by themselves or others) but which was later accepted?
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R E L AT I VI T Y
Questions 1
E xplain what is meant by an inertial frame of reference. ( 1 mark)
2
An electron is travelling parallel to a metal wire that carries an electric current. D iscuss the nature o the orce on the electron in terms o the rame o reerence o:
3
4
5
(IB) The radioactive decay o a particular nuclide involves the release o a - particle. A betaparticle detector is placed 0.3 7 m rom the actinium source, as measured in the laboratory reerence rame. The Lorentz actor o the beta particle ater release is 4. 9. a) C alculate, or the laboratory reerence rame:
a) a proton stationary with respect to the wire
(i)
b) the moving electron.
(ii) the time taken or the - particle to reach the detector.
Two electrons are travelling directly towards one another. Each has a speed o 0.002 c relative to a stationary observer. C alculate the relative velocity o approach, as measured in the rame o reerence o one o the electrons according to the Galilean transormation. ( 2 mark)
(IB) a) D efne: (i) proper length
b) The events described in ( a) can be described in the - particles rame o reerence. For the rame o the beta particle: (i)
(ii) proper time
(IB) a) E xplain what is meant by an inertial frame of reference. b) An observer in reerence rame A measures the relativistic mass and the length o an obj ect that is at rest in their reerence rame. The observer also measures the time interval between two events that take place at one point in the reerence rame. The relativistic mass and length o the obj ect and the time interval are also measured by a second observer in reerence rame B . B is moving at constant velocity relative to A.
Muons are accelerated to a speed o 0.95 c as measured in the reerence rame o the laboratory. They are counted by detector 1 and any muons that do not decay are counted by detector 2 . The distance between detector 1 and 2 is 1 3 70 m. v = 0.95 c muons detector 2 1370 m
c) (i) Hal the number o muons pass through detector 2 as pass through detector 1 in the same given time. in the laboratory rame
(ii) in the rest rame o the muons. (iii) D etemine the separation o the counters in the muon rest rame. c) Use your answers in ( b) to explain what is meant by the terms time dilation and length contraction. ( 1 1 marks)
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State whether the observer in rame B measures the quantities as being larger, the same size or smaller than when measured in rame A.
(ii) C ompare the density o the obj ect measured in rame B with the same measurement made in A. ( 7 marks)
b) D etermine, the hal- lie o the muons (i)
state the speed o the detector
(ii) calculate the distance travelled by the detector. ( 7 marks)
6
detector 1
the speed o the -particle
7
Some students are marooned on a planet carrying out feld work when it becomes clear that the star o the planet system is about to become a supernova. A spaceship is despatched to rescue them and the students can be beamed
QUESTION S
aboard the spaceship without the need or it to change speed. The spacetime diagram shows the rame o the star and planet and the rame o the spaceship. The star and the planet do not move relative to each other. The star becomes a supernova at spacetime point S .
ct
ct
ip ra m
e
x
a ce
time axis o planet rame
sh
S fnishing line
xi s
o
sp
a) D iscuss who wins the race in the reerence rame o:
ti m
ea
(i)
S
ce sp a
a xi
s
p o s
ac
ip esh
ra
me
(iii) the reeree. b) Discuss whether there is agreement about the result o the race. ( 1 0 marks)
x planet
space axis o planet rame
a) C opy the diagram to scale and on it identiy: (i)
the spacetime point at which the spaceship arrives at the planet
(ii) the spacetime point o the star when the spaceship arrives at the planet. b) D iscuss whether the spaceship arrives at the planet beore or ater the supernova: (i)
in the rame o the spaceship
(ii) in the rame o the planet. c) D iscuss when the IB students see the supernova. ( 9 marks)
8
F
(ii) S
9
star
F fnishing line
Two athletes compete in a race. Athlete S is slow and is awarded a handicap so that he has less ar to run than athlete F. The spacetime diagram or the race is shown or the rame o reerence o the reeree who is stationary relative to the fnishing lines.
(IB) a) D efne rest mass. b) An electron o rest mass m 0 is accelerated through a potential dierence V. Explain why, or large values o V, the equation 1 __ m 0 v2 = eV cannot be used to determine the 2 speed v o the accelerated electron. c) D etermine the mass equivalence o the change in kinetic energy o the electron when V is 5 MV. ( 7 marks)
1 0 (IB) a) A charged particle o rest mass m 0 and carrying charge e, is accelerated rom rest through a potential dierence V. Deduce that eV = 1 + _2 m 0c where is the Lorentz actor and c is the speed o light in ree space. b) C alculate the speed attained by a proton accelerated rom rest through a potential dierence o 5 00 MV. ( 5 marks)
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R E L AT I VI T Y 1 1 A proton is accelerated rom rest through a potential dierence o 2 . 0 1 0 9 V. C alculate, in MeV c1 the fnal momentum o the proton. ( 3 marks) 1 2 (IB) a) D istinguish between the rest massenergy o a particle and its total energy.
acceleration
initial direction of light beam
initial direction of light beam
b) The rest mass o a proton is 93 8 MeV c2 . State the value o its rest massenergy. c) A proton is accelerated rom rest. D etermine the potential dierence through which it must be accelerated to reach a speed o 0.98c, as measured by a stationary observer in the laboratory reerence rame. ( 9 marks)
acceleration
base
(i)
base
Describe the path taken by the light beam as observed by the inertial observer.
(ii) Describe the path taken by the light beam as observed rom within the spaceship. (iii) Explain the dierence between the paths.
1 3 (IB) a) In both the special and general theories o relativity, Einstein introduced the idea o spacetime. D escribe, with a diagram, what is meant by spacetime with reerence to a particle that is ar rom other masses and is moving with a constant velocity. b) Explain how the general theory o relativity accounts or the gravitational attraction between the E arth and an orbiting satellite.
b) Explain how this dierence relates to the equivalence principle. ( 7 marks)
1 5 (IB) a) S tate the principle o equivalence. b) A spacecrat is initially at rest on the surace o the Earth. Ater accelerating away rom Earth into deep space, it then moves with a constant velocity. A spring balance supports a mass rom the ceiling.
c) D escribe what is meant by a black hole. d) Estimate the radius o the S un necessary or it to become a black hole. ( 1 2 marks)
1 4 (IB) a) A spaceship in a gravity- ree region o space accelerates uniormly with respect to an inertial observer, in a direction perpendicular to its base. A narrow beam o light is initially directed parallel to the base. D iagram 1: View with D iagram 2: View with respect to respect to inertial observer observer in ship
0
0
0
(a)
(b)
(c)
The diagrams show the readings on the spring balance at dierent stages o the motion. Identiy, with an explanation, the reading that will be obtained when the spaceship is: (i)
at rest on the Earths surace
(ii)
moving away rom Earth with acceleration
(iii) moving at constant velocity in deep space.
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QUESTION S c) The spacecrat now accelerates in deep space such that the acceleration equals that o ree all at the E arths surace. State and explain which reading would be observed on the spring balance. ( 1 0 marks)
1 8 (IB) On 2 9 March 1 91 9, an experiment by Eddington carried out during a total eclipse provided evidence to support Einsteins general theory. The diagram below (not to scale) shows the relative position o the Sun, Earth and a star S on this date.
1 6 (IB) The gravitational feld o a black hole warps spacetime.
S
sun
earth
orbit path of earth about sun
Eddington measured the apparent position o the star and six months later, he again measured the position o the star rom Earth. a) (i) (ii)
D escribe what is meant by the centre and the surface o a black hole. D efne the Schwarzchild radius.
(iii) C alculate the Schwarzchild radius or a star that has a mass 1 0 times that o the S un ( solar mass = 2 .0 1 0 30 kg) . b) In 1 979, astronomers discovered two very distant quasars separated by a small angle. Examination o the images showed that they were identical. O utline how these observations give support to the theory o general relativity. ( 9 marks)
1 7 One prediction rom the principle o equivalence is the eect known as gravitational lensing. a) S tate the principle of equivalence. b) Use the principle to explain gravitational lensing. ( 5 marks)
a) S tate why the experiment was carried out during a total solar eclipse. b) Explain why the position o the star was measured again six months later. c) C opy the diagram and draw the path o a ray o light rom S to the E arth as suggested by the general theory. d) Explain how Einsteins theory accounts or the path o the ray. e) Label the apparent position o the star as seen rom Earth. ( 6 marks)
1 9 (IB) a) (i)
D escribe, with reerence to spacetime, the nature o a black hole.
(ii) D efne, with reerence to ( a) ( i) , the S chwarzchild radius. (iii) A star that has a mass o 4.0 1 0 31 kg evolves into a black hole. C alculate the S chwarzchild radius o the black hole, stating any assumption that you make.
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A
R E L AT I VI T Y b) A spacecrat approaches the black hole in ( a) ( iii) . I it were to continue to travel in a straight line it would pass within 1 000 km o the black hole. (i) S uggest the eect the black hole would have on the motion o the spacecrat. (ii) Explain gravitational attraction in terms o the warping o space-time by matter. ( 1 0 marks) 2 0 Alan and B renda are in a spaceship that is moving with constant speed. C lose to Alan is a light source fxed to the oor o the spaceship. B oth Alan and B renda measure the same value or the requency o the light emitted by the source.
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Alan
Brenda
The spaceship begins to accelerate. a) Explain why B renda observes the source close to Alan to emit light o a lower requency during the acceleration. b) O utline how the situation described in ( a) leads to the idea o gravitational redshit. ( 5 marks)
B E N G I N E E RI N G PH YS I CS Introduction Engineering physics covers some o the key topics that you would expect to meet on an undergraduate engineering course. These are quite diverse topics covering rotational motion, thermodynamics, uid mechanics, and orced vibrations and
resonance. The mathematical content o these topics is naturally restricted to be in line with a preuniversity course; however, those students with a strong mathematical inclination will fnd much in these sub-topics to stimulate urther research.
B.1 Rigid bodies and rotational dynamics Understanding Torque Moment o inertia
Applications and skills Calculating torque or single orces and couples Solving problems involving moment o inertia,
Rotational and translational equilibrium Angular acceleration
Equations o rotational motion or uniorm
angular acceleration Newtons second law applied to angular motion Conservation o angular momentum
Nature of science Extended objects The work covered until now has almost invariably used the model o an object being represented by a single particle o zero dimensions. This model works well or many situations but alls short when orces are not applied along a line passing through the centre o mass o the object. In such cases an extended object can still be modelled; however, objects o dierent shape will not always move in the same way when orces are applied to them. Using simpliying models, engineers were able to design and manuacture sophisticated machines that had signifcant impact during the industrial revolution o the late eighteenth to early nineteenth century.
torque, and angular acceleration Solving problems in which objects are in both rotational and translational equilibrium Solving problems using rotational quantities analogous to linear quantities Sketching and interpreting graphs o rotational motion Solving problems involving rolling without slipping
Equations Torque equation: = Fr sin Moment o inertia: I = mr2 Newtons second law for rotational motion:
= I Relationship between angular requency and requency: = 2f Equations o motion or constant angular acceleration: = i + t 2 = 2i + 2 = i t + _1 t2 2 Angular momentum: L = I Rotational kinetic energy: E K = _1 I 2 2 rot
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B
E N G I N E E R I N G P H YS I C S
Introduction Note
The angular acceleration is diferent rom both centripetal acceleration ( a c) and tangential acceleration ( a t) this is the rate o change o the linear tangential velocity with time.
As an object moving in a circle undergoes angular acceleration, its tangential acceleration increases as does its centripetal acceleration.
The tangential acceleration o the body is related to the angular acceleration and the radius o the circle (r) by the relationship: at = r
This is because vt = r at
v This means that = _______ = ____r rt where vt is the tangential velocity o the object. Thus the linear quantity (velocity or acceleration) is the angular quantity multiplied by the radius.
This sub- topic is very closely related to the linear mechanics and circular motion that you studied in Topics 2 and 6. There are quantities that are directly analogous to the linear quantities o displacement, velocity, acceleration, orce, etc. With the knowledge that you have already absorbed and an understanding that a rigid body is an extension o a point obj ect you will be able to solve most rotational dynamics problems.
Uniorm motion in a circle You will remember that or a body moving around a circle with a constant linear speed ( or angular speed) there needs to be a means o providing a centripetal orce. This can be a contact orce, such as tension or riction ( or a component o these orces) , or a orce provided by a feld, such as Newtons law o gravitation or C oulombs law. In all cases the orce ( or component) providing the centripetal orce will be given by: mv2 2 F= _ r = m r These terms were defned in Topic 6. Remember that is known as the angular velocity or angular requency and is related to requency f by the equation: = 2 f
Angular acceleration Let us consider an obj ect moving in the circle so that it is given an angular acceleration and its angular speed increases. We defne angular acceleration () as the rate of change of angular sp eed with time. = _ t is measured in radian per second squared ( rad s 2 )
Worked example The diagram shows a hula- hoop ( a large plastic ring) rolling with constant angular speed along a horizontal surace prior to rolling down a uniorm inclined plane. When it reaches a second horizontal surace it, again, moves with a constant angular speed.
hula-hoop
A
S ketch graphs to show the variation with time o a) the angular velocity and b) the angular acceleration o the hula- hoop.
B
Figure 1
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C
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Solution a) The graph shows the hula- hoop travelling with constant angular velocity along A and C . It has a greater value along C since it has now undergone angular acceleration. As B is o constant gradient, the angular acceleration is constant here. b) The second graph shows zero angular acceleration throughout A and C and a constant angular acceleration along B . You should compare these graphs with those for a point object moving along a frictionless surface.
When the hula-hoop is travelling at the higher angular velocity it covers the same distance in a shorter time
/arbitrary units
C B A t/arbitrary units /arbitrary units
B
A
C
t/arbitrary units
Figure 2
Equations of motion In Topic 2 we met the our equations o motion under constant linear acceleration: v = u + at 1 s = ut + __ at 2 2
v2 = u 2 + 2as (v + u)t s=_ 2 In rotational dynamics there are our analogous equations that apply to a body moving with constant angular acceleration: = i + t 1 t2 = it + __ 2
2 = 2i + 2 ( i + )t =_ 2 Here i = initial velocity (in rad s 1 ) , = fnal velocity (in rad s 1 ) , = angular displacement (in rad) or sometimes just " angle" , = angular acceleration (in rad s2 ) , t = time taken or the change o angular speed (in s) .
Note The IB Physics Data Booklet does not include the last of these rotational equations it is probably the most straightforward of the equations because it simply equates two ways of expressing the average angular speed.
Worked example
Solution
A wheel is rotated rom rest with an angular acceleration o 8. 0 rad s 2 . It accelerates or 5 .0 s.
a) = i + t = > = 0 + 8.0 5 . 0 = 40.0 rad s 1 ( i + )t ( 0 + 40.0) 5 . 0 b) = _ = __ = 1 00 rad 2 2 Each revolution makes an angle o 2 radian 1 00 so the number o revolutions = ___ 2
D etermine: a) the angular speed b) the number o revolutions that the wheel has rotated through.
= 1 5 .9 revolutions.
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Moment of inertia The moment o inertia ,I, o a body is the rotational equivalent o the role played by mass in linear dynamics. In a similar way to the inertial mass o an object being a measure o its opposition to a change in its linear motion, the moment o inertia o an object is its resistance to a change in its rotational motion. Objects such as ywheels that need to retain their rotational kinetic energy are designed to have large moments o inertia as shown in fgure 3. The moment o inertia o an obj ect depends on the axis about which it is rotated. For a particle ( a single point) o mass m rotating at a distance r about an axis, the moment o inertia is given by: Figure 3
A pair o fywheels with much o the mass distributed to be around the perimeter.
I = mr2 Moments o inertia are scalar quantities and are measured in units o kilogram metres squared ( kg m 2 ) . For an obj ect consisting o more than one point mass, the moment o inertia about a given axis can be calculated by adding the moments o inertia or each point mass.
m
r
I = mr2
axis o rotation
This summation is a mathematical abbreviation or m 1 r1 2 + m 2 r2 2 + m 3 r1 3 + . . . + m nrn2 when the obj ect is made up o n point masses. The moment o inertia o a simple pendulum o length l and mass m is given by Ipendulum = ml2 while that o a simple dumb-bell consisting o two masses m connected by a light rod o length l, when rotating about the centre o the rod, will be
Figure 4 Moment o inertia
o a
point mass.
( )
l I dumb-bell = m _ 2
2
( )
l + m _ 2
2
( )
l = 2m _ 2
2
1 ml2 =_ 2
axis o rotation into plane o the page light string
light connecting rod l/2
l
point masses o value m
bob o mass m l/2
Note With linear motion there is just one single mass value or an object. In rotational motion, the moment o inertia is also a unction o the position o the axis o rotation. This means that there are an infnite range o possible moments o inertia or any one object.
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simple pendulum
Figure 5 Moment o inertias o a
m axis o rotation simple dumb-bell into plane o page
simple pendulum and dumb-bell.
In each o these cases we assume that the mass is a point mass and that the string and the connecting rod are so light that their masses can be ignored. For a more complicated structure we need to use integral calculus in order to derive ormulae or moments o inertia. In IB D iploma Programme physics questions, any equations that need to be used or moments o inertia will be provided in the question.
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Torque Torque , , can be defned as shown in fgure 6. C onsider orce F acting at point P on an obj ect. The direction o F is such that it makes an angle to the radius o the circle in which the obj ect rotates. The torque will be given by: = Frsin r sin
This is sometime called the orce multiplied by the " arm o the lever" . Torques are also called " moments" , but to avoid conusion between " moment" and " momentum" we advise sticking to the term torque. They are ( pseudo) vector quantities with the direction perpendicular to the plane o the circle in which the obj ect rotates. Imagine your right hand gripping the axis about which the obj ect rotates so that the fngers curl round the axis, when your thumb is " up" it will be pointing in the direction o the torque ( as shown in fgure 7) . This rule also applies to the directions o angular displacement , , angular velocity , , and angular acceleration , , which, like torque, are all vector quantities. The maximum torque that a orce can apply to a body is when the orce is perpendicular to the arm o the lever. In this case = 90 and so sin = 1 . This means that = Fr
F radius (r)
P
Figure 6 Defnition
o torque.
F applied orce radius rom axis
r
torque direction
Couples A coup le consists of a p air of equal and op p osite forces that do not act in the same straight line ( see fgure 7) . This combination o orces produces a torque that causes an obj ect to undergo angular acceleration without having any translational acceleration.
d
F P d
Figure 7
Direction o torque.
-F
P -F
(a)
F (b)
Figure 8 Two examples o couples.
The torque o the couple about the pivot P in both fgure 8( a) and ( b) is equal to Fd ( that is the product o one o the orces and the perpendicular distance between them) .
A single resultant orce acting through the centre o mass o an object will produce translational acceleration. This means that the object will move in a straight line. An object in ree-all is an example o this as (ignoring air resistance) the pull o gravity is the only orce acting on it.
The same orce, but acting through a point displaced rom the centre o mass o the object, will produce a combination o both linear and angular acceleration. This means that the object will move in a helical path.
Note Torque is measured in units of newton metre (N m) but dont confuse this with the unit of work or energy (the joule). In the case of torque the force is perpendicular to the direction in which the object moves but in the case of work the force and direction moved are the same. Torque is a vector quantity while work is a scalar.
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E N G I N E E R I N G P H YS I C S Figure 9 shows a orce acting at the top o an obj ect the eect o this orce can be split into the ( i) same orce acting at the centre o mass o the obj ect ( producing translation) and ( ii) a couple ( producing rotation) . 1 1 The couple consists o __ F at the top and - __ F at the bottom o the obj ect. 2 2 This is a general p rincip le that can be app lied to any force not acting through the centre of mass of an obj ect. A couple acting on an obj ect will produce angular acceleration with no linear motion at all. This means that the object will move in a circular path ( as shown in fgure 8) . 1 2F
F
=
+ F
centre of mass
-1F 2 translation Figure 9
rotation
Force acting through point displaced from centre of mass.
Newtons frst law or angular motion rotational equilibrium We saw in Topic 2 that a body in (translational) equilibrium does not accelerate but remains in a state o rest or travels with uniorm velocity. This means that there can be no resultant orce acting on the body in line with Newtons frst law o motion. For an object to be in rotational equilibrium there can be no external resultant torque acting on it it will then remain in a state o rest or continue to rotate with constant angular velocity. For rotational motion Newtons frst law may be stated as: An object continues to remain stationary or to move at a constant angular velocity unless an external torque acts on it. When the obj ect is in rotational equilibrium: total clockwise torque = total anticlockwise torque. This statement is oten called the " principle o moments" .
Newtons second law or angular motion angular acceleration Again, in Topic 2 we considered the simpler statement o Newtons second law o ( translational) motion equating orce to the product o mass and acceleration. The rotational equivalent o Newtons second law relates the angular acceleration and torque on a body o moment o inertia. = I
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In IB D ip loma Programme Physics questions, the axis to which the torque is ap p lied and the axis about which the moment of inertia is taken will always be the same ( although there are quite simple ways o dealing with situations when this is not the case) .
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Newtons third law for rotational motion You can probably guess that this version o the law says that action torque and reaction torque are equal and opposite this pair o torques, like action and reaction orces or linear motion, act on dierent bodies. I body A applies a torque to body B , then body B applies an equal and opposite torque to body A.
Angular momentum Linear momentum ( p) is a vector quantity defned as being the product o mass and velocity. Angular momentum ( L) is the rotational equivalent to this and is defned as being the product o a bodys moment o inertia and its angular velocity; this, too, is a vector. L = I Angular momentum is measured in units o kg m 2 rad s 1 .
Worked example Solution
A couple, consisting o two 4.0 N orces, acts tangentially on a wheel o diameter 0.60 m. The wheel starts rom rest and makes one complete rotation in 2.0 s. C alculate:
1 t2 = > 2 = 0 + _ 1 2 .0 2 a) = it + _ 2 2 4 = _ = = 3 .1 4 rad s 2 4. 0 b) torque o couple = Fd = 4.0 0.60 = 2.4 N m 2 .4 _ 2 as = I = > I = _ = 3 .1 4 = 0.76 kg m
a) the angular acceleration b) the moment o inertia o the wheel.
The conservation of angular momentum In linear dynamics we fnd the total ( linear) momentum o a system remains constant providing no external orces act on it. In rotational dynamics, the total angular momentum o a system remains constant providing no external torque acts on it. Figure 1 0 shows a small mass being gently dropped onto a reely spinning disc. The addition o the mass increases the combined moment o inertia o the disc and the mass and so the angular velocity o the system now alls in order to conserve the angular momentum. small mass
r
spinning disc Figure 10 An
example of the conservation of angular momentum.
The conservation o angular momentum has many applications in physics, such as the speeding up o an ice skater when pirouetting with decreasing angular momentum, or a gymnast putting in turns or twists by changing the distribution o mass.
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Rotational kinetic energy You probably eel confdent about the analogies between linear and rotational dynamics, so it should not surprise you that the relationship or angular kinetic energy is 1 E K = __ I 2 2 ro t
This, like all energies, is a scalar quantity and could, in principle, be ound by adding the ( translational) kinetic energies o all the particles making up a rotating obj ect. Two more useul rotational analogies are given in the table below.
Quantity
Linear equation
Angular equation
work
W = Fs
W =
power
P = Fv
P =
Rolling and sliding
R + v
v R
R - v Figure 11
Disc rolling on a fat
surace.
I an obj ect makes a perectly rictionless contact with a surace it is impossible or the obj ect to roll it simply slides. When there is riction the obj ect can roll; as the point o contact between the rolling body and the surace along which it rolls is instanteously stationary, the coefcient o static riction should be used in calculations involving rolling. The point o contact must be stationary because it does not slide. Figure 1 1 shows a disc o radius R rolling along a at surace such that its centre o mass has a velocity v. Each point on the perimeter o the disc will have a tangential velocity = R. This means that the top o the disc will have total velocity o R + v and the point o contact with the surace will have a velocity o R - v. As the disc is not slipping, the bottom o the disc has zero instantaneous velocity and so R = v. This means that the top o the disc will have an instantaneous velocity = 2 v. The total kinetic energy o a body that is rolling without slipping will 1 1 be = __ I 2 + __ mv2 . When an obj ect rolls down a slope so that it loses a 2 2 vertical height h, the loss o gravitational potential energy will become the total kinetic energy. This gives us 1 1 mgh = __ I 2 + __ mv 2 2 2
Worked example A solid ball, o radius o 45 mm, rolls down an inclined plane o length 2 .5 m. The sphere takes a time o 6.0 s to roll down the plane. Assume that the ball does not slip. The moment o inertia, I, o a solid ball o 2 mass M and radius R is given by I = __ MR 2 5 a) C alculate the velocity o the sphere as it reaches the end o the plane.
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b) C alculate the angular velocity o the sphere as it reaches the end o the plane. c) D etermine the angle o inclination o the plane. d) C omment on whether the assumption that the ball does not slip was appropriate in this instance.
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Solution a) For the ball starting at rest rolling along the slope or time t the ( translational) equations o motion give: 1 s = __ at2 and v = at 2
Substituting the values or s and t 2s _ 2 2 .5 a=_ = = 0.1 4 m s 2 t2 6.0 2 v = at = 0.1 4 6.0 = 0.84 m s 1
substituting or the moment o inertia equation we have: 2 MR2 _ 5 1 __ 2 _ + M Mgh = 2 v R2
(
)
C ancelling M and R2 gives
(
)
2 + 1 =_ 7 2 1 gh = __ v2 _ v 2 5 10 v 7 So h = __ __ g where h is the height o the end 10 o the inclined plane. 2
0.84 b) The angular speed = _vr = _______ = 1 8.7 45 1 0 1 1 9 rad s 3
__1 mv
1 c) rom the equations mgh = __ I 2 + 2 and v = R
(
1 2 _ mgh = __ v I2 + m 2 R
7 0.84 2 h = _ _ = 0.05 m 10 9.81
2
2
0.05 sin = _ 2.5
)
(
)
0.05 = sin 1 _ = 1 .1 2 .5
d) This angle is very small and so there is no problem with assuming that the ball rolls without slipping.
Investigate! Measuring the moment o inertia o a fywheel A fywheel is mounted so that it can rotate on a horizontal axle (the axle orming part o the fywheel) . A mass, suspended by a string, is wound round the axle o a fywheel ensuring that it does not overlap. When the mass is released, the gravitational potential energy o mass is converted into the linear kinetic energy o the mass + the rotational kinetic energy o the fywheel + work done against the rictional orces (all these values are taken when the string loses contact with the axle) . 1 1 I 2 + n W mgh = _ mv2 + _ 1 2 2 The symbols here have their usual meaning with n 1 being the number o turns o string and W the work done against rictional orces during each revolution. When the string disengages, the rotational kinetic energy makes a urther n 2 rotations beore it comes to rest so the rotational kinetic energy o the fywheel must be equal to n 2 W or 1 I 2 _ I 2 = n 2 W making W = _ 2 n2 2 2
n I 1 1 This means mgh = __ mv 2 + __ I 2 + ____ 2 2 2n 1
or
n 1 1 mgh = __ mv 2 + __ I 2 1 + __ n 2 2
(
1
2
)
2
v h In addition to this = vr and __ = __ t 2
Measure the radius r o the axle o the system using vernier or digital callipers.
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Attach a mass hanger to one end o the piece o string and make a loop at the other end so that, as the mass hanger j ust touches the oor, the loop slips o the peg in the axle.
Loop the string over the peg, wind up the string or a whole number o revolutions n 1 and measure the height h o hanger above the oor. Make sure the windings do not overlap. view rom ront
view rom side
axle
peg r axle
fywheel axis o rotation
slotted masses mass hanger
h
Figure 12
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Measuring the moment o inertia o a fywheel.
Adj ust the number o slotted masses on the hanger.
Measure the time t rom the moment o release until the hanger reaches the oor.
C ount the number o revolutions n 2 the ywheel makes ater the string comes o the peg until it comes to rest.
Repeat several times and calculate mean values rom which a value or the moment o inertia o the ywheel and axle I can be deduced.
C onsider how the data could be used graphically to fnd a value or I.
B . 2 TH E RM O D YN AM I CS
B.2 Thermodynamics Understandings The frst law o thermodynamics The second law o thermodynamics Entropy
Applications and skills Describing the frst law o thermodynamics as a
Cyclic processes and pV diagrams Isovolumetric, isobaric, isothermal, and
adiabatic processes Carnot cycle Thermal eciency
Nature of science Dierent viewpoints o the second law o thermodynamics Thermodynamics is an area o physics which, through dierent scientifc eras, has been shaped by a combination o practice and theory. When Sadi Carnot wrote his treatise about heat engines, he was a believer in the caloric standpoint yet his ideas were suciently developed to inuence the development o machines in the industrial revolution. Clausius, Boltzmann, Kelvin, and Gibbs were all responsible or dierent statements o the second law o thermodynamics a law that has a undamental impact on whether or not a process, allowed by the frst law o thermodynamics, can actually occur.
statement o conservation o energy Explaining sign convention used when stating the frst law o thermodynamics as Q = U + W Solving problems involving the frst law o thermodynamics Describing the second law o thermodynamics in Clausius orm, Kelvin orm and as a consequence o entropy Describing examples o processes in terms o entropy change Solving problems involving entropy changes Sketching and interpreting cyclic processes Solving problems or adiabatic processes or 5 _
monatomic gases using p V 3 = constant Solving problems involving thermal eciency
Equations First law o thermodynamics: Q = U + W
3 2 Q Entropy change: S = _ T Equation o state or adiabatic change: Internal energy: U = _ nRT
5 _
p V 3 = constant (or monatomic gases) Work done when volume changes at constant pressure: W = pV useul work done Thermal eciency: = __________________________ energy input Tc o l d Carnot eciency: Ca rn o t = 1- _ Th o t
Introduction When using thermodynamics, there is a convention that we talk about the body that we are interested in as being the system. Everything else that may have an impact on the system is known as the surroundings. The system is separated from the surroundings by a boundary or wall. Everything including the system and the surroundings is called the universe.
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surroundings universe
system
the fask represents the boundary in this case Figure 1
Thermodynamic system and surroundings.
As we saw in Topic 3 , James Joule showed that work done on a system or energy transerred to the system because o temperature dierences result in the same outcome: the internal energy o that system increases. The nature o the way the system changes and how that energy reveals itsel depends on whether or not the phase o the substance changes. When there is no change o state the most apparent eect is the increase in the mean random kinetic energy o the particles; when there is a change o state the increase in the potential energy is the most signifcant eect. Remember that the internal energy of a system is the total of the potential energy and the random kinetic energy of all the particles making up the system. We oten simpliy discussion by considering the system to be an ideal gas. In this case the internal energy is entirely kinetic and we can use the relationship derived in Sub-topic 3 .2 . This showed that, or n moles o an ideal gas, the internal energy U is related to the absolute temperature T 3 by the equation U = __ nRT, where R is the universal molar gas constant. 2
The frst law o thermodynamics The internal energy o a system may change as any combination o ( i) doing work on the system ( or allowing the system to do work on the surroundings) and (ii) transerring energy to or rom the system as a result o a dierence in temperature. Saying this amounts to stating the conservation o energy. There are a variety o versions o the equation or the frst law o thermodynamics and each has its merits they will all be sel-consistent and understandable but you will need to make sure that you read the defnition o each o the terms in the equation. The preerred version o the equation or the IB Physics syllabus is written as: Q = U + W When each o these quantities is p ositive:
Q represents the energy transerred from the surroundings to the system because the surroundings are at a higher temperature than the system
U represents the increase in the internal energy o the system ( this is not simply a change, it is an increase)
W represents the work done by the system as it expands and pushes back the surroundings.
When each o these quantities is negative:
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Q represents the energy transerred from the system to the surroundings because the system is at a higher temperature than the surroundings
B . 2 TH E RM O D YN AM I CS
U represents the decrease in the internal energy o the system
W represents the work done on the system as the surroundings compresses it.
Worked example A systems internal energy alls by 2 00 J as a result o energy transer and work being done. The system does 5 00 J o work on the surroundings.
Solution
a) S tate and explain whether the system is at a higher or lower temperature than its surroundings.
b) Using the frst law o thermodynamics Q = U + W, U must be negative and W must be positive.
b) C alculate the amount o energy transerred causing the reduction in internal energy.
a) The system must gain energy in order to be able to do this amount o work and so its temperature must be below that o the surroundings.
Q = - 2 00 + 5 00 = 3 00J S o 3 00 J are transerred to the system rom the surroundings.
Nature of science Human metabolism and the frst law Let us consider a human body as a thermodynamic system. When we eat, our internal energy increases because we are taking in ood in the orm o chemical potential energy. When there is no energy transer because o temperature dierences this must be a process which is related to work although we will not discuss the biochemistry here. S o, using the frst law with no energy transer: 0 = U + W As U is positive this means that W is negative. The chemical energy in the ood we eat does three main things: it allows us to do work on
our surroundings, it allows us to transer energy to the ( usually) cooler surroundings and the remainder is stored in our bodies ( as at) . With a well- adj usted, balanced diet the build up o at and change in internal energy is zero and the ood we eat allows us to stay warm and do work and be active. S o the ideal system is to make the net change in the internal energy o our bodies zero eat too much and we will build up at ( U is positive) , eat too little and we will lose weight ( U is negative) . In reality, human metabolism is much more complicated than this, but the frst law o thermodynamics does represent a model that has many more applications than the gases we will now ocus on.
Using the frst law or ideal gases When discussing the changes that can be made to the state o the gas, it is usual to consider an ideal gas enclosed in a cylinder by a moveable piston. The gas represents the system, the cylinder and the piston represent the boundaries or walls. Everything else becomes the surroundings.
Calculating the work done in an isobaric change An isobaric change is one which occurs at constant pressure. Consider an ideal gas at a pressure p enclosed in a cylinder o cross-sectional area A. When the gas expands it pushes the piston a distance x so that the volume
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E N G I N E E R I N G P H YS I C S o the gas increases by V (= Ax) . When energy is supplied to the gas rom the surroundings the pressure can remain constant at value at p. The work done by the gas on the surroundings during this expansion will be W. The orce F o the gas on piston = pA. This means that the work done during expansion = Fx = pAx = pV.
piston
W = p V V= A X
X
area A
Figure 3 shows a p- V graph or an isobaric change. The area under the graph is equal to the work done. The arrow on the line connecting the end points AB shows that the gas is expanding and doing work on the surroundings an arrow in the opposite direction would show a compression.
volume V
ideal gas = system
cylinder Figure 2
Applying the frst law o thermodynamics to this, we get Q = U + pV. This could result in any number o possibilities but they must be consistent with this equation; Q must be positive or an expansion and negative or a compression. For example, supplying 2 0 J to the gas could increase the internal energy by 1 9 J and allow the gas to do 1 J or work or increase the internal energy by 1 9.1 and do 0.9 J o work etc.
Work done by an ideal gas.
p area under line = work done by gas on surroundings = p V A
p
V The equation o state or an isobaric change is __ = constant. T
B
Work done for non-isobaric changes
p
V Figure 3
This will be a positive quantity because the system is doing work on the surroundings. When the gas is compressed W will be negative.
When changes do not occur at constant pressure we can still calculate the work done rom the area under a p- V graph. We can make the assumption that the pressure will be unchanged over a small change in volume and, thereore, we approximate the overall change to a series o small isobaric changes as shown in fgure 4.
V
Work done in an isobaric change.
The area o the frst constant pressure rectangle = p 1 V1 = W1 p
A
The area o the second rectangle would be p 2 V2 = W2 , etc.
area of isobaric rectangle = p 1 V1 = work done
p1 p2
Thereore, the total work done = n Wn = n p n Vn or n rectangles this is the area under the curve. S o or any p- V graph the area under the curve will give the work done this depends on the path taken and not j ust on the end points.
B
Isothermal changes V1
V
V2
Figure 4 Work done in
a non-isobaric change.
Note Isothermal changes normally take place very slowly and the boundary between the system and the surrounding must be a good conductor of energy.
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Isothermal changes are those that occur resulting in the internal energy of the system staying constant. The internal energy o an ideal gas consists o the sum o the mean random kinetic energies o the particles o gas. The mean kinetic energy is proportional to the temperature o the gas. An isothermal change, thereore, means that there is no change in the temp erature of the system. Using the frst law o thermodynamics Q = U + W with U = 0, this leaves Q = W. As there is no change in internal energy, the energy transerred to the system because o a temperature dierence between the system and the surroundings Q will allow the system to do work W on the surroundings. As W = pV this can only mean that the gas is expanding the direction arrow on the graph should go rom A to B .
B . 2 TH E RM O D YN AM I CS
The other possibility is that - Q = - W so the energy transerred rom the system to the surroundings is equal to the work done on the system by the surroundings. In this case the gas is being compressed the direction arrow on the graph should go rom B to A.
p
We saw in Topic 3 that when the temperature does not change pV = constant. This is the equation o the line on a p- V graph or an isothermal change. The lines are known as isotherms. As shown in fgure 6, changes at higher temperatures will always produce isotherms that are urther rom the origin than those at lower temperatures. For a given volume the pressure will always increase in moving rom a low temperature isotherm to one at higher temperature.
A
B area = work done V
Adiabatic changes
Figure 5 Work done in
This is the name given to a change in which no energy is transferred between the system and the surroundings. This does not mean that the system and the surroundings are always at the same temperature, although that could be so, but it is more likely that there is a wellinsulated barrier between them. Adiabatic changes usually happen very quickly, which means that there is no time or the energy to transer.
an isothermal change.
p T1 > T2
Applying the frst law o thermodynamics to a system undergoing an adiabatic change gives:
T1
Q = U + W but, as Q = 0, this can mean either U = - W ( an increase in internal energy occurs because o work being done on the system) or - U = W ( a decrease in internal energy occurs because the system is doing work on the surroundings) . For an ideal monatomic gas the equation or an adiabatic change takes the orm:
T2 V Figure 6 Isotherms at diferent temperatures.
Note
5 _
p V 3 = constant 5 _
You dont need to derive the equations or adiabatic changes. The exponent or V does vary i the gas molecules are diatomic (two atom molecules) or polyatomic (three or more molecules in the atom) . The value is actually the ratio o the principal molar specifc heats but, as this is not going to be examined, we leave you to do urther research on this.
5 _
This can be written as p 1 V1 3 = p 2 V2 3 when there is a constant temperature. As the gas is an ideal gas the equation p 1 V1 p 2 V2 _ = _ T1 T2 also applies to an adiabatic change. 5 5 _ _ p 1 V1 p 2 V2 D ividing p 1 V1 3 = p 2 V2 3 by _ = _ T1 T2 gives 2 _
2 _
p isotherms
T1 V1 3 = T2 V2 3 In the p- V equation or an adiabatic change V is raised to a power o greater than one. This means that or an adiabatic change the line will be steeper than that or an isothermal change as shown in fgure 7. The area under the adiabatic change, as or all other changes, will be the work done. The direction o the arrow will determine whether work is done on or by the gas.
adiabatic process
Isovolumetric changes These changes occur at constant volume and, thereore, mean that no work can be done by or on the system. Q = U + W = U + p V
work done V Figure 7
Adiabatic and isothermal changes.
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With no change in the volume, the frst law o thermodynamics becomes Q = U ( when the energy transerred to the system increases the internal energy) or - Q = - U ( when the energy transerred rom the system decreases the internal energy) .
B
Figure 8 shows an isovolumetric change on a p- V graph. In this case the change shows an increase in temperature ( moving to a higher isotherm on an isothermal graph) .
A V Figure 8 Isovolumetric change.
p The equation o state or an isovolumetric change is __ = constant. T
Worked example a) D istinguish between an isothermal process and an adiabatic process as applied to an ideal gas. b) An ideal gas is held in a cylinder by a moveable piston and energy is supplied to the gas such that the gas expands at a constant pressure o 1 .5 1 0 5 Pa. The initial volume o the cylinder is 0. 040 m 3 and its fnal volume is 0.1 2 m 3 . The total energy supplied to the gas during the process is 7.5 1 0 3 J. ( i) S tate and explain the type o change that the gas undergoes. ( ii) D etermine the work done by the gas. ( iii) C alculate the change in internal energy o the gas.
Solution
transerred because o the temperature dierence between the gas and the surroundings and work done one on the other. An adiabatic process is one in which there is no energy exchanged between the system and the surroundings. This means that changes in the internal energy (and hence temperature) occurs because o work done by or on the ideal gas. b)
( i) As the pressure does not change the gas undergoes an isobaric expansion. ( ii) Work done = area under the p- V graph or the change ( = p V) = 1 .5 1 0 5 ( 0.1 2 - 0.04) = 1 .2 1 0 4 J ( iii) Using the frst law o thermodynamics Q = U + W
a) An isothermal process is one that takes place at constant temperature (and constant internal energy) so there is an interchange o energy
U = Q - W because only 7.5 1 0 3 J is supplied to the gas and 1 2 .0 1 0 3 J o work is done, the internal energy must all by 4.5 1 0 3 J
Cycles and engines hot reservoir T1 Q1
heat engine
work = Q 1 - Q 2
Q2 cold reservoir T2
Figure 9
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The principle of a heat engine.
Work can be converted into internal energy eectively through rictional orces. Work done by riction is usually undesirable because it increases the temperature o the system ( good) and the surroundings ( bad) . The reverse process o continuously converting energy into work is more difcult to achieve, but it can be done using a heat engine that operates through a cycle o changes. In 1 82 4 the French physicist and engineer, S adi C arnot, published the frst description o a heat engine; in this he described what has come to be known as the " C arnot cycle" . The principle o a heat engine is to take in energy at a high temperature, rej ect energy at a low temperature and use the remainder o the energy to do work on the system as illustrated in fgure 9. We can think o the system as being a gas enclosed in a cylinder with a rictionless moveable piston. When energy Q 1 is supplied to the gas rom a hot reservoir at temperature T1 , the gas expands and moves the piston, doing work. This will stop as soon as the gas pressure is equal to that o the surroundings. The gas now needs to be returned to its original state beore it can do urther work. This can only happen i some o the energy (Q 2 ) that was initially absorbed is rejected to a cold reservoir at lower temperature T2 .
B . 2 TH E RM O D YN AM I CS
The thermal efciency o the heat engine will be given by: Q1 - Q2 useul work done W = __ = _ = _ Q1 Q1 energy input
p A
In the C arnot cycle we imagine a completely riction- ree engine that is able to take a gas through a cycle o two isothermal and two adiabatic changes as shown in fgure 1 0.
B T1
S tarting at point A where the gas is at its highest temperature T1 , it expands isothermally to B by absorbing energy Q 1 . The internal energy o the gas does not change so all the energy absorbed is doing work.
D Q2
T2
C
V
The gas now expands adiabatically to C where the temperature alls to T2 . No energy is now being absorbed but the gas still does work on the surroundings by losing some internal energy. D uring the expansion AB C the area under the two expansion curves gives the work done on the surroundings.
T1 > T2
Q1
Figure 10
The Carnot cycle.
Note
At C the gas now needs work being done on it so it is compressed isothermally to D and rej ects energy Q 2 . The internal energy does not change so the work done on the gas is all rej ected as energy. Finally, the gas is urther compressed adiabatically rom D back to A. The work done on the gas is all used to increase the internal energy in returning the gas to temperature T1 . D uring the compression C D A the area under the two compression curves gives the work done by the surroundings on the gas. The area enclosed by the curve is the net work done by the gas on the surroundings in one cycle. The C arnot heat engine is said to be reversible. This is a theoretical concept in which, at any part in the cycle, the system can be returned to a previous state without any energy transerence this must be done infnitely slowly and means that the system returns exactly to its initial state at the end o the cycle. It can be shown that, or the C arnot cycle ( and all reversible heat engines) , that the thermal efciency is given by T1 - T2 T = _ = 1 - _2 T1 T1
Temperatures used here must be in kelvin.
This is the equation or the maximum eciency o a heat engine.
The maximum eciency is increased by raising the temperature o the hot reservoir and/or by lowering the temperature o the cold reservoir.
The maximum eciency can never equal 100% as this would mean that the cold reservoir was at absolute zero or else the hot reservoir was at an infnitely high temperature neither o these requirements is possible.
In practice thermodynamic cycles can be achieved but none will be more ecient than the Carnot cycle.
This is written in the IB Physics S yllabus as Tcold Carnot = 1 - _ Thot
Worked example A quantity o an ideal gas is used as the working substance o a heat engine. The cycle o operation o the engine is shown in the p- V graph opposite. C hange C A is isothermal.
p I10 5 Pa A
12.0
B
8.0
The temperature o the gas at A is 3 00 K. a) D uring the change AB the change in internal energy o the gas is 7.2 kJ.
C
4.0
( i) C alculate the temperature, at B , o the gas. ( ii) D etermine the amount o energy transerred during change AB .
0
0
2
4
6
V I10 -3 m 3
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The increase in internal energy = 7.2 kJ so the energy transerred to the gas must equal 4.8 + 7.2 = 1 2 . 0 kJ.
b) S tate why, or the change B C , the change in the internal energy o the gas is numerically the same as that in AB . c) C alculate: ( i) the net work done in one cycle ( ii) the efciency.
Solution a)
b) The gas undergoes the same change in temperature and, as the gas is ideal, this means that the change in internal energy depends solely on the temperature. c)
pV ( i) For an ideal gas __ = constant T V V As pressure is constant __ = __ = >TB T T A
B
A
B
V T 6.0 1 0 300 = ______ = _____________ = 900 K V 2.0 1 0 B
-3
A
-3
A
( ii) W = p V = 1 2 .0 1 0 5 ( 6.0 1 0 - 3 - 2 .0 1 0 - 3 ) = 4. 8 kJ
hot reservoir T1
( i) We need to fnd the area enclosed by the cycle. Each large square is equivalent to 1 .0 1 0 - 3 2 .0 1 0 5 J = 2 00 J Estimate that there are 1 4 large squares, making a total 2 800 J or 2 .8 kJ. W 2 .8 ( ii) = _ = _ = 0.2 3 or 2 3 % 12 Q1
Heat pumps and refrigerators Heat pumps and rerigerators act in a similar manner to a heat engine working in reverse. They take in energy Q 2 at a low temperature T2 , do work W on the working substance and rej ect energy Q 1 at the high temperature T1 . Although they are very similar in their working the heat pump is designed to add energy to the high temperature reservoir ( or example, the room being heated by extracting energy rom the ground) whilst the rerigerator is designed to remove energy rom the low temperature reservoir ( the cool box) . Air conditioning units are another example o heat pumps.
Q1 refrigerator or heat pump W = Q1 - Q2 Q2 cold reservoir T2 Figure 11
The principle of a heat pump and refrigerator.
Worked example The diagram shows the relationship between the pressure p and the volume V o the working substance o a rerigerator or one cycle o its operation. The working substance is a volatile liquid which is made to vaporize and condense.
S uggest the reason why both the changes rom C D and AB are isothermal, isobaric changes. b) S tate during which process o the cycle energy is absorbed rom the cold reservoir and during which process energy is transerred to the hot reservoir.
p B
a) The working substance at point C o the cycle is entirely in the liquid phase.
A
c) S tate how the value o the work done during one cycle may be determined rom the pV diagram. C
D V
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Solution a) B oth changes are isobaric and isothermal because there is no pressure or temperature change. Each o the changes occurs because o a change o phase o the working substance. From C to D the liquid vaporizes and rom A to B the vapour condenses.
b) Energy is absorbed during C to D as the liquid needs energy to vaporize and it is ejected during A to B in order or the vapour to condense. c) The area enclosed by the cycle will always indicate the net amount o work in this case it is work done on the system (working substance) .
The second law of thermodynamics Although this is a undamental law that has its origins in practical experiences, the second law o thermodynamics can be stated in a number o dierent ways. The frst law o thermodynamics equates work to energy, the second law deals with the circumstances in which energy can be converted to work. Each o the statements is equivalent to the others and communicates an expression o the impracticability o reversing real thermodynamic processes. The C lausius version o the second law can be stated as: It is imp ossible to transfer energy from a body at a lower temp erature to one at higher temp erature without doing work on the system. The Kelvin ( or Kelvin-Planck) version states: It is imp ossible to extract energy from a hot reservoir and transfer this entirely into work. I the second law was not true it would be possible to power ships by energy extracted rom the sea this cannot be done because there needs to be a cold reservoir into which the dierence between the energy extracted and the work done would be rej ected. When, ater a time, we return to ull cup o coee we do not expect to fnd it at a higher temperature than when it was made. Energy passes rom the hot coee to the cooler room until the two bodies are at the same temperature i we wanted the coee to heat up we would have to transer energy to it using a heating coil or else do work on it by stirring it rapidly! The internal energy o an obj ect is related to the random motion o the molecules o the obj ect. B y trying to convert this internal energy into work we are trying to convert random motion into something more ordered. It is impossible to do this because we cannot take control over the individual motion o a colossal number o molecules.
Entropy A third version o the second law o thermodynamics involves the concept o entrop y S. This quantity can be defned ( or a reversible change) in terms o the equation Q S = _ T
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E N G I N E E R I N G P H YS I C S S is the increase in entropy
Note
Q is the energy absorbed by the system
T is always positive so when energy is absorbed by a system and Q is positive there will be a positive change in entropy an increase. When energy is rejected by the system the entropy will decrease.
For an adiabatic change Q = 0 and so S = 0
A substance taken through a complete reversible cycle will undergo no change in T2
= entropy as = 1- ____ T Q2
Q1
1
1
1
Entropy is a scalar quantity and has units o j oule per kelvin ( J K 1 )
Worked example 0.2 0 kg o ice at 0 C melts. The specifc latent heat o usion o water is 3 .3 1 0 5 J kg 1 . C alculate the change in entropy o the ice as it melts.
Solution E nergy needed to melt the ice = mL = 0.2 0 3 .3 1 0 5 = 6.6 1 0 4 J Q 6. 6 1 0 4 S = _ = _ = 2 .4 1 0 2 J K 1 T 2 73
Q2
1- ____ this means ____ = ____ T T Q
T is the temperature in kelvin at which this occurs.
2
All heat engines reject energy to the surroundings and generate an overall increase in entropy of the universe.
Entropy as a measure of disorder We have now seen that or real processes the entropy o the universe increases in act this is another way o stating the second law o thermodynamics frst suggested by B oltzmann. Real processes always degrade the energy, i.e. change the energy rom being localized to being more spread out. I some hot water is mixed with cold water in a completely insulated container there is no loss o energy, however, the opportunity to use the energy to do work is now restricted by having cold water. It is not a sensible proposition to separate the most energetic molecules to produce some hot water and some cold water rom the mixture the energy has changed rom the localized situation in the hot water molecules to a situation where it is spread- out amongst all o the molecules. Imagine having 1 0 coins all placed heads up on the table. The coins are picked up and shaken beore being returned to the table, without looking to see where they are placed; some coins will, thereore, have heads up but others will have heads down. The coins have moved rom an ordered 1 10 1 state to a disordered state. There is a small likelihood ( __ = ____ that 2 ) 1 024 the coins would be replaced with all 1 0 heads up. B y increasing the number o coins rom 1 0 to 1 00 it decreases the likelihood o them all being heads up to 1 in 1 .3 1 0 30, and, this really isnt likely to happen! The coin experiment mirrors nature in that a system does not naturally become more ordered. We have now seen that the entropy o a system naturally increases with the disorder o the system. This is not coincidental since it can be shown that entropy is a measure of the disorder of a system.
(
Figure 12
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Ten coins all heads up.
)
B . 2 TH E RM O D YN AM I CS
TOK The arrow of time Many scientists have discussed increasing entropy as representing the "arrow o time". Because the disorder o a large system will increase with time, fnding such a system with increased order would be equivalent to time going backwards. Perpetually increasing entropy is
contrary to Newton's laws o motion in which the change o state o a system is equally predictable whether we go orwards or backwards in time. Does this statement o the Second law prevent the possibility o time travel?
Worked example
Solution
a) S tate what is meant by an increase in entropy of a system.
a) When the entropy increases, there is an increase in the degree of disorder in the system.
b) S tate, in terms of entropy, the second law of thermodynamics.
b) The total entropy of the universe increases.
c) When a chicken develops inside an egg, the entropy of the egg and its contents decreases. E xplain how this observation is consistent with the second law of thermodynamics.
c) E ntropy of the surroundings must increase more than the decrease of entropy in the developing egg. The energy generated by the biochemical processes within the egg becomes more spread out as a consequence of some passing into the surroundings.
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B.3 Fluids and fuid dynamics (AHL) Understandings Density and pressure
Applications and skills Determining buoyancy orces using
Buoyancy and Archimedes principle Pascals principle
Hydrostatic equilibrium The ideal uid
Streamlines The continuity equation
The Bernoulli equation and the Bernoulli efect Stokes law and viscosity Laminar and turbulent ow and the Reynolds
number
Archimedes principle Solving problems involving pressure, density and Pascals principle Solving problems using the Bernoulli equation and the continuity equation Explaining situations involving the Bernoulli efect Describing the rictional drag orce exerted on small spherical objects in laminar uid ow Solving problems involving Stokes law Determining the Reynolds number in simple situations
Equations Buoyancy orce: B = V g Pressure in a uid: p = p 0 + gd Continuity equation: Av =constant The Bernoulli equation: ___ 1 v2 + gz + p = constant 2
Stokes law: FD = 6rv
vr Reynolds number: R = _
Nature of science Fluids in motion The study o the transportation o mass is very important in medicine and engineering. Knowledge o how the velocity and pressure changes throughout a moving uid is vital in our understanding o many physical processes, including how blood ows around the body, how aircrat y, and how jet engines operate. The visualization o uid ow can be extremely appealing and has been used extensively in the visual arts. 570
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L)
Introduction The gaseous and liquid states o matter are j ointly known as fuids substances that can ow and take the shape o their containers. There are important dierences between gases and liquids:
It is easy to compress a gas but liquids, like solids, are almost incompressible.
Gases are not restricted by a surace but liquids are you can have hal a cup o coee but not hal a cup o air! The most energetic molecules may, however, be able to break through the surace to orm a vapour above the liquid.
Although this topic is called uids and uid dynamics, it is almost entirely restricted to idealized uids those that cannot be compressed, are non-viscous, and ow in a steady manner. These properties are best represented by a low viscosity liquid, such as water, and not by a ' sticky' one, such as oil ( which has much internal riction) .
Static fuids density and pressure The density o a substance is given by the ratio o the mass m o the substance to the volume V o the substance: m = _ V Pressure is the ratio o the perpendicular contact orce acting on a surace to the area o the surace: F p=_ A Although the direction o orce is at right angles to the surace on which it acts, pressure is a scalar quantity and acts in all directions; with the orce in newton and the area in metres squared, pressure is measured in pascal ( Pa) . In a uid, pressure is ound to increase with depth and, at a given depth, the pressure is ound to be equal in all directions. This results in a perpendicular orce acting on any surace at a given depth. I this was not the case, any pressure dierences would cause the liquid to ow until the pressure was constant. C onsider a cylinder o height h and base area A in a uid o density as shown in fgure 2 the top o the cylinder is at the surace o the uid. The orces acting on the bottom o the cylinder are the weight W o the column o liquid above it acting downwards and the pressure orce pA rom the liquid acting upwards. These are in equilibrium so
orce F p= F A area A Figure 1
Defnition o pressure.
cross-sectional area A
h
W
uid o density
W = pA W = mass o cylinder gravitational feld strength = mg, this means that W = Ahg = pA and thereore pA
p = hg In a liquid that is not in a sealed container the atmospheric pressure p 0 will also act on the liquid surace. This makes the total pressure p at depth h become hg + p 0 p = hg + p 0
Figure 2
Pressure in a uid.
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Nature of science Measuring density of immiscible liquids using a U-tube When two liquids that do not mix are added to the limbs o a U-tube, their levels settle in a way that is dependent on their relative densities. At the level XY in fgure 3 the pressure will be constant, so
liquid A o density A
p 0 + h A A g = p 0 + h B B g C ancelling p 0 and g gives
atmospheric pressure = p 0
liquid B o density B
h A A = h B B When we know the density o one liquid we are then able to calculate that o the second liquid by taking a ratio o the heights. This experiment is an example o hydrostatic equilibrium.
X
Y
Figure 3 Hydrostatic equilibrium to measure density.
Pascals principle This principle says that the pressure applied at one point in an enclosed fuid under equilibrium conditions is transmitted equally to all parts o the fuid. The principle allows hydraulic systems to operate. For example, the hydraulic jack is used to raise heavy objects such as cars.
Fx Fy Ax
Ay
pressure throughout fuid = p x Figure 4 Hydraulic jack.
A small orce Fx applied to a piston o small cross-sectional area A x produces a pressure Px in a liquid such as oil. This pressure is communicated through the liquid so that it acts on a second, larger piston o cross-sectional area A y. This will then produce a larger orce Fy so that Fy Fx px = _ = _ Ax Ay The device cannot ampliy the energy, so the distance moved by Fx must be ar more than that moved by Fy in order to conserve energy ( work being orce distance) .
pxA fuid o density
Archimedes principle cylinder o cross-sectional area A
h
(p x + p x) A Figure 5 Archimedes principle.
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When you dive into a swimming pool you quickly eel the buoyancy orce o the water acting on you. The magnitude o this upward orce is expressed by Archimedes' principle; this says that or an obj ect wholly or p artially immersed in a fuid there will be an up ward buoyancy orce acting on the obj ect which is equal to the weight o fuid that the obj ect disp laces. The buoyancy orce is oten known as the up thrust. We can deduce this relationship by considering a cylinder o height h and cross- sectional area A ully immersed in a liquid o density . The pressure is p x at the top and ( p x + p x) at the bottom. The latter is higher because o the extra depth o uid. The orces on the top and bottom o the cylinder will be p xA and ( p x + p x) A. We ignore the atmospheric pressure because it acts on both the top and the bottom o the cylinder.
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L) The buoyancy orce B will, thereore, be the dierence between these orces B = ( p x + p x ) A - p xA = p xA p x = hg making B = hgA
Note When an object is foating:
The buoyancy orce acting on the object must equal the weight o the object.
The buoyancy orce will equal the weight o fuid that is displaced by the submerged portion o the foating object.
V = hA And so B = Vg, which is the weight o the uid displaced by the cylinder. To emphasize that this applies to a uid, the IB Physics data booklet shows this equation as B = V g
Worked example A block o wood o mass 5 0 kg is completely immersed in water. The density o the wood is 1 .1 1 0 3 kg m 3 and that o the water 1 .0 1 0 3 kg m 3 a) C alculate the resultant orce on the block. b) When the block is cut into planks and made into a boat it oats, displacing 0.1 5 m 3 o water when empty. C alculate the buoyancy orce acting on the boat.
Solution a) The weight o the block = mg = 5 0 9.81 = 4.90 1 0 2 N B uoyancy orce = weight o water displaced = B = V g m 50 _ 3 V = volume o block = _ = 1 .1 1 0 3 = 0. 045 m B = V g = 1 .1 1 0 3 0.045 9. 81 = 4.86 1 0 2 N The overall orce is, thereore, 4.90 1 0 2 - 4.86 1 0 2 = 4 N downwards b) The weight o water displaced by the boat = 0.1 5 1 .0 1 0 3 9.81 = 1 .5 1 0 3 N S o B = 1 .5 1 0 3 N ( around three times the original value)
Fluid dynamics There are many instances o uid ow. Examples include the smoke rom a fre, hot water in a central heating system and wind spinning around in a tornado. Fluid dynamics also deals with solid objects moving through stationary uids. An ideal uid oers no resistance either to a solid moving through it or it moving through or around a solid object. An ideal uid is said to be non-viscous. We will start our discussion o uid dynamics by considering a non-viscous, incompressible uid in a stream tube.
Streamline (or laminar) fow Ideal uids ow in a very predictable way. We can represent the paths taken by particles within a uid using the concept o streamlines. In
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E N G I N E E R I N G P H YS I C S streamline fow, the motion o a particle passing a particular point is identical to the motion o all the particles that preceded it at that point. Streamlines will be close together when the particles move quickly and urther apart when they move more slowly. A group o streamlines is known as a stream tube. Fluid never crosses the surace o a stream tube. Figure 6 shows the streamlines in a tube that narrows and widens again.
Figure 6 Streamlines in
a tube that narrows.
The continuity equation With steady fow the mass o fuid entering one end o a stream tube must be equal to that leaving the other end. The stream tube may be a physical tube or pipe or it may be a series o streamlines within the fuid.
Ay vy vx
Ax
Y X Figure 7
Streamlines at the boundary of a stream tube.
For a length o stream tube the fuid enters at X through an area o cross- section A x at velocity vx and leaves at Y through an area o crosssection A y with velocity vy. In a short time, t, the fuid leaving X will travel a distance vx t thus meaning that a volume A xvx t and a mass A xvx t enter the tube in this time. In the same time a mass A yvy t will leave the stream tube through Y. As there cannot be any discontinuities in an ideal liquid these masses must be equal so A xvx t = A yvy t meaning that when t is cancelled rom both sides A xvx = A yvy We see rom this that Av = constant This is known as the continuity equation and the product Av is known as the volume fow rate ( or, sometimes, j ust fow rate) . Volume fow rate is measured in units o m 3 s 1 .
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B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L)
The Bernoulli equation This equation is a generalized equation which deals with how a uid is able to both speed up and rise to a higher level as it passes through a stream tube. The equation is derived rom the conservation o energy and, although the derivation is shown below, you will not need to repeat this in an IB Physics examination. In your data booklet the B ernoulli equation is written as 1 __ v 2 + gz + p = constant 2
Here is the uid density, v is its speed, g the gravitational feld strength, z the height above a chosen level, and p the pressure at that height. You will notice in the equation that, i was replaced by m in each o the frst two terms, we would have kinetic energy and potential energy. m As density is equal to __ , multiplying the B ernoulli equation by V V would give 1 __ mv 2 + mgz + pV = dierent constant 2
This has now turned the B ernoulli equation into the conservation o energy because it implies that the kinetic energy + gravitational potential energy ( z is height here) + work done ( remember the frst law o thermodynamics) remains the same. The equation in the B ernoulli orm expresses each o the terms as an energy density ( energy per unit volume) and is measured in J m 3 . This equation also tells us that, or any point in a continuous steady ow, the total o all the quantities will remain the same: i the pressure changes then one or more o the other terms must also change.
Derivation of the Bernoulli equation y Fy = p y A y Y Y' vy
x
hy
Fx = p x A x hx
X
X' vx
Figure 8 The Bernoulli equation.
C onsider a uid entering and leaving a pipe that becomes wider and higher. It is obvious that the pressure at the inlet must be greater than the pressure at the outlet. When the inlet pressure is p x the orce Fx at the inlet is p xA x and that at the outlet is p yA y. In a short time, t, the uid entering the pipe at X moves a short distance x rom X to X at velocity vx. This means that the work done on the uid is Fx x = p xA x x. In the same time, the same mass o uid moves a distance y in moving rom Y to Y and the work done will be Fy y = p yA y y.
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E N G I N E E R I N G P H YS I C S The net work done will be p xA x x - p yA y y As this happens in the same time, t, and because equal masses must mean equal volumes or an ideal fuid o constant density, this gives A x x = A yy = V thus net work done = V( p x- p y ) 1 the change o kinetic energy or this mass = __ m ( vy 2 - vx 2 ) 2
and the change in potential energy = mg( h y - h x ) O verall, or the conservation o energy: work done = gain in kinetic energy + gain in gravitational potential energy so 1 V( p x - p y ) = __ m v 2 - vx2 ) + mg ( h y - h x ) 2 ( y
D ividing by V and collecting terms or x and y we get 1 1 vx2 + gh x = p y + __ vy2 + gh y p x + __ 2 2
This is the B ernoulli equation. In the IB Physics data booklet the symbol z is used or heights.
Applications of the Bernoulli equation From the B ernoulli equation it ollows that, when a fuid speeds up, there must be a decrease in either the pressure or the gravitational potential energy or both o these. When the fow is horizontal there can be no change in the gravitational potential energy, so there must be a reduction in pressure. Aerooils, used on aircrat wings and mounted on racing cars, are shaped so that the air fows aster over the more curved surace than over the fatter surace in most cases it is the aerooil that moves, but the relative eect is identical. The aster moving air produces a lower pressure and less orce acts on that section o the aerooil this causes the aerooil to be orced upwards ( a lit) or downwards ( a down thrust) depending on which way it is positioned. The eect o the shape o the aerooil causes the streamlines to be closer together on the curved side and to be urther apart on the fatter side. longest path, high velocity streamlines
direction o air fow
shorter path, low velocity Figure 9
The aerooil.
Venturi tubes When a tube narrows, the speed o the fuid increases at the narrow part o the tube beore returning to the original value at the wider part. This means that the pressure is higher at the edges o the tube and lower
576
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L) in the centre. This is the principle o the Venturi gauge, which is used or measuring uid speeds as demonstrated in fgure 1 0. The higher air pressure has a greater eect on the right limb o the manometer. This pressure dierence can be calibrated to measure relative uid speeds and ow rates. narrow high speed wide low speed
lower pressure
higher pressure
height proportional to pressure diference
Figure 10 Venturi gauge.
Pitot static tubes Pitot static tubes are used or measuring the velocity o a uid o density . The tubes X and Y must be on the same streamline. The opening o tube X is perpendicular to the direction o uid ow but the opening o tube Y is parallel to the ow. The uid can then ow into the opening o tube Y. The two tubes must be ar enough apart or tube X not to aect the velocity at Y. When there is steady ow, no uid particles can move rom one streamline to another and so none enter tube X. Tube X measures the static pressure p x. Particles will enter tube Y where their kinetic energy will be brought to zero. Pitot tube
static tube
hy
hx ow direction velocity = v X Figure 11
Y
owing liquid
Pitot static tubes.
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E N G I N E E R I N G P H YS I C S Applying the B ernoulli equation to the streamline along XY gives 1 vx2 = p y p x + __ 2
which gives ________
vx =
_2 ( p
y
- px )
As p x = h xg + p 0 and p y = h yg + p 0 the velocity equation becomes vx =
X
_________
2 g( h y -
hx )
For an idealized uid this will be the velocity along the streamline XY. This arrangement is the one used or measuring the velocity o liquids, but it can be adapted to measure the ow o gases. A Pitot tube is installed on aircrat wing to measure the speed o the aircrat relative to the air.
hx
Y
Flow out of a container You may be amiliar with the leaking can demonstration. Figure 1 2 shows liquid pressure increasing with depth. This is another application o the B ernoulli equation. The line XY represents a streamline j oining water at the surace to a hole in the side o the can. Each o these two points will be open to the atmosphere and so they will both be at atmospheric pressure. This means that p x = p y = p 0 . Figure 12
The B ernoulli equation becomes:
Leaking can demonstration.
1 1 __ vx2 + gh x = __ vy2 + gh y 2 2
Here h x is the height above Y and h y = 0. I we assume the container is wide enough and the holes small enough or the velocity o the water surace to approximate to zero, the equation reduces to 1 gh x = __ vy2 2
C ancelling and re- arranging the equation gives ____
vy = 2 gh x
This confrms that the velocity is related to the height o water above opening. Viscosity and turbulence have been neglected.
Worked example X
tank 12.0 m
The diagram shows a large water tank, open to the atmosphere. The tank eeds a pipe o crosssectional area 4.0 1 0 3 m 2 which, in turn, eeds a narrow tube o cross- sectional areas 5 .0 1 0 4 m 2 . The narrow tube is sealed with a closed valve. The water surace in the tank is 1 2 .0 m above the centre o the two tubes. density o water = 1 .00 1 0 3 kg m 3
Y
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Z
valve
atmospheric pressure = 1 . 0 1 0 5 Pa
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L)
a) C alculate the pressure at Y. b) C alculate the pressure at Z. c) The valve is now opened so that water leaves the system. The pressure at the open valve is now atmospheric.
Note That this is not an equation given in the IB Physics data booklet. I you were tested on it, the question would need to be structured in order to lead you through the development o this equation.
( i) D etermine the velocity o the water at Z. ( ii) C alculate the velocity o the water at Y. ( iii) C alculate the pressure at Y.
( ii) Using the continuity equation A yvy = A zvz
Solution
4.0 1 0 3 vy = 5 . 0 1 0 4 1 5 .3
a) and b) The pressure at Y and Z will be equal because they are at the same level. The pressure will be p = p 0 + gd = 1 .0 1 0 5 + ( 1 . 00 1 0 3 9. 81 1 2 . 0) = 2 .1 8 1 0 5 Pa 2 .2 1 0 5 Pa
vy = 1 .9 m s 1
c) ( i) Assuming the surace area is large enough or the kinetic energy at X to be zero we have shown that ___ _____________ v = 2 gh = 2 9. 81 1 2 . 0 = 1 5 .3 m s 1
( iii) Using the B ernoulli equation between Y and Z ( they are on the same level so there will be no dierence in gravitational potential energy) . 1 1 vz2 = p y + __ vy2 p z + __ 2 2 1 and as p z = p 0 then p y = p 0 + __ v 2 - vy 2 ) 2 ( z 1 py = 1 .0 1 05 + __ 1.00 1 03 ( 1 5.3 2 - 1.9 2 ) 2
= 2 .2 1 0 5 Pa
Viscosity o fuids We have seen that ideal uids are non-viscous. Real uids do, o course, have viscosity with some uids being more viscous than others. For a viscous uid in laminar ow, each layer impedes the motion o its neighbouring layers. In a pipe, the layers adjacent to the inner walls o the tube are stationary while the layers in the centre o it travel at the highest speed. The viscosity o uids is highly temperature dependent, with most liquids becoming less viscous at higher temperatures and gases becoming more viscous at higher temperatures. The viscosity o a uid is measured in terms o its coefcient o viscosity ( at a given temperature) . This is a quantity measured in units o pascal second ( Pa s) .
Stokes law When a sphere, o radius r, alls slowly through a viscous uid it pulls cylindrical layers with it. Under these conditions o streamline ow, the sphere experiences a viscous drag orce FD . This is given by FD = 6rv where v is the velocity o the sphere and is the coefcient o viscosity o the uid. I the sphere is contained in a tube, the tube would need to be very wide or S tokes equation to apply. When a small metal ball is released in oil, the ball will initially accelerate until the drag orce ( plus the buoyancy orce) is equal to the weight o the ball the ball will then travel at its terminal speed so 4 6rvt + __ r3 g = mg 3
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E N G I N E E R I N G P H YS I C S or 4 4 r3 g = __ r3 g 6rvt + __ 3 3
and 2 r 2 g( - ) vt = __ 9 Here vt is the balls terminal speed, is the density o the uid and is the density o the ball.
Investigate!
metal ball
Measuring the coecient o viscosity o oil
0
Using the apparatus shown in fgure 1 3 you can investigate Stokes equation.
Use six or more balls o dierent radii ( but made o the same material) small balls should reach the terminal velocity in a ew centimetres i you use a viscous liquid such as engine oil or glycerol.
light gate 10 20 metre ruler 30 40
You will need to know the densities o the liquid and the material o the balls.
50 seconds
You will need to make sure that the balls reach their terminal velocity use a long, wide transparent cylinder.
60
oil
I you use steel balls, a magnet can be used to remove them rom the oil.
80
Think about ways o releasing the balls consistently. Think how you are going to measure the diameter o balls accurately.
timer
70
90 light gate 100 Figure 13
Stokes law investigation.
Turbulent fow laminar
turbulent
Figure 14 Laminar and
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turbulent fow.
At low velocities uids ow steadily and in layers that do not mix because the uid is viscous the layers travelling closest to the walls will move slowest as shown in fgure 1 4. When the uid velocity is increased or obstacles proj ect into the uid the ow becomes turbulent. The ow is no longer laminar and the particles in the dierent layers mix with each other. This causes the smooth streamlines, seen during laminar ow, to break up and orm eddies and vortices. It is not easy to predict when the rate o ow is sufciently high to cause the onset o turbulence. A quantity, known as the Reynolds number R, gives a practical " rule o thumb" that can be used to predict whether the ow is great enough to become turbulent. This is a dimensionless quantity which is calculated rom the equation: vr R=_
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L) For a uid v is velocity, r the radius o the pipe ( or other dimension or a river, length o a plate etc.) , the density and the coefcient o viscosity. Using this defnition a Reynolds number that is less than 1 000 is taken to represent laminar ow. I the Reynolds number is greater than 1 000, it does not mean that the ow will be turbulent because there is a transition stage between the uid being laminar and becoming turbulent. It is generally accepted that, using this defnition o the Reynolds number, a value o above 2 000 will be turbulent.
Worked example O il ows through a pipe o diameter 3 0 mm with a velocity o 2 .5 m s 1 . The oil has viscosity 0. 3 0 Pa s and density o 890 kg m 3 a) D etermine whether or not this ow is laminar. b) C alculate the maximum velocity at which the oil will ow through the pipe and still remain laminar.
Solution C oefcient o viscosity is oten j ust called " viscosity" . vr 2 .5 1 5 1 0 3 890 ___ a) R = _ = 111 = 0.3 0 This value is below 1 000 so the ow is laminar. R 1 000 0. 3 0 __ 1 b) v = _ r = 1 5 1 0 3 890 = 2 2 m s
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E N G I N E E R I N G P H YS I C S
B.4 Forced vibrations and resonance (AHL) Understandings Natural requency o vibration
Applications and skills Qualitatively and quantitatively describing
Q actor and damping Periodic stimulus and the driving requency Resonance
Nature of science Risk assessment and resonance Resonance can be a useul phenomenon and, as discussed in this sub-topic, it is utilized in many walks o lie. Despite this, the possibility o one system interacting with another has its drawbacks loud music with heavy bass notes or the noise during construction work can carry over long distances and can, thereore, impinge upon peoples lives. Most nations promote the inclusion o resonance hazards during risk assessment applicable when carrying out building or maintenance work. For example, devising saety measures such as incorporating sound insulation in wall cavities or installing double or triple glazing to shield rom the efects o unwanted sounds rom outside.
examples o under-, over-, and criticallydamped oscillations Graphically describing the variation o the amplitude o vibration with driving requency o an object close to its natural requency o vibration Describing the phase relationship between driving requency and orced oscillations Solving problems involving Q actor Describing the useul and destructive efects o resonance
Equations Quality actor equations energy stored Q = 2 ______ energy dissipated per cycle energy stored Q = 2 resonant requency ___ power loss
Introduction All mechanical systems, and some electrical systems, will vibrate when they are set in motion. We have seen this in S ub- topics 4.1 and 9.1 when studying simple harmonic motion. More intricate examples o oscillations, that are not simple harmonic, include the motion o a tall building in the wind, the shaking o an unbalanced washing machine as its drum spins, or the vibrations caused by a car engine misfring. In some instances, the vibrations are useul but, oten, the unction o the system will beneft rom being damped when the amplitude o oscillation is restricted. D ampers are increasingly being incorporated in tall buildings to restrict potential earthquake damage.
Free vibrations When a mechanical system is displaced rom its rest position and allowed to vibrate, without any external orces being applied, it will oscillate at its natural requency f0 . S uch vibrations are called free vibrations.
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B . 4 F O R C E D V I B R AT I O N S A N D R E S O N A N C E ( A H L )
Oscillations and damping At a given time the amplitude o an oscillation depends on how much the system is damped as well as the size o the driving orce. This happens when an oscillating system experiences a resistive orce which causes the amplitude o the object oscillating to decay. The resistive orce acts in the opposite direction to the motion o the system and it increases with the speed o the oscillation. This means that the damping orce is a maximum when the system passes through its equilibrium position as it will have a maximum speed at this point ( and it will be zero at the maximum displacement at which point the system momentarily stops) . In opposing the damping orce, the system must do work and this reduces the energy that it stores causing the amplitude to decay. The resistive nature o the orce slows the oscillator down which increases the time period ( although this eect is negligible when a system is lightly damped) . Figure 1 shows how the displacement o an oscillator varies with time when there is light damping. This is also known as under-damp ing. The envelope o the curve takes an exponential shape when the damping or resistive orce is proportional to the speed; this means the ratio o the amplitudes at hal- period intervals is a constant value 0.25 0.2 1 0.1 8 ____ ____ etc. ) . ( ____ 0.21 0.1 8 0.1 5
Investigate! Iteratation with damped SHM
( the symbols having their usual meaning and b being the damping actor . .. a number between 0 and 1 ) .
The term bv should now be subtracted rom each acceleration value in the spreadsheet iteration discussed in the SHM spreadsheet investigate! in sub-topic 9.1 (the second box in the ow chart is modifed to a n + 1 = - kxn - bvn) .
You should try out dierent values or b in order to j udge the impact that it has on the damping o the shm.
An example o this spreadsheet is provided on the webpage.
displacement/m 0.25 0.20 0.15
The equation or a velocity dependent S HM can be written as ma = - kx - bv
0.10 0.05 0
2
4
6
8
10
12
14
16
18
20 t/s
-0.05 -0.10 -0.15 -0.20
Figure 1
Under-damped vibration.
Although the amplitude decreases with time, the requency and time period o the oscillator are approximately constant. Heavier damping may completely stop a system rom oscillating. When the system is displaced and it returns to its equilibrium position without overshooting it, the system is " heavily damped" . A system returning to its equilibrium position in the shortest possible time is said to be critically damped, while a system taking longer than this to return to the equilibrium position is over-damped. Critical damping is necessary in many mechanical systems using this type o damping in car suspension systems avoids oscillations on uneven suraces; such oscillations may lead to a loss o control and a very uncomortable ride. For a similar reason, fre doors in buildings are oten ftted with automatic closers that are critically damped.
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E N G I N E E R I N G P H YS I C S amplitude
over-damped critically damped time
under-damped Figure 2
Variation o amplitude with time or diferent degrees o damping.
These curves can be modelled by adding a velocity- dependent damping term to the simple harmonic motion equation. Figure3
Child being pushed on
amplitude o the orced vibration
swing.
zero damping
light damping heavy damping natural requency driving requency
Figure 4 Efect o damping on
resonance.
Note
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For zero damping the amplitude would be infnite, however, this could not happen in practice because infnite amplitude would require an infnite amount o energy to be supplied! As the damping increases the peak moves slightly to the let o the natural requency. Increased damping reduces the sharpness o the resonance peak.
Forced vibrations When an external orce acts on a mechanical system, the orce may have its own requency o vibration, which may aect the motion o the mechanical system. A simple example o this is a child on a swing. When the child is let to his own devices he will only be able to swing at a particular requency which is a property o the childswing system. Pushing the child provides an external orce that causes the motion o the swing to change. The amplitude o the swing will only increase i the pushes are applied at a rate matching the swing requency. When the applied requency does not match the swing requency the amplitude might decrease. Forced vibrations are those that occur when a regularly changing external orce is applied to a system resulting in the system vibrating at the same requency as the orce.
Resonance When a mechanical system is orced to oscillate by a driving orce that has the same requency as the natural requency o the mechanical system, it will vibrate with maximum amplitude. This is called resonance. When the requency o the driving orce becomes closer to the natural requency o the system the amplitude o the oscillation will be greater. This can be seen in fgure 4, which shows the variation with the driving requency o the systems amplitude. The degree o damping alters the systems amplitude response. Resonance occurs in many physical systems ranging rom tuning radios to the tidal eects o the Moon and the unction o lasers.
Q factor The Q or " quality" actor is a criterion by which the sharpness o resonance can be assessed. It is defned by the relationships: energy stored Q = 2 ___ energy dissipated per cycle energy stored = 2 resonant requency __ power loss This is an arbitrary defnition with the 2 being included so that using the equation with a real system becomes simplifed! In the equations or many rotating or oscillating systems the 2 actor will cancel ...making estimation o Q actors straightorward. The two defnitions are equivalent relationships:
B . 4 F O R C E D V I B R AT I O N S A N D R E S O N A N C E ( A H L ) writing Es or energy stored, Ed or the energy dissipated per cycle, f0 or the resonant requency, T0 or the resonant period, P or power loss ( = power dissipated) and E or the energy lost in time t. The second relationship becomes Es E 1 _ _ Q = 2 f0 __ = 2 P T0 _ E t E t s = 2 _ _ T0 E s
tim e co n side re d t The actor __ = ________________ = n ( number o oscillations) T tim e o r o n e o scillatio n 0
Es
S o Q = 2 n __ E E E = 2 __ = 2 __ E __ E s
n
s
d
e ne rgy sto re d or in words Q = 2 ___________________ e ne rgy dissip ate d p e r cycle
This is the frst relationship. The Q actor is a numerical quantity and has no unit. A system with a high Q actor is lightly damped and will continue vibrating or many oscillations as the energy dissipated per cycle will be small. As a rule o thumb, the Q actor is approximately the number o oscillations that the system will make beore its amplitude decays to zero (without urther energy input) . The larger the Q actor, the sharper the resonance peak on the graph o amplitude against driving requency. A high quality actor means a low loss o energy. For light damping Q = m/b where b is the damping actor relating the resistive orce Fr to the speed v: Fr = - bv S ome typical values o Q actors are:
Oscillator
Q factor
critically damped door
0.5
loaded test tube oscillating in water
10
mass on spring
50
simple pendulum
200
oscillating quartz crystal
30 000
Worked example
Solution
An electrical pendulum clock has a period o 1 . 0 s. An electrical power supply o 2 5 mW maintains its constant amplitude. As the pendulum passes its equilibrium position it has kinetic energy o 40 mJ.
a) The pendulum has a requency o 1 .0 Hz.
a) E xplain how these quantities apply to the Q actor relationship. b) C alculate the Q actor or the pendulum clock.
As it is storing 40 mJ, the rate o energy supplied must equal the rate at which energy is lost i.e., the power is supplied at a rate o 25 mW. energy stored b) Q = 2 resonant requency __ power loss 40 1 0 3 = 2 1 _ 2 5 1 0 3 =1 0
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B
E N G I N E E R I N G P H YS I C S The Q factor is an especially important quantity for electrical oscillations transmitting radio waves. When selecting radio and television stations it is essential that the transmitting and receiving circuits are tuned to the same resonance frequency.
Barton's pendulums Figure 5 shows an arrangement of pendulums that can be used to investigate resonance. The apparatus is known as B artons pendulums named after the B ritish physicist, E dwin B arton.
2
1
0
3 5
4 paper cones
Figure 5 Bartons pendulums.
brass bob
As the diagram shows, the set up consists of a number of paper cone pendulums of varying lengths. All are suspended from the same string as the brass bob that acts as the " driver" pendulum. When the driver pendulum is displaced from its rest position and released, it forces all the paper cone pendulums to oscillate with the same frequency, but with different amplitudes. This is an example of forced oscillations. The " cone" that has the same length as the driver pendulum has the greatest amplitude because it has the same natural frequency as the driver pendulum. In calculating the period using the equation for the period of a simple pendulum, the " equivalent length" of each pendulum should be taken from the centre of the bob to the horizontal broken line shown on the diagram. C areful observation shows that:
cone 3 ( which has the same length as the driver pendulum) always lags behind the driver pendulum by one quarter of period ( equivalent to 90 or __ radian) 2
the shorter cones ( 1 and 2 ) are almost in phase with the driver pendulum
the longer pendulums ( 4 and 5 ) are almost in anti- phase ( 1 80 or radian out of phase) with the driver pendulum.
Figure 6 shows the variation, with forcing frequency, of the phase lag for a pair of pendulums. The length of the driver pendulum string is adjusted when it is shorter than the driven pendulum the forcing frequency is higher and the driven pendulum will lag behind the driver by radian, etc. phase lag/rad driver leads by half a period lighter damping heavier damping /2 driver leads by quarter of a period
driver and driven in phase 0 f0 Figure 6 Phase relationship for the displacement of a
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driver frequency
forced vibration.
B . 4 F O R C E D V I B R AT I O N S A N D R E S O N A N C E ( A H L )
Nature of science Examples of resonance In nature, resonance is a very common phenomenon having examples in virtually all areas o physics. Further examples include:
The human voice uses resonance to produce loud sounds rom a relatively weak source the vocal cord.
The sound box o musical instruments amplifes the energy, causing air in the box to resonate at the same requency.
O zone in the stratosphere has natural requencies o vibration that match the requency o ultraviolet and absorb this radiation.
Microwave cookers emit electromagnetic waves that match the natural requency o water molecules in ood.
The optical cavities in lasers set up standing waves or light in order to produce coherent beams.
MRI ( magnetic resonance imaging) scans use resonating protons in atoms to provide vital inormation about body cells.
Resonance can have drawbacks as well as benefts, or example:
Vibration set up by soldiers marching across bridges has meant that they are told to " break step" . In June 2 000, Londons Millennium Footbridge ( fgure 7) was ound to sway alarmingly rom the resonances set by pedestrians crossing it. Fitting dampers solved the problem but ailed to prevent the bridge being known as " the wobbly bridge" .
The Tacoma Narrows Bridge in Washington State, USA, collapsed in November 1 940 as a result o cross winds matching its natural requency.
Figure 7 The London Millennium Footbridge.
Feedback can be heard at rock concerts a loud howling sound is produced when microphones or pickups are too close to loudspeakers, giving an uncontrolled amplifcation o the sound.
The clamped hacksaw blade will have a natural requency that depends on the length projecting.
The amplitude o the end o the blade can be determined using a vertically clamped millimetre scale ( not shown) .
This apparatus can be used to measure the variation o amplitude with ( a) supply requency f ( b) proj ecting length L.
Vibrations in machinery can cause vibrations in nearby obj ects overtaking mirrors on lorries that are idling can be seen to vibrate with large amplitude.
Investigate! Resonance of a hacksaw blade Apparatus similar to that shown in fgure 8 can be used to investigate the resonance o a hacksaw blade.
An electromagnet is powered by a signal generator that behaves as a variable requency ac supply.
The requency can be varied by adj usting the signal generator.
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B
E N G I N E E R I N G P H YS I C S
The general orm o the relationship may be x0 = kfn or x0 = kL n
This suggests a loglog graph should generate a straightline graph.
at both ends. The magnet is positioned somewhere near one end o the wire.
Melde' s string oers a similar investigation when a long metallic wire carries a current in a magnetic feld ( see S ub- topic 4.5 ) :
An alternating small current is set up in the wire.
B y changing the tension in the wire the natural requency o the wire will approach the requency o the AC current.
When resonance is reached, standing waves will appear along the wire, the harmonic depending on the conditions.
Take care because the wire can become red-hot!
The wire is placed between the poles o a strong U- shaped magnet and is clamped
signal generator wooden blocks
hacksaw blade
G-clamp
electromagnet
laboratory jack
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Figure 8 Resonance of a hacksaw blade.
QUESTION S
Questions 1
An obj ect o mass 3 .0 kg is attached to a string which is wrapped round a thin uniorm disc o radius 0.2 5 m and mass 7. 0 kg. The disc rotates about a horizontal axis passing through its centre. Assume the rotation o the disc is rictionless.
7 kg
(b)
(a)
3
The angular speed o a rotating disc is increased rom 2 0 rad s - 1 to 85 rad s - 1 in a time o 6.0 s by a constant torque. The disc has mass o 8.0 kg and radius is 0.3 5 m. C alculate: a) the work done by the torque in this time b) the average power applied to the disc during this time. ( The moment o inertia o a disc is given by the equation in question 1 . ) ( 7 marks)
C alculate: a) the angular acceleration o the disc when the mass is released rom rest b) the angular acceleration i the orce were applied by pulling the cord with a constant orce o 3 0 N. The moment o inertia I o a thin uniorm disc o mass M and radius R is given by 1 I = __ MR 2 ( 5 marks) 2 2
A skater has moment o inertia 2 .85 kg m 2 with her arms outstretched. She rotates with an initial angular speed o 2 .0 rad s - 1 as shown in the diagram. B y bringing in her arms the skaters moment o inertia reduces to 1 .5 kg m 2 . For the skater, calculate:
4
( IB) The graph shows the variation with volume o the pressure o a fxed mass o gas when it is compressed adiabatically and also when the same sample o gas is compressed isothermally. 7.0
C
6.0 pressure/ 10 5 Pa
3.0 kg
5.0
B
4.0 3.0
a) her fnal angular speed b) the change in her rotational kinetic energy.
A
2.0
( 5 marks)
1.0
2.0
3.0
4.0
5.0
6.0
Volume /10 -3 m 3
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B
E N G I N E E R I N G P H YS I C S a) S tate and explain whether line AB or AC represents the isothermal compression.
O n a copy o the diagram above: ( i) draw arrows to show the direction o the changes
b) O n a copy o the graph, shade the area that represents the dierence in work done in the adiabatic change and in the isothermal change.
(ii) label with the letter A an isobaric change ( iii) label with the letter B the change during which energy is transerred to the working substance because o a temperature dierence.
c) D etermine the dierence in work done, as identifed in ( b) . d) Use the frst law o thermodynamics to explain the change in temperature during the adiabatic compression. ( 9 marks)
5
c) Use data rom the diagram in ( b) to estimate the work done during one cycle o the working substance. d) ( i) B y reerence to entropy change, state the second law o thermodynamics.
(IB) An ideal gas at an initial pressure o 4.0 1 0 5 Pa is expanded isothermally rom a volume o 3.0 m3 to a volume o 5.0 m 3 .
( ii) The cycle o the working substance in ( b) reduces the temperature inside the rerigerator. E xplain how your statement in ( d) ( i) is consistent with the operation o a rerigerator. ( 1 1 marks)
a) C alculate the fnal pressure o the gas. b) O n graph paper, sketch a graph to show the variation with volume V o the pressure p during this expansion. 7
c) Use the sketch graph in ( b) to: ( i) estimate the work done by the gas during this process ( ii) explain why less work would be done i the gas were to expand adiabatically rom the same initial state to the same fnal volume. ( 7 marks) 6
( i) the volume rate o ow in m 3 s 1 through the tube ( ii) the speed o the water in the narrower part o the tube.
p I10 5 Pa
b) The diagram below shows the pressure volume ( p- V) changes or one cycle o the working substance o a rerigerator.
590
a) C alculate:
(IB) a) S tate what is meant by isobaric change.
13 12 11 10 9 8 7 6 5 4 3 2 1 0
A horizontal tube o cross-sectional area 3 .0 1 0 4 m 2 narrows to 1 .6 1 0 4 m 2 . Water ows through the tube at a speed o 0. 40 m s 1 at the wider part.
b) Explain how the ow o water in the pipe is accelerated. ( 6 marks) 8
The diagram shows a venturi meter which is used to measure the ow rate o low-density liquid through a horizontal pipe. When the liquid is owing, the mercury manometer is used to measure the dierence between the pressure o the liquid in the pipe at the entrance to the venturi meter and that in the throat o the meter: entrance pipe
throat
exit pipe liquid fow
0
0.2
0.4
0.6 0.8 1.0 V I10 -3 m 3
1.2
1.4
1.6
mercury manometer
80 mm
QUESTION S a) In one measureme nt the die rence in levels o the me rcury in the manome ter arms is 8 0 mm. C alculate the pressure diere nce betwe en the liquid in the e ntrance to the ve nturi meter and that in the thro at, e xpre ssing your answer in pascals. density o me rcury = 1 . 4 1 0 4 kg m - 3 b) the corss- sectional areas o the pipe and venturi meter throat are 4.0 1 0 - 2 m 2 and 1 .0 1 0 - 2 m 2 respectively. D etermine the ratio o the speed o the liquid owing in the throat o the venturi meter to its speed in the entrance pipe. c) The liquid has a density o 8. 0 1 0 2 kg m - 3. ( i)
Use the B ernoulli equation to estimate the speed o the liquid in the horizontal pipe.
( ii) Hence estimate the ow rate o the liquid in kg s - 1 . ( 1 0 marks) 9
a) ( i) Use the continuity equation to show that an incompressible liquid moving rom a wider pipe to a narrower pipe must increase its velocity. ( ii) S tate B ernoullis relation or the ow o an incompressible inviscid uid along a horizontal stream line. b) Figure 1 shows a cross-section through a simple laboratory flter pump. B
( i)
Explain how a ow o water rom A to C through the pump produces a partial vacuum at D .
( ii) C alculate the velocity o the water emerging rom the nozzle B . ( iii) C alculate the rate, in m 3 s - 1 , at which water ows through the pump. ( density o water = 1 000 kg m - 3 ) ( 1 3 marks) 1 0 A spring or which the extension is directly proportional to the weight hung on it is suspended vertically rom a fxed support. When a weight o 2 . 0 N is attached to the end o the spring the spring extends by 5 0 mm. A mass o 0 . 5 0 kg is attached to the lower end o the unloaded spring. The mass is pulled down a distance o 2 0 mm rom the equilibrium position and then released. a) ( i) S how that the time period o the simple harmonic vibrations is 0. 70 s. ( ii) O n a sheet o graph paper sketch the displacement o the mass against time, starting rom the moment o release and continuing or two oscillations. S how appropriate time and distance scales on the axes. b) The massspring system described in part (a) is attached to a support that can be made to vibrate vertically with a small amplitude. Describe the motion o the massspring system with reerence to requency and amplitude when the support is driven at a requency o ( i) 0.5 Hz
( ii) 1 .4 Hz ( 8 marks)
A
C D
1 1 The Millennium Footbridge in London was discovered to oscillate when large numbers o pedestrians were walking across it.
Figure 1
The Pressure at C is at atmospheric pressure ( 1 00 kPa) and the water at C is moving very slowely. The Pressure at D is 45 kPa. Nozzle B has a diameter o 2 . 0 mm. C alculate
a) What name is given to this kind o physical phenomenon? b) E xplain the conditions that would cause this phenomenon become particularly hazardous?
591
B
E N G I N E E R I N G P H YS I C S c) Suggest two measures which engineers might adopt in order to reduce the size of the oscillations of a bridge. ( 7 marks) 1 2 a)
A forced vibration could show resonance. E xplain what is meant by: ( i) forced vibrations ( ii) resonance.
592
b) ( i) E xplain what is meant by damping. ( ii) What effect does damping have on resonance? ( 5 marks)
C I M AG I N G Introduction D ata collection in science often depends on using our senses to make an observation. S ometimes the image we perceive needs to be enhanced in some way. For example, doctors
may need to see inside the human body in order to make a diagnosis. This topic deals with the physics of imaging in both visual and non- visual contexts.
C.1 Introduction to imaging Understanding Thin lenses
Applications and skills Describing how a curved transparent interace
Converging and diverging lenses Converging and diverging mirrors
Ray diagrams Real and virtual images Linear and angular magnifcation
Spherical and chromatic aberrations
Nature of science Virtual images cannot be ormed directly on a
screen. The technique o ray tracing allows the position and size o a virtual image or a virtual object to be inerred. This is an example o deductive logic in action.
modifes the shape o an incident waveront Identiying the principal axis, ocal point, and ocal length o a simple converging or diverging lens on a scaled diagram Solving problems involving not more than two lenses by constructing scaled ray diagrams Solving problems involving not more than two curved mirrors by constructing scaled ray diagrams Solving problems involving the thin lens equation, linear magnifcation, and angular magnifcation Explaining spherical and chromatic aberrations and describing ways to reduce their eects on images
Equations 1 =_ 1 +_ 1 Thin lens equation: _ v u f 1 Power o a lens: P = _ f
h ho Angular magnifcation: M = _i o
v
i Linear magnifcation: m = _ = - _ u
Magnifcation o a magniying glass:
D + 1; M D M n e a rp o i n t = _ =_ in f n ity f f
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Converging and diverging mirrors Topic 4 introduced the rules o reection at plane suraces. In this section we look at what happens when suraces are no longer at. A curved reecting surace allows the rays to be manipulated, leading to images o various types ormed by mirrors. The shapes o convex and concave suraces are shown in fgure 2 . Mirrors made with these shapes are commonly ound in a number o situations: as mirrors used or makeup or shaving where a magnifed image is needed, or as a way or seeing a wide angle o view.
Figure 1 Concave and convex mirrors and their refections.
We will mainly consider mirror suraces that have a spherical profle. Figure 2 shows what happens when light is incident on both convex and concave reectors that have been ormed rom the surace o a hollow sphere. The line that goes through the centre o the mirror surace at 90 to the surace is the p rincip al axis. concave mirror
convex mirror
Q
principal axis
C
F
P
C
F
Q f ocal length R radius o curvature (a)
ocal length
(b)
Figure 2 When rays that are parallel and close to the principal axis are incident on a concave mirror then they all reect through a point known as the principal focus F. F is also known as the focal p oint o the mirror. It is easy to see why this has to be the case. The second law o reection tells us that the angle o incidence at the mirror is equal to the angle o reection. The surace o the mirror is part o a sphere (drawn as a circle in the two-dimensional diagram) and thereore the line joining the centre o the sphere C to the mirror at the point where the light ray is incident is the normal to the surace. This defnes the angle o incidence i. The angle o reection r must equal i. I the incident ray is close to the principal axis then, ater reection, the ray will cross the principal axis at a point F that is hal way rom the pole P o the mirror to the centre o the sphere. The distance rom F to the pole is known as the focal length f. When the distance between the principal axis and the incident ray is small, f is hal the radius o curvature o the mirror. We shall look at what happens when incident rays lie well away rom the principal axis later. For now, our assumption is always that the distance between rays and the axis is small. When a convex mirror is used ( fgure 2 ( b) ) the position is very similar, except that this time the reected ray appears to have come from the ocus. There is the same relationship between f and the radius o the mirror as beore.
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Investigate! Images from a mirror These experiments are intended to introduce you to the images ormed by a mirror. The mirrors that will work best or these experiments have a ocal length o about 1 5 2 5 cm. You need to know the approximate ocal length o each mirror beore you begin.
(1) Concave B egin by taking the concave mirror and looking into it with an approximate distance between your eye and the mirror that is ( a) about double the ocal length, ( b) less than the ocal length. What do you notice about the image you see? Is the image the right way up or upside down? C an you describe what happens when the eye is exactly at the ocal point? Now set the mirror on the bench using suitable stands so that the light rom an illuminated object is incident on the mirror and then refected to orm
an image on a screen. Make sure that the object mirror distance u is greater than the ocal length. Find out the range o u or which the image is larger, smaller, and the same size as the object.
(2) Convex This time simply look into the mirror. Where is the image ormed? Is it similar to the image ormed by a plane mirror, what are the dierences? lamp
concave mirror
wooden block
white paper ruler
Figure 3.
To explain the positions o images ormed by curved mirrors we use the technique o drawing a ray diagram. A ray diagram establishes the relationship between an obj ect and its image that results when rays rom the obj ect are refected in the mirror. Using the idea that light is reversible we can identiy p redictable rays that we will use to construct a scaled ray diagram. Reversibility means that when light rays are refected back along their original path they will trace out the incident path but in the reverse direction.
X
Y Z
The predictable rays we will use are:
X: initially parallel to the principal axis and then going, ater refection, through the ocal point.
Y: initially through the ocal point and then going, ater refection, parallel to the principal axis.
Z: a ray through the centre o curvature that goes ater refection back along its original path.
Figure 4 Predictable rays.
Using these rays we can determine the position and the nature ( size and type) o the image ormed by a curved mirror. Remember that X, Y, and Z are close to the principal axis and the mirror is almost fat at the centre. The drawing technique uses several conventions:
The mirror is drawn in two dimensions as a straight line at 90 to the principal axis. O nly the top and bottom o the mirror are curved to indicate whether the mirror is convex or concave.
The vertical scale and the horizontal scale need not be the same. O bj ects and mirrors are sometimes only centimetres high but can
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The obj ect is represented by a vertical arrow drawn to show the largest dimension o the obj ect to scale.
At least two predictable rays are drawn rom the top o the obj ect arrow. The position o the top o the image o the arrowhead will be shown by the direction o the rays ater reection.
Arrows are drawn to show the direction in which the rays travel.
The step-by- step construction o one ray diagram each or a concave and a convex mirror is shown in sequence in fgure 5 . Concave
object C
C
C
C
Convex 1 Draw the principal axis and the mirror. Add the centre o curvature and the ocal point. Add the object to scale (vertical and horizontal scales can be diferent) .
F
F
2 Draw one ray rom the top o the object parallel to the principal axis. Ater reection the ray goes through/appears to have come rom the ocal point. Note that a line or the convex mirror is dotted because it is a construction line, not a ray.
3 Draw another ray either (a) through the ocal point and then reecting parallel to the principal axis or (b) to and rom the centre o curvature.
F
F image
object
4 Where the rays/construction lines cross is the position o the image. The nature and position o the image give inormation about it.
This image is smaller than the object (diminished) , upside down (inverted) , and real (because the rays cross) .
F
C
F
C
F
C
image F
C
This image is smaller than the object (diminished) , the right way up (erect) , and virtual (because the rays appear tohave come rom the image and do not cross) .
Figure 5 Step-by-step instructions or drawing concave mirror and convex mirror ray drawing. The concave mirror has a number o separate cases each o which depends on the position o the object. I the rays rom the top o the object converge to a single point then this marks the top o the arrowhead. This is now below the principal axis (whereas the object was above) because the image is an inversion o the object (turned upside down) . This point where the rays converge is still called the top o the image.
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The rays in these cases orm a real image. Rays that intersect ( converge) in this way can be used to orm an image on a screen. This is what was happening when you saw a picture orming on the screen in the Investigate!
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However, when the obj ect is closer than the ocal point, the fnal ray directions diverge ( move apart) and do not cross. When these diverging rays are constructed back ( notice that construction lines are always drawn as broken lines in the worked examples as they are not rays) , they orm a dierent type o image known as a virtual image. This type o image cannot be ormed on a screen but it can be seen or manipulated using another mirror or a lens. Virtual images in the Investigate! were ormed by the lens in your eye. A convex mirror by itsel can never orm a real image rom a real obj ect. As the rays always diverge, the convex mirror is called a diverging mirror. For the concave mirror, unless the obj ect is closer to the lens than the ocal point, the rays converge. C oncave mirrors are usually called converging mirrors in some books they are reerred to as converging ( concave) mirrors.
Magnifcation B ecause the diagrams or both types o mirrors are scaled, the fnal scale sizes o the obj ect and image can be used to measure the linear magnifcation m o the mirror system. hi height o image m = __ = _ height o obj ect ho
object v __ u,
Geometry shows that m is also equal to where u and v are the distances rom the mirror pole to the obj ect and the image respectively.
ho
o i
C
hi F image
In some mirror examples that we shall see later, the obj ect and image heights cannot be easily identifed. In these cases angular magnifcation M is used. M = __ where i and o are the angles i
o
subtended at the mirror by the rays rom the image o the top o the image and the rays rom the top o the obj ect, respectively.
Figure 6 Magnifcation by a mirror, linear and angular.
Aberrations in a mirror S o ar we have assumed that the rays rom the obj ect are travelling close and approximately parallel to the principal axis. When this is not the case then the spherical mirror introduces distortions ( aberrations) into the images. The principal reason or this is that at large distances rom the pole o the mirror the rays no longer reect through the ocal point. When rays parallel to the principal axis are incident over the whole mirror surace then they orm a pattern known as a caustic curve. You may have noticed this pattern orming when strong sunlight strikes the reecting surace o a circular cup o coee or another opaque liquid. The inner surace o the cup reects the light onto the liquid surace. The position o ocal point and the shape o the caustic can be clearly seen. To reduce this problem in cases where a large mirror surace is needed, a parabolic surace can be used. Rays parallel to the principal axis o a parabolic mirror always pass through the ocus and no caustic orms in these circumstances ( fgure 7( b) ) . An example is the large mirror used or an astronomical telescope.
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(a) spherical mirror
(b) parabolic mirror
Figure 7 Spherical mirror distortions and the parabolic mirror. The reversibility of light tells us that a parabolic mirror shape can also be used to generate a parallel beam of light if a light source is placed at the focus. This is often used to produce a searchlight beam that diverges very little even at large distances from the mirror. In some situations, two or more mirrors can be combined in optical instruments such as telescopes. We will give examples of these in a later section.
Worked example 1
c)
C onstruct a ray diagram for a concave mirror when: a) the obj ect is between F and C
object C
b) the obj ect is at C
F
c) the obj ect is at F d) the obj ect is between the pole of the mirror and F.
d) object
Solution a) C
2 image C
object
F
image
F
C onstruct a ray diagram for a convex mirror when the obj ect is between the pole of the mirror and the focal distance.
Solution b)
C
image
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image
object
F
object
F
C
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Converging and diverging thin lenses A mirror has only one surace and the light always travels in the air. When light enters a transparent medium it is reracted and the speed and wavelength o the light change. This change o speed makes the analysis o light passing through a curved interace into a dierent medium more complex than the reection case. Lenses usually have two interaces through which the light enters and leaves. B eore considering this double reraction in detail we will look at the eect o a single interace between two dierent media. In this example, the light travels slower in the second medium than the frst. Figure 8(a) shows parallel rays o light as they travel through a convex surace rom one medium to another o greater optical density. Parallel plane wave ronts are associated with these rays and are also shown. The centre o these waveronts meets the curved convex surace earlier than the outer edges o the wave ront. As the wave enters the medium it slows down and the wavelength o the light becomes smaller. This means that the centre o the wave is travelling at a slower speed in the second medium than the wave edges that have not yet reached the interace.
curved interace (a)
waveronts
ray ray
(b)
O
I
Figure 8 Wave ronts through a medium.
The overall eect on the wave when it has ully entered the medium is that what was a parallel wave has become curved. The rays are no longer parallel either. They meet at a ocal point. O ne interpretation o the action o a curved surace is that it adds curvature to waveronts ( in the convex case) or removes curvature ( in the concave case) . The curvature o a surace can be defned in general 1 terms as __ where R is the radius o the surace. The smaller the radius R o the surace, the more curved it is. Now we can see what happens when a real lens is used to modiy rays ( fgure 8( b) ) . There are two curved interaces and it is the cumulative eect o both interaces that determine the properties o the lens ( fgure 9) . We use a similar model to that o the mirrors to defne the parts o the lens ( fgure 1 0) . The op tical centre o the lens is the point in the lens that does not deviate rays o light passing through it. As with the mirror we defne a p rincip al axis, focal p oint and focal length. Rays parallel to the principal axis pass through the ocal point ater reraction or a convex lens. For a concave lens, the parallel rays appear to have come rom the ocal point. We shall assume that the convex or concave lenses have lens thicknesses that can be ignored compared to the distances between obj ect and image. The thin-lens theory that uses this assumption is more straightorward than when the thickness o the lens is included. We will also consider only spherical lenses, that is, those with suraces shaped so as to orm part o the surace o a sphere. Many modern lenses are designed to have an asp herical shape ( non- spherical) , such lenses, or example, spectacle lenses, can be made thinner and lighter than their spherical counterparts.
Figure 9 Rays through convex and concave lenses. rays rom infnity
optical centre
ocal point F
O
F principal axis
F
O
F principal axis
Figure 10 Defnitions o the parts o a lens.
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Investigate! Finding an approximate value for the focal length of a convex lens
We have used the idea of rays that travel parallel to the principal axis. These can be obtained (approximately) by using rays that have come from a distant object. The diagram shows how these arise. Two rays that leave the same place on an object and then travel a large distance compared with the diameter of the lens have to be travelling very close to parallel if they are both to enter the lens.
Hold your lens near a window so that rays from a distant obj ect outside ( more than about 2 0 m away will do) form an image on a sheet of card ( called a screen) held behind the lens.
The distance from the card to the lens is, approximately, the focal length. Get someone to measure this distance while you hold the lens and the card. Notice that the image is upside- down.
long distance C screen 2F
F
O
O
Figure 11
Investigate! The relationship between object and image for a thin convex lens
For this experiment you will need a convex lens of approximate focal length 0.1 5 m, an illuminated obj ect, a screen, and a ruler to measure the distance between the obj ect and the lens ( the obj ect distance) and the lens and the screen ( the image distance) . Set the obj ect about 0.4 m from the lens. Move the screen until there is a clearly focused picture of the obj ect on the screen you should expect it to be upsidedown. Measure the obj ect distance u and the image distance v and record them.
Repeat the procedure for a range of obj ect distances ( do not attempt to have the obj ect distance closer than the focal length) .
Plot your data on a graph of: 1 1 __ __ v against u
lamp (object)
lens u
screen (image) v optical bench
Figure 12
The results of this experiment can be summed up using scaled ray diagrams. Ray diagrams for both convex and concave lenses are constructed on a similar basis to those for mirrors. Additional rules are that:
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The lens is drawn as a vertical line to emphasize that it is thin. S ymbols at the top and bottom of the lens indicate whether it is convex or concave.
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The ocal length should be shown on both sides o the lens.
There are three predictable rays:
A ray rom the top o the obj ect parallel to the principal axis goes through the ocal point ater reraction by the convex lens or, in the case o a concave lens, deviates so that it appears to have passed through the virtual ocus.
A ray rom the top o the obj ect through the optical centre o the lens does not deviate.
A ray rom the top o the obj ect through the ocal point travels parallel to the principal axis ater reraction.
O nly two o these rays are needed to complete the diagram and show the position size and nature o the image. The third can be used as a check.
Rays should have an arrow to show the direction in which they travel.
(a) object
(b) image image 2F
O
F
F
2F
object object F
2F
O
F
F
F
2F image
object
2F
F
O
F
2F
image
Figure 13 Ray diagrams for a convex (converging) lens. Figure 1 3 ( a) shows the ray diagrams that match some o the cases you tested in the Investigate! When the obj ect is closer to the lens that the ocal point, a real image cannot be ormed and a virtual image results ( fgure 1 3 ( b) ) . This is the ray diagram or a magnifying glass. D iverging lenses have ewer ray diagrams than the converging lenses, the fnal image is always virtual. Such images cannot be ormed on a screen and an additional converging lens ( such as the lens in the eye) is required to orm a real image that can be proj ected ( fgure 1 4) .
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object
image
F
F
Figure 14 Ray diagram for a concave ( diverging) lens.
The lens equation The Investigate! shows that the position o the image depends both on the position o the obj ect and the ocal length o the lens. The lens equation quantifes this relationship. In curvature terms, a particular lens o ocal length f adds the same amount o curvature to any waveront passing through it. The amount 1 added is __ . This is because a lens with a small f gives more curvature to f incident parallel rays than does a lens where f is large ( the small f lens 1 1 __ bends the rays more) . The quantities __ u and v determine the curvature o the obj ect and image waveronts. Figure 1 5 shows how the symbols are defned and relate to each other. Thereore we can write: curvature o waveronts leaving lens = curvature o waveronts entering lens + curvature added by lens so 1 1 1 _ _ _ v = u + f
u
f v
Figure 15 The lens equation. However, the waveronts rom the obj ect are spreading out when they reach the lens whereas the waveronts that have let the lens to orm the image are converging. The curvatures o the obj ect waveronts are in the opposite sense to those o the image. To allow or this we should consider the u value to be negative with respect to v. This leads to the lens equation: 1 1 1 _ _ _ u + v = f
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S ome care needs to be taken with the signs given to u, v and f. The rule that applies when using the lens equation in this orm is:
Note In fact, although this will not be tested in the examination, the lens equation also applies to mirrors. In this case f is equal to 1/R.
real obj ects and images are treated as positive
virtual obj ects and images are treated as negative
the ocal length o a converging lens is positive
the ocal length o a diverging lens is negative
this is known as the real-is-p ositive sign convention; you may see texts that use other conventions.
The ability or power o the lens to add (or subtract) curvature to a waveront is: 1 power, P ( in dioptre) = __. f( in metre) A strong lens has a small ocal length and adds a large curvature to any waveront passing through it. The power o a lens is measured in diop tres ( not an S I unit) . A converging lens with a ocal length o 5 0 cm ( 0.5 m) has a power o +2 D ; the lens is said to be a positive lens. A diverging lens o ocal length 1 2 .5 cm has a power o 8D ; the lens is said to be a negative lens. I you wear eye glasses or contact lenses you may have seen these units used on your lens prescription. The defnitions o linear and angular magnifcation are identical to those or curved mirrors. height o image v m = __ = _ u height o obj ect i M=_ o where the o and i are the angles subtended by the obj ect and the image at the lens.
Worked examples 1
An obj ect o height 5 cm is placed 1 2 cm rom a converging lens o power + 1 0 D . C alculate:
c) The positive sign tells us that the image is real, 60 = 5 times and is thereore magnifed by __ 12 2 5 cm high. The image is upsidedown and on the opposite side o the lens to the obj ect.
a) the ocal length o the lens b) the position o the image c) The nature o the image.
Solution 1 a) f = _ = 0.1 0 m = 1 0 cm D 1 1 1 1 1 1 _ _ _ _ _ b) _ u + v = f so v = f - u 1 1 2 1 _ _ _ _ v = 1 0 - 1 2 = 1 20 ; v = +60 cm rom the lens
2
An obj ect o height 5 cm is placed 8. 0 cm rom a converging lens o ocal length + 1 0 cm. C alculate the position and nature o the image.
Solution 1 = _ 1 - _ 1 = _ 1 1 -2 _ - _ = _; v = - 40 cm v u 8 10 80 f The image is ormed 40 cm rom the lens; it is virtual, magnifed 4 times, and is the right way up. It orms on the same side as the obj ect. This is the magniying glass arrangement.
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3
C alculate the position and nature o an image placed 1 2 cm rom a diverging lens o ocal length 2 4 cm.
Although not asked or in the question, the ray diagram is given as an example o a more complicated arrangement. frst lens
Solution
second lens
36 1 =_ 1 - _ 1 =_ 1 1 -1 _ - _ = -_ = _ v u -24 +1 2 24 1 2 8 f The image is ormed 8 cm rom the lens on the same side as the obj ect. It is diminished with 8 a magnifcation o ( ) __ = 0.67. The image is 12 upright and virtual. 4
a)
fnal image f1
f2
A converging lens o ocal length + 1 5 cm is 2 5 cm rom an obj ect. C alculate the position o the image.
b) Another converging lens also o ocal length + 2 5 cm is placed 1 8 cm rom the lens on the image side. C alculate the new position o the image.
Solution 1 = _ 1 - _ 1 =_ 1 1 a) _ - _; v = 3 7. 5 cm. v u 15 25 f b) The new u is 3 7.5 1 8 = 1 9. 5 cm rom the lens. This is a virtual obj ect or the second lens and so will have a negative sign in the lens equation. 1 1 1 _ _ _ v = 2 5 - - 1 9.5 ; v = + 1 1 cm A real image orms 1 1 cm rom the second ( f = 2 5 cm) lens.
f1
f2 intermediate (virtual) object
The frst lens and its image is drawn frst ( green lines) the rays beyond the position o the second will not orm so are drawn dashed as construction lines. Then the second lens is added together with its ocal points (red lines) . There are a number o possible predictable rays that can be drawn. One shown here is the ray that is aimed at the optical centre o the second lens this ray will not deviate as it goes through the lens and must also go through the top o the virtual image. The other ray or the second lens comes rom the top o the original object, through f1 and then must go parallel to the principal axis ater being reracted by the lens. Ater passing through the second lens, it must go through f2 and defnes the top o the fnal real image.
The simple magnifying glass A single convex lens used as a magniying glass is the oldest optical instrument recorded. Around 2 400 years ago, Aristophanes magnifed small obj ects using spherical asks flled with water like those used by present- day chemists. We have already drawn the ray diagram or the magniying glass, but because o its importance, both alone and in conj unction with other lenses, it merits a section o its own. The human eye has a near p oint distance D that is the closest distance at which the eye can ocus on an obj ect without strain. For a normal eye this distance is taken to be 2 5 cm ( although it varies greatly rom individual to individual) . Thereore, the best we can do to see the fne detail in an obj ect is to hold it 2 5 cm away rom the eye. The magniying glass helps us to improve on this.
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h
D without lens image at near point
image at infnity
h f
object at f
image f
object
F
f with lens
Figure 16 Image ormed by a magniying glass at near point and infnity O ne way ( and the least tiring method or prolonged viewing) is to position the obj ect at the ocus o the converging lens and thereore orm the image at infnity where it can be viewed with relaxed eye muscles ( fgure 1 6) . I f or the lens is smaller than 2 5 cm the obj ect will need to be closer than the near point with the eye muscles relaxed. In such circumstances, the eye should be as close to the lens as possible. In this case, the angle subtended by the unaided eye is the size o the h h image __ whereas the angle subtended by the aided eye is __ so the D f angular magnifcation M is h _ f D _ = _. f h _ D S o a magniying glass with ocal length 0. 1 m will give a magnifcation o 2 .5 . However this is not the best that the magniying glass can do. Another way to use the lens, but more tiring because the muscles are not relaxed, is to place the eye close to the lens as beore and to adj ust the position o the obj ect so that the image orms at the near point. The obj ect can now be much closer to the eye than would be comortable without the lens.
u
h' _
u The angular magnifcation now changes to M = __ where u, h' and h are h _ defned in fgure 1 7. D
The lens equation gives 1 1 1 _ =_ + _ u -D f D - u D D D __ S o _____ = __ or __ u u - 1= f f
S o or this setting o the magniying glass D M= _+ 1 f
h' G
h image f
object D
Figure 17 Angular magnifcation or a magniying glass.
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Note You may see that some books give the equation as M = ___D - 1. This results when the sign o D is not introduced during the proo but is used in the fnal calculation. Using the numerical examples above together with the equation with the negative sign, when the image is at infnity is M 2.5 and when the image is at the near point it is 3.5. So the magnitude o the magnifcation still remains larger or the second case.
For a magniying glass with ocal length 0.1 0 m and a near point o 0.2 5 m, the magnifcation is 3 . 5 .
Spherical and chromatic aberrations Chromatic aberrations The absolute reractive indices o the materials used or making lenses vary with wavelength. Dierent colours travel through the lenses at dierent speeds and thus orm ocal points at dierent distances rom the lens. crown int
chromatic aberration
Worked example A magniying glass has a ocal length o 7.5 cm. It is used by a person with a near point o 2 5 cm to view an obj ect. C alculate the angular magnifcation o the obj ect when the image is at: a) infnity b) the near point.
Solution 25 D; M = _ a) M = _ = 3.3 7. 5 f b) M = infnity angular magnifcation + 1 = 4.3
achromatic doublet
Figure 18 Chromatic aberration and a way to correct it. This problem is known as chromatic aberration and it leads to the appearance o colour ringes around the image. The best place to view the image is at the intersection o the green rays. This is the point known as the circle of least confusion. A circle drawn around all the rays here is at its smallest. One common way to correct or chromatic aberrations is to use a doublet lens, one o the lenses is converging with positive power and the other diverging with a negative and smaller magnitude power. The lenses produce chromatic aberrations in opposite senses. Over the whole visible range, the colour perormance o the combined lens is better than with one lens alone.
Spherical aberrations Like mirrors, lenses also suer rom aberrations caused through the geometry o the spherical lens. As fgure 1 9 shows, rays that are ar rom the optical centre are brought to a ocus closer to the convex lens than those near the lens centre. This is sp herical aberration and leads to an obj ect consisting o a square grid distorting as shown. This is known as barrel distortion.
Figure 19 Rays that produce spherical aberration and the efect on a square grid.
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C .1 I N TRO D U CTI O N TO I M AG I N G The easiest way to cure spherical aberration is to reduce the aperture ( diameter) o the lens perhaps by putting an obstacle with a hole cut in the centre over the lens. O course, this will reduce the amount o light energy arriving at the image position and will make the image appear less bright.
Nature of science Same travel times edge ray O
centre ray
shorter physical path but spends more time in lens
short time in lens longer physical path I same time to go from O to I
TOK Sign conventions what is their efect? There are two common sign conventions used in optics: the real-is-positive convention used here and the New Cartesian convention. Earlier we used a conventional current o positive charges in electricity. To what extent, i any, do sign conventions afect our understanding and use o science?
Figure 20 There is another interpretation o how lenses work. We have regarded the lens as reracting light rays or as adding curvature to or subtracting curvature rom a waveront. Think about two rays going through a converging lens ( fgure 2 0) . O ne goes through the centre o the lens where it is thickest. The other goes through the edge o the lens where there is not much glass. The image orms at the place on the other side o the lens where the time taken by the light to take these two paths is the same. The ray through the centre travels arther than the other ray in the glass, and because the speed o the light is slower in the glass, this ray spends longer in the glass than the edge ray. The edge ray has a longer overall path. What is special about the image position is that it is the ( unique) place where all rays rom the obj ect take the same time to reach their respective places on the image. This has to be the case given our waveront interpretation. The lens keeps the wave together so that all parts meet at the same time at the same place on the image. Taking the same time and having the same eective path distance is what makes a lens a useul thing to manipulate rays. This is an example o Fermats principle o least time.
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C.2 Imaging instrumentation Understanding Optical compound microscopes
Applications and skills Constructing and interpreting ray diagrams
Simple optical astronomical reracting
telescopes Simple optical astronomical reecting telescopes Single-dish radio telescopes Radio intererometry telescopes Satellite-borne telescopes
Nature of science Optical instruments have been developed over the centuries to improve our ability to observe very distant objects in the sky and very small objects on Earth. Instruments to improve our vision were used in the time o the Egyptians. Authors described combinations o lenses in mediaeval Europe and it is certain that the knowledge o optics had been known in the Arab world since at least 1000 years BP. This line o improvement in instrumentation is clear in optics.
o optical compound microscopes at normal adjustment Solving problems involving the angular magnifcation and resolution o optical compound microscopes Investigating the optical compound microscope experimentally Constructing or completing ray diagrams o simple optical astronomical reracting telescopes at normal adjustment Solving problems involving the angular magnifcation o simple optical astronomical telescopes Investigating the perormance o a simple optical astronomical reracting telescope experimentally Describing the comparative perormance o Earth-based telescopes and satellite-borne telescopes
Equations f0 Magnifcation o astronomical telescope M = _ fe
Optical compound microscope Although the convex lens used as a magniying glass provides magnifcation, small ocal lengths are required to produce large D magnifcations ( M = __ ) . Lenses with small ocal length have large f surace curvatures and this can give rise to so much aberration that eatures in the image cannot be seen clearly. The compound microscope is designed to produce a virtual magnifed image o a small obj ect. The term compound means that it is composed o more than one lens: an obj ective lens, close to the obj ect, with a very short ocal length fo , and an eyep iece lens into
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which the user looks As with all optical systems, you should try to understand the unction o each component in the microscope:
eye lens O
The obj ective orms a real, highly magnifed image o the obj ect at a position that is closer to the eyepiece than fe . The eyepiece then acts as a magniying glass the obj ectives real image as its own obj ect.
The fnal image, viewed by the observers eye, is virtual and very highly magnifed.
The fnal image should not be closer to the eye than the near point D and when the image is ormed here then the microscope is said to be in normal adj ustment. Many experienced microscope users do not ocus the microscope in ull normal adj ustment, however. It is oten much more convenient or the fnal image to be ocused at the plane o the bench on which the microscope stands. A notebook next to the microscope on the bench will then also be in ocus or note keeping or drawing and can be viewed by the other open eye.
objective lens fo
fe I1 O D
Figure 1 Compound microscope.
The ray diagram ( see fgure 1 ) shows how the microscope in normal adj ustment orms its image. When learning about microscopes and telescopes, try to understand the role o the separate elements as this will make it easier or you to draw the ray diagrams in examinations. It is a good idea to devise a strategy or drawing the diagram as it is easy to end up with a fnal image that is too large or the paper or o the page. Figure 3 ( on page 61 1 ) shows one route to achieve a good sketch. Notice that the vertical size o the lenses does not necessarily reect their actual sizes in the instrument itsel. The eyepiece and obj ective apertures are usually very small in diameter, but are shown large on the diagram so that it appears that the rays actually go through them. In practice, with real microscopes the lenses are small to minimise the aberrations and so that the lenses can be ground very accurately. As usual, i the ray diagram is drawn to scale, then measurements can be taken o the obj ect and image scaled sizes to establish the overall magnifcation. L For the obj ective lens the magnifcation is equal to __ where L is the f o
length in the microscope tube between the obj ective ocal point and the eyepiece ocal point ( so that the total physical length o the microscope D tube is fo + L + fe ) . The angular magnifcation o the eyepiece is __ as f e
it is a magniying glass ( remember that D is the near point distance) . The angular magnifcation o the complete microscope is given by the product o the magnifcation o the two lenses acting separately and this is approximately DL _ fo fe
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Investigate! Make a microscope
Take two converging lenses, with ocal lengths perhaps 5 cm and 1 5 cm. D etermine their ocal lengths by using rays o light rom a distant obj ect and measuring the approximate ocal length between the lens and a piece o card.
Figure 2
Fix the short ocal length ( obj ective) lens securely to the end o a short ruler ( length 3 0 cm or so) using modelling clay or similar material. Set up the ruler so that it is a short distance rom an obj ect a well- illuminated piece o graph paper makes a good obj ect or this experiment.
Fix the other ( eyepiece) lens so that when you look into it, you see a magnifed virtual obj ect.
B y comparing the size o the grid on the obj ect with the size o the grid on the image, estimate the magnifcation o the microscope.
D oes your microscope approach the theoretical value o the magnifcation? The approximations assume that fo and fe are very small compared with D and L. Is this true in this case?
Resolution of a microscope All light that goes through an aperture such as a lens is diracted. Point obj ects become image disks as a result. I the image disks o two adj acent point obj ects overlap, it is difcult to tell them apart and the two images are not resolved. This is a signifcant problem in microscopy.
Nature of science Electron microscopes E lectron microscopes are used to image the smallest obj ects. The electrons are accelerated and as a result have extremely small wavelengths, much less that that o light. Glass lenses cannot be used to ocus the electrons but magnetic and electrostatic felds can, and these are used to bend the paths o the electrons to produce a ocusing eect. Look at some web sites that explain how the various types o electron microscope work. Try to understand how the felds replace the lenses o the compound optical microscope.
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Rayleigh suggested that two images were j ust resolved when the minimum o the diraction pattern o one image coincided with the central maximum o the pattern o the other. This means that when the images are j ust resolved, the angle subtended at the eye by rays rom each image is given by: sin = 1 .2 2 _ d where is the wavelength o the light and d is the diameter o the aperture (in this case circular) . For the microscope the aperture diameter is the eective diameter o the lenses. This is known as the Rayleigh criterion. You can fnd more discussion o the Rayleigh criterion in Topic 9. The criterion is oten modifed or use with microscopes and the eective aperture is replaced by N the numerical aperture o the lens. This is equal to n sin where n is the reractive index o the medium in which the lens is placed (n = 1 in the case o most microscopes that are in the air) and is the hal angle o the maximum cone o light that can enter the lens. The criterion becomes: sin = 1 .2 2 _ N ( You will not need to use this equation in the examination.)
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First, draw the principal axis. Add the lenses a scaled distance apart i you have been given dimensions. Then draw the intermediate image use an intermediate image height that will allow the fnal image to be on the page.
principal axis
Work backwards rom the intermediate image to orm the object to the let o the objective. Either choose the ocal lengths yoursel or use the scaled values i they have been provided. Dont draw rays to the right o the intermediate image yet.
objective lens eyepiece lens
object
fo
fe
fe
fo
Finally, draw the rays to the right o the intermediate image. These do not deviate between the lenses, but ater going through the eyepiece they must appear to have come rom the top o the virtual image. To complete the diagram add the arrows to all rays (not construction lines) and label all ocal points and both lenses.
virtual image
Figure 3 Strategy or drawing the compound microscope diagram. The equations suggest that the resolution o a microscope can be improved by:
using a short wavelength or the light ( sometimes microscopists use ultraviolet radiation or taking photographs o specimens to improve the resolution)
using as a wide an aperture or the obj ective as is consistent with reducing aberrations ( the design o the fnal eye piece lens is fxed by the average size o the eye)
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using a liquid of high refractive index between the specimen and the obj ective lens. The liquid obj ective allows a wider cone of light rays to be collected by the lens and this extra information improves the resolution of the image. Typically an oil of refractive index around 1 . 6 is used in special microscopes that have oil- immersion obj ectives.
Astronomical refracting telescopes Figure 4 An astronomical reector.
O bj ects in the night sky have always fascinated astronomers, and telescopes for viewing the sky were, like the microscope, developed early in the history of science. objective eyepiece
fo , fe
fo
fe
Begin by drawing the principal axis, the lenses and the common ocal points Add an intermediate image. Do not make it too small allow enough room or the construction later.
objective eyepiece
fo , fe
fo
fe
Draw construction lines to show you the fnal direction o the rays when they have been reracted by the eyepiece. These are construction lines so should be dashed.
objective eyepiece
fo , fe
fo
fe
The frst ray should be the ray that ends at the top o the intermediate image and goes through the optical centre. This ray is not deviated at the objective. Ater passing through the eyepiece the ray must travel parallel to the construction lines.
objective eyepiece
fo
fo , fe
o
image at
fe
e
Choose two more rays parallel to the frst ray beore they enter the objective. Ater passing through the objective they must go to the top o the intermediate image. Ater passing through the eyepiece they must go parallel to the frst ray (and the construction lines) . Finally, add urther construction lines to show the direction o the fnal image. Then label the diagram and add the ray arrow
Figure 5 Telescope ray diagram and drawing sequence.
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Also like the microscope, the astronomical reracting telescope uses two converging lenses. The obj ective lens has a long ocal length fo and an eyep iece lens that has a short ocal length fe . The telescope is designed primarily to ocus the light rom very distant obj ects and in its normal adj ustment is set up or this purpose only. Light rom infnity (that is, parallel rays) rom the object are ocused by the objective lens at its ocal point. The ocal points o the objective and eyepiece lenses are coincident (at the same place) inside the telescope and thereore the eyepiece treats the image ormed by the objective as though it is a real object placed at fe. We know rom earlier that this leads to parallel rays emerging rom the eyepiece and these then enter the eye o the observer. The angle subtended by the rays rom the top o the image and the principal axis is much greater than the angle subtended by the rays rom the top o the object. The image is magnifed. The ray directions show that the viewed image is also upsidedown, but this is not an issue or astronomers (maps o the Moon are oten inverted to take account o this) . The angular magnifcation o the telescope ollows directly by looking at the incident and emerging rays and the angles, and e, they make with the principal axis
Note In astronomy the term power is sometimes used in place o angular magnifcation. The maximum practical power or any given astronomical telescope is limited by the need to produce a high resolution image. This in itsel is determined by the atmospheric conditions.
o fo M= _= _ fe e
Investigate! (fe about 1 01 5 cm) . The combined total o fo + fe should not exceed the length o the ruler.
Making a telescope
Measure and record the ocal lengths o both lenses.
Fix the eyepiece lens at the end o the ruler with modelling clay as beore.
Holding or clamping the ruler horizontally, fx the objective lens a distance equal to fo + fe rom the eyepiece while looking through the eyepiece. You may need to move the objective orward and backwards along the ruler a short distance to get the optimum ocus or your own eye.
View a distant obj ect ( say a brick wall) and estimate how much larger one brick is when viewed through the telescope compared with using your naked eye.
D oes your telescope approach the theoretical angular magnifcation give above?
Figure 6 The basic method is similar to that o the microscope in that the instrument is constructed on a ruler or optical bench. This time a metre ruler is required together with a long ocal length objective (fo about 5 0 cm or longer) and a short ocal length eyepiece
Reracting telescopes are requently used by amateur astronomers and also by proessional astronomers or some applications. However, the largest reracting telescopes made have objective apertures up to 2 m and this represents the limit o the technology or grinding the glass lens. It is also difcult to support the lens given that it needs to be at the end o a long tube. Mirrors can be made much larger and thereore collect more light energy meaning that more distant objects can be viewed.
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Worked example 1
A small obj ect is placed 3 0 mm rom the obj ective lens o a compound microscope in normal adj ustment. A real intermediate image is ormed 1 5 0 mm rom the obj ective lens. The eyepiece lens has a ocal length o 75 mm and orms a virtual image at the near point. For the observer, the near point distance is 3 00 mm. C alculate the overall magnifcation o the telescope.
2
An astronomical telescope in normal adj ustment has an obj ective ocal length o 1 5 0 cm and an eyepiece ocal length o 5 .0 cm. C alculate: a) the angular magnifcation o the telescope b) the overall length o the telescope.
Solution f 1 50 a) M = __ = ___ = 3 0 ( angular magnifcation 5 f o
Solution 1 50 Magnifcation o the obj ective = ___ = 5 . 30 5 1 1 1 1 ____ __ ___ For the eyepiece __ = __ u + - 300 ; u = 300 ; 75 u = 60 mm. S o magnifcation o eyepiece 300 = 5 . = ___ 60
e
and linear are the same or the telescope in normal adj ustment) b) O verall length = fo + fe = 1 .5 5 m
O verall magnifcation = 5 5 = 2 5
Astronomical refecting telescopes The frst known reecting telescope was completed by Newton in 1 668. The basis o the telescope is straightorward. Rays rom the top and bottom o a distant obj ect are reected by a concave mirror and come to a ocus at the ocal point where an image o the obj ect orms. An observer positioned at the ocal point will see an image in the mirror. Rays rom the top and bottom tend to conuse the diagram. The convention in many reecting telescope ray diagrams is to show only the rays parallel to the principal axis and you may well see this in some books. B ecause the rays are parallel or at very small angles to the principal axis, either spherical mirrors or parabolic mirrors can be used in the reecting telescopes. For astronomical use, parabolic is chosen so that the spherical aberrations discussed in the earlier section are eliminated.
prime
prime focus primary mirror
Figure 7 Picture of astronomical telescope and basic ray diagram.
As with compound microscopes, the angular resolution o a reecting telescope is given by sin = 1 .2 2 _ d where d is the diameter o the aperture. So a large aperture or the mirror is an advantage as this increases the resolving power o the instrument. There are many advantages o reecting telescopes over their reracting equivalents:
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There is one surace which is usually ground to shape and then coated with a reective metallic surace. The rays o light do not have to pass through a number o layers o glass. The telescope can thereore in principle have no chromatic aberration.
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The mirror only has to be supported rom one side. This makes engineering large reecting telescopes more straightorward.
With only one surace to grind, the telescopes are cheaper or a comparable quality o image.
O nly one surace has to be made perect.
There are also some disadvantages however:
The mirror surace is vulnerable to damage and needs to be cleaned ( unless covered which introduces aberration) .
The optics can easily get out o alignment i the support or the mirror shits in some way.
In addition, this simple design means that the observers head or a camera blocks some o the light travelling along the tube thereby reducing some o the benefts o using a large mirror. For this reason, various arrangements are used to transer the image rom inside to outside the main tube o the telescope. There are a number o dierent design variants or achieving this, some o which use lenses and other devices to produce a high- quality image. In this course we will examine only two simple systems: the newtonian telescope and the cassegrain telescope. B oth are named or their inventors.
Newtonian mounting
newtonian mounting
cassegrain mounting
Figure 8 Newtonian and cassegrain mirror systems. In this system a small secondary plane mirror diverts the rays rom inside to outside the tube. Plane mirrors o high quality can be produced that introduce little or no distortion to the image. This is one o the cheapest designs o telescope and is very popular with amateur astronomers, some o whom grind their own mirrors and build their telescopes rom scratch.
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Cassegrain mounting Replacing the plane mirror with a hyperbolic secondary mirror enables the rays to be sent through a hole in the primary mirror surace. This means the observer can be in line with the telescope and can look in the observing direction. A principal advantage o this mounting is that the ocal length o the telescope can be eectively longer than the tube itsel. However, the secondary mirror inside the tube cuts o some o the incoming light, and the principal mirror has lost some o its reecting surace. In general, all telescope types have specifc observational advantages and disadvantages.
Radio telescopes Many o the principles o the reecting telescopes are carried over into the single- dish radio telescopes. These instruments are intended to collect electromagnetic signals in the radio region that originate rom astronomical obj ects. rays from distant object
F
Figure 9 Radio telescopes. A parabolic dish has the property that all rays parallel to the principal axis are brought to a ocus at the same point. I a radio antenna (aerial) is placed at this point then it will collect all the ocused radio waves that arrive simultaneously. The larger the area o the dish, the greater the power collected, so in principle the designer should aim or as large an area as possible. Another advantage o a large dish diameter is that, as with the optical reecting telescopes, the resolution o the telescope improves ( in the Rayleigh criterion equation decreases) as the diameter becomes larger. Resolution is a particularly important actor as the wavelengths used by the telescopes are those o the radio segment o the electromagnetic spectrum. These wavelengths are o the order o centimetres to tens o metres and are much larger than those o the visible light used with optical instruments ( order o 1 0 7 m) . S o a telescope working at radio wavelengths needs to have an aperture at least 1 00 000 times greater than its optical counterpart to have the same resolution. Figure 10 The Arecibo Observatory, Puerto Rico.
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This raises engineering problems or a dish that has to be steered to point at particular obj ects in the sky. There is the difculty o moving the dish and problems associated with the dish deorming rom the ideal parabolic shape under its own weight. S ome radio telescopes get over this problem by building the dish into a cavity in the ground.
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The disadvantage is that the dish can no longer be steered. Arecibo O bservatory in Puerto Rico is a good example o this ( fgure 1 0) . S ome exibility can be engineered into the system by moving the antenna, which in this telescope is suspended at the ocus and can be moved on a suspended railway track that can be seen in the photograph. The Arecibo telescope is 3 00 m across but a larger telescope o 5 00 m diameter, also in a natural crater, is being constructed in Pingtang C ounty, Guizhou Province, south-west C hina. The C hinese telescope is designed so that the panels that make up the dish are moveable to allow the principal axis o the telescope to be steered within limits. B y contrast, the largest dish telescope in the world at present is the Green B ank 1 00 m diameter telescope in West Virginia USA ( fgure 1 1 ) .
Interferometer telescopes To overcome the inherent design problems o a dish and the inability to steer a telescope built in a crater, a recent trend is to construct intererometer telescopes. from source
Figure 11 The Green Bank Telescope, West Virginia, USA.
baseline
beam combiner
Figure 12 Interferometer telescope. S uch telescopes come in various ormats with two or more radio telescopes combined, but the basic idea behind all o the arrangements is that signals rom a source are combined in the individual components o the telescope to produce a total signal. The signals can be combined in such a way that the whole telescope is equivalent to a single dish with an eective diameter equal to a baseline B. The baseline is the dimension across the individual dishes that make up the telescope. The resolution o such an array is sin = _. B O ne ormat that is requently used is a series o small steerable dishes. These are individually o low cost and do not have the engineering problems associated with large dishes. The signals are combined using computers to give a fnal signal.
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I M AG I N G Another approach is to use a very long baseline. Signals rom steerable dishes situated in dierent countries can be combined to give a network, the baseline o which can approach the size o the planet. For example the E uropean Very- long- baseline intererometry Network ( EVN) is a combination o 1 2 telescopes and can, itsel, be combined with intererometers in the UK and the US A to make a global network that achieves a very high resolution. There are plans to build a square kilometre array ( S KA) that will combine telescopes in Australia, S outh Arica and other countries in the southern hemisphere with a baseline o the order o 3 000 km. The telescope is due to be completed by 2 02 4. O ther intererometer ormats are possible and these at present include linear arrays o two long antennae or arrangements o antennae built in the shape o a plus sign.
Nature of science Discoveries in radio astromony Not everyone has to be a proessional. S ome branches o science are known or the number o discoveries made by amateurs. Astronomy is one o these. The frst radio telescope was constructed in 1 931 by a radio engineer named Karl Jansky who worked in the B ell Telephone Laboratories in the US. He studied sources o static noise in radio signals and discovered that some o the static originated outside the Earth. His telescope was a mobile collection o radio antennae (aerials) . The frst true dish radio telescope was built by a North American amateur radio enthusiast named Grote Reber (call sign W9GFZ) who was inspired by Janskys
work. He spent considerable time (rom 1 93338) constructing a series o parabolic dish reectors that were eventually sensitive enough to reproduce Janskys results. Reber and Jansky were the frst true pioneers o radio astronomy. Today, amateur astronomers have excellent telescopes coupled to digital cameras. This makes it possible or them to contribute to science in a signifcant way. Proessional astronomers cannot view all areas o the sky night ater night. For example, amateur astronomers discover deep sky supernovae in distant galaxies and alert proessional observers who then make observations using larger telescopes.
Earth and satellite-borne telescopes Recently, high-quality telescopes have been placed in satellite orbit above the Earth or on spacecrat that are aimed away rom the Earth into the S olar S ystem. O bservational astronomy carried out on the surace is subj ect to a number o limitations, as described overlea:
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When we look at the night sky with the naked eye, the stars appear to twinkle. This is because our observations are made through tens o kilometres o air. The atmosphere introduces brightness variations and position errors into the images o the stars through local short- term changes in the air density caused by heating and convection eects. A common way to reduce this problem is to build optical telescopes on mountain tops in places where the atmosphere is relatively stable.
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Some recent telescopes use up-to- date techniques to remove light variation using adaptive optics. When viewing a distant dim star the optics make reerence also to light rom a nearby bright star that is assumed to have a constant intensity and position. Parts o the telescope mirror suraces are then moved rapidly to correct or the distortion o the light, which is assumed to be the same or both reerence and observed stars. However, observations that are made rom a satellite platorm above the atmosphere do not need these corrections.
TOK What do we see?
Stars emit radiation right across the electromagnetic spectrum, and all this inormation is o value to astronomers. Many wavelengths rom the stars are absorbed by the atmosphere ( fgure 1 3 ) and cannot be detected at the Earths surace. The only way to measure these is through the use o a telescope or sensor mounted on a satellite. X- rays are a particularly important region o the spectrum to astronomers and orbiting satellites are almost the only way to study emissions at these wavelengths rom stars ( fgure 1 3 ) .
ultraviolet X-ray blocked by atmosphere
visible infra-red blocked window by atmosphere
radio window long-wavelength radio blocked
10 cm
There is an increasing problem o electromagnetic pollution rom artifcial sources o radiation associated with cities on Earth. Optical telescopes need to be constructed in increasingly remote areas, and radio telescopes also need to be placed in places where the radio spectrum is relatively uncluttered.
1 m
Any optical or radio instrument extends our senses beyond their design limits. This can be in terms of wavelengths to which we are not normally sensitive, or in terms of magnifying beyond what we would normally see. To what extent do the images we see or interpret represent true reality?
opaque
1 km
100 m
10 m
1m
1 cm
1 mm
100 m
10 m
100 nm
10 nm
1 nm
0.1 nm
transparent
wavelength/m
Figure 13 Absorption of the electromagnetic spectrum by the atmosphere.
An Earth- based telescope, whether radio or optical, can either be fxed or steerable. A fxed telescope can only view the parts o the sky that move in ront o it. A steerable instrument can view ( within the limits o the local horizon) the whole o the hemisphere on which it is centred. Similarly, some satellite telescopes are fxed so that they observe certain sections o the sky. O ther sky- survey satellites allow the entire sky to be mapped over the period o the satellites lietime.
The expense o placing an observing satellite into orbit and the costs o building surace- bound telescopes is so great that international collaboration is common in astronomy with research groups rom around the world booking observation time on the instruments.
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C.3 Fibre optics Understandings
Applications and skills
Structure o optic fbres
Solving problems involving total internal reection
Step-index fbres and graded-index fbres Total internal reection and critical angle Waveguide and material dispersion in
optic fbres Attenuation and the decibel (dB) scale
Nature o science Modern communications rely heavily on the use o fbre optics. A relatively simple piece o science has led through applied science to a transormation o all types o communications across the globe.
and critical angle in the context o fbre optics Describing how waveguide and material dispersion can lead to attenuation and how this can be accounted or Solving problems involving attenuation Describing the advantages o fbre optics over twisted pair and coaxial cables
Equations 1 critical angle equation n = _ sin c I _ attenuation (dB) = 10 log I0
Structure and use o optic fbres The concepts o total internal reection and critical angle were introduced in Sub- topic 4. 4. These ideas are used in the modern technology o fbre optics ( Figure 1 ) which began to be developed or communication purposes in the early 1 970s. However, the idea o sending light along a light pipe was frst demonstrated by C olladon and B abinet as early as 1 82 0. Nowadays, under-sea fbres link nations with international collaboration and agreement on common standards or the transmission o the inormation. cladding core
cladding
Figure 1 Total internal reection in a fbre.
Communications and the advantages o optic fbres The principal methods using physical links or communication ( as opposed to the radiation o electromagnetic waves as in radio) are:
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Twisted p air. In this technique developed by Alexander B ell, the wires connecting one operator with another were twisted together ( hence the name) . External electrical signals can induce an em in both wires and this em is to some extent cancelled out by the
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twisting. However, the noise cancellation is not necessarily complete and i two pairs o twisted- pair cables are next to each other then signals rom one pair can be induced in the other cable pair giving cross-talk and a lack o security. Twisted- pair cables are generally used or low- requency applications.
C oaxial cable. This cable is constructed with a central core conductor insulated rom a metallic shield that is earthed at zero potential relative to the changes in the signal voltage ( but which carries the return current) . The cable is completed with a tough protective outer cover. S uch cable is generally used or radiorequency signals. You may well have seen an example o it carrying the ( very weak) signals rom your tv antenna or satellite dish to the tv itsel. This type o cable rej ects outside electrical noise as the earth shield acts as the surace o a conducting shell. Little noise rom outside can distort the weak signal travelling along the cable and equally the signal itsel cannot radiate signifcantly beyond the earthed conductor. However, the cable is bulky and expensive. O p tic fbres. In this technique the signal to be transmitted down the fbre modulates an electromagnetic wave ( usually at visible or inra- red wavelengths) . This modulated wave is shone along a very thin glass fbre. The thickness o the fbre is so small that the light strikes the walls at angles ( much) greater than the critical angle and so the light propagates along the fbre. E lectrical noise does not aect the passage o the light through the fbre. There are wavelength windows in optic fbre glasses at which the loss o signal with distance ( the attenuation o the glass) is very low. The fbre is surrounded by a cladding material with a reractive index smaller than that o the core in order to ensure that the critical angle condition is maintained.
twisted pair
plastic jacket metallic shield dielectric insulator centre core co-axial cable
optic fbre
cladding core cladding
Figure 2 Physical communication links.
D igital signals are transmitted along the optic fbres. S uch signals consist o a sequence o changes between two states; these might be on/o, variations between two fxed requencies, or abrupt changes in the phase o the signal. For our discussions we will assume a simple model where the light is either on or o, so that a series o light pulses is transmitted down the cable. The advantage o optic fbres over conductors are:
Immunity to electromagnetic intererence unlike most electrical cables.
Low attenuation loss over long distances in glass compared to metal conductors.
B roader bandwidth ( total range o usable requencies) in the glass compared to wires. This means that one fbre can carry millions o telephone conversations.
Glass is an insulator and does not suer rom the inherent problems o a electrical- cable telephone system where electrical connections to earth ( shorts) are a problem.
Very small diameter so that compact bundles o fbres can be constructed thus increasing the capacity o the system even more or the same physical space as an electrical cable.
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Greater security to signal detection rom outside the fbre ( fbre tapping) . S ome optical fbres have been designed with a dual core and are said to be completely secure to outside tapping.
The main types o fbre are:
S tep -index fbre in which there is an abrupt change in the reractive index at the interace between core and cladding.
Graded-index fbre in which there is a gradual reduction in the reractive index rom the centre to the outside o the core.
S ingle-mode fbre, which has a diameter o only a ew times the wavelength o the light used in it. At such small diameters, the rules o geometrical optics that we use or our discussion o optic fbres break down and wave theories are required. We are not going to consider this type o fbre urther in this course.
input power
We will look at the behaviour o step- and graded- index fbres in more detail later.
Attenuation and dispersion time
O ptic fbres have very low loss. I sea water had the same optical properties as glass used in fbres, then the details at the bottom o the 1 1 km deep Mariana Trench, o the coast o Japan, would be clearly visible rom the ocean surace. Nevertheless, eventually the signal weakens ( attenuates) so that it needs to be amplifed beore it can continue its j ourney. The device that carries out this re- amplifcation is known as a rep eater.
output power
At the point where the signal enters the cable, the pulse will have an abrupt on/o change in intensity. time
Figure 3 Attenuation with time in an optical fbre. The time axes are to the same scale.
Note This defnition uses the intensity (or power) o the signal. The intensity I o the signal is related to its amplitude A by I A 2 . So
The repeater needs to do two things: re- shape the pulse into its original square ormat and also boost the amplitude o the signal. We will discuss the reasons or the loss o amplitude and the dispersion later, but or the moment we concentrate on how attenuation is measured.
A attenuation = 10 log10 _____
The signal amplitude can all by many orders o magnitude beore it needs to be boosted. To deal easily with large ratio changes a logarithmic scale called the B el scale is used. In this scale, the attenuation in bels ( B ) is defned to be the logarithm to base 1 0 o the ratio o the intensity ( or power) o a signal to a reerence level o the signal.
This can be re-written as
So
attenuation = 10 log10 ____II 0
translates to 2
A 20
A attenuation = 20 log10 ____ A 0
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Figure 3 shows how the power o the pulse varies with time beore and ater passing along the cable: the power is reduced ( by a very large actor) so that the total energy in the pulse ( the area under the power time curve) is reduced also. This is attenuation; energy has been lost to the cable. Separate disp ersion eects change the shape and cause the pulse to spread out as it travels along the fbre. I two pulses that were initially separate overlap through dispersion, then the receiving system cannot disentangle them rom each other. D ispersion imposes an upper limit on the rate at which a particular fbre can transmit inormation.
I attenuation in bel = log 1 0 _ I0
C. 3 FI B RE O PTI CS where I is the attenuated ( output) power level o the signal and I0 is the input intensity level ( the subscript is a zero not a letter o) . This means I that 1 0 B is equal to a power ratio __ o 1 0 1 0 , which is a large ratio. As I power ratios tend to be smaller than this, it is more usual to quote power ratios in decibels, where I attenuation in decibel ( dB ) = 1 0 log 1 0 _ I0
( ) 0
A change in power o 1 0 dB is equivalent to a power ratio o 1 0 times. A change o 3 dB is equivalent to a change by a actor o 2 in the power.
Link
Attenuation per unit length dB/100 m Coaxial cable 15300 or requencies up to about 1 GHz depending on signal requency and cable design Twisted pair 550 or requencies up to about 300 MHz Optic fbre 0. 02 at 10 1 4 Hz
The table shows some typical values or attenuations in optic fbres, twisted cable, and coaxial cable.
Worked example
Solution
A signal o power 7.5 mW is input to a fbre that has an attenuation loss o 3 .0 dB km 1 . The signal needs to be amplifed when the power has allen to 1 .5 1 0 1 8 W. C alculate the distance required between amplifers in the system.
7 .5 1 0 Power loss = 1 0 log 1 0 ( ________ ) = 1 57 dB 1 .5 1 0 -3
-18
The fbre loses 3 dB every kilometre so another 1 57 amplifer will be required in ___ = 52 km. 3
The requencies most used or fbre optics are in the range 0. 81 .5 m. Attenuations in these ranges arise rom two principal causes:
Absorp tion This loss arises rom the chemical composition o the glass and rom any impurities that remain in it ater manuacture. Glass absorbs some inra- red wavelengths strongly and these wavelengths have to be avoided or transmission.
S cattering Rayleigh scattering ( named ater the B ritish physicist Lord Rayleigh, who developed the ideas o image resolution) is the main scattering loss. It is caused by small variations in reractive index o the glass introduced during manuacture. Rayleigh scattering accounts or about 95 % o the attenuation.
inrared absorption total absorption loss/dB km -1
Grace Hopper and her nanoseconds The US computer scientist Grace Hopper was amous as one o the frst people to construct a compiling language or computers. She was also amous or her lectures on computer science. To illustrate the problem o communication at high speed she would hand out lengths o wire 30 cm long. This is the distance travelled by light in a vacuum in one nanosecond. She used this to link the speeds and sizes o computers in the minds o her audience. To what extent does the fnite speed o light put limitations on our ability to carry out computations aster and aster?
rayleigh scattering
1.5
TOK
1.0
0.5
0 1.0
1.2
1.4 wavelength/m
Figure 4 Attenuation mechanisms in optic fbres.
1.6
1.8
Figure 5
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I M AG I N G Light in the core interacts with the molecules o the glass. I the light continues in the general orward direction there is no attenuation, but i the light is scattered in directions other than rom which it came, then attenuation occurs. This depends very strongly on the size o the small variations in density, etc. in the glass and also on the wavelength o the light. The loss due to Rayleigh scattering is proportional to the 4 and is greatest at short wavelengths. The graph ( fgure 4) shows how the two eects o absorption and scattering contribute to the overall attenuation o a typical fbre or wavelengths greater than 1 um.
Nature of science Why is the sky blue? Rayleigh scattering is the reason why the sky is blue. B ecause the blue and violet light rom the S un have shorter wavelengths than the other colours, they are preerentially scattered out o the direct beam by the gas molecules in the air compared to longer wavelengths. We see the S un as having a yellow disk with a blue sky. At
cladding acceptance cone
Two other eects contribute to distortions o the light as it passes through the fbre:
core
sunset, the S uns rays pass through a thicker atmosphere and the presence o particles in the air scatters longer wavelengths too, so the S un and the clouds now appear red. In act, this is a simplistic explanation o a complex phenomenon. I you want to know more, investigate both Rayleigh and Mie scattering.
cladding
Material disp ersion This is similar to the chromatic aberration we met earlier in S ubtopic C .1 . B oth occur because the reractive index o the material used or the lens or fbre depends on wavelength. Reractive index is w ave sp e e d in vacu u n equal to the ratio o: _________________________ . w ave sp e e d in tran sm itting m e diu m
Figure 6 Dispersion.
The reractive index o glass decreases as the wavelength increases and so the wave speed in the glass also increases with increasing wavelength ( red light travels aster than violet) . Using light with a wide range o colours ( a large bandwidth) means that the long wavelength light will reach the end o the fbre ahead o the shorter wavelengths leading to a spreading o the pulse. The answer is to restrict the wavelengths but this, at the same time, means that ewer wavelengths can be used and so there may be ewer channels available or communication.
Waveguide (or modal) disp ersion Even i monochromatic light is used, a large diameter step-index optic fbre will still be aected by waveguide dispersion. Rays o light that propagate along the fbre can travel by dierent routes depending on their initial angle o incidence at the end o the fbre. C ompare a ray that travels along the central axis o the fbre with one that is at a large angle to the axis and reects many times.
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C. 3 FI B RE O PTI CS
Although the scale o this drawing ( fgure 7) is unrealistic ( because, to scale, it has ar too large a diameter or any real optic fbre) , you should see that the large- angle ray takes a much longer path than the central ray to travel rom one end o the fbre to the other. The large-angle ray will arrive later and the time dierence will appear as an increase in the pulse width. fbre cross-section
fbre n
distance
fbre n
distance
Figure 7 Graded-index fbres. To make a correction or this eect, graded- index fbres are used. As described earlier the reractive index o the core is not constant. It has a high index in the centre o the core and a decreasing index towards the corecladding interace. The speed o the light is slowest in the centre with the speed increasing towards the cylinder wall. This means that large-angle rays travel more quickly than central axis rays during the periods when they are close to the wall. C ombining this with a smaller overall diameter o core reduces ( but does not eliminate) the waveguide dispersion eect.
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C.4 Imaging the body (AHL) Understanding Detection and recording o X-ray images in
medical contexts Generation and detection o ultrasound in medical contexts Medical imaging techniques (magnetic resonance imaging) involving nuclear magnetic resonance (NMR)
Nature of science Decisions made by a physician can involve an assessment o risk. The doctor will use some imaging techniques in the knowledge that the ionizing radiations involved can harm the patient. The real question is whether the techniques lead to a possible overall beneft.
Applications and skills Explaining eatures o X-ray imaging, including
attenuation coecient, hal-value thickness, linear/mass absorption coecients, and techniques or improvements o sharpness and contrast Solving X-ray attenuation problems Solving problems involving ultrasound acoustic impedance, speed o ultrasound through tissue and air, and relative intensity levels Explaining eatures o medical ultrasound techniques, including choice o requency, use o gel, and the dierence between A and B scans Explaining the use o gradient felds in NMR Explaining the origin o the relaxation o proton spin and consequent emission o signal in NMR Discussing the advantages and disadvantages o ultrasound and NMR scanning methods, including a simple assessment o risk in these medical procedures
Equations
I1 I0
Attenuation (dB) : L I = 10 log _ Intensity: I = I 0 e - x
Linear absorption: x 1 /2 = ln2 Acoustic impedence: Z = c
Introduction Medicine and physics combine in the world o medical diagnosis and treatment. D octors have come to rely on technology rom developments in physics. There has been a revolution in the methods that enable a doctor to visualize the interior o a patients body.
X-ray images in medicine Rontgen discovered X-rays in 1 895 and, within one month o the publication o his original scientifc paper, the radiation had been used or medical purposes. Nowadays, X- radiation is used both diagnostically and therapeutically. Here we concentrate on the use o X- rays to generate an image o the interior o the body that will inorm doctors in making their diagnoses.
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C. 4 I M AG I N G TH E B O D Y ( AH L) The basic principles behind the imaging are that X-rays are shone through part or all o the body. Materials such as bone absorb some o the X-rays or scatter them out o the beam, so that a reduced intensity emerges rom the patient. Other, soter materials such as muscle and tissue do not absorb or scatter the rays so well. The X-rays are then incident on a photographic plate or a sensor. The plate is developed or a computer computes the data rom the sensor, in both cases an image on flm or monitor is produced. Where the intensity o the X-rays is high (not absorbed by the tissue) , the plate is exposed (darkened) or the sensor detects the arrival o many photons. When signifcant absorption has occurred, ewer photons leave the body and the plate is not exposed to the same degree. As X-rays image are traditionally viewed as negatives, this means that the photograph is darkened where the rays are not absorbed and remains transparent (white) where the radiation has been absorbed by the body.
Figure 1 X-ray images.
The X- rays themselves are produced when electrons travelling at high speed are decelerated in a heavy- metal target such as tungsten. The electrons are frst accelerated in a vacuum through tens o thousands o volts in an electric feld and then strike the target. O n colliding with the target, electrons are slowed down rapidly and as a result lose energy to internal energy o the target ( 99% o the incident energy) and as energy in the orm o X- ray photons ( 1 % ) . Given the large amount o internal energy, the X- ray tubes need to be cooled and the anode target is oten rotated to avoid hot spots developing on it. D etails o the generation o the X- rays are not required or the examination. X- rays are attenuated as they pass through material, and this means that some o the photons are removed. O thers change direction ( so that they can no longer be considered part o the beam) . The principal mechanisms or removing photons or changing their direction are:
C oherent scattering A process similar to the scattering mechanisms described in the previous sub- topic on fbre optics. It is the predominant mechanism or low- energy X-ray photons up to energies o about 3 0 keV.
Photoelectric effect This is identical to the eect described in Topic 1 2 . The incoming electron has enough energy to remove an inner-shell electron rom an atom. When other electrons rom higher energy states lose energy to occupy the inner shell, a photon o light is emitted. This mechanism is most important in the energy range 3 01 00 keV. The photoelectric scattering is particularly important as it provides contrast on the image between tissue and bone.
C omp ton scattering In this mechanism, which occurs at energies generally greater than those used in diagnostic X-rays, the high-energy X-ray photon ejects an outer-shell electron rom an atom and, as a result, a photon o lower energy moves o in a dierent direction to the original photon. This scattering mode is o principal importance in therapeutic X-ray medicine where X-rays o high energy are involved.
Pair p roduction As explained in Topic 1 2, at high energies in excess o 1 MeV, a photon can interact with a nucleus to produce an electronpositron pair.
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I M AG I N G Attenuation also occurs when the intensity o the X- ray beam decreases with distance rom the X- ray tube as the beam diverges. For the photons in a monochromatic beam o X- rays, the chance o an individual photon being scattered or absorbed is related to the probability o this photon interacting with an atom in the material being X- rayed. Suppose a photon has a 1 0% chance o removal by a particular thickness o material ( fgure 2 ) .
81 000 66 000 53 000 43 000 90 000 73 000 60 000 48 000
100 000 photons
1
2
3
4
5
6
7
8
4 5 thickness
6
7
8
100 000
80 000
60 000 hal thickness 40 000
20 000
0 0
1
2
3
Figure 2 Absorption efects lead to a hal thickness. Then i 1 00 000 photons are incident on material o this thickness, 90 000 ( 90% o 1 00 000) will remain in the beam when it leaves the material. I the material were twice as thick, then 81 000 ( 90% o 90 000) will remain ater this urther thickness. This is an identical argument to that used in radioactive decay ( except that there the argument related the time taken or the ensemble o atoms to decay to the probability o decay or one atom) . However, this similarity means that, like radioactivity, the intensity o the X-rays obeys an exponential relationship and we can state that: I = I0 e - x l
where I is the intensity o the attenuated beam, I0 is the intensity o the incident beam, x is the thickness o the material and l is the linear absorp tion coefcient measured in units o metre 1 . Although the linear absorption coefcient ollows directly rom the way the absorption probability is defned, it is not the most convenient absorption coefcient to use as it depends on the density o the absorbing material. As an example, compare water vapour ( steam) with ice. Ice will scatter more eectively than steam because it has a greater density and thereore the X- ray photons are more likely to encounter
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C. 4 I M AG I N G TH E B O D Y ( AH L)
The relationship between the linear absorption coefcient l and the mass absorption coefcient m is l m = _ where is the density o the material. The units o mass absorption coefcient are m 2 kg 1 . As a consequence, when the mass absorption coefcient is used, the appropriate intensity equation is I = I0 e -
intensity
lead
distance
soft x-rays
lead intensity
a water molecule every centimetre o travel in ice than in steam. It is more convenient to quote a single mass absorp tion coefcient which is density independent and depends only on the element or compound that is absorbing the X- rays.
x
m
distance
soft x-rays
This equation is only strictly correct when the beam is monochromatic because values o the absorption coefcient vary with the energy o the X-ray photons. Highly penetrating radiation has a short wavelength (about 0.01 nm) and the absorption coefcients are small at these wavelengths; such radiation said to be hard. Long wavelength photons (about 1 nm) are termed sot X-rays and the absorption coefcients are much larger.
Figure 3
In the same way that radioactive materials have a defned hal- lie, so or X-rays the thickness o material required to halve the intensity is called the hal thickness x _1 . 2
Again, ollowing radioactive decay: ln 2 ln 2 _ x 1 /2 = _ l or x 1 /2 = m The value o l may be determined rom the gradient o a graph o ln I against x. Frequently the beam will pass through a number o layers o dierent materials all with dierent thicknesses and absorption coefcients. This is easy to treat mathematically i the layers have parallel plane interaces. S uppose that layer 1 has a thickness x' and a linear absorption coefcient l' and that layer 2 has a thickness x'' and a linear absorption coefcient l''. Then the intensity when the beam leaves layer 1 , I1 is I1 = I0 e - 'x' l
and the intensity when the beam leaves layer 2 , I2 is I2 = I1 e - ''x'' l
Thereore x
I2 = I0 e - ' x' e - '' x'' l
x!
l
layer 1 layer 2
or I2 = I0 e - ( ' x' + '' x'') l
l
For each layer add the product o the linear absorption coefcient and the layer thickness and then use the exponential unction to fnd the fnal intensity o the beam. When the mass absorption coefcients are known the equation becomes: I2 = I0 e m - (' '
m
x' + '' m'' x'')
I2 = I0 e -(1 x + !1 x!)
I1 = I0 e -1 x
IO
l
l !
Figure 4
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Worked examples 1
A sample o a metal o density 4000 kg m 3 has a hal thickness o 1 2 mm with a particular wavelength o radiation. C alculate, or the metal, the:
2
a) linear absorption coefcient
Solution
C alculate the thickness o muscle tissue that will reduce the intensity o a certain X-radiation by a actor o 1 0 3 , assuming that the linear absorption coefcient or muscle is 0.035 cm 1 .
b) mass absorption coefcient.
I1 = I0 e - 'x' l
Solution
S o 0.001 = e - 0.035 x
ln 2 -1 a) l = _ x1 /2 = 58 m
ln 0.001 = - 0.03 5 x 6.9 x = _ = 1 97 cm 0.03 5 About 2 m o muscle tissue is required to reduce the intensity by 1 000 .
l 58 _ -2 2 -1 b) m = _ = 4000 = 1 .4 1 0 m kg
aluminium flter
collimator
Improving the image Collimation Figure 5 shows a common arrangement used to take diagnostic X- ray images. A number o eatures are used to enhance both the sharpness and contrast o the image ormed on the photographic plate:
The X- ray beam is fltered by passing it through a thin plate o aluminium. This selectively removes low- energy photons because the absorption coefcient is much greater at low energies. These low- energy photons would simply be absorbed by the patient ( adding to the radiation dose) and would serve no useul imaging purpose. The plate also reduces the intensity o the beam somewhat, but overall the penetrating power o the beam increases and the X- rays become harder.
The beam as it leaves the tube is very divergent. It is collimated by passing through a series o channels in lead plates. O - axis rays are absorbed by the lead and do not reach the patient. This produces a narrower, more parallel beam. This is benefcial because photons scattered rom the o- axis angles tend to blur the photographic image.
As or the X- rays that reach the patient: some do not interact, some are absorbed and disappear rom the system, and others are scattered. Again, the scattering leads to blurring on the image. A grid system o lead plates below the patient, arranged parallel to the beam, is used to absorb the scattered rays but allows the on- axis rays to reach the imaging system.
The X-rays fnally arrive at the flm cassette and this contains two uorescent screens, one each side o the flm. As the X- rays interact with these screens they cause the materials in the screens to uoresce and emit light that improves the blackening o the photographic negative. Many o the X- rays would not contribute to the image without this arrangement.
lead plate uorescent material
flm flm cassette
Figure 5 Contrast techniques in X-ray imaging.
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Contrast S ome tissues in the body are hard to distinguish on the photographs without enhancements to the contrast. Heavy elements with large absorption coefcients can be introduced into the body to help with this. B arium and bismuth are examples o elements used. Patients with stomach disorders may be asked to drink barium sulate ( B aS O 4 ) , which coats the lining o the stomach and makes its outline very clear on the image. Iodine can be introduced intravenously to produce clear images o the cardiovascular system. O ther ways to enhance detection include using charge- coupled sensors in place o flm together with computer systems to orm electronic images viewed on monitors. The advantage o X- ray imaging is that it is a quick and inexpensive technique, costing ar less than an MRI or C T scan. The X- ray machines can be highly portable, meaning that patients may not have to travel to a central location or a scan. However, unlike ultrasound and MRI techniques, X- rays involve ionizing radiation and this presents a risk to both the patient and the radiographer. Techniques to reduce the X- ray intensity and exposure time are constantly being improved to reduce the risks. Nevertheless this exposure needs to be kept in perspective. The average chest or dental X- ray provides a much smaller dose o radiation to the body than a commercial intercontinental ight at 1 2 km above the Earth.
Computed tomography C omputer ( computerized) tomography ( C T scan) , also known as computed axial tomography ( C AT scan) was introduced during the 1 970s. It gives a much greater range o grey scales to the image and provides an axial scan an image o a slice through the patient. X- rays are incident on the area o the patient under investigation and the scattered and direct photons are detected by a series o small detectors placed in an arc around the patient. Ater the frst exposure the X- ray tube and the detectors move around the patient taking exposure ater exposure until an entire range covering 3 60 around the body has been made ( fgure 6) . The detection method uses scintillation counters that respond to the presence o a photon by emitting a ash o light, which is then detected by a photomultiplier that produces an enhanced electric current or each ash that occurs.
X-ray tube beam of X-rays
A computer builds up the inormation rom each detector and each exposure to produce a complete image o the slice o the patient up to a ew tens o millimetres thick. I urther slices are required, the patient can be moved under computer control to a new position and the process is repeated. The sensitivity o the system is very high, and many eatures can be seen and interpreted by doctors. A disadvantage o the C T scan, however, is that the cumulative dose to the patient is high, and there is a greater risk o damage to the patient than with a normal X- ray. However, exposures rom the scanners are dropping all the time as improved detection techniques are developed.
scattered X-rays
series of rotating X-ray detectors
Figure 6 CT scanning.
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Ultrasound in medicine Generation Ultrasound is a sound wave generated at a requency above that at which humans hear. The lower requency limit o ultrasound is taken to be 2 0 kHz. The range o requencies used in medicine is rom 2 to 2 0 MHz. Like electromagnetic radiation, ultrasound waves can be absorbed by matter, and reected and reracted when they meet a boundary between two media. The usual rules or reection and reraction o light at an interace also apply to ultrasound. B rothers Pierre and Paul- Jacques C urie frst observed the piezoelectric eect that is used to generate the ultrasound. They ound that when a quartz crystal is deormed, it produces a small em between opposite aces o the crystal. C onversely, applying a potential dierence across the crystal causes it to deorm and, under the right circumstances, to vibrate at high requencies. This property o piezoelectricity was ound to be shared by other materials besides quartz including some synthetic ceramic materials. These newer materials are now used in preerence to quartz or producing ultrasound. To make the scan a piezoelectric transducer ( which can both emit and receive the ultrasound) is placed in contact with the skin. A gel is used between the transducer and skin to prevent a large loss o energy that would occur at the transducerair and airskin interaces. A single pulse o the ultrasound is transmitted into the tissues and the system then waits or the reected wave ( an echo) to return rom each interace inside the patient. There are a number o ways to display the scan but the simplest display ( known as an A- scan) is to show a graph o reected signal strength against time. Figure 7( a) shows a typical A-scan with the various echo returns labelled. A knowledge o the speed o the ultrasound and the time or return enables the size and location o an organ to be determined ( remembering that, like all echoes, the ultrasound has to travel to the interace and return) . refections at boundary
probe
transducer movement organ 3 signal strength b) B scan
4 2 3 X
time
5 scan Y 4
time
2
scan X
Figure 7 A and B scans.
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1
1 time a) A scan
6
gel
bone
signal strength
abdomen wall
5 6 Y
patient
signal strength
C. 4 I M AG I N G TH E B O D Y ( AH L) A more complex type o scan is known as the B - scan ( fgure 7( b) ) . This requires a computer to combine a series o scans produced by the transducer. The operator rocks the transducer backwards and orwards to illuminate the internal suraces and the computer builds up a slice image ( in a similar way to the C T scan) through the patient by combining the signals rom a whole series o A-scans. The transmitted ultrasound that passes through the patient is not used in ultrasound diagnostic techniques, so it is o maj or importance to minimize the absorption o the ultrasound by tissues and to maximize the energy in the returned echo. To compare the ultrasound perormance o dierent types o material and tissue we use the acoustic impedance o the medium. This is a number that indicates how easy it is to transmit ultrasound through a particular material. The acoustic impedance Z o a material is ound to depend on the speed o sound c in the material and the density o the material: Z = c The S I unit or acoustic impedance is kg m 2 s 1 ; this is sometimes abbreviated to the non- SI unit the rayl ( named or Lord Rayleigh) but we will use the SI unit here. The table gives some typical values or speeds, densities and Z values or various tissues, though it should be remembered that c and thereore Z depend on the ultrasound requency being used.
Medium
Bone Soft tissue Liver Muscle Water Air at 15C
Speed / m s 1
Density / kg m 3
4100 1500 1550 1600 1480 340
1900 1050 1070 1080 1000 1.21
Acoustic impedance / 10 6 kg m 2 s 1 7.8 1.6 1.7 1.7 1.5 4.1 10 4
The proportion o the incident wave energy that is reected at an interace depends on dierences between the acoustic impedances o the two media. It can be shown that the ratio o the initial intensity I0 to the reected intensity Ir is: Ir ( Z2 - Z1 ) 2 _ = _2 I0 ( Z2 + Z1 ) where Z1 and Z2 are the acoustic impedances o the frst material and the second material respectively. The resolution o the image is o great importance in ultrasound imaging. The obvious approach might be to aim or the highest requency possible as this will give the shortest wavelength and the best resolution. However, the attenuation o the wave also increases markedly with requency as does the resolution itsel ( typically about 2 mm) i the requency is taken beyond a certain maximum. Generally the doctor has to accept a compromise, but requencies in the range 2 5 MHz are used or most applications.
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I M AG I N G Not all medical ultrasound use leads to an image in the conventional sense. There are many other diagnostic and therapeutic uses or ultrasound. These include the detection o blood ow and measurement o its speed using Doppler shits, and recent innovations where microbubbles o gas are introduced to enhance the image o blood vessels (in a similar way to the injection o contrast enhancers in X-radiography) . The advantages o ultrasound include:
It is a non- invasive technique.
It is relatively quick and inexpensive.
There are no known harmul eects.
It is o particular value in imaging sot tissues.
However:
Image resolution can be limited.
Ultrasound does not transmit through bone.
Ultrasound cannot image the lungs and the digestive system as these contain gas which strongly reects at the interace with tissue.
Nuclear magnetic resonance (NMR) in medicine Magnetic resonance imaging ( MRI scans) are medical diagnostic tools that use the phenomenon o nuclear magnetic resonance ( NMR) . In use, a patient is placed in a strong uniorm magnetic feld produced by an electromagnet. Other magnetic felds around the patient are varied and signals emitted rom the tissues are detected, measured and transormed into an image using a computer. The images produced have good resolution and good contrast or parts o the body that contain large proportions o water (and hence hydrogen nuclei) . The resolution or NMR techniques depends on the resonance requency o the signal, so the resolution is proportional to the magnetic feld strength. At the time o writing, a typical resolution or NMR imaging is 2 mm. MRI is a technique preerred or diagnosis o brain and central nervous system disorders. An explanation o MRI comes in two parts: the basic NMR eect itsel and then a description o the way it is modifed or medical use. top precession S
S N
N
S
high energy state
S conventional current loop high energy state
B
N
proton precession
N
Figure 8 Larmor precession.
The basic NMR efect
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Protons possess the properties o charge and spin and this leads them to behave as small magnets. The situation is analogous to charges moving around a circular loop o wire producing a magnetic feld that acts through the centre o the loop.
C. 4 I M AG I N G TH E B O D Y ( AH L)
Under normal conditions, hydrogen- rich materials such as water have the proton spins arranged randomly to give no overall magnetic feld. However, when a strong magnetic feld is imposed on the material the spins o the protons, and thus their magnetic felds, line up with the imposed magnetic feld.
Normally the protons line up in the lowest energy state, but i a radio- requency ( r) feld o an appropriate requency is applied to the system, some o the protons will ip into the opposite high energy state. This is analogous to a bar magnet ipping into a state where the north-seeking pole o the bar magnet is next to the northseeking pole o the feld source ( see fgure 8) .
The appropriate requency to cause the ip is known as the Larmor requency and, crucially or MRI, is directly proportional to the strength o the magnetic feld in which the material is placed. The Larmor requency in Hz is 4. 2 6 1 0 7 B where B is the magnetic feld strength in T.
What happens to the spinning protons when the r is applied can be thought o in terms o a spinning childs top. A top precesses around its spin axis in response to the Earths gravitational feld. The protons can be thought o as similarly precessing at the Larmor requency.
As the protons precess, the changing direction o the magnetic feld that they produce induces an em in a coil o wire nearby.
When the r feld is switched o, the protons in the high- energy state relax ( return) to the low- energy state.
MRI modifcations The process is essentially the same as the basic phenomenon. The patient is placed in a strong uniorm magnetic feld. However an additional gradient feld is added to the strong feld.
i Larmor requency is 64.8 MHz then signal must be coming rom here
Larmor requency = 64.5 MHz 1.54
65.6
1.52
64.8
1.50
63.9
Larmor re uenc /MHz
uniorm feld + gradient feld/T
patient
distance
Figure 9 A specifc Larmor requency comes rom a unique slice in the patient.
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TOK Henry Thoreau said It's not what you look at that matters, it's what you see. To what extent do you agree with this comment in the context of imaging inside the body?
The gradient feld is designed so that the total feld ( uniorm + gradient) varies linearly across the patient. The Larmor requency itsel depends linearly on B and thereore the Larmor requency varies predictably with position in the body.
The r feld is switched on and proton precession occurs. The relaxation o the protons ater the r feld has been removed leads to an electromagnetic signal that contains inormation about the number o protons emitting at each Larmor requency. This inormation is recovered rom the signal allowing a computer to plot the inormation spatially on an image.
MRi produces excellent images or diagnostic purpose without exposing the patient to radiation ( whether rom X- rays or radio- isotopes) . The energy o the photons in MRI are well below the 1 eV levels that correspond to molecular bonds and also are below the energies o X- ray photons. However, MRi is not entirely without risk. Factors involved in the risk include:
Strong magnetic felds. S ome patients are unsuitable or an MRI scan i they have, or example, a knee or hip j oint replacement that would distort the magnetic feld or a heart pacemaker that could be severely aected by currents induced in it when the strong magnetic felds change. There is no known risk rom a strong magnetic feld by itsel.
Radio-requency (r) felds. These can lead to local heating in the tissues o the patient.
Noise. The large changes in magnetic feld strength within the scanner cause parts o it to attempt to change shape during the scan. This can give rise to high intensities o sound that patients can fnd disturbing.
Claustrophobia. A strong uniorm feld is difcult to produce and can only be maintained over a small volume o space. This means that, typically, the patient is scanned while in a small- diameter tunnel, which some people fnd uncomortable.
Additionally, there may be elements o discomort or some patients in that a scan can take up to 90 minutes to complete; lying still in a confned space may prove difcult or some. In particular, young children and babies need to be sedated as they cannot understand the need to remain still.
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QUESTION S
Questions 1
Four rays o light rom O are incident on a thin concave ( diverging) lens. The focal points o the lens are labelled F. The lens is represented by the straight line XY.
( i)
O n a copy o the diagram, draw rays to locate the position o the image o the obj ect ormed by the lens.
( ii) Explain whether the image is real or virtual.
X
b) A convex lens o ocal length 62 .5 cm is used to view an insect o length 8.0 mm that is crawling on a table. The lens is held 5 0 mm above the table.
F
( i)
F
C alculate the distance o the image rom the lens.
( ii) C alculate the length o the image o the insect. ( 8 marks) Y
a) D efne the term focal point.
3
b) O n a copy o the diagram: ( i)
complete the our rays to locate the position o the image ormed by the lens
(IB) a) A parallel beam o light is incident on a convex lens o ocal length 1 8 cm. The light is ocused at point X as shown below.
( ii) show where the eye must be placed in order to view the image. c) S tate and explain whether the image is real or virtual.
X
d) The ocal length o the lens is 5 0. 0 cm. C alculate the linear magnifcation o an obj ect placed 75 . 0 cm rom the lens. e) Hal o the lens is now covered such that only rays on one side o the principal axis are incident on the lens. D escribe the eects, i any, that this will have on the linear magnifcation and the appearance o the image. (1 4 marks) 2
P
S tate the value o the distance PX. b) A diverging lens o ocal length 2 4 cm is now placed 1 2 cm rom the convex lens as shown below.
(IB) a) The diagram shows a small obj ect O represented by an arrow placed in ront o a converging lens. The ocal points o the lens are labelled F.
X P
12 cm F
O
F
c) ( i) ( ii)
Explain why point X acts as a virtual obj ect or the diverging lens. C alculate the position o the image as produced by the diverging lens.
L
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C
I M AG I N G d) A lens combination, such as a diverging and a convex lens, is reerred to as a telephoto lens. Suggest why a telephoto lens is considered to have a longer ocal length than that o a single convex lens. ( 7 marks) 4
( ii) Using the completed ray diagram above, derive an expression in terms o O and E or the angular magnifcation o an astronomical telescope. Assume that the fnal image is at infnity. c) When speciying an astronomical telescope, the diameter o the obj ective lens is requently quoted. S uggest a reason or quoting the diameter. ( 8 marks)
(IB) The diagram below shows the image o a square grid produced by a lens that does not cause spherical aberration. 6
(IB) A compound microscope consists o two convex lenses o ocal lengths 1 . 2 0 cm ( lens A) and 1 1 . 0 cm ( lens B ) . The lenses are separated by a distance o 2 3 .0 cm as shown below. ( The diagram is not drawn to scale.) O
1.30 cm
a) O n a copy o the diagram, draw the shape o the image when produced by a lens that causes spherical aberration.
lens A f = 1.20 cm
b) D escribe one way in which spherical aberration can be reduced. ( 4 marks)
5
23.0 cm
lens B f = 11.0 cm
An obj ect O is placed 1 .3 0 cm rom lens A. An image o O is ormed 1 5 .6 cm rom A.
(IB)
a) This image orms an obj ect or lens B . C alculate the obj ect distance or lens B .
The diagram below shows two lenses arranged so as to orm an astronomical telescope. The two lenses are represented as straight lines.
b) C alculate the distance rom lens B o the image as produced by lens B .
objective lens
c) C alculate the magnifcation o the microscope. ( 5 marks)
eye lens
7
(IB) a) S tate one cause o attenuation and one cause o dispersion in an optical fbre.
focal length fO
focal length fE
The ocal lengths o the obj ective lens and o the eye lens are O and E respectively. Light rom a distant obj ect is shown ocused in the ocal plane o the obj ective lens. The fnal image is to be ormed at infnity. a) C omplete the ray diagram to show the ormation o the fnal image. b) ( i)
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S tate what is meant by angular magnifcation.
b) An optical fbre o length 5 .4 km has an attenuation per unit length o 2 .8 dB km 1 . The signal power input is 80 mW. ( i)
C alculate the output power o the signal.
( ii) In order or the power o the output signal to be equal to the input power, an amplifer is installed at the end o the fbre. State the gain, in decibels ( dB ) , o the amplifer at the end o the fbre.
QUESTION S c) The signal to noise ratio ( S NR) , in dB , is defned as Psignal
SNR = 1 0 log ____ where Psignal and Pnoise are P noise
the powers o the signal and noise respectively. The S NR o the signal in ( b) beore amplifcation was 2 0 dB . C alculate the S NR ater amplifcation. ( 7 marks) (IB)
d) Annotate your sketch- graph to show the hal-value thickness x 1 /2 . e)
The variation with time t o the output power rom the optic fbre is shown in diagram 2 .
( i)
output power
1 0 a)
State the name o one o the mechanisms responsible or the attenuation o diagnostic X-rays in matter. ( 6 marks)
S tate and explain one situation, in each case, where the ollowing diagnostic techniques would be used.
0
0
t diagram 1
0
0
( i)
attenuation o the signal
( ii)
signal noise.
t diagram 2
b) The duration ( time width) o the signal increases as it travels along the optic fbre.
( ii)
( ii) Ultrasound
b) Apart rom health hazards, explain why dierent means o diagnosis are needed. ( 6 marks)
a) State and explain the eature o the graphs that shows that there is:
( i)
S tate two reasons or this increased time duration. S uggest why this increase in the width o the pulse sets a limit on the requency o pulses that can be transmitted along an uninterrupted length o optic fbre. ( 7 marks)
1 1 B eam energies o about 3 0 keV are used or diagnostic X- rays. This results in good contrast on the radiogram because the most important attenuation mechanism is not simple scattering. a)
( i)
I IO
attenuation coefcient
( ii) hal-value thickness. c)
The attenuation coefcient at 3 0 keV varies with the atomic number Z as attenuation coefcient Z3
The data given below list average values o the atomic number Z or dierent biological materials.
Biological material fat muscle bone
State what is meant by X-ray quality.
A parallel beam o X-rays o intensity I0 is incident on a material o thickness x as shown below. The intensity o the emergent beam is I.
Outline the most important attenuation mechanism that is taking place at this energy.
b) E xplain what is meant by:
(IB) a)
X- rays
( iii) Nuclear magnetic resonance
The scales are the same on both diagrams.
9
Draw a sketch-graph to show how intensity I varies with distance x.
The variation with time t o the input power to an optic fbre is shown in diagram 1 .
input power
8
c)
( i)
Atomic number Z 5.9 7.4 13.9
C alculate the ratio: attenuation coefcient or bone ____ attenuation coefcient or muscle
( ii) Suggest why X-rays o 3 0 keV energy are useul or diagnosing a broken bone, but a dierent technique must be used or examining a at- muscle boundary.
x
( 1 3 marks) b) D efne hal-value thickness.
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I M AG I N G 1 2 (IB)
d
a) State and explain which imaging technique is normally used: ( i)
O
to detect a broken bone
ultrasound transmitter and receiver
( ii) to examine the growth o a etus. The graph below shows the variation o the intensity I o a parallel beam o X-rays ater it has been transmitted through a thickness x o lead.
b)
The pulse strength o reected pulses is plotted against time t where t is the time elapsed between the pulse being transmitted and the time that the pulse is received.
15 10 5 0
0
(i)
2
4
6 x/mm
8
10
12
Use the graph to estimate the hal-value thickness x 1 /2 or this beam in lead.
c)
b) The diagram below shows an ultrasound transmitter and receiver placed in contact with the skin.
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25 50 75 100 125 150 175 200 225 250 275 300 t/s
Identiy the origin o the reected pulses A, B and C and D .
The above scan is known as an A- scan. State one way in which a B -scan diers rom an A- scan.
d) State one advantage and one disadvantage o using ultrasound as opposed to using X-rays in medical diagnosis. ( 1 0 marks)
1 5 S tate and explain the use o: a)
S tate a typical value or the requency o ultrasound used in medical scanning.
C
( ii) The mean speed in tissue and muscle o the ultrasound used in this scan is 1 . 5 1 0 3 m s 1 . Using data rom the above graph, estimate the depth d o the organ beneath the skin and the length l o the organ O .
( 1 1 marks)
E xplain which photon energy would be most suitable or obtaining a sharp picture o a broken leg. ( 2 marks)
D
B
( i)
( iii) A second metal has a hal-value thickness x 1 /2 or this radiation o 8 mm. C alculate what thickness o this metal is required to reduce the intensity o the transmitted beam by 80% .
1 3 The attenuation o X-rays depends not only on the nature o the material through which they travel but also on the photon energy. For photons with energy o about 3 0 keV, the halfvalue thickness o muscle is about 5 0 mm and or photons o energy 5 keV, it is about 1 0 mm.
A
0
( ii) Determine the thickness o lead required to reduce the intensity transmitted to 20% o its initial value.
1 4 a)
l
The scan is to estimate the depth d o the organ labelled O and also to fnd its length, l.
pulse strength/relative units
I/arbitrary units
20
layer of skin and fat
a barium meal in X-ray diagnosis
b) a gel on the skin during ultrasound imaging c)
a non-uniorm magnetic feld superimposed on a much larger constant feld in diagnosis using nuclear magnetic resonance. ( 7 marks)
D A S T R O P H YS I C S Introduction Astrophysics probes some of the most fundamental questions that humanity has sought to answer since the dawn of civilization. It links the experimental discipline of astronomy with the theoretical understanding of everything in the universe cosmology. It offers insights into the universe and provides answers, albeit tentatively, to its size, age, and content. Astrophysicists
can theorize on the life cycles of stars and gain an appreciation of how the universe looked at the dawn of time. The weakest of the four fundamental forces, gravity, comes into its own on an astronomic scale. It provides the mechanism to attach planets to stars, stars to other stars (in galaxies) , and galaxies to other galaxies (in clusters and super clusters) .
D.1 Stellar quantities Understandings Objects in the universe The nature o stars Astronomical distances Stellar parallax and its limitations Luminosity and apparent brightness
Nature of science When we look upwards, away rom the Earth, on a dark clear night we see many points o light in the sky. Some o these, such as the planets, artifcial satellites, and aircrat, are visible because they are reecting light rom the Sun. Others, such as stars and galaxies, are visible because o the light that they emit. The light rom distant stars and galaxies has travelled truly astronomical distances to reach us and we are able to construct a historical account o space rom analysing this. However, the light rom the dierent sources will have travelled varying distances to reach us and have been emitted at a range o times. This is analogous to looking at a amily photograph containing a mixture o many generations o a amily at dierent stages in their lives. It is a credit to humanity that we are able to conjecture so much about space when we can only justiy our reasoning with circumstantial evidence. In this way astrophysics mirrors orensic science we can never obtain evidence at the actual time an event happens.
HEAD A_UND
Applications and skills O B J TE XT_UND Identiying objects in the universe Qualitatively describing the equilibrium
between pressure and gravitation in stars Using the astronomical unit (AU) , light year (ly) and parsec (pc) Describing the method to determine distance to stars through stellar parallax Solving problems involving luminosity, apparent brightness, and distance
Equations
1 parsec defnition: d (parsec) = ___ p (arc-second)
luminosity equation: L = AT
4
L apparent brightness equation: b = __ 2 4 d
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Introduction In this frst sub-topic we rapidly move outwards rom our solar system to the remainder o our galaxy and beyond. In doing this we briey examine some o the obj ects that make up the universe. We then consider the range o units that we use or astronomical measurements. We end by considering how we can estimate the distance and luminosity o relatively near stars by treating them as black-body radiators and using their apparent brightness.
Objects that make up the universe The solar system comets
Pluto asteroids
Mercury Earth
Venus Mars Sun
Saturn
Neptune
Jupiter Uranus
(not drawn to scale)
Figure 1 The solar system. The solar system is a collection o planets, moons, asteroids, comets, and other rocky obj ects travelling in elliptical orbits around the S un under the inuence o its gravity. The S un is a star ormed, we believe, rom a giant cloud o molecular hydrogen gas that gravitated together, orming clumps o matter that collapsed and heated up. A gas disc around the young, spinning S un evolved into the planets. It is thought that the planets were ormed about 4.6 1 0 9 years ago ( see fgure 1 ) .
Figure 2 Thermal emission from the young star Fomalhaut and the debris disc surrounding it.
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The high temperature close to the S un permitted only those compounds with high condensation temperatures to remain solid, gradually accreting ( sticking together) particles to orm the our terrestrial planets: Mercury, Venus, Earth, and Mars. Further away rom the S un the Gas Giants or j ovian planets comprising o Jupiter, S aturn, Uranus, and Neptune were ormed rom cores o rock and metal and an abundance o ice. The huge quantity o ice meant that these planets became very large and produced strong gravitational felds that captured the slow moving hydrogen and helium. Pluto used to be called the ninth planet but, in 2 006, it was downgraded to a dwar planet. The planets move in elliptical orbits round the Sun with only Mercury occupying a plane signifcantly dierent to that o the other planets. Further out rom the S un, beyond Neptune, is the Kuiper belt. This is similar to the asteroid belt but much larger; it is the source o short- period comets and contains dwar planets ( including Pluto) . The Kuiper belt is set to be the next rontier o exploration in our solar system.
D .1 S TE LL AR Q U AN TI TI E S
S ix o the planets have moons orbiting them: Mars has two moons, while Jupiter has at least 5 0 acknowledged moons with several provisional ones that might be asteroids captured by its gravitational feld. About 4.5 billion years ago, the Earths moon is believed to have been ormed rom material ej ected when a collision occurred between a Mars- size obj ect and the Earth. Asteroids are rocky obj ects orbiting the S un with millions o them contained in solar orbit in the asteroid belt situated between Mars and Jupiter. Those o size less than 3 00 km have irregular shape because their gravity is too weak to compress them into spheres. S ome o the asteroids, such as C eres with a diameter o about 1 0 6 m, are large enough to be considered as minor planets. C omets are irregular obj ects a ew kilometres across comprising rozen gases, rock, and dust. O bservable comets travel around the S un in sharply elliptical orbits with periods ranging rom a ew years to thousands o years. As they draw near to the S un the gases in the comet are vaporized, orming the distinctive comet tail that can be millions o kilometres long and always points away rom the S un.
Figure 3 Comet Hale-Bopp.
Stars gas and radiation pressure Like the S un, all stars initially orm when gravity causes the gas in a gravity nebula to condense. As the atoms move towards one another, they lose gravitational potential energy that is converted into kinetic energy. This raises the temperature o the atoms which then orm a protostar. When the mass o the protostar is large enough, the temperature and pressure at the centre will be sufcient or hydrogen to use into helium, with the release o very large amounts o energy the star has ignited. Ignition produces emission o radiation rom the core, producing a radiation pressure that opposes the inward gravitational orces. When this is balanced the star is in a state o hydrostatic equilibrium and will remain stable or up to billions o years because it is on the main sequence. As the hydrogen is used up the star will eventually undergo changes that will move it rom the main sequence. D uring these changes the colour o the star alters as its surace temperature rises or alls and it will change size accordingly. The original mass o material in the star determines Figure 4 Hydrostatic equilibrium. how the star will change during its lietime.
Groups of stars D espite the difculties in assessing whether stars exist singly or in groups o two or more, it is thought that around fty per cent o the stars nearest to the S un are part o a star system comprising two or more stars. B inary stars consist o two stars that rotate about a common centre o mass. They are important in astrophysics because their interactions allow us to measure properties that we have no other way o investigating. For example, careul measurement o the motion o the stars in a binary system allows their masses to be estimated. A stellar cluster is a group o stars that are positioned closely enough to be held together by gravity. S ome clusters contain only a ew dozen stars while others may contain millions. All o the stars in a star cluster were ormed at the same time rom the same nebula. The Pleiades ( fgure 5 ) is a stellar cluster o about 5 00 stars that can be seen with the naked eye;
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A S T R O P H YS I C S this is an example o an op en cluster. O pen clusters consist o up to several hundred stars that are younger than ten billion years and may still contain some gas and dust. They are located within our galaxy, the Milky Way, and so lie within a single plane. Globular clusters contain many more stars and are older than eleven billion years and, thereore, contain very little gas and dust. There are 1 5 0 known globular clusters lying j ust outside the Milky Way in its galactic halo. Globular clusters are essentially spherically shaped.
Figure 5 The Pleiades.
A constellation, unlike a stellar cluster, is a pattern ormed by stars that are in the same general direction when viewed rom the E arth. They are more signifcant historically than physically as many ancient societies attributed them with religious importance. Today, the patterns made by the constellations are helpul to astronomers in locating areas o the sky or telescopic study. Naturally, some o the stars in a constellation are much closer to the E arth than others. B ecause o proper motion they will appear very dierent in, say, ten thousand years time. S uch stars are not held together by gravity.
Nebulae
Figure 6 The Crab Nebula.
Regions o intergalactic cloud o dust and gas are called nebulae ( singular is nebula) . As all stars are born out o nebulae, these regions are known as stellar nurseries. There are two dierent origins o nebulae. The frst origin o nebulae occurred in the matter era around 3 8 0 000 years ater the B ig B ang. D ust and gas clouds were ormed when nuclei captured electrons electrostatically and produced the hydrogen atoms that gravitated together. The second origin o nebulae is rom the matter which has been ej ected rom a supernova explosion. The C rab Nebula, shown in fgure 6 , is a remnant o such a supernova. O ther nebulae can orm in the fnal, red giant, stage o a low mass star such as the S un.
Galaxies
Figure 7 The Andromeda galaxy with two smaller satellite galaxies.
A galaxy is a creation o stars, gas, and dust held together by gravity and containing billions o stars. The Milky Way contains about 3 1 0 1 1 stars and, probably, at least this number o planets. S ome galaxies exist in isolation but the maj ority o them occur in groups known as clusters that have anything rom a ew dozen to a ew thousand members. The Milky Way is part o a cluster o about 3 0 galaxies called the Local Group which includes Andromeda ( fgure 7) and Triangulum. Regular clusters consist o a concentrated core and are spherical in shape. Irregular clusters also exist, with no apparent shape and a lower concentration o galaxies within them. S ince the launch o the Hubble Space Telescope it has been observed that even larger structures, called superclusters, orm a network o sheets and flaments; approximately 90% o galaxies can be ound within these. In between the clusters there are voids that are apparently empty o galaxies. The Milky Way and Andromeda are members o the most common class o galaxies spiral galaxies. These are characterized by having a discshape with spiral arms spreading out rom a central galactic bulge that contains the greatest density o stars. It is increasingly speculated that, at the centre o the galactic bulge, there is a black hole. The spiral arms
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D .1 S TE LL AR Q U AN TI TI E S
contain many young blue stars and a great deal o dust and gas. O ther galaxies are elliptical in shape, being ovoid or spherical these contain much less gas and dust than spiral galaxies; they are thought to have been ormed rom collisions between spiral galaxies. Irregular galaxies are shapeless and may have been stretched by the presence o other massive galaxies the Milky Way appears to be having this eect on some nearby dwar galaxies.
Astronomical distances Resulting rom the huge distances involved in astronomical measurements, some unique, non- S I units have been developed. This avoids using large powers o ten and allows astrophysicists to gain a eel or relative sizes and distances. The light year (ly) : The speed o light is one o the most undamental constants in physics; all inertial observers measure light as travelling at the same speed ( see Option A or a background to this) . This property can be used to defne a set o units based on the speed o light. For example, the distance travelled by light in one minute is called a light minute. As it takes light approximately 8 minutes to travel rom the Sun to the Earth, the distance between them is 8 light minutes. The light year (ly) is a more commonly used unit and is the distance travelled by light in one year. 15
1 ly = 9.46 1 0 m The astronomical unit (AU): This is the average distance between the Sun and the Earth. It is really only useul when dealing with the distances o planets rom the Sun. 1 AU = 1 .5 0 1 0 1 1 m 8 light minutes The parsec (pc): This is the most commonly used unit o distance in astrophysics. 1 pc = 3.26 ly = 3.09 1 0 1 6 m D istances between nearby stars are measured in pc, while distances between distant stars within a galaxy will be in kiloparsecs (kpc) , and those between galaxies in megaparsecs (Mpc) or gigaparsecs (Gpc) . The distances used in astronomy are truly enormous, and this has meant that a variety o indirect methods have been developed or their measurement. The method used to measure the distance o an astronomical object rom the Earth is dependent on its proximity.
Stellar parallax O n Earth, surveyors measure distances by using the method o triangulation. A known or measured length is taken as a baseline. The angles that a distant obj ect makes at either end o the baseline are then measured ( using a theodolite) . From these angles the distance o the obj ect can be calculated. When measuring the distance o a star rom the Earth ( in parsec) a similar technique is employed. Parallax is based on the act that nearby obj ects will appear to cross distant obj ects when viewed rom dierent positions. This can be seen rom inside a moving car when ence posts by the roadside appear to speed by but distant hills hardly move. As the Earth orbits the Sun, the stars that are quite close to us appear to move across the distant fxed stars as shown in fgure 8.
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fxed stars in background
July image
parallax angle p
star or which distance is being measured
position o Earth sun January image in July baseline = diameter o Earths orbit position o = 2AU Earth in January
Figure 8 Stellar parallax. When the image o a nearby star is recorded in both January and July the star appears to have moved across the fxed stars in the background. Using the equation 1 d= _ p gives the distance d in parsecs when p is the parallax angle in arcsecond. This simple relationship is used or defning the parsec: when a star is at a distance o 1 pc rom the Earth the parallax angle given by the equation will be one arcsecond. In a circle there are 3 60 degrees. In every degree there are 60 arcminutes and 60 arcseconds in every arcminute. Thus 1 arcsecond is very small 1 being j ust ____ o a degree. 3600 There is a limit to the distance that can be measured using stellar parallax parallax angles o less than 0.01 arcsecond are difcult to measure rom the surace o the Earth because o the absorption and scattering o light by the atmosphere. Turbulence in the atmosphere also limits the resolution because it causes stars to twinkle. Using the 1 parallax equation, gives a maximum range o d = ____ = 1 00 pc. 0.01
Figure 9 The Gaia mission.
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In 1 989, the satellite Hipparcos ( an acronym or High Precision Parallax C ollecting S atellite) was launched by the European Space Agency ( ES A) . B eing outside the atmosphere, Hipparcos was able to measure the parallaxes o 1 1 8 000 stars with an accuracy o 0.001 arcsecsond ( to distances 1 000 pc) ; its mission was completed in 1 993. Gaia ( fgure 9) , Hipparcoss successor, was launched in 2 01 3 and is charged with the task o producing an accurate three- dimensional map showing the positions o about a billion stars in the Milky Way. This is about one per cent o the total number o stars in the galaxy! Gaia is able to resolve a parallax angle o 1 0 microarcsecond measuring stars at a distance o 1 00 000 pc. Amateur astronomers taking images will be working at approximately 1 arcsecond per pixel.
D .1 S TE LL AR Q U AN TI TI E S
Luminosity and apparent brightness of stars In Sub-topic 4.3 we looked at the intensity o a wave at a point distance r rom the source. Intensity was defned as the power emitted by a source divided by the area o the sphere over which the energy is spread equally. P I = _2 4 r For stars P is called the luminosity ( L) o the star and represents the total energy emitted by the star per second in watt. I in this context is the apparent brightness ( b) and is measured in watts per square metre ( W m 2 ) ; the distance rom the star is usually denoted by d. This version o the equation is written as: L b = _2 4 d The luminosity o a star is a very important quantity in establishing the nature o a star. Hence, the equation is particularly useul because, although we are not in a position to measure the luminosity o the star, we can measure its apparent brightness (or example, by using a telescope and a charge-coupled device) . I we also calculate the distance o the star we can then work out its luminosity. Alternatively, when we know the luminosity o a star (because it is similar to other stars o known luminosity) we are able to use the measurement o its apparent brightness to estimate the stars distance.
Worked example a) For a star, state the meaning o the ollowing terms: ( i) Luminosity ( ii) Apparent brightness. b) The spectrum and temperature o a certain star are used to determine its luminosity to be approximately 6. 0 1 0 31 W. The apparent brightness o the star is 1 . 9 1 0 - 9 W m 2 . These data can be used to determine the distance o the star rom Earth. C alculate the distance o the star rom Earth in parsec.
Solution a) ( i) The luminosity is the total power emitted by the star. ( ii) The apparent brightness is the incident power per unit area received at the surace o the Earth. L b) Using b = _2 we rearrange to get 4 d ___
d=
L = ___ 4 b
__________
6.0 1 0 = 5 .0 ____________ 4 1 . 9 1 0 31
-9
1 0 19 m
As 1 pc = 3 .2 6 ly and 1 ly = 9. 46 1 0 1 5 m then 1 pc = 3 .2 6 9.46 1 0 1 5 m
c) D istances to some stars can be measured by using the method o stellar parallax.
= 3 .08 1 0 1 6 m. 5 .0 1 0 or 1 62 3 pc 5 .0 1 0 1 9 m is ________ 3 .08 1 0 1 600 pc 19
( i) O utline this method. (ii) Modern techniques enable the measurement rom Earths surace o stellar parallax angles as small as 5 .0 1 0 3 arcsecond. Calculate the maximum distance that can be measured using the method o stellar parallax.
16
c)
( i) The angular position o the star against the background o fxed stars is measured at six month intervals. The distance d is then 1 ound using the relationship d = __ p ( this is shown in fgure 6.) 1 ( ii) d = _______ = 2 00 pc. 5 1 0 -3
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Black-body radiation and stars
Worked example Some data or the variable star B etelgeuse are given below. Average apparent brightness = 1 .6 1 0 7 Wm 2 Radius = 790 solar radii EarthB etelgeuse separation = 1 3 8 pc The luminosity o the S un is 3 .8 1 0 26 W and it has a surace temperature o 5 800 K. a) C alculate the distance between the Earth and B etelgeuse in metres. b) D etermine, in terms o the luminosity o the Sun, the luminosity o B etelgeuse. c) C alculate the surace temperature o B etelgeuse.
In Sub-topic 8.2 we considered black bodies as theoretical objects that absorb all the radiation that is incident upon them. B ecause there is no reection or re-emission they appear completely black as their name suggests. Such bodies would also behave as perect emitters o radiation, emitting the maximum amount o radiation possible at their temperature. All objects at temperatures above absolute zero emit black-body radiation. This type o radiation consists o every wavelength possible but containing dierent amounts o energy at each wavelength or a particular temperature. Although stars are not perect black-bodies they are capable o emitting and absorbing all wavelengths o electromagnetic radiation. Figure 1 0 shows the black-body radiation curves or the Sun, the very hot star Spica, and the cold star Antares. B ecause each o the stars will produce dierent intensities, the curves have been normalized by dividing the intensity emitted at a given wavelength by the maximum intensity that the star yields this means the vertical scale has no unit and the maximum value it can take is 1 .00. The maximum intensity o radiation emitted by the Sun has a wavelength o just over 5 00 nm making it appear yellow; the peak intensity or Spica is in the ultraviolet region, but there is sufcient intensity in the blue region or it to appear blue; the peak or Antares is in the near inra-red but, with plenty o red light emitted, it appears to be red to the naked eye.
a) As 1 pc = 3 .1 1 0 1 6 m, 1 3 8 pc = 1 3 8 3 .1 1 0 1 6 = 4.3 1 0 1 8 m 2 L b) b = ____ L = 4 d b 2
4 d
= 4[4.3 1 0 ] 1 .6 1 0 18 2
= 3 .7 1 0
31
-7
3 .7 1 0 31 ________ = 9.7 1 0 4. 26
So Betelgeuse has a luminosity o 9.7 1 0 4 L Sun c) As L = 4 R T , by taking ratios we get 2
4
2
4
L Sun 4 R S u n T S u n _______ = _______________ 2 4 L 4 R B e te lge u s e T B e te lge u s e
_________
Sun
=
4
4
2
L B e te lge u se R S u n _________ L S u n R B e te lge u se 2
_______
9.8 1 04 _______ = 0. 63 7902
TBetelgeuse = 0.63 5 800 K = 3 700 K
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1.00 0.75
the Sun (5800 K)
0.50
Antares (3400 K)
0.25 0 0
500
1000 wavelength/nm
1500
2000
Figure 10 Black-body radiation curves for three stars.
3.8 1 0
T B e te lge u s e ______ = T
Spica (23,000 K)
W
D ividing by the luminosity o the S un gives
B e te lg u e u s e
normalized intensity
Solution
For a star the S teanB oltzmann law is written as L = AT4 where L is the luminosity in watt, A the surace area o the star in square metres, and T the temperature in kelvin. is the S teanB oltzmann constant = 5 .67 1 0 8 Wm 2 K 4. When we assume that a star is spherical we can use this equation in the orm: L = 4 R 2 T4 where R is the radius o the star. We can see that the luminosity o a star depends on its temperature and its size ( measured here by its surace area) . In the next sub- topic we will see how the balance between temperature and surace star size is used to categorize star type.
D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N
D.2 Stellar characteristics and stellar evolution Understandings Stellar spectra HertzsprungRussell (HR) diagram Massluminosity relation or main sequence stars
Applications and skills Explaining how surace temperature may be
Cepheid variables Stellar evolution on HR diagrams Red giants, white dwars, neutron stars, and
black holes Chandrasekhar and OppenheimerVolko limits
Nature of science In 1859, the physicist Gustav Kirchho and chemist Robert Bunsen worked at the University o Heidelberg in Germany. Having developed the frst spectroscope, they repeated Foucaults experiment o passing sunlight through a bright sodium ame to fnd the absorption lines seen in the solar spectrum. They then repeated their experiment with other alkali metals and were able to show that the solar absorption spectra were the reverse o emission spectra. Kirchho went on to provide strong evidence or the presence o iron, magnesium, sodium, nickel, and chromium in the atmosphere o the Sun. Their experiments paved the way or our understanding o many o the properties o stars.
obtained rom a stars spectrum Explaining how the chemical composition o a star may be determined rom the stars spectrum Sketching and interpreting HR diagrams Identiying the main regions o the HR diagram and describing the main properties o stars in these regions Applying the mass luminosity relation Describing the reason or the variation o Cepheid variables Determining distance using data on Cepheid variables Sketching and interpreting evolutionary paths o stars on an HR diagram Describing the evolution o stars o the main sequence Describing the role o mass in stellar evolution
Equations Wiens law: m a x T = 2.9 10
-3
mK luminosity-mass relationship: L M 3 .5
Introduction The magnitude o the distance between stars is unimaginable with the nearest star (ignoring the Sun) being so distant that, even with the astest rocket ever built, it would take almost a hundred thousand years or us to reach it. The curiosity o the human race knows no bounds and we have striven to fnd out all we can about stars their lie cycles and their distances rom the Earth. In this sub-topic we investigate how it is possible to estimate the distance o stars more than 1 00 pc rom the Earth, and explore the lives o stars o dierent mass, radius, temperature, and colour.
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Stellar spectra In Sub- topic 7.1 we considered emission and absorption spectra occurring as electrons make transitions between energy levels. S uch spectra provide important inormation about the chemical composition, density, surace temperature, rotational and translational velocities o stars. When we observe the spectrum o stars using a spectrometer we fnd that nearly all stars show a continuous spectrum which is crossed by dark absorption lines; some stars also show bright emission lines. S eeing absorption lines across a continuous stellar spectrum tells us that the stars have a hot dense region ( to produce the continuous spectrum) surrounded by cooler, low- density gas ( to produce the absorption lines) . In general, the density and temperature o a star decreases with distance rom its centre. B ecause its temperature is so high, a stars core has to be composed o high- pressure gases and not o molten rock, unlike the cores o some planets.
Composition of stars Absorption o certain wavelengths is apparent when we observe the intensitywavelength relationship or stars. The smooth theoretical black- body curve is modifed by absorption dips as can be seen in fgure 1 . The graph shows the variation with wavelength o the intensity or the Sun and Vega ( which is much hotter than the S un having a surace temperature o 9600 K) . In the case o Vega, the cooler hydrogen in the stars outer layers ( the photosphere) absorbs the photons emitted by hydrogen. The clear pattern between the wavelengths absorbed and those o the visible part o the hydrogen absorption spectrum shown in fgure 2 strongly suggests that Vegas photosphere is almost entirely made up o hydrogen.
intensity/arbitrary units
1.5
Vega
1
Sun
0.5
0
-0.5 400
450
500
550
600 650 wavelength/nm
Figure 1 Intensitywavelength relation for Vega and the Sun.
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700
750
800
D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N The absorption lines correspond to those o hydrogen shown in the spectrum o fgure 2 . hydrogen absorption spectrum
400 nm
700 nm
Figure 2 Atomic hydrogen absorption spectrum. It also implies that the transitions occur in agreement with the B almer series rom level n = 2 to higher levels. O n the other hand, the S un has some o the visible hydrogen lines. B ecause the Sun is cooler than Vega, many o its hydrogen atoms are in the ground state producing absorption lines that are in the ultraviolet region o the spectrum ( corresponding to transitions rom level n = 1 in the Lyman series) . S tars that are even hotter than Vega tend not to produce hydrogen absorption lines in the visible spectrum because their high temperatures mean that the hydrogen in the photosphere is ionized ( and thereore has no electron to become excited by the absorption o a photon) .
Nature of science Spectral classes S tars were originally categorized in terms o the strength o the hydrogen absorption lines with the stars producing the darkest absorption lines being called type A and those with successively weaker lines type B and C etc. Within the last century astronomers recognized that the line strength depended on temperature o the stars rather than their composition. Subsequently, the
whole system o classifcation was revised with the result that many categories were abandoned and many others reordered as shown in the table below. This orm o star classifcation will not be included in your IB D iploma Programme Physics examinations, but is included here or historical interest and to indicate how many scientists have a reluctance to abandon a well-loved system!
Main sequence star properties Colour
Class Solar masses Solar diameters
Nature Oof science 20100
bluest
Spectral classes bluish B
420
1225 412
S tars were originally categorized in terms o the blue-white A 24 1.54 strength o the hydrogen absorption lines with the stars producing the darkest absorption lines white F 1.052 1.11.5 being called type A and those with successively weaker lines type etc. Within the last yellow-white G B , C0.81.05 0.851.1 century astronomers recognized that the line orange depended K on temperature 0.50.8 0.60.85 strength o the stars rather than their composition. Subsequently, the
red
M
0.080.5
0.10.6
Temperature/K Prominent lines 40 000
ionized helium
18 000
neutral helium, neutral hydrogen
whole system o classifcation was revised with 10result 000 thatneutral the manyhydrogen categories were abandoned and many others reordered as shown in the 7000 neutral hydrogen, ionized calcium table below. This orm o star classifcation will not5500 be included in your IB Physics neutral hydrogen, strongestexaminations, ionized calcium but is included here or historical interest and to 4000 how neutral metals (calcium, , ionized to indicate many scientists have iron) a reluctance calcium abandon a well-loved system!
3000
molecules and neutral metals
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A S T R O P H YS I C S Using the absorption spectrum to determine the chemical elements in a particular star is not easy as the spectra o many elements ( dierently ionized) are superimposed on one another. Although the lines present in an absorption spectrum tell us quite a lot about the temperature o the photosphere ( as described above) , they are difcult to interpret in terms o the abundance o elements. The movement o the gas atoms in the star causes the photons o light emitted to undergo both red and blue D oppler shits. In hotter stars, the atoms move aster than in cooler stars and, thereore, the D oppler broadening is more pronounced. The rotation o the stars themselves means that the light reaching us will come rom dierent parts o the star one edge moving towards us, the other moving away rom us and the central region rotationally stationary. This, o course, adds to the thermal D oppler broadening.
Wiens displacement law and star temperature B y treating a star as a black body it is possible to estimate its surace temperature using Wiens law as discussed in S ub- topic 8.2 . S tars range in surace temperature rom approximately 2 000 K to 40 000 K. The temperature T o the star in kelvin is given by: 2 .9 1 0 - 3 mK T = __ max where max is the wavelength in metres at which the black- body radiation is o maximum intensity and T is in kelvin.
Worked example a) Explain the term black-body radiation. The diagram is a sketch graph o the blackbody radiation spectrum o a certain star.
Solution a) B lack-body radiation is that emitted by a theoretical perect emitter or a given temperature. It includes all wavelengths o electromagnetic waves rom zero to infnity.
intensity
intensity
b) and c)
b) C opy the graph and label its horizontal axis. c) O n your graph, sketch the black-body radiation spectrum or a star that has a lower surace temperature and lower apparent brightness than this star. d) The star B etelgeuse in the O rion constellation emits radiation approximating to that emitted by a black-body radiator with a maximum intensity at a wavelength o 0. 97 m. C alculate the surace temperature o B etelgeuse.
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wavelength
The red line intensity should be consistently lower and the maximum shown shited to a longer wavelength. -3
2 .9 1 0 mK 2 .9 1 0 - 3 d) T = __ = __ max 0. 97 1 0 - 6 = 2 .99 1 0 3 K 3 000K
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Cepheid variables
luminosity/arbitary units
C epheid variables are extremely luminous stars that undergo regular and predictable changes in luminosity. B ecause they are so luminous it means that very distant C epheids can be observed rom the Earth. In 1 784, the periodic pulsation o the supergiant star, D elta C ephei, was discovered by the E nglish amateur astronomer, John Goodricke. This star has a period o about 5 .4 days as seen in fgure 3 .
period 5.4 days 4
6 8 10 12 time ( days) Figure 3 Lum inositytime relationship for Delta Cephei.
In 1 908, the American astronomer, Henrietta Leavitt, was working at the Harvard C ollege Observatory. She published an article describing the linear relationship between the luminosity and period o pulsation or 25 stars, all practically the same distance rom the Earth, in the Small Magellanic C loud a galaxy that orbits the Milky Way. This relationship allows us to estimate the luminosity o a given star by measuring its period o pulsation. It is now known that there are many more o these variable stars collectively known as Cepheids. The period o these stars varies between twelve hours and a hundred days. Although the period is regular it is not sinusoidal and it takes less time or the star to brighten than it does to ade.
C epheid variable stars are known as standard candles because they allow us to measure the distances to the galaxies containing C epheid variable stars. The distances, d, o the C epheids can be calculated rom the apparent brightnessluminosity equation: L b = _2 . 4 d The apparent brightness is measured using a telescope and C C D and the luminosity is calculated rom a measurement o the period o the C epheid.
10 3
type II (W vignis) Cepheids
10 2 RR Lyrae
Figure 4 shows the relative luminosityperiod relationship or C epheid stars. The relative luminosity is the ratio o the luminosity o the star to that o the S un. The S uns luminosity is conventionally written as L C epheid stars are stars that have completed the hydrogen burning phase and moved o the main sequence ( see later or an explanation o this) . The variation in luminosity occurs because the outer layers within the star expand and contract periodically. This is shown diagrammatically in fgure 5 . I a layer o an element loses hydrostatic equilibrium ( between the gas and radiation pressure and that due to gravity) and is pulled inwards by gravity ( 1 ) , the layer becomes compressed and less transparent to radiation ( 2 ) ; this means that the temperature inside the layer increases, building up the internal pressure ( 3 ) and causing the layer to be pushed outwards ( 4) . D uring expansion the layer cools, becoming less dense ( 5 ) and more transparent, allowing radiation to escape and letting the pressure inside all ( 6) . S ubsequently the layer alls inwards under gravity ( 1 ) and the cycle repeats causing the pulsation o the radiation emitted by the star.
type I (classical) Cepheids
10 4
L /L
2
1 0.5
1
3 5 10 period/days
30 50
100
Figure 4 Relative luminosity period relationships for Cepheid stars.
2
3
4
1
6
5
Figure 5 Cycle in a Cepheid star.
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TOK Women in science, technology, and engineering Over the ages there have been ew women scientists, and physics has been no exception. Astronomy, however, is one o the areas in which women have shone. Many women have made signifcant contributions. For example, in astrophysics, we have the German British astronomer Caroline Herschel (who discovered several comets), Agnes Clerk, Maria Mitchell, Annie Jump Cannon, and Henrietta Leavitt. Even today, there continues to be ewer women ollowing careers in science, and ewer opportunities or them to gain promotion to senior roles. Why should women be so under- represented in science, technology, and engineering?
Worked example a) D efne ( i) luminosity ( ii) apparent brightness.
Solution
b) State the mechanism or the variation in the luminosity o the C epheid variable.
a) ( i) Luminosity is the total power radiated by a star.
The variation with time t, o the apparent brightness b, o a C epheid variable is shown below. 1.3
A
b/10 -10 W m -2
1.2
( ii) Apparent brightness is the power rom a star received by an observer on the Earth per unit area o the observers instrument o observation. b) O uter layers o the star expand and contract periodically due to interactions o the elements in a layer with the radiation emitted.
1.1 1.0
c) ( i) The radius is larger ater two days (point A) because, at this time the luminosity is higher and so the stars surace area is larger.
0.9 B 0.8 0
1
2
3
4 5 6 time/da s
7
8
9
10
Two points in the cycle o the star have been marked A and B . c) ( i) Assuming that the surace temperature o the star stays constant, deduce whether the star has a larger radius ater two days or ater six days. ( ii) Explain the importance o C epheid variables or estimating distances to galaxies. d) ( i) The maximum luminosity o this C epheid variable is 7.2 1 0 29 W. Use data rom the graph to determine the distance o the C epheid variable. ( ii) C epheids are sometimes reerred to as standard candles. Explain what is meant by this.
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(ii) Cepheid variables show a regular relationship between period o variation o the luminosity and the luminosity. By measuring the period the luminosity can be calculated and, by L using the equation b = _ , the distances 4 d 2 to the galaxy can be measured. This assumes that the galaxy contains the Cepheid star. L 7.2 1 0 29 d) ( i) b = _2 thus 1 . 2 5 1 0 1 0 = _ 4 d 4 d 2 _______________
d=
7.2 1 0 __ 4 1 .2 5 1 0
d = 2 .1 4 1 0
29
-10
19
m
( ii) A standard candle is a light source o known luminosity. Measuring the period o a Cepheid allows its luminosity to be estimated. From this, other stars in the same galaxy can be compared to this known luminosity.
D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N
Hertzsprung-Russell (HR) diagram We saw in S ub- topic D .1 that the luminosity o a star is proportional both to its temperature to the ourth power and to its radius squared. C learly, large hot stars are the most inherently bright, but how do other combinations o temperature and radius compare to these? In the early 1 900s, two astronomers, Ej nar Hertzsprung in D enmark and Henry Norris Russell in America, independently devised a pictorial way o illustrating the dierent types o star. B y plotting a scattergram o the luminosities o stars against the stars temperatures, clear patterns emerged. These scattergrams are now known as Hertzsp rung-Russell ( HR) diagrams. In general, cooler red stars tend to be o relatively low luminosity, while hotter blue stars tend to be o high luminosity. With high temperature conventionally drawn to the let o the horizontal axis, the maj ority o stars create a diagonal stripe which goes rom top let to bottom right this is known as the main sequence. A small number o stars do not ollow the main sequence pattern but, instead, orm island groups above and below the main sequence. The vertical axis is commonly modifed to show the ratios o star luminosity to that o the S un ( denoted as L ) as shown in fgure 6 in this case the axis is logarithmic and has no unit. The temperature axis is also logarithmic and doubles with every division rom right to let. The HR diagram shows the position o many stars o dierent ages; during the lietime o a star its position will move on the diagram as its temperature and luminosity changes. We know rom black-body radiation that the luminosity depends on the size o a star and its temperature. Small, cool stars will be dim and be positioned to the bottom right o the diagram rom Wiens law we know they will be red. Large, hot stars will be o high luminosity and blue or blue-white in colour thus placing them at the top let o the diagram. A modifcation o the HR diagram to include the dierent star classes is shown in fgure 7. Main sequence stars are ordinary stars, like the S un, that produce energy rom the usion o hydrogen and other light nuclei such as helium and carbon. Nearly 90% o all stars ft into this category. 10 6 10 6 10 4
10 4
cool bright
hot bright
supergiants
10 2 L star L
10
2
L star L
10 -2 10 -4
giants 10 0
Sun
10 0
main sequence Sun
10 2 main sequence hot dim 40 000
20 000
cool dim 10 000 T/K
5000
Figure 6 HertzsprungRussell diagram.
2500
10 4 40 000
white dwars 20 000
10 000 5000 surace temperature/K
2500
Figure 7 HertzsprungRussell diagram showing the diferent classes o stars.
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A S T R O P H YS I C S Red giants are cooler than the Sun and so emit less energy per square metre o surace. However, they have a higher luminosity, emitting up to 1 00 times more energy per second than the S un. This means that they must have a much greater surace area to be able to emit such large energies. They, thereore, have a much larger diameter than the S un making them giant stars. S up ergiant stars are gigantic and very bright. A supergiant emitting 1 00 000 times the energy per second and at the same temperature o the Sun must have a surace area 1 00 000 times larger. This leads to a diameter that is over 3 00 times the diameter o the S un. O nly about 1 % o stars are giants and supergiants. White dwarfs are the remnants o old stars and constitute about 9% o all stars. Although they were very hot when they fnally stopped producing energy, they have a relatively low luminosity showing them to have a small surace area. These very small, hot stars are very dense and take billions o years to cool down.
Massluminosity relation for main sequence stars Not all main sequence stars are the same as the S un some are smaller and cooler while others are larger and hotter. High mass stars have shorter lietimes a star with a mass o 1 0 times the solar mass might only live 1 0 million years compared to the expected lietime o around 1 0 billion years or the Sun. O bservations o thousands o main sequence stars have shown there to be a relationship between the luminosity and the mass. For such stars this takes the orm L M 3.5 where L is the luminosity in W ( or multiples o the S uns luminosity, L ) and M is the mass in kg ( or multiples o the S uns mass, M ) . Because mass is raised to a positive power greater than one, this means that even a slight dierence in the masses o stars results in a large dierence in their luminosities. For example, a main sequence star o 1 0 times the mass o the Sun has a luminosity o (1 0) 3.5 3200 times that o the Sun. For a star to be stable it needs to be in hydrostatic equilibrium, where the pressure due to the gravitational attraction o inner shells is equalled by the thermal and radiation pressure acting outwards. For a stable star o higher mass there will be greater gravitational compression and so the core temperature will be higher. Higher temperatures make the usion between nuclei in the core more probable giving a greater rate o nuclear reaction and emission o more energy; thus increasing the luminosity. The mass o a star is undamental to the stars lietime those with greater mass have ar shorter lives.
Stellar evolution Formation of a star We saw in S ub- topic D .1 that the initial process in the ormation o a star is the gravitational attraction o hydrogen nuclei. The loss o potential energy leads to an increase in the gas temperature. The gas becomes
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denser and, when the protostar has sufcient mass, the temperature becomes high enough or nuclear usion to commence. The star moves onto the main sequence where it remains or as long as its hydrogen is being used into helium this time occupies most o a stars lie. E ventually when most the hydrogen in the core has used into helium the star moves o the main sequence.
The fate of stars All stars collapse when most o the hydrogen nuclei have used into helium. Gravity now outweighs the radiation pressure and the star shrinks in size and heats up. The hydrogen in the layer surrounding the shrunken core is now able to use, raising the temperature o the outer layers which makes them expand, orming a giant star. Fusion o the hydrogen adds more helium to the core which continues to shrink and heat up, orming heavier elements including carbon and oxygen. The very massive stars will continue to undergo usion until iron and nickel (the most stable elements) are ormed. What happens at this stage depends on the mass o the star.
A. Sun-like stars For stars like the S un o moderate mass ( up to about 4 solar masses) the core temperature will not be high enough to allow the usion o carbon. This means that, when the helium is used up, the core will continue to shrink while still emitting radiation. This blows away outer layers orming a planetary nebula around the star. When the remnant o the core has shrunk to about the size o the Earth it consists o carbon and oxygen ions surrounded by ree electrons. It is prevented rom urther shrinking by an electron degeneracy p ressure. Paulis exclusion principle prevents two electrons rom being in the same quantum state and this means that the electrons provide a repulsive orce that prevents gravity rom urther collapsing the star. The star is let to cool over billions o years as a white dwarf. Such stars are o very high density o about 1 0 9 kg m 3 . Figure 8 shows S irius B , the companion star o S irius A, and the frst white dwar to be identifed.
Figure 8 Sirius A and B.
The probable uture or the Sun is shown as the purple line on the HR diagram fgure 9. 10 6 core star shrinks. ejected gases thin and orm planetary nebula
10 4
yellow giant
10 2
burning helium in core
L star L
core star cools
10 0
10 2
no uel available or burning, star cools
10 4 40 000
star ejects outer layers red supergiant 12.2 billion years helium in core ignites: helium fash red giant
12 billion years burning hydrogen Sun now 4.5 billion years in shell around core burning hydrogen in core
main sequence white dwar
20 000
10 000 5000 surace temperature/K
2500
Figure 9 HertzsprungRussell diagram showing the Suns path.
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B. Larger stars
Figure 10 Supernova 1987A with the right hand image the region of the sky taken just before the event
The core o a star that is much bigger than the S un undergoes a dierent evolutionary path rom that o a S un-like star. When such stars are in the red giant phase, the core is so large that the resulting high temperature causes the usion o nuclei to create elements heavier than carbon. The giant phase ends with the star having layers o elements with proton numbers that decrease rom the core to the outside ( much like layers in an onion see S ub-topic D . 4, fgure 8) . The dense core causes gravitational contraction which, as or lighter stars, is opposed by electron degeneracy pressure. E ven with this pressure, massive stars cannot stabilize. The C handrasekhar limit stipulates that it is impossible or a white dwar to have a mass o more than 1 . 4 times the mass o the Sun. When the mass o the core reaches this value the electrons combine with protons to orm neutrons emitting neutrinos in the process. The star collapses with neutrons coming as close to each other as in a nucleus. The outer layers o the star rush in towards the core but bounce o it in a huge explosion a supernova. This blows o the outer layers and leaves the remnant core as a neutron star. In a quantum mechanical process similar to that o electron degeneracy, the neutrons provide a neutron degeneracy p ressure that resists urther gravitational collapse. The O p p enheimerVolkoff limit places an upper value on a neutron star or which neutron degeneracy is able to resist urther collapse into a black hole. This value is currently estimated at between 1 . 5 and 3 solar masses. life cycle of a low-mass star
protostar
main sequence star
red giant
helium burning star
double-shell burning red giant
planetary nebula
white dwarf
ssupernova pernova
neutron star or black hol hole
life cycle of a high-mass star
protostar
blue main sequence star
red supergiant
helium multiple-shell hell burning burning superupersupergiant gaint
Figure 11 The evolution of Sun-like and more massive stars.
Black holes B lack holes are discussed in more detail in Sub-topic A.5 but here we discuss their importance in astrophysics. It is not possible to orm a neutron star having a mass greater than the OppenheimerVolko limit instead the remnant o a supernova orms a black hole. Nothing can escape rom a black hole including the astest known particles, photons. For this reason it is impossible to see a black hole directly but their existence can be strongly inerred by the ollowing.
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The X- rays emitted by matter spiralling towards the edge o a black hole and heating up. X-ray space telescopes, such as NASAs C handra, have observed such characteristic radiation.
Giant j ets o matter have been observed to be emitted by the cores o some galaxies. It is suggested that only spinning black holes are sufciently powerul to produce such j ets.
The unimaginably strong gravitational felds have been seen to inuence stars in the vicinity, causing them to eectively spiral. A black hole has been detected in the centre o the Milky Way and it has been suggested that there is a black hole at the centre o every galaxy.
Worked example A partially completed HertsprungRussell ( HR) diagram is shown below.
b) State and explain the change in the luminosity o the Sun that occurs between positions S and I. c) E xplain, by reerence to the C handrasekhar limit, why the fnal stage o the evolutionary path o the Sun is at F.
I
d) O n the diagram, draw the evolutionary path o a main sequence star that has a mass o 3 0 solar masses.
luminosity
Solution a) Most o the S uns hydrogen has used into helium. S
b) B oth the luminosity and the surace area increase as the Sun moves rom S to I. c) White dwars are ound in region F o the HR diagram. Main sequence stars that end up with a mass under the C handrasekhar limit o 1 . 4 solar masses will become white dwars.
F
temperature
The line indicates the evolutionary path o the S un rom its present position, S , to its fnal position, F. An intermediate stage in the S uns evolution is labelled by I.
d) The path must start on the main sequence above the Sun. This should lead to the super red giant region above I and either stop there or curve downwards towards and below white dwar in the region between F and S.
a) S tate the condition or the S un to move rom position S .
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D.3 Cosmology Understandings
Applications and skills
The Big Bang model Cosmic microwave background (CMB) radiation Hubbles law
Describing both space and time as originating
The accelerating universe and redshit (Z) The cosmic scale actor (R)
Nature of science Cosmology and particle physics When we look at astronomical objects we see them as they were in the past. The Sun is always viewed as it was 8 minutes earlier and the most distant galaxies appear as they were more than 10 billion ( 10 1 0 ) years ago. The urther away the galaxy, the urther back in time we are looking. Cosmology is a way o studying the history o the universe. Particle physicists study the universe in a diferent way, but they have the same objectives. They use particle accelerators, such as the Large Hadron Collider (LHC) or the Relativistic Heavy Ion Collider (RHIC) , to attempt to recreate events that mimic conditions in the very early universe.
with the Big Bang Describing the characteristics o the CMB radiation Explaining how the CMB radiation is evidence or a hot big bang Solving problems involving z, R, and Hubbles law Estimating the age o the universe by assuming a constant expansion rate
Equations ___vc the redshit equation: z = _______ 0
relation between redshit and cosmic scale actor: R z = _____ -1 R0
Hubbles law: v = H 0 d 1 age o the universe estimate: T ______ H 0
Introduction S ir Isaac Newton believed that the universe was infnite and static. In his model o the universe he argued that the stars would exert equal gravitational attractions on each other in all directions, and this would provide a state o equilibrium. In 1 82 3 , the German astronomer, Heinrich O lbers, suggested that Newtons view o an infnite universe conicted with observation; when we look up into the night sky we see darkness but, in an infnite universe, we should be able to see a star in every direction and, thereore, the night sky should be uniormly bright on a cloudless night. In 1 848, the America author, E dgar Allan Poe, suggested that the universe, although infnite, was simply not old enough to have allowed the light to travel rom the most ar- ung regions to reach us. Although Poe was a naive scientist, he had suggested a theory that was a orerunner o the B ig B ang model. In this sub- topic we will consider how the ( inationary) B ig B ang model has developed into the most probable explanation o the beginning o the universe.
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The redshift and Hubbles law In the 1 92 0s, the American astronomer, Edwin Hubble, was working at the Mount Wilson O bservatory in C aliornia. He set out to fnd an additional way o gauging the distance o remote galaxies that would supplement using C epheid stars. Following on rom the work o others Hubble started to compare the spectra o distant galaxies with their E arth-bound equivalents. He ound that the spectra rom the galaxies invariably appeared to be redshited in line with the D oppler eect. Such consistent results could only mean that all the galaxies were moving away rom the Earth. Figure 1 shows the typical absorption spectra or a range o sources.
Note
In the case of infra-red or
microwaves the radiation already has wavelengths that are longer than red light. For these, the term redshift is ambiguous because redshift changes their wavelengths to even longer wavelengths and not towards the wavelength of red light.
very distant galaxy
distant galaxy
Stars within the Milky Way
and, therefore, relatively close to the Earth might actually be moving towards us and could show a blueshift this is simply a local phenomenon.
nearby galaxy nearby star labratory reference 500
600
700
wavelength/nm
Figure 1 Redshifted absorption spectra. For optical spectra the wavelengths are moved towards the red end o the spectrum. The shit applies to all waves in the spectrum, so the absorption lines in the spectrum can be seen to have shited. In addition to recognizing the consistent redshit, Hubble showed that the urther away the galaxy the greater the redshit. To do this he used the standard candles available to him, C epheid variables. Although his data had large uncertainties ( see fgure 2 ) he suggested that the recessional speed o a galaxy is proportional to its distance rom Earth. Hubbles law is written as
velocity of recession/km s -1
400
1000
500
0 0
1 distance/Mpc
2
Figure 2 Recreation of Hubbles original 1929 data.
where v is the velocity o recession and d is the distance o the galaxy ( both measured rom Earth) , H 0 is the Hubble constant. v is usually being measured in km s 1 and d in Mpc, H 0 is usually measured in km s 1 Mp c 1 . From the gradient o the graph o Hubbles data you will see that it gives a value or H 0 o about 5 00 km s 1 Mpc 1 . Now that we have more reliable data we can see that Hubbles intuition was valid but, by using 1 3 5 5 galaxies, data give a modern value or H 0 that is much closer to 70 km s 1 Mpc 1 . However, it can be seen in fgure 3 that there is still uncertainty in the value o the Hubble constant.
velocity of recession/km s -1
v = H0d 10000
5000
0 0
50 distance/Mpc
100
Figure 3 More recent velocity distance plot.
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Worked example a) The value o the Hubble constant H 0 is accepted by some astronomers to be in the range 60 km s 1 Mpc 1 to 90 km s 1 Mpc 1 . ( i) S tate and explain why it is difcult to determine a precise value o H 0 . ( ii) S tate one reason why it would be desirable to have a precise value o H 0 . b) The line spectrum o the light rom the quasar 3 C 2 73 contains a spectral line o wavelength 75 0 nm. The wavelength o the same line, measured in the laboratory, is 660 nm. Using a value o H 0 equal to 70 km s 1 Mpc 1 , estimate the distance o the quasar rom Earth.
determine how ar away they are. This is because o the difculty o both locating a standard candle, such as fnding a Cepheid variable within the galaxy, and the difculties o accurately measuring its luminosity. ( ii) Having a precise value o H 0 would allow us to gain an accurate value o the rate o expansion o the universe and to determine an accurate value to distant galaxies. It would also allow us to determine a more reliable value or the age o the universe. b) From the D oppler shit equation ( see Topic 9 and later in this topic) ___ _vc 0
Solution a) ( i) The Hubble constant is the constant o proportionality between the recessional velocity o galaxies and their distance rom Earth. The urther galaxies are away (rom Earth) the more difcult it is to accurately
0
TOK
The Big Bang model and the age of the universe
Values for H 0
Hubbles conclusion that the galaxies are moving urther apart provides compelling evidence that they were once much closer together. According to the S tandard Model, about 1 3 . 7 billion years ago the universe occupied a space smaller than the size o an atom. At that instant the entire universe exploded in a B ig B ang, undergoing an immense expansion in which both time and space came into being. 32 S tarting o at a temperature o 1 0 K, the universe rapidly cooled so that one second ater the B ig B ang it had allen to 1 0 1 0 K. In the time since the B ig B ang, the universe has continued to cool to 2 .7 K. In this time there has been an expansion o the abric o space the universe has not been expanding into a vacuum that was already there. As the galaxies move apart the space already between them becomes stretched. This is what we mean by expansion and why we believe the redshit occurs. The space through which the electromagnetic radiation travels is expanding and it stretches out the wavelength o the light. The urther away the source o the light, the greater space becomes stretched, resulting in a more stretched- out wavelength and increasing the cosmological redshift not to be conused with local redshit due to the D oppler eect.
There is still much debate about the value o the Hubble constant. Using better data has brought Hubbles original value down by a actor o almost 10 but the nature o measurement o intergalactic distances means that there are inevitable uncertainties. In December 2012, NASAs Wilkinson Microwave Anisotropy Probe (WMAP) gave a value or H 0 o (69.32 0.80) km s 1 Mpc 1 but, in March 2013, the Planck Mission provided a value o (67.80 0.77) km 1 s 1 Mpc . With two very credible scientifc projects providing values that barely overlap at their extreme values, should we believe scientists claims?
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= 90 1 0 9 m 90 1 0 - 9 v = 3 1 0 8 __ 660 1 0 - 9 7 = 4.1 1 0 m s 1 4 4. 1 1 0 v _ d = __ = = 5 90 Mpc H 70
C osmologists have spent a great deal o time considering what happened between the time that the B ig B ang occurred and the present day. One o the key questions is how did the universe appear in the distant past? Hubbles law certainly suggests that the galaxies were closer together than they are now and, logically, it ollows that there must have been a point in time when they were all in the same place that time being the B ig B ang. It is possible to estimate the age o the universe using Hubbles law.
D . 3 CO SM O LO G Y
Assuming that Hubbles law has held true for all galaxies at all times, the light from the most distant star ( at the edge of the observable universe) has taken the age of the universe to travel to us. If the light was emitted immediately after the time of the B ig B ang, the space between the galaxy and the Earth must have expanded at slightly less than the speed of light for the light to have j ust reached us. This makes the recessional speed of the galaxy ( almost) that of the speed of light, c. The distance that light has travelled from the galaxy = c T where T is the age of the universe v = H0 d c H 0 cT 1 T _ H0 Using a value for H 0 of 70 km s 1 Mpc 1 from the IB Physics data booklet 1 ly = 9.46 1 0 1 5 m and 1 pc = 3 .2 6 ly so 1 Mpc = 1 0 6 9.46 1 0 1 5 3 . 2 6 3 .1 1 0 22 m 3 .1 1 0 22 1 _ = 4.4 1 0 1 7 s = _ H0 70 1 0 3 10 1 . 4 1 0 yr ( or 1 4 billion years) The derivation of the age of the universe equation assumes that the galaxy and the Earth are moving at a relative constant speed of c and that there is nothing in their way to slow them down. In Sub-topic D.5 we consider a range of possibilities for the continued expansion of the universe.
The importance of the cosmic microwave background (CMB) Until the 1 960s there were two competing theories of the origin of the universe. One was the B ig B ang theory and the other was known as the steady state theory. One aspect of the B ig B ang theory is that it suggested a very high temperature early universe that cooled as the universe expanded. In 1 948, Gamow, Alpher, and Herman predicted that the universe should show the spectrum of a black-body emitter at a temperature of about 3 K. 5 In the Big B ang model, at approximately 4 1 0 years after the formation of the universe, the temperature had cooled to about 3000 K and the charged ion matter was able to attract electrons to form neutral atoms. This meant that space had become transparent to electromagnetic radiation, allowing radiation to escape in all directions (previously, when matter was ionic, it had been opaque to radiation) . The expansion of the universe has meant that each of the photons emitted at this time has been shifted to a longer wavelength that now peaks at around 7 cm in the microwave region of the spectrum. At earlier times the photons would have been much more energetic and of far shorter wavelengths, peaking in the visible or ultraviolet region of the electromagnetic spectrum. The C MB in the sky looks essentially the same in all directions ( it is isotropic) and does not vary with the time of day; this provides compelling support for the B ig B ang model. With the discovery of C MB , the advocates of the steady state theory were forced to concede to the strength of evidence. The Wilkinson Microwave Anisotropy
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A S T R O P H YS I C S Probe ( WMAP) was able to show that, on a fner scale, there are some uctuations in the isotropy o the C MB ; this indicates the seeds o galaxies in the early universe.
Figure 4 The CMB sky.
Figure 4 is an image produced by the European Space Agencys Planck mission team showing an all-sky image o the inant universe created rom 1 5 .5 months o data. The radiation is essentially uniorm but shows tiny variations in temperature ( 400 000 M ) . Given such mass, it is no surprise that stars orm in clusters in these regions individual stars have a mass between 0.1 M and 1 5 0 M . In even cooler regions o space, with higher gas densities o 1 0 1 1 particles per cubic metre and temperatures o 1 0 K, the Jeans mass alls to approximately 5 0 M . With these masses, short-lived giant stars are ormed. The atmospheric number density close to the surace o the Earth is o the order o 1 02 5 molecules per cubic metre. B ut, with the high temperature and low mass o gas, you need not worry that a star will orm in the atmosphere!
Nuclear fusion We have seen that the energy generated by a star is the result o thermonuclear usion reactions that take place in the core o the star. The process that occurs during the main sequence is known as hydrogen burning; here hydrogen uses into helium. For Sun-like stars the process advances through the p rotonproton chain but stars o greater than our solar masses undergo a series o reactions known as the C NO cycle.
stage 1
ra ma
p p
stage 3
stage 2 ga m ga m ma
p n p n
y
ra y
ga p p n
a mm
p np
p
total reaction
ra y
p np
p p n pn p
p p p p
gamma ray p n pn
gamma ray
Figure 2 The protonproton chain.
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A S T R O P H YS I C S The protonproton chain has three stages:
1 1
H + 11 H 21 H + 01 e + 00
[two protons ( hydrogen- 1 nuclei) fuse into a hydrogen- 2 nucleus ( plus a positron and a neutrino) ]
1 1
H + 21 H 32 He +
[a third proton fuses with the hydrogen-2 to form a helium-3 nucleus + a gamma-ray photon]
3 2
He + 32 He 42 He + 11 H + 11 H
[two helium- 3 nuclei fuse to produce helium- 4 and two hydrogen- 1 nuclei]
Thus, in order to produce a helium nucleus, four hydrogen nuclei are used in total ( six are used in the fusion reactions and two are generated) . p
12 6
The C NO process occurs in the larger stars with a minimum core temperature of 2 1 0 7 K and involves six stages:
C
1 1
13 7
1 1
H+
13 6
C
14 7
N+
[carbon- 1 3 fuses with proton to give nitrogen- 1 4 + a gamma- ray photon]
1 1
H+
14 7
N
15 8
O+
[nitrogen- 1 4 fuses with proton to give unstable oxygen- 1 5 + a gamma- ray photon]
15 8
1 1
gamma ray 13 7
N
neutrino e+
13 6
p
C
positron gamma ray
14 7
p
N gamma ray
15 8
O
15 7
N
12 6
13 6
N
15 7
O
H+
C
15 7
13 7
N+
C + 01 e + 00
N + 01 e + 00
N
12 6
C + 42 He
neutrino e+
p
H+
positron 4 2
He
[proton fuses with carbon-1 2 to give unstable nitrogen-1 3 + a gamma-ray photon] [nitrogen- 1 3 undergoes positron decay into carbon-1 3 ]
[oxygen- 1 5 undergoes positron decay into nitrogen- 1 5 ] [nitrogen- 1 5 fuses with proton to give carbon-1 2 ( again) and helium- 4]
Again, four protons are used to undergo the fusion process; carbon- 1 2 is both one of the fuels and one of the products. Two positrons, two neutrinos and three gamma- ray photons are also emitted in the overall process. The fusing of hydrogen into helium takes up the maj ority of a stars lifetime and is the reason why there are far more main sequence stars than those in other phases of their life- cycle.
Figure 3 CNO chain.
Fusion after the main sequence Note The CNO cycle is part of the main sequence and no heavy elements are synthesized in this process.
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We saw in Sub-topic D.2 that, once the hydrogen in a stars core is used up, the core converts into helium. The lack of radiation pressure causes the core to shrink and heat up. The hot core instigates the fusion and expansion of the hydrogen surrounding it, causing the red giant phase. Eventually, the cores rising temperature becomes high enough to make the star move off the main sequence. Helium now fuses into the unstable beryllium which then fuses with a further helium nucleus to produce carbon and then oxygen:
4 2
He + 42 He 84B e [two helium nuclei fuse to produce unstable beryllium- 8]
4 2
He + 84B e
4 2
He +
12 6
C
12 6
16 8
C
[a helium nucleus quickly fuses with beryllium- 8 to produce carbon-1 2 ]
O
[a further helium nucleus fuses with carbon- 1 2 to produce oxygen- 1 6]
D . 4 S TE LL AR PRO CE SS E S AH L
You can see rom these reactions that nucleosynthesis (the production o dierent nuclides by the usion o nuclei) is not a simple process. Eventually, both the carbon and oxygen produced will undergo usion and orm nuclei o silicon, magnesium, sodium, and so on until iron-56 is reached. This element represents one o the most stable o all the elements (nickel-62 is the most stable nuclide but is ar less abundant in stars than iron-56) . These nuclides have the highest binding energy per nucleon. Energy cannot be released by urther usions o these elements to produce even heavier nuclides, instead energy will need to be taken in to allow usion to occur. I nickel-62 is the most stable nuclide then how are heavier nuclides made? This is where neutron capture comes in. Neutrons, being uncharged, do not experience electrostatic repulsion and can approach so close to nuclei that the strong nuclear orce is able to capture them. The capture o a neutron increases the nucleon number by one and so does not produce a new element, just a heavier isotope o the original element (isotopes o X in fgure 4) . The newly created isotope will be excited and decay to a less energetic sibling by emitting a gamma-ray photon. The neutron in the newly ormed nucleus might be stable or it could decay into a proton, an electron and antineutrino (by negative beta decay) . This raises the proton number o the nucleus by one and produces the nucleus o a new element (Y in fgure 4) . The new nuclide will initially be excited and releases a gamma-ray photon in decaying into a less energetic nuclide. The hal-lie o beta decay depends solely on the nature o the particular parent nuclide. Whether or not there is sufcient time or a nucleus to capture a urther neutron depends on the density o neutrons bombarding the nuclei. -particle neutron antineutrino
target A ZX
neutron capture
excited nucleus A+1 Z X*
A+1 ZX
beta decay
gamma radiation A+1 Z+1 Y
A+1 Z+1 Y*
gamma radiation
Figure 4 Neutron capture. In a massive star, heavy nuclides ( up to bismuth- 2 09) can be produced by slow neutron capture or the s-p rocess. These stars provide a airly small neutron ux as a by- product o carbon, oxygen, and silicon burning. This means that there is time or nuclides to undergo beta decay beore urther neutron captures build up their nucleon number producing successively heavier isotopes o the original element.
ep Wn
In rapid neutron capture, or the r-process, there is insufcient time or beta decay to occur so successively heavier isotopes are built up very quickly, one neutron at a time. Type II supernovae produce a very high neutron ux and orm nuclides heavier than bismuth-209 in a matter o minutes (and well beore there is any likelihood o beta decay occurring) . This is something
neutrino interacts with neutron p + en+
Figure 5 Neutrino interacts with neutron.
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A S T R O P H YS I C S that does not appear to happen in the massive stars. There is also a high neutrino ux in a supernova and this has the eect o causing neutrons to convert into protons through the weak interaction, orming new elements. This weak interaction is shown in the Feynman diagram, fgure 5.
Worked example a) S uggest why nuclear usion processes inside stars can only synthesize elements with a nucleon number less than 63 . b) O utline how heavier elements could be produced by stars.
Solution a) Energy is released when the binding energy per nucleon increases. The binding energy per nucleon is a maximum at nucleon number 62
(nickel) and so urther usions would require energy to be supplied. This means that, in a star producing heavier elements, usion is no longer energetically avourable. b) With a strong neutron ux, nuclei can absorb neutrons. This can either be a slow process in massive stars, which are capable o producing nuclei no more massive than bismuth- 2 09, or it can be a rapid process in a supernova, that is able to produce still more massive nuclei.
Lifetimes of main sequence stars It might appear that the more matter a star contains the longer it can exist. However, this takes no account o stars having dierent luminosities and, in act, the more massive a star is, the shorter its lietime! Massive stars need higher core temperatures and pressures to prevent them rom collapsing under gravity. This means that usion proceeds at a aster rate than in stars with lower mass. Thereore, massive stars use up their core hydrogen more quickly and spend less time on the main sequence than stars o lower mass. In S ub- topic D .2 we saw that the luminosity o a main sequence star is related to its mass by the relationship: L M 3.5 Luminosity is the total energy E released by the star per unit time while hydrogen is being used or E L=_ t While usion occurs, the energy emitted is accompanied by a loss o mass. This will amount to a proportion o the total star mass during its lietime so a star o mass M loses a mass M. Using Einsteins mass 2 energy relationship E = mc , this makes the energy emitted during the hydrogen burning phase o the stars lie E = Mc 2 . Its average luminosity will be given by: Mc2 L=_ where is the lietime o the star. This can be rearranged to give: Mc 2 =_ L and because L M 3.5
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D . 4 S TE LL AR PRO CE SS E S AH L we can deduce that M _ M 3.5 or M - 2.5 It is useful to compare the mean lifetime of a star with that of the S un and so, using the normal notation for solar quantities, we have:
( )
M _ _ = M
- 2.5
The Sun is expected to have a lifetime of approximately ten billion years ( = 1 0 1 0 years) . A bigger star of, say, 1 0 solar masses would start with ten times the S uns hydrogen but it would have a much shorter lifetime given by
(
1 0M star _ = _ 10 M 10
)
- 2.5
= 3 . 2 1 0 -3 7
This gives star = 3 .2 1 0 years. We can see from this that although it has ten times the Suns mass it only lives for 0.3% of the lifetime of the Sun.
Note You will not be tested on this relationship in your IB Physics examinations, but it is included here to help explain the basic ideas, as these might be tested.
Figure 6 is a HR diagram showing the masses and lifetimes of a number of stars. This HR diagram is for interest only and there is no need to try to learn it; however, the various regions should make sense to you. 10 6 60 m Sun 30 m Sun 10 sol ar r adi i 10 4
lifetime 10 7 yrs 1s ola r ra diu s
10 2 0 .1
L star L 10
sol
ar r
adi
us
0
1 0 2
sol
ar r
adi
10 2 1 0 3
sol
ar r
us
adi
10 4
40 000
10 2
sol
10 3 ar r
adi
i
Deneb Rigal
sol
ar r
adi
i
Supergiants
Spica 10 m Sun
Betelgeuse Canopus Antares Bellatrix 6 m Sun Polaris ma Giants i n s Achernar lifetime e q u Arcturus Aldebaran e n c 3 m Sun 10 8 yrs e Vega Pollux Sirius 1.5 m Sun Procyon lifetime 1 m Sun 10 9 yrs Sun
Sirius b
us
Ceti lifetime 0.3 m Sun 10 10 yrs white lifetime 0.1 m Sun dwarfs 10 11 yrs Barnards star Ross 128 Wolf 359 Proxima centauri Procyon b
20 000
10 000 T/K
Dx cancri
5 000
2500
Figure 6 HR diagram.
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Worked example a) Explain why a star having a mass o 5 0 times the solar mass would be expected to have a lietime o many times less than that o the Sun. b) B y reerring to the massluminosity relationship, suggest why more massive stars will have shorter lietimes.
Solution a) The more massive stars will have much more nuclear material ( initially hydrogen) . Massive stars have greater gravity so equilibrium is reached at a higher temperature at which the outward pressure due to radiation and the hot gas will balance the inward gravitational
pressure. This means that usion proceeds at a aster rate than in stars with lower mass meaning that the nuclear uel becomes used up ar more rapidly. b) As the luminosity o the star is the energy used per second, stars with greater luminosity are at higher temperatures and will use up their uel in shorter periods o time. The luminosity o a star is related to its mass by the relationship L M 3.5 . Thereore, increasing the mass raises the luminosity by a much larger actor which in turn means the temperature is much higher. At the higher temperature the uel will be used in a much shorter time.
Supernovae We now delve into supernovae a little more because o their importance to astrophysics. Supernovae are very rare events in any given galaxy but, because there are billions o galaxies, they are detected quite regularly. They appear as very bright stars in positions that were previously unremarkable in brightness. The advent o high-powered automated space telescopes has meant that astrophysicists are no longer dependent on observing supernovae as random events within or close to the Milky Way. Until recently amateur astronomers had discovered more supernovae than the proessionals. With several automated surveys, such as the C atalina Real-Time Survey, more and more supernovae are being detected so that the number detected in the last ten years is greater than the total detected beore ten years ago. Supernovae are classifed as being Type I or Type II in terms o their absorption spectra (Type I have no hydrogen line but Type II do. This is because Type I are produced by old, low-mass stars and Type II by young, massive stars.) . Type I supernovae are subcategorized as Ia, Ib, Ic, etc. depending on other aspects o their spectra.
Figure 7 An artists impression of the accretion of stellar matter leading to a Type Ia supernova. The white dwarf is the star on the right, while the left-hand star is a red giant.
Type Ia supernovae result rom accretion o matter between two stars in a binary star. O ne o the stars is a white dwar and the other is either a giant star or a smaller white dwar. The ormation o these supernovae show up as a rapid increase in brightness ollowed by a gradual tapering o. Type II supernovae have been discussed briey in Sub- topic D .2 and consist o single massive stars in the fnal stages o their evolution. These classes o supernovae produce light curves with dierent characteristics as shown in fgure 9.
Type Ia supernovae These are very useul to astrophysicists as they always emit light in a predictable way and behave as a standard candle or measuring the distance o the galaxy in which the supernova occurs. Given the immense density o the material within a white dwar, the gravitational feld is unimaginably strong and attracts matter rom the companion star. When the mass o the
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D . 4 S TE LL AR PRO CE SS E S AH L
growing white dwar exceeds the Chandrasekhar limit o 1 .4 solar masses, the star collapses under gravity. The usion o carbon and oxygen into nickel generates such radiation pressure that the star is blown apart, reaching a luminosity o 1 0 1 0 times that o the Sun. Because the chain reaction always occurs at this mass we know how bright the supernova actually is and, by comparing it with the apparent brightness observed on Earth, we can estimate the distance o the supernovas galaxy up to distances o 1 000 Mpc. Ater the explosion the ejected material continues to expand in a shell around the remnant or thousands o years until it mixes with the interstellar material giving the potential to orm a new generation o stars.
Type II supernovae Ater approximately 1 0 million years (or stars o 81 0 solar masses) all the hydrogen in the core has converted into helium and hydrogen usion can now only continue in a shell around the helium core. The core undergoes gravitational collapse until its temperature is high enough or usion o helium into carbon and oxygen. This phase lasts or about a million years until the cores helium is exhausted; it will then contract again under gravity, causing it to heat up and allowing the usion o carbon into heavier elements. It takes about 1 0 thousand years until the carbon is exhausted. This pattern continues, with each heavier element lasting or successively shorter lengths o time, until silicon is used into iron-56 taking a ew days (see fgure 8) . At this point the star is not in hydrostatic equilibrium because there is now little radiation pressure to oppose gravity. On reaching the Chandrasekhar limit o 1 .4 M , electron degeneracy pressure is insufcient to oppose the collapse and the star implodes producing neutrons and neutrinos. The implosion is opposed by a neutron degeneracy pressure that causes an outward shock wave. This passes through the outer layers o the star causing usion reactions to occur. Although this process lasts just a ew hours it results in the heavy elements being ormed. As the shock wave reaches the edge o the star, the temperature rises almost instantly to 20 000 K and the star explodes, blowing material o as a supernova. non-burning hydrogen
10 9
hydrogen fusion helium fusion carbon fusion
relative 10 8 luminosity (L/L ) 10 7
Type Ia Type II
oxygen fusion neon fusion magnesium fusion
10 6
0 100 200 300 days after maximum brightness
Figure 9 Light curves for class Ia and II supernovae.
silicon fusion iron ash
Figure 8 The onion model of a massive star before it goes supernova.
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A S T R O P H YS I C S Type Ia and Type II supernovae can be distinguished by observers on Earth rom the manner in which the stars emit light. The Type Ia emits light up to 1 0 1 0 times the luminosity o the Sun; this rapidly reaches a maximum and then gradually tails o over six months or so. The Type II emits light up to 1 0 9 times the luminosity o the Sun; however, the burst alls a little beore reaching a slight plateau where it stays or some days beore alling away more rapidly. The light curves or these supernovae are shown in fgure 9.
Worked example a) Outline the dierence between a Type Ia and a Type II supernova. b)
( i) What is meant by a standard candle? ( ii) Explain how a Type Ia supernova can be used as a standard candle.
Solution a) A Type Ia supernova results rom a white dwar in a binary star system accreting material rom its companion giant or smaller white dwar star. Eventually, when the mass o the white dwar passes the Chandrasekhar limit, the star collapses and is blown apart as a supernova by the huge temperature generated. Type II supernovae occur in stars having masses between 8 and 50 times the mass o the Sun. When the core has changed into inert iron and nickel, with no urther usion occurring, gravity collapses the core. Again, on reaching the Chandrasekhar limit o 1 .4 M , electron degeneracy pressure is insufcient to oppose the collapse and the star implodes producing neutrons and neutrinos. The implosion is opposed by a neutron degeneracy pressure that causes an outward shock wave blowing away the surrounding material as a supernova. b)
( i) A standard candle is a star o known luminosity that, when compared with its apparent brightness, can be used to calculate its distance. ( ii) As all Type Ia supernova occur when the mass reaches the C handrasekhar mass they are all ( essentially) o the same peak luminosity. B y measuring the apparent brightness, the distance o the supernova ( and the galaxy in which it is a L member) can be calculated (rom b = ____ ). 4d2
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D . 5 FU R TH E R CO S M O LO G Y AH L
D.5 Further cosmology AHL Understanding The cosmological principle Rotation curves and the mass of galaxies Dark matter
Applications and skills Describing the cosmological principle and its
Fluctuations in the CMB The cosmological origin of redshift
Critical density Dark energy
Nature of science How constant is the Hubble constant? ...or when is a constant not a constant? The Hubble constant H0 is a quantity that we have used in Subtopic D.3 to estimate the age of the universe. This is a very important cosmological quantity, indicating the rate of expansion of the universe. The current value is thought to be around 70 km s 1 Mpc 1 , but it has not always been this value and will not be in the future. So it is a constant in space but not in time and it would be more appropriately named the Hubble parameter. The zero subscript is used to indicate that we are talking about the present value of the constant in general use we should omit the subscript. In the IB Physics course, H0 is used to indicate all values of the Hubble constant.
role in models of the universe Describing rotation curves as evidence for dark matter Deriving rotational velocity from Newtonian gravitation Describing and interpreting the observed anisotropies in the CMB Deriving critical density from Newtonian gravitation Sketching and interpreting graphs showing the variation of the cosmic scale factor with time Describing qualitatively the cosmic scale factor in models with and without dark energy
Equations velocity of rotating galaxies: v =
____
4G r _________ 3
3H2 critical density of universe: c = _______ 8G
Introduction In Sub-topic D .3 we saw that the B ig B ang should correspond to the simultaneous appearance o space and time (spacetime) . Hubbles law and the expansion o the universe tell us that the observable universe is certainly larger than it was in the past and that it can be traced back to something smaller than an atom, containing all the matter and energy currently in the universe. There is no special place in the universe that would be considered to be the source o the B ig B ang, it expanded everywhere in an identical manner. It is not possible to use the B ig B ang model to speculate about what is beyond the observable universe it should not be thought o as expanding into some sort o vacuous void. In this sub-topic we will consider ways in which the universe might continue to expand and look at possible models or fat, open, and closed universes.
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Nature of science Philosophy or cosmology? Although there is substantial evidence leading us to believe in the B ig B ang model nobody actually knows what instigated the B ig B ang. We have explained that this was the beginning o space and time and so we cannot ask what happened beore the B ig B ang? There are a
number o theories that reect on why the B ig B ang occurred such as uctuations in gravity or quantum uctuations but these theories stimulate other questions such as what caused this? As o yet these theories cannot be put to the test.
The Cosmological Principle Buoyed up by the success o his general theory o relativity in 1 91 5, Einstein sought to extend this theory to explain the dynamics o the universe or cosmos. In order to make headway, because he recognized that this would be a complex matter, he made two simpliying assumptions these have subsequently been shown to be essentially true on a large scale:
the universe is homogenous
the universe is isotropic
The frst o these requirements simply says that the universe is the same everywhere which, when we ignore the lumpiness o galaxies, it is. The second requirement is that the universe looks the same in all directions. Although this may not seem too dierent rom the homogeneity idea it actually is! It really says that we are not in any special place in the universe and ties in with the theory o relativity that says there is no special universal reerence rame. Imagine that we are positioned towards the edge o a closed universe and we look outward there would be a limited number o galaxies to send photons to us. I we look inward there would be an immense number o galaxies to send us photons. The two situations would appear very dierent. The isotropicity says that this isnt the case. Jointly, these two prerequisites o Einsteins theory are known as the Cosmological Principle. These assumptions underpin the Big Bang cosmology and lead to specifc predictions or observable properties o the universe. Figure 1 shows an image produced by the Automated Plate Measurement ( APM) Galaxy survey o around 3 million galaxies in the S outhern Hemisphere sky. The image shows short- range patterns but, in line with the cosmological principal, on a large scale the image shows no special region or place that is dierent rom any other. Using the cosmological principle and the general theory o relativity it can be shown that matter can only distort spacetime in one o three ways. This is conventionally shown diagrammatically by visualizing the impact o the third dimension on a at surace; however, the ourth dimension o time is also involved and this makes visualization even more complex.
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Figure 1 APM Galaxy survey image.
The at surace can be positively curved into a spherical shape o a fnite size. This means that, by travelling around the surace o the sphere, you could return to your original position or, by travelling through the universe, you could return to your original position in spacetime.
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The at surace can be negatively curved like the shape o a saddle and have an infnite size. In this universe you would never return to the same point in spacetime.
The surace could also remain at and infnite as given in our everyday experiences. Again, you would never return to the same position in spacetime.
Knowledge o the amount o matter within the universe is essential when determining which model is applicable. There is a critical density ( c) o matter that would keep the universe at and infnite this density would provide a gravitational orce large enough to prevent the universe running away but j ust too little to pull it back to its initial state. With less than the critical density the universe would be open and infnite. With greater density than the critical value the universe would be closed and fnite with gravity pulling all matter back to the initial state o spacetime. The critical density o matter appears to be no greater than ten particles per cubic metre and current research suggests that the average density is very close to this critical value.
Figure 2 The visualizations o closed, open, and fat universes.
The implications of the density of intergalactic matter As can be seen in fgure 5 in Sub-topic D .3, theory suggests that, ater the initial inationary period ollowing the B ig B ang, the rate o expansion o the universe has been slowing down. Instrumental to the ate o the universe is the uncertainty about how much matter is available to provide a strong enough gravitational orce to reverse the expansion and cause a gravitational collapse. As discussed in Sub-topic D .3 , data rom Type Ia supernovae has suggested that the universe may actually be undergoing an accelerated expansion caused by mysterious dark energy. We can derive a relationship or the critical density using Newtonian mechanics: Imagine a homogenous sphere o gas o radius r and density . A galaxy o mass m at the surace o the sphere will be moving with a recessional speed v away rom the centre o the sphere along a radius as shown in fgure 3. B y Hubbles law the velocity o the galaxy is given by: v = H0 r The total energy o the galaxy is the sum o its kinetic energy and its gravitational potential energy ( relative to the centre o mass o the sphere o gas) . ET = E K + E P Mm 1 ET = __ mv2 - G _ 2 r
r
v
m
remembering that potential energy is always negative or obj ects separated by less than infnity. The mass M is that o the sphere o gas is given by 4 M = _ r3 3 4 __ r3 m 3 _ ET = 1 m( H0 r) 2 - G _ r 2
Figure 3 Critical density or the universe.
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A S T R O P H YS I C S The galaxy will continue to move providing that it has sufcient kinetic energy, thus making ET positive. In the limit ET = 0 this gives
Note
This equation contains
4 __ r3 cm 3 1 _ m( H0 r) 2 = G _ r 2
nothing but constants. Any value for the critical density of the material of the universe is dependent on how precisely the Hubble constant can be determined.
where c is the critical density o matter. S impliying this equation gives 3 H0 2 c = _ 8G
A more rigorous derivation
of this equation requires the use of general relativity.
The cosmic scale factor and time The ratio o the actual density o matter in the universe ( ) to the critical density is called the density parameter ( indicated in fgure 2 ) and is given the symbol 0 . 0 = _ c There are three possibilities ( shown in fgure 4) or the ate o the universe, depending on the density parameter o the universe: 1
I 0 = 1 ( or = c) the density must equal the critical density and must be the value or a at universe in which there is j ust enough matter or the universe to continue to expand to a maximum limit. However, the rate o expansion would decrease with time. This is thought to be the least likely option.
2
I 0 < 1 ( or < c) the universe would be open and would continue to expand orever.
3
I 0 > 1 ( or > c) then the universe would be closed. It would eventually stop expanding and would then collapse and end with a B ig C runch.
An accelerated expansion o the universe ( shown by the red line on fgure 4) might be explained by the presence o dark energy. This oers an interesting and, increasingly likely, prospect.
cosmic scale actor (R)
4
In S ub- topic D .3 we considered the cosmic scale actor ( R) . This is essentially the relative size, or radius, o the universe. Figure 4 shows how R varies with time or the dierent density parameters. Each o the models gives an 0 value that is based on the total matter in the universe. An explanation or the accelerated universe depends on the concept o the ( currently) hypothetical dark energy outweighing the gravitational eects o baryonic and dark matter.
accelerated universe
3
0< 1 0 =1
2 1
0> 1 0
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now 10 time/10 9 years
20
30
Figure 4 Variation o R or diferent density parameters.
The cosmic scale factor and temperature The wavelength o the radiation emitted by a galaxy will always be in line with the cosmic scale actor (R) . So, as space expands, the wavelength will expand with it. We know rom Wiens law that the product o the maximum intensity wavelength and the temperature is a constant. Assuming that the spectrum o a black body retains its shape during the expansion this means that Wiens law has been valid rom the earliest
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times ollowing the B ig B ang. Thus, the wavelength and the cosmic scale actor are both inversely proportional to the absolute temperature. 1 1) T _ ( and T _ R
Worked example The diagram below shows the variation o the cosmic scale actor R o the universe with time t. The diagram is based on a closed model o the universe. The point t = T is the present time.
b)
R
R
T
T
t
a) Explain what is meant by a closed universe. b) O n a copy o the diagram, draw the variation based on an open model o the universe. c) Explain, by reerence to your answer to b) , why the predicted age o the universe depends upon the model o the universe chosen. d) ( i) What evidence suggests that the expansion o the universe is accelerating? ( ii) What is believed to be the cause o the acceleration?
Solution a) A closed universe is one that will stop expanding at some uture time. It will then start to contract due to gravity.
t
The graph should start at an early time ( indicating an older universe) and touch the closed universe line at T. It should show curvature but not fatten out as a fat universe would do. c) We only know the data or the present time so all curves will cross at T. B y tracing the curve back to the time axis, we obtain the time or the B ig B ang. This extrapolation will give a dierent time or the dierent models. d) ( i) The redshit rom distant type Ia supernovae has suggested that the expansion o the universe is now accelerating. ( ii) The cause o this is thought to be dark energy something o unknown mechanism but opposing the gravitational attraction o matter ( both dark and baryonic) .
Evidence for dark matter Let us imagine a star o mass m near the centre o a spiral galaxy o total mass M. In this region the average density o matter is . The star moves in a circular orbit with an orbital velocity v and radius r. B y equating Newtons law o gravitation to the centripetal orce we obtain Mm mv2 G_ = _ 2 r r C ancelling m and r M 2 G_ r = v
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A S T R O P H YS I C S In terms o the density and taking the central hub to be spherical, this gives 4 __ r3 3 _ G r = v2
_____
This means that v =
G4 r _ 3
or
v = constant r
From this we can see that the velocity is directly proportional to the radius.
What i the star is in one o the less densely populated arms o the galaxy? In this case we would expect the star to behave in a similar manner to the way in which planets rotate about the S un. The galaxy would behave as i its total mass was concentrated at its centre; the stars would be ree to move with nothing to impede their orbits. This gives
Figure 5 The spiral galaxy M81.
M 2 G_ r = v and so 1_ v _ r When the rotational velocity is plotted against the distance rom the centre o the galaxy, we would expect to see a rapidly increasing linear section that changes to a decaying line at the edge o the hub. This is shown by the broken line in fgure 6. What is actually measured ( by measuring the speed rom the redshits o the rotating stars) is the upper observed line. This is surprising because this at rotation curve shows that the speed o stars, ar out into the region beyond the arms o the galaxy, are moving with essentially the same speed as those well inside the galaxy. O ne explanation or this eect is the presence o dark matter orming a halo around the outer rim o the galaxy ( as shown in fgure 6) . This matter is not normal luminous or baryonic matter and emits no radiation and, thereore, its presence cannot be detected. v/km s -1 observed 100
expected from luminous disc
50
dark matter halo
5 luminous matter
Figure 6 Dark matter halo surrounding a galaxy.
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Figure 7 The rotation curve for the spiral galaxy M33.
10
r/kpc
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In fgure 8, the experimental curve has been modelled by assuming that the halo adds sufcient mass to that o the galactic disc. This maintains the high rotational speeds well away rom the galactic centre. 200
Vcir /km s -1
150 halo 100
50 disc
0
0
10
20
30
40
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radius/kpc
Figure 8 distribution of dark matter in NGC 3198.
O ther evidence or the presence o dark matter comes rom:
the velocities o galaxies orbiting each other in clusters these galaxies emit ar less light than they ought to in relation to the amount o mass suggested by their velocities
the gravitational lensing eect o radiation rom distant obj ects ( such as quasars) because the radiation passes through a cluster o galaxies it becomes much more distorted than would be expected by the luminous mass o the cluster
the X-ray images o elliptical galaxies show the presence o haloes o hot gas extending well outside the galaxy. For this gas to be bound to the galaxy, the galaxy must have a mass ar greater than that observed up to 90% o the total mass o these galaxies is likely to be dark matter.
At the moment no one knows the nature o dark matter but there are some candidates:
MAC HO s are MAssive C ompact Halo O bj ects that include black holes, neutron stars, and small stars such as brown dwars. These are all high density ( compact) stars at the end o their lives and might be hidden by being a long way rom any luminous obj ects. They are detected by gravitational lensing, but it is questionable whether or not there are sufcient numbers o MAC HO s to be able to provide the amount o dark matter thought to be in the universe.
WIMPs are Weakly Interacting Massive Particles subatomic particles that are not made up o ordinary matter (they are non-baryonic) . They are weakly interacting because they pass through ordinary, baryonic, matter with very little eect. Massive does not mean big, it means that these particle have mass (albeit very small mass) . To produce the amount o mass needed to make up the dark matter there would need to be
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A S T R O P H YS I C S unimaginably large quantities o WIMPs. In 1 998, neutrinos with very small mass were discovered and these are possible candidates or dark matter; other than this the theory depends on hypothetical particles called axions and neutralinos that are yet to be discovered experimentally.
Dark energy In 1 998, observations by the Hubble S pace Telescope ( HS T) o a very distant supernova showed that the universe was expanding more slowly than it is today. Although nobody has a defnitive explanation o this phenomenon its explanation is called dark energy. ES As Planck mission has provided data that suggest around 68% o the universe consists o dark energy ( while 2 7% is dark matter leaving only 5 % as normal baryonic matter see fgure 9) . dark matter 22.7%
dark matter 26.8%
ordinary matter 4.5%
ordinary matter 4.9%
dark energy 72.8%
dark energy 68.3%
before Planck
after Planck
Figure 9 The mass/energy recipe for the universe.
It has been suggested that dark energy is a property o space and so, with the expansion o the universe as space expands, so too does the amount o dark energy, i.e. more dark energy coming into existence along with more space. This orm o energy would subsequently cause the expansion o the universe to accelerate. Nobody knows i this model is viable. It is possible that an explanation or the accelerated expansion o the universe requires a new theory o gravity or a modifcation o Einsteins theory such a theory would still need to be able to account or all the phenomena that are, at the moment, correctly predicted by the current theory.
Nature of science The Dark Energy Survey (DES) With dark energy being one o the most upto- the- minute research topics in the whole o science, it is unsurprising that international collaboration is prolifc. The D E S is a survey with the aim o gaining the best possible data or the rate o expansion o the universe. This is being carried out by observing around 3 0 0 0 distant supernovae, the most distant o which exploded when the universe was about hal
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its current size. Using the Victor M. B lanco Telescope at C erro Tololo Inter- American O bservatory ( C TIO ) in C hile, 1 2 0 collaborators rom 2 3 institutions in 5 countries are using a specially developed camera to obtain images in the near inra- red part o the visible spectrum. The survey will take fve years to complete and will add to the sky- based research o missions such as WMAP and Planck.
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Anisotropies in the CMB In S ub- topic D . 3 we discussed the importance o the cosmic microwave background with regard to the B ig B ang. In this model, the universe came into being almost 1 3 . 8 billion years ago when its density and temperature were both very high ( this is oten reerred to as the hot Big Bang) . S ince then the universe has both expanded and cooled. The Planck satellite image in S ub- topic D . 3 , fgure 4 showed that, although the C MB is essentially isotropic, there are minute temperature uctuations; these variations are called anisotrop ies. In the early 1 9 9 0s, the C osmic B ackground E xplorer ( C O B E ) satellite provided the frst evidence o these anisotropies but, with the launch o NAS As WMAP in 2 001 and the E S As Planck satellite in 2 009 , the resolution has been improved dramatically. B oth these missions have shown signifcant low- level temperature uctuations. It is thought that these uctuations appear as the result o tiny, random variations in density, implanted during cosmic ination the period o accelerated expansion that occurred immediately ater the B ig B ang. When the universe was 3 8 0 000 years old and became transparent, the radiation emitted rom the B ig B ang was released and it has travelled outwards through space and time, including towards the E arth. This radiation was in the red part o the electromagnetic spectrum when it was released, but its wavelength has now been stretched with the expansion o the universe so that it corresponds to microwave radiation. The pattern shown in the variation demonstrates the dierences present on the release o the radiation: uctuations that would later grow into galaxies and galaxy clusters under the inuence o gravity. The inormation extracted rom the Planck sky map shows isotropy on a large scale but with a lack o symmetry in the average temperatures in opposite hemispheres o the sky. The S tandard Model suggests that the universe should be isotropic but, given these dierences, this appears not to be the case. There is also a cold spot ( circled in S ub- topic D . 3 , fgure 4) extending over a patch o sky and this is larger than WMAP had previously shown. The concepts o dark matter and dark energy have already been added to the S tandard Model as additional parameters. The evidence rom the C MB anisotropies may require urther tweaks to the theory or even a maj or re- think because o this. The Planck data also identiy the Hubble constant to be 67. 1 5 km s 1 Mpc 1 ( signifcantly less than the current standard value in astronomy o around 1 00 km s 1 Mpc 1 ) . The data imply that the age o the universe is 1 3 . 82 billion years. O ver the next ew years this value may well be modifed because o additional data being collected by this mission and rom orbiting telescopes including the James Webb space telescope ( fgure 1 0) and the j oint NAS A/E S A E uclid mission. There has, arguably, never been a more productive time in the history o cosmology.
Figure 10 The James Webb space telescope artists impression.
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Questions 1
b) O utline why:
( IB)
( i) white dwar stars cannot have a greater mass than 1 .4 M
a) The star Wol 3 5 9 has a parallax angle o 0.41 9 arcsecond. ( i)
( ii) it is possible or a main sequence star with a mass equal to 8 M to evolve into a white dwar. ( 6 marks)
D escribe how this parallax angle is measured.
( ii) C alculate the distance in light- year rom Earth to Wol 3 5 9. ( iii) S tate why the method o parallax can only be used or stars at a distance less than a ew hundred parsecond rom E arth. b) The ratio ap p are nt b righ tne ss o Wo l 3 5 9 ______________________ is 3 .7 1 0 1 5 . ap p are nt b rightn e ss o the S u n
4
(IB) a)
D efne luminosity.
b) The sketch- graph below shows the intensity spectrum or a black body at a temperature o 6000 K.
( 1 1 marks)
2
The average intensity o the Suns radiation at the surace o the E arth is 1 .3 7 1 0 3 W m - 2 . C alculate ( a) the luminosity and ( b) the surace temperature o the S un. The mean separation o the Earth and the S un = 1 .5 0 1 0 1 1 m, radius o the S un = 6.96 1 0 8 m, S teanB oltzmann constant = 5 .67 1 0 - 8 W m 2 K 4. ( 4 marks)
3
(IB) The diagram below is a ow chart that shows the stages o evolution o a main sequence star such as the Sun. ( Mass o the S un, the solar mass = M )
main sequence star mass M
red giant nebula
planetary nebula
white dwar
a) C opy nad complete the boxes below to show the stages o evolution o a main sequence star that has a mass greater than 8 M . main sequence star mass > 8M
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0 0
wavelength
O n a copy o the axes, draw a sketch-graph showing the intensity spectrum or a black body at 8000 K. c) A sketch o a HertzsprungRussell diagram is shown below.
luminosity
lu m ino sity o Wo l 3 5 9 ________________ is 8.9 1 0 4. lu m ino sity o the S u n
intensity
Show that the ratio
temperature
C opy the diagram above and identiy the: ( i)
main sequence ( label this M)
( ii) red giant region ( label this R) ( iii) white dwar region ( label this W) . d) In a HertzsprungRussell diagram, luminosity is plotted against temperature. Explain why the diagram alone does not enable the luminosity o a particular star to be determined rom its temperature. ( 8 marks)
QUESTION S 5
7
(IB)
a) In an observation of a distant galaxy, spectral lines are recorded. Spectral lines at these wavelengths cannot be produced in the laboratory. Explain this phenomenon.
The diagram below shows the grid of a HertzsprungRussell ( HR) diagram on which the positions of the S un and four other stars A, B , C and D are shown. 10 6
A
b) Describe how Hubbles law is used to determine the distance from the Earth to distant galaxies.
B
10 4
c) Explain why Hubbles law is not used to measure distances to nearby stars or nearby galaxies (such as Andromeda) . (6 marks)
luminosity (L) (Sun L = 1)
10 2 Sun
1
8
10 -2
C D
00 40 0 30 0 00
00
a) Describe any differences between this galaxy and the Milky Way.
50
60
0
00
80
00 10
00
0
10 -6
surface temperature (T/K)
Hubbles law predicts that NGC 5 1 2 8 is moving away from Earth.
a) Name the type of stars shown by A, B, C, and D.
b) (i) State Hubbles law.
b) Explain, using information from the HR diagram and without making any calculations, how astronomers can deduce that star B is larger than star A.
(ii) State and explain what experimental measurements need to be taken in order to determine the Hubble constant. c) A possible value for the Hubble constant is 68 km s 1 Mpc 1 . Use this value to estimate:
c) Using the following data and information from the HR diagram, show that star B is at a distance of about 700 pc from Earth.
(i) the recession speed of NGC51 28 (ii) the age of the universe.
Apparent brightness of the S un = 1 .4 1 0 3 W m 2 9
Apparent brightness of star B = 7.0 1 0 8 W m 2
(IB) O ne of the most intense radio sources is the Galaxy NGC 5 1 2 8. Long exposure photographs show it to be a giant elliptical galaxy crossed by a band of dark dust. It lies about 1 .5 1 0 7 light years away from Earth.
10 -4
25
(IB)
(1 0 marks)
a) D escribe what is meant by a nebula. b) E xplain how the Jeans criterion applies to star formation. (3 marks)
Mean distance of the S un from Earth = 1 .0 AU 1 parsec = 2 .1 1 0 5 AU
6
1 0 O utline how hydrogen is fused into helium in: ( 1 1 marks)
(IB)
a) stars of mass similar to that of the S un b) stars of mass greater than ten solar masses. (6 marks)
a) State what is meant by cosmic microwave background radiation. b) D escribe how the cosmic microwave background radiation provides evidence for the expanding universe. ( 5 marks)
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Explain what is meant by neutron capture.
( ii) Write a nuclear equation to show nuclide A capturing a neutron to become nuclide B . b) O utline the dierence between s and r processes in nucleosynthesis. ( 8 marks)
1 2 Explain why the lietimes o more massive main sequence stars are shorter than those o less massive ones. ( 4 marks)
a) D escribe the observational evidence in support o an expanding universe. b) Explain what is meant by the term critical density o the universe. c) D iscuss the signifcance o comparing the density o the universe to the critical density when determining the uture o the universe. ( 6 marks)
1 6 (IB)
1 3 B riey explain the roles o electron degeneracy, neutron degeneracy and the C handrasekhar limit in the evolution o a star that goes supernova. ( 6 marks)
a) Recent measurements suggest that the mass density o the universe is likely to be less than the critical density. S tate what this observation implies or the evolution o the universe in the context o the B ig B ang model.
1 4 (IB)
b) ( i)
a) E xplain the signifcance o the critical density o matter in the universe with respect to the possible ate o the universe. The critical density c o matter in the universe is given by the expression: 3 H0 2 c = _ 8G where H0 is the Hubble constant and G is the gravitational constant. An estimate o H0 is 2 .7 1 0 1 8 s 1 . b) ( i) C alculate a value or c. ( ii) Using your value or H0 , determine the equivalent number o nucleons per unit volume at this critical density. ( 5 marks)
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1 5 (IB)
O utline what is meant by dark matter.
( ii) Give two possible examples o dark matter. ( 5 marks)
I N TERN AL ASSESSM EN T Introduction In this chapter you will discover the important role of experimental work in physics. It guides you through the expectations and requirements of an independent investigation called the
internal assessment ( IA) . The investigation itself is likely to occur late in your second year, so you do not need to read this chapter until your teacher advises you to.
Advice on the internal assessment Understanding theory and experiment
Applications and skills to appreciate the interrelationship o theory
internal assessment requirements internal assessment guidance
internal assessment criteria
Nature of science Empirical evidence is a key to objectivity in science. Evidence is obtained by observation, and the details o observation are embedded in experimental work. Theory and experiment are two sides o the same coin o scientifc knowledge.
and experiment ability to plan your internal assessment understand teacher guidance appreciate the ormal requirements o an internal assessment to be critically aware o academic honesty
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I N T E R N AL A S S E S S M E N T
Theory and experiment The sciences use a wide variety o methodologies and there is no single agreed scientifc method. However, all sciences are based on evidence obtained by experiment. E vidence is used to develop theories, which then orm laws. Theories and laws are used to make predictions that can be tested in experiments. Science moves in a cycle that moves between theory and experiment. O bservations inorm theory. However, the refnement o a theory and improvements in instrumentation re-ocus on more observation. E xperimentation allows us to have confdence that a theory is not merely pure speculation. C onsider a amous analogy used by Albert Einstein o a man trying to understand the mechanism o a pocket watch. In the ollowing quote, Einstein illustrates that our scientifc knowledge can be tested against reality. He shows that we can confrm or deny a theory by experiment, but we can never know reality itsel. There is a continual dance between theory and experiment. Physical concepts are ree creations o the human mind, and are not, however it may seem, uniquely determined by the external world. In our endeavor to understand reality we are somewhat like a man trying to understand the mechanism o a closed watch. He sees the ace and the moving hands, even hears it ticking, but he has no way o opening the case i he is ingenious he may orm some picture o the mechanism which could be responsible or all the things he observes, but he may never be quite sure his picture is the only one which could explain his observations. He will never be able to compare his picture with the real mechanism and he cannot even imagine that possibility o the meaning o such a comparison. B ut he certainly believes that, as his knowledge increases, his picture o reality will become simpler and simpler and will explain a wider and wider range o his sensuous impressions. He may also believe in the existence o the ideal limit o knowledge and that it is approached by the human mind. He may call this ideal limit the objective truth. Albert Einstein and Leopold Infeld, The Evolution of Physics.
The internal assessment requirements E xperimental work is not only an essential part o the dynamic o scientifc knowledge it also plays a key role in the teaching and learning o physics. Experimental work should be an integral and regular part o your physics lessons consisting o demonstrations, hands- on group work, and individual investigations. It may include computer simulations, mathematical models, and online databases resources. It is only natural then that time should be allocated to you in order to ormulate, design, and implement your own physics experiment. You will produce a single investigation that is called an internal assessment. This means that your teacher will assess your report using IB criteria, and the IB will externally moderate your teachers assessment. Your investigation will consist o:
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selecting an appropriate topic
researching the scientifc content o your topic
defning a workable research question
adapting or designing a methodology
obtaining, processing, and analysing data
appreciating errors, uncertainties, and limits o data
writing a scientifc report 61 2 A4 pages long
receiving continued guidance rom your teacher.
AD VI CE O N TH E I N TE R N AL ASS E SS M E N T
Planning and guidance Ater your teacher introduces the idea o an internal assessment investigation, you will have an opportunity to discuss your investigation topic with your teacher. Through dialogue with your teacher you can select an appropriate topic, defne a workable research question, and begin by doing research into what is already known about your topic. You will not be penalized or seeking your teachers advice. It is your teachers responsibility to provide you with a clear understanding o the IA expectations, rules, and requirements. Your teacher will:
provide you with continued guidance at all stages o your work
help you ocus on a topic, then a research question, and then an appropriate methodology
provide quidance as you work and read a drat o your report, making general suggestions or improvements or completeness.
Your teacher will not, however, edit your report nor give you a tentative grade or achieving level or your report until it is fnally completed. Once your report is completed and ormally submitted you are not allowed to make any changes. Your teacher is responsible or your guidance, making sure you understand the IA expectations, and that your work is your own. As the student it is your resp onsibility to appreciate the meaning o academic honesty, especially authenticity and the respect o intellectual property. You are also responsible or initiating your research question with the teacher, seeking help when in doubt, and demonstrating independence o thought and initiative in the design and implementation o your investigation. You are also responsible or meeting the deadlines set by your teacher.
The internal assessment report There is no prescribed ormat or your investigation report. However, the IA criteria encourage a logical and j ustifed approach, one that demonstrates personal involvement and exhibits sound scientifc work. The style and orm o your report or the IA investigation should model a scientifc journal article. You should be amiliar with a number o high school level physics journal articles. For example, journals like the B ritish Physics Education publication ( http://iopscience.iop.org) or the American The Physics Teacher publication (http:// tpt.aapt.org) oten have articles that are appropriate or high school work. Moreover, many o these articles can provide good ideas or an investigation. There is no prescribed narrative mode, and your teacher will direct you to the style that they wish you to use. However, because a report describes what you have done, it is reasonable to write in the past tense. D escriptions are always clearer to understand i you avoid the use o pronouns (usually ' it' ) and reer specifcally to the relevant noun ( ' the wire' , ' the ammeter' , ' a digital caliper' , etc.) .
Academic honesty The IB learner profle ( page iii) describes IB students as aspiring to develop many qualities, including that o being principled. This means that you act with integrity and honesty, with a strong sense o airness and j ustice, and that you take responsibility or your actions and their consequences. The IA is your responsibility, and it is your work. Plagiarism and copying others work is not permissible. You must
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I N T E R N AL A S S E S S M E N T clearly distinguish between your own words and thoughts and those o others by the use o quotation marks ( or other methods like indentation) ollowed by an appropriate citation that denotes an entry in the bibliography. Although the IB does not prescribe reerencing style or in- text citation, certain styles may prove most commonly used; you are ree to choose a style that is appropriate. It is expected that the minimum inormation included is: name o author, date o publication, title o source, and page numbers as applicable.
Types of investigations Ater you have covered a number o physics syllabus topics and perormed a number o hands-on experiments in class, you will be required to research, design, perorm, and write up your own investigation. This project, known as an internal assessment, will count or 2 0% o your grade, so it is important that you make the most o your opportunity to do well. You will have 1 0 hours o class time, be able to consult with your teacher at all stages o your work, and research and write your report out o class. Your IA investigation may not be used as part o a physics extended essay. The variety and range o possible investigations is large, you could choose rom:
Traditional hands-on exp erimental work. You may want to measure the acceleration o gravity using Atwood' s approach, or determine the gas constant using standard B oyles law equipment.
D atabase investigations. You may obtain data rom scientifc websites and process and analyse the inormation or your investigation. Perhaps you fnd a pattern in the ebb and ow o the ocean tides, or process inormation on global warming, or use an astronomical database to confrm Keplers law.
S p readsheet. You can make use o a spreadsheet with data rom any type o investigation. You can process the data, graph the results, even design a simple model to compare textbook theory with your experimental values.
Simulations. It may not be easible to perorm some investigations in the classroom, but you may be able to fnd a computer simulation. The data rom a simulation could then be processed and presented in such a way that something new is revealed. Perhaps you might determine the universal gravitation constant through a simulation (an experiment too sensitive to perorm in most school laboratories) . Or you might investigate the eect o air resistance on projectile motion.
C ombinations o the above are also possible. The subj ect matter o your investigation is up to you. It may be something within the syllabus, something you have already studied or are about to study, or it can be within or outside the syllabus. The depth o understanding should be, however, commensurate with the course you are taking. This means that your knowledge o IB Physics ( either S L or HL) will be sufcient to achieve maximum marks when assessed.
The assessment criteria Your IA consists o a single investigation with a report 61 2 pages long. The report should have an academic and scholarly presentation, and demonstrate scientifc rigor commensurate with the course. There is the expectation o personal involvement, an understanding o physics, and that the study is set within a known academic context. This means you need to research your topic and fnd out what is already known about it. There are six assessment criteria, ranging in weight rom 82 5 % o the total possible marks. Each criterion reects a dierent aspect o your overall investigation.
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AD VI CE O N TH E I N TE R N AL ASS E SS M E N T
Criterion
Points
Weight
Personal engagement
02
8%
Exploration
06
25%
Analysis
06
25%
Evaluation
06
25%
Communication
04
17%
024
100%
Total
The IA grade will count or 2 0% o your total physics grade. The criteria are the same or standard and higher level students. We will now consider each criterion in detail. PE RS O NAL E NGAGE ME NT. This criterion assesses the extent to which you engage with the investigation and make it your own. Personal engagement may be recognized in dierent attributes and skills. These include thinking independently and/or creatively, addressing personal interests, and presenting scientifc ideas in your own way. For maximum marks under the personal engagement criterion, you must provide clear evidence that you have contributed signifcant thinking, initiative, or insight to your investigation: that you take the responsibility or ownership o your investigation. Your research question could be based upon something covered in class or an extension o your own interest. For example, you may be a keen music student and your teacher may have demonstrated resonance with a wine glass whilst studying sound. You could have been ascinated with this phenomenon and decide to design and perorm an investigation on the resonance o a wine glass. Personal signifcance, interest, and curiosity are expressed here. You may also demonstrate personal engagement by showing personal input and initiative in the design, implementation, or presentation o the investigation. Perhaps you designed an improved method or measuring the timing o a bouncing ball or devised an interesting method or the analysis o data. You are not to simply perorm a cookbook-like experiment. The key here is to be involved in your investigation, to contribute something that makes it your own. E XPLO RATIO N. This criterion assesses the extent to which you establish the scientifc context or your work, state a clear and ocused research question, and use concepts and techniques appropriate to the course you are studying. Where appropriate, this criterion also assesses awareness o saety, environmental, and ethical considerations. For maximum marks under the exploration criterion, your topic must be appropriately identifed and you must describe a relevant and ully ocused research question. B ackground inormation about your investigation must be appropriate and relevant, and the methodology must be suitable to address your research question. Moreover, or maximum marks, your research must identiy signifcant actors that may inuence the relevance, reliability, and sufciency o your data. Finally, your work must be sae and it must demonstrate a ull awareness o relevant environmental and ethical issues. The key here is your ability to select, develop, and apply appropriate methodology and produce a scientifc work.
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I N T E R N AL A S S E S S M E N T ANALYS IS . This criterion assesses the extent to which your report provides evidence that you have selected, processed, analysed, and interpreted the data in ways that are relevant to the research question and can support a conclusion. For maximum marks under the analysis criterion, your investigation must include sufcient raw data to support a detailed and valid conclusion to your research question. Your processing o the data must be carried out with sufcient accuracy. Moreover, your analysis must take a ull and appropriate account o experimental uncertainty. Finally, or maximum marks, you must correctly interpret your data, so that completely valid and detailed conclusions to the research questions can be deduced. The key here is to make an appropriate and j ustifed analysis o your data that is ocused on your research question. EVALUATIO N. This criterion assesses the extent to which your report provides evidence of evaluation of the investigation and results with regard to the research question and the wider world. For maximum marks under the evaluation criterion, you must describe a detailed and j ustifed conclusion that is entirely relevant to the research question, and ully supported by your analysis o the data presented. You should make a comparison to the accepted scientifc context i relevant. The strengths and weakness o your investigation, such as the limitations o data and sources o uncertainty, must be discussed and you must provide evidence o a clear understanding o the methodological issues involved in establishing your conclusion. Finally, to earn maximum marks or evaluation, you must discuss realistic and relevant improvements and possible extensions to your investigation. The key here is dierent rom the analysis criterion. The ocus o evaluation is to incorporate the methodology and to set the results within a a wider scientifc context while making reerence to your research topic. CO MMUNIC ATIO N. This criterion assesses whether the investigation is presented and reported in a way that supports effective communication of the investigations focus, process, and outcomes. For maximum marks under the communication criterion, your report must be clear and easy to ollow. Although your writing does not have to be perect, any mistakes or errors should not hamper the understanding, ocus, process, and outcomes o your investigation. Your report must be well structured and ocused on the necessary inormation, the process and outcomes must be presented in a logical and coherent way. Your text must be relevant and avoid wandering o onto tangential issues. Your use o specifc physics terminology and conventions must be appropriate and correct. Graphs, tables, and images must all be well presented. Your lab report should be 61 2 pages long. E xcessive length ( beyond 1 2 pages) will be penalized under the communication criterion. The key here is to demonstrate a concise, logical, and articulate report, one that is easy to ollow and is written in a scientifc context. This is not an assessed criterion but nevertheless is likely to be key to the ulflment o a successul IA. In conclusion, the IA represents a unique opportunity or you to take ownership o your physics learning by investigating something that matters to you. It is a chance or you to work independently and to ollow your own scientifc instincts. True, you should heed the advice and experience o your teacher and be guided so that you don' t go o down a blind alley; however you should be prepared to research your topic independently and approach your teacher ull o ideas and suggestions. Experience suggests that those students brim ull o proposals are likely to be successul, providing they stick to the physics skills and principles that have been encouraged to develop throughout the course.
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INDEX Page numbers in italics refer to questions. absorption 623 linear absorption coefficient 628 mass absorption coefficient 629 absorption spectra 271 3 Fraunhofer lines 272 accelerating universe 660, 665 acceleration 27, 334, 46 acceleration and displacement 1 1 5, 1 1 81 9 force, mass and acceleration 47 suvat/kinematic equations of motion 27, 368 accuracy 9 acoustic impedance Z 633 addition 8, 1 2 adiabatic changes 559, 563 air resistance 59 albedo 329, 3423 alpha particles 267, 273, 274, 2745 scattering of alpha particles 291 2, 4936 alternating current (ac) 1 94 ammeters 1 95, 21 4, 21 6 ampere (A) 3, 1 72, 1 84 definition of the ampere 238 amplitude 1 1 7, 1 25 interference by division of amplitude 373 analogue meters 1 94 angle of incidence 1 46 angular displacement 245, 246 angular magnification 593, 597 angular momentum 475, 483, 555 angular motion 554 angular speed 245, 2478, 354 angular velocity 245, 2478 anistrophies 683 approximations 1 Archimedes principle 570, 5723 assessment criteria 6901 analysis 692 communication 692 evaluation 692 exploration 691 2 personal engagement 691 asteroids 643 astronomical distances 641 , 6456 astronomical reflecting telescopes 61 41 5 cassegrain mounting 61 6 Newtonian mounting 61 5 astronomical refracting telescopes 61 21 4 astronomical unit (AU) 645 astrophysics 641 , 6846 astronomical distances 641 , 6456 luminosity and apparent brightness of stars 6478 objects that make up the universe 6423 stars 6435 atomic physics 267, 268, 3046 absorption of radiation 267, 2801 absorption spectra 271 3 background radiation 267, 27980 Bohr model 4834 emission spectra 2701 energy levels 267, 2689 half-life 267, 2767 nuclear radiation and safety 274 patterns in physics 270 probability of decay 267, 276 radioactive decay 267, 2736 transitions between energy levels 267, 26970 atoms 1 05 attenuation 6246 Avrogado constant 1 00, 1 023 Bartons pendulums 586 baryons 290, 295, 297 Becquerel, Henri 267 Bel scale 6223 Bernoulli equation 570, 575
applications 576 derivation 5756 flow out of a container 5789 Pitot static tubes 5778 Venturi tubes 5767 beta particles 267, 2734, 274, 275 Big Bang 660, 675 cosmic microwave background [CMB] 6634 redshift equation and the cosmic scale factor 6645 Type Ia supernovae 665 binary stars 643 black holes 5402, 5589 black-body radiation 329, 3367 black-body radiation and stars 648 building a theory 3401 emission spectrum from a black body 3379 grey bodies and emissivity 329, 33940 StefanBoltzmann law 329, 339 Wiens displacement law 329, 339 Bohr model 483 energies in the Bohr orbits 4834 Bohr, Niels 270, 483 Boyle, Robert 1 01 , 1 31 Brahe, Tycho 258 bridge circuits 4534 brightness, apparent 641 , 647-8 calculations 8, 1 21 4 adding and subtracting vectors 203 candela (cd) 3 capacitance 455, 457 capacitance of a parallel-plate capacitor 459 unit of capacitance 456 capacitors 455, 4567 capacitance of a parallel-plate capacitor 459 capacitors in practice 45961 charging a capacitor 46870 charging a capacitor with a constant current 4578 combining capacitors in parallel and series 4623 energy stored in a capacitor 458 Carnot cycle 559, 564-5 CAT scans (computed axial tomography) 631 CCDs (charge-coupled devices) 380 centripetal acceleration 245, 24950 centripetal force 245, 250 amusement park rides 2523 banking a bike 2534 centripetal or centrifugal? 251 2 how does speed change when motion is in a vertical circle? 2567 investigating how F varies with m, v and r 2501 moving in a vertical circle 2546 satellites in orbit 252 turning a car 253 Cepheid variables 649, 653-4 CERN 303, 492 Chandrasekhar limit 649, 658 charge 1 69, 405 charge and field 1 701 charge carrier drift speed 1 69, 1 846, 238 charge carriers 1 82 charge density 1 85 electric current 1 84 electronic or elementary charge 1 72 measuring and defining charge 1 725 moving charge 1 803 positive test charge 1 77 radial fields 1 77 charged parallel plates 3945, 407, 455 chromatic aberrations 606, 61 4 circle of least confusion 606 circuit diagrams 1 92, 1 93 circuit conventions 1 94
circuit symbols 1 934 constructing practical circuits from a diagram 1 956 practical measurements of current and potential difference 1 945 circular motion 245, 246, 2656 angular displacement 245, 246 angular speed 245, 2478 angular velocity 245, 2478 centripetal acceleration 245, 24950 centripetal force 245, 2507 circular motion and SHM 3545 linking angular and linear speeds 245, 2489 period and frequency 245, 248 radians or degrees 2467 climate change 329, 3489 international perspective 349 clock synchronization 521 collaboration 1 00, 303 collisions 734, 7980 elastic collisions 73, 767 inelastic collisions 73, 78 comets 643 communications and the advantages of optic fibres 620 coaxial cable 621 optic fibres 621 2 twisted pair 6201 compressions 1 26 conducting spheres 1 7980 conduction 91 , 1 81 conduction in gases and liquids 1 82 conduction in metals 1 81 2 models of conduction 1 83 conductors 1 789, 2302, 2368 conservation laws 532 conservation of charge 21 3 conservation of energy 61 , 63, 1 47, 21 3 conservation of linear momentum 73, 745 conservation of momentum 73, 745 constellations 644 convection 91 , 329, 3323 convection currents 332 convection in the Earth 333 modelling convection 3345 sea breezes 333 why the winds blow 3334 Copenhagen interpretation 486 Copernicus 258 Cosmic Background Radiation 1 33 cosmic microwave background [CMB] 660, 6634 anistrophies in the CMB 683 discovery of CMB 664 redshift equation and the cosmic scale factor 6645 cosmic scale factor [R] 660 cosmology 660, 6756 anistrophies in the CMB 683 Big Bang and the age of the universe 6625 cosmic scale factor and temperature 6789 cosmic scale factor and time 678 cosmological principle 6769 dark energy 6823 evidence for dark matter 67982 implications of the density of intergalactic matter 6778 redshift and Hubbles law 661 2 Coulombs law 1 69, 1 73, 1 77 coulombs 1 69, 1 72, 456 critical angle 1 45, 1 4850 CT scans (computer tomography) 631 current balances 238 damping 5834 dark energy 682 Dark Energy Survey {DES} 682
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INDEX dark matter 67982 MACHOs 681 WIMPs 681 2 database investigations 690 de Brglie wavelength 4801 decibels 623 deflection effect 540 degraded energy 31 4 density 570, 571 , 572 diatomic gases 1 03 dielectric materials 455, 4601 diffraction 1 45, 1 51 2 diffraction and resolution 377 diffraction and the satellite dish 3789 electron diffraction 481 2 single-slit diffraction 3647 diffraction grating 367, 371 2 appropriate wavelengths for effective dispersion 372 grating spacing and number of lines per mm 372 using a spectrometer 373 digital meters 1 95, 21 4, 21 6 diodes 2001 diode bridges 439, 4534 dioptres 603 direct current [dc] 1 69, 1 94 discharging a capacitor 4638 discrete energy 267 dispersion 6224 material dispersion 624 waveguide (or modal) dispersion 6245 displacement 27, 2830 acceleration and displacement 1 1 5, 1 1 81 9 angular displacement 245, 246 displacementdistance graphs 1 267 displacementtime graphs 1 27 relationship between displacement, velocity and acceleration 355 Wiens displacement law 329, 339, 654 distance 27, 2830 forcedistance graphs 61 , 65 division 8, 1 3 Doppler effect 381 astronomy 385 light 3845 measuring rate of blood flow 386 moving observer and stationary source 3823 radar 3856 sound waves 381 3 double-slit interference 1 524, 367, 368 intensity variation with the double-slit 3689 drag force 59 drift speed 1 69, 1 846, 238 Earth 262, 329, 333 energy balance in surfaceatmosphere system 329, 3423, 3479 gases in the atmosphere 41 9 greenhouse effect 329, 3437 orbit 341 2 Edison, Thomas 1 98, 453 efficiency 61 , 72 Einstein, Albert 48, 61 , 63, 79, 1 34, 284, 285, 340, 475, 507, 508, 51 2, 51 7, 688 Einsteins photoelectric equation 478 equivalence principle 5345 general theory of relativity 534, 538, 539, 540, 5423 elastic potential energy 61 , 702 electric cells 21 71 8, 2404 anodes and cathodes 21 8 capacity 2201 discharge 21 7, 221 2 electromotive force (emf) 21 7, 2236 internal resistance 21 7, 2236 lead-acid cells 21 920 Leclanch cells 21 81 9 power supplied by a cell 226 primary and secondary cells 21 7, 21 8221 , 223 recharging secondary cells 223 terminal potential difference 21 7, 221 , 222,
694
224, 225 electric current 1 69, 1 836, 2404 charge 1 84 chemical effect 1 92, 21 7 circuit diagrams 1 92, 1 936 conventional and electron current 1 88 conventional current 1 87, 2301 direction rules 2301 , 238, 431 eddy currents 446 Edison vs Westinghouse 453 effects 1 923 full-wave rectification 439, 451 2 half-wave rectification 439, 4501 heating effect 1 92, 21 1 1 2 high-voltage direct-current transmission (HVDC) 449 Kirchhoffs laws 1 92, 21 21 5 magnetic effect 1 92, 227 magnets 229 mechanism for electric current 1 81 2 rectifying ac 450 resistance 1 92, 1 96202 resistivity 1 92, 2021 2 Wheatstone bridge circuits 4534 Wien bridge circuits 454 electric fields 1 69, 1 75, 2404, 392, 406, 426 charge moving in magnetic and electric fields 4223 charged parallel plates 407 charges moving in electric fields 421 2 close to a conductor 1 789 comparison of gravitational and electric fields 4245 conducting sphere 1 7980 electric field strength 1 778, 392 electric field strength and potential gradient 4089 electric field strength and surface charge density 409 graphical interpretations of electric field strength and potential 41 01 1 plotting electric fields 1 76 potential difference 391 , 393 potential inside a hollow conducting charged sphere 41 21 3 electric lighting 1 98 electrical breakdown 1 82 electromagnetic force 290, 2989, 300 electromagnetic induction 427, 428, 471 4 applications 438 electromagnetic force 430 Lenzs law 427, 4302 magnetic flux 427, 4325 making a current 42830 electromagnetism 227, 2404 catapult field 235 Flemings left-hand rule 235, 4301 force between bar magnet field and currentcarrying wire 2346 force on a current-carrying conductor 2369 forces between two current-carrying wires 2334 magnetic field due to current in conductors 2302 magnetic field patterns 2279 motor effect 235 electromotive force [emf] 1 89, 21 7, 2236, 427 changing fields and moving coils 436 coil moving 437 inducing an emf 42830 magnetic field changes 4378 straight wire moving in uniform field 437 electron microscopes 61 0 electrons 2923 charge carriers 1 82 conduction electrons 1 82 electron degeneracy pressure 657 electron diffraction 481 2 free electrons 331 electronvolts 1 69, 1 901 electrostatics 1 701 , 391 , 392, 393 field between parallel plates 3945, 407
linking electrostatics and gravity 41 2 emissivity 329, 33940 empirical versus theoretical models 1 1 2 energy density 308, 31 1 1 2 energy sources 307, 3502 fossil fuels 308, 31 7 nuclear fuel 31 722 primary sources 308, 3091 0, 31 722 pumped storage 308, 3246 renewable and non-renewable energy sources 307, 3089 Sankey diagrams 308, 31 41 6 secondary sources 308 solar energy 308, 3268 specific energy and energy density 307, 31 1 13 types of energy sources 3091 1 wind generators 3224 energy transfers 61 doing work 61 , 635 efficiency 72 elastic potential energy 702 energy forms and transfers 623 energy moving between GPE and KE 6870 KE and GPE 667, 678 power 61 , 656 energy-momentum relation 531 engineering physics 549, 58992 engines 5647 entropy 559, 5678 entropy as a measure of disorder 5689 equilibrium position 1 1 6 equipotentials 391 , 396, 3989 field and equipotential in gravitation 399401 equivalence principle 5345 errors 8 error bars 8, 1 4 propagation of errors 1 7 random errors 8, 1 01 1 systematic errors 8, 91 0, 1 1 zero error 9 escape speed 405, 41 71 9 estimation 1 , 78 event horizon 541 exchange particles 290, 299300 experiment 688, 690 explosions 73, 80 exponential decay 466, 470 falsifiability 77 Faraday, Michael 427, 4356, 439 Faradays law of induction 427, 435 farads 456 Feynman diagrams 290, 3001 fibre optics 620 attenuation and dispersion 6224 structure and use of optic fibres 6202 field lines 391 , 394, 395 field and equipotential in gravitation 399401 field between a point charge and a charged plate 399 field due to a charged sphere 399 field due to a co-axial conductor 399 field due to a single point charge 398 field due to two point charges 399 measuring potential in two dimensions 3957 fields 391 , 392, 393, 405 comparison of gravitational and electric fields 4245 concept map for field theory 424 edge effects 395 field strength 392 radial fields 398 fluid dynamics 570, 573 continuity equation 570, 574 streamline or laminar flow 570, 5734 fluid resistance and terminal speed 27, 59 maximum speed of a car 601 skydiving 5960 fluids 570, 571 Archimedes principle 570, 5723 Bernoulli equation 570, 5759
measuring density of immiscible liquids using a U-tube 572 measuring the coefficient of viscosity of oil 580 Pascals principle 570, 572 static fluids density and pressure 571 Stokes law 57980 turbulent flow 5801 viscosity of fluids 579 flux 427, 4324, 4345 focal length 594, 599 focal point 594, 599 forces 44, 391 Aristotle and the concept of force 445 conservative forces 68 forcedistance graphs 61 , 65 forcetime graphs 73, 834 force, mass and acceleration 47 forces and inverse-square law behaviour 405, 4067 forces between charged objects 1 723 free-body force diagrams 44, 4951 identifying force pairs 44, 48 Newtons first law of motion 44, 456 Newtons second law of motion 44, 467, 1 24 Newtons third law of motion 44, 489, 1 24 representing forces as vectors 44, 53 resultant force 44, 52 translational equilibrium 44, 524 triangle of forces 53 fossil fuels 308, 309, 31 1 Fourier synthesis 1 58, 353 Franklin, Benjamin 1 70, 1 79, 1 88, 238 free-body force diagrams 44, 4951 frequency 1 1 5, 1 1 7, 1 23, 1 25, 354 angular speed or frequency [w] 354 Larmor frequency 635 natural frequency 1 47 period-frequency relationship 1 1 5, 245, 248 threshold frequency 477 friction 58 coefficients of friction 44, 55, 56, 57 dynamic friction 44, 56, 57 solid friction 44, 545 static friction 44, 556, 57 values for pairs of surfaces 56 full scale deflections (fsd) 1 94 full-wave rectification 439, 451 2 fusion (melting) 97 GM tubes 279, 502 galaxies 641 , 6445 Galileo Galilei 44, 45, 258 Galilean relativity and Newtons postulates 507, 5081 0 Galilean transformations 5091 0 gamma rays 1 33, 267, 2745 gases 91 , 1 00, 1 1 31 4, 1 69 Boltzmann constant 1 00, 1 09, 1 1 0 Boyles law 1 01 , 1 034 Brownian motion 1 06 Charless law 1 02, 1 045 differences between real and ideal gases 1 00, 1 1 1 1 2 diffusion 1 06 first law of thermodynamics 563 gas laws 1 01 5 ideal gas equation 1 00, 1 02, 1 078, 1 09, 1 1 0 kinetic model of an ideal gas 73, 1 00, 1 07, 1 1 2, 329 MaxwellBoltzmann distribution 1 1 01 1 molar mass 1 00, 1 03 mole and Avrogado constant 1 00, 1 023 pressure 1 00, 1 02 third gas law 1 02, 1 05 total internal energy of an ideal gas 1 09 gauge bosons 299 Geiger, Johannes 290, 291 2, 492 generators 439 alternating current [ac] generators 43941 measuring alternating currents and voltages 4434
modelling an ac generator 441 3 transformers 439, 4449 global warming 329, 3489 GPS (global positioning system) 521 , 537 gradient fields 6356 graphs 1 4 adding and subtracting vectors 203 calculating gradient of a graph 32 describing motion with graphs 27, 31 , 346 displacementdistance graphs 1 267 displacementtime graphs 1 27 drawing graphs manually 1 6 false origin 1 6 forcedistance graphs 61 , 65 forcetime graphs 73, 834 graphical interpretations of electric field strength and potential 41 01 1 linearizing graphs 1 7 SHM (Simple Harmonic Motion) 1 1 921 gravitation 245, 2656 field and equipotential in gravitation 399401 Newtons law of gravitation 257, 2601 , 263, 264, 402 universal gravitational constant 261 gravitational field strength 257, 258, 392 defining gravitational field strength 25960 field strength at a distance r from a point mass, M 261 2 field strength at a distance r from the centre outside a sphere of mass M 262 g and the acceleration due to gravity 260 inside the Earth 262 linking orbits and gravity 2623 gravitational fields 391 , 392, 405, 406, 426 comparison of gravitational and electric fields 4245 escaping the Earth 41 71 9 orbiting Earth 41 51 7 potential 41 31 5 potential energy 391 , 393 potential inside a planet 41 5 gravitational force 2989, 300 gravitational potential energy (GPE) 61 , 6870 gravitational time dilation 539 greenhouse effect 329, 3434 modelling climate balance 3457 why greenhouse gases absorb energy 3445 hadrons 290, 296 Large Hadron Collider (LHC) 492, 660 half-wave rectification 439, 4501 hard radiation 629 harmonics 1 58, 1 60 harmonics in pipes 1 623 harmonics on strings 1 61 Hawking, Stephen 542 heat pumps 5667 Heisenberg uncertainty principle 475, 4878, 489 HertzsprungRussell [HR] diagram 649, 6556 Higgs boson 290, 303 Hooke, Robert 70, 71 , 1 30 Hookes law 702 Hopper, Grace 623 Hubbles law 660, 661 2, 6623 Huygens, Christiaan 1 30, 1 34, 1 45 hydrostatic equilibrium 570 imaging 593, 63740 converging and diverging mirrors 593, 5948 converging and diverging thin lenses 593, 599606 imaging instrumentation 608 real images 593, 596 spherical and chromatic aberrations 593, 6067, 61 4 virtual images 593, 597 impulse 73, 823 inertia 45 inertial frame of reference 508, 51 0 infra-red radiation 1 32 interference 1 45, 367 constructive and destructive interference 1 52
double-slit interference 1 524, 3689 interference by division of amplitude 373 measuring wavelength of laser light using double slit 1 56 measuring wavelength of microwaves using double slit 1 57 multiple-slit interference 367, 36971 path difference and the double-slit equation 1 546 interferometer telescopes 61 71 8 internal assessment requirements 6889 academic honesty 68990 assessment criteria 6902 internal assessment report 689 planning and guidance 689 types of investigations 690 internal energy 91 , 95 invariant quantities 51 721 inverse-square relationships 1 38, 1 71 , 2601 forces and inverse-square law behaviour 405, 4067 isobaric change 559, 561 2 work done for non-isobaric changes 562 isothermal changes 559, 5623 isotopes 268 isovolumetric changes 559, 5634 Jeans criterion 6667 joule (J) 63, 1 69 kelvin (K) 3, 339 Kepler, Johannes 258, 260, 264 kilogram (kg) 3 kinetic energy (KE) 61 , 667, 6870 KE and momentum 80 Kirchhoffs laws 21 21 4 ideal and non-ideal meters 21 41 6 Large Hadron Collider (LHC) 492, 660 Larmor frequency 635 lasers 1 56 length contraction 5267 lenses 3745, 593, 599 aspherical shape 599 converging and diverging thin lenses 599 doublets lens 606 eyepiece lenses 6089, 61 3 finding approximate value for focal length of a convex lens 600 lens equation 6024 magnifying glasses 601 , 6046 objective lenses 608, 61 3 relationship between object and image for a thin convex lens 6002 same travel times 607 thin-lens theory 599 Lenzs law 427, 4302 leptons 290, 294, 297 light 1 45 interface between two media 1 47 polarization of light 1 42 reversibility of light 1 48 linear magnification 593, 597 light year (ly) 645 liquids 91 Lorentz transformations 51 3, 51 41 6 inverse Lorentz transformation 51 5 Lorentz factor 51 4 luminosity 641 , 6478 MACHOs (Massive Compact Halo Objects) 681 magic numbers 283 magnetic field lines 228 magnetic field patterns 2279 magnetic fields 227 charge moving in magnetic and electric fields 4223 charges moving in magnetic fields 41 921 magnetic field due to current in conductors 2302 magnetic field strength 2378 magnetic flux 427
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INDEX flux density 4324 flux linkage 427, 4345 magnetic forces 227 force between bar magnet field and currentcarrying wire 2346 force on a current-carrying conductor 2369 forces between two current-carrying wires 2334 magnetic poles 2289 magnets 229, 232, 2346, 236 hard and soft magnetic materials 447 magnifying glasses 601 , 6046 magnitude, orders of 1 , 7 main sequence stars 649, 655, 656 lifetimes of main sequence stars 6702 Maluss law 1 33, 1 423 Marsden, Ernest 290, 291 2, 492 mass 46 force, mass and acceleration 47 inertial and gravitational mass 46 massspring system 361 , 362 matter 290 states of matter 91 matter waves 475, 4801 electron diffraction 481 2 Maxwell, James 1 77, 507, 51 2 measurements 1 4, 246 uncertainties in measurement 91 1 mechanics 27, 8790 medical imaging 626 computed tomography (CT) 631 nuclear magnetic resonance (NMR) in medicine 626, 6346 ultrasound in medicine 6324 X-ray images in medicine 62631 Meldes string 1 5960 melting (fusion) 97 Mendeleev, Dmitri 282 mesons 290, 295, 297 metabolism 561 metallic bond 1 81 2 meters 21 41 6 metre (m) 3, 63 metric multipliers 1 , 7 microscopes electron microscopes 61 0 make a microscope 61 0 normal adjustment 609 optical compound microscopes 6089 resolution of a microscope 61 01 2 microwaves 1 323, 1 57 Minkowski diagrams see spacetime diagrams mirrors 594 aberrations in a mirror 597598 caustic curves 597 images from a mirror 595 linear and angular magnification 593, 597 pole P 594 modelling 1 00, 224 mole (mol) 3 mole and Avrogado constant 1 00, 1 023 moment of inertia 549, 552, 55758 momentum 734 collisions and changing momentum 745 impulse 73, 823 KE and momentum 80 momentum and safety 73, 856 momentum and sport 73, 86 Newtons second law of motion 73, 845 relativistic momentum 529, 5301 two objects colliding 78 two objects when energy is gained 79 two objects with different masses 778 two objects with same mass 767 momentum conservation 745 helicopters 82, 85 recoil of gun 80 rocketry 82, 85 water hoses 81 2 Moon 343, 346, 41 9 motion 27 describing motion with graphs 27, 31 , 346
696
Newtons laws of motion 44, 456, 467, 489 objects and motion 445, 45 suvat/kinematic equations of motion 27, 368 MRI scans 6356 multiple-slit interference 367, 36971 multiplication 8, 1 3 muons 293 nanoseconds 623 near point 604 nebulae 644 Neumanns equation 427, 435 neutrinos 293, 499 neutron capture 669 r-process 66970 s-process 669 neutron stars 494 neutrons 293 neutron degeneracy pressure 658 thermal neutrons 31 8 newton (N) 4, 63 Newton, Isaac 44, 1 34, 1 45, 258, 61 4, 660 Newtons law of gravitation 257, 2601 , 263, 264, 402 Newtons laws of motion 44, 456, 467, 489, 73, 845, 1 24, 51 0 Principa Mathemetica 509 night vision 1 33 nuclear fusion 282, 2867 CNO cycle 667, 668 protonproton chain 6678 star formation 6678, 66870 nuclear interactions 73 nuclear magnetic resonance (NMR) in medicine 626, 634 basic NMR effect 6345 MRI modifications 6356 nuclear physics 267, 3046, 476, 493, 5046 alternative nuclear binding energy plot 289 energy levels in the nucleus 4689 fission chain reaction 2889 law of radioactive decay 5005 mass and energy units for nuclear changes 285 mass defect and nuclear binding energy 282, 2845 nuclear fission 2878 nuclear fusion 282, 2867 nuclides 274, 2823 Rutherford scattering and the nuclear radius 492, 4936 unified atomic mass unit 282, 2834 using electrons of higher energies 4968 variation of nuclear binding energy per nucleon 2856 nuclear power 31 722 fast breeder reactors 320 safety issues 321 society and nuclear power 321 2 nucleosynthesis 668, 671 nuclides 274, 282 patterns for stability 2823 Ohms law 1 92, 1 98200 OppenheimerVolkoff limit 649, 658 optic fibres 621 2 graded-index fibre 621 single-mode fibre 621 step-index fibre 621 optical centre 599 optical compound microscopes 6089 orbital energy 405 orbital motion 405 orbital speed 405, 41 51 6, 41 8 orbits 2634 satellites of Jupiter 264 oscillations 1 1 5, 1 648, 353 describing periodic motion 1 1 61 8 isochronous oscillations 1 1 61 8 oscillations and damping 5834 pair annihilation 293, 475, 489
pair production 293, 475, 4889 pair production and the Heisenberg uncertainty principle 489 X-rays 627 paradigm shifts 507, 51 2 parsec (pc) 645 particle acceleration 532 particle accelerators 531 particle physics 267, 290, 291 4, 3046, 660 annihilation and pair production 293 classification of particles the Standard Model 2948 conservation rules 2978 exchange particles 290, 299300 Feynman diagrams 290, 3001 fundamental forces 290, 2989, 300 international collaboration 303 particle properties 290 strangeness 290, 301 2 Pascals principle 570, 572 path difference 1 546 pendulums 35860, 586 period 1 1 5, 1 1 6, 1 23 period-frequency relationship 1 1 5 periodic motion 1 1 5, 1 1 61 8 Periodic Table 282 phase and phase difference 1 21 2 in phase 1 21 phase change 91 molecular exchange of phase change 99 photoelectric effect 475 demonstration of the photoelectric effect 4767 Einsteins photoelectric equation 478 explanation of the photoelectric effect 477 gold leaf experiment 4778 Millikans photoelectric experiment 479 wave theory and the photoelectric effect 480 X-rays 629 photons 269, 475, 5345 photovoltaic cells 308, 327 pitch 1 63 Planck, Max 340, 476 polarization 1 34, 1 41 2, 1 44 Maluss law 1 33, 1 423 partial polarization 1 41 polarimeters 1 43 polarization of light 1 42 Polaroids 1 42, 1 43 positrons 292 potential 405 graphical interpretations of electric field strength and potential 41 01 1 potential and potential energy 405, 41 3 potential inside a hollow conducting charged sphere 41 21 3 potential difference [pd] 1 69, 1 868, 1 90, 393, 405 electrical potential 391 , 404 electromotive force [emf] 1 89 gravitational potential 391 , 401 4 measuring pd 1 945 power, current and pd 1 8990 using a potential divider to give a variable pd 2091 1 potential gradient 405, 4089 electric field strength and potential gradient 4089 potentiometers 1 96 potters kiln 338 power 61 , 656, 439 power dissipation 1 92 power stations 308 baseload stations 325 control rods 31 9 enriched fuel 31 8 fossil fuels 31 7 heat exchangers 31 9 nuclear fuel 31 722 thermal power stations 31 21 3 precision 1 0, 1 1 pressure 570, 571
electron degeneracy pressure 657 gases 1 00, 1 02 neutron degeneracy pressure 658 principal axis 594, 599 principal focus 594 principle quantum number 484 projectile motion 27 falling (moving vertically) 3941 , 42 moving horizontally 423 proper length 5201 proper time 51 820 pumped storage hydroelectric systems 308, 3246 Q factor 5846 quantities 1 3 fundamental and derived quantities 2 invariant quantities 51 721 scalar quantities 1 81 9 vector quantities 1 81 9, 1 920, 21 , 223 quantum 269 quantum mechanics 484 quantum physics 476, 5046 Heisenberg uncertainty principle 4878 matter waves 4807 pair production and annihilation 48890 photoelectric effect 47680 quantum tunnelling 475, 4901 quarks 290, 2945, 500 quark confinement 290, 295 radar 1 33 radio telescopes 61 61 7 discoveries in radio astronomy 61 8 radio waves 1 32 radioactive decay 267, 2734, 4991 alpha decay 2745 decay constant and half-life 501 decay in height of water column 5034 gamma ray emission 2756 measuring long half-lives 5023 measuring radioactive decay 279 measuring short half-lives 501 modelling radioactive decay 2789 negative beta decay 275, 498 nuclide nomenclature 274 positron decay 275 raising to a power 8, 1 1 4 rarefactions 1 26 ratios 1 ray diagrams 593, 5957, 6002 predictable rays 595 Rayleigh criterion 340, 376, 377, 61 0, 61 6 Rayleigh scatter 6234 RayleighJeans law 476 rays 1 345 real-is-positive 603 red giants 649, 656 redshift 660, 661 2 cosmological redshift 662 gravitational redshift 5379 redshift equation and the cosmic scale factor 6645 reflection 1 45, 1 46 total internal reflection 1 45, 1 4850 refraction 1 45, 1 46 reflection and refraction of waves 1 46 refractive index 1 46, 1 47, 1 50 refrigerators 5667 Relativistic Heavy Ion Collider [RHIC] 660 relativistic mechanics 529 particle acceleration 532 photons 5323 relativistic momentum 529, 5301 total energy and rest energy 52930 relativity 507, 54448 applications of general relativity to the universe as a whole 5423 bending of light 53940 charges and currents a puzzle 51 01 2 clock synchronization 521 equivalence principle 5345
Galilean relativity and Newtons postulates 507, 5081 0 GPS (global positioning system) 521 , 537 gravitational redshift 5369 gravitational time dilation 537 invariant quantities 51 721 Lorentz transformations 51 3, 51 41 6 Maxwell and electromagnetism 507, 51 2 reference frames 507, 58 Schwarzchild black holes 5402 two postulates of special relativity 51 31 4 velocity addition 51 61 7 renewable energy 3089 repeaters 624 resistance 1 92, 1 967 diodes and thermistors 2001 how resistance depends on size and shape 202 measuring internal resistance of a fruit cell 225 Ohms law 1 92, 1 98200 resistance of metal wire 1 97 variation of resistance of a lamp filament 1 99 resistivity 1 92, 2023 combining resistors 2045 complicated networks 207 heating effect equations 21 1 1 2 potential dividers 208 practical resistors 204 resistivity of pencil lead 203 resistors in parallel 205, 2067 resistors in series 205, 2056 thermal and electric resistivity 331 2 using a potential divider to give a variable pd 2091 1 using a potential divider with sensors 2089 variable resistor or potential divider? 21 1 resistorcapacitor [RC] series circuits 455 resolution 376, 377, 38790 diffraction and resolution 377 resolution equation 378 resolution in a CCD (charge-coupled device) 380 resolvance of diffraction gratings 379 resonance 584 examples of resonance 587 resonance of a hacksaw blade 58788 rest mass 528 rheostats 21 0 root mean square (rms) 444 rotational dynamics 549, 550 angular acceleration 5501 equations of motion 551 uniform motion in a circle 550 Rutherford, Ernest 290, 291 2, 493 Sankey diagrams 308, 31 41 6 satellite dishes 3789 satellites 41 51 6, 41 71 9 geostationary orbit 41 7 geosynchronous satellites 41 61 7 GPS (global positioning system) 521 , 537 orbit shapes 41 81 9 polar orbit 41 6 scalars 1 8, 239 scalar quantities 1 81 9 scattering 291 2, 492, 493, 6234 deviations from Rutherford scattering 495 electron diffraction 4956 method of closest approach 4934 nuclear density 494 why is the sky blue? 624 Schrdingers equation 4856 Schwarzchild black holes 5402 Schwarzchild radius 541 scientific notation 1 , 57 second (s) 3 serendipity 44, 664 SHM (Simple Harmonic Motion) 1 1 5, 1 1 81 9, 353, 354 angular speed or frequency [] 354 circular motion and SHM 3545 energy changes in SHM 1 223, 353, 3623
graphing SHM 1 1 922 iteration with damped SHM 583 massspring system 361 , 362 modelling SHM with a spreadsheet 3557 relationship between displacement, velocity and acceleration 355 SHM equation and 2 357 simple pendulums 35860 velocity equation 353, 358 SI units 1 3 capital or lower case? 3 fundamental and derived SI units 1 , 34, 1 84 sign conventions 607 significant figures 1 , 45 simulations 690 single-slit diffraction 364 graph of intensity against angle 364 single slit with monochromatic and white light 3667 single-slit equation 3656 small angle approximation 365 Snells law 1 45, 1 46, 1 47 soap films 3756 solar constant 329, 341 2 solar energy 308 solar heating panels 3267 solar photovoltaic panels 3278 solar system 6423 solids 91 sound waves 1 23 bell jar experiment 1 301 measuring speed of sound 1 2930 spacecraft 41 9 Earths rotation 41 9 slingshots 41 9 spacetime diagrams 5223 simultaneity 5245 time dilation and length contraction 5257 twin paradox 52728 spacetime interval 51 71 8 time travel 51 8 specific energy 308, 31 1 1 2 specific heat capacity 91 , 957 specific latent heat 91 , 979 spectrometers 373 Bainbridge mass spectrometer 423 speed 27, 30 instantaneous and average values 323 terminal speed 27, 60 spherical aberrations 6067 spreadsheet models 34, 3557 spreadsheets 692 springs 70, 1 1 71 8, 1 245 extension 71 2 massspring system 361 , 362 spring constant 71 stars 641 , 643 binary stars 643 black-body radiation and stars 648 Cepheid variables 649, 6534 composition of stars 6502 fate of stars 657 formation of a star 6567 groups of stars 6434 HertzsprungRussell [HR] diagram 649, 6556 Jeans criterion for star formation 6667 larger stars 658 luminosity and apparent brightness of stars 641 , 64749 main sequence stars 649, 655, 656, 6702 massluminosity relation for main sequence stars 649, 656 spectral classes 651 stellar evolution 649, 65659 stellar parallax 641 , 6456 stellar spectra 649, 6502 Sun-like stars 657 Wiens displacement law and star temperature 652 stellar clusters 643 globular clusters 644 open clusters 6434
697
INDEX stellar evolution 649, 65659 stellar parallax 641 , 6456 stellar processes 666 fusion after the main sequence 66872 Jeans criterion for star formation 6667 nuclear fusion 666, 66769 supernovae 6724 stellar spectra 649, 6501 strangeness 290 conservation of strangeness 301 2 strong nuclear force (strong interaction) 290, 2989, 300 subtraction 8, 1 2, 1 3 Sun 329, 341 2, 660 supergiant stars 656 supernovae 6724, 673 Type Ia supernovae 665, 666, 6723 Type II supernovae 666, 67374 superposition 1 58 symbols 2 symmetry 290, 296 telescopes astronomical reflecting telescopes 61 41 6 astronomical refracting telescopes 61 21 4 Earth and satellite-borne telescopes 61 81 9 interferometer telescopes 61 71 8 making a telescope 61 3 radio telescopes 61 61 8 temperature 91 , 92 absolute temperature 91 , 945 conversion from Celsius to kelvin 91 temperature and energy transfer 92 thermometers 934 tesla (T) 237 theory 688 thermal energy transfer 91 , 32930, 3502 atomic vibration 331 convection 91 , 329, 3325 thermal and electric resistivity 331 2 thermal conduction 329, 3301 thermal physics 91 , 1 1 31 4 thermal radiation 91 , 329, 335 black and white surfaces 3356 radiators 336 saucepans 336 thermistors 93, 94, 2001 thermodynamics 55960 cycles and engines 559, 5647 first law of thermodynamics 559, 5604 second law of thermodynamics 559, 56769 thermal efficiency 559 thermometers 934 thin film interference 367, 3734 coating of lenses 3745 vertical soap films 3756 waves reflected by the film 374 waves transmitted through the film 374 thin lenses see lenses tidal effects 538 time constant 455, 467 time dilation 5256, 537 torque 549, 553 angular momentum 555 conservation of angular momentum 555 couples 5534 Newtons first law for angular motion 554 Newtons second law for angular motion 554 Newtons third law for rotational motion 555 rolling and sliding 5567 rotational kinetic energy 556 total energy and rest energy 52930 trajectories 40, 523 transformers 439, 4446 choosing core material 447 choosing wire in the coils 447 core design 447 efficiency of transformer 446 laminating the core 4467 step-up and step-down transformers 446 transformer action 4478 transformer rule 446
698
transformers in action 4489 twin paradox 52728 ultrasound 628, 6346 ultraviolet catastrophe 476 ultraviolet radiation 1 32 uncertainties 8, 246 absolute, fractional and percentage uncertainty 8, 1 1 1 2 gradient and intercepts 8, 1 41 5 propagation of uncertainties 8, 1 2 zone of uncertainty 1 4 unified atomic mass unit 282, 2834 binding energy 284 values 1 vaporization 989 vectors 1 8, 239 adding vector quantities at right angles 21 adding vector quantities not at right angles 223 combination and resolution 1 8 horizontal and vertical components of vector A 18 representing vector quantities 1 920 resolving vectors 21 2 resultant vectors 20 scale drawing [graphical] approach 20 subtraction of vectors 23 vector quantities 1 81 9 velocity 27, 30 angular velocity 245, 2478 relationship between displacement, velocity and acceleration 355 velocity addition 51 61 7 velocity equation 353, 358 vibrations 582 forced vibrations 584 free vibrations 582 voltmeters 1 95, 21 4, 21 6 watt (W) 1 90 Watt, James 66 wave function 475 wave phenomena 353, 38790 waves 1 23, 1 24, 1 456, 1 648 boundary conditions 1 58, 1 63 comparison of travelling waves and stationary waves 1 58, 1 63 critical angle and total internal reflection 1 48 50 describing waves 1 256 displacementdistance graphs 1 267 displacementtime graphs 1 27 electromagnetic waves 1 23, 1 24, 1 31 3 HuygensFresnel principle 1 37 intensity 1 378 longitudinal waves 1 23, 1 25 measuring the refractive index 1 50 mechanical waves 1 24 motion of wavefronts 1 357 nodes and antinodes 1 589 plane polarized waves 1 41 polarization 1 34, 1 41 2 reflection and refraction of waves 1 46 refractive index and Snells law 1 46, 1 47 reversibility of light 1 48 standing waves 1 58 standing waves in pipes 1 61 3 standing waves on strings 1 5861 superposition 1 34, 3940 transverse waves 1 23, 1 245 travelling waves 1 23, 1 245 wave equation 1 23, 1 289 wave theory and the photoelectric effect 480 waveparticle duality 480, 482 wavefronts and rays 1 345 wavelength 1 23, 1 25 weak nuclear force (weak interaction) 290, 2989, 300, 301 white dwarfs 649, 656, 657 Wiens displacement law 329, 339, 652
WIMPs (Weakly Interacting Massive Particles) 681 2 wind generators 308, 3224 women scientists 654 work 61 , 634 rate of doing work 66 work done against resistive force 645 work function 477 worldlines 5227 X-rays 1 33 coherent scattering 627 collimation 630 Compton scattering 627 computed tomography (CT) 631 contrast 631 pair production 627 photoelectric effect 627 soft X-rays 629 X-ray images in medicine 62630 zone of stability 282
PH YSI CS
2 0 1 4 ED I TI O N
Supporting the latest syllabus at SL and HL, this 2014 edition was developed with the IB to most closely embody the IB way of teaching. The holistic approach to all aspects of the syllabus, including Nature of Science, encourages an inquiring, active approach to learning whilst integrated worked examples support exceptional achievement. Oxford course books are the only DP resources developed with the IB. This means that they are: The most accurate match to IB specifications Written by expert and experienced IB examiners and teachers Packed with accurate assessment support, directly from the IB Truly aligned with the IB philosophy, challenging learners with fresh and topical TOK Experiments strengthen practical understanding, cementing comprehension
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Unrivalled support for the new concept-based approach to learning
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Authors David Homer Michael Bowen-Jones
2 0 1 4 ED I TI O N
PH YSI CS C O U R S E C O M PA N I O N
David Homer Michael Bowen-Jones
Great Clarendon Street, Oxford, OX2 6DP, United Kingdom Oxford University Press is a department of the University of Oxford. It furthers the Universitys objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries Oxford University Press 2014 The moral rights of the authors have been asserted First published in 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above. You must not circulate this work in any other form and you must impose this same condition on any acquirer British Library Cataloguing in Publication Data Data available 978-0-19-839213-2 1 3 5 7 9 10 8 6 4 2 Paper used in the production of this book is a natural, recyclable product made from wood grown in sustainable forests. The manufacturing process conforms to the environmental regulations of the country of origin. Printed in Great Britain
Adrian Hillman/Shutterstock; p420: sciencephotos / Alamy; p427: Zigzag Mountain Art/Shutterstock; p445: Ziga Cetrtic/Shutterstock; p461: Shutterstock; p475: Kamioka Observatory, ICRR (Institute for Cosmic Ray Research), The University of Tokyo; p481: ANDREW LAMBERT PHOTOGRAPHY/SCIENCE PHOTO LIBRARY; p486: Paul Noth/Condenast Cartoons; p499: Kamioka Observatory, ICRR (Institute for Cosmic Ray Research), The University of Tokyo; p507: MIKKEL JUUL JENSEN / SCIENCE PHOTO LIBRARY; p540: NASA, ESA, and STScI; p549: PERY BURGE/SCIENCE PHOTO LIBRARY; p552: Ricardo high speed carbon-fibre flywheel; p570: Shutterstock; p584: iStock; p587: Travelpix Ltd/Getty Images; p593: Allison Herreid/Shutterstock; p594: GIPhotoStock/Science Source; p599a: DAVID PARKER/SCIENCE PHOTO LIBRARY; p599b: DAVID PARKER/SCIENCE PHOTO LIBRARY; p612: Donald Joski/Shutterstock; p614: Image courtesy of Celestron; p616: Israel Pabon/Shutterstock; p617: PETER BASSETT/ SCIENCE PHOTO LIBRARY; p620: Edward Kinsman/Getty Images; p623: US AIR FORCE/SCIENCE PHOTO LIBRARY; p627: iStock; p631: GUSTOIMAGES/ SCIENCE PHOTO LIBRARY; p641: ROBERT GENDLER/SCIENCE PHOTO LIBRARY; p642: ALMA/NAOJ/NRAO/EUROPEAN SOUTHERN OBSERVATORY/ NASA/ESA HUBBLE SPACE TELESCOPE/SCIENCE PHOTO LIBRARY; p643: Alan Dyer, Inc/Visuals Unlimited/Corbis; p644a: CHRIS COOK/SCIENCE PHOTO LIBRARY; p644b: ESO/NASA; p644c: GALEX, JPL-Caltech/NASA; p646: C. CARREAU/EUROPEAN SPACE AGENCY/SCIENCE PHOTO LIBRARY; p654: ROYAL ASTRONOMICAL SOCIETY/SCIENCE PHOTO LIBRARY; p657: NASA/ CXC/ SAO/NASA; p658: ROYAL OBSERVATORY, EDINBURGH/SCIENCE PHOTO LIBRARY; p664: ESA and the Planck Collaboration; p665: GSFC/ NASA; p667: ESO/VISTA/J. Emerson; p676: GSFC/NASA; p677: NASA/WMAP Science Team; p677: NATIONAL OPTICAL ASTRONOMY OBSERVATORIES/ SCIENCE PHOTO LIBRARY; p680: NASA/JPL-Caltech/S.Willner (HarvardSmithsonian CfA); p683: NASA/SCIENCE PHOTO LIBRARY; p689: Dorling Kindersley/Getty Images Artwork by Six Red Marbles and OUP
Acknowledgements
The authors and publisher are grateful for permission to reprint extracts from the following copyright material:
The publishers would like to thank the following for permissions to use their photographs:
P651 Nick Strobel, table Main Sequence Star Properties from www. astronomynotes.com, reprinted by permission.
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P499 John Updike, Telephone Poles and Other Poems from Cosmic Gall, (Deutsch, 1963), copyright 1959, 1963 by John Updike, reprinted by permission of Alfred A. Knopf, an imprint of the Knopf Doubleday Publishing Group, a division of Random House LLC, and Penguin Books Ltd, all rights reserved. Sources: P472 Albert Einstein Considerations concerning the fundaments of theoretical physics, Science 91:492 (1940) p112 Neutrino faster than light scientist resigns, BBC News 2013 BBC
Contents 1 Measurements and uncertainties Measurements in physics Uncertainties and errors Vectors and scalars
1 8 18
2 Mechanics Motion Forces Work, energy, and power Momentum
27 44 61 73
91 1 00
4 Oscillations and waves O scillations Travelling waves Wave characteristics Wave behaviour S tanding waves
D Astrophysics
S imple harmonic motion S ingle- slit diraction Intererence Resolution The D oppler eect
353 3 64 3 67 3 76 3 81
S tellar quantities S tellar characteristics and stellar evolution C osmology S tellar processes Further cosmology
3 91 405
Internal assessment
10 Fields (AHL) D escribing felds Fields at work
11 Electromagnetic induction (AHL)
3 Thermal physics Temperature and energy changes Modelling a gas
9 Wave phenomena (AHL)
115 1 23 1 34 1 45 1 58
Electromagnetic induction Power generation and transmission C apacitance
The interaction o matter with radiation Nuclear physics
5 Electricity and magnetism
A Relativity The beginnings o relativity Lorentz transormations S pacetime diagrams Relativistic mechanics General relativity
1 92 21 7
Index
693
42 7 43 9 45 5
475 492
5 07 513 522 529 534
B Engineering physics 2 45 257
7 Atomic, nuclear, and particle physics D iscrete energy and radioactivity Nuclear reactions The structure o matter
(with thanks to Mark Headlee for his assistance with this chapter) 687
227
6 Circular motion and gravitation C ircular motion Newtons law o gravitation
65 9 660 666 675
12 Quantum and nuclear physics (AHL)
E lectric felds Heating eect o an electric current E lectric cells Magnetic eects o electric currents
1 69
641
Rigid bodies and rotational dynamics Thermodynamics Fluids and uid dynamics Forced vibrations and resonance
5 49 559 5 70 5 82
C Imaging 2 67 2 82 2 90
Introduction to imaging Imaging instrumentation Fibre optics Imaging the body
5 93 608 62 0 62 6
8 Energy production E nergy sources Thermal energy transer
3 07 329
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Course book defnition
The IB Learner Profle
The IB D iploma Programme course books are resource materials designed to support students throughout their two- year D iploma Programme course o study in a particular subj ect. They will help students gain an understanding o what is expected rom the study o an IB D iploma Programme subj ect while presenting content in a way that illustrates the purpose and aims o the IB . They reect the philosophy and approach o the IB and encourage a deep understanding o each subj ect by making connections to wider issues and providing opportunities or critical thinking.
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iv
Inquirers: They develop their natural curiosity. They acquire the skills necessary to conduct inquiry and research and snow independence in learning. They actively enj oy learning and this love o learning will be sustained throughout their lives. Knowledgeable: They explore concepts, ideas, and issues that have local and global signifcance. In so doing, they acquire in-depth knowledge and develop understanding across a broad and balanced range o disciplines. Thinkers: They exercise initiative in applying thinking skills critically and creatively to recognize and approach complex problems, and make reasoned, ethical decisions. C ommunicators: They understand and express ideas and inormation confdently and creatively in more than one language and in a variety o modes o communication. They work eectively and willingly in collaboration with others. Princip led: They act with integrity and honesty, with a strong sense o airness, j ustice and respect or the dignity o the individual, groups and communities. They take responsibility or their own action and the consequences that accompany them. O p en-minded: They understand and appreciate their own cultures and personal histories, and are open to the perspectives, values and traditions o other individuals and communities. They are accustomed to seeking and evaluating a range o points o view, and are willing to grow rom the experience. C aring: They show empathy, compassion and respect towards the needs and eelings o others. They have a personal commitment to service, and to act to make a positive dierence to the lives o others and to the environment. Risk-takers: They approach unamiliar situations and uncertainty with courage and orethought, and have the independence o spirit to explore new roles, ideas, and strategies. They are brave and articulate in deending their belies.
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C ollusion is defned as supporting malpractice by another student. This includes:
allowing your work to be copied or submitted or assessment by another student
duplicating work or dierent assessment components and/or diploma requirements.
O ther orms o malp ractice include any action that gives you an unair advantage or aects the results o another student. Examples include, taking unauthorized material into an examination room, misconduct during an examination and alsiying a C AS record.
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Using your IB Physics Online Resources What is Kerboodle? Kerboodle is an online learning platorm. I your school has a subscription to IB Physics Kerboodle O nline Resources you will be able to access a huge bank o resources, assessments, and presentations to guide you through this course.
What is in your Kerboodle Online Resources? There are three main areas or students on the IB Physics Kerboodle: planning, resources, and assessment.
Resources There a hundreds o extra resources available on the IB Physics Kerboodle O nline. You can use these at home or in the classroom to develop your skills and knowledge as you progress through the course. Watch videos and animations o experiments, difcult concepts, and science in action. Hundreds o worksheets read articles, perorm experiments and simulations, practice your skills, or use your knowledge to answer questions. Look at galleries o images rom the book and see their details close up. Find out more by looking at recommended sites on the Internet, answer questions, or do more research.
Planning B e prepared or the practical work and your internal assessment with extra resources on the IB Physics Kerboodle online. Learn about the dierent skills that you need to perorm an investigation. Plan and prepare experiments o your own. Learn how to analyse data and draw conclusions successully and accurately.
One of hundreds of worksheets.
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Practical skills presentation.
Assessment C lick on the assessment tab to check your knowledge or revise or your examinations. Here you will fnd lots o interactive quizzes and examstyle practice questions. Formative tests: use these to check your comprehension, theres one auto-marked quiz or every sub-topic. E valuate how confdent you eel about a sub-topic, then complete the test. You will have two attempts at each question and get eedback ater every question. The marks are automatically reported in the markbook, so you can see how you progress throughout the year. Summative tests: use these to practice or your exams or as revision, theres one auto- marked quiz or every topic. Work through the test as i it were an examination go back and change any questions you arent sure about until you are happy, then submit the test or a fnal mark. The marks are automatically reported in the markbook, so you can see where you may need more practice. Assessment practice: use these to practice answering the longer written questions you will come across when you are examined. These worksheets can be printed out and perormed as a timed test.
Don't forget! You can also fnd extra resources on our ree website www.oxfordsecondary.co.uk/ib-physics Here you can fnd all o the answers and even more practice questions. vii
Introduction Physics is one o the earliest academic disciplines known i you include observational astronomy, possibly the oldest. In physics we analyse the natural world to develop the best understanding we can o how the universe and its constituent parts interrelate. Our aim as physicists is to develop models that correspond to what is observed in the laboratory and beyond. These models come in many orms: some may be quantitative and based on mathematics; some may be qualitative and give a verbal description o the world around us. But, whatever orm the models take, physicists must all agree on their validity beore they can be accepted as part o our physical description o the universe. Models used by physicists are linked by a coherent set o principles known as concepts. These are overarching ideas that link the development o the subject not only within a particular physical topic (or example, orces in mechanics) but also between topics (or example, the common mathematics that links radioactive decay and capacitor discharge) . In studying physics, take every opportunity to understand a new concept when you meet it. When the concept occurs elsewhere your prior knowledge will make the later learning easier. This book is designed to support your learning o physics within group 4 o the IB Diploma Programme. Like all the disciplines represented in this subject group it has a thorough basis in the acts and concepts o science, but it also draws out the nature o science. This is to give you a better understanding o what it means to be a scientist, so that you can, or example, identiy shortcomings in scientifc topics presented to you in the media or elsewhere. Not everyone taking IB Physics will want to go on to be a physicist or engineer, but all citizens need to have an awareness o the importance o science in modern society. The structure o this book needs an explanation; all o the topics include the ollowing elements:
Understanding The specifcs o the content requirements or each sub- topic are covered in detail. C oncepts are presented in ways that will promote enduring understanding.
Investigate! These sections describe practical work you can undertake. You may need to modiy these
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experiments slightly to suit the apparatus in your school. These are a valuable opportunity to build the skills that are assessed in IA ( see page 687) .
Nature of science These sections help you to develop your understanding by studying a specifc illustrative example or learning about a signifcant experiment in the history o physics. Here you can explore the methods o science and some o the knowledge issues that are associated with scientifc endeavour. This is done using careully selected examples, including research that led to paradigm shits in our understanding o the natural world.
Theory of Knowledge These short sections have articles on scientifc questions that arise rom Theory o knowledge. We encourage you draw on these examples o knowledge issues in your TOK essays. O course, much o the material elsewhere in the book, particularly in the nature o science sections, can be used to prompt TOK discussions.
Worked example These are step-by-step examples o how to answer questions or how to complete calculations. You should review them careully, preerably ater attempting the question yoursel.
End -of-Topic Questions At the end o each topic you will fnd a range o questions, including both past IB Physics exam questions and new questions. Answers can be ound at www. oxordsecondary.co.uk/ib- physics Authors do not write in isolation. In particular, our ways o describing and explaining physics have been honed by the students we have been privileged to teach over the years, and by colleagues who have challenged our ways o thinking about the subj ect. O ur thanks go to them all. More specifcally, we thank Jean Godin or much sound advice during the preparation o this text. Any errors are, o course, our responsibility. Last but in no sense least, we thank our wives, Adele and B renda, or their ull support during the preparation o this book. We could not have completed it without their understanding and enormous patience. M Bowen-Jones D Homer
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M EASU REM EN TS AN D U N CERTAI N TI ES
Introduction This topic is different from other topics in the course book. The content discussed here will be used in most aspects of your studies in physics. You will come across many aspects of this work
in the context of other subj ect matter. Although you may wish to do so, you would not be expected to read this topic in one go, rather you would return to it as and when it is relevant.
1.1 Measurements in physics Understanding Fundamental and derived SI units Scientifc notation and metric multipliers Signifcant fgures Orders o magnitude Estimation
Applications and skills Using SI units in the correct ormat or all
required measurements, fnal answers to calculations and presentation o raw and processed data Using scientifc notation and metric multipliers Quoting and comparing ratios, values, and approximations to the nearest order o magnitude Estimating quantities to an appropriate number o signifcant fgures
Nature of science In physics you will deal with the qualitative and the quantitative, that is, descriptions o phenomena using words and descriptions using numbers. When we use words we need to interpret the meaning and one person's interpretation will not necessarily be the same as another's. When we deal with numbers (or equations) , providing we have learned the rules, there is no mistaking someone else's meaning. It is likely that some readers will be more comortable
with words than symbols and vice-versa. It is impossible to avoid either methodology on the IB Diploma course and you must learn to be careul with both your numbers and your words. In examinations you are likely to be penalized by writing contradictory statements or mathematically incorrect ones. At the outset o the course you should make sure that you understand the mathematical skills that will make you into a good physicist.
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Quantities and units Physicists deal with p hysical quantities, which are those things that are measureable such as mass, length, time, electrical current, etc. Quantities are related to one another by equations such as m which is the symbolic orm o saying that density is the ratio = __ V o the mass o an obj ect to its volume. Note that the symbols in the equation are all written in italic ( sloping) onts this is how we can be sure that the symbols represent quantities. Units are always written in Roman ( upright) ont because they sometimes share the same symbol with a quantity. S o m represents the quantity mass but m represents the unit metre . We will use this convention throughout the course book, and it is also the convention used by the IB .
Nature of science The use of symbols
Greek
The use o Greek letters such as rho () is very common in physics. There are so many quantities that, even using the 5 2 Arabic letters (lower case and capitals) , we soon run out o unique symbols. Sometimes symbols such as d and x have multiple uses, meaning that Greek letters have become just one way o trying to tie a symbol to a quantity uniquely. O course, we must consider what happens when we run out o Greek letters too we then use Russian ones rom the C yrillic alphabet.
alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu
Russian
nu ksi omicron pi rho sigma tau upsilon phi chi psi omega
Fundamental quantities are those quantities that are considered to be so basic that all other quantities need to be expressed in terms o them. m In the density equation = __ only mass is chosen to be undamental V ( volume being the product o three lengths) , density and volume are said to be derived quantities. It is essential that all measurements made by one person are understood by others. To achieve this we use units that are understood to have unambiguous meaning. The worldwide standard or units is known as S I Systme international dunits. This system has been developed rom the metric system o units and means that, when values o scientifc quantities are communicated between people, there should never be any conusion. The SI defnes both units and prefxes letters used to orm decimal multiples or sub- multiples o the units. The units themselves are classifed as being either undamental ( or base) , derived, and supplementary. There are only two supplementary units in S I and you will meet only one o these during the D iploma course, so we might as well mention them frst. The two supplementary units are the radian ( rad) the unit o angular measurement and the steradian ( sr) the unit o solid angle . The radian is a useul alternative to the degree and is defned as the angle subtended by an arc of a circle having the same length as the radius,
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1 . 1 M E A S U R E M E N T S I N P H YS I C S as shown in fgure 1 . We will look at the radian in more detail in S ub- topic 6 . 1 . The steradian is the three- dimensional equivalent o the radian and uses the idea o mapping a circle on to the surace o a sphere.
r r 1 rad r
Fundamental and derived units In S I there are seven fundamental units and you will use six o these on the D iploma course ( the seventh, the candela, is included here or completeness) . The undamental quantities are length, mass, time, electric current, thermodynamic temperature, amount o substance, and luminous intensity. The units or these quantities have exact defnitions and are precisely reproducible, given the right equipment. This means that any quantity can, in theory, be compared with the undamental measurement to ensure that a measurement o that quantity is accurate. In practice, most measurements are made against more easily achieved standards so, or example, length will usually be compared with a standard metre rather than the distance travelled by light in a vacuum. You will not be expected to know the defnitions o the undamental quantities, but they are provided here to allow you to see j ust how precise they are.
Figure 1 Defnition o the radian.
Figure 2 The international prototype kilogram.
metre (m) : the length o the path travelled by light in a vacuum during 1 o a second. a time interval o _________ 2 9 9 7 9 2 45 8 kilogram (kg) : mass equal to the mass o the international prototype o the kilogram kept at the B ureau International des Poids et Mesures at S vres, near Paris. second (s) : the duration o 9 1 92 63 1 770 periods o the radiation corresponding to the transition between the two hyperfne levels o the ground state o the caesium-1 3 3 atom. amp ere (A) : that constant current which, i maintained in two straight parallel conductors o infnite length, negligible circular cross- section, and placed 1 m apart in vacuum, would produce between these conductors a orce equal to 2 1 0 7 newtons per metre o length. 1 o the thermodynamic temperature o the kelvin (K) : the raction _____ 2 73 .1 6 triple point o water.
mole (mol) : the amount o substance o a system that contains as many elementary entities as there are atoms in 0.01 2 kg o carbon1 2 . When the mole is used, the elementary entities must be specifed and may be atoms, molecules, ions, electrons, other particles, or specifed groups o such particles. candela (cd) : the luminous intensity, in a given direction, o a source that emits monochromatic radiation o requency 5 40 1 0 1 2 hertz and 1 that has a radiant intensity in that direction o ___ watt per steradian. 683 All quantities that are not undamental are known as derived and these can always be expressed in terms o the undamental quantities through a relevant equation. For example, speed is the rate o change o distance with s respect to time or in equation orm v = ___ (where s means the change in t distance and t means the change in time) . As both distance (and length) and time are undamental quantities, speed is a derived quantity.
TOK Deciding on what is fundamental Who has made the decision that the undamental quantities are those o mass, length, time, electrical current, temperature, luminous intensity, and amount o substance? In an alternative universe it may be that the undamental quantities are based on orce, volume, requency, potential dierence, specifc heat capacity, and brightness. Would that be a drawback or would it have meant that humanity would have progressed at a aster rate?
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Note I you are reading this at the start o the course, it may seem that there are so many things that you might not know; but, take heart, Rome was not built in a day and soon much will come as second nature. When we write units as m s 1 and m s 2 it is a more efective and preerable way to writing what you may have written in the past as m/s and m/s2 ; both orms are still read as metres per second and metres per second squared.
The units used or undamental quantities are unsurprisingly known as undamental units and those or derived quantities are known as derived units. It is a straightorward approach to be able to express the unit o any quantity in terms o its undamental units, provided you know the equation relating the quantities. Nineteen undamental quantities have their own unit but it is also valid, i cumbersome, to express this in terms o undamental units. For example, the S I unit o pressure is the pascal ( Pa) , which is expressed in undamental units as m 1 kg s 2 .
Nature o science Capitals or lower case? Notice that when we write the unit newton in ull, we use a lower case n but we use a capital N or the symbol or the unit unortunately some word processors have deault setting to correct this so take care! All units written in ull should start with a lower case letter, but those that have been derived in honour o a scientist will have a symbol that is a capital letter. In this way there is no conusion between the scientist and the unit: Newton reers to Sir Isaac Newton but newton means the unit. Sometimes units are abbreviations o the scientists surname, so amp (which is a shortened orm o ampre anyway) is named ater Ampre, the volt ater Volta, the arad, Faraday, etc.
Example o how to relate undamental and derived units The unit o orce is the newton ( N) . This is a derived unit and can be expressed in terms o undamental units as kg m s 2 . The reason or this is that orce can be defned as being the product o mass and acceleration or F = ma. Mass is a undamental quantity but acceleration is not. Acceleration is the rate o change o velocity or a = v where v t represents the change in velocity and t the change in time. Although time is a undamental quantity, velocity is not so we need to take another step in defning velocity in undamental quantities. Velocity is the rate o change o displacement ( a quantity that we will discuss later in the topic but, or now, it simply means distance in a given direction) . s So the equation or velocity is v = ___ with s being the change in t displacement and t again being the change in time. D isplacement ( a length) and time are both undamental, so we are now in a position to put N into undamental units. The unit o velocity is m s 1 and these are already undamental there is no shortened orm o this. The units o acceleration will thereore be those o velocity divided by time and so m s will be ____ which is written as m s 2 . So the unit o orce will be the unit s o mass multiplied by the unit o acceleration and, thereore, be kg m s 2 . This is such a common unit that it has its own name, the newton, ( N kg m s 2 a mathematical way o expressing that the two units are identical) . So i you are in an examination and orget the unit o orce you could always write kg m s 2 ( i you have time to work it out! ) .
___
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Figure 3 Choosing fundamental units in an alternative universe.
Signifcant fgures C alculators usually give you many digits in an answer. How do you decide how many digits to write down or the fnal answer?
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1 . 1 M E A S U R E M E N T S I N P H YS I C S
S cientists use a method o rounding to a certain number o signifcant fgures ( oten abbreviated to s. .) . S ignifcant here means meaningul. C onsider the number 84 072 , the 8 is the most signifcant digit, because it tells us that the number is eighty thousand and something. The 4 is the next most signifcant telling us that there are also our thousand and something. Even though it is a zero, the next digit, the 0, is the third most signifcant digit here. When we ace a decimal number such as 0. 002 45 , the 2 is the most signifcant digit because it tells us that the number is two thousandth and something. The 4 is the next most signifcant, showing that there are our ten thousandths and something. I we wish to express this number to two signifcant fgures we need to round the number rom three to two digits. I the last number had been 0.00244 we would have rounded down to 0.0024 and i it had been 0.00246 we would have rounded up to 0.0025 . However, it is a 5 so what do we do? In this case there is equal justifcation or rounding up and down, so all you really need to be is consistent with your choice or a set o fgures you can choose to round up or down. Oten you will have urther digits to help you, so i the number had been 0.002 45 1 and you wanted it rounded to two signifcant fgures it would be rounded up to 0.0025 .
Some rules or using signifcant fgures
A digit that is not a zero will always be signifcant 345 is three signifcant fgures (3 s..) .
Zeros that occur sandwiched between non- zero digits are always signifcant 3 405 ( 4 s.. ) ; 1 0.3 405 ( 6 s..) .
Non- sandwiched zeros that occur to the let o a non- zero digit are not signifcant 0.3 45 ( 3 s.) ; 0. 03 4 ( 2 s.. ) .
Zeros that occur to the right o the decimal point are signifcant, provided that they are to the right o a non- zero digit 1 .03 4 ( 4 s..) ; 1 .00 ( 3 s..) ; 0. 3 45 00 ( 5 s. .) ; 0. 003 ( 1 s..) .
When there is no decimal point, trailing zeros are not signifcant ( to make them signifcant there needs to be a decimal point) 400 ( 1 s..) ; 400. ( 3 s..) but this is rarely written.
Scientifc notation O ne o the ascinations or physicists is dealing with the very large ( e.g. the universe) and the very small ( e.g. electrons) . Many physical constants ( quantities that do not change) are also very large or very small. This presents a problem: how can writing many digits be avoided? The answer is to use scientifc notation. The speed o light has a value o 299 792 45 8 m s 1 . This can be rounded to three signifcant fgures as 300 000 000 m s 1 . There are a lot o zeros in this and it would be easy to miss one out or add another. In scientifc notation this number is written as 3.00 1 0 8 m s 1 (to three signifcant fgures) . Let us analyse writing another large number in scientifc notation. The mass o the Sun to our signifcant fgures is 1 989 000 000 000 000 000 000 000 000 000 kg ( that is 1 989 and twenty- seven zeros) . To convert
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S this into scientifc notation we write it as 1 .989 and then we imagine moving the decimal point 3 0 places to the let ( remember we can write as many trailing zeros as we like to a decimal number without changing it) . This brings our number back to the original number and so it gives the mass o the Sun as 1 . 989 1 0 30 kg. A similar idea is applied to very small numbers such as the charge on the electron, which has an accepted value o approximately 0.000 000 000 000 000 000 1 602 coulombs. Again we write the coefcient as 1 .602 and we must move the decimal point 1 9 places to the right in order to bring 0.000 000 000 000 000 000 1 602 into this orm. The base is always 1 0 and moving our decimal point to the right means the exponent is negative. We can write this number as 1 .602 1 0 1 9 C . Apart rom avoiding making mistakes, there is a second reason why scientifc notation is preerable to writing numbers in longhand. This is when we are dealing with several numbers in an equation. In writing the value o the speed o light as 3 .00 1 0 8 m s 1 , 3 . 00 is called the coefcient o the number and it will always be a number between 1 and 1 0. The 1 0 is called the base and the 8 is the exponent. There are some simple rules to apply:
When adding or subtracting numbers the exponent must be the same or made to be the same.
When multiplying numbers we add the exponents.
When dividing numbers we subtract one exponent rom the other.
When raising a number to a power we raise the coefcient to the power and multiply the exponent by the power.
Worked examples In these examples we are going to evaluate each o the calculations. 1
1 .40 1 0 6 + 3 .5 1 0 5
S o we write this product as: 7.8 1 0 3
Solution
4
These must be written as 1 . 40 1 0 6 + 0.3 5 1 0 6 so that both numbers have the same exponents.
Solution
They can now be added directly to give 1 .75 1 0 6 2
3 .7 1 0 5 2 .1 1 0 8
Solution
So we write this product as: 7. 8 1 0 3
4.8 1 0 5 _ 3 .1 1 0 2
The coefcients are divided and the exponents are subtracted so we have: 4.8 3 . 1 = 1 .5 48 ( which we round to 1 .5 ) And 5 2 = 3
The coefcients are multiplied and the exponents are added, so we have: 3.7 2.1 = 7.77 (which we round to 7.8 to be in line with the data something we will discuss later in this topic) and: 5 + 8 = 1 3 13
3 .7 1 0 5 2 .1 1 0 8
Solution Again the coefcients are multiplied and the exponents are added, so we have: 3 .7 2 . 1 = 7. 8
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Here the exponents are subtracted ( since the 8 is negative) to give: 5 8 = 3
This makes the result o the division 1 .5 1 0 3 5
( 3 .6 1 0 7 ) 3
Solution We cube 3 .6 and 3 .6 3 = 46.7 And multiply 7 by 3 to give 2 1 This gives 46.7 1 0 21 , which should become 4.7 1 0 22 in scientifc notation.
1 . 1 M E A S U R E M E N T S I N P H YS I C S
Metric multipliers (prefxes)
Factor 10 24 10 21 10 18 10 15 10 12 10 9 10 6 10 3 10 2 10 1 10 -1 10 -2 10 -3 10 -6 10 -9 10 -12 10 -15 10 -18 10 -21 10 -24
S cientists have a second way o abbreviating units: by using metric multipliers ( usually called prefxes ) . An S I prefx is a name or associated symbol that is written beore a unit to indicate the appropriate power o 1 0. S o instead o writing 2 . 5 1 0 1 2 J we could alternatively write this as 2 . 5 TJ ( teraj oule) . Figure 4 gives the 2 0 S I prefxes these are provided or you as part o the data booklet used in examinations.
Orders of magnitude An important skill or physicists is to understand whether or not the physics being considered is sensible. When perorming a calculation in which someones mass was calculated to be 5 000 kg, this should ring alarm bells. S ince average adult masses ( weights) will usually be 6090 kg, a value o 5 000 kg is an impossibility. A number rounded to the nearest power o 1 0 is called an order of magnitude. For example, when considering the average adult human mass: 6080 kg is closer to 1 00 kg than 1 0 kg, making the order o magnitude 1 0 2 and not 1 0 1 . O course, we are not saying that all adult humans have a mass o 1 00 kg, simply that their average mass is closer to 1 00 than 1 0. In a similar way, the mass o a sheet o A4 paper may be 3 . 8 g which, expressed in kg, will be 3 . 8 1 0 3 kg. S ince 3 . 8 is closer to 1 than to 1 0, this makes the order o magnitude o its mass 1 0 3 kg. This suggests that the ratio o adult mass to the mass o a piece o paper (should 10 you wish to make this comparison) = ____ = 1 0 2 ( 3) = 1 0 5 = 1 00 000. In 10 other words, an adult human is 5 orders o magnitude ( 5 powers o 1 0) heavier than a sheet o A4 paper.
Name yotta zetta exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto zepto yocto
Symbol
d c m n p f a z y
Figure 4 SI metric multipliers.
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Estimation E stimation is a skill that is used by scientists and others in order to produce a value that is a useable approximation to a true value. E stimation is closely related to fnding an order o magnitude, but may result in a value that is more precise than the nearest power o 1 0. Whenever you measure a length with a ruler calibrated in millimetres you can usually see the whole number o millimetres but will need to 1 estimate to the next __ mm you may need a magniying glass to help 10 you to do this. The same thing is true with most non- digital measuring instruments. S imilarly, when you need to fnd the area under a non- regular curve, you cannot truly work out the actual area so you will need to fnd the area o a rectangle and estimate how many rectangles there are. Figure 5 shows a graph o how the orce applied to an obj ect varies with time. The area under the graph gives the impulse ( as you will see in Topic 2 ) . There are 2 6 complete or nearly complete yellow squares under the curve and there are urther partial squares totalling about our ull squares in all. This gives about 3 0 ull squares under the curve. E ach curve has an area equivalent to 2 N 1 s = 2 N s. This gives an estimate o about 60 N s or the total impulse.
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S In an examination, estimation questions will always have a tolerance given with the accepted answer, so in this case it might be ( 60 2 ) N s.
12 These two partial squares may be combined to approximate to a whole square, etc.
force/N
10 8 6 4 2 0 0
1
2
3
4
5
6
7
8
time/s
Figure 5
1.2 Uncertainties and errors Understanding Random and systematic errors Absolute, ractional, and percentage uncertainties Error bars Uncertainty o gradient and intercepts
Nature of science In Sub-topic 1.1 we looked at the how we defne the undamental physical quantities. Each o these is measured on a scale by comparing the quantity with something that is precisely reproducible. By precisely reproducible do we mean exact? The answer to this is no. I we think about the defnition o the ampere, we will measure a orce o 2 10 7 N. I we measure it to be 2.1 10 7 it doesnt invalidate the measurement since the defnition is given to just one signifcant fgure. All measurements have their limitations or uncertainties and it is important that both the measurer and the person working with the measurement understand what the limitations are. This is why we must always consider the uncertainty in any measurement o a physical quantity.
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Applications and skills Explaining how random and systematic errors
can be identifed and reduced Collecting data that include absolute and/or ractional uncertainties and stating these as an uncertainty range (using ) Propagating uncertainties through calculations involving addition, subtraction, multiplication, division, and raising to a power Using error bars to calculate the uncertainty in gradients and intercepts
Equations Propagation o uncertainties: I: y = a b then: y = a + b ab I: y = _____ c y
______ ______ ______ a c b then: ______ y = a + b + c I: y = a n y ______ a then: _______ y = n a
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
Uncertainties in measurement Introduction No experimental quantity can be absolutely accurate when measured it is always subj ect to some degree o uncertainty. We will look at the reasons or this in this section. There are two types o error that contribute to our uncertainty about a reading systematic and random.
Systematic errors As the name suggests, these types o errors are due to the system being used to make the measurement. This may be due to aulty apparatus. For example, a scale may be incorrectly calibrated either during manuacture o the equipment, or because it has changed over a period o time. Rulers warp and, as a result, the divisions are no longer symmetrical. A timer can run slowly i its quartz crystal becomes damaged ( not because the battery voltage has allen when the timer simply stops) .
Figure 1 Zero error on digital calliper.
When measuring distances rom sealed radioactive sources or lightdependent resistors ( LD Rs) , it is hard to know where the source is actually positioned or where the active surace o the LD R is. The zero setting on apparatus can drit, due to usage, so that it no longer reads zero when it should this is called a zero error. Figure 1 shows a digital calliper with the j aws closed. This should read 0.000 mm but there is a zero error and it reads 0.01 mm. This means that all readings will be 0.01 mm bigger than they should be. The calliper can be reset to zero or 0.01 mm could be subtracted rom any readings made. O ten it is not possible to spot a systematic error and experimenters have to accept the reading on their instruments, or else spend signifcant eort in making sure that they are re-calibrated by checking the scale against a standard scale. Repeating a reading never removes the systematic error. The real problem with systematic errors is that it is only possible to check them by perorming the same task with another apparatus. I the two sets give the same results, the likelihood is that they are both perorming well; however, i there is disagreement in the results a third set may be needed to resolve any dierence. In general we deal with zero errors as well as we can and then move on with our experimentation. When systematic errors are small, a measurement is said to be accurate.
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Nature of science Systematic errors Uncertainty when using a 300 mm ruler may be quoted to 0.5 mm or 1 mm depending on your view o how precisely you can gauge the reading. To be on the sae side you might wish to use the larger uncertainty and then you will be sure that the reading lies within your bounds.
Figure 2 Millimeter (mm) scale on ruler.
You should make sure you observe the scale rom directly above and at right angles to the plane o the ruler in order to avoid parallax errors.
Figure 4 Analogue scale.
Figure5 Digital scale.
The digital ammeter in fgure 5 gives a value o 0.27 A which should be recorded as (0.27 0.01 ) A. In each o these examples the uncertainty is quoted to the same p recision ( number o decimal places) as the reading it is essential to do this as the number o decimal places is always indicative o precision. When we write an energy value as being 8 J we are implying that it is ( 8 1 ) J and i we write it as 8. 0 J it implies a precision o 0. 1 J.
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7 8
9 10
The meter in fgure 4 shows an analogue ammeter with a airly large scale there is justifcation in giving this reading as being (40 5 ) A.
Figure 3 Parallax error.
Random errors Random errors can occur in any measurement, but crop up most requently when the experimenter has to estimate the last signifcant fgure when reading a scale. I an instrument is insensitive then it may be difcult to j udge whether a reading would have changed in dierent circumstances. For a single reading the uncertainty could well be better than the smallest scale division available. B ut, since you are determining the maximum possible range o values, it is a sensible precaution to use this larger precision. D ealing with digital scales is a problem the likelihood is that you have really no idea how precisely the scales are calibrated. C hoosing the least signifcant digit on the scale may severely underestimate the uncertainty but, unless you know the manuacturers data regarding calibration, it is probably the best you can do. When measuring a time manually it is inappropriate to use the precision o the timer as the uncertainty in a reading, since your reaction time is likely to be ar greater than this. For example, i you timed twenty oscillations o a pendulum to take 1 6. 2 7 s this should be recorded as being ( 1 6.3 0.1 ) s. This is because your reaction time dominates the precision o the timer. I you know that your reaction time is greater than 0.1 s then you should quote that value instead o 0.1 s.
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1 . 2 U N CE R TAI N TI E S AN D E R R O R S
The best way o handling random errors is to take a series o repeat readings and fnd the average o each set o data. Hal the range o the values will give a value that is a good approximation to the statistical value that more advanced error analysis provides. The range is the largest value minus the smallest value. Readings with small random errors are said to be p recise ( this does not mean they are accurate, however) .
Worked examples 1
In measuring the angle o reraction at an airglass interace or a constant angle o incidence the ollowing results were obtained ( using a protractor with a precision o 1 o ) :
T/C
Solution The mean o these values is 45 . 4 and the range is ( 47 44) = 3 . Hal the range is 1 .5 . How then do we record our overall value or the angle o reraction? S ince the precision o the protractor is 1 , we should quote our mean to a whole number ( integral) value and it will round down to 45 . We should not minimize our uncertainty unrealistically and so we should round this up to 2 . This means that the angle o reraction should be recorded as 45 2 . 2
The diagram below shows the position o the meniscus o the mercury in a mercury-in- glass thermometer.
1
2
3
4
5
6
7
8
9
10
Express the temperature and its uncertainty to an appropriate number o signifcant fgures.
45 , 47, 46, 45 , 44 How should we express the angle o reraction?
0
Solution The scale is calibrated in degrees but they are quite clear here, so it is reasonable to expect a precision o 0.5 C . The meniscus is closer to 6 than to 6.5 ( although that is a j udgement decision) so the values should be recorded as ( 6.0 0. 5 ) C . Remember the measurement and the uncertainty should be to the same number o decimal places. 3
A student takes a series o measurements o a certain quantity. He then averages his measurements. What aspects o systematic and random uncertainties is he addressing by taking repeats and averages?
Solution S ystematic errors are not dealt with by means o repeat readings, but taking repeat readings and averaging them should cause the average value to be closer to the true value than a randomly chosen individual measurement.
Absolute and fractional uncertainties The values o uncertainties that we have been looking at are called absolute uncertainties. These values have the same units as the quantity and should be written to the same number o decimal places. Dividing the uncertainty by the value itsel leaves a dimensionless quantity (one with no units) and gives us the fractional uncertainty. Percentaging the ractional uncertainty gives the percentage uncertainty.
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Worked example C alculate the absolute, ractional, and percentage uncertainties or the ollowing measurements o a orce, F:
range = ( 2 .8 2 .5 ) N = 0. 3 N, giving an absolute uncertainty o 0.1 5 N that rounds up to 0.2 N
2 .5 N, 2 .8 N, 2 .6 N
We would write our value or F as ( 2 .6 0.2 ) N
Solution
0.2 = 0.077 and the the ractional uncertainty is ___ 2 .6 percentage uncertainty will be 0.077 1 00% = 7.7%
2 .5 N + 2 .8 N + 2 .6 N
mean value = ________________ = 2 .63 N, this is 3 rounded down to 2.6 N
Propagation of uncertainties O ten we measure quantities and then use our measurements to calculate other quantities with an equation. The uncertainty in the calculated value will be determined rom a combination o the uncertainties in the quantities that we have used to calculate the value rom. This is known as p rop agation of uncertainties. There are some simple rules that we can apply when we are propagating uncertainties. In more advanced treatment o this topic we would demonstrate how these rules are developed, but we are going to ocus on your application o these rules here ( since you will never be asked to prove them and you can look them up in a text book or on the Internet i you want urther inormation) . In the uncertainty equations discussed next, a, b, c, etc. are the quantities and a, b, c, etc. are the absolute uncertainties in these quantities. 11
Addition and subtraction
12
This is the easiest o the rules because when we add or subtract quantities we always add their absolute uncertainties. When a = b + c
or a = b c
then a = b + c
13
In order to use these relationships dont orget that the quantities being added or subtracted must have the same units.
14
S o i we are combining two masses m 1 and m 2 then the total mass m will be the sum o the other two masses.
15
m 1 = ( 2 00 1 0) g and m 2 = ( 1 00 1 0) g so m = 3 00 g and m = 2 0 g meaning we should write this as:
16
m = ( 3 00 2 0) g
17 18
We use subtraction more oten than we realise when we are measuring lengths. When we set the zero o our ruler against one end o an object we are making a judgement o where the zero is positioned and this really means that the value is (0.0 0.5) mm.
19
A ruler is used to measure a metal rod as shown in fgure 6. The length is ound by subtracting the smaller measurement rom the larger one. The uncertainty or each measurement is 0.5 mm.
20
Larger measurement = 1 95 .0 mm S maller measurement = 1 1 8. 5 mm
12
Figure 6 Measuring a length.
Length = ( 76.5 1 .0) mm as the uncertainty is 0. 5 mm + 0.5 mm
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
Nature of science Subtracting values When subtraction is involved in a relationship you need to be particularly careful. The resulting quantity becomes smaller in size (because of subtraction) , while the absolute uncertainty becomes larger (because of addition) . Imagine two values that are subtracted: b = 4.0 0.1 and c =3.0 0.1 . We wont concern ourselves with what these quantities actually are here.
If a = b c then a = 1 .0 and since a = b + c then a = 0.2 We have gone from two values in which the percentage uncertainty is 2.5 % and 3.3% respectively to a calculated value with uncertainty of 20%. Now that really is propagation of uncertainties!
Multiplication and division When we multiply or divide quantities we add their fractional or percentage uncertainties, so: when a = bc or a = __bc or a = __bc a b c ___ ___ then ___ a = b + c
There are very few relationships in physics that do not include some form of multiplication or division. m We have seen that density is given by the expression = __ where V m is the mass of a sample of the substance and V is its volume. For a particular sample, the percentage uncertainty in the mass is 5 % and for the volume is 1 2 % .
The percentage uncertainty in the calculated value of the density will therefore be 1 7% . If the sample had been cubical in shape and the uncertainty in each of the sides was 4% we can see how this brings about a volume with uncertainty of 1 2 % : For a cube the volume is the cube of the side length ( V = l 3 = l l l) V so ___ = V
__ + __ + __ = 4% + 4% + 4% = 1 2 % l l
l l
l l
This example leads us to:
Raising a quantity to a power V l From the cube example you might have spotted that ___ = 3 __ V l
This result can be generalized so that when a = b n ( where n can be a positive or negative whole, integral, or decimal number)
a b ___ then ___ a = n b
The modulus sign is included as an alternative way of telling us that the uncertainty can be either positive or negative.
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Worked example The period T o oscillation o a mass __ m on a spring, m having spring constant k is T = 2 __ k Dont worry about what these quantities actually mean at this stage. The uncertainty in k is 1 1 % and the uncertainty in m is 5 % . C alculate the approximate uncertainty in a value or T o 1 .2 0 s.
Solution First lets adj ust the equation a little we can write it as
( )
m T = 2 __ k
1 __ 2
()
which is o the orm a = 2 __bc
n
Although we will truncate , we can really write it to as many signifcant fgures as we wish and so the percentage uncertainty in , as in 2 will be zero.
Using the division and power relationships: a b c m k T 1 ___ 1 ___ ___ ___ ___ ___ __ __ a = n b + n c or here T = 2 m + 2 k
so the percentage uncertainty in T will be hal that in m + hal that in k. This means that the percentage error in T = 0.5 5 % + 0.5 1 1 % = 8% I the measured value o T is 1 .2 0 s then the 8 absolute uncertainty is 1 .20 ___ = 0. 096 This 1 00 rounds up to 0.1 0 and so we quote T as being ( 1 . 2 0 0.1 0) s. Remember that the quantity and the uncertainty must be to the same number o decimal places and so the zeros are important, as they give us the precision in the value.
Drawing graphs speed/m s 1
3.0 2.0 1.0 0 0
0.1
0.2 0.3 time/s
0.4
Figure 7 Error bars.
An important j ustifcation or experimental work is to investigate the relationship between physical quantities. O ne set o values is rarely very revealing even i it can be used to calculate a physical constant, such as dividing the potential dierence across a resistor by the current in the resistor to fnd the resistance. Although the calculation does tell you the resistance or one value o current, it says nothing about whether the resistance depends upon the current. Taking a series o values would tell you i the resistance was constant but, with the expected random errors, it would still not be defnitive. B y plotting a graph and drawing the line o best ft the pattern o results is ar easier to spot, whether it is linear or some other relationship.
Error bars In plotting a point on a graph, uncertainties are recognized by adding error bars. These are vertical and horizontal lines that indicate the possible range o the quantity being measured. S uppose at a time o ( 0. 2 0.05 ) s the speed o an obj ect was ( 1 .2 0.2 ) m s 1 this would be plotted as shown in fgure 7.
The value could lie anywhere inside this rectangle
speed/m s 1
3.0
This means that the value could possibly be within the rectangle that touches the ends o the error bars as shown in fgure 8. This is the zone of uncertainty or the data point. A line o best ft should be one that spreads the points so that they are evenly distributed on both sides o the line and also passes through the error bars.
2.0 1.0 0 0
0.1
0.2 0.3 time/s
Figure 8 Zone of uncertainty.
14
0.4
Uncertainties with gradients Using a computer application, such as a spreadsheet, can allow you to plot a graph with data points and error bars. You can then read o the gradient and the intercepts rom a linear graph directly. The application
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
will automatically draw the best trend line. You can then add the trend lines with the steepest and shallowest gradients that are j ust possible while still passing through all the error bars. S tudents quite commonly, but incorrectly, use the extremes o the error bars that are urthest apart on the graphs. Although these could be appropriate, it is essential that all the trend lines you draw pass through all o the error bars. In an experiment to measure the electromotive orce (em) and internal resistance o a cell, a series o resistors are connected across the cell. The currents in and potential dierences across the resistors are then measured. A graph o potential dierence, V, against current, I, should give a straight line o negative gradient. As you will see in Topic 5 the em o a cell is related to the internal resistance r by the equation: = I( R + r) = V + Ir This can be rearranged to give V = Ir So a graph o V against I is o gradient r (the internal resistance) and intercept (the em o the cell) . The table on the right shows a set o results rom this experiment. With a milliammeter and voltmeter o low precision the repeat values are identical to the measurements given in the table. The graph o fgure 9 shows the line o best ft together with two lines that are j ust possible. EMF and Internal Resistance 1.8
I 5/mA
V 0.1/V
15
1.5
20
1.4
1.2
25
1.4
1.0
30
1.3
0.8
35
1.2
0.6
50
1.1
0.4
55
0.9
0.2
70
0.8
85
0.6
90
0.5
1.6
V = 0.013I + 1.68 V = 0.0127I + 1.64 V = 0.0153I + 1.78
1.4
V/V
0.0 0
20
40
60
80
100
120
140
I/mA
Figure 9 A graph o potential diference, V, against current, I, or a cell.
Converting rom milliamps to amps, the equations o these lines suggest that the internal resistance (the gradient) is 1 3.0 and the range is rom 1 2.7 to 1 5.3 (= 2.6 ) meaning that hal the range = 1 .3 . This leads to a value or r = ( 1 3 . 0 1 . 3 ) . The intercept on the V axis o the line o best ft = 1 .68 which rounds to 1 .7 V ( since the data is essentially to 2 signifcant fgures) . The range o the j ust possible lines gives 1 .6 to 1 . 8 V ( when rounded to two signifcant fgures) . This means that = ( 1 .6 0. 1 ) V.
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Nature of science Drawing graphs manually O ne o the skills expected o physicists is to draw graphs by hand and you may well be tested on this in the data analysis question in Paper 3 o the IB D iploma Programme physics examination. You are also likely to need to draw graphs or your internal assessment.
Try to look at your extreme values so that you have an idea o what scales to use. You will need a minimum o six points to give you a reasonable chance o drawing a valid line.
Use scales that will allow you to spread your points out as much as possible ( you should fll your page, but not overspill onto a second sheet as that would damage your line quality and lose you marks) . You can always calculate an intercept i you need one; when you dont
include the origin, your axes give you a false origin ( which is fne) .
Use sensible scales that will make both plotting and your calculations clear-cut (avoid scales that are multiples o 3, 4, or 7 stick to 2, 5, and 1 0) .
Try to plot your graph as you are doing the experiment i apparently unusual values crop up, you will see them and can check that they are correct.
B eore you draw your line o best ft, you need to consider whether or not it is straight or a curve. There may well be anomalous points ( outliers) that you can ignore, but i there is a defnite trend to the curve then you should opt or a smooth curve drawn with a single line and not sketched artistically!
Figure 1 0 shows some o the key elements o a good hand- drawn graph. C alculating the gradients on the graph is very useul when checking values. best straight line
line that is just possible
gradient of best straight line (59.6 42.0) 10 6 m 3
58
=
Second just possible line should be added for a real investigation it has been missed out here so that you can clearly see the values on the two lines
(398 288) K
= 1.60 10 7 m 3 K1 56 gradient of just possible line (58.6 42.0) 10 6 m 3
54
=
(398 286) K = 1.48 10 7 m 3 K1
V/10 6 m 3
52
All values on V axis have been divided by 10 6 and are in m 3
Use a large gradient triangle to reduce uncertainties
50 48 46 outlier ignored for best straight line
44 False origin neither line has been forced through this point
Uncertainties in temperature are too small to draw error bars
42 40 260
16
Figure 10 Hand-drawn graph.
280
300
320
340 T/K
360
380
400
1 . 2 U N CE R TAI N TI E S AN D E R R O R S
Linearizing graphs
Note
Many relationships between physical quantities are not directly proportional and a straight line cannot be obtained simply by plotting one quantity against the other. There are two approaches to dealing with non- proportional relationships: when we know the orm o the relationship and when we do not.
Relationships can never be indirectly proportional this is a meaningless term since it is too vague. Consequently the term proportional means the same as directly proportional.
1 (or a gas held I we do know the orm o the relationship such as p __ _ V at constant temperature) or T l (or a simple pendulum) we can plot a graph o one quantity against the power o the other quantity to obtain a straight-line origin graph. An alternative or the simple pendulum is to plot a graph o T 2 against l which will give the same result.
This topic is dealt with in more detail and with many further examples on the website.
You should think about the p rop agation of errors_ when you consider the relative merits of plotting T against l or T 2 against l. The following discussion applies to HL examinations, but offers such a useful technique that SL students may wish to utilize it when completing IAs or if they undertake an Extended Essay in a science subject.
By taking logs o this equation we obtain log y = log k + n log x which we can arrange into log y = n log x + log k and is o the orm y = m x + c. This means that a graph o log y against log x will be linear o gradient n and have an intercept on the log y axis o log k.
N/arbitrary units
I we dont know the actual power involved in a relationship, but we suspect that one quantity is related to the other, we can write a general relationship in the orm y = kxn where k and n are constants.
expontential shape 45 30 25 20 15 10 5 0
N = 32e -0.039t
0
N = N0 e - t by taking logs to the base e we get lnN = lnN0 - t ( where lnN is the usual way o writing log e N) .
40 60 t/arbitrary units
80
100
linear form of same data
B y plotting a graph o lnN against t the gradient will be - t and the intercept on the lnN axis will be lnN0 . This linearizes the graph shown in fgure 1 1 producing the graph o fgure 1 2 . A linear graph is easier to analyse than a curve. C apacitors also discharge through resistors using the same general mathematical relationship as that used or radioactive decay.
20
Figure 11
In N
This technique is very useul in carrying out investigations when a relationship between two quantities really is not known. The technique is also a useul way o dealing with exponential relationships by taking logs to base e, instead o base 1 0. For example, radioactive nuclides decay so that either the activity or the number o nuclei remaining alls according to the same general orm. Writing the decay equation or the number o nuclei remaining gives:
4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0
In N = 0.0385t + 3.4657
0 10 20 30 40 50 60 70 80 90 100 t/arbitrary units
Figure 12
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1.3 Vectors and scalars Understanding
Applications and skills
Vector and scalar quantities
Solving vector problems graphically and
Combination and resolution of vectors
algebraically.
Equations The horizontal and vertical components of vector A: A H = A cos A V = A sin A
AV
u AH
Nature of science All physical quantities that you will meet on the course are classified as being vectors or scalars. It is important to know whether any quantity is a vector or a scalar since this will affect how the quantity is treated mathematically. Although the concept of adding forces is an intuitive application of vectors that has probably been used by sailors for millennia, the analytical aspect of it is a recent development. In the
Philosophi Naturalis Principia Mathematica, published in 1687, Newton used quantities which we now call vectors, but never generalized this to deal with the concepts of vectors. At the start of the 19th century vectors became an indispensible tool for representing three-dimensional space and complex numbers. Vectors are now used as a matter of course by physicists and mathematicians alike.
Vector and scalar quantities S calar quantities are those that have magnitude ( or size) but no direction. We treat scalar quantities as numbers ( albeit with units) and use the rules o algebra when dealing with them. D istance and time are both scalars, as is speed. The average speed is simply the distance divided by the time, so i you travel 80 m in 1 0 s the speed will always be 8 m s 1 . There are no surprises.
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Vector quantities are those which have both magnitude and direction. We must use vector algebra when dealing with vectors since we must take into account direction. The vector equivalent o distance is called displacement ( i. e., it is a distance in a specifed direction) . The vector equivalent o speed is velocity ( i.e. , it is the speed in a specifed direction) . Time, as we have seen, is a scalar. Average velocity is defned as being displacement divided by time.
1 . 3 VE CTO RS AN D S C AL ARS D ividing a vector by a scalar is the easiest operation that we need to do involving a vector. To continue with the example that we looked at with scalars, suppose the displacement was 80 m due north and the time was, again, 1 0 s. The average velocity would be 8 m s 1 due north. S o, to generalize, when we divide a vector by a scalar we end up with a new vector that has the direction o the original one, but which will be o magnitude equal to that o the vector divided by that o the scalar.
Commonly used vectors and scalars Vectors
Scalars
Comments
orce (F)
mass (m)
displacement (s)
length/distance (s, d, etc.)
velocity (v or u)
time (t)
momentum (p)
volume (V)
acceleration (a)
temperature (T)
gravitational feld strength (g)
speed (v or u)
velocity and speed oten have the same symbol
electric feld strength (E)
density ( )
the symbol or density is the Greek rho not the letter p
magnetic feld strength (B)
pressure (p)
area (A)
energy/work (W, etc.)
F displacement used to be called space now that means something else!
the direction o an area is taken as being at right angles to the surace
power (P) current (I)
with current having direction you might think that it should be a vector but it is not (it is the ratio o two scalars, charge and time, so it cannot be a vector) . In more advanced work you might come across current density which is a vector.
resistance (R) gravitational potential (VG ) electric potential (VE ) magnetic ux ()
the subscripts tell us whether it is gravitational or electrical ux is oten thought as having a direction it doesnt!
Representing vector quantities A vector quantity is represented by a line with an arrow.
The direction the arrow points represents the direction o the vector.
The length o the line represents the magnitude o the vector to a chosen scale.
When we are dealing with vectors that act in one dimension it is a simple matter to assign one direction as being positive and the opposite direction as being negative. Which direction is positive and which negative really doesnt matter as long as you are consistent. S o, i one orce acts upwards on an obj ect and another orce acts downwards, it is a simple matter to fnd the resultant by subtracting one rom the other.
5.0 N this vector can represent a force of 5.0 N in the given direction (using a scale of 1 cm representing 1 N, it will be 5 cm long)
Figure 1 Representing a vector.
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M E A S U R E M E N TS AN D U N C E R TAI N T I E S Figure 2 shows an upward tension and downward weight acting on an obj ect the upward line is longer than the downward line since the obj ect is not in equilibrium and has an upward resultant.
tension = 20 N upwards
Adding and subtracting vectors
resultant = 5 N upwards
When adding and subtracting vectors, account has to be taken o their direction. This can be done either by a scale drawing ( graphically) or algebraically.
weight = 15 N downwards
Figure 2 Two vectors acting on an object.
V2
V1
V1
V
V2
Figure 3 Two vectors to be added.
Figure 4 Adding the vectors.
Scale drawing (graphical) approach Adding two vectors V1 and V2 which are not in the same direction can be done by orming a parallelogram to scale.
Make a rough sketch o how the vectors are going to add together to give you an idea o how large your scale needs to be in order to fll the space available to you. This is a good idea when you are adding the vectors mathematically too.
Having chosen a suitable scale, draw the scaled lines in the direction o V1 and V2 ( so that they orm two adj acent sides o the parallelogram) .
C omplete the parallelogram by drawing in the remaining two sides.
The blue diagonal represents the resultant vector in both magnitude and direction.
Worked example
resultant
36
17 0 180 160 0 1 5 0 2 0 10 30
40
30 40 150 14 0
80 90 10 0 1 1 0 1 70 20 6 0 1 1 0 10 0 90 80 7 0 13 60 0 0 0 5 12 50 0 13
0
D ont orget that the vector must have a magnitude and a direction; this means that the angle is j ust as important as the size o the orce.
8.7 N
6N
14
Solution
Scale 10 mm represents 1.0 N
0 10 20 180 17 0 1 60
Two orces o magnitude 4.0 N and 6.0 N act on a single point. The orces make an angle o 60 with each other. Using a scale diagram, determine the resultant orce.
4N length of resultant = 87 mm so the force = 8.7 N angle resultant makes with 4 N force = 36
20
6N
1 . 3 VE CTO RS AN D S C AL ARS
Algebraic approach Vectors can act at any angle to each other but the most common situation that you are going to deal with is when they are at right angles to each other. We will deal with this frst. v2
Adding vector quantities at right angles Pythagoras theorem can be used to calculate a resultant vector when two perpendicular vectors are added ( or subtracted) . Assuming that the two vector quantities are horizontal and vertical but the principle is the same as long as they are perpendicular.
v1 v1
Figure 5 shows two perpendicular velocities v1 and v2 ; they orm a parallelogram that is a rectangle. ______
The magnitude o the resultant velocity = v 21 + v 22
The resultant velocity makes an angle to the horizontal given by
( )
v2
( )
v1 v1 1 _ tan = _ v2 so that = tan v2
Notice that the order o adding the two vectors makes no dierence to the length or the direction o the resultant.
Worked example A walker walks 4.0 km due west rom his starting point. He then stops beore walking 3 .0 km due north. At the end o his j ourney, how ar is the walker rom his starting point?
Solution
Figure 5 Adding two perpendicular vectors.
sketch
3 km
______
2 resultant = 4 2 + 3 = 5 km
angle = tan
1
()
3 = 3 6.9 __ 4
4 km
B eore we look at adding vectors that are not perpendicular, we need to see how to resolve a vector i.e. split it into two components.
Resolving vectors We have seen that adding two vectors together produces a resultant vector. It is sensible, thereore, to imagine that we could split the resultant into the two vectors rom which it was ormed. In act this is true or any vector it can be divided into components which, added together, make the resultant vector. There is no limit to the number o vectors that can be added together and, consequently, there is no limit to the number o components that a vector can be divided into. However, we most commonly divide a vector into two components that are perpendicular to one another. The reason or doing this is that perpendicular vectors have no aect on each other as we will see when we look at proj ectiles in Topic 2 .
Fsin
F
Fcos
Figure 6 Resolving a force.
The orce F in fgure 6 has been resolved into the horizontal component equal to F cos and a vertical component equal to F sin . ( The component opposite to the angle used is always the sine component.)
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Worked example An ice- hockey puck is struck at a constant speed o 40 m s 1 at an angle o 60 to the longer side o an ice rink. How ar will the puck have travelled in directions a) parallel and b) perpendicular to the long side ater 0.5 s?
vx = 40 cos 60 = 2 0 m s 1 distance travelled ( x) = vxt = 2 0 0.5 = 10 m b) Resolving parallel to shorter side: vx = v sin 60
Vy = 40 sin 60
Solution
vy = 40 cos 60 = 3 4. 6 m s 1 distance travelled ( y) = vyt = 3 4.6 0.5 = 17 m
40 ms 1
Just to demonstrate that resolving is the reverse o adding the components we can use Pythagoras theorem to add together our two components giving: _________
60
total speed = 2 0 2 + 3 4. 6 2 = 3 9. 96 m s 1 long side of rink
Vx = 40 cos 60
a) Resolving parallel to longer side:
as the value or vy was rounded this gives the expected 40 m s 1
vx = v cos 60
Adding vector quantities that are not at right angles
V1 = V1 sin 1
You are now in a position to add any vectors.
V1
Resolve each o the vectors in two directions at right angles this will oten be horizontally and vertically, but may be parallel and perpendicular to a surace.
Add all the components in one direction to give a single component.
Add all the components in the perpendicular direction to give a second single component.
C ombine the two components using Pythagoras theorem, as or two vector quantities at right angles.
1
V1x = V1 cos 1
V2y = V2 sin 2
V1 and V2 are the vectors to be added.
V2
Total x component Vx = V1 x + V2x = V1 cos 1 - V2 cos 2 V2x = V2 cos
22
Each vector is resolved into components in the x and y directions. Note that since the x component o V2 is to the let it is treated as being negative ( the y component o each vector is in the same direction . . . so upwards is treated as positive) .
2 2
Figure 7 Finding the resultant of two vectors that are not perpendicular.
Total y component Vy = V1 y + V2y = V1 sin 1 + V2 sin 2 Having calculated Vx and Vy we can fnd the resultant by using ______
Pythagoras so V =
V2x +
V2y
and the angle made with the horizontal = tan
( ).
Vy 1 __ Vx
1 . 3 VE CTO RS AN D S C AL ARS
Worked example Magnetic felds have strength 2 00 mT and 1 5 0 mT respectively. The felds act at 2 7 to one another as shown in the diagram.
Solution The 1 5 0 mT feld is horizontal and so has no vertical component. Vertical component o the 2 00 mT feld = 2 00 sin 2 7 = 90.8 mT
200 mT
(not drawn to scale)
This makes the total vertical component o the resultant feld. Horizontal component o the 2 00 mT feld = 2 00 cos 2 7 = 1 78.2 mT
27 150 mT
C alculate the resultant magnetic feld strength.
Total horizontal component o resultant feld = ( 1 5 0.0 + 1 78.2 ) mT = 3 2 8.2 mT ____________
Resultant feld strength 90.8 2 + 32 8.2 2 = 340 mT The resultant feld makes an angle o: tan 1
(_____) = 1 5 with the1 5 0 mT feld. 90.8 3 2 8.2
Subtraction of vectors S ubtracting one vector rom another is very simple you j ust orm the negative o the vector to be subtracted and add this to the other vector. The negative o a vector has the same magnitude but the opposite direction. Lets look at an example to see how this works: S uppose we wish to fnd the dierence between two velocities v1 and v2 shown in fgure 8. V2 V1 V2 V1 V2
V1
V1
Figure 8 Positive and negative vectors.
V1
In fnding the dierence between two values we subtract the frst value rom the second; so we need v1 . We then add v1 to v2 as shown in fgure 9 to give the red resultant. The order o combining the two vectors doesnt matter as can be seen rom the two versions in fgure 9. In each case the resultant is the same it doesnt matter where the resultant is positioned as long as it has the same length and direction it is the same vector.
V2 V1
V2
Figure 9 Subtracting vectors.
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1
M E A S U R E M E N TS AN D U N C E R TAI N T I E S
Questions 1
E xpress the ollowing units in terms o the S I undamental units.
5
Write down the order o magnitude o the ollowing ( you may need to do some research) .
a) newton ( N)
a) the length o a human oot
b) watt ( W)
b) the mass o a y
c) pascal ( Pa)
c) the charge on a proton
d) coulomb ( C )
d) the age o the universe
e) volt ( V)
e) the speed o electromagnetic waves in a vacuum
( 5 marks)
( 5 marks) 2
Express the ollowing numbers to three signifcant fgures.
6
a) 2 5 7.5 2 b) 0.002 3 47 c) 0.1 783 d) 7873 e) 1 .997 ( 5 marks)
Without using a calculator estimate to one 2 4.9 signifcant fgure the value o _. 480 b) When a wire is stretched, the area under the line o a graph o orce against extension o the wire gives the elastic potential energy stored in the wire. Estimate the energy stored in the wire with the ollowing characteristic:
C omplete the ollowing calculations and express your answers to the most appropriate number o signifcant fgures.
12 10 force/N
3
a)
a) 1 .3 4 3 .2 1 .3 4 1 0 2 b) _ 2 .1 1 0 3 c) 1 .87 1 0 2 + 1 . 97 1 0 3
6 4 2
d) ( 1 .97 1 0 5 ) ( 1 . 0 1 0 4)
0
e) ( 9.47 1 0 2 ) ( 4.0 1 0 3 )
0
( 5 marks)
4
8
Use the appropriate metric multiplier instead o a power o ten in the ollowing. a) 1 .1 1 0 4 V b) 4.2 2 1 0 4 m c) 8.5 1 0 1 0 W
1
2
3
4 5 6 extension/mm
7
8
( 4 marks) 7
The grid below shows one data point and its associated error bar on a graph. The x- axis is not shown. S tate the y- value o the data point together with its absolute and percentage uncertainty.
d) 4.2 2 1 0 7 m 5.0
e) 3 .5 1 0 1 3 C ( 5 marks)
4.0 3.0 2.0 1.0
24
( 3 marks)
QUESTION S 8
A ball alls reely rom rest with an acceleration g. The variation with time t o its displacement s 1 is given by s = __ gt2 . The percentage uncertainty 2 in the value o t is 3 % and that in the value o g is 2 % . C alculate the percentage uncertainty in the value o s. ( 2 marks)
9
The volume V o a cylinder o height h and radius r is given by the expression V = r2 h. In a particular experiment, r is to be determined rom measurements o V and h. The percentage uncertainty in V is 5 % and that in h is 2% . C alculate the percentage uncertainty in r. ( 3 marks)
1 0 ( IB) At high pressures, a real gas does not behave as an ideal gas. For a certain range o pressures, it is suggested that or one mole o a real gas at constant temperature the relation between the pressure p and volume V is given by the equation pV = A + Bp
where A and B are constants.
In an experiment, 1 mole o nitrogen gas was compressed at a constant temperature o 1 5 0 K. The volume V o the gas was measured or dierent values o the pressure p. A graph o the product pV against p is shown in the diagram below.
c) p was measured to an accuracy o 5 % and V was measured to an accuracy o 2 % . D etermine the absolute error in the value o the constant A. ( 6 marks) 1 1 ( IB) An experiment was carried out to measure the extension x o a thread o a spiders web when a load F is applied to it. 9.0 8.0 7.0 6.0 F/10 2 N 5.0 4.0 3.0 2.0 1.0 0.0 0.0
thread breaks at this point
1.0
2.0
3.0 4.0 x/10 2 m
5.0
6.0
a) C opy the graph and draw a best-ft line or the data points. b) The relationship between F and x is o the orm F = kxn S tate and explain the graph you would plot in order to determine the value n. c) When a load is applied to a material, it is said to be under stress. The magnitude p o the stress is given by F p= _ A
13
where A is the cross- sectional area o the sample o the material.
12 pV/10 2 Nm
Use the graph and the data below to deduce that the thread used in the experiment has a greater breaking stress than steel.
11
B reaking stress o steel = 1 .0 1 0 9 N m 2 10 0
5
10 15 p/ 10 6 Pa
20
a) C opy the graph and draw a line o best ft or the data points. b) Use your graph to determine the values o the constants A and B in the equation pV = A + Bp
Radius o spider web thread = 4.5 1 0 6 m d) The uncertainty in the measurement o the radius o the thread is 0. 1 1 0 6 m. D etermine the percentage uncertainty in the value o the area o the thread. ( 9 marks)
25
1
M E A S U R E M E N TS AN D U N C E R TAI N T I E S 1 2 A cyclist travels a distance o 1 2 00 m due north beore going 2 000 m due east ollowed by 5 00 m south-west. D raw a scale diagram to calculate the cyclists fnal displacement rom her initial position. ( 4 marks)
1 3 The diagram shows three orces P, Q, and R in equilibrium. P acts horizontally and Q vertically.
R
a) C alculate the resultant velocity o the boat relative to the bank o the river. b) The river is 5 0 m wide. C alculate the displacement rom its initial position when the boat reaches the opposite bank. ( 7 marks)
1 5 A car o mass 85 0 kg rests on a slope at 2 5 to the horizontal. C alculate the magnitude o the component o the cars weight which acts parallel to the slope.
P
( 3 marks)
Q
When P = 5 .0 N and Q = 3 . 0 N, calculate the magnitude and direction o R. ( 3 marks)
26
1 4 A boat, starting on one bank o a river, heads due south with a speed o 1 .5 m s 1 . The river ows due east at 0.8 m s 1 .
2
M E CH AN I CS
Introduction Everything moves. The Earth revolves on its axis as it travels around the S un. The S un orbits within the Milky Way. Galaxies move apart. Motion and its causes are important to a study o physics. We begin by defning the meaning
o everyday terms such as distance, velocity, acceleration and go on to develop models or motion that will allow us to predict the uture motion o an obj ect.
2.1 Motion Understanding Distance and displacement Speed and velocity Acceleration
Applications and skills Determining instantaneous and average values
Graphs describing motion Equations o motion or uniorm acceleration
Projectile motion
Fluid resistance and terminal speed
Nature of science An understanding o motion lies at the heart o physics. The areas o the subject are linked by the concepts o movement and the orces that produce motion. Links are used in a creative way by scientists to illuminate one part o the subject by reerence to insights developed or other topics. The study o motion also relies on careul observation o the world so that accurate models o motion can be developed.
or velocity, speed, and acceleration Solving problems using equations o motion or uniorm acceleration Sketching and interpreting motion graphs Determining the acceleration o ree-all experimentally Analysing projectile motion, including the resolution o vertical and horizontal components o acceleration, velocity, and displacement Qualitatively describing the efect o uid resistance on alling objects or projectiles, including reaching terminal speed
Equations Kinematic equations o motion: v = u + at 1 2 s = ut + ___ at 2 2 2 v = u + 2as (v + u) t s = _____________ 2
27
2
M E C H AN I C S
Distance and displacement Your j ourney to school is unlikely to take a completely straight line from home to classroom. How do we describe j ourneys when we turn a corner and change direction? The map shows the j ourney a student makes to get to school together with the times of arrival at various points on the way. To keep life simple, the j ourney has no hills.
A
home
bus stop
B school bus stop
Figure 1 Journey to school.
journey leg
time
distance for leg / m
leave home
08.10.00
0
walk to bus stop
08.20.15
800
bus arrives at stop
08.24.30
0
bus arrives near school
08.31.10
2400
walk from bus to school
08.34.00
200
The total length of this j ourney is 3 .4 km including all the twists and turns. This is the distance travelled. As we saw in Topic 1 , distance is a scalar quantity; it has no direction. If the student walks home by exactly the same route, then the distance travelled going home is the same as going to school.
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2 .1 M OTI ON
Nature of science How long is a piece of string? Distance can be measured in any appropriate unit o length: metres, miles, and millimetres are all common. Some countries and some proessions use alternatives. Surveyors use chains, the English-speaking world once used measures o length called rods, poles, and perches all related to agricultural measurements. Astronomers use light years (a unit o length, not time) and a measure called simply the astronomical unit. Sometimes in everyday lie it is a question o using a convenient unit rather than the correct metric version. In your exam, however, lengths will be in multiples and sub-multiples o the metre or in a wellrecognised scientifc unit such as the light year.
Journeys can be defned by the starting point ( A) and fnishing point ( B ) without saying anything about the intermediate route. A vector is drawn that starts at A and fnishes at B . This straight line rom start to fnish has, as a vector, magnitude, and direction. This vector measurement is known as the disp lacement. As you saw in Sub- topic 1 . 3 , when you give the details o a displacement you must always give two pieces o inormation: the magnitude ( or size) o the quantity together with its unit, and also the direction o the vector. This direction can be written in a number o ways. O ne way is as a heading such as N3 5 E meaning that the fnishing direction is at an angle o 3 5 clockwise rom north. I you are a sailor or a keen orienteer, you may have other ways o measuring the direction o displacement. The displacement o the students j ourney to school ( AB ) is also shown on the diagram. The displacement is the vector that connects home to school. This time the j ourney rom school to home is not the same as the trip to school. It has the same vector length, but the direction is the opposite to that o the outward route.
Nature of science Moving in 3D D isplacement has been described here in terms o a j ourney in a at landscape. D oes a change in level alter things? O nly one thing changes, and that is the number o pieces o inormation required to speciy the fnal position relative to the start. Three pieces o inormation are now required: the magnitude plus its unit, the heading, and the overall change in height during the j ourney. S peciying motion in three dimensions requires three numbers or coordinates, you are already
amiliar with the idea o a coordinate rom drawing and using graphs. There is exibility in how the three numbers can be chosen. You may have seen three- dimensional graphs with three axes each at 90 to the others, in this case coordinate numbers are given that relate to the distance along each axis. Another option is to use polar coordinates ( fgure 2 ) : where a distance and two angles are required,
29
2
M E C H AN I C S
one angle is the bearing rom North, the other the angle up or down rom the horizontal needed to look directly at the obj ect above (or below) us.
(r, , )
r
In some circumstances, even the distance itsel may not be required. S ailors use latitude and longitude when they are navigating. They stay on the surace o the sea and this is ee ctively a constant distance rom the centre o the E arth.
Worked examples 1
A cyclist travels 1 6 km in 70 minutes. Calculate, in m s 1 , the speed o the cyclist.
Solution 70 minutes is 60 70 = 42 00 s, 1 6 km is 1 6 000 m. The quantities are now in the units required by the question. The speed o the cyclist 1 6 000 is _____ = 3 .8 m s 1 . 42 0 0
2
The speed o light in a vacuum is 3 .0 1 0 8 m s 1 . A star is 2 2 light years rom Earth ( 1 light year is the distance travelled by light in one year) . C alculate the distance o the star rom Earth in kilometres.
Solution Light travels 3 .0 1 0 8 m in 1 s. S o in a year it travels 3 .0 1 0 8 3 65 2 4 60 60 = 9.5 1 0 1 5 m. The distance o the star rom the Earth is 22 9.5 1 0 1 5 = 2 .1 1 0 1 7 m. The question asks or an answer in kilometres, so the distance is 2 .1 1 0 1 4 km.
30
Figure 2 Polar coordinates.
Speed and velocity In j ust the same way that there are scalar and vector measures o the length o a j ourney, so there are two ways o measuring how quickly we cover the ground. The frst o these is a scalar quantity sp eed, which is defned as distan ce trave lle d o n the j o u rn e y _______________________ . Units or speed, amiliar to you already, may tim e take n o r the j o u rne y
include metre per second ( m s 1 ) and kilometres per hour ( km h 1 ) , but any accepted distance unit can be combined with any time unit to speciy speed. Velocity is the vector term and, j ust as or displacement, a magnitude and direction are required. Examples might be 4.2 m s 1 due north or 5 5 km h 1 at N2 2 .5 E.
Nature of science Measuring speed? To measure speed it is necessary to measure the the distance travelled (using a ruler) and time taken (using a clock) . The trick is in a good choice o ruler, clock and method or recording the measurements! A 30 cm ruler and a wrist watch will be fne or a biologist measuring the speed at which an earthworm moves. B ut i we want to measure the speed o a 1 00 m sprinter, then a measured distance on the ground, a good stop watch and a human observer is barely good enough. Even then the observer has to be careul to watch the smoke rom the starting pistol and not to wait or the sound o the gun. I you need to measure the speed o a soccer ball during a penalty kick, then stop watch-plus-human is no longer adequate to make a valid measurement. Perhaps a video camera that takes rames at a known rate (the clock) and a scale near the path o the ball visible on the picture (the ruler) is needed now? Move up to measuring the speed o a jet aircrat and the equipment needs to change again. C hoosing the right equipment or the task in hand is all part o the job o the working science student.
2 .1 M OTI ON
Describing motion with a graph I The use o two variables, distance and time, to calculate speeds and velocities means that things become more complicated. Figure 1 showed a map o the j ourney the student makes to school. Part o this j ourney is on oot, part is by bus. It is unlikely that the student will travel at the same speed all the time, as the bus will travel aster than the student walks. I we want to display data in a visual way, a graph o distance against time is one o the most common approaches. 3500 distance travelled/m
3000 2500 2000 1500 1000 500 0
500
1000 time elapsed/s
1500
Figure 3 Distancetime graph.
The distance travelled by the student is plotted on the vertical axis while the time since the beginning o the j ourney is plotted horizontally ( the clock times o the j ourney have been translated into elapsed times since the start o the j ourney or this graph) . The dierent regions o the graph are identifed and you should confrm that they are matched correctly. Notice how the gradient o the graph changes or the dierent parts o the j ourney: small values o the gradient or the walking sections o the j ourney, horizontal ( zero gradient) or stationary at the bus stop, and steep or the bus j ourney. What will the graph or the j ourney home rom school look like, assuming that the time or each segment o the j ourney is the same as in the morning?
distance travelled/m
0
distance travelled on bus = 2400 m
time taken or bus journey = 400 s
Inormation can easily be extracted rom this graph. The gradient o the graph is the speed. Add the overall direction to this speed and we have the velocity too. For the frst walk to the bus stop the distance was 800 m and the time taken 800 was 61 5 s. The constant walking speed was thereore ___ , which is 61 5 1 1 .3 m s .
speed = 2400 = 6 ms 1 400
1000 time elapsed/s
Figure 4 Distancetime graph rom fgure 3 with gradients.
2 400 The gradient o the bus journey is ____ (this is marked on the graph) and so 400 1 the speed was 6.0 m s .
O course, even this j ourney with its changes is simplistic. Real j ourneys have ew straight lines, so we must introduce some ways to handle rapidly varying speeds and velocities.
31
2
M E C H AN I C S
Instantaneous and average values The bus driver knows how ast the bus in the students j ourney is travelling because it is displayed on the speedometer. This is the instantaneous sp eed, as it gives the value o the speed at the moment in time at which the speed is determined. 3500
distance travelled/m
3000
2500
1500
Nature of science Calculating a gradient of a graph Taking the gradient o a graph is actually a way o averaging results. As discussed in Topic 1 , a straightline graph drawn through points that have some scatter makes this obvious. The value o the gradient ch an ge in the valu e o n the y- axis . is the _______________________ change in th e valu e o n the x- axis The unit associated with the gradient y- axis u nit is the ________ . D ont orget to x- axis u nit quote the unit every time you write down the value o the gradient.
use as large a tangent line as possible, change in distance = (2500 500) m change in time = (1300 900) s = 5.0 m s 1 gradient = 2000 400
2000 a more realistic graph for the bus
1000
tangent to graph at t = 1000 s
500 1000 time elapsed/s
Figure 5 Instantaneous speed.
The instantaneous speed is also the gradient o the distancetime graph at the instant concerned. Figure 5 shows how this is calculated when 1 000 s into the whole journey to school. The original red line or the bus rom fgure has been replaced by a green line that is more realistic or the motion o a real bus the speed varies as the driver negotiates the trafc. You will requently be asked to calculate gradients on physics graphs. Make sure that you can do this accurately by using a transparent ruler. From a mathematical point o view we can describe the instantaneous speed as the rate of change of position with respect to time. ds A mathematician will write this as __ , where s is the distance and t is the dt s time. You may also have seen ___ , where the symbol means change in. t change in distance s So ___ is just shorthand or ____________ . change in tim e t
There is however another useul measure o speed. This is the average speed and is the speed calculated over the whole the journey without regard to variations in speed. So as an equation this is distan ce trave lle d o ve r w ho le j o u rne y average speed = __________________________ tim e take n o r w ho le j o u rne y
In terms o the distancetime graph, the average speed is equal to the gradient o the straight line that j oins the beginning and the end o the time interval concerned. S o, or the part o the students j ourney up to the moment when the bus arrives at the stop, the distance travelled is 800 m, the time taken is 870 s ( including the wait at the stop) so the average speed is 0.92 m s 1 .
32
2 .1 M OTI ON
E verything said here about average and instantaneous speeds can also reer to average and instantaneous velocities. Remember, o course, to include the directions when quoting these measurements.
Worked example Solution
distance/m
The graph shows how the distance run by a boy varies with the time since he began to run.
a)
100 90 80 70 60 50 40 30 20 10 0
( i) The boy runs at constant speed so the graph is straight rom 0 1 0 s. The gradient o the straight line is 48 __ = 4.8 m s 1 . 10
S o the instantaneous speed at 5 .0 s is 4.8 m s 1 . ( ii) Again the boy is running at a constant speed, but this time slower than in the frst 1 0 s. 0
5
10
15
20 time/s
25
30
35
C alculate: a) the instantaneous speed at ( i) 5 .0 s ( ii) 2 0 s
The speed rom 1 0 s to 3 0 s is: ( 90 48) 42 _ = 2 .1 m s 1 . = __ 20 ( 3 0 1 0) The instantaneous speed at 2 0 s is 2 .1 m s 1 . b) The total distance travelled in 3 0 s is 90 m, so 90 the average speed is __ = 3 .0 m s 1 . 30
b) the average speed or the whole 3 0 s run.
Acceleration In real j ourneys, instantaneous speeds and velocities change requently. Again we need to develop a mathematical language that will help us to understand the changes. The quantity we use is acceleration. Acceleration is taken to be a vector, but sometimes we are not interested in the direction and then write the magnitude o the acceleration meaning the size o the acceleration ignoring direction. The defnition is: acceleration =
change in velocity ___________________ tim e taken or the change
m s and this means that the units o acceleration are ____ which is usually s 1
written as m s 2 ( or sometimes you will see m/s 2 ) . It is important to understand what acceleration means, not j ust to be able to use it in an equation. I an obj ect has an acceleration o 5 m s 2 then or every second it travels, its velocity increases in magnitude by 5 m s 1 in the direction o the acceleration vector. As an example: the Japanese N700 train has a quoted acceleration o 0.72 m s 2 . Assume that this is a constant value (very unlikely) . One second ater starting rom rest, the speed o the train will be 0.72 m s 1 . One second later (at 2 s rom the start) the speed will be 0.72 + 0.72 = 1 .44 m s 1 . At 3 s it will be 2.1 6 m s 1 and so on. Each second the speed is 0.72 m s 1 more.
Worked example How many seconds will it take the N700 to reach its maximum speed o 3 00 km h 1 on the S anyo Shinkansen route?
Solution 3 00 km h 1 = 83 .3 m s 1 Time taken to reach the maximum speed: 83 .3 _ = 1 1 6 s, 0. 72 just under 2 minutes.
33
2
M E C H AN I C S
Nature of science Spreadsheet models One powerful way to think about acceleration (and other quantities that change in a predictable way) is to model them using a spreadsheet. The examples here use a version of Microsoft Excel but any computer spreadsheet can be used for this. This is a spreadsheet model for the N700 train. The value of the acceleration is in cell B 1 . C ells A4 to A2 9 give the time in increments of 5 s; the computed speed at each of these times is in B
C
D
E
F
G
H
I
J
0.72 speed/ms -1 0 3.6 7.2 10.8 14.4 18.0 21.6 25.2 28.8 32.4 36.0 39.6 43.2 46.8 50.4 54.0 57.6 61.2 64.8 68.4 72.0 75.6 79.2 82.8 86.4 90.0
100 90 80 70 60 50 40 30 20 10 0
speed /m s -1
A 1 acceleration/s 2 time/s 3 4 0 5 5 6 10 7 15 8 20 9 25 10 30 11 35 12 40 13 45 14 50 15 55 16 60 17 65 18 70 19 75 20 80 21 85 22 90 23 95 24 100 25 105 26 110 27 115 28 120 29 125
cells B 4 to B 2 9. The speed is calculated by taking the change in time between the present cell and the one above it, and then multiplying by the acceleration (the acceleration is written as $B $1 so that the spreadsheet only uses this cell and does not drop down a cell every time the new speed is calculated) . Finally, the spreadsheet plots speed against time showing that the graph is a straight line and that the acceleration is uniform.
0
20
40
60
80 time/s
100
120
140
Describing motion with a graph II D istancetime plots lead to a convenient display of speed and velocity changes. Plots of speed ( or velocity) against time also help to display and visualize acceleration. The data table and the graph show a j ourney with various stages on a bicycle. From the start until 1 0 s has elapsed, the bicycle accelerates at a uniform rate to a velocity of + 4 m s 1 . The positive sign means that the velocity is directed to the right.
34
2 .1 M OTI ON
5 4 3 2 1 velocity /m s -1 0 1 0 2 3 4 5
20
40
60
80
100
120
140
time/s Figure 6 Velocitytime graph
for the bicycle.
From 1 0 s to 45 s the cycle moves at a constant velocity o + 4 m s 1 and at 45 s the cyclist applies the brakes so that the cycle stops in 5 s. The cycle is then stationary or 1 0 s. From then on the velocity is negative meaning that the cycle is travelling in the opposite direction. The pattern is similar, an acceleration to 3 m s 1 , a period o constant velocity and a deceleration to a stop at 1 20 s. As beore, the gradient o the graph has a meaning. The gradient o this velocitytime graph gives the magnitude ( size) o the acceleration and its sign ( direction) as well. From 45 s to 5 0 s the velocity goes rom 4 m s 1 to 0 and so the f nal sp e e d - in itial sp e e d ____ ( 0 4) = 0. 8 m s 2 . From 90 s to 1 2 0 s acceleration is __________________ 5 tim e take n
3 the magnitude o the acceleration is __ = 0.1 m s 2 . We need to take care 30 with the sign o the acceleration here. B ecause the cycle is moving in the negative direction and is slowing down, the acceleration is positive (as is the gradient on the graph) this simply means that there is a orce acting to the right, that is, in the positive direction which is slowing the cycle down. We will discuss how orce leads to acceleration later in this topic.
The area under a velocitytime graph gives yet more inormation. It tells us the total displacement o the moving object. The way to see this is to realize that the product o velocity time is a displacement (and that product o speed time is a distance) . The units tell you this too: when the units are m e tre multiplied the seconds in _____ second cancel to leave metre only. se co nd
velocity/m s -1
In the case o a graph with uniorm acceleration, the areas, and hence the displacements (distances) are straightorward to calculate. Divide the graph into right-angled triangles and rectangles and then work out the areas or each individual part. This is shown in fgure 7. 5 area = 1 10 4 = 20 m 2 4 3 area = 12 5 4 = 10 m 2 1 0 0 40 60 80 100 120 20 1 2 area = 35 4 = 140 m 3 4 time/s
Figure 7
140
Velocity-time graph broken down into areas.
35
2
M E C H AN I C S The individual areas are shown on the diagram and or the motion up to a time o 60 s the area is 1 70 m; the area rom the 60 s time to the end is 1 2 0 m. As usual, the negative sign indicates motion in the opposite direction to the original. As discussed in Topic 1 , when the velocitytime graph is curved, you will need to: ( i) estimate the number o squares ( ii) assess the area ( distance) or one square, and fnally ( iii) multiply the number o squares by the area o one square. This will usually give you an estimate o the overall distance. Figure 8 gives an example o how this is done. There are about 85 squares between the x-axis and the line. ( You may disagree slightly with this estimate, but that is fne there is always an allowance made or this.)
speed/m s -1
E ach o the squares is 2 s along the time axis and 0.5 m along the speed axis. S o the area o one square is equivalent to 1 .0 m o distance. The total distance travelled is 85 m ( or, at least, somewhere between 80 and 90 m) . 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0
10
20 time/s
30
40
Figure 8 A more difcult area to estimate.
The suvat equations of motion
36
symbol
quantity
s
displacement/distance
u
initial (starting) velocity/speed
v
fnal velocity/speed
a
acceleration
t
time taken to travel the distance s
The graphs o distancetime and speedtime lead to a set o equations that can be used to predict the value o the parameters in motion. They also help you to understand the connection between the various quantities in your study o motion. A note about symbols: rom now on we will use a consistent set o symbols or the quantities. The table gives the list. I you read down the list o symbols they spell out suvat and the equations are sometimes known by this name, another name or them is the kinematic equations o motion. The derivation o the equations o uniormly accelerated motion begins rom a simple graph o speed against time or a constant acceleration rom velocity u to velocity v in a time t.
2 .1 M OTI ON
v
speed
gradient = area =
(v u) t
1 (v u) t 2
u area = u t 0
0
t time
Figure 9 Deriving the frst two equations o motion.
The acceleration is the gradient o the graph: change in speed __ time taken or change The change in speed is v u, the time taken is t. Thereore, vu a= _ t and re arranging gives
v = u + at
frst equation o motion
The area under the speedtime graph rom 0 to t is made up o two parts, the lower rectangle, area and the upper right- angled triangle, area . 1 base height = _ 1 t( v u) area = _ 2 2 area = base height = ut 1 ( v u) t = ut + _ 1 ( at) t s = total area = area + area = ut + _ 2 2 So
1 at2 s = ut + _ 2
second equation o motion
The frst equation has no s in it; the second has no v. There are three more equations, one with a missing t and one with a missing a. There is one equation that has a missing u, but this is not oten used. To eliminate t rom the frst and second equations, re arrange the frst in terms o t: vu t= _ a This can be substituted into the second equation:
(
vu vu 1 _ _ s=u_ a + 2 a a
)
2
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2
M E C H AN I C S and 1 1 1 1 as = u( v u) + _ ( v u) 2 = uv u 2 + _ v2 + _ u 2 _ 2 uv 2 2 2 2 which gives 2 as = v2 u 2 or
v2 = u 2 + 2as
third equation of motion
The derivation o the fnal equation is let to you as an exercise:
( )
v+ u s = _________ t 2
fourth equation of motion
There are two ways to approach this proo. O ne way is to think about v+ u the meaning o the speed that corresponds to ____ and then to recognize 2 the time at which this speed occurs in the motion. The second way is to take the third equation and amalgamate it with the frst.
Remember! These equations only apply if the acceleration is uniform. In other words, acceleration must not change during the motion.
You will not be expected to remember these proos or the equations themselves (which appear in the data booklet) , but they do illustrate how useul graphs and equations can be when solving problems in kinematics.
Worked examples 1
A driver o a car travelling at 2 5 m s 1 along a road applies the brakes. The car comes to a stop in 1 5 0 m with a uniorm deceleration. C alculate a) the time the car takes to stop, and b) the deceleration o the car.
Solution a) O ne way to answer kinematic equation questions is to begin by writing down what you do and dont know rom the question. 1
s = 1 5 0 m; u = 2 5 m s ; v = 0 ; a = ? ; t = ? To work out t, the ourth equation is required: v+ u t s = ____ 2
( )
( )
2 which rearranges to t = ____ s v+ u
Substituting the values gives
( )
2 t = __ 1 50 = 2 6 = 1 2 s 25
b) To fnd a the equation v2 = u 2 + 2 as is best. Substituting: 0 = 252 + 2 a 1 50
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25 25 25 a = _ = _ = 2 .1 m s 2 . 3 00 12 The minus sign shows that the car is decelerating rather than accelerating. 2
A cyclist slows uniormly rom a speed o 7.5 m s 1 to a speed o 2.5 m s 1 in a time o 5.0 s.
C alculate a) the acceleration, and b) the distance moved in the 5 .0 s.
Solution a) s = ?; u = 7.5 m s 1 ; v = 2 .5 m s 1 ; a = ? ; t = 5 .0 s Use v = u + at and thereore 2 .5 = 7.5 + a 5 .0 5 .0 so, a = ___ = 1 .0 m s 2 5 .0
The negative sign shows that this is a deceleration. 1 b) s = ut + __ at2 so 2 1 s = 7.5 5 .0 __ 1 .0 5 .0 2 2 = 3 7.5 1 2 .5 = 25 m
2 .1 M OTI ON
Projectile motion Falling freely When an obj ect is released close to the Earths surace, it accelerates downwards. We say that the orce o gravity acts on the obj ect, meaning that it is pulled towards the centre o the E arth. E qually the obj ect pulls with the same orce on the Earth in the opposite direction. Not surprisingly, with small obj ects, the eect o the orce on the E arth is so small that we do not notice it. In Topic 6, we shall look in more detail at the eects o gravity but or the moment we assume that there is a constant acceleration that acts on all bodies close to the surace o the Earth. The acceleration due to gravity at the Earths surace is given the symbol g. The accepted value varies rom place to place on the surace, so that at Kuala Lumpur g is 9.776 m s 2 whereas at Stockholm it is 9. 81 8 m s 2 . Reasons or the variation include variations in the shape o the E arth ( it is not a perect sphere, being slightly attened at the poles) and eects that are due to the densities o the rocks in dierent locations. The dierent tangential speeds o the Earth at dierent latitudes also have an eect. It is better to buy gold by the newton at the equator and sell it at the North Pole than the other way round!
Investigate! Measuring g Alternative 1
There are a number o ways to measure g. This method uses a data logger to collect data. One o the problems with measuring g by hand is that the experiment happens quickly. Manual collection o the data is difcult.
An ultrasound sensor is mounted so that it senses obj ects below it.
S et the logging system up to measure the speed o the obj ect over a time o about 1 s. S et a sensible interval between measurements.
S witch on the logger system, and drop an obj ect vertically that is large enough or the sensor to detect.
The output rom the system should be a speedtime graph that is a straight line; it may be that the loggers sotware can calculate the gradient or you.
You could extend this experiment by testing obj ects o dierent mass but similar size and shape to confrm a suggestion by Galileo that such dierences do not aect the drop.
Ultrasound sensor
Figure 10 Ultrasound sensor.
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M E C H AN I C S
Alternative 2
There are other options that do not involve a data logger.
This system measures the time o ight t o the sphere rom the contacts to the trapdoor.
Measure the distance h rom the bottom o the sphere to the top o the trapdoor ( you might think about why these are the appropriate measurement points) .
A possible way to carry out the experiment is to measure t or one value o h with, o course, a ew repeat measurements or the
electromagnet
ruler
height of fall = h
to switch and timer
1 gt2 with u = 0 to same h. Then use h = ut + __ 2
trap-door
calculate g.
Figure 11 Trap door method.
A magnetic feld holds a small steel sphere (such as a ball bearing) between two metal contacts. The magnetic feld is produced by a coil o wire with an electric current in it. When the current is switched o, the feld disappears and the sphere is released to all vertically.
As the sphere leaves the metal contacts, a clock starts. The clock stops when the sphere opens a small trapdoor and breaks the connection between the terminals o a timing clock or computer. ( The exact details o these connections will depend on the equipment you have.)
This is a one- o measurement that is prone to error. C an you think o some reasons why? O ne way to reduce the errors is to change the vertical distance h between the sphere and trapdoor and to plot a graph o h against t2 . g The gradient o the graph is __ . I you observe 2 an intercept on the h- axis, what do you think it represents?
Alternative 3
There is a urther method that involves taking a video or a multiash image o a alling object and analysing the images to measure g. You will see an example o such an analysis later in this sub-topic.
What goes up must come down displacement
Perhaps you have seen a toy rocket flled with water under pressure and fred vertically upwards? O r you may have thrown a ball vertically, high into the air?
time
Figure 12a Displacementtime for ball thrown vertically.
maximum height
distance
0 0
40
time
Figure 12b Distancetime for ball thrown vertically.
Ater the ball has been released, as the pull o gravity takes eect, the ball slows down, eventually stopping at the top o its motion and then alling back to Earth. I there was no air resistance the displacement time graph would look like fgure 1 2 a. Remember that this is a graph o vertical displacement against time, not the shape o the path the ball makes in the air which is called the traj ectory. The ball is going vertically up and then vertically down to land in the same spot rom where it began. A distancetime graph would look dierent ( fgure 1 2 b) , it gives similar inormation but without the direction part o the displacement and velocity vectors. Make sure that you understand the dierence between these graphs. The suvat equations introduced earlier can be used to analyse this motion. The initial vertical speed is u, the time to reach the highest point is t, the highest point is h and the acceleration o the rocket is g. The
2 .1 M OTI ON
sign o g is negative because upwards is the positive direction. As the acceleration due to gravity is downwards, g must have the opposite sign. The kinematic equations are printed again but with dierences to reect the vertical motion to the highest point: 0 = u gt which comes rom v = u + at 1 1 h = ut __ gt2 which comes rom s = ut + __ at2 2 2
0 = u 2 2 gh which comes rom v2 = u 2 + 2 as I you want to fnd out the time or the entire motion ( that is, up to the highest point and then back to E arth again) , it is simply 2 t.
Reminder! Try to remember this crucial point about the signs in the equations when you answer questions on vertical motion: upwards is +ve and downwards is ve. Something else that students forget is that at the top of the motion the vertical speed of the rocket is zero.
Worked examples 1
A student drops a stone rom rest at the top o a well. She hears the stone splash into the water at the bottom o the well 2.3 s ater releasing the stone. Ignore the time taken or the sound to reach the student rom the bottom o the well. a) C alculate the depth o the well. b) C alculate the speed at which the stone hits the water surace. c) Explain why the time taken or the sound to reach the student can be ignored.
Solution a) The acceleration due to gravity g is 9.8 m s 2 . u = 0; t = 2 .3 s. s= s=
1 ut + __ at2 and thereore 2 1 9.8 2 .3 2 0 + __ 2
= 26 m b) v = u + at; v = 0 + 9. 8 2 . 3 = 2 3 m s 1 c) The speed o sound is about 300 m s 1 and so the time to travel about 25 m is about 0.08 s. Only about 4% o the time taken or the stone to all. 2
A hot-air balloon is rising vertically at a constant speed o 5 .0 m s 1 . A small obj ect is released rom rest relative to the balloon when the balloon is 3 0 m above the ground. a) C alculate the maximum height o the obj ect above the ground. b) C alculate the time taken to reach the maximum height. c) C alculate the total time taken or the obj ect to reach the ground.
Solution a) The obj ect is moving upwards at + 5 .0 m s 1 when it is released. The acceleration due to gravity is 9. 8 m s 2 . When the object is released it will continue to travel upwards but this upward speed will decrease under the inuence o gravity. When it reaches its maximum height it will stop moving and then begin to all. 0 5 v2 = u 2 + 2 as and s = _______ = + 1 .3 m 2 9 . 8 2
This shows that the obj ect rises a urther 1 .3 m above its release point, and is thereore 3 1 .3 m above the ground at the maximum height. 0 5 b) v = u + at; t = ____ = + 0.5 1 s ( the plus sign 9 . 8 shows that this is 0.5 1 s ater release)
c) Ater reaching the maximum height ( at which point the speed is zero) the obj ect alls with the acceleration due to gravity. s = 3 1 .3 m ; u = 0; v = ?; a = 9.8 m s 2 ; t = ? 1 at2 , 3 1 . 3 = 0 0.5 9.8 t2 . Using s = ut + __ 2
( Notice that s is 3 1 .3 m as it is in the opposite direction to the upwards + direction.) This gives a value or t o 2 . 5 3 s. The positive value is the one to use. Think about what the negative value stands or. S o the total time is the 0.5 1 s to get to the maximum height together with the 2 .5 3 s to all back to Earth. This gives a total o 3.04 s which rounds to 3.0 s. Notice that, in this example, i you carry the signs through consistently, they give you inormation about the motion o the obj ect.
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Moving horizontally There is little that is new here. We are going to assume that the surace o the E arth is large enough or its surace to be considered at ( Topic 6 will go into more details o what happens in reality) and that there is no riction. Gravity acts vertically and not in the horizontal direction. This will be important when we combine the horizontal and vertical motions later. B ecause the horizontal acceleration is zero, the suvat equations are simple. For horizontal motion:
the horizontal velocity does not change
the horizontal distance travelled is horizontal speed time for the motion.
Putting it all together
projected to the right
dropped
A student throws a ball horizontally. Figure 1 3 shows multiple images o the ball every 0 . 1 0 s as it moves through the air. The picture also shows, or comparison, the image o a similar ball dropped vertically at the same moment as the ball is thrown. What do you notice about both images? It is obvious which o the two balls was thrown horizontally. C areul examination o the images o this ball should convince you that the horizontal distance between them is constant. Knowing the time interval and the distance scale on the picture means that you can work out the initial ( and unchanging) horizontal speed.
dropped
Figure 13 Multi-fash images o two alling objects.
projected to the right
The images tell us about the vertical speeds too. The clue here is to concentrate on the ball that was dropped vertically. The distance between images (strictly, between the same point on the ball in each image) is increasing. The distance s travelled varies with time t rom release as s t2 . I t doubles then s should increase by actor o 4. D oes it look as though this is what happens? C areul measurements rom this fgure ollowed by a plot o s against t2 could help you to confrm this. The two motions, horizontal and vertical, are comp letely indep endent of each other. The horizontal speed continues unchanged ( we assumed no air resistance) while the vertical speed changes as gravity acts on the ball. This independence allows a straightorward analysis o the motion. The horizontal and the vertical parts o the motion can be split up and treated separately and then recombined to answer questions about the velocity and the displacement or the whole o the motion. The position is summed up in fgure 1 4. At two positions along the traj ectory, the separate components o velocity are shown and the resultant ( the actual velocity including its direction) is drawn.
Figure 14 Horizontal and vertical speed components.
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Real- lie situations do not always begin with horizontal motion. A wellaimed throw will proj ect the ball upwards into the air to achieve the best range ( overall distance travelled) . B ut the general principles above still allow the situation to be analysed.
2 .1 M OTI ON
vertical speed = 0 at maximum height
at maximum height: 0 = u sin - gt
gt 2 and 2 0 = u 2 sin 2 - 2gh h = (u sin ) t -
horizontal speed is constant if air resistance negligible
displacement
vertical acceleration is -g
range: = 2t (u cos ) t is the time to maximum height
h g only acts on vertical speed initial velocity u at to horizontal initial horizontal component = u cos initial vertical component = u sin
time time for whole motion is twice time to maximum height
Figure 15 Projectile motion.
S tudy fgure 1 5 careully and apply the ideas in it to any proj ectile problems you need to solve.
Worked examples 1
An arrow is fred horizontally rom the top o a tower 3 5 m above the ground. The initial horizontal speed is 3 0 m s 1 . Assume that air resistance is negligible. C alculate:
c) To calculate the velocity, the horizontal and vertical components are required. The horizontal component remains at 3 0 m s 1 . The vertical speed is calculated using v = u + at and is 0 + 9.8 2 .67 = 2 6. 2 m s 1 . _________
The speed is 3 0 2 + 2 6.2 2 = 3 9. 8 m s 1 which rounds to 40 m s 1 .
a) the time or which the arrow is in the air b) the distance rom the oot o the tower at which the arrow strikes the ground c) the velocity at which the arrow strikes the ground.
Solution a) The time taken to reach the ground depends on the vertical motion o the arrow. At the instant when the arrow is fred, the vertical speed is zero. The time to reach the ground can be ound 1 using s = ut + __ at2 2 2 35 t2 = _, so t = 2 .67 s or 2 .7 s to 2 s. . 9.8 b) The distance rom the oot o the tower depends only on the horizontal speed.
The angle at which the arrow strikes the 2 6.2 ground is tan ____ = 41 30 2
An obj ect is thrown horizontally rom a ship and strikes the sea 1 .6 s later at a distance o 3 7 m rom the ship. C alculate: a) the initial horizontal speed o the obj ect b) the height o the obj ect above the sea when it was fred thrown.
Solution a) The obj ect travelled 3 7 m in 1 .6 s and the 37 horizontal speed was ___ = 2 3 m s 1 . 1 .6 1 b) Use s = ut + __ at2 to calculate s above the sea. 2
s = 0 + 0. 5 9.8 1 .6 2 = 1 2 .5 m
s = ut = 3 0 2 .67 = 80. 1 m = 80 m.
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M E C H AN I C S
2.2 Forces Understanding
Applications and skills
Objects as point particles
Representing orces as vectors
Free-body diagrams
Sketching and interpreting ree-body diagrams
Translational equilibrium
Describing the consequences o Newtons frst
Newtons laws o motion Solid riction
Nature of science The application o mathematics enhances our understanding o orce and motion. Isaac Newton was able to use the work o earlier scientists and to ormalize it through the use o the calculus, a orm o which he developed or this purpose. He made many insights during his scientifc thinking within the topic o orce and motion, but also beyond it. The story o the alling apple, whether true or not, illustrates the importance o serendipity in science and the requirement that the creative scientist can orm links that go beyond what exists already.
law or translational equilibrium Using Newtons second law quantitatively and qualitatively Identiying orce pairs in the context o Newtons third law Solving problems involving orces and determining resultant orce Describing solid riction (static and dynamic) by coecients o riction
Equations Newtons second law: F = ma static riction equation: F s R dynamic riction equation: F = d R
Introduction We depend on forces and their effects for all aspects of our life. Forces are often taught in most elementary physics courses as though they are pushes or pulls, but forces go well beyond this simple description. Forces can change the motion of a body and they can deform the shapes of bodies. Forces can act at a distance so that there is no contact between obj ects or between a system that produces a force and the obj ect on which it acts.
TOK Aristotle and the concept of force Discussions about the concept o what is meant by a orce go back to the dawn o scientifc thought. Aristotle, a Greek philosopher who lived about 2300 years ago, had an overarching view o the world (called an Aristotelian cosmology) and he can be regarded as being an important actor in the development o science. The German philosopher Heidegger wrote that there would have been no Galileo without Aristotle beore him.
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Despite his importance to us, however, we would not have regarded Aristotle as a scientist in any modern sense. For one thing he is not known to have perormed experiments to veriy his ideas; some o his ideas seem very odd to us today. Aristotle believed in the nature o all objects including living things. He believed that all objects had a natural state which was to be motionless on the surace o the Earth and that all
2 . 2 FO R CE S
objects, i let alone, would try to attain this state. Then he distinguished between natural motion in which, or example, heavy objects all downwards and unnatural or orced motion in which the objects need to have a orce continually applied i they are to remain anywhere other than their natural state. Unortunately or those learning physics this is a very persuasive idea because we know intuitively that i we want to hold something in our hand, our muscles have to keep working in order to do this.
There are many other examples o Aristotelian thought and how later scientists moved our thinking orward. But it is important to remember the contribution that Aristotle made to science, even i some o his ideas are now overturned. What do you think students in 50, 100, or even a 1000 years will make o the physics o our century?
Newtons laws of motion Newtons frst law B y the time o Galileo, scientists had begun to realise that things were not as simple as the Greek philosophers such as Aristotle had thought. They were coming to the view that moving obj ects have inertia, meaning a resistance to stopping and that, once in motion, obj ects continue to move. Galileo carried out an experiment with inclined planes and spheres. In act, this may have been a thought experiment this was oten the way orward in those days but in any event it is easy to see what Galileo was trying to suggest.
(a)
(b)
(c)
Figure 1 Galileos thought experiment.
In the frst experiment ( a) , the two arms o the inclined plane are at the same angle and the sphere rolls the same distance up the slope as it rolled down ( assuming no energy losses) . In the second experiment ( b) , the second arm is at a lower angle than beore but the sphere rolls up this plane to the same height as that rom which it was released. Galileo then concluded that i the second plane is horizontal ( c) , the sphere will go on rolling or ever because it will never be able to climb to the original release height. Newton included this idea in his frst law o motion that says: An obj ect continues to remain stationary or to move at a constant velocity unless an external orce acts on it. Galileo had suggested that the sphere on his horizontal plane went on orever, but it was Newton who realised that there is more to say than this. Unless something rom outside applies a orce to change it, the velocity o an obj ect ( both its speed and its direction) must remain the same.
45
2
M E C H AN I C S This was directly opposed to Aristotles view that a force had to keep pushing constantly at a moving obj ect for the speed to remain the same.
Newtons second law The next step is to ask: if a force does act on an obj ect, in what way does the velocity change? Newton proposed that there was a fundamental equation that connected force and rate of change of velocity. This is contained in his second law of motion. This law can be written in two ways, one way is more complex than the other and we will look at the more complicated form of the law later in this topic. Newtons second law, in its simpler form, says that Force = mass acceleration As an equation this can be written F = ma The appropriate S I units are force in newtons ( N) , mass in kilograms ( kg) , and acceleration in metres per second 2 ( m s 2 ) . As discussed in Topic 1 the newton is a derived unit in S I. We can represent it in terms of fundamental units alone as kg m s 2 . Two things arise from this equation:
Mass is a scalar, so there will not be a change in the direction of the acceleration if we multiply acceleration by the mass. The direction of the force and the direction of the acceleration must be the same. So, applying a force to a mass will change the velocity in the same direction as that of the force.
One way to think about the mass in this equation is that it is the ratio of the force required per unit of acceleration for a given object. This helps us to standardize our units of force. If an object of mass 1 kg is observed to accelerate with an acceleration of 1 m s 2 then one unit of force (1 N) must have acted on it.
Nature of science Inertial and gravitational mass W hat e xactly do we m e an b y m ass? The m as s that we u s e in Ne wto ns s e co nd law o f m o tio n is ine rtial m as s . Ine rtial m as s is the p ro p e rty that p e rm its an o b j e ct to re s is t the e ffe cts o f a fo rce that is trying to change its m o tio n. In o the r p arts o f this b o o k, m as s is u s e d in a diffe re nt co nte xt. We talk ab o u t the we ight o f an o b j e ct and we kno w that this we ight aris e s fro m the gravitatio nal attractio n b e twe e n the
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m as s o f the o b j e ct and the m as s o f the E arth. The two m as s e s in this cas e are gravitatio nal m as s e s and are the re s p o ns e o f m atte r to the e ffe cts o f gravity. It has b e e n e xp e rim e ntally ve rifie d that we ight is p ro p o rtio nal to m as s to b e tte r than 3 p arts in 1 0 1 1 . It is a p o s tu late o f E ins te ins ge ne ral the o ry o f re lativity that ine rtial m as s and gravitatio nal m as s are p ro p o rtio nal.
2 . 2 FO R CE S
Investigate! Force, mass, and acceleration card
break the light beam to calculate, frst, the average speed at each gate and then ( using the distance apart o the gates) , the acceleration o the cart. The kinematic equations are used or this. ( This is why you must pull the cart with a constant orce so that the acceleration is uniorm.)
light gate
cart
Repeat the experiment with two, three, and possibly our elastic threads, all identical, all extended by the same amount. This means that you will be using one unit o orce ( with one thread) , two units ( with two threads) and so on.
Plot a graph o calculated acceleration against number o orce units. Is your graph straight? D oes it go through the origin? Remember that this experiment has a number o possible uncertainties in j udging the best- ft line.
Experiment 1: force and acceleration
The idea in this experiment is to measure the acceleration o the cart ( o constant mass) when it is towed by dierent numbers o elastic threads each one extended by the same amount. The timing can be done using light gates and an electronic timer as shown here. Alternatively it can be done using a data logger with, say, an ultrasound sensor and the cart moving away rom the sensor. Practise accelerating the cart with one elastic thread attached to the rear o the cart. The thread( s) will have to be extended by the same amount or each run. A convenient point to j udge is the orward end o the cart. Make sure that your hand clears the light gates i you are using this method. Your hand needs to move at the same speed as the cart so that the thread is the same length throughout the run.
Experiment 2: mass and acceleration
The setup is essentially the same as or Experiment 1 , except this time you will use a constant orce ( possibly two elastic threads is appropriate) .
C hange the mass o your cart (some laboratory carts are specially designed to stack, one on top o another) and measure the acceleration with a constant orce and varying numbers o carts.
This time Newtons second law predicts that mass should be inversely proportional to acceleration. Plot a graph o acceleration 1 against ____ m ass . Is it a straight line?
2
An aircrat o mass 3 .3 1 0 5 kg takes o rom rest in a distance o 1 .7 km. The maximum thrust o the engines is 83 0 kN. C alculate the take- o speed.
The card on the cart needs to be o a known length so that you can use the time taken to
Worked examples 1
A car with a mass o 1 5 00 kg accelerates uniormly rom rest to a speed o 2 8 m s 1 ( about 1 00 km h 1 ) in a time o 1 1 s. C alculate the average orce that acts on the car to produce this acceleration.
Solution
8.3 1 0 The acceleration o the aircrat is _______ = 3.3 1 0 2 2 .5 1 m s .
vu 28 The acceleration = a = _ = _ = 2 .5 4 m s 2 . t 11 Force = ma = 1 5 00 2 . 5 4 = 3 . 8 kN.
v2 = u 2 + 2 as so v = 0 + 2 2 . 5 1 1 700 which leads to v = 92 .4 m s 1 .
5
5
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TOK
Newtons third law
But are they really laws?
Newtons third law of motion is sometimes written in such a way that the true meaning o the law has to be teased out. The law can be expressed in a number o equivalent ways.
The essential question is: can Newtons laws of motion be proved? The answer is that they cannot; strictly speaking they are assertions as Newton himself recognized. In his famous Principia (written in Latin as was a custom in those days) he writes Axiomata sive leges motus, or the axioms or laws of motion. However, they appear to be an excellent set of rules that allow us to predict most of the motion that we undertake. They remained unchallenged as a theory for about 200 years until the two theories of relativity were formulated by Einstein at the turn of the twentieth century. Essentially (said Einstein) the rules that Newton proposed do not always apply to, for example, motion that is very fast. However, for the modest speeds at which humans travel, the rules are reliable to a high degree and are certainly good enough for our needs most of the time.
O ne common way to write Newtons third law o motion is: E very action has an equal and op p osite reaction. The frst point to make is that the words action and reaction really mean action orce and reaction orce. S o, once again, Newton is reerring directly to the eects o orces. A second point is that the actionreaction pair must be o the same type. S o a gravitational action orce must correspond to a gravitational reaction. It could not, or example, reer to an electrostatic orce. The law suggests that orces must appear in pairs, but in thinking about a particular situation it is important to identiy all possible orce pairs and then to pair them up correctly. Take, as an example, the situation o a rubber ball resting on a table. At frst glance the obvious action orce here is the weight o the ball, that is, the gravitational pull the E arth exerts on the sphere. This orce acts downwards and i the table were not there the ball would accelerate downwards according to Newtons second law. It would all to the oor. What is the reaction orce here? Given that action orce and reaction orce must pair up like or like, the reaction must be the gravitational orce that the ball exerts on the Earth. This is o exactly the same size as the pull o the Earth on the ball, but is in the opposite direction. What prevents the ball accelerating downwards to the oor? A orce must be exerted by the table on the ball and i there are no accelerations happening, then this orce is equal and opposite to the downwards gravitational pull o the E arth on the ball. B ut the upwards table orce is not the reaction orce to the balls weight we have already seen that this is the gravitation pull on the E arth. The origin o the table orce is the electrostatic orces between atoms. As the ball lies on the table, it deorms the horizontal surace very slightly, rather like what happens when you push downwards with a fnger on a metre ruler suspended by supports at its end the ruler bends in a spring- like way to provide a resistance to the orce acting downwards. The dent in the table surace is the response o the atoms in the table to the weight lying on it. Remove the ball and the surace will return to being at. This upwards orce that returns the table to the horizontal is pushing upwards on the ball. There is a corresponding downwards orce rom the deormed ball ( the ball will become slightly attened as a response to the gravitational pull) . S o here is the second actionreaction pair, between two orces that are electrostatic in origin. In summary, there are our orces in this situation: the weight o the ball, the upwards spring- like orce o the table, and the reactions to these orces which are the pull o the ball on the E arth and the downwards push o the deormed ball on the table.
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2 . 2 FO R CE S
To ( literally) get a eeling or this, take a one- metre laboratory ruler and suspend it between two lab stools as suggested earlier. Press down gently with your fnger in the centre o the ruler so that it becomes curved. The ruler will bend, you will be able to eel it resisting your eorts to deorm it too ar. Remove your fnger and the ruler will return to its original shape. In explaining Newtons third law in a particular example, you must remember to emphasise the nature, the size, and the direction o the orce you are describing. Another common example is that o a rocket in space. S tudents sometimes write that .. .by Newtons third law, the rocket pushes on the atmosphere to accelerate but this shows a poor understanding o how the propulsion actually works. First, o course, the rocket does not push on anything. The act that a rocket can accelerate in space where there is no atmosphere proves this. What happens inside the rocket is that chemicals react together producing a gas with a very high temperature and pressure. The rocket has exhaust nozzles through which this gas escapes rom the combustion chamber. At one end o the chamber inside the rocket the gas molecules rebound o the end wall and exert a orce on it, as a result they reverse their direction. In principle, the rebounding molecules could then travel down the rocket and leave through the nozzles. S o there is an action- reaction pair here, the orce orwards that the gas molecules exert on the chamber ( and thereore the rocket) and the orce that the chamber exerts on the gas molecules. It is the frst o these two orces that accelerates the rocket. I the chamber were completely sealed and the gas could exert an equal and opposite orce at the back o the rocket, then the orward orce would be exactly countered by the backwards orce and no acceleration would occur.
Figure 2 Chemical rockets in action.
Later we shall interpret this acceleration in a dierent way. B ut the explanation given here will still be correct at a microscopic level. Think about the ollowing situations and discuss them with ellow students.
A freman has to exert considerable orce on a fre hose to keep it pointing in the direction that will send water to the correct place.
There is a suggestion to power space travel to deep space by ej ecting ions rom a spaceship.
A sailing dinghy moves orward when the wind blows into the sails.
Free-body force diagrams As a vector quantity, a orce can be represented by an arrow that gives both the scaled length and the direction o the orce. In simple cases where ew orces act this works well, but as the situations become more complex, diagrams that show all the arrows can become complicated. O ne way to avoid this problem is to use a free-body force diagram to illustrate what is happening. The rules or a ree- body diagram or a body are simple, as ollows.
49
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M E C H AN I C S
The diagram is or one body only and the orce vectors are represented as arrows.
O nly the orces acting on the body are considered.
The orce ( vector) arrows are drawn to scale originating at a point that represents the centre o mass o the body.
All orces have a clear label.
Procedure for drawing and using a free-body diagram
B egin by sketching the general situation with all the bodies that interact in the situation.
S elect the body o interest and draw it again removed rom the situation.
D raw, to scale, and label all the orces that act on this body due to the other bodies and orces.
Add the orce vectors together ( either by drawing or calculation) to give the net orce acting on the body. The sum can be used later to draw other conclusions about the motion o the obj ect.
Examples of free-body diagrams 1
A ball falling freely under gravity with no air resistance. situation diagram ball
pull of Earth on ball
pull of ball on Earth
Earth
free-body diagram for the ball ball acceleration of ball pull of Earth on ball
Figure 3 Ball falling freely under gravity.
This straightorward situation speaks or itsel. There are two orces acting: the Earth pulling on the ball and the ball pulling on the Earth. For the ree-body diagram o the ball we are only interested in the frst o these. So the ree-body diagram is a particularly simple one, showing the object and one orce. Notice that the ball is not represented as a real obj ect, but as a point that reers to the centre o mass o the object.
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2 . 2 FO R CE S 2
The same ball resting on the ground.
general situation
free-body diagram for the ball force on ball from ground
acceleration = 0
ball
ball
ground
force of Earth on ball
Figure 4 Ball resting on the ground.
As we saw earlier, our orces act in this case:
the weight o the ball downwards
the reaction o the Earth to this weight
the upwards orce rom the ground
the reaction o the ball to this spring-like orce.
A ree- body diagram ( fgure 4) defnitely helps here because, by restricting ourselves to the orces acting on the ball, the our orces reduce to two: the weight o the ball and the upwards orce on it due to the deormation o the ground. These two orces are equal and opposite. The net resultant ( vector sum) o the orces is zero and there is thereore no acceleration. 3
An object accelerating upwards in a lift.
The weight o the object is downwards and the size o this orce is the same as though the object had been stationary on the Earths surace. However, the upwards orce o the oor o the lit on the object is now larger than the weight and the resultant orce o the two has a net upwards component. The object is accelerated upwards as you would expect. For the lit, there is an upward orce in the lit cable and a downwards weight o the lit is downwards together with the weight o the object. The resultant orce is upwards and is equal to the orce in the cable less the weight orces o lit and obj ect. general situation
free-body diagram for the mass lift cable
free-body diagram for the lift
acceleration of lift upwards force exerted by lift on object
acceleration of object lift
acceleration of object upwards
object
object
force exerted by cable on lift
force exerted by earth on lift (weight of lift)
force exerted by Earth on object force exerted by object on lift
Figure 5 Mass in a lift accelerating upwards.
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M E C H AN I C S
Translational equilibrium
orce 1
orce 2
(a)
orce components add vertically
When an obj ect is in translational equilibrium it is either at rest or moving at a constant velocity ( not j ust constant speed in this case) . Translational here means moving in a straight line. ( There is also rotational equilibrium where something is at rest or rotating at a constant angular speed; this is discussed in option B ) . Newtons frst and second laws remind us that i there is no change o velocity then there must be zero orce acting on the object. This zero orce in many cases is the resultant (addition) o more than one orce. In this section we will examine what equilibrium implies when there is more than one orce. The simplest case is that o two orces. I they are equal in size and opposite in direction, they will cancel out and be in equilibrium ( Figure 6( a) ) .
orce components cancel horizontally (b)
Figure 6 Two equal orces in diferent directions.
I the orces are equal in size but not in the same direction then equilibrium is not possible. Horizontally, in fgure 6( b) , the two components o the orce vectors are still equal and opposite and we could expect that there would be no change in this direction. B ut vertically the two vector components point in the same direction so that overall there will be an unbalanced vertical orce and an acceleration acting on the obj ect on which these orces act. This gives a clue as to how we should proceed when there are three or more orces. free-body diagram
situation diagram
T1
gb
a la
nc
e
T2
sp
ri n
mg
string ring
weight
T2 sin T1
T2 cos horizontally
mg
vertically
Figure 7 Three orces acting on a ring.
Figure 7 shows a situation diagram and a ree- body diagram or a ring on which three orces act. For equilibrium, in whatever direction we resolve the orces, all three components must add up to zero in this direction.
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2 . 2 FO R CE S
Figure 7 shows that horizontal and vertical are two good directions with which to begin, because two orces are aligned with these directions and one disappears in each direction chosen. Thus, vertical orce mg has no component in the horizontal direction and horizontal orce T1 has no part to play vertically. B ut whichever direction is chosen, i there is to be no resultant orce and consequently no acceleration, then all the orces must cancel. There is one more consequence o this idea. Figure 8 shows the orces drawn, as usual, to scale and in the correct direction ( in red) . The orces can be moved, as shown by the green arrows, into a new arrangement ( shown in black) . What is special here is that the three orces orm a closed triangle where all the arrows meet.
T2 Move the mg vector vertically upwards to begin at the start of T1 .
Move T2 sideways to the end of T1 . T1 mg
The three vectors now form a closed triangle. The three forces are in translational equilibrium.
Figure 8 Making a triangle for forces.
Algebraically, this must be true because we know (ignoring the directions o the orces) that T1 = T2 cos ( horizontally) and mg = T2 sin ( vertically) So T12 = T22 cos 2 and ( mg) 2 = T22 sin 2 Adding these together gives T12 + ( mg) 2 = T22 ( sin 2 + cos 2 ) and thereore as sin 2 + cos 2 = 1 T12 + ( mg) 2 = T22 This is equivalent to Pythagoras theorem where x = T1 2 , y = ( mg) 2 and z = T2 2 so that x2 + y 2 = z2 . S o, because the sum o the squares o two vector lengths equals the square o the third vector length, the three must ft together as a right- angled triangle. We have a right- angled triangle ormed by the scaled lengths and directions o the three orce vectors. This important fgure is known as a triangle of forces. I you can draw the vectors or a system in this way, then the system must be in translational equilibrium.
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Investigate! Three forces in equilibrium T2
Your arrangement or hanging the spring balances will depend on the resources in your school.
S elect a mass to suit the sensitivity o your spring balances.
Place a vertical drawing board with a large sheet o paper pinned to it behind the balances so that you can mark the position o the strings. When these are marked, use a protractor to measure the angles.
B egin with a simple case, say when T1 is horizontal so that 1 = 90.
The knot must be stationary and when this is true:
T1
2
1
string
mg
I a point is in equilibrium under the infuence o three orces, then the three orces must orm a closed triangle. To veriy this, use a mass and two spring balances with a knot halway along the string between the two spring balances to act as the point obj ect.
T1 /N
1 /
T2 /N
2 /
mg/N
T1 cos 1 = T2 cos 2 ( resolving horizontally)
T1 sin 1 + T2 sin 2 = mg ( resolving vertically)
Construct this table to enable you to veriy that both equations are true or every case you set up.
T1 cos 1 /N
T1 sin 1 /N
T2 cos 2 /N
T2 sin 2 /N
Case 1 For each case check that the equations are correct within experimental error.
Solid friction Friction is the orce that occurs between two suraces in contact. I you live in a part o the world where there is snow and ice then you will know that when the riction between your shoes and the ice disappears it can be a good thing ( or skiing) or a bad thing ( or alling over) .
Investigate! Solid friction
spring balance
In this experiment you use a spring balance to measure the orce that acts between two suraces. It gives some unexpected results. weight
Set up the platorm on rollers with a suitable weight and a spring balance. A string connects the spring balance to the weight. Ensure that the weight is attached securely to the block.
Pull the platorm using the winch system with an increasing orce observing, at the same time, the reading on the balance.
rollers
winch system
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platform
2 . 2 FO R CE S
Note the maximum value o the orce as measured on the balance.
Note the way the size o the orce changes when the platorm begins to move.
Is this orce constant? O r does it uctuate?
C hange the size o the weights on the platorm. How does the total weight aect the orce measured on the balance?
C hange the type o surace at the bottom o the weights. You might use abrasive ( emery) paper or cloth in dierent abrasive grades pinned to the platorm.
Investigate other changes: does lubrication at the bottom o the weights aect the riction?
I you carried out the Investigate! S olid riction experiment you may have ound that:
The orce on the balance increases as you pull harder and harder, but the platorm does not begin to move relative to the weights immediately.
Eventually the platorm suddenly begins to move at a particular value o orce and at this instant the orce, shown by the balance, drops to a new lower value.
This new value is then maintained as the platorm moves steadily.
You may observe stick-slip behaviour where the platorm alternately sticks and then j umps to a new sticking position. This behaviour is associated with two values o riction, but this may be too difcult to observe unless you get very suitable suraces.
The riction orces depend on the magnitude o the weights.
The rictional orces that occur in this experiment are described empirically as:
static riction ( when there is no relative movement between the suraces)
dynamic riction ( when there is relative movement) .
As the pulling orce increases but without any slip happening, the riction is said to be static (because there is no motion) . Eventually however, the pulling orce will exceed the value o the static riction and the suraces will start to move. As the movement continues, the riction orce drops to a new value, lower than the maximum static value. This new lower value is known as the dynamic riction. B oth static and dynamic riction orces are highly dependent on the two suraces concerned.
Static friction The static riction orce F is ound to be given empirically by F s R where F is the rictional orce exerted by the surace on the block. R is the normal reaction o the surace on the block, this equals the weight o the block as there is no vertical acceleration. The symbol s reers to the coefcient o static riction. The less than or equal to symbol indicates that the static riction orce can vary rom zero up to a maximum value. B etween these limits F is
55
2
M E C H AN I C S equal to the pull on the block. O nce the pull on the block is equal to F, then the block is j ust about to move. O nce the pull exceeds F then the block begins to slide ( fgure 9) . For these larger orces, the riction operating is in the dynamic regime.
pull on block
pull of Earth on block
friction force exerted by surface on block
reaction force of surface on block
Figure 9 Friction force acting on an object.
Dynamic friction Dynamic riction only applies when the suraces move relative to each other. The riction drops rom its maximum static value (there is an explanation or this below) and remains at a constant value. This value depends on the total reaction orce acting on the surace but (according to simple theory) is not thought to depend on the relative speed between the two suraces. So, or dynamic riction F = dR where d is the coefcient o dynamic riction. The values o s and d vary greatly depending on the pair o suraces being used and also the condition o the suraces ( or example, whether lubricated or not) . A ew typical values are given in the table. I you want to investigate a wider range o suraces, there are many sources o the coefcient values on the Internet search or C oefcients o riction . Each riction coefcient is a ratio o two orces (F and R) and so has no units.
Surface 1
Surface 2
s
d
glass
metal
0.7
0.6
rubber
concrete
1.0
0.8
rubber
wet tarmac
0.6
0.4
rubber
ice
0.3
0.2
metal
metal (lubricated)
0.15
0.06
Values o s and d or p airs o suraces It is possible or the coefcients to be greater than 1 or some surace pairs. This reects the act that or these suraces the riction is very strong and greater than the weight o the block. Remember that the suraces are being pulled sideways by a horizontal orce whereas the reaction orce is vertical so we are not really comparing like with like in these empirical rules or riction.
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Investigate! Friction between a block and a ramp One way to measure the static coefcient o riction between two suraces in a school laboratory is to use the two suraces as part o a ramp system.
O ne surace is the top o the ramp, the other is the base o the block.
Resolving at 90 to plane, N = mg cos ; resolving along the plane, F = mg sin F S o tan = _ = s N S tart with the ramp horizontal, and then gradually raise one end until the block starts to slip.
free-body diagram of just the block
a block on a ramp
N
Ff
N
Ff
m
At this moment o slip, measure the angle o the ramp. The tangent o this angle is equal to the coefcient o static riction.
2
A skier places a pair o skis on a snow slope that is at an angle o 1 . 7 to the horizontal. The coefcient o static riction between the skis and the snow is 0.02 5 .
mg
mg
Worked examples 1
A box is pushed across a level oor at a constant speed with a orce o 2 80 N at 45 to the oor. The mass o the box is 5 0 kg.
F 45
D etermine whether the skis will slide away by themselves.
Solution reaction force of surface on ski
C alculate:
weight component down slope
friction force
1.7
a) the vertical component o the orce
weight of ski
b) the weight o the box c) the horizontal component o the orce d) the coefcient o dynamic riction between the box and the oor.
Solution a) the vertical component is 2 80 sin 45 = 1 98 N b) the weight o the box = mg = 5 0 9.8 = 490 N c) the horizontal component o the orce = 2 80 cos 45 = 1 98 N d) the vertical component o the orce exerted by the oor on the box = 490 + 1 98 N = 688 N the riction orce = the horizontal component ( the box is travelling at a steady speed) , so 1 98 d = _ = 0. 2 9 688
C all the weight o the skis W. The component o weight down the slope = W sin 1 .7 The reaction orce o the surace on the ski = W cos 1 .7. Thereore, the maximum riction orce up slope = S W cos 1 .7. The skis will slide i S W cos 1 .7 < W sin 1 .7 in other words, i S < tan 1 .7. The value o tan 1 .7 is 0.02 96 and this is greater than the value o S , which is 0.02 5 , so the skis will slide away.
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Origins of friction Friction originates at the interace between the two materials, in other words at the surace where they meet. The actual causes o riction are still being investigated today and the explanation given here is highly simplifed and a historical one. Leonardo da Vinci mentions riction in his notebooks and some o the next scientifc writings about riction appear in works by Guillaume Amontons rom around 1 700. moving to right
(a)
moving to left
liquid lubricant layer
(b)
Figure 10 How solid friction arises.
One model or solid riction suggests that what are very smooth suraces to us are not smooth at all. At the atomic level, they are actually ull o peaks and troughs o atoms (fgure 1 0(a) ) . When static riction occurs and two suraces are at rest relative to each other, then the atomic peaks rest in the troughs, and it needs a certain level o orce to deorm or break the peaks sufciently or sliding to begin. Once relative motion has started (so that dynamic riction occurs) , then the top surace rises a little above the deormed peaks. Less orce is now required to keep the motion going. B ecause the irregularities on the surace are very small and o atomic size, even the small orces applied in our lab experiments cause large stresses to act on the peaks. The peaks then deorm like a sot plastic irrespective o whether the material is hard steel or something much soter. Moving suraces are oten coated with a lubricant to reduce wear due to riction ( fgure 1 0( b) ) . The lubricant flls the space between the two suraces and either prevents the peaks and troughs o atoms rom touching or reduces the amount o contact. In either event, the atoms rom the suraces do not interact as much as beore and the riction orce and the coefcient are reduced. The origins o these riction orces are bound up in the complex electronic properties o the materials that make up the suraces. However, this simple theory should give you some understanding o riction as well as an awareness that the bulk materials we perceive on the macroscopic scale arise rom microscopic properties that are operating at the atomic level. What is important to realize is that the two equations or static and dynamic riction do not arise rom a study o the interatomic orces between the suraces; they are derived purely rom experiments with bulk materials. The results are empirical not theoretical.
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2 . 2 FO R CE S
Fluid resistance and terminal speed The assumption that air resistance is negligible is oten unrealistic. An obj ect that travels through a uid ( a liquid or a gas) is subj ect to a complex process in which, as the obj ect travels through the material, the uid is stirred up and becomes subj ect to a drag force. The process is complex, so even ater introducing some simple assumptions about the resistance o a uid, we will still not be able to give a complete analysis. In energy terms, the action o air resistance is to transer some o the energy o the moving body into the uid through which it is moving. S ome uids absorb this energy better than others: swimming through water is much more tiring than running in the air. For your IB D iploma Programme physics exams, you only need to describe the eects o uid resistance without going into the mathematics.
Skydiving In 2 01 2 , Felix B aumgartner j umped saely rom a height o 3 9 km above New Mexico to reach a top speed o 1 3 42 km h 1 aster than the speed o sound. A skydive rom more usual heights will not take place at such high speeds, usually up to about 2 00 km h 1 . The dierence is due to the variation in the resistance o the air at dierent heights above the Earth.
drag force mg drag force
mg
mg body released from rest
forces on body during acceleration
forces on body at terminal speed
Figure 11 Forces acting on skydiver.
The diagram shows the orces acting on a skydiver. The weight o the diver acts vertically downwards and is eectively constant ( because there is little change in the Earths gravitational feld strength at the height o the dive) . The air resistance orce acts in the opposite direction to the motion o the diver and or a diver alling vertically, this will be vertical too. O ther orces acting on the diver include the upwards buoyancy caused by the displacement o air by the diver.
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2
M E C H AN I C S When the skydiver initially leaves the aircrat, the divers weight acts downwards and because the vertical speed is almost zero, there is almost no air resistance. Air resistance increases as the speed increases so that as the diver goes aster and aster the resistance orce becomes larger and larger. The net orce thereore decreases and consequently the acceleration o the diver downwards also decreases. E ventually the weight orce downwards and the resistance orce upwards are equal in magnitude and, o course, opposite in direction. At this point there is no longer any acceleration and the diver has reached a constant rate o all known as the terminal sp eed. The graph o vertical speed against time is as shown in fgure 1 2 . parachute opens
vertical speed
accelerates to terminal speed
greater drag force so terminal speed smaller
(not to scale)
lands
time
Figure 12 Speedtime for a parachute jump.
Eventually the skydiver opens the parachute. Now, because o the large surace area o the parachute envelope, the upwards resistive orce is much larger than beore and is greater than the weight. As a result, the directions o the net orce and acceleration are also upwards; the vertical velocity decreases in magnitude. Once again a balance will be reached where the upward and downward orces are equal and opposite but at a much lower speed than beore (about 1 2 m s 1 or a landing) and the diver reaches the ground saely.
Maximum speed of a car The top ( maximum) speed o a motor car is determined by a number o actors. The most obvious o these is the maximum orce that the engine can exert through the tyres on the road surace. B ut, as with the skydiver, this is not the only orce acting. There is a considerable drag on the vehicle due to the air and this drag orce increases markedly as the speed o the car becomes larger. Typically, when the speed doubles the drag orce will increase by at least a actor o our. There is a maximum power that the car engine can produce. When the car accelerates and the speed increases towards the maximum, the power dissipated in riction also increases. When the maximum energy output o the engine every second is completely used in overcoming the energy losses, then the car cannot accelerate urther and has reached its maximum speed.
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2 . 3 W O R K , E N E R G Y, A N D P O W E R
Worked example 1
C alculate the upward force acting on a skydiver of mass 80 kg who is falling at a constant speed.
horizontal force of 3 0 N. The mass of P is 2 .0 kg and the mass of Q is 4.0 kg.
Solution
Q
The weight of the skydiver is 80 9. 8 = 784 N. B ecause the skydiver is falling at constant speed ( i.e. terminal speed) the upwards drag force is equal to the downwards weight. The upward force is 784 N. 2
P and Q are two boxes that are pushed across a rough surface at a constant velocity with a
P
F
S tate the resultant force on box Q.
Solution Q is moving at constant speed. S o the resultant force on this box must be 0 N.
2.3 Work, energy, and power Understanding Principle o conservation o energy Kinetic energy Gravitational potential energy
Applications and skills Discussing the conservation o total energy
Elastic potential energy Work done as energy transer
Power as rate o energy transer Eciency
Nature of science The theory o conservation o energy allows many areas o science to be understood at a undamental level. It allows the explanation o natural phenomena but also means that scientists can predict the outcome o a previously unknown efect. The conservation o energy also demonstrates that paradigm shits occur in science: the interchangeability o mass and energy as predicted by Einstein is an example o this.
within energy transormations Sketching and interpreting orce distance graphs Determining work done including cases where a resistive orce acts Solving problems involving power Quantitatively describing eciency in energy transers
Equations work: W = Fs cos 1 kinetic energy: EK = ___ mv2 2
1 k(x) 2 elastic potential energy: Ep = ___ 2
change in gravitational potential energy:
Ep = mgh power = Fv useul power out useul workout = _________________________ eciency = ______________________ total work in total power in
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M E C H AN I C S
Introduction This sub- topic looks at the physics of energy transfers. The importance of energy is best appreciated when it moves or transfers between different forms. Then we can make it do a useful j ob for us. What is important is that you learn to look below the surface of the bald statement: electrical energy is converted into internal energy and ensure that you can talk about the detailed changes that are taking place.
Energy forms and transfers Energy can be stored in many different forms. S ome of the important ones are listed in the table below.
Energy
Nature of energy associated with...
kinetic (gravitational) potential electric/magnetic chemical nuclear elastic (potential) thermal (heat)
the motion o a mass the position o a mass in a gravitational feld charge owing atoms and their molecular arrangements the nucleus o an atom an object being deormed a change in temperature or a change o state
mass
conversion to binding (nuclear) energy when nuclear changes occur mechanical waves in solids, liquids, or gases
vibration (sound)
light
photons o light
Notes sometimes the word gravitational is not used
related to a mass change by E = mc2 The word potential is not always used A change o state is a change o a substance between phases, i.e. solid to liquid, or liquid to gas. This is reerred to as energy transerred as a result o temperature dierence in line with the IB Guide. The colloquial term heat is usually acceptable when reerring to situations involving conservation o energy situations.
the amount o sound energy transerred is almost always negligible when compared with other energy orms sometimes called radiant energy another orm o electric/magnetic
Energy can be transferred between any of its forms and it is during such transfers we see the effects of energy. For example, water can fall vertically to turn the turbine of a hydroelectric power station and drive a generator. Lots of things are happening here. The water molecules are attracted by the Earth and accelerate downwards through the pipe. Their momentum is transferred to the blades of the turbine which rotates and turns the coils in the generator. As a result of the coils turning, electrons are forced to move and there is an electric current. Overall, this chain of physical processes can be summed up as the conversion of gravitational potential energy of the water into an electrical form. Another example of an energy transfer is that of an animal converting stored chemical energy in the muscles leading to kinetic and gravitational potential energy forms with some of the chemical energy also appearing as frictional losses.
62
2 . 3 W O R K , E N E R G Y, A N D P O W E R Learn to recognise the physical (and sometimes chemical) processes that are going on in the system. It is easy to describe the changes in airly broad energy transer terms. Always try, however, to explain the eects in terms o microscopic or macroscopic interactions. Whatever the orm o the energy transer, we use a unit o energy called the joule (J) . This is in honour o James Joule (1 81 81 889) the English scientist who devoted his scientifc eorts to studying energy and its transers. O ne j oule is the energy required when a force of one newton acts through a distance of one metre. When energy changes rom one orm to another we fnd that nothing is lost ( providing that we take care to include every single orm o energy that is included in the changes) . This is known as the p rincip le of conservation of energy which says that energy cannot be created or destroyed. S ince E insteins work at the turn o the twentieth century, we now recognise that mass must be included in our table o energy orms. For most changes the mass dierence is insignifcant, but in nuclear changes it makes a maj or contribution. In some applications, such as when discussing the output o power stations, the joule is too small a unit so you will requently see energies expressed in megajoules (MJ or 1 0 6 J) or even gigajoules (GJ or 1 0 9 J) . Get used to working in large powers o ten and with prefxes when dealing with energies.
Worked example D escribe the mechanisms associated with the energy changes that occur when a balloon is blown up.
Solution Air molecules gain kinetic energy that is used to store elastic potential energy in the skin o the balloon and to make changes in the energy o the air. The air inside is able to exert more orce ( pressure) outwards on the skin o the balloon until a new equilibrium is established between the tension in the skin and the atmospheric pressure.
In some parts o physics, dierent energy units are used. These have usually arisen historically. An example o this is the electronvolt ( eV) , which is the energy gained by an electron when it is accelerated through a potential dierence o one volt. Another example is the calorie used by dieticians; this is an old unit that has lingered in the public domain or perhaps longer than it should! O ne calorie ( cal) is 4.2 J. You will learn the detail about any special units used in the course in the appropriate place in this book.
Doing work In 1 82 6, Gaspard- Gustave C oriolis was studying the engineering involved in raising water rom a ooded underground mine. He realized that energy was being transerred when the steam engines were pumping the water vertically. He described this energy transer as work done, and he recognized that the energy transerred when the pumping engines exerted a orce on a particular mass o water and lited it rom the bottom o the mine to the surace. In other words work done ( in J) = force exerted ( in N) distance moved in the direction of the force ( in m) So, when a weight o 5 N o water is lited vertically through a height o 1 5 0 m, then the work done by the engine on the water is 5 1 50 = 75 0 J. In the example o the mine, the orce and the distance moved are in the same direction ( vertically upwards) , but in many cases this will not
Figure 1 Mine steam engine.
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M E C H AN I C S be the case. A good example of this is a sand yacht, where the force F from the wind acts in one direction and the sail is set so that the yacht moves through a displacement s that is at an angle to the wind. In this case, the distances moved in the direction of the force and by the yacht are not the same. You must use the component of force in the direction of movement. sail distance moved by sand yacht, s force exerted by wind, F plan view
Figure 2 Force on sail.
In this case, work done = F cos s
Work done against a resistive force Work is done when a resistive force is operating too. C onsider a box being pushed at a constant speed in a horizontal straight line. For the speed to be constant, friction forces must be overcome. The force that overcomes the friction may not act in the direction of movement. Again, the work that will be done by the force (or the force provider) is force acting distance travelled cos (figure 3) . force F horizontal component of force = F cos
box
direction of motion
Figure 3 Resistive force.
Worked examples 1
The thrust ( driving force) of a microlight aircraft engine is 3 .5 1 0 3 N. C alculate the work done by the thrust when the aircraft travels a distance of 1 5 km.
that acts at 5 0 to the horizontal. C alculate the work done in moving the box 8.5 m. 55 N 8.5 m
Solution Work done = force distance = 3 5 00 1 5 000 = 5 .3 1 0 7 J 5 3 MJ. 2
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A large box is pulled a distance of 8. 5 m along a rough horizontal surface by a force of 5 5 N
50
2 . 3 W O R K , E N E R G Y, A N D P O W E R
Solution
distance in which the obj ect will come to rest i a net orce o 4. 0 N opposes the motion.
The component o orce in the direction o travel is 55 cos 50 = 35.4 N.
Solution
The work done = this orce component distance travelled = 3 5 .4 8.5 = 3 01 J.
There must be 2 4 J o work done to stop the motion o the obj ect. The orce acting is 4.0 so 24 __ = 6 m. 4
An obj ect moving in a straight line has an initial kinetic energy o 2 4 J. C alculate the
3
Forcedistance graphs In practice, it is rare or the orce acting on a moving obj ect to be constant. Real railway trucks or sand yachts have air resistance and other orces that lead to energy losses that vary with the speed o the obj ect or the surace over which it runs. We can deal with this i we know how the orce varies with distance. S ome examples are shown in fgure 4. total area = (number of squares) (energy represented by one square)
area = F s 0
distance moved (a)
s s
0
0
F
Worked example
distance moved (b)
The graph shows the variation with displacement d o a orce F that is applied to a toy car. C alculate the work done by F in moving the toy through a distance o 4.0 cm.
Figure 4 Forcedistance graphs.
For a constant orce ( fgure 4( a) ) , the graph o orce against distance will be a straight line parallel to the x-axis. The work done is the product o force distance ( we are assuming that = 90 in this case) ; this corresponds to the area under the graph o orce against distance. When the orce is not constant with distance moved ( fgure 4( b) ) the work done is still the area under the line, but this time you have to work a little harder by estimating the number o squares under the graph and equating each square to the energy that it represents. The product o ( energy for one square) ( number of squares) will then give you the overall work done. There is a urther example o this type o calculation later in the topic on elastic potential energy.
Power Imagine two boys, Jean and Phillipe with the same weight ( lets say 65 0 N) who climb the same hill ( 70 m high) . B ecause they have the same weight and climb the same vertical distance they both gain the same amount o gravitational potential energy. This is at the expense o the chemical energy reserves in their bodies. However, suppose Jean climbs the hill in 1 5 0 s whereas Phillipe takes 3 00 s.
6 5 4 F/N
0
one square = F s
force
force
F
3 2 1 0
0
2
4 d/10 - 2 m
6
Solution The work done is equal to the area o the triangle enclosed by the graph and 1 the axes. This is __ base 2 o the triangle height o 1 4. 0 1 0 - 2 triangle = __ 2 5 .0 = 0.1 0 J
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James Watt was a mechanical engineer who worked at the end o the 18th century and the beginning o the 19th. He did major work in improving steam engine design and even developed a way o making copies o paper documents that was used in ofces until early in the 20th century.
The obvious dierence he re is that Jean is gaining potential energy twice as ast as Phillipe because Jeans time to make the climb is hal that o Phillipe . This die rence is important when we want to compare two machines taking dierent times to carry out the same amount o work. The quantity p ower is used to measure the rate of doing work, in other words it is the number o j oules that can be converted every second. S o, power is defned as: energy transferred power = __ time taken for transfer Using the correct quantities is important here. I the energy change is in j oules, and the time or the transer measured in seconds then the power is in watts ( W) . 1 W 1 J s 1 In the example o Jean and Phillipe above, both did 45 . 5 kJ o work, but 650 70 Jeans power in climbing the hill was _______ = 3 03 W and Phillipes was 1 50 1 5 2 W because he took twice as long in the climb. The equation work done = force distance can be rearranged to give another useul expression or power. distance moved by force work done power = _ = force __ time time It is easy to see that this is the same as power = force speed. S o the power required to move an obj ect travelling at a speed v with a orce F is Fv.
Nature of science You will see other units or power used in various areas o everyday lie. As an example, the horsepower is one measure used by car manuacturers to provide customers with data about car engines. This is a unit that dates back to the beginning o the 1 8th century and was used to compare the power o an average horse with steam engines. (1 horsepower is equal to about 75 0 W in modern units.)
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Kinetic energy, KE Kinetic energy is the energy an obj ect has because o its motion; kinetic energy ( sometimes abbreviated to KE) has its own symbol, EK. O bj ects gain kinetic energy when their speed increases.
mass m
KE =
1 2
u
mu 2
KE = change in KE = 12 m(v2 u 2 )
mass m
Figure 5 Kinetic energy of mass.
1 2
mv2
v
2 . 3 W O R K , E N E R G Y, A N D P O W E R Imagine an obj ect of mass m which is at rest at time t = 0 and which is accelerated by a force F for a time T. The kinematic equations and Newtons second law allow us to work out the speed of the obj ect v at time t = T. F The acceleration a is __ m ( using Newtons second law of motion) . mv F ___ Therefore v = 0 + __ m T ( because the initial speed is zero) . S o F = T
The work done on the mass is the gain in its kinetic energy, EK , and is F s where s is the distance travelled and therefore mv vT EK = F s = _ _ T 2 (v + 0) T because s = _ 2
Worked example A car is travelling at a constant speed of 2 5 m s 1 and its engine is producing a useful power output of 2 0 kW. C alculate the driving force required to maintain this speed.
Solution power driving force = _ speed 2 0 000 _ = = 800 N 25
The work done by the force is equal to the gain in kinetic energy and is 1 EK = __ mv2 2
Remember that this is the case where the initial speed was 0. If the obj ect is already moving at an initial speed u then the change in kinetic 1 energy EK will be __ m( v2 u 2 ) . 2 There is a subtle piece of notation here. When we talk about a value of kinetic energy, we write EK, but when we are talking about a change in kinetic energy from one value to another then we should write EK where , as usual, means the change in.
Examiners tip This equation for EK is one that needs a little care. Notice where the squares are; they are attached to each individual speed. This equation is not the same as ___12 m (v u) 2 .
Worked examples 1
A vehicle is being designed to capture the world land speed record. It has a maximum design speed of 1 700 km h 1 and a fully fuelled mass of 7800 kg. C alculate the maximum kinetic energy of the vehicle.
Solution 1 700 km h 1 470 m s 1 ( this is greater than the speed of sound in air! ) 1 EK = __ mv2 = 0.5 7800 470 2 = 8.6 1 0 8 J 2
0.86 GJ 2
A car of mass 1 .3 1 0 3 kg accelerates from a speed of 1 2 m s 1 to a speed of 20 m s 1 . Calculate the change in kinetic energy of the car.
Solution 1 EK = __ m( v2 u 2 ) = 0. 5 1 3 00 ( 2 0 2 1 2 2 ) 2
= 65 0 ( 400 1 44) = 1 . 7 1 0 5 J
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Gravitational potential energy, GPE mass, m
g
h
gain in gpe( E P) = mgh
Gravitational p otential energy is the energy an obj ect has because o its position in a gravitational feld. When a mass is moved vertically up or down in the gravity feld o the E arth, it gains or loses gravitational potential energy ( GPE) . The symbol assigned to GPE is EP. O nly the initial and fnal positions relative to the surace determine the change o GPE ( assuming that there is no air resistance on the way) . For this reason, gravitational orce is among the group o orces said to be conservative, because they conserve energy ( see the Nature o S cience box below) . The work done when an obj ect is raised at constant speed through a change in height h is, as usual, equal to force distance moved. In this case the orce required is mg and work done, EP = mg h.
Figure 6 Gravitational potential energy.
As usual, the value o g is 9.8 m s 2 close to the Earths surace, but this value becomes smaller when we move away rom the E arth.
Nature of science Conservative forces There is an important dierence between orces such as gravity and rictional orces. When only gravity acts, the energy change depends on the start height and end height o the motion but not on the route taken by an obj ect to get rom start to fnish. Horizontal movement does not have to be counted i there is no riction acting. This type o orce is said to be conservative; in other words, it conserves energy. We could recover all the energy by moving the obj ect back to the start. C ontrast this with the riction orce that acts between a book and a table as the book is moved around the tables surace rom one point to another. I the book goes directly rom start to fnish a certain amount o energy will be used up to overcome the riction. B ut i the book goes by a longer route, more energy is needed ( work done = friction force distance travelled) . When it is necessary to know the exact route beore we can calculate the total energy conversion, the orce is said to be non-conservative. I we move the book back again, we cannot recover the energy in the way that we could when only gravity acted. You will not need to write about the meaning o the term conservative orce in the IB Diploma Programme physics examination.
Energy moving between GPE and KE S ometimes the use o gravitational potential energy and kinetic energy together provides a neat way to solve a problem. A snowboarder is moving down a curved slope starting rom rest ( u = 0) . The vertical change in height o the slope is h = 5 0 m. What is the speed o the snowboarder at the bottom o the slope? Assume we can ignore riction at the base o the board and the air resistance. What we must not do in this example is to use the kinematic ( suvat) equations. They cannot be applied in this case because the acceleration
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2 . 3 W O R K , E N E R G Y, A N D P O W E R o the snowboarder will not be constant suvat only works when we are certain that a does not change. Although the suvat equations give the correct answer here, they should not be used because the physics is incorrect. The fnal answer happens to be correct only because we use the start and end points and also as a consequence o the conservative nature o gravity. We are also using an average value or acceleration down the slope by assuming that the angle to the horizontal is constant. The equations would not give the correct answer i we brought riction orces into the calculation. C onservation o energy comes to our aid because ( as the riction losses are negligible) we know that the loss o gravitational potential energy as the boarder goes down the slope is equal to the gain in kinetic energy over the length o the slope. B ecause we know that the GPE change only depends on the initial and fnal positions then we do not need to worry at all about what is going on at the base o a board during the ski run.
h = 50 m
Figure 7 Mechanics of snowboarding.
_____
1 So EP = mgh = __ mv2 and v = 2gh , 2
__________
in this case v = 2 9.8 5 0 = 3 1 m s 1 ( This is a speed o about 1 1 0 km h 1 which tells you that the assumption about no air resistance and no riction is a poor one, as any snowboarder will tell you! ) Notice that the answer does not depend on the mass o the boarder; the mass term cancels out in the equations.
Investigate! Converting GPE to KE cart
D evise a way to measure the speed o the cart. You could use a smart pulley that can measure the speed as the string turns the pulley wheel, or an ultrasound sensor, or a data logger with light gates.
When the mass is released, the cart will gain speed, as the gravitational potential energy o the weight changes.
Make measurements to assess the gravitational potential energy lost ( you will need to know the vertical height through which the mass alls) and the kinetic energy gained ( you will need the fnal speed) . Notice that only the alling mass is losing GPE but both the mass and the cart are gaining kinetic energy.
C ompare the two energies in the light o the likely errors in the experiment. Is the energy conserved? Where do you expect energy losses in the experiment to occur?
smart pulley to interface clamp mass
This experiment will help you to understand the conversion between KE and GPE.
Arrange a cart on a track and compensate the track or riction. This is done by raising the let- hand end o the track through a small distance so that the cart neither gains nor loses speed when travelling down the track without the string attached.
Have a string passing over a pulley at the end o the track and tie a weight o known mass to the other end o the track.
Measure the mass o the cart too.
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Worked examples 1
A ball of mass 0.3 5 kg is thrown vertically upwards at a speed of 8. 0 m s 1 . C alculate a) the initial kinetic energy b) the maximum gravitational potential energy c) the maximum height reached.
Solution 1 1 mv2 = __ a) EK = __ 0.3 5 8 2 = 1 1 .2 J 2 2
b) At the maximum height all the initial kinetic energy will have been converted to gravitational potential energy so the maximum value of the GPE is also 1 1 . 2 J. c) The maximum GPE is 1 1 . 2 J and this is equal to mg h, so 1 1 .2 1 1 .2 ________ h = ____ m g = 0 . 3 5 9 . 8 = 3 .3 m. 2
A pendulum bob is released from rest 0.1 5 m above its rest position. C alculate the speed as it passes through the rest position.
0.15 m
Solution EK at the rest position = EP at the release position; 1 __ mv2 = mg h which rearranges to 2
___
____________
v = 2 gh = 2 9.8 0.1 5 = 1 .7 m s 1 .
Elastic potential energy The shape of a solid can be changed by applying a force to it. D ifferent materials will respond to a given force in different ways; some materials will be able to return the energy that has been stored in them when the force is removed. A metal spring is a good example of this; most springs are designed to store energy in this way in many different contexts. The materials that can return energy in this way have stored elastic p otential energy. To discuss elastic potential energy we need to know something about the properties of springs. This is an area of physics that has been studied for a long time. Robert Hooke, a contemporary of Newton, published a rule about springs that has become known as Hookes law.
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2 . 3 W O R K , E N E R G Y, A N D P O W E R
Investigate! Investigating Hookes law
support
Robert Hooke realized that there was a relationship between the load on a spring and the extension of the spring. spring
Arrange a spring of known unstretched length with a weight hanging on the end of the spring.
You will need to devise a way to measure the extension ( change in length from the original unstretched length) of the string for each of a number of different increasing weights hanging on the end of the spring.
0 5
pointer
10
scale
15 20
pan
Repeat the measurements as you remove the weights as a check.
Plot a graph of force ( weight) acting on the spring ( y- axis) against the extension ( x-axis) . This is not the obvious way to draw the graph, but it is the way normally used. Normally, we would plot the dependent variable ( the extension in this case) on the y-axis.
25 30
weight
table
For small loads acting on the spring, Hooke showed that the extension of the spring is directly proportional to the load. (For larger loads, this relationship breaks down as the microscopic arrangement of atoms in the spring changes. We shall not consider this part of the deformation, however.) Hookes rule means that the graph of force F against extension ( x) is a straight line going through the origin. In symbols the rule is F x, or F = kx where k is a constant known as the spring constant, which has units of N m 1 . The gradient of the graph is equal to k ( this is why the graph is plotted the wrong way round) . We can now relate this graph to the work done in stretching the spring. The force is not constant ( the bigger the extension, the bigger the force
force
Fmax
0
1
Area = 2 Fmax x = work done in extending spring = stored elastic potential energy
0
extension
x
Figure 8 Work done in stretching spring.
71
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Worked examples
required) but we now know how to deal with this. The work done on the spring is the area under the graph o orce against extension.
1
The area is a right-angled triangle and is equal to
A spring, o spring constant 48 N m 1 , is extended by 0.40 m. C alculate the elastic potential energy stored in the spring.
Solution 1 Energy stored = __ kx2 = 2 2 0.5 48 0.40 = 0.38 J
2
An object o mass 0.78 kg is attached to a spring o unstretched length 5 60 mm. When the obj ect has come to rest the new length o the spring is 62 0 mm. C alculate the energy stored in the spring as a result o this extension.
Solution The change in length o the spring x = 62 0 5 60 = 60 mm The tension in the spring will be equal to the weight o the obj ect = mg = 0.78 9.8 = 7.64 N The energy stored in the 1 spring = __ Fx = 0.5 2 7.64 0.06 = 0.2 3 J
1 __ force to stretch spring extension, 2 1 in symbols EP = __ Fmax x. 2 F We know rom Hookes law that F = k x, so k = ___ and thereore EP is x 1 2 also equal to __ k( x) . 2
Efciency In the Investigate! experiment where gravitational potential energy was transerred to kinetic energy, it is likely that some gravitational potential energy did not appear in the motion o the cart. Some energy will have been lost to internal energy as a consequence o riction and to elastic potential energy stored when the string changes its shape. We need to have a way to quantiy these losses. One way to do this is to compare the total energy put into a system with the useul energy that can be taken out. This is known as the efciency o the transer and can be applied to all energy transers, whether carried out in a mechanical system, electrical system or other type o transer. You can expect to meet efciency calculations in any area o physics where energy transers occur. The defnition can also be applied to power transers, because the energy change in these cases takes place in the same time or the total energy in and the useul work out. useul power out useul work out efciency = __ = __ total energy in total power
Worked example 1
An electric motor raises a weight o 1 5 0 N through a height o 7.2 m. The energy supplied to the motor during this process is 3 .5 1 0 4 J. C alculate: a) the increase in gravitational potential energy b) the efciency o the process.
Solution a) EP = 1 5 0 7. 2 = 1 080 J u se u l w o rk o u t 1 080 b) Efciency = ___________ = ____ = 0.3 1 or 3 1 % 3500 e ne rgy in
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2.4 M OM EN TUM
2.4 Momentum Understanding Newtons second law expressed as a rate of
change of momentum Impulse and forcetime graphs Conservation of linear momentum Elastic collisions, inelastic collisions, and explosions
Applications and skills Applying conservation of momentum in simple
Nature of science The concept of momentum has arisen as a result of evidence from observations carried out over many centuries. The principle of conservation of momentum is an example of a law that has universal applicability. It allows the prediction of the outcomes of physical interactions at both the macroscopic and the microscopic level. Many areas of physics are informed by it, from the kinetic theory of gases to nuclear interactions.
isolated systems including (but not limited to) collisions, explosions, or water jets Using Newtons second law quantitatively and qualitatively in cases where mass is not constant Sketching and interpreting forcetime graphs Determining impulse in various contexts including (but not limited to) car safety and sports Qualitatively and quantitatively comparing situations involving elastic collisions, inelastic collisions, and explosions
Equations momentum: p = mv Newton's second law (momentum version) : p F = ______ t
p2 kinetic energy: EK = ______ 2m
impulse: = Ft = p
Introduction Many sports involve throwing or catching a ball. C ompare catching a table-tennis ( ping-pong) ball with catching a baseball travelling at the same speed. O ne o these may be a more painul experience than the other! What is dierent in these two cases is the mass o the obj ect. The velocity may be the same in both cases but the combination o velocity and mass makes a substantial dierence. Equally, comparing the experience o catching a baseball when gently tossed rom one person to another with catching a frm hit rom a good player should tell you that changes in velocity make a dierence too.
Momentum We call the product o the mass m o an obj ect and its instantaneous velocity v the momentum p of the obj ect ( p = m v) and this quantity turns out to have ar-reaching importance in physics. First, here are some basics ideas about momentum.
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mass m
Momentum is mass velocity never mass speed.
Momentum has direction. The mass is a scalar but velocity is a vector. When mass and velocity are multiplied together, the momentum is also a vector with the same direction as the velocity. Think o the mass as scaling the velocity in other words, j ust making it bigger by a actor equal to the mass o the obj ect.
It ollows that the unit o momentum is the product o the units o mass and velocity, in other words kg m s 1 . There is a shorter alternative to this that we shall see later.
I velocity or mass is changing then the momentum must also be changing. We shall be looking at the two cases where one quantity changes while the other is held constant.
When a net resultant orce acts on an obj ect, the obj ect accelerates and the velocity must change. This means a change in momentum too. S o a net orce leads to a change in momentum.
velocity v
momentum = mv
Figure 1 Momentum and velocity.
Worked examples 1
A ball o mass 0.2 5 kg is moving to the right at a speed o 7.4 m s 1 . C alculate the momentum o the ball.
Solution p = mv = 0.2 5 7.4 = 1 .85 kg m s 2
1
to the right.
A ball o mass 0.2 5 kg is moving to the right at a speed o 7.4 m s 1 . It strikes a wall at 90 and rebounds rom the wall leaving it with a speed o 5 .8 m s 1 moving to the let. C alculate the change in momentum.
Solution From the previous example the initial momentum is 1 .85 kg m s 1 to the right. The fnal momentum is 0.25 5.8 = 1 .45 kg m s 1 to the let. So taking the direction to the right as being positive, the change in momentum = 1 .45 ( + 1 .85 ) = 3 .3 kg m s 1 to the right ( or alternatively + 3 .3 kg m s 1 to the let) .
Collisions and changing momentum You may have seen a Newtons cradle. Newton did not invent this device (it was developed in the twentieth century as an executive toy) , but it helps us to visualize some important rules relating to his laws o motion, and it certainly seems appropriate to associate his name with it.
Figure 2 Newtons cradle.
O ne o the balls ( the right- hand one in fgure 2 ) is moved up and to the right, away rom the remaining our. When released the ball alls back and hits the second ball rom the right. The right-hand ball then stops moving and the let-most ball moves o to the let. It is as though the motion o the original ball transers through the middle three which remain stationary and appears at the let-hand end. You can analyse this in terms o physics that you already know. The right-hand ball gains gravitational potential energy when it is raised. When it is released the potential energy is converted into kinetic energy as the ball gains speed. Eventually the ball strikes the next stationary one along, and a pair o actionreaction orces acts at the suraces o the two balls (you might like to consider what they are) . The stationary ball is compressed slightly by the orce acting on it and this compression moves as a wave through the middle three balls. When the compression reaches the
74
2.4 M OM EN TUM
let-hand ball the elastic potential energy associated with the compression wave is converted into kinetic energy (because there is a net orce on this sphere) and work begins to be done against the Earths gravity. The lethand ball starts to move to the let. When the speed o the ball becomes zero, it is at its highest point on the let-hand side and the motion repeats in the reverse direction. The overall result is that the end spheres appear to perorm an alternate series o hal oscillations. Another way to view this sequence is as a transer o momentum. Think about a simpler case where only two spheres are in contact (in the toy, three balls can be lited out o the way) . The right-hand sphere gains momentum as it alls rom the top o its swing. When it collides with the other sphere, the momentum appears to be transerred to the second sphere. The frst sphere now has zero momentum (it is stationary) and the second has gained momentum. What rules govern this transer o momentum? These interactions between the balls in the Newtons cradle are called collisions. This is the term given to any interaction where momentum transers. E xamples o collisions include fring a gun, hitting a ball with a bat in sport, two toy cars running into each other, and a pile driver sinking vertical cylinders into the ground on a construction site. There are many more.
Investigate! Is momentum conserved? The exact details o this experiment will depend on the apparatus you have in your school. light gate stationary cart
timing card moving cart raised to compensate for friction (see page 68)
The experiment consists o measuring the speed o a cart o known mass and then launching it at another cart also o known mass that is initially stationary. O ten there will be carts available to you o almost identical mass, as this is a particularly easy case to begin with. You will need a way to measure the velocity o the carts just beore and just ater the collision. This could be done in various ways:
using a data logger with motion sensors
using a paper tape system where a tape attached to the cart is pulled through a device that makes dots at regular time intervals on the tape
using a video camera and computer sotware
using a stop watch to measure the time taken to cover a short, known distance beore and ater the collision.
I your carts run on a track, you can allow or the riction at the cart axles and the air resistance. This is done by raising the end o the track a little so that, when pushed, a cart runs at a constant speed. The riction at the bearings and the air resistance will be exactly compensated by the component o the cart weight down the track.
For the frst part o the experiment, begin by arranging that the two carts will stick together ater colliding. This can be done in a number o ways, including using modelling clay, or a pin attached to one cart entering a piece o cork in the other, or two magnets ( one on each cart) that attract.
Make the frst ( moving) cart collide with and stick to the second ( stationary) cart.
Measure the speed o the frst cart beore the collision and the combined speed o the carts ater the collision.
Repeat the experiment a number o times and think careully about the likely errors in the results.
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The initial momentum is mass o frst cart velocity o frst cart. The fnal momentum is (mass o frst cart + mass o second cart) combined velocity o both carts.
and the momentum ater the collision are approximately equal.
What can you say about the total momentum o the system beore the collision compared with the total momentum ater the collision? You should consider the experimental errors in the experiment beore making your j udgement. I you have done the experiment careully, you should fnd that the momentum beore
Now extend your experiment to dierent cases:
where the carts do not stick together
where they are both moving beore the collision
where the masses are not the same, and so on.
You may need to alter how you measure the velocity to cope with the dierent cases.
I you carry out an experiment that compares the momentum beore with the momentum ater a collision then you should fnd ( within the experimental uncertainty o your measurements) that the total momentum in the system does not change. An important point here is that there must be no orce rom outside the system acting on the obj ects taking part in the collision. As we saw earlier, external orces produce accelerations, and this changes the velocity and hence the momentum. You might argue that gravitational orce is acting and gravity certainly is acting on the carts in the experiment but because the gravitational orce is not being allowed to do any work, the orce does not contribute to the interaction because the carts are not moving vertically. Momentum is always constant if no external force acts on the system. This is known as the p rincip le of conservation of linear momentum, the word linear is here because this is momentum when the obj ects concerned are moving in straight lines. This conservation rule has never been observed to be broken, and is one o the important conservation rules that ( as ar as we know) are true throughout the universe. In nuclear physics, particles have been proposed in order to conserve momentum in cases where it was apparently not being conserved. These particles were subsequently ound to exist. Momentum conservation is such an important rule that it is worth us considering a ew dierent situations to see how momentum conservation works. In each o these cases we will assume that the centres o the obj ects lie on a straight line so that the collision happens in one dimension.
Two objects with the same mass, one initially stationary, when no energy is lost This is known as an elastic collision and or it to happen no permanent deormation must occur in the colliding obj ects and no energy can be released as internal energy ( through riction) , sound, or any other way. The spheres in the Newtons cradle lose only a little energy every time they collide and this is why this is a reasonable demonstration o momentum eects.
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2.4 M OM EN TUM u
beore:
mass m 1
0 m s 1
TOK
mass m 2
Popper and alsifability
mass m 1
ater:
mass m 2
0 m s 1 v= u
Figure 3 Elastic collision between two identical masses.
The frst obj ect collides with the second stationary obj ect. The frst obj ect stops and remains at rest while the second moves o at the speed that the frst obj ect had beore the collision. ( Try icking a coin across a smooth table to hit an identical coin head-on.) In this case momentum is conserved because ( using an obvious set o symbols or the mass m o the obj ects and their velocities u and v) m 1 u = m2v B ecause m 1 = m 2 then u = v so the velocity o one mass beore the collision is equal to the velocity o the second mass aterwards. The 1 kinetic energy o the moving mass ( whichever mass is moving) is __ mu 2 2 and it does not change either.
Two objects with diferent masses when no energy is lost This time ( as you may have seen in an experiment or demonstration) the situation is more complicated. u1 u2 beore:
m1 m2
v1 v2 ater:
m1 m2
No-one has yet observed a case where the momentum in an isolated system is not conserved, but we should continue to look! Karl Popper, a philosopher o the twentieth century, argued that the test o whether a theory was truly scientifc was that it was capable o being alsifed. By this he meant that there has to be an experiment that could in principle contradict the hypothesis being tested. Popper argued that psychoanalysis was not a science because it could not be alsifed by experiment. Popper also applied his ideas to scientifc induction. He said that, although we cannot prove that the Sun will rise tomorrow, because it always has we can use the theory that the Sun rises in the morning until the day when it ails to do so. At that point we must revise our theory. What should happen i momentum appears not to be conserved in an experiment? Is it more sensible to look or a new theory or to suggest that something has been overlooked?
Figure 4 Two moving objects with diferent mass in an elastic collision.
Again m 1 u 1 + m 2 u 2 = m 1 v1 + m 2 v2 but this time we cannot eliminate the mass terms so easily. What we do know, though, is that kinetic energy is conserved. S o the kinetic energy beore the collision must equal the kinetic energy ater the collision ( because no energy is lost) . This means that ( summing the kinetic energies beore and ater the collision) 1 1 1 1 __ m 1 u 21 + __ m 2 u 22 = __ m 1 v21 + __ m 2 v22 2 2 2 2
77
2
M E C H AN I C S The momentum and kinetic energy equations can be solved to show that
( (
) ( ) (
) )
m1 m2 2 m2 v1 = _ u 1 + _ u 2 m1 + m2 m1 + m2 and
m2 m1 2 m1 v2 = _ u 2 + _ u 1 m1 + m2 m1 + m2 ( You will not be expected to prove these in, or memorize them or, an examination. ) There are some interesting cases when mass m 2 is stationary and is struck by m 1 :
C ase 1 : m 1 is much smaller than m 2 . Look at the v1 equation. I m 1 m 2 m then the frst term becomes roughly _____ u 1 which is u 1 ; the second m term is zero because u 2 = 0. So the small mass bounces o the large mass, reversing its direction ( shown by the minus sign) . The large mass gains speed in the orward direction (the original direction o the small mass) . The magnitude o the speed o the larger mass is roughly 2m ___ u 1 and this is a small raction o the original speed o the small m 2m mass because ___ is much smaller than 1 . m
( ) 2
2
( ) 1
2
( ) 1
2
C ase 2 : m 1 is much greater than m 2 . Here the original mass loses hardly any speed ( though it must lose a little) . The momentum lost by m 1 is given to m 2 which moves o in the same direction, but at about twice the original speed o m 1 . Look at the v1 and v2 equations and satisy yoursel that this is true.
Two objects colliding when energy is lost When a moving object collides with a stationary one and the two objects stick together, then some o the initial kinetic energy is lost. Ater the collision, there is a single object with an increased (combined) mass and a single common velocity. This is known as an inelastic collision (fgure 5) . This is a case you may have studied experimentally in the Investigate! The momentum equation this time is m 1 u 1 = ( m 1 + m 2 ) v1 m A rearrangement shows that v1 = _______ u 1 and, as we might expect, (m + m ) 1
1
2
the fnal velocity is in the same direction as beore but is always smaller than the initial velocity. u1 before:
0
m1
m2 v1
after:
78
m1 + m2
Figure 5 Inelastic collision between two masses.
As or energy loss, the incoming kinetic energy is kinetic energy is ( substituting or v1 ) m 21 1 __ __ ( m + m ) u2 1 2 2 ( m1 + m 2) 2 1 This is m 12 1 _ __ u 21 . 2 (m + m ) 1 2 ( m1 + m2) initial kinetic energy The ratio __ is _ m1 fnal kinetic energy
__1 m u 2
1
2 1
and the fnal
2.4 M OM EN TUM
Two objects when energy is gained There are many occasions when two initially stationary masses gain energy in some way. S ome laboratory dynamics carts have a way to show this. An easy way is to attach two small strong magnets to the ront o the carts so that when the carts are released ater being held together with like poles o the magnets acing each other, the magnets repel and drive the carts apart. The analysis is quite straightorward in this case. The initial momentum is zero as neither obj ect is moving. Ater the collision the momentum is m 1 v1 + m 2 v2 = 0.
before:
0
0
m1
m2
S o m 1 v1 = m 2 v2 The obj ects move apart in opposite directions. I the masses are equal the speeds are the same, with one velocity the negative o the other. I the masses are not equal then v2 m _ m 2 = v1
1 _
after:
v1
v2
m1
m2
Figure 6 Energy gained in a collision.
TOK Helpful or not? Are conservation laws helpul to scientists? On the one hand they allow the prediction o as yet untested cases, but on the other hand they may restrict the progress o science. This can happen i scientists are not prepared to challenge the status quo. In 2012 the results o an experiment suggested that neutrinos could travel aster than the speed o light.
This few in the ace o the accepted science originally proposed by Einstein. Later investigations showed that small errors in the timings had occurred in the experiments, and that there was no evidence or aster-than-light travel. Were the scientists right to publish their results so that others could test the new proposals?
Worked examples 1
A rail truck o mass 45 00 kg moving at a speed o 1 .8 m s 1 collides with a stationary truck o mass 1 5 00 kg. The two trucks couple together. C alculate the speed o the trucks immediately ater the collision.
Solution The intial momentum = 45 00 1 .8 kg m s 1 . The fnal momentum = ( 45 00 + 1 5 00) v, where v is the fnal speed. 45 0 0 1 . 8 Momentum is conserved and so v = __________ ( 45 0 0 + 1 5 0 0 ) 4. 5 1 . 8 _______ 1 = = 1 .3(5 ) = 1 .4 m s . 6.0
2
S tone A o mass 0. 5 kg travelling at 3 . 8 m s 1 across the surace o a rozen pond collides
with stationary stone B o mass 3 .0 kg. S tone B moves o at a speed o 0.65 m s 1 in the same original direction as stone A. C alculate the fnal velocity o stone A.
Solution The initial momentum is 0.5 3 .8 = 1 .9 kg m s 1 The fnal momentum is 3 .0 0.65 + 0.5 vA and this is equal to 1 .9 kg m s 1 1 .9 ( 3 .0 0.65 ) 0.05 vA = __ = _ = 0.1 kg m s 1 0.5 0.5 The fnal velocity o stone A is 0.1 kg m s 1 in the opposite direction to its original motion.
79
2
M E C H AN I C S
3 before: after:
2.5 m s 1
0 m s 1
v m s 1
1.9 m s 1
6000 kg
3000 kg
A railway truck o mass 6000 kg collides with a stationary truck o mass 3000 kg. The frst truck moves with an initial speed o 2.5 m s 1 and the second truck moves o with a speed o 1 .9 m s 1 in the same direction. C alculate:
Solution a) The initial momentum is 6000 2 .5 = 1 5 000 kg m s 1 . The fnal momentum is 6000v + 3 000 1 .9 Equating these values: 1 5 000 = 6000v + 5 700, so 6000v = 93 00 and v = 1 .5 (5 ) kg m s 1 . This is positive so the frst truck continues to move to the right ater the collision at a speed o 1 .6 kg m s 1 . 1 6000 2 .5 2 b) The initial kinetic energy = __ 2 = 1 8 75 0 J.
Ater the collision the total kinetic energy =
a) the velocity o the frst truck immediately ater the collision b) the loss in kinetic energy as a result o the collision.
1 1 __ 6000 1 .5 5 2 + __ 3 000 1 .9 2 2 2
= 72 08 + 5 41 5 = 1 2 62 3 or to 2 s. . 1 3 000 J. S o the loss in kinetic energy is 1 8 75 0 1 3 000 = 5 800 J.
Energy and momentum There is a convenient link between kinetic energy and momentum. 1 Kinetic energy, EK= __ mu 2 ; momentum, p = mu and thereore p 2 = m 2 u 2 . 2 So p2 EK = _ 2m
This is oten o use in calculations.
Applications of momentum conservation Recoil of a gun Figure 7 shows a gun being fred to trigger a snow all. This prevents a more dangerous avalanche. When the gun fres its shell, the gun moves backwards in the opposite direction to that in which the shell goes. You should be able to explain this in terms o momentum conservation. Initially, both gun and shell are stationary; the initial total momentum is zero. The shell is propelled in the orwards direction through the gun by the expansion o gas ollowing the detonation o the explosive in the shell. S o this is one o the cases discussed earlier, one in which energy is gained. The explosion is a orce internal to the system. Gas at high pressure is generated by the explosion in the chamber behind the shell. The gas exerts a orce on the interior o the shell chamber and hence a orce on the gun as well. The explosive releases energy and this is transerred into kinetic energies o both the shell and the gun.
80
Figure 7 Firing a gun to cause an avalanche.
The initial linear momentum was zero and no external orce has acted on the system. The momentum must continue to be zero and this can only be so i the gun and the shell move in opposite directions with the same magnitude o momentum. The shell will go ast because it has a small mass compared to the gun; the gun moves relatively slowly.
2.4 M OM EN TUM
Water hoses Watch a fre being extinguished by fremen using a high- pressure hose and you will see the eect o water leaving the system. O ten two or more fremen are needed to keep the hose on target because there is a large orce on the hose in the opposite direction to that o the water. This can be seen in a more modest orm when a garden hose connected to a tap starts to shoot backwards in unpredictable directions i it is not held when the tap is turned on. The cross- sectional area o the hose is greater than that o the nozzle through which the water emerges. The mass o water owing past a point in the hose every second is the same as the mass that emerges rom the nozzle every second. S o the speed o the water emerging rom the nozzle must be greater than the water speed along the hose itsel. The water gains momentum as it leaves the hose because o this increase in exit speed compared with the ow speed in the hose.
Figure 8 Fire fghting.
The momentum o the system has to be constant and so there must be a orce backwards on the end o the hose which needs to be countered by the eorts o the fremen. The kinetic energy and the momentum are being supplied by the water pump that eeds water to the fre hose or whatever originally created the pressure in the supply to the garden tap. The momentum lost by the system per second is (mass of water leaving per second) (speed at which water leaves the nozzle speed in the hose). m The mass o water lost per second is ___ and so the momentum lost per t m ___ second p is t ( v u) where v is the speed o water as it leaves the nozzle and u is the speed o the water in the hose.
v A, cross-sectional area o nozzle
v
u
one second o water
Figure 9 Water leaving the hose.
I we know the cross- sectional area A o the nozzle o the hose and the m density o the water, , then ___ can be determined. Figure 9 shows what t happens inside the hose during a one-second time interval. Every second you can imagine a cylinder o water leaving the hose; this cylinder is v long and has an area A. The volume leaving per second is thereore Av m The mass o the water leaving in one second ___ t
= (density of water) (volume of cylinder) = Av. B ecause the mass entering and leaving the nozzle per second must be the same, this means that the change o momentum in one second m = ___ ( v u) = Av( v u) . I u v then the expression simplifes t to Av2 . The hose is an example o where care is needed when looking at the whole system. C onsider what happens i the water is directed at a vertical wall. The water strikes the wall, loses all its horizontal momentum, and trickles vertically down the wall. The momentum must
81
2
M E C H AN I C S
Worked example A mass o 0.48 kg o water leaves a garden hose every second. The nozzle o the hose has a cross- sectional area o 8.4 1 0 5 m 2 . The water ows in the hose at a speed o 0.71 m s 1 . The density o water is 1 000 kg m 3 . C alculate: a) the speed at which water leaves the hose b) the orce on the hose.
Solution a) 0.48 kg o water in one second corresponds to 0 . 48 m 3 a volume o ____ 1 000 leaving the hose per second. This leaves through a nozzle o area 8.4 1 0 5 m 2 so the speed v must be 0 . 48 _____________ = 1 0 0 0 8 . 4 1 0 5
5.71 m s 1 . b) The orce on the hose = mass lost per second change in speed = 0.48 (5 .71 0.71 ) = 2.4 N
have gone into the wall, its oundations and, thereore, the ground. S o we might conclude that the E arth itsel has gained momentum and that we can speed up the Earths rotation by using a garden hose. B ut this is not true, because the water had to be given momentum originally by a pump. This gain in momentum at the pump must have given some momentum to the Earth too. The amount o momentum the E arth gained at the pump is equal and opposite to the momentum gained by the Earth when the water strikes the wall.
Rocketry Earlier in this topic we discussed the acceleration o a rocket and looked at the situation rom the perspective o Newtons second and third laws. A similar analysis is possible in terms o momentum conservation. Rockets operate eectively in the absence o an atmosphere because momentum is conserved. All rockets release a liquid or gas at high speed. The uid can be a very hot gas generated in the combustion o a solid chemical ( as in a domestic rocket) or rom the chemical reaction when two gases are mixed and burnt. O r it can be a uid stored inside the rocket under pressure. In each case, the uid escapes rom the combustion/storage chamber through nozzles at the base o the rocket. As a result the rocket accelerates in the opposite direction to the direction in which the uid is ej ected. Momentum is conserved. The rate o loss o momentum rom the rocket in the orm o high- speed uid must be equal to the rate o gain in momentum o the rocket. We shall look again at the mathematics o the rocket later in this topic.
Helicopters Helicopters are aircrat that can take o and land vertically and also can hover motionless above a point on the ground. Vertical ight is said to have been invented in C hina about 2 5 00 years beore the present, but anyone who has seen the seeds rom certain trees spiral down to the ground will realize where the original design came rom. There were many attempts to build ying machines on the helicopter principle over the centuries, but the frst commercial aircrat ew in the 1 93 0s. A helicopter uses the principle o conservation o linear momentum in order to hover. The rotating blades exert a orce on originally stationary air causing it to move downwards towards the ground gaining momentum in the process. No external orce acts and as a result there is an upward orce on the helicopter through the rotors.
Impulse Earlier we used Newtons second law o motion: F = ma where the symbols have their usual meaning. We can rearrange this equation using one o the kinematic equations: (v u) a = _____ t
Figure 10 Helicopter.
m( v u) Eliminating a gives F = _ t change in momentum which means that force = __ time taken for change or force time = change in momentum
82
2.4 M OM EN TUM
This equation gives a relationship between orce and momentum and provides a urther clue to the real meaning o momentum itsel. The equation shows that we can change momentum ( in other words, accelerate an obj ect) by exerting a large orce or a short time or by exerting a small orce or a long time. A small number o people can get a heavy vehicle moving at a reasonable speed, but they have to push or a much longer time than the vehicle itsel would take i powered by its own engine ( which produces a larger orce) .
Worked examples 1
The product o force and time is given the name imp ulse, its units are newton seconds ( N s) . Impulse is the same as change in momentum, and N s gives us an alternative to kg m s 1 as a unit or momentum. You can use whichever you preer.
An impulse o 85 N s acts on a body o mass 5 .0 kg that is initially at rest. C alculate the distance moved by the body in 2 .0 s ater the impulse has been delivered.
Solution The change in momentum is 85 kg m s 1 so that the fnal 85 speed is __ = 1 7 m s 1 . In 5 2.0 s the distance travelled is 34 m.
ch an ge in m om en tu m
There is a short-hand way to write the force = ______________ equation. tim e We have already used the convention that means change in. S o in symbols the equation becomes p F= _ t
2
where p is the symbol or momentum and t ( as usual) means time. ( I you are amiliar with dierential calculus you may also come across dp this expression in the orm F = __ , but using the equation in this orm dt will take us too ar away rom IB D iploma Programme physics. )
An impulse I acts on an object o mass m initially at rest. D etermine the kinetic energy gained by the object.
Solution
Forcetime graphs
The change in speed o the I obj ect is __ m . The obj ect is initially at rest and the gain 1 in kinetic energy is __ mv2 2 2 I 1 I ___ m _ which is __ 2 m = 2m
Up till now we have usually assumed that orces are constant and do not change with time. This is rarely the case in real lie and we need a way to cope with changes in momentum when the orce is not constant. p The equation F = ___ helps here because it suggests that we use a orce t time graph.
( )
2
force
force
area = F T
force
Fmax
F
area = 12 Fmax T 0
0
time (a)
T
0
0
time
T
0
0
(b)
time
T
(C)
Figure 11 Forcetime graphs.
I a constant orce acts on a mass, then the graph o orce, F, against time, t, will look like fgure 1 1 ( a) . The change in momentum is F T and this is the shaded area below the line. The area below the line in a orcetime graph is equal to the change in momentum. Another straightorward case that is more plausible than a constant orce is the graph shown in fgure 1 1 ( b) where the orce rises to a maximum Fmax and then alls back to zero in a total time T. The area under the 1 graph this time is __ F T and, again, this is the change in momentum. 2 max
(
)
The fnal case ( fgure 1 1 (c) ) is one where there is no obvious mathematical relationship between F and t, but nevertheless we have
83
2
M E C H AN I C S a graph o how F varies with t. This time you will need to estimate the number o squares under the graph and use the area o one square to evaluate the momentum change.
Worked examples 1
b) the change in kinetic energy over the 1 0 s o the motion.
The sketch graph shows how the orce acting on an obj ect varies with time.
200 180
15 force/N
160
0
0
5
20
25
time/s
p/kg m s 1
140 120 100 80
The mass o the obj ect is 5 0 kg and its initial speed is zero.
60 40
C alculate the fnal speed o the obj ect. 20
Solution The total area under the F t graph is 1 2 __ 1 5 5 + ( 1 5 1 5 ) = 75 + 2 2 5 2 = 3 00 N s
(
)
This is the change in momentum.
0
1
2
3
4
5 t/s
Solution a) The gradient o the graph is equal to the orce.
300 So the fnal speed is ___ = 6.0 m s 1 50
2
0
The graph shows how the momentum o an obj ect o mass 40 kg varies with time.
6
7
8
9
10
p ___ and this is t
200 The magnitude o the gradient is ___ and this 10 gives a value or orce o 2 0 N. 2
p 200 b) EK = ___ = _____ = 5 00 J 2m 2 40
C alculate, or the obj ect:
2
a) the orce acting on it
Revisiting Newtons second law p In the last sub- topic we used F = ma to show that F also equals ___ . Using t the ull expression or momentum p gives ( m v) F = _____ t
This can be written as ( using the product rule) v m F= m_+ v_ t t You may have to take this algebra on trust but, by thinking through what the two terms stand or, you should be able to understand the physics that they represent. ch an ge in velocity The frst term on the let-hand side is j ust mass ___________ which we ch an ge in tim e
84
know as mass acceleration our original orm o Newtons second law o motion. The second term on the right-hand side is something new, it is the change in m ass . instantaneous velocity __________ change in tim e
2.4 M OM EN TUM
Our frst version o Newtons second law was a simpler orm o the law than the ull change in momentum version. The new extra term takes account o what happens when the mass o the accelerating object is changing.
Rockets and helicopters again Rockets We showed that rockets are an excellent example o momentum conservation in action. The difculty in analysing the rocket example is that the rocket is always losing mass ( in the orm o gas or liquid propellant) , so m is not constant. mv v m F = ____ + ____ is our new version o Newtons second law o motion t t and, in this case, F = 0 because there is no external orce acting on the system. mv v m mv S o ____ = ____ . The two terms have quite separate meanings: ____ t
t
t
reers to the instantaneous mass o the rocket ( including the remaining v uel) and to the acceleration o this total mass ___ . The other term t is the ej ection speed o the uel and the rate at which mass is lost m rom the system ___ . S o at one instant in time, the acceleration o t the rocket
Worked example A small frework rocket has a mass o 3 5 g. The initial rate at which hot gas is lost rom the frework has been lit is 3.5 g s 1 and the speed o release o this gas rom the rear o the rocket is 1 3 0 m s 1 . C alculate the initial acceleration o the rocket.
Solution v v m a = ___ = ____ where v t mt is the release speed o the m gas, ___ is the rate o loss t o gas, and m is the mass o the rocket 1 30 3.5 a = _______ = 7. 0 m s 2 65
v v m a = ___ = ____ . t mt
The negative sign reminds us that the rocket is losing mass while gaining speed.
Helicopters With the helicopter hovering, there is a weight orce downwards and so the equation now becomes mv v m Mg = ____ + ____ where M is the mass o the helicopter t t
There is no change in the speed o the helicopter (it is hovering) so v = 0, there is however momentum gained by the air that moves downwards. This is the speed, v, the air gains multiplied by the mass o the air accelerated m every second so Mg = v___ or t weight of helicopter ( in N) = mass of air pushed downwards per second ( in kg s 1 ) speed of air downwards (in m s 1 )
Momentum and safety Many countries have made it compulsory to wear seat belts in a moving vehicle. Likewise, many modern cars have airbags that inate very quickly i the car is involved in a collision. O bviously, both these devices restrain the occupants o the car, preventing them rom striking the windscreen or the hard areas around it. B ut there is more to the physics o the air bag and the seat belt than this. O n the ace o it, someone in a car has to lose the same amount o kinetic energy and momentum whether they are stopped by the windscreen or restrained by the seat belt. What diers in these two
Figure 12 Car safety: seat belts and air bags.
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2
M E C H AN I C S cases is the time during which the loss o energy and momentum occur. Unrestrained, the time to stop will be very short and the deceleration will thereore be very large. A large deceleration means a large orce and it is the magnitude o the orce that determines the amount o damage. S eat belts and air bags dramatically increase the time taken by the occupants o the car to stop and as force time = momentum change, or a constant change in momentum, a long stopping time will imply a smaller, and less damaging, orce.
Momentum and sport This sub- topic began with a suggestion that it was less painul to catch a table-tennis ball than a baseball. You should now be able to understand the reason or the dierence. You should also realize why good technique in many sports hinges on the application o momentum change. Many sports in which an obj ect usually a ball is struck by hand, oot or bat rely on the efcient transer o momentum. This transer is oten enhanced by a ollow through, which increases the contact time between bat and ball. The player maintains the same orce but or a longer time, so the impulse on the ball will increase, increasing the momentum change as well. Think about your sport and how eective use o momentum change can help you.
Investigate! Estimating the force on a soccer ball This experiment will allow you to estimate the orce used to kick a soccer ball. It uses many o the ideas contained in this topic and is a good place to conclude our look at IB mechanics!
To measure the contact time: stick some aluminium oil to the shoe o the kicker and to the soccer ball. S et up a data logger or asttimer so that it will measure the time T or which the two pieces o oil are in contact.
To measure the change in momentum: the ball starts rom rest so all you need is the magnitude o the fnal momentum. Get the kicker to kick the ball horizontally rom a lab bench. Measure the distance s rom where the ball is kicked to where it lands. Measure the distance h rom the bottom o the ball on the bench to the oor. Use the ideas o projectile motion to calculate (i) the time t taken or the ball __ to reach the
h
s aluminum foil foot ball
[
to timer
86
The basis o the method is to measure the contact time between the oot and the ball and the subsequent change in momentum o the ball. The use o force contact time = change in momentum allows the orce to be calculated.
]
2h 1 oor h = __ gt2 and so t = __ g . Then using 2 this t, the initial speed u o the ball can be estimated as _st . Measure the mass o the ball M and thereore the change in momentum is Mu which is equal to the force on the ball T.
This method can be modifed or many sports including hockey, baseball, and gol.
QUESTION S
Questions 1
3
(IB) Christina stands close to the edge o a vertical cli and throws a stone at 1 5 m s 1 at an angle o 45 to the horizontal. Air resistance is negligible.
A marble is projected horizontally rom the edge o a wall 1 .8 m high with an initial speed V. A series o fash photographs are taken o the marble and combined as shown below. The images o the marble are superimposed on a grid that shows the horizontal distance x and vertical distance y travelled by the marble.
P
15 m s 1 O
(IB)
The time interval between each image o the marble is 0.1 0 s.
Q
0
0.50
x/m 1.0
1.5
2.0
0 25 m
0.50
y/m
sea
Point P on the diagram is the highest point reached by the stone and point Q is at the same height above sea level as point O . C hristinas hand is at a height o 2 5 m above sea level.
1.50
a) At point P on a copy o the diagram above draw arrows to represent:
2 .0
( i) the acceleration o the stone ( label this A)
Use data rom the photograph to calculate a value o the acceleration o ree all. ( 3 marks)
( ii) the velocity o the stone ( label this V) . b) D etermine the speed with which the stone hits the sea. ( 8 marks) 2
1.0
4
( IB)
(IB)
A cyclist and his bicycle travel at a constant velocity along a horizontal road.
Antonia stands at the edge o a vertical cli and throws a stone vertically upwards.
a) ( i) S tate the value o the resultant orce acting on the cyclist.
The stone leaves Antonias hand with a speed v = 8.0 m s 1 . The time between the stone leaving Antonias hand and hitting the sea is 3 . 0 s. Assume air resistance is negligible.
( ii) C opy the diagram and draw labelled arrows to represent the vertical orces acting on the bicycle.
a) C alculate: ( i) the maximum height reached by the stone ( ii) the time taken by the stone to reach its maximum height. b) D etermine the height o the cli. ( 6 marks)
87
2
M E C H AN I C S ( iii) E xplain why the cyclist and bicycle are travelling at constant velocity.
1.4 1.2
b) The total mass o the cyclist and bicycle is 70 kg and the total resistive orce acting on them is 40 N. The initial speed o the cycle is 8.0 m s 1 . The cyclist stops pedalling and the bicycle comes to rest.
P/kW
1
0.4 0 0
( ii) Estimate the distance taken by the bicycle to come to rest rom the time the cyclist stops pedalling.
0.5
1
1.5
2 v/m s -1
2.5
3
3.5
For a steady speed o 2 .0 m s 1 : a) use the graph to determine the power o the boats engine
( iii) State and explain one reason why your answer to b) ( ii) is an estimate.
b) calculate the rictional (resistive) orce acting on the boat. (3 marks)
( 1 3 marks)
A car o mass 1 000 kg accelerates on a straight fat horizontal road with an acceleration a = 0.30 m s 2 . The driving orce T on the car is opposed by a resistive orce o 5 00 N. C alculate T. (3 marks)
0.6 0.2
( i) C alculate the magnitude o the initial acceleration o the bicycle and rider.
5
0.8
8
(IB) The graph shows the variation with time t o the speed v o a ball o mass 0.5 0 kg that has been released rom rest above the Earths surace. 25
A crane hook is in equilibirium under the action o three orces as shown in the diagram. 3.8 kN
20 v/m s -1
6
15 10 5
T2
60
30
0
T1
0
2
4
6
8
10
t/s
The orce o air resistance is not negligible. a) S tate, without any calculations, how the graph could be used to determine the distance allen. C alculate T1 and T2 .
7
( 4 marks)
b) ( i)
C opy the diagram and draw and label arrows to represent the orces on the ball at 2 .0 s.
(IB) A small boat is powered by an outboard motor o variable power P. The graph below shows the variation with speed v o the power P or a particular load.
ball at t = 2.0s
Earths surface
( ii) Use the graph to show that the acceleration o the ball at 2 .0 s is approximately 4 m s 2 .
88
QUESTION S
( iii) C alculate the magnitude o the orce o air resistance on the ball at 2 . 0 s.
5.0 m s 1
( iv) State and explain whether the air resistance on the ball at t = 5 .0 s is smaller than, equal to, or greater than the air resistance at t = 2 . 0 s.
B
A immediately before collision
c) Ater 1 0 s the ball has allen 1 90 m. ( i)
v
S how that the sum o the potential and kinetic energies o the ball has decreased by about 800 J. ( 1 4 marks)
B
A immediately after collision
9
(IB) A bus is travelling at a constant speed o 6.2 m s 1 along a section o road that is inclined at an angle o 6.0 to the horizontal. 6.2m s 1
The mass o truck A is 800 kg and the mass o truck B is 1 2 00 kg. a) ( i) C alculate v. ( ii) C alculate the total kinetic energy lost during the collision. b) S uggest where the lost kinetic energy has gone. ( 6 marks)
6.0
a) ( i)
1 1 (IB) D raw a labelled sketch to represent the orces acting on the bus.
( ii) S tate the value o net orce acting on the bus. b) The total output power o the engine o the bus is 70 kW and the efciency o the engine is 3 5 % .
Large metal bars are driven into the ground using a heavy alling obj ect. object mass = 2.0 10 3 kg
C alculate the input power to the engine. c) The mass o the bus is 8.5 1 0 3 kg. D etermine the rate o increase o gravitational potential energy o the bus.
bar mass = 400 kg
d) Using your answer to c ( and the data in b) , estimate the magnitude o the resistive orces acting on the bus. ( 1 2 marks) The alling obj ect has a mass 2 000 kg and the metal bar has a mass o 400 kg. 1 0 (IB) Railway truck A moves along a horizontal track and collides with a stationary truck B . The two j oin together in the collision. Immediately beore the collision, truck A has a speed o 5 .0 m s 1 . Immediately ater collision, the speed o the trucks is v.
The obj ect strikes the bar at a speed o 6.0 m s 1 . It comes to rest on the bar without bouncing. As a result o the collision, the bar is driven into the ground to a depth o 0.75 m. a) D etermine the speed o the bar immediately ater the obj ect strikes it. b) D etermine the average rictional orce exerted by the ground on the bar. ( 7 marks)
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2
M E C H AN I C S 1 2 (IB) crane
An engine for a spacecraft uses solar power to ionize and then accelerate atoms of xenon. After acceleration, the ions are ej ected from the spaceship with a speed of 3 .0 1 0 4 m s 1 . xenon ions = 3.0 10 4 m s 1 speed
5.8 m
wall
spaceship mass = 5.4 10 2 kg
The mass of one ion of xenon is 2 .2 1 0
25
kg metal ball
a) The original mass of the fuel is 81 kg. D etermine how long the fuel will last if the engine ej ects 77 1 0 1 8 xenon ions every second.
To knock a wall down, the metal ball of mass 3 5 0 kg is pulled away from the wall and then released. The crane does not move. The graph below shows the variation with time t of the speed v of the ball after release.
2
b) The mass of the spaceship is 5 .4 1 0 kg. D etermine the initial acceleration of the spaceship. The graph below shows the variation with time t of the acceleration a of the spaceship. The solar propulsion engine is switched on at time t = 0 when the speed of the spaceship is 1 .2 1 0 3 m s 1 .
The ball makes contact with the wall when the cable from the crane is vertical. a) For the ball j ust before it hits the wall use the graph, to estimate the tension in the cable. The acceleration of free fall is 9.8 m s 2 .
10.0
b) D etermine the distance moved by the ball after coming into contact with the wall.
a/10 5 m s 2
9.5
c) C alculate the total change in momentum of the ball during the collision of the ball with the wall. ( 7 marks)
9.0 8.5
3.0
1.0
2.0
3.0 4.0 t/ 10 7 s
5.0
6.0
c) Explain why the acceleration of the spaceship increases with time. d) Using data from the graph, calculate the speed of the spaceship at the time when the xenon fuel has all been used. ( 1 5 marks)
1 3 (IB) A large metal ball is hung from a crane by means of a cable of length 5 .8 m as shown below.
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2.0 v/m s 1
8.0 0.0
cable
1.0
0.0 0.0
0.2
0.4
0.6
0.8 t/s
1.0
1.2
1.4
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T H E R M A L P H YS I C S
Introduction In this topic we look at at thermal processes resulting in energy transfer between obj ects at different temperatures. We consider how the energy transfer brings about further temperature changes and/or changes of state or phase.
We then go on to look at the effect of energy changes in gases and use the kinetic theory to explain macroscopic properties of gases in terms of the behaviour of gas molecules.
3.1 Temperature and energy changes Understanding Temperature and absolute temperature Internal energy Specifc heat capacity
Applications and skills The three states o matter are solid, liquid,
Phase change Specifc latent heat
Nature of science Evidence through experimentation Since early humans began to control the use o fre, energy transer because o temperature dierences has had signifcant impact on society. By controlling the energy ow by using insulators and conductors we stay warm or cool o, and we prepare lie-sustaining ood to provide our energy needs (see fgure 2) . Despite the importance o energy and temperature to our everyday lives, conusion regarding the dierence between thermal energy or heat and temperature is commonplace. The word heat is used colloquially to mean energy transerred because o a temperature dierence, but this is a throwback to the days in which scientists thought that heat was a substance and dierent rom energy.
and gas Solids have fxed shape and volume and comprise o particles that vibrate with respect to each other Liquids have no fxed shape but a fxed volume and comprise o particles that both vibrate and move in straight lines beore colliding with other particles Gases (or vapours) have no fxed volume or shape and move in straight lines beore colliding with other particles this is an ideal gas Thermal energy is oten misnamed heat or heat energy and is the energy that is transerred rom an object at a higher temperature by conduction, convection, and thermal radiation
Equations Conversion rom Celsius to kelvin:
T(K) = (C) + 273 Specifc heat capacity relationship: Q = mcT Specifc latent heat relationship: Q = mL
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Introduction From the 1 7th century until the end o the 1 9th century scientists believed that heat was a substance that owed between hot and cold obj ects. This substance travelling between hot and cold obj ects was known as phlogiston or caloric and there were even advocates o a substance called rigoric that owed rom cold bodies to hot ones. In the 1 840s James Joule showed that the temperature o a substance could be increased by doing work on that substance and that doing work was equivalent to heating. In his paddle wheel experiment ( see fgure 1 ) he dropped masses attached to a mechanism connected to a paddle wheel; the wheel churned water in a container and the temperature o the water was ound to increase. Although the caloric theory continued to have its supporters it was eventually universally abandoned. The unit calorie which is sometimes used relating to ood energy is a residual o the caloric theory, as is the use o the word heat as a noun.
2 kg
when masses fall they turn the axle and cause the paddle wheels to churn up the water this raises the water temperature
2 kg
thermometer paddles
water
Figure 1 Joules paddle wheel experiment.
Temperature and energy transfer You may have come across temperature described as the degree o hotness o an obj ect. This is a good starting point since it relates to our senses. A pot o boiling water eels very hot to the touch and we know instinctively that the water and the pot are at a higher temperature than the cold water taken rom a rerigerator. The relative temperature o two obj ects determines the direction in which energy passes rom one obj ect to the other; energy will tend to pass rom the hotter obj ect to the colder obj ect until they are both at the same temperature ( or in thermal equilibrium) . The energy owing as a result o conduction, convection, and thermal radiation is what is oten called heat . Figure 2 Dinner is served!
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Temp erature is a scalar quantity and is measured in units o degrees C elsius ( C ) or kelvin ( K) using a thermometer.
3 . 1 T E M P E R AT U R E A N D E N E R G Y C H A N G E S
Nature of science Thermometers Most people are amiliar with liquid- in- glass thermometers in which the movement o a column o liquid along a scale is used to measure temperature. Thermometers are not j ust restricted to this liquid- in- glass type; others use the expansion o a gas, the change in electrical resistance o a metal wire, or the change in em ( electromotive orce) at the j unction o two metal wires o dierent materials. A thermometer can be constructed rom any obj ect that has a property that varies with temperature ( a therm om etric p rop erty) . In the case o a liquid- in- glass thermometer the thermometric property is the expansion o the liquid along a glass capillary tube. The liquid is contained in the bulb which is a reservoir; when the bulb is heated the liquid expands, travelling along the capillary. S ince the bore o the capillary is assumed to be constant, as the volume o the liquid increases with temperature so does the length o the liquid. S uch thermometers are simple but not particularly accurate. Reasons or the inaccuracy could be that the capillary may not be uniorm, or its cross- sectional area may vary with temperature, or it is difcult to make sure that all the liquid is at the temperature o the obj ect being investigated. Glass is also a relatively good insulator and it takes time or thermal energy to conduct through the glass to the liquid, making
the thermometer slow to respond to rapid changes in temperature. Liquid- in- glass thermometers are oten calibrated in degrees C elsius, a scale based on the scale reading at two fxed p oints, the ice point and the steam point. These two temperatures are defned to be 0 C and 1 00 C respectively ( although C elsius actually used the steam point or 0 and ice point or 1 00) . The manuacturer o the thermometer assumes that the length o the liquid in the capillary changes linearly with temperature between these two points, even though it may not actually do so. This is a undamental assumption made or all thermometers, so that thermometers only agree with each other at the fxed points between these points they could well give dierent values or the same actual temperature. D igital thermometers or temperature sensors have signifcant advantages over liquid- in- glass thermometers and have largely taken over rom liquid- in- glass thermometers in many walks o lie. The heart o such devices is usually a thermistor. The resistance o most thermistors alls with temperature ( they are known as ntc or negative temperature coefcient o resistance thermistors) . S ince the thermistor is usually quite small, it responds very quickly to temperature changes. Thermistor thermometers are usually ar more robust than liquid- in- glass ones. steam point
ice point melting ice
ask steam boiling water
unnel beaker
fnding the ice point
fnding the steam point
Figure 3 Calibration o a liquid-in glass thermometer at the ice point and steam point.
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Investigate! Calibrating a thermistor against an alcohol-in-glass thermometer This investigation can be perormed by taking readings manually or by using a data logger. Here we describe the manual way o perorming the experiment. The temperature can be adjusted either using a heating coil or adding water at dierent temperatures (including some iced water, perhaps) .
Plot a graph o resistance/ against temperature/C ( since this is a calibration curve there can be no systematic uncertainties and any uncertainties will be random being sufciently small to be ignored in the context o this investigation) .
A multimeter set to ohms or an ohm-meter is connected across the thermistor (an ammeter/ voltmeter method would be a suitable alternative but resistance would need to be calculated) .
The graph shown is typical or a ntc thermistor.
S uggest why, on a calibration curve, systematic uncertainties are not appropriate.
Would the designers o a digital thermometer assume a linear relationship between resistance and temperature?
Would you expect the reading on a digital thermometer to correspond to that on a liquid- in- glass thermometer?
The thermistor is clamped so that it lies below the water surace in a S tyrooam ( expanded polystyrene) container next to an alcohol-inglass thermometer. O btain pairs o values o readings on the multimeter and the thermometer and record these in a table.
thermistor thermometer rubber seals cardboard square Styrofoam cups
resistance of themistor/
calibration of thermistor
2.0 10 4 1.5 10 4 1.0 10 4 0.5 10 4 0
water
0 multimeter
50
100
temperature/C
Absolute temperature The C elsius temperature scale is based on the ice point and steam point o water; by defnition all thermometers using the C elsius scale agree at these two temperatures. B etween these fxed points, thermometers with dierent thermometric properties do not all agree, although dierences may be small. The absolute temp erature scale is the standard S I temperature scale with its unit the kelvin ( K) being one o the seven S I base units. Absolute temp erature is defned to be zero kelvin at absolute zero ( the temperature at which all matter has minimum kinetic energy) and 2 73 . 1 6 K at the triple point o water ( the unique temperature and pressure at which water can exist as liquid water, ice, and water vapour) . D ierences in absolute temperatures exactly correspond to those in C elsius temperatures ( with a temperature dierence o 1 C being identical to 1 K) . For this reason
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3 . 1 T E M P E R AT U R E A N D E N E R G Y C H A N G E S
it is usual to write the units o temperature dierence as K but not C ( although you are unlikely to lose marks in an examination or making this slip) . To convert rom temperatures in degrees C elsius to absolute temperatures the ollowing relationship is used: T( K) = ( C ) + 2 73 where T represents the absolute temperatures and the temperature in degrees C elsius. Other aspects o the absolute temperature scale will be considered in Sub-topic 3.2.
Internal energy S ubstances consist o particles ( e. g., molecules) in constant random motion. As energy is transerred to a substance the separation o the particles could increase and they could move aster. When particles move urther apart ( or closer to other neighbouring particles) the potential energy o the particles increases. As they move aster their random kinetic energy increases. The internal energy o a substance is the total o the p otential energy and the random kinetic energy o all the p articles in the substance. For a solid, these two orms o energy are present in roughly equal amounts; however in a gas the orces between the particles are so small that the internal energy is almost totally kinetic. We will discuss this urther in the S ub-topic 3 .2 .
Worked example A temperature o 73 K is equivalent to a temperature o A. 3 46 C . C . + 73 C .
B . 2 00 C . D . + 2 00 C .
Solution S ubstituting values into the conversion equation: 73 = + 2 73 so = 73 2 73 = 200 making the correct option B.
Note: Students oten conuse the terms internal energy and thermal energy. Avoid using the term thermal energy talk about internal energy and the energy transerred because o temperature dierences, which is what most people mean by thermal energy.
Specifc heat capacity When two dierent obj ects receive the same amount o energy they are most unlikely to undergo the same temperature change. For example, when 1 000 J o energy is transerred to 2 kg o water or to 1 kg o copper the temperature o each mass changes by dierent amounts. The temperature o the water would be expected to increase by about 0. 1 2 K while that o copper by 2 . 6 K. The two masses have dierent heat cap acities. You may think that it is hardly a air comparison since there is 2 kg o water and 1 kg o copper; and you would be right to think this! I we chose equal masses o the two substances we would discover that 1 kg o water would increase by 0. 2 4 K under the same conditions.
solid
liquid
In order to be able to compare substances more closely we defne the sp ecifc heat cap acity (c) o a substance as the energy transerred to 1 kg o the substance causing its temp erature to increase by 1 K. The defning equation or this is Q c= _ m T where Q is the amount o energy supplied to the obj ect o mass m and causing its temperature to rise by T. When Q is in J, m in kg, and T in K, c will be in units o J kg 1 K 1 . You should check that the data provided show that water has a specifc heat capacity o approximately 4.2 1 03 J kg 1 K1 and copper a value o approximately 380 J kg1 K1 .
gas
Figure 4 Motion of molecules in solids, liquids, and gases.
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Worked example A piece o iron o mass 0. 1 3 3 kg is placed in a kiln until it reaches the temperature o the kiln. The iron is then quickly transerred to 0. 476 kg o water held in a thermally insulated container. The water is stirred until it reaches a steady temperature. The ollowing data are available. Specifc heat capacity o iron = 45 0 J kg 1 K 1 Specifc heat capacity o water
= 4.2 1 0 3 J kg 1 K1
Initial temperature o the water = 1 6 C
b) C alculate the increase in internal energy o the water as the iron cools in the water. c) Use your answers to b) and c) to determine .
Solution a) Using Q = mc T or the iron Q = 0.1 33 45 0 ( 45 ) = 60 ( 45 ) b) Using Q = mc T or the water Q = 0.476 42 00 (45 1 6) = 5 .8 1 0 4 J c) Equating these and assuming no energy is transerred to the surroundings 60 ( 45 ) = 5 .8 1 0 4
Final temperature o the water = 45 C The specifc heat capacity o the container and insulation is negligible.
60 2 700 = 5 . 8 1 0 4
a) S tate an expression, in terms o and the above data, or the energy transer o the iron in cooling rom the temperature o the kiln to the fnal temperature o the water.
60 = 6.1 1 0 4 6. 1 1 0 4 = _ 60 = 1 000 C ( or 1 01 0 C to 3 s.. )
Investigate! Estimating the specifc heat capacity o a metal This is another investigation that can be perormed manually or by using a data logger. Here we describe the manual way o perorming the experiment. A metal block o known mass is heated directly by a heating coil connected to a low-voltage electrical supply. Channels in the block allow the coil to be inserted and also a thermometer or temperature probe. The power supplied to the block is calculated rom the product o the current (I) in the heating coil and the potential dierence (V) across it. The supply is switched on or a measured time (t) so that the temperature changes by at least 1 0 K. When the temperature has risen sufciently the power is switched o, but the temperature continues to be monitored to fnd the maximum temperature rise (it takes time or the thermometer to reach the same temperature as the surrounding block) .
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Thermal energy transerred to the block in time t = VIt
The heater itsel, the temperature probe and the insulation will all have heat capacities that will mean that there is a urther unknown term to the relationship: which should become VIt = ( mc + C) T where C is the heat capacity o everything except the block which will undergo the same temperature change as the block since they are all in good thermal contact. This term may be ignored i the mass o the block is high ( say, 1 kg or so) .
As the block is heated it will also be losing thermal energy to the surroundings and so make the actual rise in temperature lower
3 . 1 T E M P E R AT U R E A N D E N E R G Y C H A N G E S
that might be predicted. You are not expected to calculate cooling corrections in your IB calculations but it is important to recognize that this happens. A simple way o compensating or this is to cool the apparatus by, say 5 K, below room temperature beore switching on the power supply. I you then allow its temperature to rise to 5 K above room temperature, then you can assume that the thermal energy gained rom the room in reaching room temperature is equal to that lost to the room when the apparatus is above room temperature.
This method can be adapted to measure the specifc heat capacity o a liquid, but you must remember that the temperature o the container will also rise.
B to immersion heater circuit A 0100 C thermometer
aluminium block (calorimeter)
TOK
immersion heater
The development o understanding o energy transer due to temperature diferences insulating board
Specifc latent heat The temperature o a block o ice in a reezer is likely to be well below 0 C . I we measure the temperature o the ice rom the time it is taken rom the reezer until it has all melted and reached room temperature we would note a number o things. Initially the temperature o the ice would rise; it would then stay constant until all o it had melted then it would rise again but at a dierent rate rom beore. I we now put the water into a pan and heat it we would see its temperature rise quickly until it was boiling and then stay constant until all o the water had boiled away. These observations are typical o most substances which are heated sufciently to make them melt and then boil. Figure 5 shows how the temperature changes with time or energy being supplied. Energy is continually being supplied to the ice but there is no temperature change occurring during melting or boiling. The energy required to achieve the change o phase is called latent heat; the word latent meaning hidden. In a similar way to how specifc heat capacity was defned in order to compare equal masses o substances, physicists defne:
sp ecifc latent heat o usion ( melting) as the energy required to change the phase o 1 kg o substance rom a solid to a liquid without any temperature change
The term specifc heat capacity is a throwback to the caloric days when energy was thought to be a substance. A more appropriate term would be to call this quantity specifc energy capacity but specifc heat capacity is a term that has stuck even though scientists now recognize its inappropriateness. Are there other cases where we still use an old-ashioned term or concept because it is simpler to do so than to change all the books? Has the paradigm shit believing that heating was a transer o substance into being a transer o energy been taken on by society or is there still conusion regarding this?
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sp ecifc latent heat o vap orization ( boiling) as the energy required to change the phase o 1 kg o liquid into a gas without any temperature change.
The equations or these take the orm Q L= _ m
co
ol s
where Q is the energy supplied to the obj ect o mass which causes its phase to change without any temperature change. When Q is in J, m in kg, L will be in units o J kg 1 .
molecules
boils
wa
s
gas
rm
ols
freezes wa
co
temperature
co
ol s
rm
s
condenses
liquid
solid
wa
rm
s
melts
time
Figure 5 Change of phase (not to scale) . The graph shows how the temperature o a substance changes with time. The portions o the graph that are at indicate when there is no temperature change and so the substance is changing phase.
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This graph assumes that energy is supplied by a constant power source and so, since energy is the product o power and time, a graph o temperature against energy transerred to the substance will take the same shape as this.
I you are told the rate at which energy is supplied ( i.e. the power supplied) then rom the gradients o the temperaturetime graph you can work out the heat capacity o the liquid, the solid and the gas.
From the time o each horizontal section o the graph you can also calculate the specifc heat capacity o the solid and liquid phases.
When a gas transers energy to another obj ect at a constant rate we see the temperature o the gas drop until it reaches the boiling point ( condensing point is the same temperature as this) at which time the graph becomes at. Again as the liquid transers energy away the temperature will all until the melting ( or solidiying) temperature is reached when the graph again becomes at until all the substance has solidifed. Ater this it will cool again.
C ooling is the term generally used to mean alling temperature rather than loss o energy, so it is important not to say the substance is cooling when the temperature is constant ( at parts o the graph) .
3 . 1 T E M P E R AT U R E A N D E N E R G Y C H A N G E S
Worked example A heater is used to boil a liquid in a pan or a measured time. The ollowing data are available. Power rating o heater = 2 5 W Time or which liquid is boiled = 6. 2 1 0 2 s Mass o liquid boiled away = 4.1 1 0 2 kg Use the data to determine the specifc latent heat o vaporization o the liquid.
Solution Using Q = Pt or heater: Q = 2 5 6.2 1 0 2 = 1 5 5 00 J Using Q = mL or the liquid: 1 5 5 00 = 4. 1 1 0 2 L 1 5 5 00 L=_ = 3 . 8 1 0 5 J kg 1 4.1 1 0 2
Molecular explanation of phase change The transer o energy to a solid increases its internal energy this means that the mean random kinetic energy o the molecules increases ( as the vibrations and speeds increase) and the intermolecular potential energy increases ( as the molecules move urther apart) . Eventually some groups o molecules move ar enough away rom their neighbours or the inuence o their neighbours to be reduced chemists would oten use the model o the intermolecular bonds being broken. When this is happening the energy supplied does not increase the mean random kinetic energy o the molecules but instead increases the potential energy o the molecules. E ventually the groups o molecules are sufciently ree so that the solid has melted. The mean speed o the groups o molecules now increases and the temperature rises once more the molecules are breaking away rom each other and j oining together at a constant rate. The potential energy is not changing on average. As the liquid reaches its boiling point the molecules start moving away rom each other within their groups. Individual molecules break away and the potential energy once more increases as energy is supplied. Again a stage is reached in which the mean kinetic energy remains constant until all the molecules are separated rom each other. We see a constant temperature ( although much higher than during melting) . The energy is vaporizing the liquid and only when the liquid has all vaporized is the energy re-applied to raising the gas temperature.
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3.2 Modelling a gas Understanding
Applications and skills
Pressure
Equation o state or an ideal gas
Equation o state or an ideal gas Kinetic model o an ideal gas
Kinetic model o an ideal gas Boltzmann equation
Mole, molar mass, and the Avogadro constant Diferences between real and ideal gases
Mole, molar mass, and the Avogadro constant Diferences between real and ideal gases
Equations Pressure: p = ___AF Number o moles o a gas as the ratio o number
_____
o molecules to Avogadros number: n = NN A Equation o state or an ideal gas: pV = nRT Pressure and mean square velocity ___ o ___ 1 2 molecules o an ideal gas: p = 3 c The mean kinetic energy o ideal gas molecules: R E K = ___32 kB T = ___32 ____ T N mean
A
Nature of science Progress towards understanding by collaboration and modelling Much o this sub-topic relates to dealing with how scientists collaborated with each other and gradually revised their ideas in the light o the work o others. Repeating and oten improving the original experiment using more reliable instrumentation meant that generalizations could be made. Modelling
the behaviour o a real gas by an ideal gas and the use o statistical methods was groundbreaking in science and has had a major impact on modern approaches to science including the quantum theory where certainty must be replaced by probability.
Introduction O ne of the triumphs of the use of mathematics in physics was the successful modelling of the microscopic behaviour of atoms to give an understanding of the macroscopic properties of a gas. Although the S wiss mathematician D aniel B ernoulli had suggested that the motion of atoms was responsible for gas pressure in the mid-1 700s, the kinetic theory of gases was not widely accepted until the work of the S cottish physicist, James C lerk-Maxwell and the Austrian, Ludwig B oltzmann working independently over a hundred years later.
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3 . 2 M O D E LLI N G A G AS
The gas laws The gas laws were developed independently experimentally between the mid-seventeenth century and the start o the nineteenth century. An ideal gas can be defned as one that obeys the gas laws under all conditions. Real gases do not do this but they approximate well to an ideal gas so long as the pressure is little more than normal atmospheric pressure. With the modern apparatus available to us it is not difcult to veriy the gas laws experimentally. As a consequence o developing the vacuum pump and thus the hermetically- sealed thermometer, the Irish physicist Robert B oyle was able to show in 1 662 that the pressure o a gas was inversely proportional to its volume. B oyles experiment is now easily repeatable with modern apparatus. B oyles law states that or a fxed mass o gas at constant temp erature the p ressure is inversely p rop ortional to the volume. The French experimenter Edm Mariotte independently perormed a similar experiment to that o B oyle and was responsible or recognizing that the relationship only holds at a constant temperature so with modern statements o the law, there is much j ustifcation or calling this law Mariottes law. 1 ( at constant temperature) or Mathematically this can be written as p __ V pV = constant ( at constant temperature) .
A graph o pressure against volume at constant temperature ( i.e. , a B oyles law graph) is known as an isothermal curve ( iso meaning the same and thermo relating to temperature) . Isothermal curves are shown in fgure 2 ( b) on p1 04.
TOK Boyles impact on scientifc method Robert Boyle was a scientifc giant and experimental science has much to thank him or in addition to his eponymous law. In the mid-1600s where philosophical reasoning was preerred to experimentation, Boyle championed perorming experiments. Boyle was most careul to describe his experimental techniques to allow them to be reproduced by others this gave reliability to experimental results and their interpretation. At this time, many experimenters were working independently. The conusion in correct attribution o gas laws to their discoverer may have been less involved had others ollowed Boyle's lead. His rapid and clear reporting o his experimental work and data avoided the secrecy that was common at the time; this was advantageous to the progress o other workers. Although Boyle was apparently not involved in the ensuing quarrels, there was great debate regarding whether Boyle or the German chemist Henning Brand discovered the element phosphorus. This is largely because Brand kept his discovery secret while he
pursued the philosophers stone in an attempt to convert base metals such as lead into gold. Is rapid or requent publication usually a case o a scientist's selpromotion or is the practice more oten altruistic?
Figure 1 Robert Boyle.
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T H E R M AL P H YS I C S The French physicist Jacques C harles around 1 787 repeated the experiments o his compatriot Guillaume Amontons to show that all gases expanded by equal amounts when subj ected to equal temperature 1 changes. C harles showed that the volume changed by ___ o the volume 273 at 0 C or each 1 K temperature change. This implied that at 2 73 C , the volume o a gas becomes zero. C harless work went unpublished and was repeated in 1 802 by another French experimenter, Joseph GayLussac. The law that is usually attributed to C harles is now stated as or a fxed mass o gas at constant p ressure the volume is directly p rop ortional to the absolute temp erature. Mathematically this law can be written as V T ( at constant pressure) V constant ( at constant pressure) . or __ T = Amontons investigated the relationship between pressure and temperature but used relatively insensitive equipment. He was able to show that the pressure increases when temperature increases but ailed to quantiy this completely. The third gas law is stated as or a gas o fxed mass and volume, the pressure is directly proportional to the absolute temperature. This law is sometimes attributed to Amontons and oten (incorrectly) to Gay-Lussac or Avogadro and sometimes it is simply called the pressure law. Maybe it is saest to reer to this as the third gas law! Mathematically this law can be written as p T ( at constant volume) p or __ = constant( at constant volume) . T The Italian physicist, C ount Amadeo Avogadro used the discovery that gases expand by equal amounts or equal temperature rises to support a hypothesis that all gases at the same temperature and pressure contain equal numbers o particles per unit volume. This was published in a paper in 1 81 1 . We now state this as the number o p articles in a gas at constant temp erature and p ressure is directly p rop ortional to the volume o the gas. Mathematically this law can be written as n V ( at constant n pressure and temperature) or __ = constant ( at constant pressure and V temperature) . S ince each o the gas laws applies under dierent conditions the constants are not the same in the our relationships. The our equations can be combined to give a single constant, the ideal gas constant, R. C ombining the our equations gives what is known as the equation o state o an ideal gas: pV _ = R or pV = nRT nT When pressure is measured in pascal ( Pa) , volume in cubic metre ( m 3 ) , temperature in kelvin ( K) , and n is the number o moles in the gas, then R has the value 8. 3 1 J K 1 mol 1 .
The mole and the Avogadro constant The mole, which is given the symbol mol, is a measure o the amount o substance that something has. It is one o the seven S I base units and is defned as being the amount o substance having the same
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3 . 2 M O D E LLI N G A G AS
number of p articles as there are neutral atoms in 1 2 grams of carbon-1 2 . O ne mole o a gas contains 6.02 1 0 23 atoms or molecules. This number is the Avogadro constant NA. In the same way that you can reer to a dozen roses ( and everyone agrees that a dozen is the name or 1 2 ) you can talk about three moles o nitrogen gas which will mean that you have 1 8.06 1 0 23 gas molecules. The mole is used as an alternative to expressing quantities in volumes or masses.
Molar mass As we have j ust seen, the mole is simply a number that can be used to count atoms, molecules, ions, electrons, or roses ( i you wanted) . In order to calculate the molar mass o a substance ( which diers rom substance to substance) we need to know the chemical ormula o a compound ( or whether a molecule o an element is made up rom one or more atoms) . Nitrogen gas is normally diatomic ( its molecules have two atoms) , so we write it as N 2 . O ne mole o nitrogen gas will contain 6.02 1 0 2 3 molecules but 1 2 .04 1 0 23 atoms. As one mole o nitrogen atoms has a mass o approximately 1 4.01 g then a mole o nitrogen molecules will have a mass o 2 8.02 g this is its molar mass. The chemical ormula or water is, o course, H 2 O , so one mole o water molecules contains two moles o hydrogen atoms and one mole o oxygen atoms. 1 mol o hydrogen atoms has a mass o 1 .00 g and 1 mol o oxygen atoms has a mass o 1 6.00 g, so 1 mol o water has a mass o ( 2 1 .00 g) + 1 6. 00 g = 1 8.00 g. The mass o water is 1 8.00 g mol 1 .
Worked example C alculate the percentage change in volume o a fxed mass o an ideal gas when its pressure is increased by a actor o 2 and its temperature increases rom 3 0 C to 1 2 0 C .
Solution S ince n is constant the equation o state can be written p 1 V1 p 2 V2 _ =_ T1 T2
V and so the We are trying to obtain the ratio __ V 2 1
V p T equation can be rearranged into the orm __ = ___ V p T 2
1
2
1
2
1
V2 p 2 = 2 p 1 and T1 = 3 03 K and T2 = 3 93 K, so __ V 1
393 = ______ = 0.65 or 65 % 2 303
This means that there is a 3 5 % reduction in the volume o the gas.
Investigate! Verifying the gas laws experimentally Boyles law
Pressure is changed using the pump.
This pushes dierent amounts o oil into the vertical tube, which changes the pressure on the gas ( air) .
C hanges are carried out slowly to ensure there is no temperature change.
A graph o pressure against volume gives a curve known as an isothermal ( line at constant temperature) .
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Isothermals can be plotted over a range o dierent temperatures to give a series o curves such as those in fgure 2 ( b) .
A graph o pressure against the reciprocal o 1 volume __ should give a straight line passing V through the origin.
This investigation could be monitored automatically using a data logger with a pressure sensor.
Charless law
Pressure is kept constant because the capillary tube is open to the surrounding atmosphere ( provided the atmospheric pressure does not change) .
C hange the temperature by heating ( or cooling/icing) the water.
Assuming that the capillary is totally uniorm ( which is unlikely) the length o the column o air trapped in the tube will be proportional to its volume.
S tir the water thoroughly to ensure that it is at a constant temperature throughout.
0 5 10
column o trapped gas
15 20 25
thermometer
30 35 40
oil column
capillary tube
Bourdon gauge
45
valve
L
rubber bands
50
air trapped by liquid index
55 60
ootball pump
(a)
pressure
increasing temperature This is your value or absolute zero
(b) volume
Figure 2 (a) Boyles law apparatus and (b) graph.
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Your results are going to be in the region rom 0 C up to 50 C, but i you draw your graph on the scale shown here it will not give a very reliable value or absolute zero it is better to use the page rom 0 C up to 50 C and to use the method o similar triangles to fnd absolute zero
L/mm
200 temperature/
0
Figure 3 Charles law apparatus and graph.
50
3 . 2 M O D E LLI N G A G AS
Leave a reasonable length o time in between readings to allow energy to conduct to the air in the capillary when this happens the length o the capillary no longer changes.
A graph o length o the air column against the temperature in C should give a straight line, which can be extrapolated back to absolute zero.
To choose larger ( and thereore better) scales you can use similar triangles to calculate absolute zero.
line, which can be extrapolated back to absolute zero.
To choose larger ( and thereore better) scales you can use similar triangles to calculate absolute zero.
Bourdon gauge
thermometer 250 ml round-bottomed fask containing air water bath
The third gas law
There are many variants o this apparatus which could, yet again, be perormed using a data logger with temperature and pressure sensors.
The gas ( air) is trapped in the glass bulb and the temperature o the water bath is changed.
The pressure is read rom the B ourdon gauge ( or manometer or pressure sensor, etc.) . Leave a reasonable length o time in between readings to allow energy to conduct to the air in the glass bulb when this happens the pressure no longer changes. A graph o pressure against the temperature in o C should give a straight
pressure This is your value or absolute zero
200 temperature/C
0
50
Figure 4 Third gas law apparatus and graph.
Nature of science Evidence for atoms The concept o atoms is not a new one but it has only been accepted universally by scientists over the past century. Around 400 B C E the Greek philosopher D emocritus theorized the existence o atoms named rom ancient Greek: meaning without division. He suggested that matter consisted o tiny indivisible but discrete particles that determined the nature o the matter o which they comprised. E xperimental science did not become ashionable until the 1 7th century and so D emocritus theory remained unproven. S ir Isaac Newton had a limited view o the atomic nature o matter but believed that all matter was ultimately made o the same substance. In the 1 9th century the chemist, John D alton, was the frst to suggest that the individual elements were made o dierent atoms and could be combined in fxed ratios. Yet even D alton and his successors were only able to iner the presence o atoms rom chemical reactions. Less explicit evidence (yet still indirect) came rom experiments with diusion and B rownian motion.
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The microscopic interpretation of gases Difusion The smell o cooked ood wating rom a barbecue is something that triggers the ow o stomach j uices in many people. This can happen on a windless day and is an example o diusion o gases. The atomic vapours o the cooking ood are being bombarded by air molecules causing them to move through the air randomly. The bromine vapour experiment shown in fgure 5 is a classic demonstration o diusion. B romine ( a brown vapour) is denser than air and sinks to the bottom o the lower right- hand gas j ar. This is initially separated rom the upper gas j ar by a cover slide. As the slide is removed the gas is gradually seen to fll the upper gas j ar ( as seen in the j ars on the let) ; this is because the air molecules collide with the bromine atoms. I this was not the case we would expect the bromine to remain in the lower gas j ar.
Brownian motion
Figure 5 Bromine difusion.
In 1 82 7, E nglish botanist, Robert B rown, frst observed the motion o pollen grains suspended in water. Today we oten demonstrate this motion using a microscope to see the motion o smoke particles suspended in air. The smoke particles are seen to move around in a haphazard way. This is because the relatively big and heavy smoke particles are being bombarded by air molecules. The air molecules have momentum, some o which is transerred to the smoke particles. At any instant there will be an imbalance o orces acting on each smoke particle giving it the random motion observed. The experiment o fgure 6( a) shows how smoke in an air cell is illuminated rom the side. E ach smoke particle scatters light in all directions and so some reaches the microscope. The observer sees the motion as tiny specks o bright light that wobble around unpredictably as shown in fgure 6( b) .
these specks o light are the smoke particles scattering light
microscope
convex lens
window
the random motion o a smoke particle showing how it moves linearly in between collisions with air molecules
smoke
lamp air cell (a)
Figure 6 The smoke cell.
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(b)
3 . 2 M O D E LLI N G A G AS
Kinetic model of an ideal gas
z
The kinetic theory of gases is a statistical treatment o the movement o gas molecules in which macroscopic properties such as pressure are interpreted by considering molecular movement. The key assumptions o the kinetic theory are:
c
A gas consists o a large number o identical tiny particles called molecules; they are in constant random motion.
L
m cx
The number is large enough or statistical averages to be made.
Each molecule has negligible volume when compared with the volume o the gas as a whole.
At any instant as many molecules are moving in one direction as in any other direction.
The molecules undergo perectly elastic collisions between themselves and also with the walls o their containing vessel; during collisions each momentum o each molecule is reversed.
There are no intermolecular orces between the molecules between collisions ( energy is entirely kinetic) .
The duration o a collision is negligible compared with the time between collisions.
Each molecule produces a orce on the wall o the container.
The orces o individual molecules will average out to produce a uniorm pressure throughout the gas ( ignoring the eect o gravity) .
y
x L
L
Figure 7 Focusing on the x-component of the velocity.
Using these assumptions and a little algebra it is possible to derive the ideal gas equation: In fgure 7 we see one o N molecules each o mass m moving in a box o volume V. The box is a cube with edges o length L. We consider the collision o one molecule moving with a velocity c towards the righthand wall o the box. The components o the molecules velocity in the x, y, and z directions are cx, cy, and cz respectively. As the molecule collides with the wall elastically, its x component o velocity is reversed, while its y and z components remain unchanged. The x component o the momentum o the molecule is mcx beore the collision and mcx ater the collision. The change in momentum o the molecule is thereore mcx (mcx) = 2mcx As orce is the rate o change o momentum, the orce Fx that the 2mc right-hand wall o the box exerts on our molecule will be _____ where t t is the time taken by the molecule to travel rom the right- hand wall o the box to the opposite side and back again ( in other words it is the time between collisions with the right- hand wall o the box) . x
2 m cx _ m c 2x 2L _ Thus t = _ = = and so F m cx 2L __ L cx
What quantity is equivalent to the reciprocal of t?
Using Newtons third law o motion we see that the molecule must exert mc a orce o ___ on the right-hand wall o the box ( i.e. a orce equal in L magnitude but opposite in direction to the orce exerted by the righthand wall on the molecule) . 2 x
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3
T H E R M AL P H YS I C S S ince there are N molecules in total and they will have a range o speeds, the total orce exerted on the right- hand wall o the box m Fx = _ ( c 2x + c 2x + c 2x + ... + c 2x ) L Where c x is the x component o velocity o the frst molecule, c x that o the second molecule, etc. 1
2
3
N
1
Which assumption is related to molecules having a range o speeds?
2
With so many molecules in even a small volume o gas, the orces average out to give a constant orce. The mean value o the square o the velocities is given by
z
( c 2x + c 2x + c 2x + .. . + c 2x ) ___ c = N This means that the __ total orce on the right- hand wall o the box is Nm 2 given by Fx = ___ c x. L __ 2 x
c cz cx
1
2
3
N
Figure 8 shows a velocity vector c being resolved into components cx cy, and cz.
cy
Using Pythagoras theorem we see that c2 = c 2x + c 2y + c 2z x
It ollows that the mean values o velocity can be resolved into the mean values o its components such that __
y
Figure 8 Resolving the velocity of a molecule.
__
__
__
c 2 = c 2x + c 2y + c 2z __
c 2 is called the mean square speed o the molecules.
Which assumption is being used here?
On average there is an equal likelihood o a molecule moving in any direction as in any other direction so the magnitude o the mean components o the velocity will be the same, i.e. , __
__
__
__
__
__
__
1 2 c 2x = c 2y = c 2z , so c 2 = 3 c 2x or c 2x = __ c 3
S o our equation or the total orce on the right- hand wall becomes __ Nm 2 1 ___ F = __ c 3 L F The pressure on this wall ( which has an area A) is given by p x = __ and A A = L 2 so that X
Nm __2 Nm __2 1_ 1_ px = _ c or p x = _ c 3 3 L 3 V S ince pressure at a point in a uid ( gas or liquid) acts equally in all directions we can write this equation as Nm __2 1_ p= _ c 3 V As Nm is the total mass o the gas and the density ( Greek, rho) is the total mass per unit volume, our equation simplifes to 1 c__2 p= _ 3
Do collisions between molecules afect this result?
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B e careful not to confuse p, the pressure, with , the density. You should note that the guidance in the IB D iploma Programme physics guide says that you should understand this proo that does not mean that you need to learn it line by line but that each o the steps should make sense to you. You should then be able to answer any question asked in examinations.
3 . 2 M O D E LLI N G A G AS
Molecular interpretation of temperature Returning to the equation o an ideal gas in the orm Nm __2 1 _ p = __ c 3 V we get Nm __ pV = _c2 3 3 Multiplying each side by __ gives 2
__
__
Nm 2 3 3 ___ 1 __ pV = ( __ c ) = N __ m c2 2 2 3 2
C omparing this with the equation o state or an ideal gas pV = nRT __
3 1 nRT = N __ m c2 we can see that __ 2 2
This may look a little conusing with both n and N in the equation so lets recap: n is the number o moles o the gas and N is the number o molecules so lets combine them to give a simpler equation: __
n RT 3 ___ 1 __ = __ m c2 2 N 2
N N B ut ___ = n so __ n is the number o molecules per mole ( the Avogadro N number, NA) making the equation: A
__
3 ___ RT 1 __ = __ m c2 2 N 2 A
R Things get even easier when we defne a new constant to be ___ ; this N constant is given the symbol kB and is called the B oltzmann constant. A
S o the equation now becomes: __
3 1 __ k T = __ m c2 2 B 2
__
1 m c2 should remind you o the equation or kinetic energy and, Now __ 2 indeed, it represents the mean translational kinetic energy o the gas molecules. We can see rom this equation that the mean kinetic energy gives a measure o the absolute temperature o the gas molecules.
The kinetic theory has linked the temperature ( a macroscopic property) to the microscopic energies o the gas molecules. In an ideal gas there are no long- range intermolecular orces and thereore no potential energy components; the internal energy o an ideal gas is entirely kinetic. This means that the total internal energy o an ideal gas is ound by multiplying the number o molecules by the mean kinetic energy o the molecules 3 total internal energy of an ideal gas = __ NkB T 2
We have only considered the translational aspects o our gas molecules in this derivation and this is fne or atomic gases ( gases with only one atom in their molecules) ; when more complex molecules are considered this equation is slightly adapted using a principle called the equipartition o energies ( something not included in the IB D iploma Programme Physics guide) .
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Alternative equation o state o an ideal gas
Note When you look at the unit or Boltzmann's constant you should see a similarity with specifc heat capacity. This is like the specifc heat capacity or one mole o every monatomic gas
The gas laws produced the equation o state in the orm o pV = nRT. Now we have met the B oltzmann constant we can use an alternative orm o this equation. As we have seen n represents the number o moles o our ideal gas and R is the universal molar gas constant. I we want to work with the number o molecules in the gas ( as physicists oten do) the equation o state can be written in the orm o pV = Nk B T where N represents the number o molecules and kB is the B oltzmann constant. Thus nR = NkB and i we consider 1 mol o gas then n = 1 R and N = NA = 6.02 1 0 23 . We see rom this why kB = __ as was used N previously. A
1
1
8.3 J mol K With R = 8. 3 J K 1 mol 1 kB must = ____________ = 1 .3 8 1 0 23 J K 1 6.02 1 0 mol 23
1
Worked example Nitrogen gas is sealed in a container at a temperature o 3 2 0 K and a pressure o 1 .01 1 0 5 Pa. a) C alculate the mean square speed o the molecules. b) C alculate the temperature at which the mean square speed o the molecules reduces to 5 0% o that in a) .
Solution
__
__
3p 3 1 .01 1 0 1 ___________ a) p = __ c2 so c2 = __ = 3 1 .2 5
= 2 .5 3 1 0 5 m 2 s 2 ( notice the unit here?) __
b) c2 T so the temperature would need to be 5 0% o the original value ( i. e. it would be 1 60 K) .
mean density o nitrogen gas over the temperatures considered = 1 .2 kg m 3
Nature of science MaxwellBoltzmann distribution In the 1 85 0s James C lerk Maxwell realized that a gas had too many molecules to have any chance o being analysed using Newtons laws ( even though this could be done in principle) . With no real necessity to consider the motion o individual molecules he realized that averaging techniques could be used to link the motion o the molecules with their macroscopic properties. He recognized that he needed to know the distribution o molecules having dierent speeds. Ludwig B oltzmann had proposed a general idea about how energy was distributed among systems consisting o many particles and Maxwell developed B oltzmanns distribution to show how many particles have a particular speed in the gas. Figure 9 shows the three typical distributions
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or the same number o gas molecules at three dierent temperatures. At higher temperatures the most probable speed increases but overall there are less molecules travelling at this speed since there are more molecules travelling at higher speeds. The speeds o molecules at these temperatures can be greater than 1 km s 1 ( but they dont travel very ar beore colliding with another molecule! ) . The MaxwellB oltzmann distribution was frst verifed experimentally between 1 93 0 and 1 934 by the American physicist I F Zartman using methods devised by the German-American Otto Stern in the 1 92 0s. Zartman measured the speed o molecules emitted rom an oven by collecting them on the inner surace o a rotating cylindrical drum.
3 . 2 M O D E LLI N G A G AS
MaxwellBoltzmann speed distribution 12.00% 273 K 373 K 473 K
relative frequency
10.00% 8.00% 6.00% 4.00% 2.00% 0.00% 0
200
400
600 800 speed/m s 1
1000
1200
1400
Figure 9 MaxwellBoltzmann distribution for the speed of gas molecules.
Nature of science Real gases An ideal gas is one that would obey the gas laws and the ideal gas equation under all conditions so ideal gases cannot be liquefed. In 1 863, the Irish physician and chemist Thomas Andrews succeeded in plotting a series o pV curves or carbon dioxide; these curves deviated rom the Boyles law curves at high pressures and low temperatures. Until this time it had been believed that certain gases could never be liquefed. Andrews showed that there was a critical temperature above which the gas could not be liquefed by simply increasing the pressure. He demonstrated that or carbon dioxide the critical temperature is approximately 31 C.
The diference between ideal gases and real gases The 1 9th century experimenters showed that ideal gases are j ust that ideal and that real gases do not behave as ideal gases. Under high pressures and low temperatures all gases can be liquefed and thus become almost incompressible. This can pV be seen i we plot a graph o ___ against p or one RT mole o a real gas we would expect this plot to give a horizontal straight line o value o 1 .00 or
an ideal gas. Figure 1 0 is such a plot or nitrogen. The pressure axis is in atmospheres this is a non-SI unit that is appropriate or high pressures and quite commonly used by experimenters (e.g. 900 atm converts to 900 1 .01 1 0 5 Pa or 9.1 1 0 7 Pa) . The ideal gas line is a better ft to the real gas line at 1 000 K and better still at low pressure. 3
200 K 500 K
2 PV RT
1000 K ideal gas
1
0 300
600 P (atm)
900
Figure 10 Deviation of a real gas from an ideal gas situation.
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Thus we can see that the best approximation o a real gas to an ideal gas is at high temperature and low pressures. In deriving the ideal gas equation we made assumptions that are not true or a real gas. In particular we said that there are no intermolecular orces between the molecules in between collisions. This is not true or real gases on two accounts:
Short-range repulsive orces act between gas molecules when they approach each other thereby reducing the eective distance in which they can move reely (and so actually reducing the useul gas volume below the value o V that we considered) . At slightly greater distances the molecules will exert an attractive orce on each other this causes the ormation o small groups
o loosely attached molecules. This in turn reduces the eective number o particles in the gas. With the groups o molecules at the same temperature as the rest o the gas, they have the same translational kinetic energy and momentum as other groups o molecules or individual molecules. The overall eect o intermolecular attraction is to reduce the pressure slightly. The D utch physicist Johannes van der Waals was awarded the 1 91 0 Nobel Physics Prize or his work in developing a real gas equation, which was a modifcation o the ideal gas equation. In this equation two additional constants were included and each o these diered rom gas to gas. In the ensuing years many have tried to ormulate a single simple equation o state which applies to all gases, no one has been successul to date.
TOK Empirical versus theoretical models
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Theoretical models such as the kinetic theory o gas are based on certain assumptions. I these assumptions are not met in the real world, the model may ail to predict a correct outcome as is the case with real gases at high pressure and low temperature.
modifed when objects are moving at speeds approaching that o light. Theoretical models are not inherently better than empirical models, or vice versa. The best model, whether it is theoretical or empirical, is the model that gives the best predictions or a particular set o circumstances.
Despite the best eorts o van der Waals and others, there is no theorectical model that gives the right answer or all gases in all states. This, however, does not mean that we cannot make correct predictions about a particular gas we can use an empirical model that is based on how the gas has behaved in these circumstances previously. We can sensibly expect that i the conditions are repeated then the gas will behave in the same way again. In the 17th century, with no understanding o the nature o gas molecules, a theoretical model o gases was not possible and so Boyle, Charles, and others could only ormulate empirical laws that were repeatable with the precision that their apparatus allowed. In the 21st century we still rely on empirical methods when gases are not behaving ideally; even when the van der Waals modifcation o the equation o state or an ideal gas is used, the constants that need to be included are dierent rom gas to gas and determined experimentally. So even though it is, perhaps, aesthetically pleasing to have a one size fts all equation, the reality is that theoretical models rarely account or all circumstances. Even Newtons laws o motion must be
When a set o experimental results are not in line with theory does this mean that the theory must be abandoned?
In March 2012 the ollowing report appeared on the BBC website: The head o an experiment that appeared to show subatomic particles travelling aster than the speed o light has resigned rom his post .. Earlier in March, a repeat experiment ound that the particles, known as neutrinos, did not exceed light speed. When the results rom the Opera group at the Gran Sasso underground laboratory in Italy were frst published last year, they shocked the world, threatening to up-end a century o physics as well as relativity theory which holds the speed o light to be the Universes absolute speed limit.
The media were very quick to applaud the experiment when the initial results were published, but the scientifc community was less quick to abandon a well-established law. Which o the two groups was showing bias in this case?
QUESTION S
Questions 1
7
Two obj ects are in thermal contact. S tate and explain which o the ollowing quantities will determine the direction o the transer o energy between these obj ects? a) The mass o each obj ect.
S pecifc latent heat o usion o water = 3 .3 4 1 0 5 J kg 1
b) The area o contact between the obj ects. c) The specifc heat capacity o each obj ect. d) The temperature o each obj ect.
2
3
S pecifc latent heat o vaporization o water = 2 .2 6 1 0 6 J kg 1
( 4 marks)
a) C alculate the amount o energy needed to be removed rom the water to reeze it.
Two obj ects are at the same temperature. Explain why they must have the same internal energy. ( 2 marks)
b) C alculate the amount o energy required by the water to vaporize it. c) Explain the dierence between the values calculated in a) and b) . (6 marks)
The internal energy o a piece o copper is increased by heating. 8
a) Explain what is meant, in this context, by internal energy and heating. b) The piece o copper has mass 0. 2 5 kg. The increase in internal energy o the copper is 1 .2 1 0 3 J and its increase in temperature is 2 0 K. Estimate the specifc heat capacity o copper. ( 4 marks)
5
C alculate the amount o energy needed to raise the temperature o 3 . 0 kg o steel rom 2 0 C to 1 2 0 C . The specifc heat capacity o steel is 490 J kg 1 K 1 . ( 2 marks)
Specifc heat capacity o water = 42 00 J kg K
1
Specifc heat capacity o copper = 3 90 J kg 1 K 1 ( 2 marks)
6
A fxed mass o an ideal gas is heated at constant volume. S ketch a graph to show the variation with C elsius temperature t with pressure p o the gas? ( 3 marks)
1 0 Under what conditions does the equation o state or an ideal gas, pV = nRT, apply to a real gas? ( 2 marks)
C alculate the energy supplied to 0. 070 kg o water contained in a copper cup o mass 0.080 kg. The temperature o the water and the cup increases rom 1 7 C to 2 5 C . 1
( IB) A container holds 2 0 g o neon and also 8 g o helium. The molar mass o neon is 2 0 g and that o helium is 4 g. C alculate the ratio o the number o atoms o neon to the number o atoms o helium. ( 2 marks)
9 4
2 .0 kg o water at 0 C is to be changed into ice at this temperature. The same mass o water now at 1 00 C is to be changed into steam at this temperature.
The temperature dierence between the inlet and the outlet o an air- cooled engine is 3 0.0 K. The engine generates 7.0 kW o waste power that the air extracts rom the engine. C alculate the rate o ow o air ( in kg s 1 ) needed to extract this power.
1 1 a) ( i) Explain the dierence between an ideal gas and a real gas. (ii) Explain why the internal energy o an ideal gas comprises o kinetic energy only. b) A fxed mass o an ideal gas has a volume o 870 cm 3 at a pressure o 1 .00 1 0 5 Pa and a temperature o 20.0 C . The gas is heated at constant pressure to a temperature o 21 .0 C . C alculate the change in volume o the gas. ( 6 marks)
Specifc heat capacity o air (at constant pressure) = 1 .01 1 0 3 J kg1 K1 (3 marks)
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T H E R M AL P H YS I C S 1 2 (IB)
S tate the relation between
a) The pressure p o a fxed mass o an ideal gas is directly proportional to the kelvin temperature T o the gas, that is, p T. ( i) State the relation between the pressure p and the volume V or a change at constant temperature. ( ii) State the relation between the volume V and kelvin temperature T or a change at a constant pressure. b) The ideal gas is held in a cylinder by a moveable piston. The pressure o the gas is p , its volume is V and its kelvin 1 1 temperature is T . 1
The pressure, volume and temperature are changed to p , V and T respectively. The 2 2 2 change is brought about as illustrated below.
( i)
p 1 , p 2 , T1 and T
( ii) V1 , V2 , T and T2 c) Use your answers to b) to deduce, that or an ideal gas pV = KT, where K is a constant.
( 6 marks)
1 3 A helium balloon has a volume o 0.2 5 m 3 when it is released at ground level. The temperature is 3 0 C and the pressure 1 .01 1 0 5 Pa. The balloon reaches a height such that its temperature has allen to 1 0 C and its pressure to 0.65 1 0 5 Pa. a) C alculate the new volume o the balloon. b) S tate two assumptions that must be made about the helium in the balloon. c) C alculate the number o moles o helium in the balloon. ( 6 marks)
p 1 , V1 , T1
p 2 , V1 , T'
heated at constant volume to pressure p 2 and temperature T'
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p 2 , V2 , T2
heated at constant pressure to volume V2 and temperature T2
4 O S C I L L AT I O N S A N D WAV E S Introduction All motion is either periodic or non-periodic. In p eriodic motion an obj ect repeats its pattern o motion at fxed intervals o time: it is regular and repeated. Wave motion is also periodic and
there are many similarities between oscillations and waves; in this topic we will consider the common eatures but also see that there are dierences.
4.1 2.1 Oscillations Motion Understanding Simple harmonic oscillations Time period, requency, amplitude,
displacement, and phase diference Conditions or simple harmonic motion
Applications and skills Qualitatively describing the energy changes
taking place during one cycle o an oscillation Sketching and interpreting graphs o simple harmonic motion examples
Equations 1 Period-requency relationship: T = ___ f
Proportionality between acceleration and
displacement: a x
Nature of science Oscillations in nature Naturally occurring oscillations are very common, although they are oten enormously complex. When analysed in detail, using a slow-motion camera, a hummingbird can be seen to ap its wings at a requency o around 20 beats per second as it hovers, drinking nectar. Electrocardiographs are used to monitor heartbeats as hearts pulsate, pushing blood around our bodies at about one per second when we are resting and maybe two or three times this rate as we exert ourselves. Stroboscopes can be used to reeze the motion in engines and motors where periodic motion is essential, but too strong vibrations can be potentially very destructive. The practical techniques that have been developed combined with the mathematical modelling that is used to interpret oscillations
Figure 1 Hummingbird hovering over fower.
are very powerul tools; they can help us to understand and make predictions about many natural phenomena.
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O S C I LL AT I O N S AN D WAVE S
Isochronous oscillations
Figure 2 Pocket watch under strobe light.
A very common type o oscillation is known as isochronous (taking the same time) . These are oscillations that repeat in the same time period, maintaining this constant time property no matter what amplitude changes due to damping occur. It is the isochronicity o oscillations such as that o a simple pendulum that has made the pendulum such an important element o the clock. I we use a stroboscope (or strobe) we can reeze an isochronous oscillation so that it appears to be stationary; the strobe is made to ash at a regular interval, which can be matched to the oscillation. I the requency o the strobe matches that o the oscillating object then, when the object is in a certain position, the strobe ashes and illuminates it. It is then dark while the object completes an oscillation and returns to the same position when the strobe ashes again. In this way the object appears motionless. The lowest requency o the strobe that gives the object the appearance o being static will be the requency o the object. I you now double the strobe requency the object appears to be in two places. Figure 2 shows a swinging pocket watch which is illuminated with strobe light at 4 Hz, so the time interval between the images is 0.25 s. The watch speeds up in the middle and slows down at the edges o the oscillation.
Describing periodic motion The graph in fgure 3 is an electrocardiograph display showing the rhythm o the heart o a healthy 48-year-old male pulsing at 65 beats per minute. The pattern repeats regularly with the repeated pattern being called the cycle o the motion. The time duration o the cycle is called the time period or the period (T) o the motion. Thus, a person with a pulse o 65 beats per minute has an average heartbeat period o 0.92 s.
A
Figure 3 Normal adult male heart rhythm. Let us now compare the pattern o fgure 3 with that o fgure 4, which is a data-logged graph o a loaded spring. The apparatus used or the datalogging is shown in fgure 5 and the techniques discussed in the Investigate! section on p1 1 7. The graph or the loaded spring is a more straightorward example o periodic motion, obtained as the mass suspended on a spring oscillates up and down, above and below its normal rest position. There are just over 1 2 complete oscillations occurring in 2 0 seconds giving a period o approximately 1 .7 s. Although the two graphs are very dierent the period is calculated in the same way. As the mass passes its rest or equilibrium p osition its displacement ( x) is zero. We have seen in Topics 1 and 2 that displacement is a vector
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distance/m
x vs t centre zero 0.25 0.20 0.15 0.10 0.05 0 0.05 0.10 0.15 0.20 0.25
10 seconds
20 seconds
retort stand time
Figure 4 Loaded spring. quantity and, thereore, must have a direction. In this case, since the motion is linear or one dimensional, it is sufcient to speciy the direction as simply being positive or negative. The choice o whether above or below the rest position is positive is arbitrary but, once decided, this must be used consistently. In the graph o fgure 4 the data logger has been triggered to start as the mass passes through the rest position. In this case positive displacements are above the rest position so the mass is moving downwards when timing is started. The maximum value o the displacement is called the amp litude ( x0 ) . In the case o the loaded spring the amplitude is a little smaller than 0.2 5 m. It is much more difcult to measure the amplitude or the heart rhythm ( fgure 3 ) because o determining the rest position additionally with no calibration o the scale it is impossible to give any absolute values. The amplitude is marked as A which is approximately 3 cm on this scale. The frequency ( f ) o an oscillation is the number o oscillations completed per unit time. When the time is in seconds the requency will be in hertz ( Hz) . Scientists and engineers regularly deal with high requencies and so kHz, MHz, and GHz are commonly seen. The vibrations producing sound waves range rom about 2 0 Hz to 2 0 kHz while radio waves range rom about 1 00 kHz to 1 00 MHz. With frequency being the number o oscillations per second and p eriod the time or one oscillation you may have already spotted the relationship between these two quantities: 1 T= _ f
spring
clamp rigid card always more than 25 cm above motion sensor motion sensor mesh guard
Figure 5 Data logger arrangement.
Worked example A childs swing oscillates simple harmonically with a period o 3 .2 s. What is the requency o the swing?
Solution 1 so f = _ 1 T= _ T f 1 = _ = 0. 3 1 Hz 3 .2
Investigate! In this investigation, a motion sensor is used to monitor the position o a mass suspended rom the end o a long spring. The data logger sotware processes the data to produce a graph showing the variation o displacement with time. The apparatus is arranged as shown in fgure 5 with the spring having a period o at least 1 second.
The apparatus is set up as shown above.
The mass is put into oscillation by displacing and releasing it.
The details o the data logger will determine the setting values but data loggers can usually be triggered to start reading at a given displacement value.
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Motion sensors use ultrasound, so reections rom surroundings should be avoided.
The sotware can normally be used to plot graphs o velocity and acceleration ( in addition to displacement) against time.
S imilar investigations can be perormed with other oscillations.
Nature of Science Simple for physicists Simple does not mean easy in the context o physics! In act, mathematically (as discussed in Sub-topic 9.1 ) simple harmonic motion (SHM) is not simple. Simple may be better thought o as simplifed in that we make lie easier or ourselves by either limiting or ignoring some o the orces that act on a body. A simple pendulum consists o a mass suspended rom a string. In dealing with the simple
pendulum the rictional orces acting on it (air resistance and riction between the string and the suspension) and the act that the mass will slightly stretch the string are ignored without having any disastrous eect on the equations produced. It also means that smooth graphs are produced, such as or the loaded spring.
Simple harmonic motion We saw that the graph produced by the mass oscillating on the spring was much less complex than the output o a human heart. The mass is said to be undergoing simple harmonic motion or SHM. In order to perorm SHM an object must have a restoring orce acting on it.
The magnitude o the orce ( and thereore the acceleration) is proportional to the displacement o the body rom a fxed point.
The direction o the orce (and thereore the acceleration) is always towards that fxed point.
Focusing on the loaded spring, the fxed point in the above defnition is the equilibrium position o the mass where it was beore it was pulled down. The orces acting on the mass are the tension in the spring and the pull o gravity (the weight) . In the equilibrium position the tension will equal the weight but above the equilibrium the tension will be less and the weight will pull the mass downward; below the equilibrium position the tension will be greater than the weight and this will tend to pull the mass upwards. The dierence between the tension and the weight provides the restoring orce the one that tends to return the mass to its equilibrium position. We can express the relationship between acceleration a and displacement x as: a x which is equivalent to a = kx This equation makes sense i we think about the loaded spring. When the spring is stretched urther the displacement increases and the tension
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4 . 1 O S C I L L AT I O N S
a
increases. B ecause force = mass acceleration, increasing the orce increases the acceleration. The same thing applies when the mass is raised but, in this case, the tension decreases, meaning that the weight dominates and again the acceleration will increase. So although we cannot prove the proportionality (which we leave or HL) the relationship makes sense. The minus sign is explained by the second bullet point the acceleration is always in the opposite direction to the displacement. S o choosing upwards as positive when the mass is above the equilibrium position, the acceleration will be downwards ( negative) . When the mass is below the equilibrium position ( negative) , the ( net) orce and acceleration are upwards ( positive) . The orce decelerates the mass as it goes up and then accelerates it downwards ater it stopped. Figure 6 is a graph o a against x. We can see that a = kx. takes the orm y = mx + c with the gradient being a negative constant.
0
x 0
Figure 6 Accelerationdisplacement graph.
Graphing SHM From Topic 2 we know that the gradient o a displacementtime graph gives the velocity and the gradient o the velocitytime graph gives the acceleration at any given time. This remains true or any motion. Lets look at the displacement graph or a typical S HM. In fgure 7 the graph is a sine curve ( but i we chose to start measuring the displacement rom any other time it could j ust as easily be a cosine, or a negative sine, or any other sinusoidally shaped graph) . I we want to fnd the velocity at any particular time we simply need to fnd the gradient o the displacementtime graph at that time. With a curved graph we must draw a tangent to the curve (at our chosen time) in order to fnd a gradient. The blue tangent at 0 s has a gradient o + 2.0 cm s 1 , which gives the maximum velocity ( this is the steepest tangent and so we see the velocity is a maximum at this time) . Looking at the gradient at around 1 .6 s or 4.7 s and you will see rom the symmetry that it will give a velocity o 2 .0 cm s 1 , in other words it will be the minimum velocity ( i.e. the biggest negative velocity) . At around 0.8, 2 .4, 3 .9, and 5 .5 s the gradient is zero so the velocity is zero. 2 1.5 1
x/cm
0.5 0 0
0.5
1
2
3
4
5
6
7
t/s
1 1.5 2
Figure 7 The displacement graph for a typical SHM.
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In many situations it is helpful to think about the magnitude of the velocity i. e. speed. For SHM organize a table for an object oscillating between +A and -A. In your table show where the magnitudes of the position, velocity, acceleration and force are (a) zero and (b) maximum. Once you have done that mark in the directions of these these vectors at +A, -A and zero. By doing this you should be thinking physics before algebra.
Using these values you will see that, overall, the velocity is going to change as shown in fgure 8(b) ; it is a cosine graph in this instance. You may remember that the gradient o the velocitytime graph gives the acceleration and so i we repeat what we did or displacement to velocity, we get fgure 8( c) or the acceleration graph. You will notice that this is a reection o the displacementtime graph in the time axis i. e. a negative sine curve. This should be no surprise, since rom our defnition o simple harmonic motion we expect the acceleration to be a negative constant multiplied by the displacement. (a)
1.0 0.5 x/cm
Directions with SHM
0
0
1
2
3
4
5
6
7
4
5
6
7
4
5
6
7
t/s
0.5 1.0 (b)
2.0
v/m s 1
1.0 0
0
1
2
3 t/s
1.0 2.0 (c)
4.0
a/m s 2
2.0 0
0
1
2
3 t/s
2.0 4.0
Figure 8 The variation with time of displacement, velocity, and acceleration.
Worked examples 1
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On a sheet o graph paper sketch two cycles o the displacementtime (x - t) relationship or a simple pendulum. Assume that its displacement is a maximum at time 0 seconds. Mark on the graph a time or which the velocity is maximum (labeled A) , a time or which the velocity is zero (labeled B) and a time or which the acceleration is a maximum (labeled C) .
Solution x
A
B
C t
4 . 1 O S C I L L AT I O N S
Note
2
This is a sketch graph so no units are needed they are arbitrary.
A, B, and C are each labelled on the time axis as this is what the question asks.
There are just two cycles (two complete periods) marked since this is what the question asks.
A is where the gradient o the displacementtime graph is a maximum.
B is where the gradient o the displacement time graph is a zero.
C is where the displacement is a minimum because a = kx this means that the acceleration will be a maximum. The equation defning simple harmonic motion is a = kx.
b) Two similar systems oscillate with simple harmonic motion. The constant or system S 1 is k, while that or system S 2 is 4k. Explain the dierence between the oscillations o the two systems.
Solution a) Rearranging the equation we obtain a k = _ x S ubstituting the units or the quantities gives ms 2 unit o k = ____ m = s 2
S o the units o k are s 2 or per second squared this is the same as requency squared ( and could be written as Hz 2 ) . b) Reerring to the solution to ( a) we can see that 4 or S 2 the square o the requency would be __ times that o S 1 . This means that S 2 has 4 or twice the requency ( or hal the period) o S. That is the dierence.
a) What are the units o the constant k?
Phase and phase diference Reerring back to fgures 8 a, b, and c we can see that there is a big similarity in how the shapes o the displacement, velocity, and acceleration graphs change with time. The three graphs are all sinusoidal they take the same shape as a sine curve. The dierence between them is that the graphs all start at dierent points on the sine curve and continue like this. The graphs are said to have a phase difference. When timing an oscillation it really doesnt matter when we start timing we could choose to start at the extremes o the oscillation or the middle. In doing this the shape o the displacement graph would not change but would look like the velocity graph (quarter o a period later) or acceleration graph (hal a period later) . From this we can see that the phase dierence between the displacementtime graph and the velocitytime graph is equivalent to quarter T o a period or __ (we could say that velocity leads displacement by quarter o 4 a period) . The phase dierence between the displacementtime graph and T the accelerationtime graph is equivalent to hal a period or __ (we could say 2 that acceleration leads displacement by hal a period or that displacement leads acceleration by hal a period it makes no dierence which) . Although we have discussed phase dierence in term o periods here, it is more common to use angles. However, transerring between period and angle is not difcult: Period T is equivalent to 3 60 or 2 radians, T so __ is equivalent to 1 80 or radians 2 T and __ is equivalent to 90 or __ radians. 4 2
When the phase dierence is 0 or T then two systems are said to be oscillating in p hase.
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Worked example 1
b) The period ( 2 . 0 s) is equivalent to 2 radians, so taking ratios:
C alculate the phase dierence between the two displacementtime graphs shown in fgure 9. Give your answers in a) seconds b) radians c) degrees.
0.2 5 2 .0 ___ = ____ ( where is the phase dierence in 2
radians)
2.0
This gives a value or o 0.79 radian.
1.5
c) The period ( 2 .0 s) is equivalent to 3 60, so taking ratios:
1.0 0.5
2 .0 0.25 ___ = ____ ( where is the phase dierence in 360
0.0 0.5
0.0
1.0
2.0 time/s
3.0
4.0
This gives a value or o 45 .
1.0
2.5 2.0
2.5
1.5 displacement/cm
1.5 2.0
2.5 2.0 1.5 displacement/cm
degrees)
5.0
1.0 0.5 0.5
1.0 0.5 time/s
0.0 0.5
0.0
1.0
2.0
3.0
4.0
5.0
1.0 1.5
0.0 0.0
1.0
2.0
3.0
4.0
2.0
5.0
2.5
time/s
0.25 s
1.0 1.5
2.5
2.0
2.0
2.5
1.5
Figure 9
Solution a) It can be seen rom either graph that the period is 2 .0 s. D rawing vertical lines through the peaks o the two curves shows the phase dierence. This appears to be between 0.2 5 0.2 6 s so we will go with ( 0.2 5 0.01 ) s.
displacement/cm
displacement/cm
2.5
1.0 0.5 time/s
0.0 0.5
0.0
1.0
2.0
3.0
4.0
5.0
1.0 1.5 2.0 2.5
Figure 10
Energy changes in SHM Lets think o the motion o the simple pendulum o fgure 1 1 as an example o a system which undergoes simple harmonic motion. For a pendulum to oscillate simple harmonically the string needs to be long and to make small angle swings (less than 1 0) . The diagram is, thereore, not drawn to scale. Lets imagine that the bob just brushes along the ground when it is at its lowest position. When the bob is pulled to position A, it is at its highest point and has a maximum gravitational potential energy
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4 . 2 T R AV E L L I N G W AV E S ( GPE) . As the bob passes through the rest position B , it loses all the GPE and gains a maximum kinetic energy ( KE) . The bob now starts to slow down and move towards position C when it briefy stops, having regained all its GPE. In between A and B and between B and C the bob has a combination o KE and GPE. In a damped system over a long period o time the maximum height o the bob and its maximum speed will gradually decay. The energy gradually transers into the internal energy o the bob and the air around it. D amping will be discussed in O ption B . 4.
O point of suspension string GPE(max) = mgh KE = 0
GPE(max) = mgh KE = 0 bob C h
A h ground
B GPE = 0 KE = maximum
Figure 11 simple pendulum.
4.2 Travelling waves Understanding Travelling waves Wavelength, requency, period, and wave speed Transverse and longitudinal waves The nature o electromagnetic waves The nature o sound waves
Applications and skills Explaining the motion o particles o a medium
when a wave passes through it or both transverse and longitudinal cases Sketching and interpreting displacement distance graphs and displacement time graphs or transverse and longitudinal waves Solving problems involving wave speed, requency, and wavelength Investigating the speed o sound experimentally
Equation The wave equation: c = f
Nature of science Patterns and trends in physics One o the aspects o waves that makes it an interesting topic is that there are so many similarities with just enough twists to make a physicist think. Transverse waves have many similarities to longitudinal waves, but there are equally many dierences. An electromagnetic wave requires no medium through which to travel, but a mechanical
wave such a sound does need a medium to carry it; yet the intensity o each depends upon the square o the amplitude and the two waves use the same wave equation. Physicists enjoy patterns but at the same time they enjoy the places where patterns and trends change the similarities give confdence but the dierences can be thought-provoking.
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Introduction Whe n yo u co nside r wave s yo u p ro b ab ly imme diate ly think o the rip p le s o n the surace o a lake o r the se a. It may b e diicult to think o many e xamp le s o wave s, b u t this to p ic is inte gral to o u r ab ility to co mmu nicate and, since lie o n E arth is de p e nde nt o n the radiatio n that arrive s ro m the S u n, hu mankind co u ld no t e xist witho u t wave s. Waves are o two undamental types:
Figure 1 Ripples spreading out rom a stone thrown into a pond.
mechanical waves, which require a material medium through which to travel
electromagnetic waves, which can travel through a vacuum.
B oth types o wave motion can be treated analytically by equations o the same orm. Wave motion occurs in several branches o physics and an understanding o the general principles underlying their behaviour is very important. Modelling waves can help us to understand the properties o light, radio, sound . . . even aspects o the behaviour o electrons.
Travelling waves Figure 2 shows a slinky spring being used to demonstrate some o the properties o travelling waves.
Figure 2 Slinky spring being used to demonstrate travelling waves.
fxed end
end ree to move
With one end o the stretched spring fxed ( as in fgure 3 ( a) ) , moving the other end sharply upwards will send a pulse along the spring. The pulse can be seen to travel along the spring until it reaches the fxed end; then it reects and returns along the spring. O n reection the pulse changes sides and what was an upward pulse becomes a downwards pulse. The pulse has undergone a phase change o 1 80 or radians on reection. When the end that was fxed is allowed to move there is no phase change on reection as shown in fgure 3 ( b) and the pulse travels back on the same side that it went out.
Note You should discuss with fellow students how Newtons second and third laws of motion apply to the motion of the far end of the spring.
a
b
Figure 3 Reection o wave pulses.
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O bserving the motion o a slinky can give much insight into the movement o travelling waves. To help ocus on the motion o the spring, it is useul i one o the slinky coils is painted to make it stand out rom the others. The coils are being used to model the particles o the medium through which the wave travels they could be air or water molecules. I a table tennis ball is placed beside the spring, it shoots o at right angles to the direction o the wave pulse. This shows that this is a transverse wave pulse the direction o the pulse being perpendicular to the direction in which the pulse travels along the slinky. When the slinky is vibrated continuously rom side to side a transverse wave is sent along the spring as shown in fgure 4 overlea.
4 . 2 T R AV E L L I N G W AV E S
Figure 4 Transverse waves on a slinky spring. From observing the wave on the spring and water waves on a pond we can draw the ollowing conclusions:
A wave is initiated by a vibrating obj ect ( the source) and it travels away rom the obj ect.
The particles o the medium vibrate about their rest position at the same requency as the source.
The wave transers energy rom one place to another.
A slinky can generate a dierent type o wave called a longitudinal wave. In this case the ree end o a slinky must be vibrated back and orth, rather than rom side to side. As a result o this the coils o the spring will vibrate about their rest position and energy will travel along the spring in a direction parallel to that o the springs vibration as shown in fgure 5 .
rarefaction
compression
Figure 5 Longitudinal waves on a slinky spring.
Describing waves When we describe wave properties we need specialist vocabulary to help us:
Wavelength is the shortest distance between two points that are in phase on a wave, i. e. two consecutive crests or two consecutive troughs.
Frequency f is the number o vibrations per second perormed by the source o the waves and so is equivalent to the number o crests passing a fxed point per second.
Period T is the time that it takes or one complete wavelength to pass a fxed point or or a particle to undergo one complete oscillation.
Amp litude A is the maximum displacement o a wave rom its rest position.
These defnitions must be learned, but it is oten easier to describe waves graphically. There are two types o graph that are generally used when describing waves: displacementdistance and displacementtime graphs. S uch graphs are applicable to every type o wave.
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4
O S C I LL AT I O N S AN D WAVE S For example displacement could represent:
the displacement o the water surace rom its normal at position or water waves
the displacement o air molecules rom their normal position or sound waves
the value o the electric feld strength vector or electromagnetic waves.
Displacementdistance graphs This type o graph is sometimes called a wave profle and represents the displacement o many wave particles at a particular instant o time. O n fgure 6 the two axes are perpendicular but, in reality, this is only the case when we describe transverse waves ( when the graph looks like a photograph o the wave taken at a particular instant) . For longitudinal waves the actual displacement and distance are parallel. Figure 7 shows how the particles o the medium are displaced rom the equilibrium position when the longitudinal wave travels through the medium orming a series o compressions (where the particles are more bunched up than normal) and rarefactions (where the particles are more spread out than normal) . In this case the particles that are displaced to the let o their equilibrium position are given a negative displacement, while those to the right are allocated a positive displacement.
displacement /cm 0.5
wavelength amplitude
0 2
4
8 dist ance/cm
6
0.5 trough
crest
trough
crest
Figure 6 Transverse wave profle. 1 cm equilibrium position with no wave position o particles with wave displacement /cm
t= 0
0.5
wavelength amplitude
0 2
4
6
centre o rareaction
centre o compression
centre o rareaction
8 dist ance/cm
0.5 centre o compression
Figure 7 Longitudinal wave profle.
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centre o compression
4 . 2 T R AV E L L I N G W AV E S
It is very easy to read the amplitude and wavelength directly rom a displacementtime graph. For the longitudinal wave the wavelength is both the crest- to-crest distance and the distance between two consecutive compressions or rareactions. Using a sequence o displacementdistance graphs can provide a good understanding o how the position o an individual particle changes with time in both transverse and longitudinal waves ( fgure 8) . O n this T diagram the wave profle is shown at a quarter period ( __ intervals; 4) diagrams like this are very useul in spotting the phase dierence between the particles. P and Q are in anti- phase here (i.e. 1 80 or radians. You may wish radians out o phase) and Q leading R by 90 or __ 2 to consider the phase dierence between P and R. displacement/cm 0.5 t= 0
Q
P
0 2
4
6
8 distance/cm
6
8 distance/cm
6
8 distance/cm
6
8 distance/cm
6
8 distance/cm
R
0.5 displacement/cm 0.5 t=
T 4
Q R
0 2
4 P
0.5 displacement/cm
R
0.5 t=
T 2
P
0
2
Q
4
0.5 displacement/cm 0.5 P
t=
3T 4
0 2
4 Q
0.5
R
displacement/cm 0.5 t= T
2 0.5
Q
P
0
4 R
Figure 8 Sequence of displacement-distance graphs at ___4T time intervals.
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Displacementtime graphs A displacementtime graph describes the displacement of one particle at a certain position during a continuous range of times. Figure 9 shows the variation with time of the displacement of a single particle. Each particle along the wave will undergo this change ( although with phase difference between individual particles) . displacement/cm 0.5
Note It is common to confuse displacementdistance graphs (where crest-to-crest gives the wavelength) with displacementtime graphs (where crest-to-crest gives the period) .
0 T 2
T
2T time/s
3T 2
0.5
Figure 9 Displacementtime graphs. This graph makes it is very easy to spot the period and the amplitude of the wave.
The wave equation When a source of a wave undergoes one complete oscillation the wave it produces moves forward by one wavelength ( ) . S ince there are f oscillations per second, the wave progresses by f during this time and, therefore, the velocity ( c) of the wave is given by c = f. With f in hertz and in metres, c will have units of hertz metres or ( more usually) metres per second. You do need to learn this derivation and it is probably the easiest that you will come across in IB Physics.
Worked example
directions of wave travel
The diagram below represents the direction of oscillation of a disturbance that gives rise to a wave. M
a) Draw two copies of the diagram and add arrows to show the direction of wave energy transfer to illustrate the difference between (i) a transverse wave and (ii) a longitudinal wave. b) A wave travels along a stretched string. The diagram to the right shows the variation with distance along the string of the displacement of the string at a particular instant in time. A small marker is attached to the string at the point labelled M. The undisturbed position of the string is shown as a dotted line.
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O n a copy of the diagram: ( i) D raw an arrow to indicate the direction in which the marker is moving. ( ii) Indicate, with the letter A, the amplitude of the wave.
4 . 2 T R AV E L L I N G W AV E S
( iii) Indicate, with the letter , the wavelength o the wave.
b)
direction of wave
crest
( iv) D raw the displacement o the string a time T/4 later, where T is the period o oscillation o the wave.
A M
( v) Indicate, with the letter N, the new position o the marker.
A
c) The wavelength o the wave is 5 .0 cm and its speed is 1 0 cm s 1 . N
D etermine: ( i) the requency o the wave
( i) With the wave travelling to the let, the trough shown will move to the let and that must mean that M moves downwards.
( ii) how ar the wave has moved in a quarter o a period.
Solution
( ii) The amplitude ( A) is the height o a crest or depth o a trough.
a) ( i) The energy in a transverse wave travels in a direction perpendicular to the direction o vibration o the medium:
( iii) The wavelength () is the distance equivalent to crest to next crest or trough to next trough. ( iv) In quarter o a period the wave will have moved quarter o a wavelength to the let ( broken line curve) .
direction of energy direction of vibration
( ii) The energy in a longitudinal wave travels in a direction parallel to the direction o vibration o the medium:
trough
( v) Ater quarter o a period the marker is now at the position o the trough ( N) . c)
direction of energy
c 10 ( i) f = __ = __ = 2 . 0 Hz ( since both 5 wavelength and speed are in cm, there is no need to convert the units) 1 ( ii) T = __ = 0.5 s f
direction of vibration
B ecause the wave moves at a constant speed T s = ct = c _ = 1 0 0.1 2 5 = 1 .2 5 cm 4
Investigate! Measuring the speed of sound Here are two o the many ways o measuring the speed o sound in ree air ( i.e. not trapped in a tube) .
Method one using a fast timer
The frst microphone triggers the timer to start.
The second microphone triggers the time to stop.
When the hammer is made to strike the plate the sound wave travels to the two microphones triggering the nearer microphone frst and the urther microphone second.
B y separating the microphones by 1 m, the time delay is around 3 .2 ms.
This is a very simple method o measuring the speed o sound:
Two microphones are connected to a ast timer ( one which can measure the nearest millisecond or even microsecond) .
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2nd microphone
fast timer 1st microphone
hammer
metre ruler plate
Figure 10 Fast timer method for the speed of sound. This gives a value or the speed o sound to be 1 .0 c = _st = _______ 3 1 0 m s 1 3 .2 1 0
Connect a signal generator to a loudspeaker and set the requency to between 500 Hz and 2.0 kHz.
This should be repeated a ew times and an average value obtained.
You might think about how you could develop this experiment to measure the speed o sound in dierent media. S ound travels aster in solids and liquids than in gases can you think why this should be the case ? I the temperature in your country varie s signifcantly over the year you might try to perorm the experiment in a hot or cold corridor and compare your results.
O ne o the microphones needs to be close to the loudspeaker, with the second a metre or so urther away.
C ompare the two traces as you move the second microphone back and orth in line with the frst microphone and the speaker.
Use a ruler to measure the distance that you need to move the second microphone or the traces to change rom being in phase to changing to antiphase and then back in phase again.
The distance moved between the microphones being in consecutive phases will be the wavelength o the wave.
The speed is then ound by multiplying the wavelength by the requency shown on the signal generator.
3
Method two using a double beam oscilloscope
C onnect two microphones to the inputs o a double-beam oscilloscope ( fgure 1 1 ) .
oscilloscope signal generator
loudspeaker microphones
upper trace shifted horizontally to align with lower trace
Figure 11 Double beam oscilloscope method for the speed of sound.
Nature of Science Analogies can slow down progress You may have seen a demonstration, called the bell j ar experiment, to show that sound needs a material medium through which to travel.
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An electric bell is suspended rom the mouth o a sealed bell jar and set ringing (fgure 1 2 ) . A vacuum pump is used to evacuate the jar. When
4 . 2 T R AV E L L I N G W AV E S
there is no longer any air present in the jar, no sound can be heard. This experiment was frst perormed by Robert B oyle in 1 660 although, long beore this time, the Ancient Greeks understood that sound needed something to travel through. The frst measurements o the speed o sound were made over our hundred years ago. These experiments were based on measuring the time delay either between a sound being produced and its echo reecting rom a distant surace or between seeing the ash o a cannon when fred and hearing the bang. From the middle to the end o the seventeenth century physicists such as Robert Hooke and C hristiaan Huygens proposed a wave theory o light. In line with the model or sound, Huygens suggested that light was carried by
a medium called the luminierous ether. It was not until 1 887, when Michelson and Morley devised an experiment in an attempt to detect the ether wind, that people began to realize that electromagnetic waves were very dierent rom sound waves.
electric bell
to vacuum pump
Figure 12 The bell jar experiment.
Electromagnetic waves You may have seen the experiment shown in fgure 1 3 in which a beam o white light is dispersed into the colours o the visible spectrum by passing it through a prism. The act that what appears to be a single colour actually consists o multiple colours triggers the question what happens beyond the red and blue ends o the spectrum? The two colours represent the limit o vision o the human eye but not the limit o detection o electromagnetic radiation by the human body. Holding the back o your hand towards the S un allows you to eel the warmth o the S uns inra- red radiation. In the longer term, your hand will be subj ected to sunburn and even skin cancer caused by the higher energy ultraviolet radiation. Figure 1 4 shows the ull electromagnetic spectrum with the atmospheric windows; these are the ranges o electromagnetic waves that can pass through the layers o the atmosphere. All electromagnetic waves are transverse, carry energy, and exhibit the ull range o wave properties. They travel at 3 00 million metres per second ( 3 .00 1 0 8 m s 1 ) in a vacuum. All electromagnetic waves ( except gamma rays) are produced when electrons undergo an energy change, even though the mechanisms might dier. For example, radio waves are emitted when electrons are accelerated in an aerial or antenna. Gamma rays are dierent, they are emitted by a nucleus or by means o other particle decays or annihilation events. All electromagnetic waves consist o a time-varying electric feld with an associated time-varying magnetic feld. As the human eye is sensitive to the electric component, the amplitude o an electromagnetic wave is usually taken as the waves maximum electric feld strength. B y graphing the electric feld strength on the y-axes, we can use displacementdistance and displacementtime graphs to represent electromagnetic waves. In the ollowing discussion o the dierent areas o the electromagnetic spectrum the range o the wavelength is given but these values are not hard and ast. There are overlaps o wavelength when radiation o the same wavelength is emitted by dierent
Figure 13 Dispersion of white light using a glass prism.
Figure 14 The electromagnetic spectrum.
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O S C I LL AT I O N S AN D WAVE S mechanisms ( most notably X- rays and gamma rays) . In addition to comparing the wavelengths o the radiations it is sometimes more appropriate to compare their requencies ( or example, radio waves) or the energy o the wave photon ( or example, with X- rays and gamma rays) . The requency range is easy to calculate rom wavelengths using the equation c = f with c = 3 . 00 1 0 8 m s 1 . Photon energies will be discussed in Topic 7. Those electromagnetic waves with requencies higher than that o visible light ionize atoms and are thus harmul to people. Those with lower requencies are generally believed to be sae.
Nature of Science Infra-red radiation ( ir~1 1000 m) William Herschel discovered inra-red in 1 800 by placing a thermometer j ust beyond the red end o the spectrum ormed by a prism. Today we might do the same experiment but using an inra-red detector as a modern thermometer. Obj ects that are hot but not glowing, i.e. below 5 00 C , emit inra-red only. At this temperature obj ects become red-hot and emit red light in addition to inra-red. At around 1 000 C objects become white hot and emit the ull visible spectrum colours. Remote controllers or multimedia devices utilize inra-red as do thermal imagers used or night vision. Inrared astronomy is used to see through dense regions o gas and dust in space with less scattering and absorption than is exhibited by visible light.
Ultraviolet radiation ( uv ~ 100 400 nm) In 1 801 , Johann Ritter detected ultraviolet by positioning a photographic plate beyond the violet part o the spectrum ormed by a prism. It can also be detected using uorescent paints and inks these absorb the ultraviolet ( and shorter wavelengths) and re- emit the radiation as visible light. Absorption o ultraviolet produces important vitamins in the skin but an overdose can be harmul, especially to the eyes. The Sun emits the ull range o ultraviolet: UV-A, UVB , and UV- C . These classifcations are made in terms o the range o the wavelengths emitted ( UV- A ~3 1 5 400 nm, UV-B ~2 803 1 5 nm, and UV- C ~ 1 002 80 nm) . UV- C rays, having the highest requency, are the most harmul but, ortunately, they are almost completely absorbed by the atmosphere. UV- B rays cause sunburn and increase the risk o D NA and other cellular damage in living organisms; luckily only about 5 % o this radiation passes through the ionosphere ( this consists o layers o electrically
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charged gases between 80 and 400 km above the Earth) . Fluorescent tubes, used or lighting, contain mercury vapour and their inner suraces are coated with powders. The mercury vapour and powders uoresce when radiated with ultraviolet light. With the atmosphere absorbing much o the ultraviolet spectrum, satellites must be positioned above the atmosphere in order to utilize ultraviolet astronomy, which is very useul in observing the structure and evolution o galaxies.
Radio waves ( radio ~1 mm 100 km) James C lerk Maxwell predicted the existence o radio waves in 1 864. B etween 1 885 and 1 889 Heinrich Hertz produced electromagnetic waves in the laboratory, confrming that light waves are electromagnetic radiation obeying Maxwells equations. Radio waves are used to transmit radio and television signals. VHF ( or FM) radio waves have shorter wavelengths than those o AM. VHF and television signals have wavelengths o a ew metres and travel in straight lines ( or rays) rom the transmitter to the receiver. Long- wave radio relies on reections rom the ionosphere and diracts around obstacles on the Earths surace. S atellite communication requires signals with wavelengths less than 1 0 m in order to penetrate the ionosphere. Radio telescopes are used by astronomers to observe the composition, structure and motion o astronomic bodies. S uch telescopes are physically large, or example the Very Large Array ( VLA) radio telescope in New Mexico which consists o 2 7 antennas arranged in a Y pattern up to 3 6 km across.
Microwaves ( micro ~ 1 mm 30 cm) Microwaves are short wavelength radio waves that have been used so extensively with radar, microwave cooking, global navigation satellite
4 . 2 T R AV E L L I N G W AV E S
systems and astronomy that they deserve their own category. In 1 940, S ir John Randall and D r H A B oot invented the magnetron which produced microwaves that could be used in radar ( an acronym or radio detection and ranging) to locate aircrat on bombing missions. The microwave oven is now a common kitchen appliance; the waves are tuned to requencies that can be absorbed by the water and at molecules in ood, causing these molecules to vibrate. This increases the internal energy o the ood the container holding the ood absorbs an insignifcant amount o energy and stays much cooler. Longer wavelength microwaves pass through the Earths atmosphere more eectively than those o shorter wavelength. The C osmic B ackground Radiation is the elemental radiation feld that flls the universe, having been created in the orm o gamma rays at the time o the B ig B ang. With the universe now cooled to a temperature o 2 .73 K the peak wavelength is approximately 1 .1 mm ( in the microwave region o the spectrum) .
X-rays ( X~30 pm 3 nm) X- rays were frst produced and detected in 1 895 by the German physicist Wilhelm C onrad Roentgen. An X- ray tube works by fring a beam o electrons at a metal target. I the electrons
have sufcient energy, X- rays will emitted by the target. X-rays are well known or obtaining images o broken bones. They are also used in hospitals to destroy cancer cells. S ince they can also damage healthy cells, using lead shielding in an X- ray tube is imperative. Less energetic and less invasive X-rays have longer wavelengths and penetrate esh but not bone: such X- rays are used in dental surgery. In industry they are used to examine welded metal j oints and castings or aults. X- rays are emitted by astronomical obj ects having temperatures o millions o kelvin, including pulsars, galactic supernovae remnants, and the accretion disk o black holes. The measurement o X- rays can provide inormation about the composition, temperature, and density o distant galaxies.
Gamma rays ( < 1 pm) Gamma rays were discovered by the French scientist Paul Villard in 1 900. Gamma rays have the shortest wavelength and the highest requency o all electromagnetic radiation. They are generated, amongst other mechanisms, by naturally occurring radioactive nuclei in nuclear explosions and by neutron stars and pulsars. Only extra-terrestrial gamma rays o the very highest energies can reach the surace o the Earth the rest being absorbed by ozone in the Earths upper atmosphere.
Nature of Science Night vision Humans eyes are sensitive to the electric component o a portion o the electromagnetic spectrum known as the visible spectrum; this is what we call sight. S ome animals, in particular insects and birds, are able to see using the ultraviolet part o spectrum. However, ultraviolet consists o relatively short wavelength radiation that can damage animal tissue, yet these animals appear to be immune to these dangers. It has been speculated whether any animal eye might be adapted to be able to use the inra- red part
o the electromagnetic spectrum. As these long wavelengths have low energy it might be aretched to believe that they can be detected visually. There are animals that have evolved ways o sensing inra-red that are similar to the processes occurring in the eye. For example, the brains o some snakes are able to interpret the inra-red radiation in a way that can be combined with other sensory inormation to enable them to have a better understanding o surrounding danger or ood sources.
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4.3 Wave characteristics Understanding
Applications and skills
Waveronts and rays
Sketching and interpreting diagrams involving
Amplitude and intensity Superposition
Polarization
waveronts and rays Solving problems involving amplitude and intensity Sketching and interpreting the superposition o pulses and waves Describing methods o polarization Sketching and interpreting diagrams illustrating polarized refected and transmitted beams Solving problems involving Maluss law
Equations Relationship between intensity and amplitude:
I A2 Maluss law: I = I0 cos 2
Nature of science Imagination and physics Imagination ... is more important than knowledge. Knowledge is limited. Imagination encircles the world 1 . Einstein was amous or his gedanken thought experiments and one o the qualities that makes a great physicist is surely a hunger to ask the question what i ... ?. Mathematics is a crucial tool or the physicist and it is central to what a physicist does, to be able to quantiy an argument. However,
imagination can also play a major role in interpreting the results. Without imagination it is hard to believe that we would have Huygens principle, Newtons law o gravitation, or Einsteins theory o special relativity. To visualize the abstract and apply theory to a practical situation is a fair that is undamental to being an extraordinary physicist. Viereck, George Sylvester (October 26, 1929) . What lie means to Einstein: an interview. The Saturday Evening Post.
1
Introduction Wavefronts and rays are visualizations that help our understanding of how waves behave in different circumstances. B y drawing ray or wave diagrams using simple rules we can predict how waves will behave when they encounter obstacles or a different material medium.
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4 . 3 W AV E C H A R A C T E R I S T I C S
D efnitions o these two quantities are:
A wavefront is a surace that travels with a wave and is perpendicular to the direction in which the wave travels the ray.
A ray is a line showing the direction in which a wave transers energy and is, o course, perpendicular to a waveront.
The distance between two consecutive waveronts is one wavelength ( ) .
The motion of wavefronts O ne o the simplest ways to demonstrate the motion o waveronts is to use a ripple tank such as that shown in fgure 1 . This is simply a glass- bottomed tank that contains water, illuminated rom above. Any waves on the water ocus the light onto a screen oten placed below the tank. The bright patches result rom the crests ocusing the light and the dark patches rom the troughs deocusing the light. Using a vibrating dipper, plane or circular waves can be produced allowing us to see what happens to waveronts in dierent situations.
lamp
to power supply
elastic bands
vibrator water surface
dipper shallow water tray
wave pattern on screen
white screen
Figure 1 A ripple tank. The images shown in fgure 2 show the eect on the waveronts as they meet ( a) a plane barrier, ( b) a shallower region over a prism-shaped glass and ( c) a single narrow slit. Using waveront or ray diagrams we can illustrate how waves behave when they are reected, reracted, and diracted. We will reer to these diagrams in the sections o this topic ollowing on rom this.
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O S C I LL AT I O N S AN D WAVE S (a)
(b)
(c)
Figure 2 Reection, reraction, and difraction o waves in a ripple tank. incident wave ront incident ray
reected wave ront reected ray
i r
metal strip placed in water
Figure 3a Waveront and ray diagrams or reection o waves.
deep water
shallow water
Figure 3b Waveront and ray diagrams or reraction o waves.
Figure 3c Waveront and ray diagrams or difraction o waves by a single slit.
In fgure 3 a we can see that there is no change o wavelength and that the angle o incidence ( i) is equal to the angle o reection ( r) . In fgure 3 b we see the wave slowing down and bending as it enters the denser medium. The wavelength o the wave in the denser medium is shorter than in the less dense medium but the requency remains unchanged ( although it is not possible to tell this rom the the pattern shown in the ripple tank) . Figure 3 c shows diraction where the wave spreads out on passing through the slit but there is no change in the wavelength.
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4 . 3 W AV E C H A R A C T E R I S T I C S
Nature of Science Huygens principle O ne o the ways to predict what will happen to waveronts under dierent circumstances is to use Huygens principle. This was suggested in 1 6 7 8 by the D utch physicist C hristiaan Huygens: the waveront o a travelling wave at a particular instant consists o the tangent to circular wavelets given out by each point on the previous waveront as shown in fgure 4. In this way the waveront travels orward with a velocity c. Huygens was able to derive the laws o reection and reraction rom his principle. What the principle does not explain is why an expanding circular ( but really spherical) wave continues to expand outwards rom its source rather than travel back and ocus on the source. The French physicist Augustin Fresnel ( 1 7881 82 7) adapted Huygens principle to explain diraction by proposing the principle o superposition, discussed in S ub-topic 4.4. The HuygensFresnel principle
( as the overall principle should more accurately be called) is useul in explaining many wave phenomena.
ct secondary wavelets
ct
primary source
secondary sources
plane wavefronts
spherical wavefronts
Figure 4 Huygens principle.
The intensity of waves The loudness o a sound wave or the brightness o a light depends on the amount o energy that is received by an observer. For example, when a guitar string is plucked more orceully the string does more work on the air and so there will be more energy in the sound wave. In a similar way, to make a flament lamp glow more brightly requires more electrical energy. The energy E is ound to be proportional to the square o the amplitude A: E A2 S o doubling the amplitude increases the energy by a actor o our; tripling the amplitude increases the energy by a actor o nine, etc. Loudness is the observers perception o the intensity o a sound and brightness that o light; loudness and brightness are each aected by requency.
I = P/4r2 r
I we picture waves being emitted by a point source, S , they will spread out in all directions. This will mean that the total energy emitted will be spread increasingly thinly the urther we go rom the source. Figure 5 shows how the energy spreads out over the surace area o a sphere. In order to make intensity comparisons more straightorward it is usual to use the idea o the energy transerred per second this is the power ( P) o the source. This means that the intensity ( I) at a distance ( r)
S
Figure 5 Energy spreading out from a point source.
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O S C I LL AT I O N S AN D WAVE S I /9 I /4 I
r 2r 3r
Figure 6 Inverse square law.
rom a point source is given by the power divided by the surace area o the sphere at that radius: P I = _2 4r This equation shows that intensity has an inverse-square relationship with distance rom the point source; this means that, as the distance doubles, the intensity alls to a quarter o the previous value and when the distance the is tripled, the intensity alls to one nineth as shown in fgure 6. The S I unit or intensity is W m 2 .
Worked example At a distance o 1 5 m rom the source, the intensity o a loud sound is 2 .0 1 0 4 W m 2 . a) Show that the intensity at 1 2 0 m rom the source is approximately 3 1 0 6 W m 2 . b) D educe how the amplitude o the wave changes.
Solution P P a) Using the equation I = ___ , __ remains constant so I1 r12 = I2 r22 this 4 4 r 2
gives 2 .0 1 0 4 1 5 2 = I 2 1 2 0 2 1 5 2 = 3 .1 1 0 6 W m 2 I2 = 2.0 1 0 4 _ 2 1 20
3 10
6
Wm
2
Note In show that questions there is an expectation that you will give a detailed answer, showing all your working and that you will give a fnal answer to more signifcant fgures than the data in the question actually this is good practice or any answer! b) With the intensity changing there must be a change o amplitude. The intensity is proportional to the square o the amplitude so: __ _________ 2 I A I A 2 .0 1 0 4 _2 = _1 or _1 = _2 = _ = 8.0 2 I2 A2 I2 3 . 1 1 0 6 A2
Thus the amplitude at 1 5 m rom the source is 8.0 times that at 1 2 0 m rom the source. Another way o looking at this is to say 1 that A is proportional to __ r.
Note Because the previous part was a show that question you had all the data needed to answer this question. In questions where you need to calculate data you will never be penalized or using incorrect data that you have calculated previously. In a question that has several parts you might ail to gain a sensible answer to one o the parts. When a subsequent part requires the use o your answer as data dont give up, invent a sensible value (and say that is what you are doing) . You should then gain any marks available or the correct method.
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The principle of superposition Unlike when solid obj ects collide, when two or more waves meet the total displacement is the vector sum o their individual displacements. Having interacted, the waves continue on their way as i they had never met at all. This principle is used to explain intererence and standing waves in S ub- topics 4.4 and 4.5 respectively. For a consistent pattern, waves need to be o the same type and have the same requency and speed; the best patterns are achieved when the waves have the same or very similar amplitudes. Figure 7 shows two pulses approaching and passing through each other. When they meet, the resultant amplitude is the algebraic sum o the two amplitudes o the individual pulses.
Figure 7 Superposition of pulses. This principle applies equally well to complete waves as to pulses this is shown in fgure 8. You should convince yoursel that the green wave is the vector sum o the red and blue waves at every instant. It is equally valid to use the principle o superposition with displacementdistance graphs. 0.3
displacement/m
0.2 0.1 0 0
0.1
0.2
0.3
0.4
0.5
0.1 time/s 0.2 0.3
Figure 8 Displacementtime graph showing the superposition of waves.
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Worked examples 1
Two identical triangular pulses of amplitude A travel towards each other along a rubber cord. At the instant shown on the diagram below, point M is midway between the two pulses.
A
C
b) ( i)
E 1 F medium R
medium I
A 2
M
A
What is the amplitude of the disturbance in the string as the pulses move through M?
B
O n entering the new medium the waves refract by the same amount to give parallel wavefronts.
Solution The pulses are symmetrical, so when they meet they completely cancel out giving zero amplitude at M. 2
a)
( ii) As the wavefronts are closer in the medium R the waves are travelling more slowly.
For a travelling wave, distinguish between a ray and a wavefront.
(iii) The frequency of a wave does not change when the wave moves from one medium to another. As c = f c then __ = constant and so:
The diagram below shows three wavefronts incident on a boundary between medium I and medium R. Wavefront C D is shown crossing the boundary. Wavefront EF is incomplete.
c1 c2 1 c1 __ __ = __ __ c = 1
3 A
C
2
2
2
The graphs below show the variation with time of the individual displacements of two waves as they pass through the same point.
E F
medium I
D
medium R displacement
A1 x1
0
0
time
T
D B A 1
( i) O n a copy of the diagram above, draw a line to complete the wavefront EF. ( ii) E xplain in which medium, I or R, the wave has the higher speed. ( iii) B y taking appropriate measurements from the diagram, determine the ratio of the speeds of the wave travelling from medium I to medium R.
Solution a) A ray is a line that shows the direction of propagation of a wave. Wavefronts are lines connecting points on the wave that are in phase, such as a crest or a trough. The distance between wavefronts is one wavelength and wavefronts are always perpendicular to rays.
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A2 displacement
b)
0 x2 0
T
time
A 2
What is the total displacement of the resultant wave at the point at time T?
Solution The total displacement will be the vector sum of the individual displacements. As they are in opposite directions the vector sum will be their difference, i.e. x1 x2
4 . 3 W AV E C H A R A C T E R I S T I C S
Polarization Although transverse and longitudinal waves have common properties they reect, reract, diract and superpose the dierence between them can be seen by the property o polarization. Polarization o a transverse wave restricts the direction o oscillation to a plane perpendicular to the direction o propagation. Longitudinal waves, such as sound waves, do not exhibit polarization because, or these waves, the direction o oscillation is parallel to the direction o propagation. Figure 9 shows a demonstration o the polarization o a transverse wave on a rubber tube. vibrations go through
same plane as ence
ence with vertical rails acting as slits
vibrations stopped
ence with horizontal rails acting as slits
diferent plane
vibrations stopped at second ence
two planes at right angles
Figure 9 Polarization demonstration. Most naturally occurring electromagnetic waves are completely unpolarized; this means the electric feld vector (and thereore the magnetic feld vector perpendicular to it) vibrate in random directions but in a plane always at right angles to the direction o propagation o the wave. When the direction o vibration stays constant over time, the wave is said to be plane polarized in the direction o vibration this is the case with many radio waves which are polarized as a result o the orientation o the transmitting aerial (antenna) . Partial polarization is when there is some restriction to direction o vibration but not 1 00%. There is a urther type o polarization when the direction o vibration rotates at the same
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unpolarized incident beam
p
plane polarized refected beam
p
90 2
requency as the wave this called circular or elliptical polarization and is caused when a wave is in a strong magnetic feld but you will not be examined on this. Figure 1 0 shows how we represent polarized and unpolarized light diagramatically; the double-headed arrow represents a polarized wave, showing the plane o polarization o the wave. The crossed arrows show that the vibration has an electric feld vector in all planes and these are resolved into the two perpendicular planes shown (you may see this marked as our double-headed arrows in some texts) . These diagrams can become a little conusing when rays are added to show the direction in which the waves are travelling. Figure 1 1 shows some examples o this.
or unpolarized light
partially polarized reracted beam
Figure 11 Polarization o refected light.
polarized light
Figure 10 Representing polarized and unpolarized waves.
Polarization of light In 1 809 the French experimenter tienne-Louis Malus showed that when unpolarized light reected o a glass plate it could be polarized depending upon the angle o incidence the plane o polarization being that o the at surace reecting the light. In 1 81 2 the Scottish physicist S ir D avid B rewster showed that when unpolarized light incident on the surace o an optically denser material ( such as glass) , at an angle called the polarizing angle, the reected ray would be completely plane polarized. At this angle the reected ray and reracted ray are at right angles as shown in fgure 1 1 . Today the most common method o producing polarized light is to use a polarizing flter ( usually called Polaroid) . These flters are made rom chains o microcrystals o iodoquinine sulate embedded in a transparent cellulose nitrate flm. The crystals are aligned during manuacture and electric feld vibration components, parallel to the direction o alignment, become absorbed. The electric feld vector causes the electrons in the crystal chains to oscillate and thus removes energy rom the wave. The direction perpendicular to the chains allows the electric feld to pass through. The reason or this is that the limited width o the molecules restricts the motion o the electrons, meaning that they cannot absorb the wave energy. When a pair o Polaroids are oriented to be at 90 to each other, or crossed, no light is able to pass through. The frst Polaroid restricts the electric feld to the direction perpendicular to the crystal chains; the second Polaroid has its crystals aligned in this direction and so absorbs the remaining energy. The frst o the two Polaroids is called the p olarizer and the second is called the analyser.
Maluss law When totally plane- polarized light ( rom a polarizer) is incident on an analyser, the intensity I o the light transmitted by the analyser is directly proportional to the square o the cosine o angle between the transmission axes o the analyser and the polarizer.
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4 . 3 W AV E C H A R A C T E R I S T I C S
Figure 1 2 shows polarized light with the electric feld vector o amplitude E0 incident on an analyser. The axis o transmission o the analyser makes an angle with the incident light. The electric feld vector E0 can be resolved into two perpendicular components E0 cos and E0 sin . The analyser transmits the component that is parallel to its transmission axis, which is E0 cos . We have seen that intensity is proportional to the square o the amplitude o a wave so I0 E02 the transmitted intensity will be I which is proportional to ( E0 cos ) 2 or I E02 cos 2 2
2
E co s I = _______ cancelling E 0 2 gives Taking ratios we have __ I E
orientation of analyser E0
E 0 cos
orientation of light from polarizer
0
2
0
0
2
I = I0 cos
Figure 12 Analysing polarized light.
When = 0 ( or 1 80) I = I0 ( since cos 0 = 1 ) ; this means that the intensity o light transmitted by the analyser is maximum when the transmission axes o the two Polaroids are parallel. When = 90, I = I0 cos 2 90 = 0; this means that no light is transmitted by the analyser when the Polaroids are crossed. Figure 1 3 shows a pair o Polaroids. The let-hand image shows that, when their transmission axes are aligned, the same proportion o light passing through one Polaroid passes through both. The central image shows that when one o the Polaroids is rotated slightly, less light passes through their region o overlap. The right- hand image shows that where the Polaroids are crossed no light is transmitted.
Figure 13 Two Polaroids.
Nature of science Uses of polarization of light Polaroid sunglasses are used to reduce the glare coming rom the light scattered by suraces such as the sea or a swimming pool. In industry, stress analysis can be perormed on models made o transparent plastic by placing the model in between a pair o crossed Polaroids. As white light passes through a plastic, each colour o the spectrum is polarized with a unique orientation. There is high stress in the regions where the colours are most concentrated this is where the model ( or real obj ect being modelled) is most likely to break when it is put under stress. C ertain asymmetric molecules ( called chiral molecules) are op tically active this is the ability to rotate the plane o plane-polarized light. The angle that the light is rotated through is measured using a p olarimeter, which consists o a light source, and a pair o Polaroids. The light passes through the frst Polaroid ( the polarizer) and, initially with no sample present, the second Polaroid ( the analyser) is aligned so that no light passes through. With a sample placed between
the Polaroids, light does pass through because the sample has rotated the plane o polarization. The analyser is now rotated so that again no light passes through. The angle o rotation is measured and rom this the concentration o a solution, or example, can be ound. An easily produced, optically active substance is a sugar solution. Polarization can also be used in the recording and proj ection o 3 D flms. These consist o two flms proj ected at the same time. Each o the flms is recorded rom a slightly dierent camera angle and the proj ectors are also set up in this way. The two flms are proj ected through polarizing flters one with its axis o transmission horizontal and one with it vertical. B y wearing polarized eye glasses with one lens horizontally polarized and one vertically polarized, the viewers let eye only sees the light rom the let proj ector and the right eye light rom the right proj ector thus giving the viewer the perception o depth. In cinemas the screen needs to be metallic as non- metallic at suraces have a polarizing eect.
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Worked examples 1
Unpolarized light o intensity I0 is incident on a polarizer. The transmitted light is then incident on an analyser. The axis o the analyser makes an angle o 60 to the axis o the polarizer.
position shown, the transmission axis o the analyser is parallel to the plane o polarization o the light ( = 0) . S ketch a graph to show how the intensity I o the transmitted light varies with as the analyser is rotated through 1 80.
analyser
Solution a) In unpolarized light the electric feld vector vibrates randomly in any plane ( perpendicular to the direction o propagation) . In polarized light this vector is restricted to j ust one plane.
polarizer
unpolarized light
b) This is an application o Maluss law I = I0 cos 2
C alculate the intensity emitted by the analyser.
Solution I The frst polarizer restricts the intensity to __ . 2 Using Maluss law I = I0 cos 2 0
2
cos 60 = 0.5 so cos 2 60 = 0.2 5 I0 thus I = 0.2 5 _ = 0. 1 2 5 I0 2 a) D istinguish between polarized and unpolarized light. b) A beam o plane- polarized light o intensity I0 is incident on an analyser.
When = 0 or 1 80, cos = 1 and so cos 2 = 1 and I = I0 When = 90, cos = 0 and so cos 2 = 0 and I = 0 These are the key points to ocus on. Note that cos 2 will never become negative. There is no need to include a unit or intensity as this is a sketch graph. I0
transmission axis plane of polarization
I
incident beam
analyser
The angle between the transmission axis o the analyser and the plane o polarization o the light can be varied by rotating the analyser about an axis parallel to the direction o the incident beam. In the
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0 0
90 /
180
4 . 4 W AV E B E H AV I O U R
4.4 Wave behaviour Understanding Reection and reraction
Applications and Skills Sketching and interpreting incident, reected
Snells law, critical angle, and total
internal reection Difraction through a single-slit and around objects Intererence patterns Double-slit intererence Path diference
and transmitted waves at boundaries between media Solving problems involving reection at a plane interace Solving problems involving Snells law, critical angle, and total internal reection Determining reractive index experimentally Qualitatively describing the difraction pattern ormed when plane waves are incident normally on a single-slit Quantitatively describing double-slit intererence intensity patterns
Equations n
sin sin1
v
_________2 = ____1 1 Snells law: ____ n = v 2
2
D Intererence at a double slit: s = ______ d
Nature of science Wave or particle? In the late seventeenth century two rival theories o the nature o light were proposed by Newton and Huygens. Newton believed light to be particulate and supported his view by the acts that it apparently travels in straight lines and can travel through a vacuum; at this time it was a strongly held belie that waves needed a medium through which to travel. Huygens wave model was supported by the work o Grimaldi who had shown that light difracts around small objects and through narrow openings. He
was also able to argue that when a wave meets a boundary the total incident energy is shared by the reected and transmitted waves; Newtons argument or this was based on the particles themselves deciding whether or not to reect or transmit this was not a strong argument and the wave theory o light became predominant. In the 21st century light is treated as both a wave and a particle in order to explain the ull range o its properties.
Introduction Now we have looked at how to describe waves, we are in the position to look at wave properties or, as it is described in IB Physics, wave behaviour. You are likely to have come across some of this topic if you
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O S C I LL AT I O N S AN D WAVE S normal
incident ray
refected ray 1 1
medium 1: reractive index = n 1 medium 2: reractive index = n 2 2
have studied physics beore starting the IB D iploma Programme and we have used some o the ideas in the previous sub- topic. In S ub- topic 4.3 we studied polarization a property that is restricted to transverse waves alone. The ideas examined in this sub- topic apply to transverse waves ( both mechanical and electromagnetic) and longitudinal waves. There are many demonstrations o wave properties utilizing sound and light; however, microwaves are commonly used or demonstrating these wave properties too.
Refection and reraction o waves reracted ra
Figure 1 Refection and reraction o waves.
We have looked at much o the content o fgure 1 when we considered polarization. We will now ocus on what is happening to the rays remember we could always add waveronts at right angles to the rays drawn on these diagrams. What the ray diagrams do not show is what is happening to the wavelength o the waves we will return to this in due course. The laws o reection and reraction can be summarized in three laws as ollows: 1
The refected and reracted rays are in the same p lane as the incident ray and the normal. This means that the event o reection or reraction does not alter the plane in which the light ray travels this is not obvious because we draw ray diagrams in two dimensions but, when we use ray boxes to perorm experiments with light beams, we can confrm that this is the case.
2
The angle o incidence equals the angle o refection. The angle o incidence is the angle between the incident ray and the normal and the angle o reection is the angle between the reected ray and the normal. The normal is a line perpendicular to a surace at any chosen point. The angle o incidence and reection are both labelled as 1 on fgure 1 .
3
For waves o a p articular requency and or a chosen p air o media the ratio o the sine o the angle o incidence to the sine o the angle o reraction is a constant called the (relative) reractive index. This is called Snells law ( or D escartess law in the French speaking world) . The angle o reraction is the angle between the reracted ray and the normal. Snells law can be written as sin 1 _ = 1n2 sin 2 For light going rom medium 1 to medium 2 this way o writing Snells law has several variants and the IB course uses one that we will look at soon.
Figure 2 Use o ray box to demonstrate the laws o refection and reraction.
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When light is normal on a surace S nells law breaks down because the light passes directly through the surace.
4 . 4 W AV E B E H AV I O U R
TOK
Nature of science
Conservation of energy and waves
What happens to light at an interace between two media?
The wave and ray diagram or refection and reraction tells just part o the story. As with many areas o physics, returning to the conservation o energy is important. For the total energy incident on the interace between the two media the energy is shared between the refected wave, the transmitted wave and the energy that is absorbed the urther the wave passes through the second medium, the more o the energy is likely to be absorbed. The conservation o mass/energy is a principle in physics which, to date, has not let physicists down. Does this mean that we have proved that the principle o conservation o energy is inallible?
This is a complex process but in general terms, when charges are accelerated, or example when they are vibrated, they can emit energy as an electromagnetic wave. In moving through a vacuum the electromagnetic wave travels with a velocity o 3 .00 1 0 8 m s 1 . When the wave reaches an atom, energy is absorbed and causes electrons within the atom to vibrate. All particles have requencies at which they tend to vibrate most efciently called the natural frequency. When the requency o the electromagnetic wave does not match the natural requency o vibration o the electron, then the energy will be re- emitted as an electromagnetic wave. This new electromagnetic wave has the same requency as the original wave and will travel at the usual speed in the vacuum between atoms. This process continues to be repeated as the new wave comes into contact with urther atoms o unmatched natural requency. With the wave travelling at 3 .00 1 0 8 m s 1 in space but being delayed by the absorptionre- emission process, the overall speed o the wave will be reduced. In general, the more atoms per unit volume in the material, the slower the radiation will travel. When the requency o the light does match that o the atoms electrons the re- emission process is occurs in all directions and the atom gains energy, increasing the internal energy o the material.
Refractive index and Snells law The absolute refractive index (n) o a medium is defned in terms o the speed o electromagnetic waves as: speed o electromagnetic waves in a vacuum c n = ____ = _ v speed o electromagnetic waves in the medium
The reractive index depends on the requency o the electromagnetic radiation and, since the speed o light in a vacuum is the limit o speed, the absolute reractive index is always greater than 1 ( although there are circumstances when this is not true but that is well beyond the expectation o your IB Physics course) . For all practical purposes the absolute reractive index o air is 1 so it is not necessary to perorm reractive index experiments in a vacuum.
Worked example C alculate the angle o reraction when the angle o incidence at a glass surace is 5 5 ( reractive index o the glass = 1 .48) .
Solution As we are dealing with air and glass there is no dierence between absolute reractive index and relative reractive index.
sin = n glass S nells law gives ____ sin 1
2
sin5 5 = _ = 1 .48 sin 2 sin5 5 0.81 9 sin 2 = _ = _ = 0.5 5 3 1 .48 1 . 48 = sin 1 ( 0.5 5 3 ) = 3 3 .6 3 4
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Reversibility of light You should be able to prove to yourself that rays are reversible. Place a ray box and a glass block on a piece of paper. Mark, on the paper, the path of the beam of light emitted by the ray box as it approaches and leaves the glass block. Then place the ray box on the other side of the block and you will see that the light travels along the same path in the opposite direction. We have seen that for light travelling from medium 1 to medium 2 S nells law can be written as sin 1 _ = 1n2 sin 2 here 1 n 2 means the relative refractive index going from medium 1 to medium 2 . For light travelling in the opposite direction and since light is reversible we have sin 2 _ = 2n1 sin 1 1 It should be clear from this that 1 n 2 = ___ n 2
1
Worked example The ( absolute) refractive index of water is 1 .3 and that of glass is 1 .5 .
c)
water water
a) C alculate the relative refractive index from glass to water. b) Explain what this implies regarding the refraction of light rays. c) D raw a wavefront diagram to show how light travels through a plane interface from glass to water.
n
vac water
= 1 . 3 and n glass =
We are calculating sin va c
______ = sin w a te r
n
vac water
glass
and _____ = sin gla s s
w a te r
This means that
w ate r
glass
n
vac glass
= 1 .5
n water
sin v ac
sin gla s s sin gla s s sin v a c _____ ______ = ______ = sin sin sin va c
glass glass
Solution a) n water =
interface
vac
glass
n glass
n water
1 n water = ___ 1 .3 = 0.87 1 .5
b) With a relative refractive index less than one this means that the light travels faster in the water than the glass and therefore bends away from the normal.
As the refractive index of water is lower than that of glass the light wave speeds up on entering the water. The frequency is constant and, as = f, the wavelength will be greater in water meaning that the wavefronts are further apart. The fact that the frequency remains constant is a consequence of Maxwells electromagnetic equations something not covered in IB Physics.
The critical angle and total internal refection When a light wave reaches an interface travelling from a higher optically dense medium to a lower one the wave speeds up. This means that the wavelength of the wave increases ( frequency being constant) and
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the direction o the wave moves away rom the normal the angle o reraction being greater than the angle o incidence so as the angle o incidence increases the angle o reraction will approach 90. O p tical density is not the same as physical density, i.e. mass per unit volume, it is measured in terms o reractive index higher reractive index material having higher optical density. Light incident at any angle > c is totally refected.
Though not bent, part o the normal ray is refected. c 1
2
3
4
critical angle c
5
n1 high index material
light source
Figure 3 Light passing rom more optically dense medium to less optically dense medium. Ray 1 in fgure 3 shows a ray passing rom a more optically dense medium to a less optically dense medium normal to an interace. Most o the light passes though the interace but a portion is reected back into the original medium. Increasing the angle o incidence ( as or rays 2 and 3 ) will increase the angle o reraction and ray 4 shows an angle o incidence when the angle o reraction is 90. The angle o incidence at this value is called the critical angle ( c) . Ray 5 shows that, when the angle o incidence is larger than the critical angle, the light wave does not move into the new medium at all but is reected back into the original medium. This process is called total internal refection. It should be noted that or angles smaller than the critical angle there will always be a reected ray; although this will carry only a small portion o the incident energy. A ray box and semicircular glass block can be used to measure the critical angle or glass as shown in fgure 4. When the beam is incident on the curved ace o the block making it travel towards the centre o the at ace, it is acting along a radius and so enters the block normally and, thereore, without bending. B y moving the ray box, dierent angles o incidence can be obtained. B oth the critical angle and total internal reection can be seen. r1
r = 90 c
i1 semicircular glass slab
critical angle
i2
r2
i2 = r2
ray box
Figure 4 Use o ray box to investigate critical angle and total internal refection.
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Worked example C alculate the ( average) critical angle or a material o ( average) absolute reractive index 1 .2 .
Solution The word average is included because the reractive index and critical angle would each be dierent or dierent colours. D ont be surprised i it is let out in some questions it is implied by talking about single values. 1 1 _ sin c = _ n 1 = 1 .2 = 0.83 3 c = sin1 0.833 = 56.4 56
When a ray box emitting white light is used, the light emerging through the glass block is seen to disp erse into the colours o the rainbow. This is due to each o the colours, o which white light is comprised, having a dierent requency. The reractive index or each o the colours is dierent.
Calculating the critical angle sin S nells law gives ____ = 1n2 . sin 1
2
In order to obtain a critical angle, medium 1 must be more optically dense than medium 2 . When 1 = c then 2 = 90 so sin 2 = 1 This gives sin c = 1 n 2 1
n2 n2 = _ n1
When the less dense medium ( medium 2 ) is a vacuum or air then n 2 = 1 1 S o sin c = _ n1
Investigate! Measuring the refractive index There are several possible experiments that you could do to measure the reractive index depending on whether a substance is a liquid or a solid ( we assume the reractive index o gases is 1 although mirages are a good example o how variations in air density can aect the reraction o light) . You could research how to use real and apparent depth measurements to measure the reractive index o a liquid. The investigation outlined below will be to trace some rays ( or beams o light, really) through a glass block o rectangular cross-section.
The arrangement is similar to that shown in fgure 2 on p1 46.
Place the block on a piece o white paper and mark its position by drawing around the edges.
D irect a beam o light to enter the block near the centre o a longer side and to leave by the opposite side.
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Mark the path o the beam entering and leaving the block you will need at least two points on each beam to do this. Remove the block and use a ruler to mark the path o the beam on either side o the
block and then inside. Add arrows to these to remind you that they represent rays and to indicate which is the incident beam and which is the reracted beam.
Using a protractor, mark in and draw normals or the beam entering and leaving the block.
Remembering that light is reversible and the beam is symmetrical, measure two values or each o 1 and 2 .
C alculate the reractive index o the block using S nells law.
Estimate the experimental uncertainty on your measurements o 1 and 2 .
Note that the uncertainty in 1 and 2 is not the same as that in sin 1 and sin 2 but you sin can calculate the uncertainty in ____ ( and sin 1
2
hence n) by calculating hal the dierence sin sin and ______ between ______ sin sin 1 m ax
1 m in
2 m in
2 m ax
Repeat the experiment or a range o values o the angle o incidence.
Which o your values is likely to be the most reliable?
4 . 4 W AV E B E H AV I O U R
Difraction The frst detailed observation and description o the phenomenon that was named diffraction was made by the Italian priest Francesco Grimaldi. His work was published in 1 665 , two years ater his death. He ound that when waves pass through a narrow gap or slit ( called an aperture) , or when their path is partly blocked by an obj ect, the waves spread out into what we would expect to be the shadow region. This is illustrated by fgure 5 and can be demonstrated in a ripple tank. He noted that close to the edges the shadows were bordered by alternating bright and dark ringes. Given the limited apparatus available to Grimaldi his observations were quite extraordinary.
aperture
obstacle
obstacle
Figure 5 Difraction. Further observation o diraction has shown:
the requency, wavelength, and speed o the waves each remains the same ater diraction
the direction o propagation and the pattern o the waves change
the eect o diraction is most obvious when the aperture width is approximately equal to the wavelength o the waves.
the amplitude o the diracted wave is less than that o the incident wave because the energy is distributed over a larger area.
E xplaining diraction by a single slit is complex and you will only be asked or a qualitative description o single- slit diraction at S L however, as mentioned in Sub-topic 4. 3 , the HuygensFresnel principle gives a good insight into how the single-slit diraction pattern comes about ( see fgure 6) .
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Worked example C omplete the ollowing diagrams to show the waveronts ater they have passed through the gaps.
Figure 6 HuygensFresnel explanation o difraction.
Solution
When the width o the slit is less than or equal to the wavelength o the wave, the waves emerge rom the slit as circular waveronts. As the slit width is increased, the spreading o the waves only occurs at the edges and the diraction is less noticeable.
Plane waves travelling towards the slit behave as i they were sources o secondary wavelets. The orange dots in fgure 6 show these secondary sources within the slit. These sources each spread out as circular waves. The tangents to these waves will now become the new waveront. The central image is bright and wide, beyond it are urther narrower bright images separated by darkness. S ingle-slit diraction is urther explored or those studying HL Physics in Topic 9.
Figure 7 Single-slit difraction pattern.
Double-slit interference We briey discussed the principle o superposition in S ub- topic 4.3 . Intererence is one application o this principle. When two or more waves meet they combine to produce a new wave this is called interference. When the resultant wave has larger amplitude than any o the individual waves the intererence is said to be constructive; when the resultant has smaller amplitude the intererence is destructive. Intererence can be achieved by using two similar sources o all types o wave. It is usually only observable i the two sources have a constant phase relationship this means that although they need not emit the two sets o waves in phase, the phase o the waves cannot alter relative to one another. S uch sources will have the same requency and are said to be coherent.
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4 . 4 W AV E B E H AV I O U R
audio signal generator
speakers a
D
x L
S
L
S
S
L
S
L
S
L
microphone
S
L
S
L
S
L
L loud sound S sot sound
Figure 8 Intererence o sound waves. Intererence o sound waves is easy to demonstrate using two loudspeakers connected to the same audio requency oscillator as shown in fgure 8. Moving the microphone ( connected to an oscilloscope) in a line perpendicular to the direction in which the waves are travelling allows an equally spaced loudsot sequence to be detected. More simply, the eect can be demonstrated by the observer walking along the loudsot line while listening to the loudness. When a coherent beam o light is incident on two narrow slits very close together the beam is diracted at each slit and, in the region difracted beam rom top slit destructive intererence (trough meets trough) S constructive intererence (crest meets crest) lamp
double slit difracted beam rom bottom slit
Figure 9 Intererence o light waves.
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4
O S C I LL AT I O N S AN D WAVE S where the two diracted beams cross, intererence occurs as seen in fgure 9. A pattern o equally spaced bright and dark ringes ( shown in fgure 1 0) is obtained on a screen positioned in the region where the diracted beams overlap. When a crest meets a crest ( or a trough meets a trough) constructive intererence occurs. When a crest meets a trough destructive intererence occurs. A similar experiment to this was perormed by the talented English physicist ( and later physician) Thomas Young in 1 801 . The coherent beam is achieved by placing a single slit close to the source o light this means that the waveronts spreading rom the single slit each reach the double slit with the same phase relationship and so the secondary waves coming rom the double slit retain their constant phase relationship.
Figure 10 Fringes produced by a double-slit.
Path diference and the double-slit equation You will never be asked to derive this equation but the ideas regarding path dierence are vital to your understanding o intererence. B A2 S
d
C
A1
O
P D
Figure 11 The double-slit geometry. Figure 1 1 shows two slits ( apertures) A 1 and A 2 distance d apart. The double slit is at distance D rom a screen. O is the position o the central bright ringe ( arising rom constructive intererence) . B is the position o the next bright ringe above O ; the distance O B is the ringe spacing s. There will be another bright ringe distance s below O . The beams rom A 1 and A 2 to O will travel equal distances and so will meet with the same phase relationship that they had at A 1 and A 2 they have zero path dierence. At B the beam rom A 1 will travel an extra wavelength compared with the beam rom A 2 the path distance ( = A 1 P) equals . B ecause o the short wavelength o light and the act that D is very much larger than d, the line A 2 P is eectively perpendicular to lines A 1 B and C B ( C being the midpoint o A 1 A 2 ) . This means that the triangles A 1 A 2 P and C B O are similar triangles. A1 P BO s Taking ratios _ = _ or _ = _ D CO A1 A2 d D Rearranging gives s = _ d This gives the separation o successive bright ringes ( or bands o loud sound or a sound experiment) .
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4 . 4 W AV E B E H AV I O U R
In general or two coherent beams starting in phase, i the path dierence is a whole number o wavelengths we get constructive intererence and i it is an odd number o hal wavelengths we get destructive intererence. It must be an odd number o hal wavelengths or destructive intererence because an even number would give a whole number ( integer) and that is constructive intererence! S ummarizing this: For constructive intererence the path dierence must = n where n = 0, 1 , 2 1 where For destructive intererence the path dierence must = ( n + __ 2 ) n = 0, 1 , 2
n is known as the order o the ringe, n = 0 being the zeroth order, n = 1 the frst order, etc.
Worked examples 1
In a double-slit experiment using coherent light o wavelength , the central bright ringe is observed on a screen at point O, as shown below.
5 ___ 9 ___ 1 3 ___ , 32 , ___ , 72 , ___ , 1 12 , ___ giving a dierence is __ 2 2 2 2 total o 7 dark ringes.
2
Two coherent point sources S 1 and S 2 oscillate in a ripple tank and send out a series o coherent waveronts as shown in the diagram.
P coherent light
P O
Q
wavelength double slit screen (not to scale)
At point P, the path dierence between light arriving at P rom the two slits is 7.
S1
S2
S tate and explain the intensity o the waves at P and Q?
a) Explain the nature o the ringe at P.
Solution
b) S tate and explain the number o dark ringes between O and P.
C onsidering each waveront to be a crest; at point P two crests meet and so superpose constructively giving an amplitude which is twice that o one wave (assuming the waveronts each have the same amplitude) the intensity is proportional to the square o this so will be our times the intensity o either o the waves alone. At Q a blue crest meets a red trough and so there is cancellation occurring and there will be zero intensity.
Solution a) As the path dierence is an integral number o wavelengths there will be a bright ringe at P. b) For destructive intererence the path dierence must be an odd number o hal wavelengths, so there will be dark ringes when the path
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Investigate! Measuring the wavelength of laser light using a double slit
bright spots
laser double slit
screen
Figure 12 Measuring the wavelength of laser light using a double slit. This experiment, using a gas laser ( or laser pointer) , is a modern version o the one perormed by Young. A laser emits a highly coherent beam o light ideal or perorming this experiment. C are should be taken not to shine the laser beam or its refection into your eye should this happen, look away immediately to avoid the risk o permanent damage to your eye. O ne way to minimise eye damage is to keep sufcient light in the room so that you can still do the experiment. This means that the iris o your eye will not be ully open.
Set up the apparatus as shown in fgure 1 2 the screen should be a ew metres rom the double slit.
The double slit can be homemade by scratching a pair o narrow lines on a piece o glass painted with a blackened material or it could be a ready prepared slide.
Light rom the laser beam diracts through the slits and emerges as two separate coherent waves. B oth slits must be illuminated by the narrow laser beam; sometimes it may be necessary to use a diverging ( concave) lens to achieve this.
The intererence pattern is then proj ected onto the screen and the separation o the spots ( images o the laser aperture) is measured as accurately as possible using a metre ruler or tape measure. This is best done by measuring the distance between the urthest spots, remembering that nine spots would have a separation o 8S.
The distance rom the double slit to the screen ( D) should also be measured using metre rulers or a tape measure.
O nce your readings are taken the wavelength Sd o the light can be calculated rom = __ . D
The slit separation ( d) can be measured using a travelling microscope ( however, d is likely to be provided by a manuacturer) .
When red light is used as the source, the bright ringes are all red. When blue light is used as a source, the bright ringes are all blue. As blue light has a shorter wavelength than red light the blue ringes are closer together than the red ringes. When the source is white light the zeroth order ( central) bright ringe is white but the other ringes are coloured with the blue edges closer to the centre and the red edges urthest rom the centre. C an you explain this?
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4 . 4 W AV E B E H AV I O U R
Investigate! Measuring the wavelength of microwaves using a double slit receiver
mA milliammeter
transmitter
double slit
Figure 13 Microwave arrangement.
The double slit is adj usted so that the slits are around 3 cm apart for maximum diffraction.
Arrange both transmitter and receiver about half a metre from the slit.
Alter the position of the receiver until the received signal is at its strongest.
occur as in Young' s double-slit interference experiment with light. Look for fringes on both sides of the maximum.
Measure as accurately as you can the values for D, d, and S.
C alculate the wavelength of the microwaves. Repeat for other distances of transmitter and receiver from the slits.
Now slowly rotate the receiver until the signal is weakest.
C over up one of the slits with a book and explain the result.
Remove the book so that two slits are again available and attempt to discover if fringes
These investigations can be repeated with sound waves or radio waves given appropriate transmitters, receivers and slits of the correct dimension.
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4.5 Standing waves Understanding
Applications and Skills
The nature of standing waves
Describing the nature and formation of standing
Boundary conditions
waves in terms of superposition Distinguishing between standing and travelling waves Observing, sketching, and interpreting standing wave patterns in strings and pipes Solving problems involving the frequency of a harmonic, length of the standing wave, and the speed of the wave
Nodes and antinodes
Nature of science Fourier synthesis Synthesizers are used to generate a copy of the sounds naturally produced by a wide range of musical instruments. Such devices use the principle of superposition to join together a range of harmonics that are able to emulate the sound of the natural instrument. Fourier synthesis works by combining a sine-wave signal with sine-wave or cosine-wave
harmonics of correctly chosen amplitude. The process is named after the French mathematician and physicist Jean Baptiste Joseph, Baron de Fourier who, in the early part of the nineteenth century, developed the mathematical principles on which synthesis of music is based. In many ways Fourier synthesis is a visualization of music.
Introduction We have seen in Sub-topic 4. 2 how travelling waves transfer energy from the source to the surroundings. In a travelling wave the position of the crests and troughs changes with time. Under the right circumstances, waves can be formed in which the positions of the crests and troughs do not change in such a case the wave is called a standing wave. When two travelling waves of equal amplitude and equal frequency travelling with the same speed in opposite directions are superposed, a standing wave is formed.
Standing waves on strings Figure 1 shows two travelling waves ( coloured green and blue) moving towards each other at four consecutive times t1 , t2 , t3 , and t4. The green and blue waves superpose to give the red standing wave. As the green and blue waves move forward there are points where the total displacement ( seen on the red wave) always remains zero these are called nodes. At other places the displacement varies between a
158
4 . 5 S T A N D I N G W AV E S
maximum in one direction and a maximum in the other direction these are called antinodes. travelling waves standing wave
time t 1
time t2
time t 3
time t 4
2
standing wave
node
node
Figure 1 The formation of a standing wave. At any instant the displacement o the standing wave will vary at all positions other than the nodes. Thus a single rame shot o a standing wave would look like a progressive wave as can be seen rom the red wave in fgure 1 . When representing the standing wave graphically it is usual to show the extremes o standing waves, but over a complete time period o the oscillation the wave will occupy a variety o positions as shown by the arrow in the loop o fgure 2 . extreme positions (antinodes)
nodes
range of motion at antinode
Figure 2 Nodes and antinodes on a string.
Meldes string The apparatus shown in fgure 3 is useul in demonstrating standing waves on a string. A variant o this apparatus was frst used in the late nineteenth century by the German physicist Franz Melde. A string is strung between a vibration generator and a fxed end. When the vibration generator is
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stroboscope
fxed end o string
node
antinode
vibration generator signal generator
Figure 3 Meldes string. connected to an audio requency oscillator, the end o the string attached to the vibration generator oscillates vertically. A wave travels down the string beore undergoing a phase change o 1 80 when it reects at the fxed end. The reected wave superposes with the incident wave and (at certain requencies) a standing wave is ormed. The requency o the audio requency generator is slowly increased rom zero and eventually a requency is reached at which the string vibrates with large amplitude in the orm o a single loop the frst harmonic. I the requency is urther increased, the amplitude o the vibrations dies away until a requency o twice the frst harmonic requency is achieved in this case two loops are ormed and we have the second harmonic requency shown in fgure 3. This is an example o resonance the string vibrates with large amplitude only when the applied requency is an integral multiple o the natural requency o the string. We return to resonance in more detail in Option B. Using a stroboscope to reeze the string reveals detail about a standing wave. For example, when the ash requency is slightly out o synch with the vibration requency, it is possible to see the variation with time o the strings displacement; this will be zero at a node but a maximum at an antinode.
Note Although we oten treat the point o attachment o the string to the vibration generator as being a node, this is not really correct. The generator vibrates the string to set up the wave and thereore the nearest node to the generator will be a short distance rom the vibrator. We call this inaccuracy an end correction but oten draw a diagram showing the node at the vibrator. Because some o the travelling wave energy is transmitted or absorbed by whatever is clamping the fxed end, the reected waves will be slightly weaker than the incident waves. This means that cancellation is not complete and there will be some slight displacement at the nodes. Within each loop all parts o the string vibrate together in phase, but loops next to each other are constistently 180 o out o phase. Although strings are oten used to produce sound waves, it is not a sound wave travelling along the string it is a transverse wave. This wave travels at a speed which is determined by the characteristics o the string.
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4 . 5 S T A N D I N G W AV E S
Harmonics on strings
N=1
We have seen rom Meldes string that a string has a number o requencies at which it will naturally vibrate. These natural requencies are known as the harmonics of string. The natural requency at which a string vibrates depends upon the tension o the string, the mass per unit length and the length o the string. With a stringed musical instrument each end o a string is fxed, meaning there will be a node at either end. The frst harmonic is the lowest requency by which a standing wave will be set up and consists o a single loop. Doubling the requency o vibration halves the wavelength and means that two loops are ormed this is called the second harmonic; three times the undamental requency gives the third harmonic (see fgure 4) .
N=2
N=3
N=4
You will see rom fgure 1 that the distance between two consecutive nodes is equal to hal a wavelength in other words each loop in a standing . Figure 4 shows the frst fve harmonics o a wave wave is equivalent to __ 2 on a string o a fxed length. With the speed o the wave along the string being constant, halving the wavelength doubles the requency; reducing the wavelength by a actor o three triples the requency, etc. (the wave equation c = f applies to all travelling waves) . The dierent harmonics can be achieved either by vibrating the string at the appropriate requency or by plucking, striking or bowing the string at a dierent position although plucking at the centre is likely to produce the frst, third and fth harmonics. B y pinching the string at dierent places a node is produced so, or example, pinching at the centre o the string would produce the even harmonics. In a musical instrument several harmonics occur at the same time, giving the instrument its rich sound.
N=5
Figure 4 Harmonics on a string.
Worked example A string is attached between two rigid supports and is made to vibrate at its frst harmonic requency f.
__
1 so a quarter This means that t = T4 = __ 4f o a period has elapsed and the string has gone through quarter o a cycle to give:
The diagram shows the displacement o the string at t = 0. ( ii) In this case the wave has gone through T 1 = __ and so will have hal a period t = __ 2 2f
(
(a) D raw the displacement o the string at time 1 1 ( i) t = _ ( ii) t = _ 4f 2f (b) The distance between the supports is 1 .0 m. A wave in the string travels at a speed o 2 40 m s 1 . C alculate the requency o the vibration o the string.
Solution (a) ( i) We must remember the relationship 1 between period T and requency f is T = __ . f
)
moved rom a crest to a trough:
(b) The string is vibrating in frst harmonic mode and so the distance between the fxed ends is hal a wavelength ( __ . 2 ) S o __ = 1 . 0 m and = 2 .0 m. 2
Using c = f
c 2 40 f = __ = ___ 2 .0
= 1 2 0 Hz
Standing waves in pipes S tanding waves in a pipe dier rom standing waves on a string. In pipes the wave medium is ( usually) air and the waves themselves are longitudinal. Pipes can have two closed ends, two open ends, or one
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O S C I LL AT I O N S AN D WAVE S open and one closed; the latter two being shown in fgure 5 . The sound waves are reected at both ends o the pipe irrespective o whether they are open or closed. The variation o displacement o the air molecules in the pipe determines how we graphically represent standing waves in pipes. There is a displacement antinode at an open end ( since the molecules can be displaced by the largest amount here) and there will be a displacement node at the closed end ( because the molecules next to a closed end are unable to be displaced) . S imilar ideas to strings apply to the harmonics in pipes. S trictly speaking, the displacement antinode orms j ust beyond an open end o the pipe but this end eect can be ignored or most purposes. 1 4
1 2
1st harmonic
1st harmonic 3 4
1 2nd harmonic
3rd harmonic 5 4
3 2
3rd harmonic
5th harmonic 7 4
2 4th harmonic
7th harmonic 9 4
9th harmonic
5 2
5th harmonic
Figure 5 Harmonics in a pipe.
Harmonics in pipes The harmonic in a pipe depends on whether or not the ends o a pipe are open or closed. For a pipe o fxed length with one open and one closed end there must always be a node at the closed end and an antinode at the open end. This means that only odd harmonics are available ( the number o the harmonic is the number o hal loops in this type o pipe) . For a pipe with two open ends there must always be an antinode at each end and this means that all harmonics are achieveable ( in this case the number o loops gives the number o the harmonic) . Lets compare the requencies o the harmonics or the one open end pipe. Suppose the pipe has a length L. The wavelength ( ) o the frst c harmonic would be 4L and, rom c = f, the requency would be __ . 4L c is the speed o sound in the pipe. 3 3c 4 For the third harmonic L = __ , so = __ L and the requency = __ ( or 4 3 4L three times that o the frst harmonic) .
A harmonic is named by the ratio o its requency to that o the frst harmonic.
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4 . 5 S T A N D I N G W AV E S
Worked examples 1
The frst harmonic requency or a particular organ pipe is 3 3 0 Hz. The pipe is closed at one end but open at the other. What is the requency o its second harmonic?
Solution A named harmonic is the ratio o its requency to that o the frst harmonic so in this case the second harmonic will be 660 Hz since the second harmonic has twice the requency o the frst harmonic. 2
The frst harmonic requency o the note emitted by an organ pipe which is closed
at one end is f. What is the frst harmonic requency o the note emitted by an organ pipe o the same length that is open at both ends?
Solution The length o a pipe closed at one end in frst harmonic mode is __ ( = L ) so = 4L 4 This length will be hal the wavelength o a pipe open at both ends so L = __ so = 2 L 2 S ince the wavelength has halved the requency must double so the new requency will be 2 f.
Boundary conditions In considering both pipes and strings we have assumed reections at the ends or boundaries. In meeting the boundary o a string the wave reects ( or at least partially reects) this is known as a fxed boundary; there will be the usual phase change o 1 80 at a fxed boundary meaning that the reected wave cancels the incident wave and so orms a node. The closed ends o pipes and edges o a drumhead also have fxed boundaries. In the case o an open-ended pipe there is still a reection o the wave at the boundary but no phase change, so the reected wave does not cancel the incident wave and there is an antinode ormed the same idea applies to strips o metal vibrated at the centre, xylophones, and vibrating tuning orks. This type o boundary is called a ree boundary.
Comparison o travelling waves and stationary waves The ollowing table summarizes the similarities and dierences between travelling waves and standing waves:
Property energy transer amplitude phase
wave profle (shape) wavelength
requency
Travelling wave energy is transerred in the direction o propagation all particles have the same amplitude within a wavelength the phase is dierent or each particle propagates in the direction o the wave at the speed o the wave the distance between adjacent particles which are in phase all particles vibrate with same requency.
Standing wave no energy is transerred by the wave although there is interchange o kinetic and potential energy within the standing wave amplitude varies within a loop maximum occurs at an antinode and zero at a node all particles within a loop are in phase and are antiphase (180 out o phase) with the particles in adjacent loops stays in the same position
twice the distance between adjacent nodes (or adjacent antinodes) all particles vibrate with same requency except at nodes (which are stationary)
TOK Pitch and frequency Musical pitch is closely linked to requency but also has a psychological component in relation to music. We think o pitch as being someones perception o requency. Musical notes o certain pitches, when heard together, will produce a pleasant sensation and are said to be consonant or harmonic. These sound waves orm the basis o a musical interval. For example, any two musical notes o requency ratio 2:1 are said to be separated by an octave and result in a particularly pleasing sensation when heard. Similarly, two notes o requency ratio o 5:4 are said to be separated by an interval o a third (or strictly a pure third) ; again this interval sounds pleasing. Has music always been thought o in this way? Is the concept o consonance accepted by all societies?
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O S C I LL AT I O N S AN D WAVE S
Questions 1
3
(IB) a)
(IB) a)
A pendulum consists of a bob suspended by a light inextensible string from a rigid support. The pendulum bob is moved to one side and then released. The sketch graph shows how the displacement of the pendulum bob undergoing simple harmonic motion varies with time over one time period.
In terms of the acceleration, state two conditions necessary for a system to perform simple harmonic motion.
b) A tuning fork is sounded and it is assumed that each tip vibrates with simple harmonic motion.
d
displacement
0
0
time
O n a copy of the sketch graph:
The extreme positions of the oscillating tip of one fork are separated by a distance d.
( i) Label, with the letter A, a point at which the acceleration of the pendulum bob is a maximum.
( i)
( ii) Label, with the letter V, a point at which the speed of the pendulum bob is a maximum.
( ii) Sketch a graph to show how the displacement of one tip of the tuning fork varies with time.
b) Explain why the magnitude of the tension in the string at the midpoint of the oscillation is greater than the weight of the pendulum bob.
(iii) On your graph, label the time period T and the amplitude A. ( 8 marks) 4
2
State, in terms of d, the amplitude of vibration.
(IB) The graph below shows how the displacement x of a particle undergoing simple harmonic motion varies with time t. The motion is undamped.
( IB) a) Graph 1 below shows the variation with time t of the displacement d of a travelling ( progressive) wave. Graph 2 shows the variation with distance x along the same wave of its displacement d.
x
d/mm
0
t
a) S ketch a graph showing how the velocity v of the particle varies with time. b) Explain why the graph takes this form. ( 4 marks)
164
Graph 2 4 2 0 0.0 2 4 d/mm
0
Graph 1 4 2 0 0.0 2 4
a)
0.1
0.2
0.3
0.4
0.5
0.6 t/s
0.4
0.8
1.2
1.6
2.0
2.4 x/cm
S tate what is meant by a travelling wave.
QUESTION S
b)
Use the graphs to determine the amplitude, wavelength, frequency and speed of the wave.
wave A 3.0 2.0
( 5 marks) xA /mm
5
1.0
( IB)
0.0 0.0
2.0
4.0
6.0
8.0
t/ms 10.0
1.0
a) With reference to the direction of energy transfer through a medium, distinguish between a transverse wave and a longitudinal wave.
2.0 3.0
The graph below shows the variation with time t of the displacement xB of wave B as it passes through point P. The waves have equal frequencies.
b) A wave is travelling along the surface of some shallow water in the x- direction. The graph shows the variation with time t of the displacement d of a particle of water.
wave B 2.0
10 1.0 xB /mm
8 6
0.0
2.0
4.0
6.0
8.0
t/ms 10.0
1.0
4
2.0
2 d/mm
0.0
( i) C alculate the frequency of the waves.
0 0.0
0.05
0.1
2
0.15
0.2
0.25
0.3 t/s
( ii) The waves pass simultaneously through point P. Use the graphs to determine the resultant displacement at point P of the two waves at time t = 1 . 0 ms and at time t = 8.0 ms.
4 6 8 10
( 6 marks)
Use the graph to determine the frequency and the amplitude of the wave. c) The speed of the wave in b) is 1 5 cm s 1 . D educe that the wavelength of this wave is 2 .0 cm. d) The graph in b) shows the displacement of a particle at the position x = 0. D raw a graph to show the variation with distance x along the water surface of the displacement d of the water surface at time t = 0.070 s. ( 1 1 marks)
7
( IB) a) With reference to the direction of energy transfer through a medium, distinguish between a transverse wave and a longitudinal wave. b) The graph shows the variation with time t of the displacement d of a particular water particle as a surface water wave travels through it. 10 8 6
(IB)
4
a)
B y referring to the energy of a travelling wave, explain what is meant by: ( i) a ray
2 d/mm
6
0 0 2
0.05
0.1
0.15
0.2
0.25
0.3 t/s
( ii) wave speed. 4
b) The following graph shows the variation with time t of the displacement xA of wave A as it passes through a point P.
6 8 10
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O S C I LL AT I O N S AN D WAVE S Use the graph to determine or the wave:
The polarizer is then rotated by 1 80 in the direction shown. S ketch a graph to show the variation with the rotation angle , o the transmitted light intensity I, as varies rom 0 to 1 80. Label your sketch- graph with the letter U.
( i) the requency ( ii) the amplitude. c) The speed o the water wave is 1 2 cm s 1 . C alculate the wavelength o the wave. d) The graph in b) shows the displacement o a particle at the position x = 0.
b) The beam in a) is now replaced with a polarized beam o light o the same intensity.
S ketch a graph to show the variation with distance x along the water surace o the displacement d o the water surace at time t = 0.2 0 s.
The plane o polarization o the light is initially parallel to the polarization axis o the polarizer.
e) The wave meets a shel that reduces the depth o the water.
polarizer
wave fronts shelf direction of rotation
direction of travel of wave 30
shallow water
deep water polarization axis polarized light
The angle between the waveronts in the shallow water and the shel is 3 0. The speed o the wave in the shallow water is 1 2 cm s 1 and in the deeper water is 1 8 cm s 1 . For the wave in the deeper water, determine the angle between the normal to the waveronts and the shel. ( 1 2 marks) 8
(IB) a) A beam o unpolarized light o intensity I0 is incident on a polarizer. The polarization axis o the polarizer is initially vertical as shown.
The polarizer is then rotated by 1 80 in the direction shown. O n the same axes in a) , sketch a graph to show the variation with the rotation angle , o the transmitted light intensity I, as varies rom 0 to 1 80. ( 5 marks) 9
(IB) An orchestra playing on boat X can be heard by tourists on boat Y, which is situated out o sight o boat X around a headland. ocean X
polarizer headland Y direction of rotation
The sound rom X can be heard on Y due to A. reraction B . refection polarization axis
166
unpolarized light
C . diraction D . transmission.
QUESTION S 1 0 (IB)
( i)
A small sphere, mounted at the end of a vertical rod, dips below the surface of shallow water in a tray. The sphere is driven vertically up and down by a motor attached to the rod.
( ii) The angle between the wavefronts and the interface in region A is 60. The refractive index An B is 1 .4. D etermine the angle between the wavefronts and the interface in region B .
The oscillations of the sphere produce travelling waves on the surface of the water. rod
( iii) O n the diagram above, construct three lines to show the position of three wavefronts in region B .
water surface
sphere
a) The diagram shows how the displacement of the water surface at a particular instant in time varies with distance from the sphere. The period of oscillation of the sphere is 0.02 7 s.
With reference to a wave, distinguish between a ray and a wavefront.
c) Another sphere is dipped into the water. The spheres oscillate in phase. The diagram shows some lines in region A along which the disturbance of the water surface is a minimum. lines of minimum disturbance
1.5
displacement/mm
1.0 wavefront
0.5 distance/mm
0 0
5
10
15
0.5 1.0
( i)
1.5
Use the diagram to calculate, for the wave: ( i)
O utline how the regions of minimum disturbance occur on the surface.
( ii) The frequency of oscillation of the spheres is increased. S tate and explain how this will affect the positions of minimum disturbance.
the amplitude
( ii) the wavelength ( iii) the frequency
( 1 5 marks)
( iv) the speed. b) The wave moves from region A into a region B of shallower water. The waves move more slowly in region B . The diagram ( not to scale) shows some of the wavefronts in region A.
1 1 (IB) a) D escribe two ways in which standing waves differ from travelling waves. b) An experiment is carried out to measure the speed of sound in air, using the apparatus shown below.
60
direction of motion region A
region B
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O S C I LL AT I O N S AN D WAVE S ( i) O n a copy o the diagram, label with the letter P the position along the pipe where the amplitude o oscillation o the air molecules is the largest.
tuning fork, frequency 440 Hz tube
( ii) The speed o sound in the air in the pipe is 3 3 0 m s 1 . C alculate the length l.
tank of water
A tube that is open at both ends is placed vertically in a tank o water until the top o the tube is j ust at the surace o the water. A tuning ork o requency 440 Hz is sounded above the tube. The tube is slowly raised out o the water until the loudness o the sound reaches a maximum or the frst time, due to the ormation o a standing wave. ( i)
Explain the ormation o a standing wave in the tube.
c) Use your answer to b) ( ii) to suggest why it is better to use organ pipes that are closed at one end or producing low requency notes rather than pipes that are open at both ends. ( 8 marks)
1 3 (IB) A microwave transmitter emits radiation o a single wavelength towards a metal plate along a line normal to the plate. The radiation is reected back towards the transmitter. metal plate
( ii) S tate the position where a node will always be produced. ( iii) The tube is raised a little urther. Explain why the loudness o the sound is no longer at a maximum.
microwave transmitter microwave detector
c) The tube is raised until the loudness o the sound reaches a maximum or a second time. Between the two positions o maximum loudness the tube has been raised by 36.8 cm. The requency o the sound is 440 Hz. Estimate the speed o sound in air. ( 1 0 marks)
A microwave detector is moved along a line normal to the microwave transmitter and the metal plate. The detector records a sequence o equally spaced maxima and minima o intensity. a) Explain how these maxima and minima are ormed.
1 2 (IB) a) State two properties o a standing ( stationary) wave. b) The diagram shows an organ pipe that is open at one end.
b) The microwave detector is moved through 1 3 0 mm rom one point o minimum intensity to another point o minimum intensity. O n the way it passes through nine points o maximum intensity. C alculate the ( i) wavelength o the microwaves. ( ii) requency o the microwaves.
l
The length o the pipe is l. The requency o the undamental ( frst harmonic) note emitted by the pipe is 1 6 Hz.
168
c) D escribe and explain how it could be demonstrated that the microwaves are polarized. ( 1 1 marks)
5
E LE CTRI CI TY AN D M AG N ETI SM
Introduction Modern society uses a whole range o electrical devices rom the simplest heated metal flaments that provide light, through to the most sophisticated medical instruments and computers. D evices o increasing technical complexity are developed every day.
In this topic we look at the phenomenon o electricity, and what is meant by charge and electric current. We consider the three eects that can be observed when charge ows in an electric circuit.
5.1 Electric felds Understanding Charge Electric feld Coulombs law
Applications and skills Identiying two species o charge and the
Electric current Direct current (dc)
Potential dierence (pd)
Nature of science Electrical theory resembles the kinetic theory o gases in that a theory o the microscopic was developed to explain the macroscopic observations that had been made over centuries. The development o this subject and some o the byways that were taken make this a ascinating study. We should remember the many scientists who were involved. It is a tribute to them that they could make so much progress when the details o the microscopic nature o electronic charge were unknown to them.
direction o the orces between them Solving problems involving electric felds and Coulombs law Calculating work done when charge moves in an electric feld in both joules and electronvolts Identiying sign and nature o charge carriers in a metal Identiying drit speed o charge carriers Solving problems using the drit-speed equation Solving problems involving current, potential dierence, and charge
Equations q
current-charge relationship: I = ______ t q1 q2 ________ Coulomb's law: F = k r2
1 the coulomb constant: k = ________ 4 0
W potential dierence defnition: V = ____ q
conversion o energy in joule to electron-volt: W(eV) W(J) __________ e
electric ield strength: E = ___qF drit speed: I = nAvq
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E L E C T R I C I T Y AN D M AG N E T I S M
Charge and feld Simple beginnings Take a plastic comb and pull it through your hair. Aterwards, the comb may be able to pick up small pieces o paper. Look closely and you may see the paper being thrown o shortly ater touching the comb. S imilar observations were made early in the history o science. The discovery that obj ects can be charged by riction ( you were doing this when you drew the comb through your hair) is attributed to the Greek scientist Thales who lived about 2 600 years ago. In those days, silk was spun on amber spindles and as the amber rotated and rubbed against its bearings, the silk was attracted to the amber. The ancient Greek word or amber is ( electron) . In the 1 700s, du Fay ound that both conductors and insulators could be electrifed ( the term used then) and that there were two opposite kinds o electrifcation . However, he was unable to provide any explanation or these eects. Gradually, scientists developed the idea that there were two separate types o charge: positive and negative. The American physicist, B enj amin Franklin, carrying out a amous series o experiments ying kites during thunderstorms, named the charge on a glass rod rubbed with silk as positive electricity . The charge on materials similar to ebonite ( a very hard orm o rubber) rubbed with animal ur was called negative . O ne o Franklins other discoveries was that a charged conducting sphere has no electric feld inside it ( the feld and the charges always being outside the sphere) . Joseph Priestley was able to deduce rom this that the orce between two charges is inversely proportional to the square o the distance between the charges. At the end o the nineteenth century J. J. Thomson detected the presence o a small particle that he called the electron. Experiments showed that all electrons have the same small quantity o charge and that electrons are present in all atoms. Atoms were ound to have protons that have the same magnitude o electronic charge as, but have opposite charge to, the electron. In a neutral atom or material the number o electrons and the number o protons are equal. We now assign a negative charge to the electron and positive to the proton. There are only these two species o charge.
Explaining electrostatics E xperiments also show that positively charged obj ects are attracted to negatively charged obj ects but repelled by any other positively charged obj ect. The possible cases are summed up in fgure 1 . There can also be an attraction between charged and uncharged obj ects due to the separation o charge in the uncharged obj ect. In fgure 1 ( c) the electrons in the uncharged sphere are attracted to the positively charged sphere ( sphere A) and move towards it. The electrons in sphere B are now closer to the positives in sphere A than the fxed positive charges on B . S o the overall orce is towards sphere A as the orce between two charges increases as the distance between them decreases.
170
5.1 ELECTRI C FI ELD S +
+ +
+
+
+
+
+
+
+
+
+
+ +
+ +
+
opposite charges attract (a)
like charges repel (b) +
+ + +
+
+
+
+
+
+
+
-
+
neutral
+ +
+
charge separation means that attraction occurs (c)
Figure 1 Attractions between charges.
We now know that the simple electrostatic eects early scientists observed are due only to the movement o the negatively charged electrons. An object with no observed charge has an exact balance between the electrons and the positively charged protons; it is said to be neutral. Some electrons in conducting materials are loosely attached to their respective atoms and can leave the atoms to move rom one object to another. This leaves the object that lost electrons with an overall positive charge. The electrons transerred give the second object an overall negative charge. Notice that electrons are not lost in these transers. I 1 000 electrons are removed rom a rod when the rod is charged by a cloth, the cloth will be let with 1 000 extra electrons at the end o the process. Charge is conserved; the law o conservation o charge states that in a closed system the amount o charge is constant. When explaining the eects described here, always use the idea o surplus o electrons or a negative charge, and describe positive charge in terms o a lack ( or defcit) o electrons. Figure 2 shows how an experiment to charge a metal sphere by induction is explained.
+ + +
+ + + + + +
connected + to Earth + +
+ +
+ +
+
+
conducting sphere insulating stand
6 electrons travel to Earth
charged rod separates charge
remove charged rod
sphere is earthed and electrons are repelled to Earth
Earth connection broken, rod removed, and electrons re-distributed to leave sphere positive after rearrangement
TOK Inverse-square laws Forces between charged objects is one o several examples o inverse-square laws that you meet in this course. They are o great importance in physics. Inverse-square laws model a characteristic property o some felds, which is that as distance doubles, observed eects go down by one quarter. Mathematics helps you to learn and conceptualize your ideas about the subject. When you have learnt the physics o one situation (here, electrostatics) then you will be able to apply the same rules to new situations (or example, gravitation, in the next topic) . Is the idea o feld a human construct or does it reect the reality o the universe?
Figure 2 Charging by induction.
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E L E C T R I C I T Y AN D M AG N E T I S M
Measuring and defning charge The unit o charge is the coulomb ( abbreviated to C ) . C harge is a scalar quantity. The coulomb is defned as the charge transp orted by a current o one amp ere in one second. Measurements show that all electrons are identical, with each one having a charge equal to 1 .6 1 0 1 9 C ; this undamental amount o charge is known as the electronic (or elementary) charge and given the symbol e. C harges smaller than the electronic charge are not observed in nature. 1 2 ( Quarks have ractional charges that appear as __ e or __ e; however, 3 3 they are never observed outside their nucleons.) In terms o the experiments described here, the coulomb is a very large unit. When a comb runs through your hair, there might be a charge o somewhere between 1 pC and 1 nC transerred to it.
Forces between charged objects In 1 785 , C oulomb published the frst o several Mmoires in which he reported a series o experiments that he had carried out to investigate the eects o orces arising rom charges. He ound, experimentally, that the orce between two point charges 1 separated by distance r is proportional to __ thus confrming the earlier r theory o Priestley. Such a relationship is known as an inverse-square law. 2
Investigate! Forces between charges
These are sensitive experiments that need care and a dry atmosphere to achieve a result.
Take two small polystyrene spheres and paint them with a metal paint or colloidal graphite, or cover with aluminum oil. S uspend one rom an insulating rod using an insulating ( perhaps nylon) thread. Mount the other on top o a sensitive top- pan balance, again using an insulating rod.
172
insulating rod
+
insulating support
C harge both spheres by induction when they are apart rom each other. An alternative charging method is to use a laboratory high voltage power supply. Take care, your teacher will want to give you instructions about this. B ring the spheres together as shown in fgure 3 and observe changes in the reading on the balance.
sensitive top-pan balance
Figure 3
5.1 ELECTRI C FI ELD S
The distance d moved by the sphere depends on the orce between the charged spheres. The distance r is the distance between the centres o the spheres.
Vary d and r making careul measurements o them both.
d sideways force on ball r = distance between balls
1 Plot a graph o d against __ . An experiment r perormed with care can give a straight- line graph. 2
d r
Figure 4
Another method is to bring both charged spheres together as shown in fgure 4.
For small deections, d is a measure o the orce between the spheres ( the larger the orce the greater the distance that the sphere is moved) whereas r is the distance between sphere centres.
Nature of science Scientists in C oulombs day published their work in a very dierent way rom scientists today. C oulomb wrote his results in a series o books called Mmoires. Part o C oulombs original Mmoire in which he states the result is shown in fgure 5 .
Figure 5
Later experiments confrmed that the orce is proportional to the product o the size o the point charges q 1 and q 2 . C ombining C oulombs results together with these gives q1 q2 F _ r2 where the symbol means is proportional to. The magnitude o the orce F between two point charges o charge q 1 and q 2 separated by distance r in a vacuum is given by kq 1 q 2 F= _ r2 where k is the constant o proportionality and is known as C oulombs constant. In act we do not always quote the law in quite this mathematical orm. The constant is requently quoted dierently as 1 k=_ 4 0 The new constant 0 is called the permittivity o ree space ( ree space is an older term or vacuum . The 4 is added to rationalize electric and
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5
E L E C T R I C I T Y AN D M AG N E T I S M magnetic equations in other words, to give them a similar shape and to retain an important relationship between them ( see the TO K section on page 1 77) . S o the equation becomes 1 q1 q2 F= __ 4 0 r2 When using charge measured in coulombs and distance measured in metres, the value o k is 9 1 0 9 N m 2 C 2 . This means that 0 takes a value o 8.85 4 1 0 1 2 C 2 N 1 m 2 or, in undamental units, m 3 kg 1 s 4 A 2 . The equation as it stands applies only or charges that are in a vacuum. I the charges are immersed in a dierent medium ( say, air or water) then the value o the permittivity is dierent. It is usual to 1 as the 0 subscript amend the equation slightly too, k becomes ___ 4 in 0 should only be used or the vacuum case. For example, the permittivity o water is 7. 8 1 0 1 0 C 2 N 1 m 2 and the permittivity o air is 8. 85 49 1 0 1 2 C 2 N 1 m 2 . The value or air is so close to the ree- space value that we normally use 8. 85 1 0 1 2 C 2 N 1 m 2 or both. The table gives a number o permittivity values or dierent materials.
Material paper rubber water graphite diamond
Permittivity / 10 12 C2 N 1 m 2 34 62 779 106 71
This equation appears to say nothing about the direction o the orce between the charged obj ects. Forces are vectors, but charge and ( distance) 2 are scalars. There are mathematical ways to cope with this, but or point charges the equation gives an excellent clue when the signs o the charges are included. force on A due to B
charge A
charge B
force on B due to A
r positive direction
Figure 6 Force directions.
We take the positive direction to be rom charge A to charge B ; in fgure 6 that is rom let to right. Lets begin with both charge A and charge B being positive. When two positive charges are multiplied q q together in ____ , the resulting sign o the orce acting on charge B due to r charge A is also positive. This means that the direction o the orce will be assigned the positive direction ( rom charge A to charge B ) : in other words, let to right. This agrees with the physics because the charges are repelled. I both charges are negative then the answer is the same, the charges are repelled and the orce is to the right. 1
2
2
174
5.1 ELECTRI C FI ELD S
I, however, one o the charges is positive and the other negative, then the product o the charges is negative and the orce direction will be opposite to the let- to- right positive direction. S o the orce on charge 2 is now to the let. Again, this agrees with what we expect, that the charges attract because they have opposite signs.
Worked examples 1
Two point charges o + 1 0 nC and 1 0 nC in air are separated by a distance o 1 5 mm.
The charges are attracted along the line j oining them. ( D o not orget that orce is a vector and needs both magnitude and direction or a complete answer.)
a) C alculate the orce acting between the two charges. b) C omment on whether this orce can lit a small piece o paper about 2 mm 2 mm in area.
Solution a) It is important to take great care with the prefxes and the powers o ten in electrostatic calculations. The charges are: +1 0 1 0 9 C and 1 0 1 0 9 C. The separation distance is 1 .5 1 0 2 m (notice how the distance is converted right at the outset into consistent units) . 8
8
(+1 1 0 ) (1 1 0 ) So F = ___ 4 0 (1 .5 1 0 2 ) 2 = 4.0 1 0 3 N
b) A sheet o thin A4 paper o dimensions 2 1 0 mm by 2 97 mm has a mass o about 2 g. S o the small area o paper has a mass o about 1 .3 1 0 7 kg and thereore a weight o 1 .3 1 0 6 N. The electrostatic orce could lit this paper easily. 2
Two point charges o magnitude + 5 C and + 3 C are 1 .5 m apart in a liquid that has a permittivity o 2 .3 1 0 1 1 C 2 N 1 m 2 . C alculate the orce between the point charges.
Solution 6
6
(+5 1 0 ) (+3 1 0 ) F = __________________ = 2 3 mN; 4 2 . 3 1 0 ( 1 . 5 ) 1 1
2
a repulsive orce acting along the line j oining the charges.
Electric felds Sometimes the origin o a orce between two objects is obvious, an example is the riction pad in a brake rubbing on the rim o a bicycle wheel to slow the cycle down. In other cases there is no physical contact between two objects yet a orce exists between them. Examples o this include the magnetic orce between two magnets and the electrostatic orce between two charged objects. Such orces are said to act at a distance. The term feld is used in physics or cases where two separated obj ects exert orces on each other. We say that in the case o the comb picking up the paper, the paper is sitting in the electric feld due to the comb. The concept o the feld is an extremely powerul one in physics not least because there are many ideas common to all felds. As well as the magnetic and electrostatic felds already mentioned, gravity felds also obey the same rules. Learn the underlying ideas or one type o feld and you have learnt them all.
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Mapping felds
Investigate! Plotting electric felds the method) . This experiment allows patterns to be observed or electric felds.
+
Put some castor oil in a Petri dish and sprinkle some grains o semolina ( or grits) onto the oil. Alternatives or the semolina include grass seed and hairs cut about 1 mm long rom an artist paint brush.
Take two copper wires and bend one o them to orm a circle just a little smaller than the internal diameter o the Petri dish. Place the end o the other wire in the centre o the Petri dish.
C onnect a 5 kV power supply to the wires take care with the power supply!
O bserve the grains slowly lining up in the electric feld.
S ketch the pattern o the grains that is produced.
Repeat with other wire shapes such as the our shown in fgure 7( c) .
(a)
semolina castor oil
(b)
(c)
Figure 7 At an earlier stage in your school career you may have plotted magnetic feld patterns using iron flings ( i you have not done this there is an Investigate! in S ub- topic 5 .4 to illustrate
In the plotting experiment, the grains line up in the feld that is produced between the wires. The patterns observed resemble those in fgure 8. The experiment cannot easily show the patterns or charges with the same sign.
+ + +
+ + + + + + + +
+
176
+
The idea o feld lines was frst introduced by Michael Faraday ( his original idea was o an elastic tube that repelled other tubes) . Although feld lines are imaginary, they are useul or illustrating and understanding the nature o a particular feld. There are some conventions or drawing these electric feld patterns.
+
Figure 8 Electric feld patterns.
The lines start and end on charges o opposite sign.
An arrow is essential to show the direction in which a positive charge would move ( i. e. away rom the positive charge and towards the negative charge) .
Where the feld is strong the lines are close together. The lines act to repel each other.
The lines never cross.
The lines meet a conducting surace at 90.
5.1 ELECTRI C FI ELD S
Electric feld strength
TOK
As well as understanding the feld pattern, we need to be able to measure the strength o the electric feld. The electric feld strength is defned using the concept o a p ositive test charge. Imagine an isolated charge Q sitting in space. We wish to know what the strength o the feld is at a point P, a distance r away rom the isolated charge. We put another charge, a positive test charge o size q, at P and measure the orce F that acts on the test charge due to Q. Then the magnitude o the electric feld strength is defned to be F E= _ q
So why not use k? James Maxwell, working in the middle o the nineteenth century, realized that there was an important connection between electricity, magnetism and the speed o light. In particular he was able to show that the permittivity o ree space 0 (which relates to electrostatics) and the permeability o ree space 0 (which relates to electromagnetism) are themselves connected to the speed o light c:
electric orce +q r
test charge, P
+Q
________ 1 2 = c 0
Figure 9 Defnition o electric feld strength.
0
This proves to be an important equation. So much so that, in the set o equations that arise rom the SI units we use, we choose to use 0 in all the electrostatics equations and 0 in all the magnetic equations.
The units o electric feld strength are N C 1 . ( Alternative units, that have the same meaning, are V m 1 and will be discussed in Topic 1 1 .) E lectric feld strength is a vector, it has the same direction as the orce F ( this is because the charge is a scalar which only scales the value o F up or down) . A ormal defnition or electric feld strength at a p oint is the orce p er unit charge exp erienced by a small p ositive p oint charge p laced at that p oint.
However, not all unit systems choose to do this. There is another common system, the cgs system (based on the centimetre, the gram and the second, rather than the metre, kilogram and second) . In cgs, the value o the constant k in Coulombs law is chosen to be 1 and the equation q q . I the appears as F = ________ r numbers are dierent, is the physics the same?
C oulombs law can be used to fnd how the electric feld strength varies with distance or a point charge. Q is the isolated point charge and q is the test charge, so 1 Qq F= __ 4 0 r2 F E=_ q Thereore 1 Qq 1 _ E= __ q 4 0 r2
1
so
2
2
1 Q E = _ _2 4 0 r The electric feld strength o the charge at a point is proportional to the charge and inversely proportional to the square o the distance rom the charge. I Q is a positive charge then E is also positive. Applying the rule that r is measured rom the charge to the test charge, then i E is positive it acts outwards away rom the charge. This is what we expect as both charge and test charge are positive. When Q is negative, E acts towards the charge Q. The feld shape or a point charge is known as a radial feld. The feld lines radiate away ( positive) or towards ( negative) the point charge as shown in fgure 1 0.
+
Q
Figure 10 Radial felds or positive and negative point charges.
Q
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Worked examples 1
8. 5 1 0 4 = 1 .2 1 0 6 N C 1 b) E = __ 4 0 ( 2 .5 ) 2
C alculate the electric feld strengths in a vacuum
The feld direction is towards the negative charge.
a) 1 .5 cm rom a + 1 0 C charge 2
b) 2 .5 m rom a 0.85 mC charge
Solution a) B egin by putting the quantities into consistent units: r = 1 .5 1 0 2 m and Q = 1 . 0 1 0 5 C .
An oxygen nucleus has a charge o + 8e. C alculate the electric feld strength at a distance o 0.68 nm rom the nucleus.
Solution
1 .0 1 0 = + 4. 0 1 0 8 N C 1 . Then E = ____________ 4 ( 1 . 5 1 0 )
The charge on the oxygen nucleus is 8 1 .6 1 0 1 9 C ; the distance is 6.8 1 0 1 0 m.
The feld direction is away rom the positive charge.
1 .3 1 0 E = _______________ = 2.5 1 0 1 0 N C 1 away rom 4 ( 6 . 8 1 0 )
5
2 2
0
1 8
1 0 2
0
the nucleus. The electric feld strengths can be added using either a calculation or a scale diagram as outlined in Topic 1 . feld due to -q
+q feld
-q feld
-q
+Q
gives test charge
net electric feld
feld due to +Q
feld due to 1 and 2 feld due to 1
+2Q charge 3 feld due to 2 test charge
charge 2 +Q
gives
gives
=
net electric feld
charge 1 +Q feld due to 3
feld due to 3
Figure 11 Vector addition o electric felds.
This vector addition o feld strengths ( fgure 1 1 ) can give us an insight into electric felds that arise rom charge confgurations that are more complex than a single point charge.
Close to a conductor Imagine going very close to the surace o a conductor. Figure 1 2 shows what you might see. First o all, i we are close enough then the surace will appear at ( in j ust the same way that we are not aware o the curvature o the Earth until we go up in an aircrat) . S econdly we would see that all the ree electrons are equally spaced. There is a good reason or this: any one electron has orces acting on it rom the other electrons. The electron will accelerate until all these orces balance out and it is in equilibrium, or this to happen they must be equally spaced.
178
Now look at the feld strength vectors radiating out rom these individual electrons. Parallel to the surace, these all cancel out with each other so there is no electric feld in this direction. ( This is the same as saying that any one electron will not accelerate as the horizontal feld strength is
5.1 ELECTRI C FI ELD S
TOK
surace
But does the test charge aect the original feld? I we wanted to use the defnition o the strength o an electric feld practically by using a test charge, we would have to take care. Just as a thermometer alters the temperature o the object it measures, so the test charge will exert a orce on the original charge (the one with the feld we are trying to measure) and may accelerate it, or disturb the feld lines. This means in practice that a test charge is really an imaginary construct that we use to help our understanding. In practice we preer to measure electric feld strengths using the idea that it is the potential gradient, in other words the change in the voltage divided by the change in distance: the larger this value, the greater the feld strength. We will leave urther discussion o this point to Topic 11 but it is ood or thought rom a theory o knowledge perspective: a measurement that we can think about but not, in practice, carry out. The German language has a name or it: gedankenexperiment thought experiment. How can a practical subject such as a science have such a thing?
perpendicular to surace add
surace
parallel to surace cancel
leaving electric feld only perpendicular to surace
Figure 12 Close to conducting suraces.
zero) . Perpendicular to the surace, however, things are dierent. Now the feld vectors all add up, and because there is no feld component parallel to the surace, the local feld must act at 90 to it. S o, close to a conducting surace, the electric feld is at 90 to the surace.
Conducting sphere This can be taken a step urther or a conducting sphere, whether it is hollow or solid. Again, the ree electrons at the surace are equally spaced and all the feld lines at the surace o the sphere are at 90 to it. The consequence is that the feld must be radial, just like the feld o an isolated point charge. So, to a test charge outside the sphere, the feld o the sphere appears exactly the same as that o the point charge. Mathematical analysis shows that outside a sphere the feld indeed behaves as though it came rom a point charge placed at the centre o the sphere with a charge equal to the total charge spread over the sphere. ( Inside the sphere is a dierent matter, it turns out that there is no electric feld inside a sphere, hollow or solid, a result that was experimentally determined by Franklin.)
Worked example Two point charges, a +25 nC charge X and a +1 5 nC charge Y are separated by a distance o 0.5 m. a) C alculate the resultant electric feld strength at the midpoint between the charges.
c) Calculate the magnitude o the electric feld strength at the point P on the diagram. X and Y are 0.4 m and 0.3 m rom P respectively. P
b) C alculate the distance rom X at which the electric feld strength is zero.
0.4 m X
0.3 m Y
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2 .5 1 0 c) PX = 0.4 m so EX at P is ________ 4 0 . 4 8
Solution
2
0
2 .5 1 0 8 a) EA = __2 = 3 600 N C 1 4 0 0. 2 5
= 1 400 N C 1 along XP in the direction away rom X.
1 .5 1 0 8 EB = __2 = 2 2 00 N C 1 4 0 0. 2 5
1 .5 1 0 PY = 0.3 m so EY at P is ________ 4 0 . 3 8
2
The feld strengths act in opposite directions, so the net electric feld strength is ( 3 600 2 2 00) = 1 400 N C 1 ; this is directed away rom X towards Y.
0
= 1 5 00 N C 1 along PY in the direction towards Y.
b) For E to be zero, EA = EB and so 2 .5 1 0 8 1 . 5 1 0 8 _ = __2 2 4 0 d 4 0 ( 0. 5 d)
P net e
thus
le c t ric
f e ld
2
d 2 .5 _ = _ 1 .5 ( 0.5 d) 2 or d _ = ( 0.5 d)
X
____
2 .5 = 1 .3 _ 1 .5
Y
d = 0.65 1 .3 d The magnitude o the resultant electric feld ___________ strength is 1 400 2 + 1 5 00 2 = 2 1 00 N C 1
2 .3 d = 0.65 d = 0.2 8 m
( The calculation o the angles was not required in the question and is let or the reader.)
Moving charge Investigate! Moving charge around pong) ball with a painted metal surace rom a long insulating thread so that it is midway between the plates.
insulating thread metal plate
C onnect the plates to a high- voltage power supply with a sensitive ammeter or light- beam galvanometer in the circuit.
When the supply is turned on, the ball should shuttle between the plates (it may need to be kickstarted by making it touch one o the plates) .
Notice what happens on the scale o the galvanometer. E very time the ball touches a plate there is a deection on the meter. Look at the direction and size o the deections do they vary?
insulating handle coated ball
A ammeter
+
high voltage (approx. 5 kV) supply
180
Figure 13 Arrange two metal plates vertically and about 2 0 cm apart. S uspend a table tennis ( ping-
5.1 ELECTRI C FI ELD S
In the previous Investigate! the power supply is connected to the plates with conducting leads. Electrons move easily along these leads. When the supply is turned on, the electrons soon distribute themselves so that the plate connected to the negative supply has surplus electrons, and the other plate has a defcit o electrons, becoming positive. As the ball touches one o the plates it loses or gains some o the electrons. The charge gained by the ball will have the same sign as the plate and as a result the ball is almost immediately repelled. A orce acts on the ball because it is in the electric feld that is acting between the plates. The ball accelerates towards the other plate where it transers all its charge to the new plate and gains more charge. This time the charge gained has the sign o the new plate. The process repeats itsel with the ball transerring charge rom plate to plate. The meter is a sensitive ammeter, so when it deects it shows that there is current in the wires leading to the plates. C harge is moving along these wires, so this is evidence that:
an electric current results when charge moves
the charge is moved by the presence o an electric feld.
A mechanism for electric current The shuttling ball and its charge show clearly what is moving in the space between the plates. However, the microscopic mechanisms that are operating in wires and cells are not so obvious. This was one o the maj or historical problems in explaining the physics o electricity. E lectrical conduction is possible in gases, liquids, solids, and a vacuum. O particular importance to us is the electrical conduction that takes place in metals.
Conduction in metals The metal atoms in a solid are bound together by the metallic bond. The ull details o the bonding are complex, but a simple model o what happens is as ollows. When a metal solidifes rom a liquid, its atoms orm a regular lattice arrangement. The shape o the lattice varies rom metal to metal but the common eature o metals is that as the bonding happens, electrons are donated rom the outer shells o the atoms to a common sea o electrons that occupies the entire volume o the metal. positive ions +
+ + +
electrons entering metal
+ +
+
+
+ +
+
+
electrons leaving metal
+ metal rod
Figure 14 Conduction by free electrons in a metal.
Figure 1 4 shows the model. The positive ions sit in fxed positions on the lattice. There are ions at each lattice site because each atom has now lost an electron. O course, at all temperatures above absolute zero
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E L E C T R I C I T Y AN D M AG N E T I S M they vibrate in these positions. Most o the electrons are still bound to them but around the ions is the sea o ree electrons or conduction electrons; these are responsible or the electrical conduction. Although the conduction electrons have been released rom the atoms, this does not mean that there is no interaction between ions and electrons. The electrons interact with the vibrating ions and transer their kinetic energy to them. It is this transer o energy rom electrons to ions that accounts or the phenomenon that we will call resistance in the next sub-topic. The energy transer in a conductor arises as ollows: electric feld
low potential ()
(+) high potential
drit direction
Figure 15
In a metal in the absence o an electric feld, the ree electrons are moving and interacting with the ions in the lattice, but they do so at random and at average speeds close to the speed o sound in the material. Nothing in the material makes an electron move in any particular direction. However, when an electric feld is present, then an electric orce will act on the electrons with their negative charge. The defnition o electric feld direction reminds us that the electric feld is the direction in which a positive charge moves, so the orce on the electrons will be in the opposite direction to the electric feld in the metal ( fgure 1 5 ) . In the presence o an electric feld, the negatively charged electrons drit along the conductor. The electrons are known as charge carriers. Their movement is like the random motion o a colony o ants carried along a moving walkway.
Conduction in gases and liquids Electrical conduction is possible in other materials too. S ome gases and liquids contain ree ions as a consequence o their chemistry. When an electric feld is applied to these materials the ions will move, positive in the direction o the feld, negative the opposite way. When this happens an electric current is observed. I the ele ctric ield is strong e nough it can, itsel, lead to the creation o ions in a gas or liquid. This is known as electrical b reakd o w n. It is a common eect during electrical storms when lightning moves be twe en a charge d cloud and the E arth. You will have see n such conduction in neon display tub es or luorescent tub es use or lighting.
182
5.1 ELECTRI C FI ELD S
Nature of science Models o conduction The model here is o a simple ow o ree electrons through a solid, a liquid, or a gas. B ut this is not the end o the story. There are other, more sophisticated models o conduction in solids that can explain the dierences between conductors, semiconductors and insulators better than the ow model here. These involve the electronic band theory which arises rom the interactions between the electrons within individual atoms and between the atoms themselves. Essentially, this band model proposes that electrons have to adopt dierent energies within the substance and that some groups o energy levels ( called band gaps) are not permitted to the electrons. Where there are wide band gaps,
electrons cannot easily move rom one set o levels to another and this makes the substance an insulator. Where the band gap is narrow, adding energy to the atomic structure allows electrons to j ump across the band gap and conduct more reely this makes a semiconductor, and you will later see that one o the semiconductor properties is that adding internal energy allows them to conduct better. In conductors the band gap is o less relevance because the electrons have many available energy states and so conduction happens very easily indeed. Full details o this theory are beyond IB Physics, but i you have an interest in taking this urther, you can fnd many reerences to the theory on the Internet.
Electric current When charge ows in a conductor we say that there is an electric current in the conductor. C urrent is measured in amp res, the symbol or the unit is A. O ten, in the E nglish speaking world, the accent is omitted. C urrent is linked to ow o charge in a simple way. point P
Tip It is not good practice to write or say that current fows, what is fowing in the circuit is the electric charge. The movement o this charge is what we call current and it is best to write that there is a current in the circuit, or in a component as appropriate.
one coulomb o electron charge
Figure 16 Charge fow leading to current.
Imagine a block o electrons with a total charge o one coulomb moving along a conductor. An observer at point P is watching these electrons move along the conductor. I all the electrons in the block move past the point in one second then, the current is one ampere. I it takes twice as long ( 2 s) or the block to pass, then the current is hal and is 0.5 A. I the block takes 0.1 s to pass the observer, then the current is 1 0 A. Mathematically total charge that moved past a point electric current, I = ____ time taken for charge to move past the point
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5
E L E C T R I C I T Y AN D M AG N E T I S M or in symbols Q I= _ t The ampere is a fundamental unit defned as part o the S I. Although it is explained here in terms o the ow o charge, this is not how it is defned. The SI defnition is based on ideas rom magnetism and is covered in Sub- topics 1 .1 and 5 .4.
Nature of science Another physics link This link between owing charge and current is a crucial one. E lectrical current is a macroscopic quantity, transer o charge by electrons is a microscopic phenomenon in every sense o the word. This is another example o a link in physics between macroscopic observations and inerences about what is happening on the smallest scales.
It was the lack o knowledge o what happens inside conductors at the atomic scale that orced scientists, up to the end o the nineteenth century, to develop concepts such as current and feld to explain the eects they observed. It also, as we shall see, led to a crucial mistake.
Worked examples 1
In the shuttling ball experiment, the ball moves between the two charged plates at a requency o 0. 67 Hz. The ball carries a charge o magnitude 72 nC each time it crosses rom one plate to the other.
b) The charge transerred is 72 nC = 7.2 1 0 8 C Each electron has a charge o 1 . 6 1 0 1 9 C , so the number o electrons involved in the 7.2 1 0 transer is ________ = 4.5 1 0 1 1 1 .6 1 0 8
1 9
2
C alculate: a) the average current in the circuit b) the number o electrons transerred each time the ball touches one o the plates.
a) C alculate the current in a wire through which a charge o 2 5 C passes in 1 5 00 s. b) The current in a wire is 3 6 mA. C alculate the charge that ows along the wire in one minute.
Solution
Solution
a) The time between the ball being at the same plate 1 1 = __ = ____ = 1 . 5 s. The time to transer 72 nC 0.67 is thereore 0.75 s.
Q 25 a) I = ___ , so the current = ____ = 1 7 mA 1 5 00 t
7.2 1 0 C urrent = _______ = 96 nA 0.75 8
b) Q = I t and t = 60 s. Thus charge that ows = 3 .6 1 0 2 60 = 2 .2 C
Charge carrier drift speed Turn on a lighting circuit at home and the lamp lights almost immediately. D oes this give us a clue to the speed at which the electrons in the wires move? In the Investigate! experiment on page 1 85 , the stain indicating the position o the ions moves at no more than a ew millimetres per second. The lower the value o the current, the slower the rate at which the ions move. This slow speed at which the ions move along the conductor is known as the drift sp eed. We need a mathematical model to confrm this observation.
184
5.1 ELECTRI C FI ELD S
Imagine a cylindrical conductor that is carrying an electric current I. The cross- sectional area o the conductor is A and it contains charge carriers each with charge q. We assume that each carrier has a speed v and that there are n charge carriers in 1 m 3 o conductor this quantity is known as the charge density. length o volume swept out in one second v cross-sectional area, A
n charge carriers per unit volume
P charge carrier q
Figure 18 A model or conduction.
Figure 1 8 shows charge carriers, each o charge q, moving past point P at a speed v. In one second, a volume Av o charge carriers passes P. The total number o charge carriers in this volume is nAv and thereore the total charge in the volume is nAvq. However, this is the charge that passes point P in one second, which is what we mean by the electric current. So I = nAvq
Investigate! potassium manganate( VII ) crocodile clip crystal
clips should be attached to leads that are connected to a low voltage power supply ( no more than 2 5 V is required in this experiment) .
+
Wet the flter paper with aqueous ammonia solution.
Take a small crystal o potassium manganate( VII) and place it in the centre o the flter paper. Ensure that the slide is horizontal.
Turn on the current and watch the crystal. You should see a stain on the paper moving away rom the crystal.
Reverse the current direction to check that the eect is not due to the slide not being horizontal.
How ast is the stain moving?
flter paper soaked in microscope slide aqueous ammonia solution
Figure 17 The speeds with which electrons move in a metal conductor during conduction are difcult to observe, but the progress o conducting ions in a liquid can be inerred by the trace they leave.
You should wear eye protection during this experiment.
Fold a piece o flter paper around a microscope slide and fx it with two crocodile clips at the ends o the slide. The crocodile
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E L E C T R I C I T Y AN D M AG N E T I S M
Worked example
Solution
1
Rearranging the equation I v= _ nAQ 2 0.65 1 0 and the area A o the wire is ( ________ ) 2 = 3 .3 1 0 7 m 2 .
A copper wire o diameter 0.65 mm carries a current o 0.2 5 A. There are 8.5 1 0 2 8 charge carriers in each cubic metre o copper; the charge on each charge carrier ( electron) is 1 .6 1 0 1 9 C . C alculate the drit speed o the charge carriers.
3
0.25 So v = _________________________ 8.5 1 0 3 .3 1 0 1 .6 1 0 28
7
1 9
= 0. 05 5 mm s 1
The example shows that the drit speed o each charge is less than onetenth o a millimetre each second. You may well be surprised by this result but it is probably o the same order as the speed you observed in the experiment with the potassium manganate( VII) . Although the electron charge is very small, the speed can also be small because there are very large numbers o ree electrons available or conduction in the metal. To see how sensitive the drit speed is to changes in the charge carrier density, we can compare the drit speed in copper with the drit speed in a semiconductor called germanium. The number o charge carriers in one cubic metre o germanium is about 1 0 9 less than in copper. So to sustain the same current in a germanium sample would require a drit speed 1 0 9 times greater than in the copper or a cross-sectional area 1 0 9 as large. The slow drit speed in conductors or substantial currents poses the question o how a lamp can turn on when there may be a signifcant run o cable between switch and lamp. The charge carriers in the cable are driting slowly around the cable. However, the inormation that the charge carriers are to begin to move when the switch is closed travels much more quickly close to the speed o light in act. The inormation is transerred when an electromagnetic wave propagates around the cable and produces a drit in all the ree electrons virtually simultaneously. So the lamp can turn on almost instantaneously, even though, or direct current, it may take an individual electron many minutes or even hours to reach the lamp itsel.
Potential diference Free electrons move in a conductor when an electric feld acts on the conductor. Later we shall see how devices such as electric cells and power supplies provide this electric feld. At the same time the power supplies transer energy to the electrons. As the electrons move through the conductors, they collide with the positive ions in the lattice and transer the energy gained rom the feld to the ions. In situations where felds act, physicists use two quantities called potential and potential difference when dealing with energy transers. Potential dierence (oten abbreviated to pd) is a measure o the electrical potential energy transerred rom an electron when it is moving between two points in a circuit. However, given the very small amount o charge possessed by each electron this amount o energy is also very small. It is better to use the much larger quantity represented by one coulomb o charge.
186
5.1 ELECTRI C FI ELD S
Potential dierence between two points is defned as the work done (energy transerred) W when one unit o charge Q moves between the points. W potential dierence = _ Q The symbol given to potential dierence is V; its unit is the J C 1 and is named the volt (symbol: V) ater the Italian scientist Alessandro Volta who was born in the middle o the eighteenth century and who worked on the development o electricity. The potential dierence between two points is one volt i one joule o energy is transerred per coulomb o charge passing between the two points. A simple circuit will illustrate these ideas: An electric cell is connected to a lamp via a switch and three leads. Figure 1 9 shows a picture o the circuit as it would look set up on the bench. +
lead
cell
conventional current
electronic current
electronic current
conventional current
lamp
Figure 19 Conventional and electronic current in a circuit.
When the switch is closed, electrons ow round the circuit. Notice the direction in which the electrons move and also that the diagram shows the direction o a conventional current. The two directions are opposite; in this case, clockwise or the electron ow and anti-clockwise or the conventional current. The reason or this dierence is explained in a later Nature o science section. You need to take care with this dierence, particularly when using some o the direction rules that are introduced later in this topic. What happens to an electron as it goes round the circuit once? The electron gains electric potential energy as it moves through the cell (this will be covered in Sub-topic 5.3) . The electron then leaves the cell and begins to move through the connecting lead. Leads are designed so that they do not require much energy transer to allow the electrons through (we say that they have a low electrical resistance) and so the potential dierence rom one end o the lead to the other is small. The electron moves through the switch which also gains little energy rom the charge carrier. Ater moving through another lead the electron reaches the lamp. This component is dierent rom others in the circuit, it is deliberately designed so that it can gain much o the electrical potential energy rom
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E L E C T R I C I T Y AN D M AG N E T I S M the electrons as they pass through it. The metal lattice in the flament gains energy and as a result the ions vibrate at greater speeds and with greater amplitudes. At these high temperatures the flament in the lamp will glow brightly; the lamp will be lit. In potential dierence terms, the pds across the leads and the switch are small because the passage o one coulomb o charge through them will not result in much energy transer to the lattice. The pd across the lamp will be large because, or each coulomb going through it, large amounts o energy are transerred rom electrons to the lattice ions in the flament raising its temperature.
Nature of science Conventional and electron currents In early studies o current electricity, the idea emerged that there was a ow o electrical uid in wires and that this ow was responsible or the observed eects o electricity. At frst the suggestion was that there were two types o uid known as vitreous and resinous. B enjamin Franklin (the same man who helped drat the US D eclaration o Independence) proposed that there was only one uid but that it behaved dierently depending on the circumstances. He was also the frst scientist to use the terms positive and negative. What then happened was that scientists assigned a positive charge to the uid thought to be moving in the wires. This positive charge was said to ow out o the positive terminal o a power supply ( because the charge was repelled) and went around the circuit re- entering the power
supply through the negative terminal. This is what we now term the conventional current. In act, we now know that in a metal the charge carriers are electrons and that they move in the opposite direction, leaving a power supply at the negative terminal. This is termed the electronic current. You should take care with these two currents and not conuse them. You may ask: why do we now not simply drop the conventional current and talk only about the electronic current? The answer is that other rules in electricity and magnetism were set up on the assumption that charge carriers are positive. All these rules would need to be reversed to take account o our later knowledge. It is better to leave things as they are.
Worked examples 1
A high efciency LED lamp is lit or 2 hours. C alculate the energy transer to the lamp when the pd across it is 2 40 V and the current in it is 5 0 mA.
Solution 2 hours is 2 60 60 = 72 00 s. The charge transerred is It = 7.2 1 0 3 5 0 1 0 3 = 3 60 C Work done = charge pd = 3 60 2 40 = 86 400 J 2
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A cell has a terminal voltage o 1 .5 V and can deliver a charge o 460 C beore it becomes discharged.
a)
C alculate the maximum energy the cell can deliver.
b) The current in the cell never exceeds 5 mA. Estimate the lietime o the cell.
Solution W a) Potential dierence, V = __ q so W = qV = 460 1 .5 = 690 J
b) The current o 5 mA means that no more than 5 mC ows through the cell at any time. 46 0 S o _____ = 92 000 s ( which is about 2 5 hours) 0.005
5.1 ELECTRI C FI ELD S
Electromotive force (emf) Another important term used in electric theory is electromotive force ( usually written as em or brevity) . This term seems to imply that there is a orce involved in the movement o charge, but the real meaning o em is connected to the energy changes in the circuit. When charge ows electrical energy can go into another orm such as internal energy ( through the heating, or Joule, eect) , or it can be converted from another orm ( or example, light ( radiant energy) in solar ( photovoltaic) cells) . The term em will be used in this course when energy is transerred to the electrons in, or example, a battery. ( O ther devices can also convert energy into an electrical orm. Examples include microphones and dynamos.) The term potential difference will be used when the energy is transerred from the electrical orm. So, examples o this would be electrical into heat and light, or electrical into motion energy. The table shows some o the devices that transer electrical energy and it gives the term that is most appropriate to use or each one.
Device Cell Resistor Microphone Loudspeaker Lamp Photovoltaic cell Dynamo Electric motor
chemical electrical sound converts electrical energy electrical from light kinetic electrical
electrical internal electrical sound into light (and internal) electrical electrical kinetic
pd or emf? emf pd emf pd pd emf emf pd
Power, current, and pd We can now answer the question o how much energy is delivered to a conductor by the electrons as they move through it. S uppose there is a conductor with a potential dierence V between its ends when a current I is in the conductor. In time t the charge Q that moves through the conductor is equal to It. The energy W transerred to the conductor rom the electrons is QV which is ( I t) V. S o the energy transerred in time t is W = IV t e n e rgy W The electrical power being supplied to the conductor is _____ = __ and t tim e thereore
electrical power P = IV Alternative orms o this expression that you will fnd useul are I = P and V = __ I
__P V
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E L E C T R I C I T Y AN D M AG N E T I S M The unit of p ower is the watt ( W) 1 watt ( 1 W) is the power developed when 1 J is converted in 1 s, the same in both mechanics and electricity. So another way to think o the volt is as the power transerred per unit current in a conductor.
Worked examples 1
A 3 V, 1 .5 W flament lamp is connected to a 3 V battery. C alculate:
a distance o 1 .5 m in 7 s. The motor is 40% efcient. C alculate the average current in the motor while the load is being raised.
a) the current in the lamp b) the energy transerred in 2400 s.
Solution 1 .5 P a) Electrical power, P = IV, so I = __ = ___ = 0.5 A V 3
b) The energy transerred every second is 1 .5 J so in 2 400 s, 3 600 J. 2
An electric motor that is connected to a 1 2 V supply is able to raise a 0. 1 0 kg load through
Solution The energy gained is mgh = 0.01 g 1 .5 = 0.1 47 J The power output o the motor must be 0 . 1 47 = _____ = 0.02 1 W 7 0.021 P The current in the motor = __ = _____ = 1 .8 mA. V 12 Since the motor is 40% efcient the current will be 4.5 mA.
Nature of science A word about potential The use o the term potential dierence implies that there is something called potential which can dier rom point to point. This is indeed the case.
energy; this is described as a loss o potential. Positive charges move rom points o high potential to low potential i they are ree to do so.
An isolated positive point charge will have feld lines that radiate away rom it. A small positive test charge in this feld will have a orce exerted on it in the feld line direction and, i ree to do so, will accelerate away rom the original charge. When the test charge is close, there is energy stored in the system and we say that the system ( the two charges interacting) has a high potential. When the test charge is urther away, there is still energy stored, but it is smaller because the system has converted energy into the kinetic energy o the charge it has done work on the charge. So to move the test charge away rom the original charge transers some o the original stored
Negative charges on the other hand move rom points o low potential to high. You can work this through yoursel by imagining a negative test charge near a positive original charge. This time the two charges are attracted and to move the negative charge away we have to do work on the system. This increases the potential o the system. I charges are ree they will all towards each other losing potential energy. In what orm does this energy re- appear? We shall return to a discussion o potential in Topic 1 0. From now on Topic 5 only reers to potential dierences.
The electronvolt Earlier we said that the energy possessed by individual electrons is very small. I a single electron is moved through a potential dierence o W equal to 1 5 V then as V = __ , so W = QV and the energy gained by this Q 1 9 electron is 1 5 1 . 6 1 0 J = 2 . 4 1 0 1 8 J. This is a very small amount and involves us in large negative powers o ten. It is more convenient to defne a new energy unit called the electronvolt ( symbol eV) .
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5.1 ELECTRI C FI ELD S
This is defned as the energy gained by an electron when it moves through a potential dierence o one volt. An energy o 1 eV is equivalent to 1 .6 1 0 1 9 J. The electronvolt is used extensively in nuclear and particle physics. B e careul, although the unit sounds as though it might be connected to potential dierence, like the j oule it is a unit o energy.
Worked examples 1
An electron, initially at rest, is accelerated through a potential dierence o 1 80 V. C alculate, or the electron: a) the gain in kinetic energy
Solution (a) The electron gains 1 80 eV o energy during its acceleration. 1 eV 1 .6 1 0 1 9 J so 1 80 eV 2 .9 1 0 1 7 J 1 (b) The kinetic energy o the election = __ mv2 and 2 31 the mass ___ o the electron is 9.1 1 0 kg.
_________
___ = ___________ = 8.0 1 0 2 E ke me
2 2 . 9 1 0 1 7 9 . 1 1 0 3 1
In a nuclear accelerator a proton is accelerated rom rest gaining an energy o 2 5 0 MeV. E stimate the fnal speed o the particle and comment on the result.
Solution
b) the fnal speed.
So v =
2
6
m s 1 .
The energy gained by the proton, in j oules, is 4.0 1 0 1 1 J. ___
2 E ke
_________
2 4. 0 1 0 As beore, v = ___ = ___________ , but using a m 1 .7 1 0 value or the mass o the proton this time. p
1 2
2 7
The numerical answer or v = 2 .2 1 0 8 m s 1 . This is a large speed, 70% o the speed o light. In act the speed will be less than this as some o the energy goes into increasing the mass o the proton through relativistic eects rather than into the speed o the proton.
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5.2 Heating efect o an electric current Understanding
Applications and skills
Circuit diagrams
Drawing and interpreting circuit diagrams
Kirchhos laws
Indentiying ohmic and non-ohmic
Heating eect o an electric current and its
consequences Resistance Ohms law Resistivity Power dissipation
Nature o science Peer review a process in which scientists repeat and criticize the work o other scientists is an important part o the modern scientifc method. It was not always so. The work o Ohm was neglected in England at frst as Barlow, a much-respected fgure in his day, had published contradictory material to that o Ohm. In present-day science the need or repeatability in data collection is paramount. I experiments or other fndings cannot be repeated, or i they contradict other scientists work, then a close look is paid to them beore they are generally accepted.
conductors through a consideration o the VI characteristic graph Investigating combinations o resistors in parallel and series circuits Describing ideal and non-ideal ammeters and voltmeters Describing practical uses o potential divider circuits, including the advantages o a potential divider over a series variable resistor in controlling a simple circuit Investigating one or more o the actors that aect resistivity Solving problems involving current, charge, potential dierence, Kirchhos laws, power, resistance and resistivity
Equations resistance defnition: R = ___VI
2
V electrical power: P = VI = I2 R = _______ R
combining resistors:
in series Rto ta l = R1 + R2 + R3 ... 1 1 1 1 in parallel ___________ = _______ + _______ + _______ + ... R R R R total
1
2
3
RA resisitivity defnition: = ________ l
Efects o electric current Introduction This is the frst o three sub- topics that discusses some o the eects that occur when charge ows in a circuit. The three eects are:
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heating effect, when energy is transerred to a resistor as internal energy
chemical effect, when chemicals react together to alter the energy o electrons and to cause them to move, or when electric current in a material causes chemical changes ( Sub- topic 5 . 3 )
magnetic effect, when a current produces a magnetic feld, or when magnetic felds change near conductors and induce an em in the conductor ( S ub-topic 5 .4) .
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
This sub-topic deals with the heating eect o a current ater giving you some advice on setting up and drawing electric circuits.
Drawing and using circuit diagrams At some stage in your study o electricity you need to learn how to construct electrical circuits to carry out practical tasks. This section deals with drawing, interpreting, and using circuit diagrams and is designed to stand alone so that you can reer to it whenever you are working with diagrams and real circuits. You may not have met all the components discussed here yet. They will be introduced as they are needed.
Circuit symbols A set o agreed electrical symbols has been devised so that all physicists understand what is represented in a circuit diagram. The agreed symbols that are used in the IB D iploma Programme are shown in fgure 1 . Most o these are straightorward and obvious; some may be amiliar to you already. Ensure that you can draw and identiy all o them accurately.
joined wires
wires crossing (not joined)
cell
battery
lamp
ac supply
switch
ammeter
voltmeter
A
V
galvanometer
resistor
variable resistor
potentiometer
heating element
fuse
thermistor
diode
variable power supply
transformer
ac supply
capacitor
Figure 1 Circuit symbols.
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E L E C T R I C I T Y AN D M AG N E T I S M There are points to make about the symbols.
06 V
01 A
S ome o the symbols here are intended or direct current (dc) circuits ( cells and batteries, or example) . D irect current reers to a circuit in which the charge ows in one direction. Typical examples o this in use would be a low- voltage ashlight or a mobile phone. O ther types o electrical circuits use alternating current (ac) in which the current direction is frst one way around the circuit and then the opposite. The time between changes is typically about 1 /1 00th o a second. C ommon standards or the requencies around the world include 5 0 Hz and 60 Hz. Alternating current is used in high- voltage devices ( typically in the home and industry) , where large amounts o energy transer are required: kettles, washing machines, powerul electric motors, and so on. Alternating supplies can be easily transormed rom one pd to another, whereas this is more difcult ( though not impossible) or dc.
There are separate symbols or cells and batteries. Most people use these two terms interchangeably, but there is a dierence: a battery is a collection o cells arranged positive terminal to negative the diagram or the battery shows how they are connected. A cell only contains one source o em. S ub- topic 5 .3 goes into more detail about cells and batteries.
Circuit conventions
It is not usual to write the name o the component in addition to giving its symbol, unless there is some chance o ambiguity or the symbol is unusual. However, i the value o a particular component is important in the operation o the circuit it is usual to write its value alongside it.
Particular care needs to be taken when it is necessary to draw one connecting lead over another. The convention is that i two leads cross and are j oined to each other, then a dot is placed at the j unction. I there is no dot, then the leads are not connected to each other. In the circuit in fgure 2 it is important to know the em o the cell, the data or the lamp at its working temperature, and the ull scale defections (sd) o the two meters in the circuit.
A 6 V, 300 mA
V 010 V
Figure 2 Circuit diagram.
Practical measurements o current and potential diference We oten need to measure the current in a circuit and the pd across components in the circuit. This can be achieved with the use o meters or sensors connected to computers ( data loggers) . You may well use both during the course. S chools use many types and varieties o meters and it is impossible to discuss them all here, but an essential distinction to make is between analogue and digital meters. Again, you may well use both as you work through the course. Analogue meters have a mechanical system o a coil and a magnet. When charge ows through the coil, a magnetic feld is produced that interacts with the feld o the magnet and the coil swings round against a spring. The position reached by the pointer attached to the coil is a measure o the current in the meter.
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D igital meters sample the potential dierence across the terminals o the meter ( or, or current, the pd across a known resistor inside the meter) and then convert the answer into a orm suitable or display on the meter. Ammeters measure the current in the circuit. As we want to know the size o the current in a component it is clear that the ammeter must have the same current. The ammeter needs to be in series with the circuit or component. An ideal ammeter will not take any energy rom the electrons as they ow through it, otherwise it would disturb the circuit it is trying to measure. Figure 2 shows where the ammeter is placed to measure the current. Voltmeters measure the energy converted per unit charge that ows in a component or components. You can think o a voltmeter as needing to compare the energy in the electrons beore they enter a component to when they leave it, rather like the turnstiles (bae gates) to a rail station that count the number o people (charges) going through as they give a set amount o money (energy) to the rail company. To do this the voltmeter must be placed across the terminals o the component or components whose pd is being measured. This arrangement is called parallel. Again, fgure 2 shows this or the voltmeter.
Constructing practical circuits from a diagram Wiring a circuit is an important skill or anyone studying physics. I you are careul and work in an organized way then you should have no problems with any circuit no matter how complex. As an example, this is how you might set up one o the more difcult circuits in this course.
step 1
V
A
A
B
step 2 (a) A A'
B'
step 3 V
step 4 (b)
link A A'; B B' (c)
Figure 3 Variable resistors.
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E L E C T R I C I T Y AN D M AG N E T I S M This is a potential divider circuit that it is used to vary the pd across a component, in this case a lamp. The most awkward component to use here is the potential divider itsel ( a orm o variable resistor, that is sometimes known as a p otentiometer) . In one orm it has three terminals ( fgure 3 ( b) ) , in another type it has three in a rotary ormat. The three- terminal linear device has a terminal at one end o a rod with a wiper that touches the resistance windings, and another two terminals one at each end o the resistance winding itsel. B egin by looking careully at the diagram fgure 3 ( a) . Notice that it is really two smaller circuits that are linked together: the top sub- circuit with the cell, and the bottom sub- circuit with the lamp and the two meters. The bottom circuit itsel consists o two parts: the lamp/ammeter link together with the voltmeter loop. The rules or setting up a circuit like this are:
I you do not already have one, draw a circuit diagram. Get your teacher to check it i you are not sure that it is correct.
B eore starting to plug leads in, lay out the circuit components on the bench in the same position as they appear on the diagram.
C onnect up one loop o the circuit at a time.
Ensure that components are set to give minimum or zero current when the circuit is switched on.
D o not switch the circuit on until you have checked everything.
Figure 3 ( c) shows a sequence or setting up the circuit step-by- step. Another skill you will need is that o troubleshooting circuits this is an art in itsel and comes with experience. A possible sequence is:
C heck the circuit is it really set up as in your diagram?
C heck the power supply ( try it with another single component such as a lamp that you know is working properly) .
C heck that all the leads are correctly inserted and that there are no loose wires inside the connectors.
C heck that the individual components are working b y substituting them into an alternative circuit known to be working.
Resistance We saw in S ub- topic 5 .1 that, as electrons move through a metal, they interact with the positive ions and transer energy to them. This energy appears as kinetic energy o the lattice, in other words, as internal energy: the metal wire carrying the current heats up. However, simple comparison between dierent conductors shows that the amount o energy transerred can vary greatly rom metal to metal. When there is the same current in wires o similar size made o tungsten or copper, the tungsten wire will heat up more than the
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
copper. We need to take account o the act that some conductors can achieve the energy transer better than others. The concept o electrical resistance is used or this. The resistance o a component is defned as potential dierence across the component ____ current in the component The symbol or resistance is R and the defnition leads to a well-known equation V R=_ I The unit o resistance is the ohm ( symbol ; named ater Georg Simon O hm, a German physicist) . In terms o its undamental units, 1 is 1 kg m 2 s 3 A 2 ; using the ohm as a unit is much more convenient! Alternative orms o the equation are: V = IR and I =
__V R
Investigate! Resistance of a metal wire
Take a piece o metal wire ( an alloy called constantan is a good one to choose) and
A
connect it in the circuit shown. Use a power supply with a variable output so that you can alter the pd across the wire easily.
I your wire is long, coil it around an insulator ( perhaps a pencil) and ensure that the coils do not touch.
Take readings o the current in the wire and the pd across it or a range o currents. Your teacher will tell you an appropriate range to use to avoid changing the temperature o the wire.
For each pair o readings divide the pd by the current to obtain the resistance o the wire in ohms.
V
Figure 4
current /A 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70
A table o results is shown in fgure 5 or a metal wire 1 m in length with a diameter o 0. 5 0 mm. For each pair o readings the resistance o the wire has been calculated by dividing the pd by the current. Although the resistance values are not identical ( it is an experiment with real errors, ater all) , they do give an average value or the resistance o 2 . 5 4 . This should be rounded to 2 .5 given the signfcant fgures in the data.
pd/V 0.13 0.26 0.50 0.76 1.01 1.27 1.53 1.78
resistance/ 2.55 2.60 2.50 2.53 2.53 2.54 2.55 2.54
Figure 5 Variation of pd with current for a conductor.
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Nature of science Edison and his lamp The conversion o electrical energy into internal energy was one o the frst uses o distributed electricity. Thomas Edison an inventor and entrepreneur, who worked in the US at around the end o the nineteenth century, was a pioneer o electric lighting. The earliest orms o light were provided by producing a current in a metal or carbon flament. These flaments heated up
until they glowed. Early lamps were primitive but produced a revolution in the way that homes and public spaces were lit. The development continues today as inventors and manuacturers strive to fnd more and more efcient electric lamps such as the light- emitting diodes ( LED ) . More developments will undoubtedly occur during the lietime o this book.
Ohms law Figure 6 shows the results rom the metal wire when they are plotted as a graph o V against I. 2.00 1.80 1.60 1.40 pd/V
1.20 1.00 0.80 0.60 0.40 0.20 0.00 0.00
Worked example The current in a component is 5 .0 mA when the pd across it is 6.0 V. C alculate: a) the resistance o the component b) the pd across the component when the current in it is 1 5 0 A.
Solution V 6 a) R = __ = ______ = 1 .2 k I 5 10
0.10
0.20
0.30
0.40 current/A
0.50
0.60
0.70
0.80
Figure 6 pd against current from the table.
A best straight line has been drawn through the data points. For this wire, the resistance is the same or all values o current measured. Such a resistor is known as ohmic. An equivalent way to say this is that the potential dierence and the current are proportional ( the line is straight and goes through the origin) . In the experiment carried out to obtain these data, the temperature o the wire did not change. This behaviour o metallic wires was frst observed by Georg S imon O hm in 1 82 6. It leads to a rule known as O hms law. O hms law states that the potential dierence across a metallic conductor is directly proportional to the current in the conductor providing that the physical conditions o the conductor do not change.
3
b) V = IR = 1 .5 1 0 4 1 .2 1 0 3 = 0.1 8 A
198
B y physical conditions we mean the temperature ( the most important actor as we shall see) and all other actors about the wire. B ut the temperature actor is so important that the law is sometimes stated replacing the term physical conditions with the word temperature.
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Nature of science Ohm and Barlow O hms law has its limitations because it only tells us about a material when the physical conditions do not change. However, it was a remarkable piece o work that did not fnd immediate avour. B arlow was an English scientist who was held in high respect or his earlier work and had recently published an alternative theory on conduction. People simply did not believe that B arlow could be wrong. This immediate acceptance o one scientists work over another would not necessarily happen today. Scientists use a system o peer review. Work published by one scientist or scientifc group must be set out in such a way that other scientists can repeat the experiments or collect the same data to check that there are no errors in the original work. O nly i the scientifc community as a whole can veriy the data is the new work accepted as scientifc act. Towards the end o the 2 0th century a research group thought that it had ound evidence that nuclear usion could occur at low temperatures ( so-called cold usion) . Repeated attempts by other research groups to replicate the original results ailed, and the cold usion ideas were discarded.
Investigate! Variation o resistance o a lamp flament
Use the circuit you used in the investigation on resistance o a metal wire, but repeat the experiment with a flament lamp instead o the wire.
two ways to achieve this: the frst is to reverse the connections to the power supply, also reversing the connections to the ammeter and voltmeter (i the meters are analogue) . The second way is somewhat easier, simply reverse the lamp and call all the readings negative because they are in the opposite direction through the lamp.
Your teacher will advise you o the range o currents and pds to use.
This time do the experiment twice, the second time with the charge owing through the lamp in the opposite direction to the frst. There are
Plot a graph o V (y-axis) against I (x-axis) with the origin (0,0) in the centre o the paper. Figure 7 shows an example o a VI graph or a lamp.
6 pd/V 4 2 0 60
40
0
20 2
20
40
60 current, I/mA
4 6
Figure 7
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TOK But is it a law? This rule o Ohm is always called a law but is it? In reality it is an experimental description o how a group o materials behave under rather restricted conditions. Does that make it a law? You decide.
The graph is not straight ( although it goes through the origin) so V and I are not proportional to each other. The lamp does not obey O hms law and it is said to be non-ohmic. However, this is not a air test o ohmic behaviour because the flament is not held at a constant temperature. I it were then, as a metal, it would probably obey the law. When the resistance is calculated or some o the data points it is not constant either. The table shows the resistance values at each o the positive current points.
Current/mA 20 34 41 47 52 55
There is also another aspect to the law that is oten misunderstood. Our defnition o resistance is that R = ___VI or V = IR. Ohms law states that
Resistance/ 50 59 73 85 96 109
Tip Notice that these resistances were calculated or each individual V data point using __ . They were not I evaluated rom the tangent to the graph at the current concerned. The defnition o resistance is in V V terms o __ not in terms o ___ which I I is what the tangent would give.
V I and, including the constant o proportionality k, V = kI We thereore defne R to be the same as k, but the defnition o resistance does not correspond to Ohms law (which talks only about a proportionality) . V = IR is emphatically not a statement o Ohms law and i you write this in an examination as a statement o the law you will lose marks.
The data show that resistance o the lamp increases as the current increases. At large currents, it takes greater changes in pd to change the current by a fxed amount. This is exactly what you might have predicted. As the current increases, more energy is transerred rom the electrons every second because more electrons ow at higher currents. The energy goes into increasing the kinetic energy o the lattice ions and thereore the temperature o the bulk material. B ut the more the ions vibrate in the lattice, the more the electrons can collide with them so at higher temperatures even more energy is transerred to the lattice by the moving charges. O ther non- ohmic conductors include semiconducting diodes and thermistors; these are devices made rom a group o materials known as semiconductors.
Investigate! Diodes and thermistors
200
You can easily extend the Investigate! lamp experiment to include these two types o device. There are some extra practical points however:
The diode will require a protecting resistor in series with it ( 1 00 is usually appropriate) as the current can become very large at even quite small potential dierences; this will cause the diode to melt as large amounts o energy are transerred to it. The resistor limits the current so that the diode is not damaged.
The thermistor also needs care, because as it heats up its resistance decreases and i too much current is used, the thermistor can also be destroyed.
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
The results o these experiments and some others are summarized in the IV graphs in fgure 8. I
I
0
0
V
0
I
ohmic conductor
0
flament lamp
I
V
semiconducting diode
I
0 0
0
V
0
V neon tube
0 0
V thermistor
Figure 8 IV graphs or various conductors.
Semiconducting diodes S emiconducting diodes are designed only to allow charges to ow through them in one direction. This is seen clearly in the graph. For negative values o V there is actually a very small current owing in the negative direction but it is ar less than the orward current. The nature o semiconductor material also means that there is no signifcant current in the orward direction until a certain orward pd is exceeded.
Thermistors Thermistors are made rom one o the two elements that are electrical semiconductors: silicon and germanium. There are several types o thermistor, but we will only consider the negative temperature coefcient type ( ntc) . As the temperature o an ntc thermistor increases, its resistance alls. This is the opposite behaviour to that o a metal. S emiconductors have many ewer ree electrons per cubic metre compared with metals. Their resistances are typically 1 0 5 times greater than similar samples o metals. However, unlike in a metal, the charge density in semiconductors depends strongly on the temperature. The higher the temperature o the semiconductor, the more charge carriers are made available in the material. As the temperature rises in the germanium:
The lattice ions vibrate more and impede the movement o the charge carriers. This is exactly the same as in metals and also leads to an increase in resistance.
More and more charge carriers become available to conduct because the increase in temperature provides them with enough energy to break away rom their atoms. This leads to a large decrease in resistance.
The second eect is much greater than the frst and so the net eect is that conduction increases ( resistance alls) as the temperature o the semiconductor rises.
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Investigate! How resistance depends on size and shape
You will need a set o wires made rom the same metal or metal alloy. The wires should have a circular cross- section and be available in a range o dierent diameters. You will also need to devise a way to vary the length o one o the wires in the circuit.
Use the circuit in fgure 4 on p1 97 to answer the ollowing questions:
How does the resistance R o one o the wires vary with length l?
How does the resistance R o the wires vary with diameter d when the wires all have the same length?
Try to make things easy or your analysis. D ouble and halve the values o length to see what dierence this makes to the resistance is there an obvious relationship? The diameter o the wire may be more difcult to test in this way, but a graph o resistance against diameter may give you a clue.
O nce you have an idea what is going on, you may decide that the best way to answer these questions is to plot graphs o R against l, and 1 R against __ d 2
Resistivity The resistance o a sample o a material depends not only on what it is made o, but also on the physical dimensions ( the size and shape) o the sample itsel. The graphs suggested in the Investigate! should give straight lines that go through the origin and can be summed up in the ollowing rule. The resistance o a conductor is:
proportional to its length l
inversely proportional to its cross-sectional area A ( which is itsel proportional to d2 ) .
So l R _ A This leads to a defnition o a new quantity called resistivity. Resistivity is defned by RA =_ l The unit o resistivity is the ohm-metre ( symbol m) . Take care here: this is ohm metre. It is not ohm metre 1 a mistake requently made by students in examinations. The meaning o ohm metre 1 is the resistance o one metre length o a particular conductor, which is a relevant quantity to know, but is not the same as resistivity. Resistivity is a quantity o considerable use. The resistance o a material depends on the shape o the sample as well as what it is made rom. Even a constant volume o a material will have values o resistance that depend on the shape. However, the value o the resistivity is the same or all samples o the material. Resistivity is independent o shape or size j ust like quantities such as density ( where the size o a material has been
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
removed by the use o volume) or specifc latent heat ( where the value is related to unit mass o the material) .
Worked examples 1
A uniorm wire has a radius o 0.1 6 mm and a length o 7.5 m. C alculate the resistance o the wire i its resistivity is 7.0 1 0 7 m.
2
Solution Unless told otherwise, assume that the wire has a circular cross-section. area o wire = ( 1 .6 1 0 4) 2 = 8. 04 1 0 8 m 2 RA =_ l l 7.0 1 0 7 7.5 and R = _ = __ = 65 A 8.04 1 0 8
C alculate the resistance o a block o copper that has a length o 0.01 2 m with a width o 0.75 mm and a thickness o 1 2 mm. The resistivity o copper is 1 .7 1 0 8 m.
Solution The cross-sectional area o the block is 7.5 1 0 4 1 . 2 1 0 2 = 9.0 1 0 6 m 2 The relevant dimension or the length is 0.01 2 m, so l 1 .7 1 0 0.01 2 R = _ = _____________ = 0.02 3 m 9.0 1 0 A 8
6
Investigate! Resistivity of pencil lead pencil
D etermine the resistance o the lead.
Measure the length o pencil lead between the crocodile clips.
Measure the diameter o the lead using a micrometer screw gauge or digital callipers and calculate the area o the lead.
Using the data, calculate the resistivity o the lead. The accepted value o the resistivity o graphite is about 3 1 0 5 m but you will not expect to get this value given the presence o clay in the lead as well.
Take the experiment one step urther with a challenge. Use your pencil to uniormly shade a 1 0 cm by 2 cm area on a piece o graph paper. This will make a graphite resistance flm on the paper. Attach the graphite flm to a suitable circuit and measure the resistance o the flm. Knowing the resistance and the dimensions o your shaded area should enable you to work out how thick the flm is. ( Hint: in the resistivity equation, the length is the distance across the flm, and the area is width o the flm thickness o the flm. )
V
A
Figure 9 Graphite is a semi- metallic conductor and is a constituent o the lead in a pencil. Another constituent in the pencil lead is clay. It is the ratio o graphite to clay that determines the hardness o the pencil. This experiment enables you to estimate the resistivity o the graphite. Take a B grade pencil ( sometimes known as #1 grade in the US ) and remove about 1 . 5 cm o the wood rom each end leaving a cylinder o the lead exposed. Attach a crocodile ( alligator) clip frmly to each end. Use leads attached to the crocodile clips to connect the pencil into a circuit in order to measure the resistance o the lead. Expect the resistance o the pencil lead to be about 1 or the purpose o choosing the power supply and meters.
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Practical resistors Resistors are o great importance in the electronics and electrical industries. They can be single value ( fxed) devices or they can be variable. They can be manuactured in bulk and are readily and cheaply available. Resistors come in dierent sizes. Small resistors can have a large resistance but only be able to dissipate ( lose) a modest amount o energy every second. I the power that is being generated in the resistor is too large, then its temperature will increase and in a worst case, it could become a thermal use! Resistors are rated by their manuacturers so that, or example, a resistor could have a resistance o 2 70 with a power rating o 0.5 W. This would ___mean that the maximum current that 0.5 = 43 mA. the resistor can saely carry is ___ 270
Combining resistors Electrical components can be linked together in two ways in an electrical circuit: in series, where the components are j oined one ater another as the ammeter, the cell and the resistor in fgure 1 0, or in p arallel as are the resistor and the voltmeter in the same fgure. A
V
Figure 10
Two components connected in series have the same current in each. The number o ree electrons leaving the frst component must equal the number entering the second component; i electrons were to stay in the frst component then it would become negatively charged and would repel urther electrons and prevent them rom entering it. The ow o charges would rapidly grind to a halt. This is an important rule to understand. In series the potential dierences ( pds) add. To see this, think o charge as it travels through two components, the total energy lost is equal to the sum o the two separate amounts o energy in the components. B ecause the charge is the same in both cases, thereore, the sum o the pds is equal to the total pd dropped across them. C omponents in parallel, on the other hand, have the same pd across them, but the currents in the components dier when the resistances o the components dier. C onsider two resistors o dierent resistance values, in parallel with each other and connected to a cell. I one o the resistors is temporarily disconnected then the current in the remaining resistor is given by the em o the cell divided by the resistance. This will also be true or the other resistor i it is connected alone. I both resistors are now connected in parallel with each other, both resistors have the same pd across them because a terminal o each resistor is connected to one o the terminals o the cell. The cell will have to supply more current than i either resistor were there alone to be precise, it has to supply the sum o the separate currents. To sum up
Currents ... In series In parallel
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...are the same ...add
Potential diferences ... ...add ...are the same
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Investigate! Resistors in series and parallel
For this experiment you need six resistors, each one with a tolerance o 5 % . Two o these resistors should be the same. The tolerance fgure means that the manuacturer only guarantees the value to be within 5 % o the nominal value nominal means the printed value. Your resistors may have the nominal value written on them or you may have to use the colour code printed on them. The code is easy to decipher. 2
You also need a multimeter set to measure resistance directly and a way to j oin the resistors together and to connect them to the multimeter.
First, measure the resistance o each resistor and record this in a table.
Take the two resistors that have the same nominal value and connect them in series. Measure the resistance o the combination. C an you see a rule straight away or the combined resistance o two resistors?
Repeat with fve o the possible combinations or connecting resistors in series.
Now measure the combined resistance o the two resistors with the same nominal value when they are in parallel. Is there an obvious rule this time?
O ne way to express the rule or combining
7 1k (5%) = 27 k (5%)
multiplier tolerance silver gold black brown red orange yellow green blue violet grey white
0 1 2 3 4 5 6 7 8 9
0.01 0.1 1 10 100 1k 10 k 100 k 1M 10 M
10% 5%
R R two resistors R 1 and R 2 in parallel is as ______ . R + R 1
1% 2%
1
2
2
Test this relationship or fve combinations o parallel resistors.
0.5%
Test your two rules together by orming combinations o three resistors such as:
These ideas provide a set o rules or the combination o resistors in various arrangements:
Resistors in series Suppose that there are three resistors in series: R1 , R2 and R3 (fgure 1 1 (a) ) . What is the resistance o the single resistor that could replace them so that the resistance o the single resistor is equivalent to the combination o three? The resistors are in series ( fgure 1 1 ( a) ) and thereore the current I is the same in each resistor. Using the defnition o resistance, the pd across each resistor, V1 , V2 and V3 is V1 = IR 1
V2 = IR 2
and
V3 = IR 3
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E L E C T R I C I T Y AN D M AG N E T I S M The single resistor with a resistance R has to be indistinguishable rom the three in series. In other words, when the current through this single resistor is the same as that through the three, then it must have a pd V across it such that V = IR R1
R2
R3 I
V1
V2
V3
V (a) series
R1 I1 R2 I I2 R3 I3 V (b) parallel
Figure 11 Resistors in series and parallel.
As this is a series combination, the potential dierences add, so V = V1 + V2 + V3 Thereore IR = IR 1 + IR 2 + IR 3 and cancelling I leads to, in series R = R1 + R2 + R3 When resistors are combined in series, the resistances add to give the total resistance.
Resistors in parallel Three resistors in parallel ( fgure 1 1 ( b) ) have the same pd V across them ( fgure 1 1 ( b) ) . The currents in the three separate resistors add to give the current in the connecting leads to and rom all three resistors, thereore I = I1 + I2 + I3 Each current can be written in terms o V and R using the defnition o resistance: V _ V V V _ = + _+ _ R R1 R2 R3
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T This time V cancels out and, in p arallel,
step 1 combine series resistors
1 1 1 _1 = _ + _+ _ R
R1
R2
R3
In parallel combinations o resistors, the reciprocal o the total resistance is equal to the sum o the reciprocals o the individual resistances.
step 2 combine parallel resistor separately
The parallel equation needs some care in calculations. The steps are:
calculate the reciprocals o each individual resistor
add these reciprocals together
take the reciprocal o the answer.
It is common to see the last step ignored so that the answer is incorrect; a worked example below shows the correct approach.
More complicated networks When the networks o resistors are more complicated, then the individual parts o the network need to be broken down into the simplest orm.
step 3 combine series resistors together
Figure 1 2 shows the order in which a complex resistor network could be worked out.
Figu re 12
Worked examples 1
Three resistors o resistance, 2 .0 , 4. 0 , and 6.0 are connected. C alculate the total resistance o the three resistors when they are connected: a) in series b) in parallel.
Solution a) In series, the resistances are added together, so 2 + 4 + 6 = 12
Solution The two resistors in parallel have a combined 3 1 1 1 1 1 resistance o __ = __ + __ = __ + __ = __ R R R 4 8 8 1
2
8 R = _ = 2 .67 3 This 2 . 67 resistor is in series with 2 .0 , so the total combined resistance is 2 .67 + 2 .0 = 4.7 . 3
Four resistors each o resistance 1 .5 are connected as shown. C alculate the combined resistance o these resistors.
b) In parallel, the reciprocals are used: 6 + 3 + 2 1 1 1 1 1 1 1 11 __ = __ + __ + __ = __ + __ + __ = ________ = __ R R R R 2 4 12 6 12 1
2
3
The fnal step is to take the reciprocal o the 12 sum, R = __ = 1 .1 11 2
2 .0 , 4.0 , and 8. 0 resistors are connected as shown. C alculate the total resistance o this combination. 4.0 2.0
Solution Two 1 .5 resistors in parallel have a resistance 1 1 1 2 given by __ = ___ + ___ = ___ . S o R = 0.75 . R 1 .5 1 .5 1 .5 Two 0. 75 resistors in series have a combined resistance o 0.75 + 0. 75 = 1 .5
8.0
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Potential divider In the section on circuit diagrams, a potential divider circuit was shown. This is a circuit commonly used with sensors and also to produce variable potential dierences. It has some advantages over the simpler variable resistor circuit even though it is more complicated to set up.
V1
R1
Rtop
R
Vin
Vin R2
Vin thermistor
V2
Vout
LDR
Vout
V (a)
(b)
(c)
Figure 13 The potential divider.
The most basic potential divider consists o two resistors with resistances R 1 and R 2 in series with a power supply. This arrangement is used to provide a fxed pd at a value somewhere between zero and the em o the power supply. Figure 1 3 ( a) shows the arrangement.
Worked example 1
A potential divider consists o two resistors in series with a battery o 1 8 V. The resistors have resistances 3 .0 and 6.0 . C alculate, or each resistor:
The two resistors have the same current in them, and the sum o the pds across the resistors is equal to the source em. So The current I in both resistors is total pd across both resistors V I = ___ = _ R total resistance As usual,
a) the pd across it b) the current in it.
Solution
The resistors are in series so
a) The pd across the 3 .0 V R 18 3 resistors _______ = _____ 3 + 6 (R + R ) in
1
1
2
= 6.0 V. The pd across the 6.0 is then 1 8 6.0 V = 1 2 V. b) The current is the same in both resistors because they are in series. The total resistance is 9. 0 and the em o the battery is 1 8 V. The current = V 18 __ = __ = 2 .0 A. R 9
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V1 = IR 1 and V2 = IR 2
R = R1 + R2 This leads to equations or the pd across each resistor R1 R2 V1 = _ Vin and V2 = _Vin ( R1 + R2 ) ( R1 + R2 )
Using a potential divider with sensors It is a simple matter to extend the fxed pd arrangement to a circuit that will respond to changes in the external conditions. S uch an arrangement might be used by a computer that can sense changes in pd and respond accordingly, or example, by turning on a warning siren i a rerigerator becomes too warm, or turning on a light when it becomes dark. Typical circuits are shown in fgures 1 3 ( b) and1 3 ( c) . In fgure 1 3 ( b) a thermistor is used instead o one o the fxed resistors.
5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Recall that when a thermistor is at a high temperature its resistance is small, and that the resistance increases when the temperature alls. Rather than calculating the values, or this example we will use our knowledge o how pd and current are related to work out the behaviour o the circuit rom frst principles. S uppose the temperature is low and that the thermistor resistance is high relative to that o the fxed-value resistor. Most o the pd will be dropped across the thermistor and very little across the fxed resistor. I you cannot see this straight away, remember the equations rom the previous section: VinR thermistor VinR resistor Vthermistor = __ and Vresistor = __ ( R thermistor + R resistor ) ( R thermistor + R resistor ) I Rthermistor R resistor ( means much greater than) then R thermistor ( R thermistor + R resistor) and Vthermistor Vin. S o the larger resistance ( the thermistor at low temperatures) has the larger pd across it. I the thermistor temperature now increases, then the thermistor resistance will all. Now the fxed- value resistor will have the larger resistance and the pds will be reversed with the thermistor having the small voltage drop across it. A voltage sensor connected to a computer can be set to detect this voltage change and can activate an alarm i the thermistor has too high a temperature. You may be asking what the resistance o the fxed resistor should be. The answer is that it is normally set equal to that o the thermistor when it is at its optimum ( average) temperature. Then any deviation rom the average will change the potential dierence and trigger the appropriate change in the sensing circuit. The same principle can be applied to another sensor device, a lightdependent resistor (LD R) . This, like the thermistor, is made o semiconducting material but this time it is sensitive to photons incident on it. When the light intensity is large, charge carriers are released in the LD R and thus the resistance alls. When the intensity is low, the resistance is high as the charge carriers now re-combine with their atoms. Look at fgure 1 3 ( c) . You should be able to explain that when the LD R is in bright conditions then the pd across the fxed resistor is high.
Using a potential divider to give a variable pd A variable resistor circuit is shown in fgure 1 4( a) . It consists o a power supply, an ammeter, a variable resistor and a resistor. The value o each component is given on the diagram. We can predict the way this circuit will behave. When the variable resistor is set to its minimum value, 0 , then there will be a pd o 2 V across the resistor and a current o 0.2 A in the circuit. When the variable resistor is set to its maximum value, 1 0 , then the total resistance in the circuit is 2 0 , and the current is 0.1 A. This means that with 0. 1 A in the 1 0 fxed resistor, only 1 V is dropped across it. Thereore the range o pd across the fxed resistor can only vary
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E L E C T R I C I T Y AN D M AG N E T I S M rom 1 V to 2 V hal o the available pd that the power supply can in principle provide. You should now be able to predict the range across the variable resistor. A 2V fxed resistor 10
2V
slider
2V 1V
variable resistor 010 0V
(a) variable resistor
V fxed resistor 10
(b) potential divider
Figure 14
The limited range is a signifcant limitation in the use o the variable resistor. To achieve a better range, we could use a variable resistor with a much higher range o resistance. To get a pd o 0 . 1 V across the fxed resistor the resistance o the variable resistor has to be about 2 00 . I the fxed resistor had a much greater resistance, then the variable resistor would need an even higher value too and this would limit the current. The potential divider arrangement ( fgure 1 4( b) ) allows a much greater range o pd to the component under test than does a variable resistor in series with the component. In a potential divider, the same variable resistor can be used but the set up is dierent and involves the use o the three terminals on the variable resistor. ( A variable resistor is also sometimes called a rheostat.) O ne terminal is connected to one side o the cell, and the other end o the rheostat resistor is connected to the other terminal o the cell. The potential at any point along the resistance winding depends on the position o the slider ( or wiper) that can be swept across the windings rom one end to the other. Typical values or the potentials at various points on the windings are shown or the three blue slider positions on fgure 1 4( b) . The component that is under test ( again, a resistor in this case) is connected in a secondary circuit between one terminal o the resistance winding and the slider on the rheostat. When the slider is positioned at one end, the ull 2 V rom the cell is available to the resistor under test. When at the other end, the pd between the ends o the resistor is 0 V ( the two leads to the resistor are eectively connected directly to each other at the variable resistor) . You should know how to set this arrangement up and also how to draw the circuit and explain its use.
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Worked examples 1
A light sensor consists o a 6. 0 V battery, a 1 800 resistor and a light- dependent resistor in series. When the LD R is in darkness the pd across the resistor is 1 . 2 V.
2
a) C alculate the resistance o the LD R when it is in darkness.
Solution
b) When the sensor is in the light, its resistance alls to 2 400 . C alculate the pd across the LD R.
Solution a) As the pd across the resistor is 1 .2 V, the pd across the LD R must be 6 1 .2 = 4. 8 V. The current in the circuit is V 1 .2 = ____ = 0.67 mA. I = __ R 1 800 The resistance o the LD R is V 4. 8 __ = ________ = 72 00 . I 0.67 1 0 3
resistan ce across L DR 2 40 0 b) The ratio o ________________ = ____ = 1 .3 3 . 1 800 resistan ce across 1 8 0 0 pd across L DR This is the same value as ___________ . For the pd across 1 8 0 0
A thermistor is connected in series with a fxed resistor and a battery. D escribe and explain how the pd across the thermistor varies with temperature.
As the temperature o the thermistor rises, its resistance alls. The ratio o the pd across the fxed resistor to the pd across the thermistor rises too because the thermistor resistance is dropping. As the pd across the fxed resistor and thermistor is constant, the pd across the thermistor must all. The change in resistance in the thermistor occurs because more charge carriers are released as the temperature rises. Even though the movement o the charge carriers is impeded at higher temperatures, the release o extra carriers means that the resistance o the material decreases.
ratio o pds to be 1 .3 3 , the pds must be 2 .6 V and 3 .4 V with the 3 .4 V across the LD R.
Investigate! Variable resistor or potential divider?
Set up the two circuits shown in fgure 1 4. Match the value o the fxed resistor to the variable resistor they do not need to be exactly the same but should be reasonably close. Add a voltmeter connected across the fxed resistor to check the pd that is available across it.
Make sure that the maximum current rating or the fxed resistor and the variable resistor cannot be exceeded.
C heck the pd available in the two cases and convince yoursel that the potential divider gives a wider range o voltages.
Heating efect equations We saw earlier that the power P dissipated in a component is related to the pd V across the component and the current I in it: P = IV The energy E converted in time t is E = IV t When either V or I are unknown, then two more equations become available: V2 P = IV = I2 R = _ R
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E L E C T R I C I T Y AN D M AG N E T I S M and V2 E = IVt = I2 Rt = _ t R These equations will allow you to calculate the energy converted in electrical heaters and lamps and so on. Applications that you may come across include heating calculations, and determining the consumption o energy in domestic and industrial situations.
Worked examples 1
C alculate the power dissipated in a 2 5 0 resistor when the pd across it is 1 0 V.
Solution 2
V2 1 02 P = _ = _ = 0.40 W R 250 A 9. 0 kW electrical heater or a shower is designed or use on a 2 5 0 V mains supply. C alculate the current in the heater.
Solution 3
9000 P = _ P = IV so I = _ = 36 A V 250 C alculate the resistance o the heating element in a 2 .0 kW electric heater that is designed or a 1 1 0 V mains supply.
Solution V2 V2 1 1 02 P = _ ; R = _ = _ = 6.1 A. R P 2 000
Kirchhos frst and second laws This section contains no new physics, but it will consolidate your knowledge and put the electrical theory you have learnt so ar into a wider physical context.
I1 I5
I4
I3
I2
Figure 15 Kirchhos frst law.
We saw that the charge carriers in a conductor move into and out o the conductor at equal rates. I 1 0 6 fow into a conductor in one second, then 1 0 6 must fow out during the same time to avoid the buildup o a static charge. We also considered what happens when current splits into two or more parts at the j unction where a parallel circuit begins. This can be taken one step urther to a situation where there is more than one incoming current at the j unction too. Figure 1 5 shows a j unction with three incoming currents and two outgoing ones. O ur rule about the incoming charge equating to the outgoing charge must apply here too, so algebraically: I1 + I2 + I3 = I4 + I5 A general way to write this is to use the sign ( meaning add up everything) , so that for any j unction
I = 0
When using the notation, remember to get the signs o the currents correct. C all any current in positive, and any current out negative.
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In words this can be written as the sum o the currents into a junction equals the sum o the currents away rom a junction or total charge fowing into a junction equals the total charge fowing away rom the junction. This important rule was frst quoted by the Prussian physicist Gustav Kirchho in 1 845 . It is known as Kirchhos frst law. It is equivalent to a statement o conservation o charge. Kirchho devised a second law which is also a conservation rule this time o energy. In any electric circuit there are sources o em ( oten a cell or a battery o cells or dc) and sinks o pd ( typically, lamps, heating coils, resistors, and thermistors) . A general rule in physics is that energy is conserved. Electrical components have to obey this too. So, in any electrical circuit, the energy being converted into electrical energy (in the sources o em) must be equal to the energy being transerred rom electrical to internal, by the sinks o pd. This is Kirchhos second law equivalent to conservation o energy. This second law applies to all closed circuits both simple and complex. In words, Kirchhos second law can be written as in a complete circuit loop, the sum o the ems in the loop is equal to the sum o the potential dierences in the loop or the sum o all variations o potential in a closed loop equals zero In symbols or any closed loop in a circuit =
IR R2
I2
E
G D
A
F R3
I1
I3
B
C R1
Figure 16 Kirchhofs second law.
Look at the circuit in fgure 1 6, it has parallel and series elements in it. A number o possible loops are drawn and analysed: Loop GABCDEG travelling anticlockwise round the loop This loop begins at the cell and goes around the circuit, through resistor R 1 and resistor R 2 fnally ending at the cell again. In this loop there is one source o em and two sinks o pd ( ignoring the leads, which we assume have zero resistance) .
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5
E L E C T R I C I T Y AN D M AG N E T I S M So ( the emf of the cell) = I1 R 1 + I2 R 2 ( the total pds aross the resistors) The direction o loop travel and the current direction are in all cases the same. We give a positive sign to the currents when this is the case. The em o the cell is driving in the same direction as the loop travel direction; it gets a positive sign as well. I the loop direction and the current or em were to be opposed then they would be given a negative sign. loop EFGE travelling clockwise round the loop This loop goes frst through resistor R 3 and the loop direction is in the same direction as the conventional current. Next the loop goes through resistor R1 but this time the current direction and the loop are dierent so there has to be a negative sign. There is no source o em in the loop so the Kirchho equation becomes 0 = I3 R 3 I2 R 2 Kirchhos frst law can be applied at point G. The total current into point G is I2 + I3 ; the total out is I1 . The application o the law is I1 = I2 + I3 . There are now three separate equations with three unknowns and these equations can be solved to work out the currents in each part o the circuit assuming that we know the value or the em o the cell and the values o the resistances in the circuit. B y setting up a series o loops it is possible to work out the currents and pds or complicated resistor networks, more complicated than could be done using the resistor series and parallel rules alone.
Ideal and non-ideal meters S o ar in this topic we have assumed ( witho ut mentioning it! ) that the me ters used in the circuits were ideal. Ideal meters have no eect on the circuits that they are me asuring. We would always want this to be true b ut, unortunate ly, the me ters we use in real circuits in the laboratory are not so ob liging and we nee d to know how to allow or this. Ammeters are placed in series with components so the ammeter has the same current as the components. It is undesirable or the ammeter to change the current in a circuit but, i the ammeter has a resistance o its own, then this is what will happen. An ideal ammeter has zero resistance clearly not attainable in practice as the coils or circuits inside the ammeter have resistance. Voltmeters are placed in parallel with the device or parts o a circuit they are measuring. In an ideal world, the voltmeter will not require any energy or its coil to move ( i it is a moving- coil type) or or its analogue to digital conversion ( i it is a digital meter) . The way to avoid current in the voltmeter is or the meter to have an infnite resistance again, not an attainable situation in practice. S ome modern digital meters can get very close to these ideals o zero ohm or ammeters and infnite ohms or voltmeters. D igital meters are used more and more in modern science.
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5 . 2 H E AT I N G E F F E C T O F A N E L E C T R I C C U R R E N T
Worked examples 1
C alculate the unknown branch current in the ollowing j unctions.
to show the technique and to convince you that it works.)
a) 3.0 V 9A
5A I1 I
3A
9.0 3.0
I2 loop 2
junction A 6.0
I3
b) loop 1 11A I
C urrent directions have been assigned and two loops 1 and 2 and j unction A defned in the diagram.
5A
3A
For loop 1
7A
3 = 3 I1 + 9I2
Solution
( the em in the loop is 3 V)
a) C urrent into the j unction = 9 + 5 A
or loop 2
C urrent out o the j unction = 3 + IA I = 9 + 5 3 = 1 1 A out o the j unction. b) Current into the junction = I + 1 1 + 3 = I + 1 4A C urrent out o the j unction = 5 + 7 = 1 2 A I = 1 2 1 4 = 2 A, so the current I is directed away rom the j unction. 2
0 = 6I3 9I2
[equation 1 ]
[equation 2]
( there is no source o em in this loop, current I2 is in the opposite direction to the loop direction 0. For j unction A I1 = I2 + I3 so
C alculate the currents in the circuit shown.
0 = 9I2 6I1 + 6I2 0 = 1 5 I2 6I1
3.0 V
And rom equation 1 6 = 6I1 + 1 8I2 I1
9.0
I2
Adding the equations
3.0
6 = 3 3 I2 I2 = 0.1 8 A
6.0 I3
Solution (This circuit can be analysed using the resistors in series and parallel equations. It is given here
S ubstituting gives: I1 = 0.45 A and I3 = 0.2 7 A
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E L E C T R I C I T Y AN D M AG N E T I S M
Nature of science How times change! D igital meters are a good way to get very close to the ideal meter. The resistance between the input terminals can be made to be very large ( 1 0 1 2 or more) or a voltmeter and to have a very small value or an ammeter. The meters themselves are
0
scale
meter terminals thread spring resistance wire
path
pivot
Figure 17
easy to read without any j udgements required about what the pointer indicates. It was not always like this. An early orm o meter was the hot-wire ammeter. This uses the heating eect o a current directly to increase the temperature o a metal wire (the current fows through the wire) . There are a number o orms o the ammeter, but in the type shown here, a spring keeps the wire under tension. As the wire expands with the increase in temperature, any point on the wire moves to the right. A pivoted pointer is attached to the insulated thread. As the wire expands, the spring pulls the thread causing the pointer to rotate about the pivot; a reading o current can now be made. The scale is usually extremely non-linear.
Worked examples 1
A 2 5 0 resistor is connected in series with a 5 00 resistor and a 6.0 V battery. a) C alculate the pd across the 2 5 0 resistor. b) Calculate the pd that will be measured across the 250 resistor i a voltmeter o resistance 1 000 is connected in parallel with it.
Solution a) The pd across the 2 5 0 resistor Vin R 1 6 250 = __ = 2 .0 V. ( R 1 + R 2 ) (250 + 500) b) When the voltmeter is connected, the resistance o the parallel combination is R1 R 2 2 5 0 1 000 R = _ = __ = 2 00 1 250 ( R1 + R2 )
216
2 00 6 V = _ = 1 .7 V 700 2
An ammeter with a resistance o 5 .0 is connected in series with a 3 .0 V cell and a lamp rated at 3 00 mA, 3 V. C alculate the current that the ammeter will measure.
Solution V 3 Resistance o lamp = __ = ___ = 1 0 . Total I 0.3 resistance in circuit = 1 0 + 5 = 1 5 . S o current V 3 in circuit = __ = __ = 2 00 mA. This assumes that R 15 the resistance o the lamp does not vary between 0.2 A and 0.3 A
5 . 3 E LECTRI C CE LLS
5.3 Electric cells Understanding Cells Primary and secondary cells Terminal potential diference
Applications and skills Investigating practical electric cells (primary
Electromotive orce em Internal resistance
and secondary) Describing the discharge characteristic o a simple cell (variation o terminal potential diference with time) Identiying the direction current ow required to recharge a cell Determining internal resistance experimentally Solving problems involving em, internal resistance, and other electrical quantities
Equation em o a cell: = I(R + r)
Nature of science Scientists need to balance their research into more and more ecient electric cells with the long-term risks associated with the disposal o the cells. Some modern cells have extremely poisonous contents. There will be serious consequences i these
chemicals are allowed to enter the water supply or the ood chain. Can we aford the risk o using these toxic substances, and what steps should the scientists take when carrying orward the research?
Introduction E lectric currents can produce a chemical effect. This has great importance in chemical industries as it can be a method for extracting ores or purifying materials. However, in this course we do not investigate this aspect of the chemical effect. O ur emphasis is on the use of an electric cell to store energy in a chemical form and then release it as electrical energy to perform work elsewhere.
Cells C ells operate as direct- current ( dc) devices meaning that the cell drives charge in one direction. The electron charge carriers leave the negative terminal of the cell. After passing around the circuit, the electrons reenter the cell at the positive terminal. The positive terminal has a higher
217
5
E L E C T R I C I T Y AN D M AG N E T I S M potential than the negative terminal so electrons appear to gain energy ( whereas positive charge carriers would appear to lose it) . The chemicals in the cell are reacting while current ows and as a result o this reaction the electrons gain energy and continue their j ourney around the circuit.
Nature of science Anodes and cathodes You will meet a naming system or anode and cathode which seems dierent or other cases in physics. In act it is consistent, but you need to think careully about it. For example, in a cathode- ray tube where the electrons in the tube are emitted rom a hot metal flament, the flament is called the cathode and is at a negative voltage. This appears to be the opposite notation rom that used or the electric cell. The reason is that the notation reers to what is happening inside the cell, not to the external circuit. The chemical reaction in the cell leads to positive
ions being generated at one o the electrodes and then owing away into the bulk o the liquid. S o as ar as the interior o the cell is concerned this is an anode, because it is an ( internal) source o positive ions. O course, as ar as the external circuit is concerned, the movement o positive charges away rom this anode leaves it negative and the electrode will repel electrons in the external circuit. As external observers, we call this the negative terminal even though ( as ar as the interior o the cell is concerned) it is an anode.
Primary and secondary cells Many o the portable devices we use today: torches, music players, computers, can operate with internal cells, either singly or in batteries. In some cases, the cells are used until they are exhausted and then thrown away. These are called p rimary cells. The original chemicals have completely reacted and been used up, and they cannot be recharged. E xamples include AA cells ( properly called dry cells) and button mercury cells as used in clocks and other small lowcurrent devices. O n the o the r hand, so me de vice s u se re charge ab le ce lls so that whe n the che mical re actio ns have inishe d, the ce lls can b e co nne cte d to a charge r. The n the che mical re actio n is re ve rse d and the o riginal che micals o rm again. Whe n as mu ch o the re - co nve rsio n as is p o ssib le has b e e n achie ve d, the ce ll is again availab le as a che mical e ne rgy sto re . Re charge ab le ce lls are kno wn as s eco nd ary cells . There many varieties o cell, here are two examples o the chemistry o a primary cell and that o a secondary cell. The Leclanch cell was invented by Georges Leclanch in 1 886. It is a primary cell and is the basis o many domestic torch and radio batteries.
218
5 . 3 E LECTRI C CE LLS
-
+
zinc rod
carbon rod
+ gas vents -
porous pot (P) ammonium chloride solution mixture of carbon and manganese dioxide
positive terminal
negative terminal
lead oxide lead insulating case
sulfuric acid
glass vessel (a)
Leclanch cell
(b)
lead-acid accumulator
Figure 1 Leclanch cell and lead-acid accumulator.
Figure 1 ( a) shows one practical arrangement o the cell. It consists o a zinc rod that orms the negative terminal. Inside the zinc is a paste o ammonium chloride separated rom manganese dioxide ( the cathode) by a porous membrane. In the centre o the manganese dioxide is a carbon rod that acts as the positive terminal or the cell. Zinc atoms on the inside surace o the case oxidize to become positive ions. They then begin to move away rom the inside o the case through the chloride paste leaving the case negatively charged. When the cell is connected to an external circuit, these electrons move around the circuit eventually reaching the carbon rod. A reaction inside the cell uses these electrons together with the components o the cell eventually orming the waste products o the cell, which include ammonia, manganese oxide, and manganese hydroxide. S econdary cells are very important today. They include the leadacid accumulator, together with more modern developments such as the nickelcadmium ( NiC d) , nickelmetal- hydride ( NiMH) , and lithium ion cells. All these types can be recharged many times and even though they may have a high initial cost ( compared to primary cells o an equivalent em and capacity) , the recharge is cheap so that, during the proj ected lietime o the cell, the overall cost is lower than that o primary cells. Although the lead- acid accumulator is one o the earliest examples o a secondary cell, it remains important ( accumulator is an old word that implies the accumulation or collection o energy into a store) . It is capable o delivering the high currents needed to start internal combustion and diesel engines, and it is reliable in the long term to maintain the current to important computer servers i the mains supply ails. The leadacid cell ( fgure 1 ( b) ) is slightly older than the Leclanch cell, and was invented in 1 85 9 by Gaston Plant. In its charged state the cell consists o two plates, one o metallic lead, the other o lead( IV ) oxide ( PbO 2 ) immersed in a bath o dilute suluric
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5
E L E C T R I C I T Y AN D M AG N E T I S M acid. D uring discharge when the cell is supplying current to an external circuit, the lead plate reacts with the acid to form lead( II) sulfate and the production of two free electrons. The liquid surrounding the plates loses acid and becomes more dilute. At the oxide plate, electrons are gained ( from the external circuit) and lead( II) sulfate is formed, also with the removal of acid from the liquid. S o the net result of discharging is that the plates convert to identical lead sulfates and the liquid surrounding the plates becomes more dilute.
Worked examples 1
Explain the difference between a primary and a secondary cell.
Solution A primary cell is one that can convert chemical energy to electrical energy until the chemicals in the cell are exhausted. Recharging the cell is not possible. A secondary cell can be recharged and the chemicals in it are converted back to their original form so that electrical energy can be supplied by the cell again. 2
A cell has a capacity of 1 400 mA h. C alculate the number of hours for which it can supply a current of 1 .8 mA.
Solution C ell capacity = current time and therefore 1 400 = 1 .8t. In this case t = 780 hours.
220
Recharging reverses these changes: electrons are forced from the positive plate by an external charging circuit and forced onto the negative plate. At the negative plate, atomic lead forms, at the positive, lead oxide is created, which restores the cell to its original state. The chargerecharge cycle can be carried out many times providing that the cell is treated carefully. However, if lead or lead oxide fall off the plates, they can no longer be used in the process and will collect at the bottom of the cell container. In the worst case the lead will short out the plates and the cell will stop operating. If the cell is overcharged ( when no more sulfate can be converted into lead and lead oxide) the charging current will begin to split the water in the cell into hydrogen and oxygen gas which are given off from the cell. The total amount of liquid in the cell will decrease, meaning that the plates may not be fully covered by the acid. These parts of the plates will then not take part in the reaction and the ability of the cell to store energy will decrease. Much industrial research effort is concentrated on the development of rechargeable cells. An important consideration for many manufacturers of electronic devices is the energy density for the battery ( the energy stored per unit volume) as this often determines the overall design and mass of an electronic device. Also a larger energy density may well provide a longer lifetime for the device.
Capacity of a cell Two cells with the same chemistry will generate the same electromotive force ( emf) as each other. However if one of the cells has larger plates than the other and contains larger volumes of chemicals, then it will be able to supply energy for longer when both cells carry the same current. The cap acity of a cell is the quantity used to measure the ability of a cell to release charge. If a cell is discharged at a high rate then it will not be long before the cell is exhausted or needs recharging, if the discharge current is low then the cell will supply energy for longer times. The capacity of a cell or battery is the constant current that it can supply for a given discharge. S o, if a cell can supply a constant current of 2 A for 2 0 hours then it said to have a capacity of 40 amp- hours ( abbreviated as 40 A h) . The implication is that this cell could supply 1 A for 40 hours, or 0.1 A for 400 hours, or 1 0 A for 4 hours. However, practical cells do not necessarily discharge in such a linear way and this cell may be able to provide a small discharge current of a few milliamps for much
5 . 3 E LECTRI C CE LLS
longer than expected rom the value o its capacity. C onversely, the cell may be able to discharge at 1 0 A or much less than 4 hours. An extreme example o this is the leadacid batteries used to start cars and commercial vehicles. A typical car battery may have a capacity o 1 00 A h. The current demand or starting the car on a cold winters day can easily approach 1 5 0 A. B ut the battery will only be able to turn the engine or a ew minutes rather than the two- thirds o an hour we might expect.
Discharging cells Investigate! Discharge of a cell
S ome cells have high capacity, and studying the discharge o such a cell can take a long time. This is a good example o how electronic data collection ( data- logging) can help an experimenter.
load resistor
Many data loggers need to be told how oten to take measurements (the sampling rate) . You will need to make some judgements about the overall length o time or which the experiment is likely to operate. Suppose you have a 1 .5 V cell rated at 1 5 00 mA h ( you can check the fgure by checking the cell specifcation on the manuacturers website) . It would seem reasonable that i the cell is going to supply a current o 2 5 0 mA, then it would discharge in a time o between 41 0 hours. You will need to set up the data logger accordingly with a suitable time between readings (the sampling interval) . D o not, however, exceed the maximum discharge current o the cell.
When the computer is set up, turn on the current in the discharge circuit and start the logging. Eventually the cell will have discharged and you can display the data as a graph. What are the important eatures o your graph?
Test other types o cell too, at least one primary and one secondary cell. Are there dierences in the way in which they discharge?
For at least one cell, towards the end o the discharge, switch o the discharge current while still continuing to monitor the terminal pd. D oes the value o the pd stay the same or does it recover? When the discharge is resumed what happens to the pd?
voltage sensor and data logger
Figure 2
The aim is to collect data to fnd how the terminal pd across the cell varies with time rom the start o the discharge. The basic circuit is shown in fgure 2 . Use a new cell to provide the current so that the initial behaviour o the cell can be seen too.
The results you obtain will depend on the types and qualities o the cells you test. A typical discharge curve looks like that in fgure 3 on p2 2 2 in which the terminal potential dierence o the cell is plotted against time since the discharge current began.
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5
E L E C T R I C I T Y AN D M AG N E T I S M 1.3
cell emf / V
1.2 midpoint voltage (1.18 volts)
1.1
1.0 end of discharge (127 minutes)
0.9 0
20
40
60 80 time (minutes)
100
120
140
Figure 3 Typical dischargetime graph for a cell.
Important features of this graph are that:
The initial terminal pd is higher than the quoted emf ( the value the manufacturer prints on the case) , but the initial value quickly drops to the rated emf ( approximately) .
For most of the discharge time the terminal pd remains more or less constant at or around the quoted emf. There is however sometimes a slow decline in the value of the terminal pd.
As the cell approaches exhaustion, the terminal pd drops very rapidly to a low level.
If the current is switched off, the terminal pd rises and can eventually reach the rated value again. However, when discharge is resumed, the terminal pd falls very quickly to the low value that it had before.
O ther experiments are possible, particularly for rechargeable cells to see the effects of repeated dischargecharge cycles on the cell. Figure 4 shows how the capacity of a NiC d cell changes as cell is taken through more and more cycles. cycle life at room temperature 5
capacity A h
4 3 2 1 0 0
222
500
1000 1500 cycle number
Figure 4 Cycle life of a rechargeable cell.
2000
2500
5 . 3 E LECTRI C CE LLS
Recharging secondary cells The clue to recharging a secondary cell is in the earlier descriptions o cell chemistry. The chemicals produce an excess o electrons at the negative terminal. D uring discharge these electrons then move through the external circuit transerring their energy as they go. When the electron arrives at the positive terminal, all o its energy will have been transerred to other orms and it will need to gain more rom the chemical store in the cell. To reverse the chemical processes we need to return energy to the cell using electrons as the agents, so that the chemical action can be reversed. When charging, the electrons need to travel in the reverse direction to that o the discharge current and you can imagine that the charger has to orce the electrons the wrong way through the cell. Part o a possible circuit to charge a 6- cell leadacid battery is shown in fgure 5 . +14 V direction o charge fow A
020 A
+ 1214 V dc discharged battery 0V
Figure 5 Charging circuit or a cell.
The direction o the charging current is shown in the diagram. It is in the opposite direction to that o the cell when it is supplying energy. An input pd o 1 4 V is needed, notice the polarity o this input. The lightemitting diode ( LED ) and its series resistor indicate that the circuit is switched on. The ammeter shows the progress o charging. When the battery is completely discharged, the charging current will be high, but as the charging level ( and thereore the em) o the battery rises, this current gradually alls. D uring the charging process the terminal voltage will be greater than the em o the cell. At ull charge the em o the battery is 1 3 .8 V. When the current in the meter is zero, the battery is ully charged.
Internal resistance and emf of a cell The materials rom which the cells are made have electrical resistance in j ust the same way as the metals in the external circuit. This internal resistance has an important eect on the total resistance and current in the circuit.
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5
E L E C T R I C I T Y AN D M AG N E T I S M
r
E r
A
B
R (a)
TOK Simple assumptions The model o a cell with a fxed internal resistance and a constant em is an example o modelling. In this case the model is a simple one that cannot be realized in practice. Another model we used earlier was to suggest that there are ideal ammeters with zero resistance and ideal voltmeters with infnite resistance. Scientists requently begin with a very simple model o a system and then explore the possibilities that this model can oer in terms o analysis and behaviour. The next step is to make the model more complicated (but without being too intricate!) and to see how much complexity is needed beore the model resembles the real system that is being modelled. Do the simplifcations and assumptions o ideal behaviour orm a suitable basis or modelling?
224
(b)
Figure 6
Figure 6( a) shows a simple model or a cell. Inside the dotted box is an ideal cell that has no resistance o its own. Also inside the box is a resistance symbol that represents the internal resistance o the cell. The two together make up our model or a real cell. The model assumes that the internal resistance is constant ( or a practical cell it varies with the state o discharge) and that the em is constant ( which also varies with discharge current) . The model cell has an em and an internal resistance r in series with an external resistance R ( fgure 6( b) ) . The current in the circuit is I. We can apply Kirchhos second law to this circuit: the em o the cell supplying energy to the circuit
=
the sum o the pds
= IR + Ir
So = IR + Ir I the pd across the external resistor is V, then = V + Ir or V = Ir It is important to realize that V, which is the pd across the external resistance, is equal to the terminal pd across the ideal cell and its internal resistance ( in other words between A and B ) . The em f is the o p en circuit p d across the term inals of a p o wer so urce in other words, the term inal p d when no current is sup p lied. The pd between A and B is less than the em unless the current in the circuit is zero. The dierence between the em and the term inal p d ( the measured pd across the terminal o the cell) is sometimes reerred to as the lost pd or the lost volts . These lost volts represent the energy required to drive the electron charge carriers through the cell itsel. O nce the energy has been used in the cell in this way, it cannot be available or conversion in the external circuit. You may have noticed that when a secondary cell is being charged, or any cell is discharging at a high current, the cell becomes warm. The energy required to raise the temperature o the cell has been converted in the internal resistance.
5 . 3 E LECTRI C CE LLS
Investigate! Measuring the internal resistance of a fruit cell The method given here works or any type o electric cell, but here we use a citrus ruit cell ( orange, lemon, lime, etc.) or even a potato or the measurement. The ions in the esh o the ruit or the potato react with two dierent metals to produce an em. With an external circuit that only requires a small current, the ruit cell can discharge over surprisingly long times.
with the equation or a straight line y = c + mx A plot o V on the y-axis against I on the x-axis should give a straight line with a gradient o r and an intercept on the V-axis o .
V
Cu
A set o results or a cell ( not a ruit cell in this case) is shown together with the corresponding graph o V against I. The intercept or this graph is 1 .2 5 V and the gradient is 2 .4 V A 1 giving an internal resistance value o 2 .4 . 1.4
Zn
lemon
terminal pd /V
1.2 1 0.8 0.6 0.4 0.2 0 0
0.1
A
Figure 7
0.2
0.3 0.4 current/A
Terminal pd / V
Current / A
1.13
0.05
1.01
0.10
0.89
0.15
0.77
0.20
0.65
0.25
0.53
0.30
0.41
0.35
To make the cell, take a strip o copper oil and a strip o zinc oil, both about 1 cm by 5 cm and insert these, about 5 cm apart deep into the ruit. You may need to use a knie to make an incision unless the oil is sti enough. C onnect up the circuit shown in fgure 7 using a suitable variable resistor. Measure the terminal pd across the ruit cell and the current in it or the largest range o pd you can manage.
0.29
0.40
0.17
0.45
C ompare the equation
0.05
0.50
V = Ir
0.5
0.6
Figure 8
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5
E L E C T R I C I T Y AN D M AG N E T I S M
Worked example 1
A cell o em 6.0 V and internal resistance 2 .5 is connected to a 7.5 resistor. C alculate: a) the current in the cell b) the terminal pd across the cell c) the energy lost in the cell when charge fows or 1 0 s.
2
A battery is connected in series with an ammeter and a variable resistor R. When R = 6.0 , the current in the ammeter is 1 .0 A. When R = 3 .0 , the current is 1 .5 A. C alculate the em and the internal resistance o the battery.
Solution Using
Solution
V = Ir
a) Total resistance in the circuit is 1 0 , so 6 current in circuit = __ = 0. 60 A 10
and knowing that V = IR gives two equations: 6 1 =1 r
b) The terminal pd is the pd across cell = IR = and
0.6 7.5 = 4.5 V c) In 1 0 s, 6 C fows through the cell and the energy lost in the cell is 1 . 5 J C 1 . The energy lost is 9.0 J.
3 1 .5 = 1 .5 r These can be solved simultaneously to give ( 6 4.5 ) = 0.5 r or r = 3 . 0 and = 9.0 V.
Power supplied by a cell
Worked example
The total power supplied by a non-ideal cell is equal to the power delivered to the external circuit plus the power wasted in the cell. Algebraically,
A battery o em 9.0 V and internal resistance 3 .0 is connected to a load resistor o resistance 6.0 . C alculate the power delivered to the external load.
P = I2 R + I2 r using the notation used earlier. The power delivered to the external resistance is 2 _ R ( R + r) 2
Solution Using the equation ______ R ( R + r) 2
2
9 leads to _____ 6 = 6.0 W (6+ 3) 2
power delivered to load resistor
2
0
r 0
load resistance R
Figure 9 Power delivered to a resistor.
Figure 9 shows how the power that is delivered to the external circuit varies with R. The peak o this curve is when r = R, in other words, when the internal resistance o the power supply is equal to the resistance o the external circuit. The load and the supply are matched when the resistances are equal in this way. This matching o supply and circuit is important in a number o areas o electronics.
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5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
5.4 Magnetic efects o electric currents Understanding Magnetic felds Magnetic orce
Applications and skills Sketching and interpreting magnetic feld patterns Determining the direction o the magnetic feld
based on current direction Determining the direction o orce on a charge moving in a magnetic feld Determining the direction o orce on a currentcarrying conductor in a magnetic feld Solving problems involving magnetic orces, felds, current and charges
Equations orce on a charge moving in a magnetic ield:
F = qvB sin orce on a current-carrying conductor in a magnetic ield: F = ILB sin
Nature of science Sometimes visualization aids our understanding. Magnetic ield lines are one o the best examples o this. Scientists began by visualizing the shape and strength o a magnetic ield through the
position and direction o the ictitious lines o orce (or ield lines) . The image proved to be so powerul that the technique was subsequently used with electric and gravitational ields too.
Introduction Electromagnetism is the third eect observed when charge moves in a circuit the electric current gives rise to a magnetic feld. B ut it was not the observation o a magnetic eect arising rom a current in a wire that began the ancient study o magnetism. Early navigators knew that some rocks are magnetic. As with the nature o electric charge, the true origins o magnetic eects remained obscure or many centuries. Only comparatively recently has an understanding o the microscopic aspects o materials allowed a ull understanding o the origins o magnetism. As with electrostatics, scientists began by using the concept o the feld to describe the behaviour o interacting magnets. However, magnetic felds dier undamentally in character rom electrostatic felds and are in some ways more complex.
Magnetic feld patterns The repulsion between the like poles o two bar magnets is amiliar to us. The orces between magnets o even quite modest strength are impressive. Modern magnetic alloys can be used to produce tiny magnets
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5
E L E C T R I C I T Y AN D M AG N E T I S M ( less than 1 cm in diameter and a ew millimetres thick) that can easily attract another erromagnetic material through signifcant thicknesses o a non-magnetic substance. At the beginning o a study o magnetism, it is usual to describe the orces in terms o felds and feld lines. You may have met this concept beore. There is said to be a magnetic feld at a point i a orce acts on a magnetic pole ( in practice, a pair o poles) at that point. Magnetic felds are visualized through the construction o feld lines. Magnetic feld lines have very similar ( but not identical) properties to those o the electric feld lines in S ub- topic 5 .1 . In summary these are:
Magnetic feld lines are conventionally drawn rom the northseeking pole to the south-seeking pole, they represent the direction in which a north- seeking pole at that point would move.
The strength o the feld is shown by the density o the feld lines, closer lines mean a stronger feld.
The feld lines never cross.
The feld lines act as though made rom an elastic thread, they tend to be as short as possible.
Nature of science Talking about poles S
A word about notation: there is a real possibility o conusion when talking about magnetic poles. This partly arises rom the origin o our observations o magnetism. When we write magnetic north pole what we really mean is the magnetic pole that seeks the geographic north pole ( fgure 1 ( a) ) . We oten talk loosely about a magnetic north pole pointing to the north pole. Misunderstandings can occur here because we also know that like poles repel and unlike poles attract. S o we end up with the situation that a magnetic north pole is attracted to the geographic north pole which seems wrong in the context o two poles repelling. In this book we will talk about north- seeking poles meaning geographic north- seeking and south- seeking poles meaning geographic south seeking . O n the diagrams, N will mean geographic north seeking, S will mean geographic south seeking.
N
(a)
N
S
(b)
S
N
(c)
Figure 1 ( b) shows the patterns or a single bar magnet and ( c) and ( d) two arrangements o a pair o bar magnets o equal strength.
N
N
(d)
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Figure 1 Magnetic feld patterns.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
In the pairs, notice the characteristic feld pattern when the two opposite poles are close ( fgure 1 ( b) ) and when the two north- seeking poles are close ( fgure 1 ( c) ) . When two northseeking poles are close ( or two south- seeking poles) , there is a position where the feld
is zero between the magnets ( called a null point) . When two opposite poles are close, the feld lines connect the two magnets. In this situation the magnets will be attracted, so this particular feld pattern implies an attraction between poles.
Investigate! Observing feld patterns o permanent magnets and electric currents
There are a number o ways to carry out this experiment. They can involve the scattering o small iron flings, observation o suspensions o magnetized particles in a special liquid, or other techniques. This Investigate! is based on the iron-fling experiment but the details will be similar i you have access to other methods.
Take a bar magnet and place a piece o rigid white card on top o it. You may need to support the card along its sides. C hoose a nonmagnetic material or the support.
Take some iron flings in a shaker ( a pepper pot is ideal) and, rom a height above the card o about 2 0 cm sprinkle flings onto the card. It is helpul to tap the card gently as the flings all onto it.
You should see the feld pattern orming as the magnetic flings all through the air and come under the inuence o the magnetic feld. S ketch or photograph the arrangement.
The iron flings give no indication o feld direction. The way to observe this is to use a plotting compass a small magnetic compass a ew centimetres in diameter that indicates the direction to which a north-seeking pole sets itsel. Place one or more o these compasses on the card and note the direction in which the north- seeking pole points.
Repeat with two magnets in a number o confgurations; try at least the two in fgure 1 .
Electric currents also give rise to a magnetic feld. However, currents small enough to be sae will only give weak felds, not strong enough to aect the flings as they all.
Figure 2 To improve the eect: cut a small hole in the centre o the card and run a long lead through it ( the lead will need to be a ew metres long) . Loop the lead in the same direction through a number o times ( about ten turns i possible) . This trick enables one current to contribute many times to the same feld pattern. You may need at least 2 5 A in total ( 2 . 5 A in the lead) to see an eect.
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Magnetic feld due to a current in conductors The magnetic feld pattern due to a current in a long straight wire is a circular pattern centred on the wire. This seems odd to anyone only used to the bar magnet pattern. Measurements show that the feld is strong close to the wire but becomes weaker urther away rom it. This should be clear in your drawings o the long wire feld pattern when the lines o orce are drawn at increased spacing as you move urther rom the wire. point o arrow indicates current leaving page towards you
tail o arrow indicates current entering page away rom you
lines o magnetic feld
south pole
north pole
Figure 3 Magnetic feld patterns around current-carrying conductors.
fngers curl around the conductor (indicating the direction o magnetic feld)
Your observations using the plotting compasses should have shown that the direction o the feld depends on the direction o the current. current ow
thumb points in direction o conventional current
Figure 4 Right-hand corkscrew rule.
230
Using the conventional current ( i.e. the direction that positive charges are moving in the wire) the relationship between the current and the magnetic feld direction obeys a right-hand corkscrew rule relationship . To remember this, hold your right hand with the fngers curled into the palm and the thumb extended away rom the fngers ( see fgure 4) . The thumb represents the direction o the conventional current and the fngers represent the direction o the feld. Another way to think o the currentfeld relationship is in terms o a
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
screwdriver being used to insert a screw. The screwdriver has to turn a right-handed screw clockwise to insert it and drive the screw orwards. The direction in which the imaginary screw moves is that o the conventional current, and the direction in which the screwdriver turns is that o the feld. Use whichever direction rule you preer, but use it consistently and remember that the rule works or conventional current. The strength o the magnetic feld can be increased by increasing the current in the conductor. The magnetic feld due to the solenoid is amiliar to you already. To understand how it arises, you need to imagine the long straight wire being coiled up into the solenoid shape taking its circular feld with it as the coiling takes place. With current in the wire, the circular feld is set up in each wire. This circular feld adds together with felds rom neighbouring turns in the solenoid. Figure 5 shows this; look closely at what happens close to the individual wires. The black lines show the feld near the wires, the blue lines show how the felds begin to combine, the red line shows the combined feld in the centre o the solenoid.
felds add together circular feld near turn consecutive turns on solenoid
Figure 5 Building up the feld pattern.
A feld runs along the hollow centre o the solenoid and then outside around the solenoid ( fgure 3 ) . O utside it is identical to the bar magnet pattern to the extent that we can assign north- and south- seeking poles to the solenoid. Again there is an easy way to remember this using an N and S to show the north- seeking and south-seeking poles. The arrows on the N and S show the current direction when looking into the solenoid rom outside. I you look into the solenoid and the conventional current is anticlockwise at the end o the solenoid closer to you then that is the end that is north- seeking. I the current is clockwise then that is the south- seeking end. The strength o the magnetic feld in a solenoid can be increased by:
increasing the current in the wire
increasing the number o turns per unit length o the solenoid
adding an iron core inside the solenoid.
anticlockwise current gives a north seeking pole
clockwise current gives a south seeking pole
Figure 6 Right hand corkscrew rule and pole direction.
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Worked example Four long straight wires are placed perpendicular to the plane o the paper at the edges o a square. The same current is in each wire in the direction shown in the diagram. D educe the direction o the magnetic feld at point P in the centre o the square.
Solution The our feld directions are shown in the diagram. The sum o these our vectors is another vector directed rom point P to the let.
A
B D B P
P
A C D
C
Nature of science But why permanent magnets? We have said nothing about the puzzle o permanent magnets or the magnetism o the Earth ( Figure 1 ( a) ) . The suggestion is that magnetism arises between charges when the charges move relative to each other. C an this idea be extended to help explain the reasons or permanent magnetism ( known as erromagnetism) ? In act, permanent magnetism is comparatively rare in the periodic table, only iron, nickel, and cobalt and alloys o these metals show it. The reason is due to the arrangement o the electrons in the atoms o these substances. Electrons are now known to have the property o spin which can be imagined as an orbiting motion around the atom. In iron, cobalt and nickel, there is a particular arrangement that involves an unpaired electron. This is the atomic origin o the moving charge that is needed or a magnetic feld to appear.
232
The second reason why iron, nickel, and cobalt are strong permanent magnets is that neighbouring atoms can co-operate and line up the spins o their unpaired electrons in the same direction. So, many electrons are all spinning in the same direction and giving rise to a strong magnetic feld. Deep in the centre o the Earth it is thought that a liquid-like metallic core containing ree electrons is rotating relative to the rest o the planet. Again, these are conditions that can lead to a magnetic feld. In which direction do you predict that the electrons are moving? However, this phenomenon is not well understood and is still the subject o research interest. Why, or example, does the magnetic feld o the Earth ip every ew thousand years? There is much evidence or this including the magnetic striping in the undersea rocks o the mid-Atlantic ridge and in the anomalous magnetism ound in some ancient cooking hearths o the aboriginal peoples o Australia.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
Forces on moving charges Force between two current-carrying wires We have used feld ideas to begin our study o magnetic eects, but these conceal rom us the underlying physics o magnetism. To begin this study we look at the interactions that arise between conductors when they are carrying electric current. +
foil strips
+
-
(a)
currents in same directions (b)
current into page
current out of page (c)
Figure 7
Figure 7( a) shows two oil strips hanging vertically. The current directions in the oils can be the same or opposite directions. When the currents are in the same direction, the strips move together due to the orce on one oil strip as it sits in the magnetic feld o the other strip. When the currents are in opposing directions, the strips are seen to move apart. You can set this experiment up or yoursel by using two pieces o aluminium oil about 3 cm wide and about 70 cm long or each conductor. The power supply should be capable o providing up to 2 5 A so take care! C onnections are made to the oils using crocodile (alligator) clips.
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5
E L E C T R I C I T Y AN D M AG N E T I S M The orces on the oil can be explained in terms o the interactions between the felds as shown in the fgure 7( b) . When the currents are in the same direction, the feld lines rom the oils combine to give a pattern in which feld lines loop around both oils. The notation used to show the direction o conventional current in the oil is explained on the diagram. Look back at fgure 1 ( d) which shows the feld pattern or two bar magnets with the opposite poles close. You know that the bar magnets are attracted to each other in this situation. The feld pattern or the oils is similar and also leads to attraction. Think o the feld lines as trying to be as short as possible. They become shorter i the oils are able to move closer together. When the currents are in opposite directions, the feld pattern changes ( fgure 7( c) ) . Now the feld lines between the oils are close together and in the same direction thus representing a strong feld. It seems reasonable that a strong magnetic feld ( like a strong electrostatic feld) represents a large amount o stored energy. This energy can be reduced i the oils move apart allowing the feld lines to separate too. Again, this has similarities to the bar magnet case but this time with like poles close together.
Force between a bar magnet feld and a current-carrying wire O ne extremely important case is the interaction between a uniorm magnetic feld and that produced by a current in a wire. Again we can start with the feld between two bar magnets with unlike poles close. In the centre o the region between the magnets, the feld is uniorm because the feld lines are parallel and equally spaced.
N
S
+
=
N
S orce on wire
(a)
(b)
Figure 8 The catapult feld.
S uppose a wire carrying a current sits in this feld. Figure 8( a) shows the arrangement and the directions o the uniorm magnetic feld and the feld due to the current. A orce acts downwards on the wire. The eect can again be explained in terms o the interaction between the two magnetic felds. The circular feld due to the wire adds to the uniorm feld due to the magnets to produce a more complicated feld. This is shown in fgure 8( b) . O verall, the feld is weaker below the wire than above it. Using our ideas o the feld lines, it is clear that the system ( the wire and the uniorm feld) can overcome this dierence in feld strength, either by attempting to move the wire downwards or by moving the magnets that cause the uniorm feld upwards. This eect
234
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
is sometimes called the catap ult feld because the feld lines above the wire resemble the stretched elastic cord o a catapult j ust beore the obj ect in the catapult is fred. This eect is o great importance to us. It is the basis or the conversion o electrical energy into kinetic energy. It is used in electric motors, loudspeakers, and other devices where we need to produce movement rom an electrical power source. It is called the motor eect. It is possible to predict the direction o motion o the wire by drawing the feld lines on each occasion when required but this is tedious. There are a number o direction rules that are used to remember the direction o the orce easily. O ne o the best known o these is due to the English physicist Fleming and is known as Flemings let-hand rule. orce (motion) along thuMb
feld along F irst fnger
F B I let hand
current along seCond fnger
Figure 9 Flemings let-hand rule.
To use the rule, extend your let hand as shown in fgure 9. Your frst ( index) fnger points in the direction o the uniorm magnetic feld and your second fnger points in the direction o the conventional current in the wire, then your thumb gives the direction o the orce acting on the wire.
The motor efect The explanation or the motor eect has been given so ar in terms o feld lines. This is, o course, not the complete story. We should be looking or explanations that involve interactions between individual charges, both those that produce the uniorm magnetic feld and those that arise rom the current in the wire. E lectrostatic eects ( as their name implies) arise between charges that are not moving. Magnetic eects arise because the sets o charges that produce the felds are moving relative to each other and are said to be in dierent rames o reerence. There have been no electrostatic eects because we have dealt with conductors in which there is an exact balance o positive and negative charges. Magnetism can be thought o as the residual eect that arises when charges are moving with respect to each other.
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Worked example Wires P and R are equidistant rom wire Q.
D escribe the direction o the orce acting on wire Q due to wires P and R.
Solution I
I
I
wire P
wire Q
wire R
Using the right-hand corkscrew rule, the feld due to wire P at wire Q is out o the plane o the paper, and the feld due to wire R at wire Q is also out o the plane o the paper.
Each wire carries a current o the same magnitude and the currents are in the directions shown.
Using Flemings let- hand rule, the orce on wire Q is in the plane o the paper and to the let.
The magnitude of the magnetic force
Investigate! Force on a current-carrying conductor current carrying lead
Arrange the apparatus as shown in the diagram.
Zero the balance so that the weight o the magnets is removed rom the balance reading.
The experiment is in two parts:
l
First, vary the current in the wire and collect data or the orce acting on the magnets, and thereore the balance, as a result. A trick to improve precision is to reverse the current in the wire and take balance readings or both directions. Then add the two together ( ignoring the negative sign o one reading) to give double the answer. D raw a graph to display your data.
Second, use two pairs o magnets side-by-side to double the length o the feld. Take care that the poles match, otherwise the orces will cancel out. This is not likely to work so well as you need to assume that the magnet pairs have the same strength. This will probably not be true. But, roughly speaking, does doubling the length o wire in the feld double the orce?
magnets on steel yoke
C
236
0
balance
Figure 10
This is an experiment that will give you an idea o the size o the magnetic orce that acts on typical laboratory currents. I carried out careully it will also allow you to see how the orce varies with the length o the conductor and the size o the current.
You will need some pairs o at magnets ( known as magnadur magnets, a sensitive top- pan balance, a power supply and a suitable long straight lead to carry the current.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
The result o experiments like the one above is that the orce acting on the wire due to the feld is proportional to:
the length ( l) o the wire
the current ( I) in the wire.
This leads us to a defnition o magnetic feld strength rather dierent rom that o electric feld strength and gravitational feld strength. We cannot defne the magnetic feld strength in terms o orce __ a single quantity because the orce depends on two quantities: current and length. Instead we defne magnetic feld strength orce acting on a current element = _____ current in the element length o the element B y element we mean a short section o a wire that carries a current. I a orce F acts on the element o length l when the current in it is I, then the magnetic feld strength B is defned by F B= _ Il
B
l
B
l
F = BIl sin
F = BIl (a)
(b)
Figure 11
The unit o magnetic feld strength is the tesla, abbreviated T and the tesla is equivalent to the undamental units kg s 2 A 1 . When a 1 metre long current element is in a magnetic feld and has a current o 1 A in it, i a magnetic orce o 1 N acts on it, then the magnetic feld strength is defned to be 1 T. The tesla can also be thought o as a shortened orm o N A- 1 m -1 . The tesla turns out to be a very large unit indeed. The largest magnetic feld strengths created in a laboratory are a ew kT and the magnetic feld o the Earth is roughly 1 0 4 T. The very largest felds are associated with some neutron stars. The feld strength can be order o 1 00 GT in such stars. Having defned B we can go on to rearrange the equation to give the orce that acts on a wire: F = BIl This applies when the feld lines, the current and the wire are all at 90 to each other as they are when you use the Fleming let-hand rule. I this is not the case (see fgure 1 1 (b) ) and the wire is at an angle to the lines then we need to take the appropriate component o I or l that is at 90 to the feld.
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5
E L E C T R I C I T Y AN D M AG N E T I S M In terms o the way the angle is defned in fgure 1 1 ( b) , the equation becomes F = BIl sin This equation is written in terms o the current in the wire. O course, the current is, as usual, the result o moving charge carriers. The equation can be changed to reect this. From S ub-topic 5 .1 we know that the current I is given by Q I= _ t where Q is the charge that ows through the current element taking a time t to do it. S ubstituting
( ) l = BQ ( _ ) sin t
Q F = B _ l sin t
The term ( _lt ) is the drift sp eed v o the charge carriers and making this substitution gives the expression F = BQv sin or the orce acting on a charge Q moving at speed v at an angle to a magnetic feld o strength B.
TOK Why direction rules? During this sub-topic we have introduced two direction rules: the corkscrew rule or magnetic felds around wires and Flemings rule or the motor eect. What is the status o these rules? Are they implicit in the way the universe operates, or are they simply indications that depend on the way we defne current direction and other undamental quantities in physics? To begin to answer this, consider what would have happened i Franklin had known that charge carriers in metals were electrons with a negative charge. What would this have changed (i anything)?
238
Notice the way that the angle is defned in the diagrams. It is the angle between the direction in which the charge is moving ( or the current direction the same thing) and the feld lines. D ont get this wrong and use cosine instead o sine in your calculations.
Nature of science The defnition o the ampere In Topic 1 you learnt that the ampere was defned in terms o the orce between two current- carrying wires. You may have thought this odd both then and when the relationship between current and charge was developed earlier in this topic. Perhaps it is clearer now. A precise measurement o charge is quite difcult. At one time it could only be achieved through chemical measurements o electrolysis and the levels o precision were not great enough to give a good value at that time. It is much easier to set up an experiment that measures the orce between two wires. The experiment can be done to a high precision using a device called a current balance ( as in the orce Investigate! above) in which the magnetic orce can be measured in terms o the gravitational orce needed to cancel it out. This is rather like oldashioned kitchen scales where the quantity being measured is j udged against a standard mass acted on by gravity.
5 . 4 M AG N E TI C E FFE CTS O F E LE CTRI C C U RR E N TS
Worked example 1
When a charged particle o mass m, charge q moves at speed v in a uniorm magnetic feld then a magnetic orce F acts on it. D educe the orce acting on a particle mass m o charge 2 q and speed 2 v travelling in the same direction in the same magnetic feld.
Solution The equation or the orce is F = BQv sin In the equation, sin and B do not change but every other quantity does, so the orce is 4F. 2
Electric currents o magnitude I and 3 I are in wires 1 and 2 as shown. wire 1
wire 2
A orce F acts on wire 2 due to the current in wire 1 . D educe the magnitude o the orce on wire 1 due to the current in wire 2 .
Solution This problem can be solved by reerence to Newtons laws o motion, but an alternative is to consider the changes in magnetic feld. I the magnetic feld at wire 2 due to wire 1 is B then the magnetic feld at wire 1 due to wire 2 is 3 B. However, the current in wire 1 is one-third o that in wire 2 . S o orce at wire 1 due to wire 2 is I F = 3 B __ l = orce in wire 2 due to wire 1 . 3
The orces are the same.
I
3I
Nature of science Vectors and their products You may have ormed the view that vectors are j ust used or scale diagrams in physics. This is not the case, vectors come into their own in mathematical descriptions o magnetic orce. There is only one way to multiply scalars together: run a relay with our stages each o distance 1 00 m and the total distance travelled by the athletes is 400 m. Multiplying two scalars only gives another scalar. Vectors, because o the added direction, can be multiplied together in two ways:
to orm a scalar product ( sometimes called dot product) where, or example, orce and displacement are multiplied together to give work done ( a scalar) that has no direction. In vector notation this is written as F s = W. The multiplication sign is the dot ( hence the name) the vectors are written with a bold ont, the scalar in ordinary ont. to orm a vector product ( sometimes called cross product) where two vectors are
multiplied together to give a third vector, thus a b = c The multiplication o qv and B to orm the vector orce F is a vector product. The charge q is a scalar but everything else in the equation is a vector. A mathematician would write F = q ( v B) to show that the vector velocity and the vector magnetic feld strength are multiplied together. The order o v and B is important. There is a vector rule or the direction o F that is consistent with our observations earlier and the sin appears when the vector multiplication is worked out in terms o the separate components o the vector. Vector notation turns out to be an essential language o physics, because it allows a concise notation and because it contains all the direction inormation within the equations rather than orcing us to use direction rules. However we do not pursue the ull theory o vectors in IB Physics.
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Questions 1
( IB) Four point charges o equal magnitude, are held at the corners o a square as shown below. 2a
The electric orce between them is + F (i.e. attractive) . The spheres are touched together and are then returned to their original separation. a) Calculate the charge on X and the charge on Y.
+Q
+Q
b) C alculate the value o the electric orce between them ater being returned to their original separation. ( 7 marks) 2a
P
4
(IB) Two charged plastic spheres are separated by a distance d in a vertical insulating tube, as shown.
Q Q
tube spheres
The length o each side o the square is 2 a and the sign o the charges is as shown. The point P is at the centre o the square. a) ( i)
d
D etermine the magnitude o the electric feld strength at point P due to one o the point charges.
( ii) O n a copy o the diagram above, draw an arrow to represent the direction o the resultant electric feld at point P. The charge on each sphere is doubled. Friction with the walls o the tube is negligible.
( iii) D etermine, in terms o Q, a and k, the magnitude o the electric feld strength at point P. ( 7 marks)
D educe the new separation o the spheres. ( 5 marks)
2
(IB) Two point charges o magnitude +2Q and Q are fxed at the positions shown below. D iscuss the direction o the electric feld due to the two charges between A and B . S uggest at which point the electric feld is most likely to be zero. ( 3 marks) Q
A
3
+2Q +
+
(IB) a)
Electric felds may be represented by lines o orce. The diagram below shows some lines o orce.
B
(IB) Two identical spherical conductors X and Y are mounted on insulated stands. X carries a charge o +6.0 nC and Y carries a charge o 2.0 nC.
+6.0 nC
conductor X
240
5
A
B
2.0 nC
insulated stands
conductor Y
( i)
State whether the feld strength at A and at B is constant, increasing or decreasing, when measured in the direction rom A towards B .
QUESTION S d) An alternative circuit or measuring the VI characteristic uses a potential divider.
( ii) E xplain why feld lines can never touch or cross. b) The diagram below shows two insulated metal spheres. Each sphere has the same positive charge.
( i)
D raw a circuit that uses a potential divider to enable the VI characteristics o the flament to be ound.
( ii) Explain why this circuit enables the potential dierence across the lamp to be reduced to 0 V. ( 1 3 marks)
+
+
7
(IB) The graph below shows the VI characteristic or two 1 2 V flament lamps A and B . lamp B
12
6
(IB) A lamp is at normal brightness when there is a potential dierence o 1 2 V across its flament and a current in the flament o 0.5 0 A.
potential diference/V
C opy the diagram and in the shaded area between the spheres, draw the electric feld pattern due to the two spheres. ( 8 marks)
lamp A
0 0
0.5 current/A
a) For the lamp at normal brightness, calculate: ( i) the power dissipated in the flament ( ii) the resistance o the flament. b) In order to measure the voltagecurrent ( VI) characteristics o the lamp, a student sets up the ollowing electrical circuit. 12 V battery
a) ( i)
1.0
E xplain why the graphs indicate that these lamps do not obey O hms law.
( ii) S tate and explain which lamp has the greater power dissipation or a potential dierence o 1 2 V. The two lamps are now connected in series with a 1 2 V battery as shown below. 12 V battery
S tate the correct positions o an ideal ammeter and an ideal voltmeter or the characteristics o the lamp to be measured. c) The voltmeter and the ammeter are connected correctly in the previous circuit. Explain why the potential dierence across the lamp ( i)
cannot be increased to 1 2 V
( ii) cannot be reduced to zero.
lamp A
lamp B
b) ( i) State how the current in lamp A compares with that in lamp B . ( ii) Use the VI characteristics o the lamps to deduce the total current rom the battery. ( iii) C ompare the power dissipated by the two lamps. ( 1 1 marks)
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E L E C T R I C I T Y AN D M AG N E T I S M 8
( IB) a)
X
E xplain how the resistance o the flament in a flament lamp can be determined rom the VI characteristic o the lamp.
2.0
b) A flament lamp operates at maximum brightness when connected to a 6.0 V supply. At maximum brightness, the current in the flament is 1 2 0 mA. ( i)
E
( i) C opy the graph in ( a) , and draw the IV characteristics or the resistor R.
C alculate the resistance o the flament when it is operating at maximum brightness.
( ii) You have available a 2 4 V supply and a collection o resistors o a suitable power rating and with dierent values o resistance. C alculate the resistance o the resistor that is required to be connected in series with the supply such that the voltage across the flament lamp will be 6.0 V. ( 4 marks) 9
R
(IB) The graph below shows the IV characteristics or component X. I/A6
( ii) D etermine the total potential dierence E that must be applied across component X and across resistor R such that the current through X and R is 3 .0 A. ( 7 marks)
1 0 (IB) A student is to measure the currentvoltage ( IV) characteristics o a flament lamp. The ollowing equipment and inormation are available.
Battery Filament lamp
4
Voltmeter
2
Ammeter 0 -8
-6
-4
-2
0
2
4
6
-2 -4
8 V/V
Potentiometer
Information emf = 3.0 V, negligible internal resistance marked 3 V, 0.2 A resistance = 30 k, reads values between 0.0 and 3.0 V resistance = 0.1 , reads values between 0.0 and 0.5 A resistance = 100
a) For the flament lamp operating at normal brightness, calculate: ( i) its resistance
-6
( ii) its power dissipation. The student sets up the ollowing incorrect circuit.
The component X is now connected across the terminals o a battery o em 6. 0 V and negligible internal resistance. a) Use the graph to determine:
V
( i) the current in component X ( ii) the resistance o component X. b) A resistor R o constant resistance 2 .0 is now connected in series with component X as shown below.
A
b) ( i) E xplain why the lamp will not light. ( ii) S tate the approximate reading on the voltmeter. E xplain your answer. ( 6 marks)
242
QUESTION S 1 1 ( IB) A particular flament lamp is rated at 1 2 V, 6.0 mA. It j ust lights when the potential dierence across the flament is 6.0 V.
a) ( i) S tate the value o the current or which the resistance o X is the same as the resistance o Y and determine the value o this resistance.
A student sets up an electric circuit to measure the IV characteristics o the flament lamp.
( ii) D escribe and suggest an explanation or the IV characteristic o conductor Y.
A
b) The two conductors X and Y are connected in series with a cell o negligible internal resistance. The current in the conductors is 0. 2 0 A.
100 k V
12 V
Use the graph to determine:
S
( i) the resistance o Y or this value o current ( ii) the em o the cell.
In the circuit, shown below, the student has connected the voltmeter and the ammeter into the circuit incorrectly.
( 8 marks)
1 3 ( IB) A cell o electromotive orce ( em) E and internal resistance r is connected in series with a resistor R, as shown below.
The battery has em 1 2 V and negligible internal resistance. The ammeter has negligible resistance and the resistance o the voltmeter is 1 00 k. The maximum resistance o the variable resistor is 1 5 . a) Explain, without doing any calculations, whether there is a position o the slider S at which the lamp will be lit. b) Estimate the maximum reading o the ammeter. ( 5 marks)
r
E
R
The cell supplies 8.1 1 0 3 J o energy when 5 .8 1 0 3 C o charge moves completely round the circuit. The current in the circuit is constant. a) C alculate the em E o the cell.
1 2 ( IB) The graph below shows the currentvoltage ( IV) characteristics o two dierent conductors X and Y. 0.50 0.45 0.40 0.35 0.30 Y X I/A 0.25 0.20 0.15 14 0.10 0.05 0.00 0.0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 . 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 V/V
b) The resistor R has resistance 6.0 . The potential dierence between its terminals is 1 .2 V. D etermine the internal resistance r o the cell. c) C alculate the total energy transer in R. d) D escribe, in terms o a simple model o electrical conduction, the mechanism by which the energy transer in R takes place. ( 1 2 marks)
( IB) A battery is connected in series with a resistor R. The battery transers 2 000 C o charge completely round the circuit. D uring this process, 2 5 00 J o energy is dissipated R and 1 5 00 J is expended in the battery. C alculate the em o the battery. ( 3 marks)
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E L E C T R I C I T Y AN D M AG N E T I S M 1 5 ( IB) A student connects a cell in series with a variable resistor and measures the terminal pd V o the cell or a series o currents I in the circuit. The data are shown in the table.
V/V 1.50 1.10 0.85 0.75 0.60 0.50
I/mA 120 280 380 420 480 520
The electric feld strength is 3 .8 1 0 5 V m 1 and the magnetic feld strength is 2 .5 1 0 2 T. C alculate the speed o the electron i the net orce acting on it due to the felds is zero. ( 3 marks)
1 8 ( IB) A straight wire lies in a uniorm magnetic feld as shown. current I magnetic feld
Use the data to determine the em and internal resistance o the cell. ( 5 marks)
1 6 ( IB) A battery is connected to a resistor as shown. The current in the wire is I and the wire is at an angle o to the magnetic feld. The orce per unit length on the conductor is F. D etermine the magnetic feld strength. ( 2 marks) 6.0 V
10
V
1 9 ( IB) A straight wire o length 0. 75 m carries a current o 3 5 A. The wire is at right angles to a magnetic feld o strength 0.05 8 T. C alculate the orce on the wire. ( 2 marks) When the switch is open the voltmeter reads 1 2 V, when the switch is closed it reads 1 1 . 6 V. a) E xplain why the readings dier. b)
( i) S tate the em o the battery. ( ii) C alculate the internal resistance o the battery.
c) C alculate the power dissipated in the battery. ( 6 marks)
1 7 ( IB) An electron enters a pair o electric and magnetic felds in a vacuum as shown in the diagram. region o magnetic feld
electron
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E
+ B
2 0 ( IB) An ion with a charge o + 3 . 2 1 0 1 9 C and a mass o 2 . 7 1 0 26 kg is moving due south at a speed o 4.8 1 0 3 m s 1 . It enters a uniorm magnetic feld o strength 4. 6 1 0 4 T directed downwards towards the ground. D etermine the orce acting on the ion. ( 4 marks)
6 CI RCU LAR M O TI O N AN D G RAVI TATI O N Introduction Two apparently distinct areas of physics are linked in this topic: motion in a circle and the basic ideas of gravitation. B ut of course they are not distinct at all. The motion of a satellite about its planet involves both a consideration of the
gravitational force and the mechanics of motion in a circle. Man cannot travel beyond the E arth without a knowledge of both these aspects of Physics.
6.1 Circular motion Understanding Period, frequency, angular displacement, and
angular velocity Centripetal force Centripetal acceleration
Nature of science The drive to develop ideas about circular motion came from observations of the universe. How was it that astronomical objects could move in circular or elliptical orbits? What kept them in place in their motion? Scientists were able to deduce that there must be a force acting radially inwards for every case of circular motion that is observed. Whether it is a bicycle going around a corner or a planet orbiting its star, the physics is the same.
Applications and skills Identifying the forces providing the centripetal
forces such as tension, friction, gravitational, O B J TE XT_UND electrical, or magnetic Solving problems involving centripetal force, centripetal acceleration, period, frequency, angular displacement, linear speed, and angular velocity Qualitatively and quantitatively describing examples of circular motion including cases of vertical and horizontal circular motion
Equations speedangular speed relationship: v = r 4 2 r v2 centripetal acceleration: a = ____ = ________ r
mv 2 2 centripetal force: F = _______ r = m r
T2
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Moving in a circle Most children take great delight in an obj ect on a string whirling in a circle though they may be less happy with the consequences when the string breaks and the obj ect hits a window! Rides at a theme park and trains on a railway are yet more examples o movement in a circle. What is needed to keep something rotating at constant speed? The choice o term (as usual in physics! ) is very deliberate. In circular motion we say that the speed is constant but not the velocity is constant.
Figure 1
A airground carousel.
Velocity, as a vector quantity, has both magnitude and direction. The obj ect on the string has a constant speed but the direction in which the obj ect is moving is changing all the time. The velocity has a constant magnitude but a changing direction. I either o the two parts that make up a vector change, then the vector is no longer constant. Whenever velocity changes (even i it is only the direction) then the object is accelerated. Understanding the physics o this acceleration is the key to understanding circular motion. B ut beore looking at how the acceleration arises we need a language to describe the motion.
Angular displacement The angle moved around the circle by an obj ect rom where its circular motion starts is known as the angular disp lacement. Unlike the linear displacement used in Topic 2 , angular displacement will not be considered to be a vector in IB Physics. Angular displacement is the angle through which the obj ect moves and it can be measured in degrees ( ) or in radians ( rad) . Radians are more commonly used than degrees in this branch o physics. I you have not met radians beore, read about the dierences between radians and degrees.
Nature of science
s
Radians or degrees C alculations o circular motion involve the use o angles. In any science you studied beore starting this course you will almost certainly have measured all angles in degrees.
r s = r
1 1 ( degree) is defned to be ___ th o the way 360 around a circle.
In some other areas o physics (including circular motion) there is an alternative measure o angle that is much more convenient, the radian. Radians are based on the geometry o the arc o a circle. 1 radian ( abbreviated as rad) is defned as the angle equal to the circumerence o an arc o a circle divided by the radius o the circle. In symbols s =_ r
246
Figure 2
Defnition o radian.
Going around the circle once means travelling around the circumerence; this is a distance o 2 r. The angle in radians subtended by the 2 r whole circle is ___ r = 2 rad. S o 3 60 = 2 = 6.2 8 rad and 1 rad = 5 7.3
6 .1 CI R CU L AR M O TI O N
S ometimes, the radian numbers are let as ractions, so 1 90 = _ _ round the circle , 2 4
( ) 1 3 0 = _ ( _ round the circle ) 6 12 and so on.
To convert other values or yoursel, use the angle in de gre e an gle in radians equation ___________ = ___________ 2 360
There are some similarities between the sine o an angle and the angle in radians. The two quantities are compared in this Nature o S cience box which shows sin and in radians. Notice that, as becomes smaller, sin ( ) and become closer together. From angles o 1 0 down to 0, the dierences between sin and are very small and in some calculations and proos we treat sin and as being equal ( this is known as the small angle approximation) . For small angles cos approximates to zero radians.
To illustrate this, here are the values o sin and in radians or our angles: 90, 45 , 1 0, and 5 . Notice how similar the sine values and the radians are or 1 0 and 5 . _ sin ( 90) = 1 .000; rad = 1 .5 71 rad 2 _ sin ( 45 ) = 0.707; rad = 0.785 rad 4 _ sin ( 1 0) = 0.1 75 ; rad = 0.1 74 rad 18 _ sin ( 5 ) = 0.087; rad = 0.087 rad 36 Finally, a practical point: S cientifc and graphic calculators work happily in either degrees or radians ( and sometimes in another type o angular measure known as grad too) . B ut the calculator has to be told what to expect! Always check that your calculator is set to work in radians i that is what you want, or in degrees i those are the units you are using. You will lose calculation marks in an examination i you conuse the calculator!
, angular speed = t t = t 2 1
Angular speed In Topic 2 we used the term speed to mean linear speed. When the motion is in a circle there is an alternative: angular sp eed, this is given the symbol ( the lower-case Greek letter, omega) .
time t2
time t 1
angular displacement average angular speed = ____ time for the angular displacement to take place
Figure 3 shows how things are defned and you will see that in symbols the defnition becomes = _ t where is the angular displacement and t is the time taken or the angular displacement.
Figure 3
Angular speed.
Nature of science Angular speed or angular velocity? You may be wondering about the distinction between angular speed and angular velocity, and whether angular velocity is a vector similar to linear velocity. The answer is that angular velocity is a vector but an unusual one. It has a magnitude equal to the angular speed, but its direction is surprising! The direction is along the axis o rotation, in other
direction of direction rotation of angular velocity vector Figure 4 Angular velocity
direction.
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words, through the centre of the circle around which the obj ect is moving and perpendicular to the plane of the rotation. The direction follows a clockwise corkscrew rule so that in this example the direction of the
angular velocity vector is into the plane of the paper. In the IB course, only the angular speed the scalar quantity is used.
Period and frequency The time taken for the obj ect to go round the circle once is known as the p eriodic time or simply the p eriod of the motion, it has the symbol T. In one period, the angular distance travelled is 2 rad. S o, 2 T= _ When T is in seconds the units of are radians per second, abbreviated to rad s 1 . If you have already studied waves in this course, you might have met the idea of time period the time for one cycle. Another quantity that is associated with T is frequency. Frequency is the number of times an object goes round a circle in unit time (usually taken to be 1 second) , so one way to express the unit of frequency would be in per second or s 1 . However, the unit of frequency is re-named after the 1 9th century physicist Heinrich Hertz and is abbreviated to Hz. There is a link between T and f so that: 1 T= _ f This leads to a link between and f = 2 f
Worked example
Solution
A large clock on a building has a minute hand that is 4.2 m long.
a) The minute hand goes round once ( 2 rad) every hour.
C alculate: a) the angular speed of the minute hand b) the angular displacement, in radians, in the time periods
O ne hour is 3 600 s angular displacement angular speed = __ time taken 2 _ = = 0.001 75 rad s 1 3 600 2 1 b) ( i) 2 0 minutes is __ of 2 , so ___ rad 3 3
( i) 1 2 noon to 1 2 .2 0
( ii) 2 .5 h is 2 2 .5 = 5 rad
( ii) 1 2 noon to 1 4.3 0. c) the linear speed of the tip of the minute hand.
c) v = r = 4.2 0.001 75 = 0.007 3 3 m s 1 = 7. 3 mm s 1
Linking angular and linear speeds S ometimes we know the linear speed and need the angular speed or vice versa. The link is straightforward: When the circle has a radius r the circumference is 2 r, and T, is the time taken to go around once. S o the linear speed of the obj ect along the edge of the circle v is 2 r v= _ T
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6 .1 CI R CU L AR M O TI O N
Rearranging the equation gives 2 r T= _ v We have j ust seen that 2 T= _ so equating the two equations or T gives 2 r _ 2 _ v = C ancelling the 2 and rearranging gives v = r Notice that, in both this equation and in the earlier equation s = r, the radius r multiplies the angular term to obtain the linear term. This is a consequence o the defnition o the angular measure.
Centripetal acceleration E arlier we showed that an obj ect moving at a constant angular speed in a circle is being accelerated. Newtons frst law tells us that, or any obj ect in which the direction o motion or the speed is changing, there must be an external orce acting. In circular motion the direction is constantly changing and so the obj ect accelerates and there must be a orce acting on it to cause this to happen. In which direction do the orce and the acceleration act, and what are their sizes? The diagram shows two points P 1 and P 2 on the circle together with the velocity vectors vold and vnew at these points. The vectors are the same length as each other because the speed is constant. However, vold and vnew point in dierent directions because the obj ect has moved round the circle by an angular distance between P 1 and P 2 . Acceleration is, as usual, change o velocity ___ time taken or the change The change in velocity is the change-o- velocity vector v that has to be added to vold in order to make it become the same length and direction as vnew. Identiy these vectors on the diagram.
v vnew
vold
P2 vnew
v
2
vold v P1
vold
2
vnew Figure 5 Proof of centripetal
acceleration direction.
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6
C I R C U L AR M O T I O N AN D G R AVI TAT I O N Notice that vold and vnew slide round the circle to meet. Where does the new vector v point? The answer is: to the centre o the circle. This is an averaging process to fnd out what the dierence is between vold and vnew hal- way between the two points. This averaging can be taken urther. The time, t, to go between P 1 and P 2 , and the linear distance around the circle between P 1 and P 2 ( which is r) are related by r t = _ v Using some trigonometry on the diagram shows that v _ = v sin _ 2 2 The size o the average acceleration a that is directed towards the centre o the circle is
( )
2 v sin ( ___ 2 ) v a = _= _ 2 r __ ___ t v
2
This can be written as sin ( ___ 2 ) v _ a= _ r ___ 2
2 sin ( 2 ) is almost exactly equal to 1 and When is very small, the ratio ______ ___ ___
2
so the instantaneous acceleration a when P 1 and P 2 are very close together is
Nature of science Linking it together Notice that some o these equations have interesting links elsewhere: mv is, or example, the magnitude o the linear momentum multiplied by . Try to be alert or these links as they will help you to piece your physics together.
v2 2 a=_ r = r = v directed to the centre o the circle. This acceleration is at 90 to the velocity vector and it points inwards to the centre o the circle.
The orce that acts to keep the obj ect moving in a circle is called the centrip etal force and this orce leads to a centrip etal acceleration. ( The origin o the word centripetal comes rom two Latin words centrum and petere literally to lead to the centre.)
Centripetal force Newtons second law o motion in its simpler orm tells us that F = ma using the usual symbols. The second law applies to the orce that provides the centripetal v 2 acceleration, so the magnitude o the orce = m __ r = m r = mv. The question we need to ask or each situation is: what orce provides the centripetal orce or that situation? The direction o this orce must be along the radial line between the obj ect and the centre o the circle. 2
Investigate! Investigating how F varies with m, v and r This experiment tests the relationship 2
v m_ r = Mg
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To do this a bung is whirled in a horizontal circle with a weight hanging rom one end o a string and mass (rubber bung) on the other end.
6 .1 CI R CU L AR M O TI O N
A paper clip is attached to the string below a glass tube. The clip is used to ensure that the radius o rotation o the bung is constant the bung should be rotated at a speed so that the paper clip j ust stays below the glass tube.
To veriy the equation you need to test each variable against the others. There are a number o possible experiments in each o which one variable is held constant (a control variable) , one is varied (the independent variable) , and the third (the dependent variable) is measured. One example is:
centripetal force apparatus r mass, m
Variation of v with r
In this experiment, m and M must be unchanged. Move the clip to change r, and or each value o r, measure v using the method given above.
Analysis: v2 _ r = constant A graph o v2 against r ought to be a straight line passing through the origin. Alternatively you could, or each experimental run, simply divide v2 by r and look critically at the answer (which should be the same each time) to see i the value is really constant. I going down this route, you ought to assess the errors in the experiment and v put error limits on your __ r value.
glass tube paper clip
string
weight (Mg) Figure 6 Centripetal
force, mass, and speed.
The tension in the string is the same everywhere ( whether below the glass tube or above in the horizontal part) . This tension is F in the equation and is equal to Mg where M is the mass o the weight ( hanging vertically) . Use a speed at which you can count the number o rotations o m in a particular time and rom this work out the linear speed v o mass m.
2
What are the other possible experimental tests?
In practice the string cannot rotate in the horizontal plane because o its own weight. How can you improve the experiment or the analysis to allow or this?
Nature of science Centripetal or centrifugal?
direction of car
When discussing circular motion, you will almost certainly have heard the term centriugal orce probably everywhere except in a physics laboratory! In this course we have spoken exclusively about centripetal orce. Why are there two terms in use? It should now be clear to you how circular motion arises: a orce acts to the centre o the circle around which the obj ect is moving. The alternative idea o centriugal orce comes rom common experience. Imagine you are in a car going round a circle at high speed. You will undoubtedly eel as i you are being fung outwards. O ne way to explain this is to imagine the situation rom the vantage point o a helicopter hovering
straight on direction at this instant
real centripetal force supplied by friction at tyres car Figure 7
Centripetal forces in a car seen from above.
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stationary above the circle around which the car is moving. From the helicopter you will see the passenger attempting to go in a straight line ( Newtons frst law) , but the passenger is orced to move in a circle through riction orces between passenger and seat. I the passenger were sitting on a riction- less seat and not wearing a seat belt, then he or she will not get the message that the car is turning. The passenger continues to move in a straight line eventually meeting the door that is turning with the car. I there were no door, what direction will the passenger take? Another way to explain this is to imagine yoursel in the car as it rotates. This is a rotating rame o reerence that is accelerating and as such cannot obey Newtons laws o motion. You instinctively think that the rotating rame is actually stationary. Thereore your tendency to go in what you believe to be a straight line actually eels like an outward orce away rom the centre o the circle ( remember the rest o the world now rotates round you, and your straight line is actually part o a circle) . Think about a cup o coee sitting on the oor o the car. I there is insufcient riction at the base o the cup, the cup will slide to the side o the car. In the inertial rame o reerence ( the Earth) the cup is trying to go in a straight line. In your rotating rame o reerence you have to invent a orce acting outwards rom the centre o the circle to explain the motion o the cup.
direction of movement observed by passenger
cup
force imagined to act on cup
centre of circle
Figure 8 Rotation
forces.
There are many examples o changing a reerence rame in physics: research the Foucault pendulum and perhaps go to see one o these ascinating pendulums in action. Look up what is meant by the C oriolis orce and fnd out how it aects the motion o weather systems in the northern and southern hemispheres. One o the tricks that physicists oten use is to change reerence rames its all part o the nature o science to adopt alternative rames o reerence to make explanations and theories more accessible. O ne last tip: D ont use explanations based on centriugal orce in an IB examination. The real orce is centripetal; centriugal orce was invented to satisy Newtons second law in an accelerated rame o reerence.
Centripetal accelerations and forces in action Satellites in orbit Figure 9 shows satellites in a circular orbit around the Earth. Why do they ollow these paths? Gravitational orces act between the centre o mass o the Earth and the centre o mass o the satellite. The direction o the orce acting on the satellite is always towards the centre o the planet and it is the gravity that supplies the centripetal orce. Figure 9
Satellites in orbit.
Amusement park rides Many amusement park rides take their passengers in curved paths that are all or part o a circle. How does circular motion provide a thrill? In the type o ride shown in fgure 1 0, the people are inside a drum that rotates about a vertical axis. When the rotation speed is large enough the people are orced to the sides o the drum and the oor drops away. The people are quite sae however because they are held against the inside o the drum as the reaction at the wall provides the centripetal orce to keep them moving in the circle. The people in the ride eel the reaction between their spine and the wall. Friction between the rider and the wall prevents the rider rom slipping down the wall.
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6 .1 CI R CU L AR M O TI O N
Reaction of wall on rider N
Figure 10
friction force weight
The rotor in action.
Turning and banking When a driver wants to make a car turn a corner, a resultant orce must act towards the centre o the circle to provide a centripetal orce. The car is in vertical equilibrium ( the driving surace is horizontal) but not in horizontal equilibrium.
Turning on a horizontal road For a horizontal road surace, the riction acting between the tyres and the road becomes the centripetal orce. The riction orce is related to the coefcient o riction and the normal reaction at the surace where riction occurs.
direction reaction, R
friction friction
centre of circle
elevation Figure 11
mg, W
plan
Car moving in a circle.
I the car is not to skid, the centripetal orce required has to be less than the rictional orce v m_ r < smg where s is the static coefcient o riction. Note that when the vehicle is already skidding the less than sign becomes an equality and the dynamic coefcient o riction should be used. 2
____
This rearranges to give a maximum speed o vmax = drg or a circle o radius r.
Banking Tracks or motor or cycle racing, and even ordinary roads or cars are sometimes banked (fgures 1 2 and 1 3) . The curve o the banked road surace is inclined at an angle so that the normal reaction orce contributes to the centripetal orce that is needed or the vehicle to go round the track
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6
N cos
C I R C U L AR M O T I O N AN D G R AVI TAT I O N at a particular speed. B icycles and motorcycles can achieve the same eect on a level road surace by leaning in to the curve. Tyres do not need to provide so much riction on a banked track compared to a horizontal road; this reduces the risk o skidding and increases saety.
N
Although you will not be asked to solve mathematical problems on this topic in your IB Physics examination, you do need to understand the principles that underpin banking.
N sin
mg Figure 12
Forces in banking.
Figure 1 2 shows orces acting on a small sphere rolling round a track. This is simplifed to a point obj ect moving in a circle to remove the complications o two or our wheels. A horizontal centripetal orce directed towards the centre o the circle is needed or the rotation. The other orces that act on the ball are the orce normal to the surace ( which is at the banking angle ) and its weight acting vertically down. The vector sum o the horizontal components o the weight and the normal orce must equal the centripetal orce. Looking at this another way, i N is the normal orce then the centripetal orce is equal to N sin
normal reaction
The normal orce resolved vertically is N cos and is, o course, equal mg sin = mg tan and opposite to mg. S o Fcentripetal = ( ____ co s ) 2
centripetal force
friction
2
v mv __ Fcentripetal = ___ r and thereore tan = gr
The banking angle is correct at a particular speed and radius. Notice that it does not depend on the mass o the vehicle so a banked road works or a cyclist and a car, provided that they are going at the same speed. weight
S ome more examples o banking:
C ommercial airline pilots y around a banked curve to change the direction o a passenger j et. I the angle is correct, the passengers will not eel the turn, simply a marginal increase in weight pressing down on their seat) .
S ome high- speed trains tilt as they go around curves so that the passengers eel more comortable.
Moving in a vertical circle
Figure 13
Cycle velodrome.
S o ar the examples have been o motion around a horizontal circle. People will queue or a long time to experience moderate ear on a airground attraction like the rollercoaster in fgure 1 4. The amount o thrill rom the ride depends on its height, speed, and also the orces that act on the riders. How is the horizontal situation modifed when the circular motion o the mass is in a vertical plane?
1 What are the forces acting when the motion is in a vertical circle? Imagine a mass on the end o a string that is moving in a vertical circle at constant speed. Look careully at fgure 1 5 and notice the way the tension in the string changes as the mass goes around.
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6 .1 CI R CU L AR M O TI O N
Figure 14 Theme park ride.
B egin with the case when the string is horizontal, at point A. The weight acts downwards and the tension in the string is the horizontal centripetal force towards the centre of the circle. The mass continues to move upwards and reaches the top of the circle at B . At this point the tension in the string and the weight both act downwards. Thus: v2 Tdown + mg = m _ r and therefore v2 Tdown = m _ r mg The weight of the mass combines with the tension to provide the centripetal force and so the tension required is less than the tension T when the string is horizontal. At C , the bottom of the circle, the tension and the weight both act vertically but in opposite directions and so v2 Tup = m _ r + mg At the bottom, the string tension must overcome weight and also provide the required centripetal force.
B Tdown
D
mg
T
T A
A mg
mg
Tup mg C Figure 15 Forces in
circular motion
in a vertical plane.
As the mass moves around the circle, the tension in the string varies continuously. It has a minimum value at the top of the circle and a maximum at the bottom. The bottom of the circle is the point where the string is most likely to break. If the maximum breaking tension of the string is Tbreak, then, for the string to remain intact, v2 Tbreak > m _ r + mg and the linear speed at the bottom of the circle must be less than ____________ r _ m ( Tbreak mg) If this seems to you to be a very theoretical idea without much practical value, think about a car going over a bridge. If you assume that the shape of the bridge is part of a circle, then there is a radius of curvature r. What is the speed at which the car will lose contact with the bridge?
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v
radius of curvature
Figure 16 Car going over a
bridge.
This is the case considered above, where the object, in this case the car, is at the top o the circle. What is the tension (in this case the orce between car and road) i the car wheels are to lose contact with the bridge? To answer this question, you might begin with a ree-body diagram. You __ should be able to show that the car loses contact at a speed equal to gr .
2 How does speed change when motion is in a vertical circle? Not all circular motion in a vertical circle is at a constant speed. As a mass moves upwards it slows as kinetic energy is transerred to gravitational potential energy ( i there is nothing to keep it moving at constant speed) . At the top o the motion the mass must not stop moving or even go too slowly, because i it did then the string would lose its tension. The motion would no longer be in a circle. The centripetal orce Fc needed to maintain the motion is v Fc = m __ r as usual, at the top o the circle, i Fc is supplied entirely by gravity then v2 Fc = mg = m _ r Just or an instant, the obj ect is in ree- all. 2
The equation can be rearranged to give __
vtop = gr
and this is the minimum speed at the top o the circle or which the motion will still be circular. The minimum speed does not depend on mass. Energy is conserved assuming that there are no losses ( or example, to internal energy as a result o air resistance as the mass goes round) . Equating the energies: kinetic energy at top + gravitational potential energy dierence between top and bottom = kinetic energy at the bottom and 1 1 mv 2 _ m v top 2 + mg( 2 r) = _ bottom 2 2 B y substituting or both tensions, Tbottom and Ttop , it is possible to show that Tbottom = Ttop + 6m You can fnd this proo on the website.
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Worked examples
Solution
1
A hammer thrower in an athletics competition swings the hammer on its chain round 7.5 times in 5.2 s before releasing it. The hammer describes a circle of radius 4.2 m and has a mass of 4.0 kg. Assume that the hammer is swung in a horizontal circle and that the chain is horizontal.
a) ( i) 7.5 revolutions = 1 5 rad
a) C alculate, for the rotation:
b) The thrower usually inclines the plane of the circle at about 45 to the horizontal in order to achieve maximum range. E ve n if the plane were horizontal, the n the we ight of the hammer would contribute to the system so that a component of the tension in the chain must allow for this. B oth assumptions are unlikely.
( i) the average angular speed of the hammer ( ii) the average tension in the chain. b) C omment on the assumptions made in this question.
1 5 angular speed = _ = 9.1 rad s 1 5.2 ( ii) Tension in the chain = centripetal force required for rotation centripetal force = mr 2 = 4.0 4.2 9.1 2 = 1 400 N
6.2 Newtons law of gravitation Understanding
Applications and skills
Newtons law o gravitation
Describing the relationship between
Gravitational feld strength
gravitational orce and centripetal orce Applying Newtons law o gravitation to the motion o an object in circular orbit around a point mass Solving problems involving gravitational orce, gravitational feld strength, orbital speed, and orbital period Determining the resultant gravitational feld strength due to two bodies
Nature of science Newtons insights into mechanics and gravitation led him to develop laws o motion and a law o gravitation. One o his motion laws and the law o gravitation are mathematical in nature, two o the motion laws are descriptive. None o these laws can be proved and there is no attempt in them to explain why the masses are accelerated under the inuence o a orce, or why two masses are attracted by the orce o gravity. Newtons ideas about motion have been subsequently modifed by the work o Einstein. The questioning and insight that leads to the development o laws are undamental to the nature o science.
Equations Mm Newtons law o gravitation: F = G ______ 2 r
F gravitational feld strength: g = ____ m
gravitational feld strength and the gravitational M constant: g = G ____ r2
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N
Nature of science Scientists from the past Three o the great names rom the history o astronomy and physics were C opernicus, Tycho B rahe, and Kepler. Their contributions were linked and ultimately led to the important gravitational work o Newton. Try to fnd out something about these astronomers. D uring the lietimes o these scientists, science was carried out in a very dierent way rom today. The realization that the Earth orbits the Sun rather than the Sun orbiting the Earth was one o the great developments in scientifc understanding.
Copernicus (Mikolaj Kopernik)
Isaac Newton Figure 1
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An astronomers portrait gallery.
Galileo Galilei and other scientists o the 1 6th century overcame cultural, philosophical and religious prejudices, and some even suered persecution or the scientifc truths they had discovered. As we study the work o these pioneers we should remember that scientists in past times were not always as ree as those today. Research the lie o Galileo ( we oten drop the second name) and explore why he and others came into conict with the Roman C atholic C hurch over their scientifc belies.
Johannes Kepler
Tycho Brahe
Galileo Galilei
6 . 2 N E W T O N S L AW O F G R AV I T AT I O N
Gravitational feld strength Like electrostatics, gravity acts at a distance and is an example o a orce that has an associated orce feld. Imagine two masses in deep space with no other masses close enough to inuence them. O ne mass ( call it A) is in the orce feld due to the second mass ( B ) and a orce acts on A. B is in the gravitational feld o A and also experiences a orce. These two orces have an equal magnitude ( even though the masses may be dierent) but act in opposite directions.
orce on A due to B
orce on B due to A
A B Figure 2
Gravitational orces between two masses.
I both masses are small, the size o a human, say, the orce o gravity is extremely weak. O nly when one o the masses is as large as a planet does the orce become noticeable to us. However, whatever the size o the mass we need a way to measure the strength o the feld to which it gives rise. Gravitational orces are the weakest o all the orces in nature and so require large amounts o mass or the orce to be elt.
large mass M
gravitational feld strength =
The strength o a gravitational feld is defned using the idea o a small test mass. This test mass has to be so small that it does not disturb the feld being measured. I the test mass is large then it will exert a orce o its own on the mass that produces the feld being measured. The test mass would then accelerate the other mass and alter the arrangement that is being measured. The gravitational orce that acts on the small test mass has both magnitude and direction. These are shown in the diagram. The test mass will accelerate in this direction i it is ree to move.
F m
orce, F small test mass, m
Figure 3
Defnition o gravitational feld strength.
Defning gravitational feld strength The concept o the small test mass leads to a defnition o the strength o the gravitational feld. I the mass o the test mass is m, and the feld is producing a gravitational orce o F on the test mass, then the gravitational feld strength ( given the symbol g) is defned as F g= _ m The units o gravitational feld strength are N kg 1 . S ince F is a vector and m a scalar, it ollows that g is a vector quantity and that its direction is that o F. A ormal defnition in words is that gravitational feld strength at a p oint is the orce p er unit mass exp erienced by a small p oint mass p laced at that p oint. This defnition requires that the test mass is not j ust small but is also an ( infnitesimally) small point in space. You might want to consider what the eect might be i the test mass has a shape and extends in space.
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N S o ar the discussion has been limited to the gravitational feld produced by one point mass. How does the situation change i there is more than one mass, excluding the test mass itsel?
resultant orce 43 = 1 unit feld strength B 3 units
feld strength A 4 units
Field strength is independent o the magnitude o the point test mass ( we divided F by m to achieve this) . S o the vector feld strengths can be added together ( fgure 4) .
g1
g2 resultant g
In IB Physics examinations you will only be asked to add the feld strengths o masses that all lie on the same straight line. B ut even in two dimensions, the addition is straightorward using the ideas o vector addition by drawing or by calculation.
g and the acceleration due to gravity Figure 4 Adding feld
vectorially.
strengths
Sometimes students are surprised that the symbol g is used or the gravitational feld strength, they think there might be a risk o conusion with g the acceleration due to gravity! However this does not happen. At the E arths surace ( using Newtons second law) acceleration due to gravity at the surace orce on a mass at the surace due to gravity = ____ size o the mass F F __ S o the acceleration = __ m but m is also the defnition o gravitational feld strength so the acceleration due to gravity = gravitational feld strength = g.
The magnitude o the gravitational feld strength ( measured in N kg 1 ) is equal to the value o the acceleration due to gravity ( measured in m s 2 ) . You should be able to show that N kg 1 m s 2 . ( The symbol means is equivalent to.)
Newtons law of gravitation an inverse-square law Isaac Newton ( 1 642 1 72 7) was a B ritish scientist who consolidated the work o others and who added important insights o his own. D uring his lie he contributed to the study o optics, mechanics and gravitation. O ne o his greatest triumphs was work on gravity that he developed using the data and ideas o the German astronomer Johannes Kepler and others about the motion o the planets. Newton realized that the gravitational orce F between two obj ects with masses M and m whose centres are separated by distance r is:
always attractive
1 proportional to __
proportional to M and m.
r2
This can be summed up in the equation Mm F _ r2 1 Laws that depend on __ are known as inverse- square. I the distance r2
between the two masses is doubled without changing mass, then the orce between the masses goes down to one-quarter o its original value. To use this equation numerically a constant o proportionality is needed and is given the symbol G,
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6 . 2 N E W T O N S L AW O F G R AV I T AT I O N GMm F= _ r2 G is known as the universal gravitational constant and it has an accepted value o 6.67 1 0 1 1 N m 2 kg2 . Gravity is always attractive so i the distance is measured from the centre o mass M to mass m then the orce on m due to M is towards M. In other words, the orce is in the opposite direction to the direction in which the distance is measured. You may see some books where a negative sign is introduced to predict this direction. The IB Diploma Programme physics course does not attribute negative signs to attractive orces, the responsibility o keeping track o the orce direction is with you.
Nature of science A universal constant? What does it mean to say that G is a universal gravitational constant? Newton did not know what the size o the constant was ( as the value was determined over a hundred years later by C avendish) . In a sense, it did not matter. Newton realized that all obj ects are attracted to the Earth the apocryphal story o him seeing a alling apple reminds us that he knew this.
The insight that Newton had was to realize that the orce o gravity went on beyond the apple tree and stretched up into the sky, to the Moon, and beyond. He realized that the Moon was alling continuously towards the E arth under the inuence o gravity and because it was also moving horizontally it was in continuous orbit.
Worked examples 1
Solution
C alculate the orce o attraction between an apple o mass 1 00 g and the E arth. Mass o Earth = 6. 0 1 0 24 kg. Radius o Earth = 6.4 Mm
GMm F= _ r 6. 7 1 0 1 1 1 . 7 1 0 27 9.1 1 0 31 = ____ ( 1 . 5 1 0 1 0 ) 2 2
Solution GMm 6.7 1 0 1 1 0.1 6.0 1 0 24 F = _ = ___ r ( 6. 4 1 0 6 ) 2
= 4. 6 1 0 48 N
2
( C ompare this with the electrostatic attraction o the electron and proton at this separation o about 1 0 8 N.)
= 1 .1 N 2
C alculate the orce o attraction between a proton o mass 1 .7 1 0 27 kg and an electron o mass 9.1 1 0 3 1 kg a when they are at a distance o 1 . 5 1 0 1 0 m apart.
Gravitational feld strengths re-visited Knowledge o Newtons law o gravitation means that the defnition o gravitational feld strength can be taken urther to help our study o real situations.
The feld strength at a distance r rom a point mass, M The simplest case is that o a single point mass M placed, as usual, a long way rom any other mass. In practice we say that the point mass is at an infnite 1 distance rom any other mass because i r is very large then __ is extremely r 1 __ small (mathematicians say that as r tends to infnity, tends to zero) . 2
r
2
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N As usual, the mass o our small test obj ect is m. This means that the magnitude o the orce F between the two masses M and m is GMm F=_ r2 so that the gravitational feld strength g is F GM _ g=_ m = r2 As beore, the direction is measured outwards from M but the orce on the point mass is in the opposite direction towards M.
The feld strength at a distance r rom the centre outside a sphere o mass M It turns out that the answer or g outside a spherical planet is exactly the same as g or the point mass j ust quoted: GM g= _ 2 r
I we are outside the sphere, all the mass acts as though it is a point mass o size M positioned at the centre o the sphere (the centre o mass) . We only need one equation or both point masses and spheres.
Nature of science And inside the Earth...? Science is all about asking questions. Outside the 1 Earth, the gravitational feld strength varies as __ . But r what happens to gravity inside the Earth? Is it zero? Is it a constant? Does it become larger and larger, reaching infnity, as we get closer to the centre? You might, at frst sight, expect this rom Newtons law. 2
this is the part that accounts or gravity
out. The second thing that happens is that only the mass inside the smaller Earth contributes to the gravitational pull. The mass o this smaller Earth varies with r3 assuming a constant density or the Earth. B ecause the gravitational orce varies 1 with __ , together these give an overall dependence r o r. The whole graph or the variation o g or a planet (inside and out) is given in fgure 6, gs is the gravitational feld strength at the surace 2
tunnel
this part does not contribute Figure 5 Journey
to the centre o the Earth.
In Journey to the centre of the Earth the novelist Jules Verne imagined going through a volcanic tunnel to the centre o the planet. Visualize his travellers halway down the tunnel as they stand on the surace o a smaller Earth defned by their present distance rom the centre. Two remarkable things happen: The frst is that the contribution rom the shell o Earth above the travellers makes no contribution to the gravitational feld; all the dierent parts o the outer shell cancel
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gravitational feld strength, g
gs
1g 4 s 1g 9 s
0
R 2R 3R distance rom centre o planet o radius R
4R
Figure 6 Gravitational feld strength inside and outside the Earth.
6 . 2 N E W T O N S L AW O F G R AV I T AT I O N
Linking orbits and gravity The gravitational orce o a planet provides the centripetal orce to keep a satellite in orbit. cannon hill
impact i Earth is fat impact i Earth is a sphere in orbit Earth
constant distance rom surace Figure 7
Newtons cannon how he thought about orbits.
Newton had an insight into this too. He used the example o a cannon on a high mountain (fgure 7) . The cannon fres its cannonball horizontally and it accelerates vertically downwards. On a at Earth, it will eventually hit the ground. Newton knew that the Earth was a sphere and that the curvature o the earth allowed the ball to travel urther beore hitting the ground. He then imagined the ball being fred at larger and larger initial speeds. E ventually the shell will travel horizontally at such a high speed that the curvature o the E arth and the curve o the traj ectory will be exactly the same. When this happens the distance between shell and surace is constant and the shell is in orbit around the E arth. What do you expect to happen to the trajectory o the cannon ball i it is fred at even greater speeds? To check your conclusions, fnd an applet on the Internet that will allow you to vary the fring speed o Newtons cannon. A good starting point or the search is applet Newton gun.
Worked examples 1
( Mass o Moon = 7.3 1 0 22 kg; radius o the Moon = 1 .7 1 0 6 m)
Solution GM g=_ r2 6.7 1 0 1 1 7.3 1 0 22 = ___ ( 1 .7 1 0 6 ) 2 = 1 .7 N kg 1
This motion o a satellite around the Earth can be analysed by combining the ideas o centripetal orce and gravitational attraction. The gravitational attraction FG provides the centripetal orce Fc , so ( ignoring signs) GM E m mv 2 _ Fc = FG = _ r = r2 where ME is the mass o the Earth, r is the distance rom the satellite to the centre o the Earth, v is the linear speed o the satellite, and m is its mass. These equations can be simplifed to ____
v=
GM _ r E
This equation predicts the speed o the satellite at a particular radius. Notice that the speed does not depend on the mass o the satellite. All satellites travelling round the Earth at the same distance above the surace have the same speed. As an exercise, use the equations on page 2 5 0 to show that GM E 2 = _ r3
C alculate the gravitational feld strength at the surace o the Moon.
2
C alculate the gravitational feld strength o the S un at the position o the Earth. (Mass o Sun = 2.0 1 0 30 kg; EarthS un distance = 1 .5 1 0 1 1 m.)
Solution GM g= _ r2 1 1 6. 7 1 0 2 .0 1 0 30 = ___ ( 1 .5 1 0 1 1 ) 2 = 6.0 mN kg 1
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N where is the angular speed, and that 4 2 r 3 T2 = _ GM E where T is the orbital period ( time for one orbit) of the satellite. This result: ( orbital period of a satellite) 2 ( orbital radius) 3 is known as Kep lers third law. It is one of the three laws that he discovered when he analysed Tycho B rahes data.
Worked examples 1
C alculate the orbital period of Jupiter about the Sun. Mass of Sun = 2.0 1 0 30 kg; radius of Jupiters orbit = 7.8 1 0 1 1 m)
Solution
2
The orbital time period of the Earth about the S un is 3 .2 1 0 7 s. C alculate the orbital period of Mars. ( radius of Earth orbit = 1 . 5 1 0 1 1 m; radius of Mars orbit = 2 .3 1 0 1 1 m)
2 3
4 r T2 = _ GM S
Solution
T = 3 .7 1 0 8 s ( about 1 2 years)
r 3M T2M _ _ = r 3E T2E ___ r 3M TM = TE _ = 6. 1 1 0 7 s ( about 1 .9 years) r 3E
Nature of science What Newton knew S everal decades before Newton began working on his ideas of gravitation, Kepler had used his own and others astronomical data to show that, for the planets, ( radius of the orbit) 3 ( time period of orbit) 2 This was an empirical result ( meaning that it came from experimental data) . Newton was able to show that this proportionality arose from his own theory; this was a theoretical result ( meaning that it came from a theory, not data) . Here are some data for the satellites of Jupiter.
Moon Io Europa Ganymede Callisto
Distance from centre of Jupiter/10 3 km 420 670 1070 1890
Orbital period/ days 1.8 3.6 7.2 16.7
Use these data to show that ( radius of the orbit) 3 ( time period of orbit) 2 3
( radiu s o f the o rb it) O r, put another way, that _______________ is a constant. ( tim e p e rio d o f o rb it) 2
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QUESTION S
Questions 1
A particle P is moving in a circle with uniorm speed. D raw a diagram to show the direction o the acceleration a and velocity v o the particle at one instant o time.
5
The S ingapore Flyer is a large Ferris wheel o radius 8 5 m that rotates once every 3 0 minutes.
P
2
State what provides the centripetal orce that causes a car to go round a bend.
3
State the centripetal orce that acts on a particle o mass m when it is travelling with linear speed v along the arc o a circle o radius r.
4
( IB) At time t = 0 a car moves o rom rest in a straight line. O il drips rom the engine o the car with one drop every 0.80 s. The position o the oil drops on the road are drawn to scale on the grid below such that 1 .0 cm represents 4.0 m. The grid starts at time t = 0.
a) C alculate the linear speed o a point on the rim o the wheel o the Flyer. b) ( i) D etermine the ractional change in the weight o a passenger on the Flyer at the top o the ride. ( ii) Explain whether the passenger has a larger or smaller apparent weight at the top o the ride. c) The capsules need to rotate to keep the oor o the cabin in the correct place. C alculate the angular speed o the capsule about its central axis.
Direction of motion 1.0 cm
6 a) ( i) S tate the eature o the diagram that indicates that the car accelerates at the start o the motion.
a) Quito in E cuador ( 1 4 minutes o arc south o the Equator)
( ii) D etermine the distance moved by the car during the frst 5 .6 s o its motion. b) The car then turns a corner at constant speed. Passengers in the car who were sitting upright eel as i their upper bodies are being thrown outwards. ( i) Identiy the orce acting on the car, and its line o action, that enables the car to turn the corner. ( ii) Explain why the passengers eel as i they are being thrown outwards.
The radius o the E arth is 6400 km. D etermine the linear speed o a point on the ground at the ollowing places on E arth:
b) Geneva in S witzerland ( 46 north o the Equator) c) the South Pole. 7
A school bus o total mass 65 00 kg is carrying some children to school. a) D uring the j ourney the bus needs to travel round in a horizontal curve o radius 1 5 0 m. The dynamic coefcient o riction between the tyres and the road surace is 0.7. Estimate the maximum speed at which the driver should attempt the turn.
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C I R C U L AR M O T I O N AN D G R AVI TAT I O N b) Later in the j ourney the driver needs to drive across a curved bridge with a radius o curvature o 75 m. Estimate the maximum speed i the bus is to remain in contact with the road. 8
A velodrome used or bicycle races has a banking angle that varies continuously rom 0 to 60. Explain how the racing cyclists use this variation in angle to their advantage in a race.
D ata needed or these questions: Radius o Earth = 6.4 Mm; Mass o E arth = 6.0 1 0 24 kg; Mass o Moon = 7.3 1 0 22 kg; Mass o S un = 2 .0 1 0 30 kg; E arthMoon distance = 3.8 1 0 8 m; S unEarth distance = 1 . 5 1 0 1 1 m; G = 6.67 1 0 1 1 N m 2 kg 2 9
D educe how the radius R o the circular orbit o a planet around a star o mass m s relates to the period T o the orbit.
1 0 A satellite orbits the Earth at constant speed as shown below.
satellite
1 1 Determine the distance rom the centre o the Earth to the point at which the gravitational feld strength o the Earth equals that o the Moon. 1 2 The ocean tides on the Earth are caused by the tidal attraction o the Moon and the S un on the water in the oceans. a) C alculate the orce that acts on 1 kg o water at the surace o the sea due its attraction by the ( i) Moon ( ii) S un. b) Optional difcult. E xplain why there are two tides every day at many coastal points on the Earth. [Hint: there are two parts to the answer, why a tide at all, and why two every day.] 1 3 There are two types o communication satellite. O ne type o communication satellite orbits over the poles at a distance rom the centre o the Earth o 7400 km; the other type is geostationary with an orbital radius o 3 6 000 km. Geostationary satellites stay above one point on the equator whereas polar- orbit satellites have an orbital time o 1 00 minutes. C alculate: a) the gravitational feld strength at the position o the polar- orbit satellite
Earth
b) the angular speed o a satellite in geostationary orbit c) the centripetal orce acting on a geostationary satellite o mass 1 .8 1 0 3 kg.
a) Explain why, although the speed o the satellite is constant, the satellite is accelerating. b) D iscuss whether or not the gravitational orce does work on the satellite.
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7 ATOMIC, NUCLEAR, AND PARTICLE PHYSICS Introduction In this topic we consider the composition o atoms. We look at extra- nuclear electrons and the nucleus and the particles o which the nucleus is composed. We see the vast array o particles that are now known to exist and how
these particles can be classifed and grouped. As is oten the case, energy plays in important role in the atom and undamental to this is the tendency or particles to be most stable when their energy is minimized.
7.1 Discrete energy and radioactivity Understanding Discrete energy and discrete energy levels Transitions between energy levels Radioactive decay
Applications and skills Describing the emission and absorption spectra
Alpha particles, beta particles, and gamma rays Hal-lie Constant decay probability
Absorption characteristics o decay particles Background radiation
o common gases Solving problems involving atomic spectra, including calculating the wavelength o photons emitted during atomic transitions Completing decay equations or alpha and beta decay Determining the hal-lie o a nuclide rom a decay curve Investigating hal-lie experimentally (or by simulation)
Equations Photon energyrequency relationship: E = hf hc Planck relationship or wavelength: = _____ E
Nature of science Unintentional discoveries Physics is scattered with examples o experimenters making signifcant discoveries unintentionally. In the case o radioactivity the French physicist Henri Becquerel, in 1896, was investigating a possible link between X-rays and phosphorescence. He had stored a sample o uranium salt (which releases light over a period o time having being exposed to light) in a drawer with a photographic plate (the orerunner to flm) wrapped in an opaque covering. Becquerel discovered that the plate had become exposed even though it had been kept in the dark. Rather
than ignore this unpredicted outcome, Becquerel sought an explanation and ound other uranium salts produced the same result suggesting that it was the uranium atom that was responsible. He went on to show that the emissions rom uranium ionize gases but dier rom X-rays in that they are deected by electric and magnetic felds. For his work, which developed as the result o a lucky accident, Becquerel was awarded a share o the Nobel Prize or Physics in 1903 (with the Curies or their work also on radioactivity) .
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TOK Assumptions in physics Topic 5 showed that protons and electrons are charged. A bigger question is why do charges behave in this way? The simple answer to this is that we dont know we do know that some objects have a property that we call charge and that charges interact at a distance through an electromagnetic feld. Yet we dont know what charges are and what the dierence between positive and negative actually is. This may seem a huge gap in our understanding. However, we use the same concept o the implied labelling o objects or properties or example, when we ask the question what is a chair?, our answers will inevitably give us a description o a chair along the lines o something we sit on, having our legs, it has a seat and back etc. Thereore, what we have done is described the properties and characteristics o chairs it remains the case that a chair is a chair. So, although we only have a uzzy idea o what electrons and charges actually are, does this prevent us rom developing eective theories that are based on poorly understood concepts?
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Introduction Although many texts take a historical tour o the development o atomic and nuclear physics, the IB Physics syllabus does not lend itsel to such a treatment. Instead, there is a ocus on the common aspects o the constituents o the atom. It is likely that you are already amiliar with the structure o an atom as consisting o a positively charged nucleus surrounded by negatively charged electrons in fxed orbits. You are probably also be aware that the nucleus consists o positively- charged protons and uncharged neutrons. We will take these aspects as read, but we will also revisit them throughout the sections o Topic 7 ( and, or those students studying HL Physics, Topic 1 2 too) .
Energy levels Returning to the nuclear model o an atom, the orbiting electrons cannot occupy any possible orbit around the nucleus. D ierent orbits correspond to dierent amounts o energy, or energy levels, and the electrons are restricted to orbits with specifc energies. Electrons change energy so that they can j ump rom one energy level to another, but they can only occupy allowed energy levels. Although you may ask what makes an allowed orbit or allowed energy level? it takes an understanding o the wave nature o electrons and quantum mechanics to attempt to explain this consistently. We return to this or HL students in Topic 1 2 , but or now it is maybe better to accept that this is how nature operates. Hydrogen is the simplest o all elements, with the hydrogen atom normally consisting o a proton bound to a single electron by the electromagnetic orce ( an attractive orce between oppositely charged particles) . There are other isotop es o hydrogen, with a nucleus including one or more neutrons. I a hydrogen atom completely loses the electron it becomes a positively- charged hydrogen ion. It is the single proton that defnes this nucleus to be hydrogen any atom with a single proton must be hydrogen. In the same way any atom having two protons is an isotope o helium, and any with three protons will be lithium, etc. A common visualization o the atom is the planetary model in which the electron orbits the proton mimicking the E arth orbiting the S un. D espite weaknesses in this model, it still gives a good introduction to energy levels in the atom. The energy levels or the hydrogen atom are shown in the table below:
Energy level ( n )
Energy/eV
1
13.58
2
3.39
3
1.51
4
0.85
5
0.54
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
The energy levels are usually shown diagrammatically as in fgure 1 . This diagram shows the energy level and the energy o an electron that occupies this level. An electron in the ground state o the hydrogen atom must have exactly 1 3 . 5 8 eV o energy; one in the second energy level must have 3 . 3 9 eV, etc. An atom with an electron occupying an energy level higher than the ground state is said to be excited; so an electron in level 2 is in the frst excited state, one in level 3 is in the second excited state, etc. The energy levels in atoms are said to be quantized which means they must have discrete fnite values. An electron in a hydrogen atom cannot have 5 . 92 eV or 1 0. 2 1 eV or any other value between 1 3 . 5 8 and 3 . 3 9 eV; it m ust have one o the values in the table.
Transitions between energy levels
Note n=5 n=4 second excited state n = 3
-3.39 eV
frst excited state n = 2
-13.58 eV
ground state n = 1
When an electron in the hydrogen atom jumps rom the ground state to the frst excited state it must gain some energy; this cannot be any randomly chosen amount o energy it must be exactly the right amount. The electron needs to gain ( 3.39) ( 1 3.5 8) = 1 0.1 9 eV o energy. The electron does not physically pass through the space between the two energy levels. B eore gaining the energy it must have 1 3.58 eV and immediately ater gaining the energy it must have 3.39 eV. The energy has to be transerred to the electron in one discrete amount; it cannot be gradually built up over time. The energy needed to excite an atom can come rom absorption o light by the atom. In order to understand this we must consider light to be a packet or quantum called a photon the rationale or this is discussed in Topic 1 2 . The energy, E, carried by a photon is related to the requency o the radiation by the equation E = hf where h is the Planck constant ( = 6.63 1 0 34 J s) and f is the requency in Hz. The wave equation c = f ( where c is the speed o electromagnetic waves in a vacuum) also applies to the photons and, by combining the two equations, we can c c relate the energy to the wavelength E = hf = h __ or = h __ . We have seen E that an electron needs to absorb 1 0.1 9 eV in order to j ump rom the ground state to the frst excited state. What, then, will be the wavelength o the electromagnetic radiation needed to do this? First we need to convert 1 0.1 9 eV into j oules. This is simply a matter o multiplying the energy in eV by the charge on an electron, so 1 0.1 9 eV = 1 0.1 9 1 .6 1 0 1 9 J ~ 1 .6 1 0 1 8 J.
When an electron has been excited in this way it is likely to be quite unstable and will very quickly (in about a nanosecond) return to a lower energy level. In order to do this it must lose this same amount o energy. This means that a photon o energy 1 0.1 9 eV and wavelength 1 20 nm must be emitted by the atom. The electron will then jump to the lower energy level.
The energies are given in units o
The energies are all negative
values this is because the potential energy o two objects is zero only when they are at infnite separation. Because there is an attractive orce between a proton and an electron the electron has been moved rom infnity until it is in its orbit. The protonelectron system has lost some energy and, since it was zero to start with, it is now negative.
Energy level 1 is the lowest energy
level having the most negative and, thereore, smallest amount o energy. It is the most stable state and is called the ground state.
-18
This radiation in the ultraviolet part o the spectrum.
Figure 1 Energy level diagram or hydrogen.
electron volt (1 eV = 1.6 10 19 J) . This is a very common unit (along with the multipliers keV and MeV) or atomic physics and it avoids having to use very small numbers. However, you might fnd that a question is set in joule.
8
c 3 .00 1 0 Now using = h __ we get = 6.63 1 0 - 34 J ________ = 1 .2 1 0 - 7 m E 1 .6 1 0 or 1 2 0 nm.
ionization
0 -0.54 eV -0.85 eV -1.51 eV
n is known as the principal
quantum number and we will return to its values in Topic 12.
For the transition rom the second energy level to the third to take place only 1 .88 eV needs to be absorbed by the atom and 0.66 eV to jump rom the third level to the ourth. When the electron is given the ull 1 3.5 8 eV it
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS is completely removed rom the nucleus (or taken to infnity) and the atom has become ionized. The energy supplied to the electron in the ground state to take it to infnity is called the (frst) ionization energy.
Worked example Singly- ionized helium ( He + ) is said to be hydrogen- like in that it only has one electron ( although it has two protons and two neutrons) . The energy levels dier rom those o hydrogen as shown in the ollowing table:
Energy level ( n )
Energy/eV
1
54.4
2
13.6
3
6.0
4
3.4
5
2.2
a) Explain why the ground state has a much higher energy level than that o hydrogen ( 1 3 .6 eV) . b) (i) D etermine the requency o the photon emitted by an electron transition rom energy level 4 to energy level 2 . (ii) State which region o the electromagnetic spectrum the emitted photon belongs to.
Solution a) The hydrogen atom has a single proton in the nucleus but the helium ion has two. This means the attractive orce between the nucleus and an electron is greater or the helium nucleus. The helium electron is more tightly bound to the nucleus and, thereore, requires more energy to remove it to infnity. b) (i) E = ( 1 3 . 6) ( 3 .4) = 1 0.2 eV ( the minus sign tells us that there is a loss o energy during the emission o photons) . 1 0.2 eV = 1 0.2 1 .6 1 0 1 9 J = 1 .6 1 0 1 8 J 1 . 6 1 0 -18 E E = h = _ = __ h 6.63 1 0 - 34 = 2 .4 1 0 1 5 Hz (ii) We have seen this energy or the hydrogen atom previously and so know that this requency corresponds to ultraviolet. You can calculate the wavelength by dividing the speed o electromagnetic waves by your known requency.
Nature of science Patterns in physics You may have noticed that the calculation or the energy o photon emitted by the helium ion in the worked example above is identical to an energy value or the hydrogen atom. This sort o coincidence is not unusual in physics and suggests that there may be a link between the energy dierences. In act in 1 88 8 the S wedish physicist Johannes Rydberg proposed a ormula or hydrogen- like atoms: singly
ionized helium, doubly ionized lithium, triply ionized beryllium, etc. based on observation o spectral line patterns emitted by these elements. Rydbergs theory, based on his practical work, utilized the idea o wave number ( the reciprocal o the wavelength) this was o undamental importance to the D anish physicist Niels B ohr in explaining quantization o energy levels in 1 9 1 3 .
Emission spectra When energy is supplied to a gas o atoms at low pressure the atoms emit electromagnetic radiation. In the laboratory the energy is usually supplied by an electrical discharge ( an electrical current passing through the gas when a high voltage is set up between two electrodes
270
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
across the gas) . I the radiation emitted by the gas is incident on the collimating slit o a spectrometer, it can then be dispersed by passing it though a diraction grating or a glass prism ( see fgure 5 ) . O bserving the spectrum through a telescope, or proj ecting it onto a screen, will give a series o discrete lines similar to those shown in fgure 2 .
Figure 2 Line emission spectra.
electron transitions for the hydrogen atom n=7 n=6 n=5 n=4
Brackett series
n=3
Paschen series
n=2
Balmer series
Figure 2 Line emission spectra.
Spectral lines appear in series in the dierent regions o the electromagnetic spectrum (inra-red, visible, and ultraviolet) . Each series o lines is dependent on the energy level that the electrons all to. In the case o hydrogen the series are named ater the frst person to discover them: in the Lyman series the electrons all to the ground state (n = 1 ) this series is in the ultraviolet region o the electromagnetic spectrum; in the Balmer series the electrons all to the frst excited level (n = 2) this is in the visible region; the Paschen series relates to n = 3 and is in the inrared region.
n=1
Lyman series
Figure 3 Electron transitions for hydrogen.
Absorption spectra The electrons in solids, liquids, and dense gases can also be excited they tend to glow when heated to a high temperature. When the emitted light is observed it is seen to consist o a spectrum o bands o colours rather than lines. In the case o solids this will give a continuous spectrum in which the colours are merged into each other and so are not discrete. This is typical o matter in which the atoms are closely packed. The neighbouring atoms are very close and this causes the energy levels in the atoms to change value. When there are many atoms, the overall energy levels combine to orm a series o very similar, but dierent, energies which makes up an energy band. When the object emitting a continuous spectrum is surrounded by a cool gas, then the continuous spectrum is modifed by the surrounding gas. The continuous spectrum is streaked by a number o dark lines. When a heated tungsten flament is viewed through hydrogen gas the absorption spectrum
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS shown in the lower part o fgure 4 can be seen. Further inspection shows that the black lines occur at exactly the same positions as the lines o the hydrogen emission spectrum. This pattern cannot simply be a coincidence.
Figure 4 Emission and absorption spectra or hydrogen.
Absorption occurs when an electron in an atom o the absorbing material absorbs a photon. The energy o this photon must be identical to the dierence between the energy levels. The material removes photons o this requency rom the continuous range o energies emitted by the light source. Naturally, this will make the absorbers atoms become unstable and they will revert to a lower energy level by emitting photons these will include the absorbed requencies but they will be emitted in random directions and not necessarily in the original direction. This will reduce the intensity o those specifc requencies in the original direction giving the black lines seen crossing the continuous spectrum in fgure 4. Absorption spectra or sodium can be demonstrated with the apparatus shown in fgure 5 . A white light source emits light which is incident on a diraction grating on the turntable o the spectrometer. A continuous spectrum is seen through the telescope or displayed on a computer monitor ( using an appropriate sensor and sotware) . A burner is used to heat sodium chloride ( table salt) . The black absorption lines o sodium should be detected in the yellow region o the continuous spectrum. ame with table salt added
white light source
telescope or sensor linked to computer
collimator burner
difraction grating
spectrometer
Figure 5 Demonstration o sodium absorption spectra.
Nature of science Fraunhofer lines
272
Figure 6 Stamp commemorating the lie and work o Fraunhoer.
The English scientist William Woolaston originally observed absorption lines in the Suns spectrum in 1 802 . However it was the German physicist, Joseph von Fraunhoer, who, in 1 81 4, built a spectrometer and invented the diraction grating with which he was able to observe and analyse these absorption lines now known as Fraunhofer lines. These lines were the frst lines in a spectrum to be observed. Fraunhoer labelled the most prominent o the lines as AK. These lines provide astronomers with a means to study the composition o a star.
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
Worked example a) The element helium was frst identifed rom the absorption spectrum o the S un.
diraction grating or a prism in order to disperse it into its component wavelengths. This gives evidence about the elements in the gases in the outer part o the Sun.
(i) Explain what is meant by the term absorption spectrum. (ii) O utline how this spectrum may be experimentally observed. b) O ne o the wavelengths in the absorption spectrum o helium occurs at 5 88 nm. (i) S how that the energy o a photon o wavelength 5 88 nm is 3 .3 8 1 0 - 1 9 J. (ii) The diagram below represents some o the energy levels o the helium atom. Use the inormation in the diagram to explain how absorption at 5 88 nm arises. (iii) Mark this transition on a copy o the energy level diagram below.
b)
hc 6.63 1 0 - 34 3 .00 1 0 8 (i) E = _ = ___ 5 88 1 0 - 9 = 3 .3 8 1 0 - 1 9 J (ii) As this is absorption, the electron is being raised to a higher energy level. The dierence between the levels must be equal to 3 .3 8 1 0 - 1 9 J and so this is between ( 5 . 80 1 0 - 1 9 J) and (2.42 1 0 -19 J) levels, i.e., (2.42 1 0 -19 J) (5 .80 1 0 - 1 9 J) = (+ ) 3.38 1 0 - 1 9 J
Note that the energy levels are given in joules and so there is no need to do a conversion rom electronvolts in this question.
0
(iii) The transition is marked in red.
0
-2.42 -3.00
-1.59 -5.80
-7.64
energy/10 -19 J
energy/10 -19 J
-1.59
-2.42 -3.00 -5.80
Solution a)
(i) An absorption spectrum consists o a continuous spectrum that has a number o absorption lines crossing it. These lines correspond to the requencies o the light in the emission spectrum o the elements within the substance that is absorbing the light. (ii) The light rom the Sun can be projected onto a screen ater passing through a
-7.64
Note that light rom the Sun needs to be fltered and should not be observed directly through a telescope.
Radioactive decay Radioactive decay is a naturally occurring process in which the nucleus o an unstable atom will spontaneously change into a dierent nuclear confguration by the emission o combinations o alpha particles, beta
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS particles, and gamma radiation. There are ewer than our hundred naturally occurring nuclides ( nuclei with a particular number o protons and neutrons) but only about 60 o these are radioactive. The trend is or these nuclei to become more stable although this may take a very long time. In radioactive decay the nuclide decaying is reerred to as the p arent and the nuclide( s) ormed as the daughter(s) . As we have discussed, the nucleus contains protons and neutrons these are jointly called nucleons. These are bound together by the strong nuclear orce, which must overcome the electrostatic repulsion between the positively charged protons. The presence o the neutrons moderates this repulsion the strong nuclear orce (which has a very short range 1 0 1 5 m) acts equally on both the protons and neutrons. For the nuclei with ew nucleons, having approximately equal numbers o protons and neutrons corresponds to nuclides being stable and not radioactive. As we will soon see, heavier nuclei need a greater proportion o neutrons in order to be stable.
Nature of science Nuclear radiation and safety Many people are apprehensive about any exposure to radiation rom radioactive sources. The danger rom alpha particles is small unless the source is ingested into the body. B eta particles and gamma rays are much more penetrating and can cause radiation burns and long term damage to D NA. Any sources used in schools are very weak but should still be treated with respect. They should always be lited with long tongs, never held near the eyes and should be kept in lead-lined boxes and stored in a locked container. D uring the 1 00 plus years since its discovery nuclear radiation has been studied extensively
and can be saely controlled. Those working with radioactive sources use radiation monitoring devices to record exposure levels. Whenever there may be concerns about nuclear radiation, exposure is limited by:
distance keeping as ar away rom a radiation source as possible
shielding placing absorbers between people and sources
time restricting the amount o time or which people are exposed.
Nuclide nomenclature This is a shorthand way o describing the composition o a nuclide. The element is described by its chemical symbol H, He, Li, B e, B , etc. The structure o the nucleus is denoted by showing both the proton number Z ( otherwise called the atomic number) and the nucleon number A ( or mass number) . This always takes the orm: AZ X so, or example, the isotopes o carbon, carbon-1 2 and carbon-1 4, are written as 1 62 C and 14 6 C . C arbon-1 2 has 6 protons and 1 2 nucleons and so it has 6 neutrons; carbon-1 4 has 6 protons and 1 4 nucleons making 8 neutrons. Isotopes are nuclides o the same element ( and so have the same number o protons) having dierent numbers o neutrons.
Alpha () decay
274
In alpha-particle decay, an unstable nuclide emits a particle o the same confguration as a helium nucleus 42 He ( having two protons and two neutrons) . Many nuclides o heavy elements decay primarily by alphaparticle emission.
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
E xamples: 238 92
U 42 He +
234 90
234 90
Th 42 He +
Th
230 88
Ra
As a consequence of the conservation of charge and massenergy, the equation must balance so that there are equal numbers of protons and nucleons on either side of it. The alpha particle is sometimes written as 42 instead of 42 He.
Negative beta ( ) decay In negative beta- particle emission, an unstable nuclide emits an electron. The emission of a beta particle does not change the nucleon number of the parent nuclide. A neutron is converted to a proton and an electron is ej ected. This decay occurs for those nuclides with too high a neutron proton ratio. The decay is accompanied by an electron antineutrino which we will discuss in Sub- topic 7. 3 . E xamples: 1 31 53
I
0 -1
234 90
Th
e+ 0 -1
1 31 54
_
Xe +
234 91
e+
_
Pa +
The negative beta particle is sometimes written as
0 -1
instead of
0 -1
e.
Again, the equation must balance. The daughter nuclide has the same nucleon number as the parent but has one extra proton meaning that its proton number increases by 1 . The antineutrino has no proton or _ nucleon number and is often written as 00 .
Positron ( + ) decay In positron or positive beta- particle emission, an unstable nuclide emits a positron. This is the antiparticle of the electron, having the same characteristics but a positive charge instead of a negative one. The emission of the positron does not change the nucleon number of the parent nuclide. A proton is converted to a neutron and a positron is ej ected. This decay occurs for those nuclides with too high a proton neutron ratio. The decay is accompanied by an electron neutrino which, again, we will discuss in S ub- topic 7.3 . E xamples: 11 6 21 11
C Na
0 +1
e+
0 +1
11 5
e+
B +
21 10
Ne +
The positron is sometimes written as
0 +1
instead of
0 +1
e.
Again, the equation balances. The daughter nuclide has the same nucleon number as the parent but has one less proton, meaning that its proton number decreases by 1 . The neutrino has no proton or nucleon number and is alternatively written as 00 .
Gamma ray emission Gamma rays are high-energy photons often accompanying other decay mechanism. Having emitted an alpha or beta particle the daughter nucleus
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS is oten let in an excited state. It stabilizes by emitting gamma photon(s) thus losing its excess energy. Examples: 60 27
Co
0 -1
e+
60 28
60 28
Ni *
60 28
Ni +
_
Ni * + +
Here the cob alt- 6 0 decays b y b e ta e mission into an excite d nicke l- 6 0 * nuclide 60 28 Ni . This is accomp anied b y a gamma photo n. The nickel- 6 0 de - excite s by e mitting a se cond gamma photo n ( o die rent e nergy rom the original pho ton) . Gamma photons have no proton or nucleo n numb er and are sometimes writte n as 00 . I we had wished to summarize the cobalt decay in one equation we could write it as: 60 27
Co
0 -1
e+
60 28
_
Ni + + 2
This tells us that overall there are two gamma photons emitted.
Worked examples 1
A nucleus o strontium-90 ( 90 38 S r) decays into an isotope o yttrium ( Y) by negative beta emission. Write down the nuclear equation or this decay.
When the iron- 5 5 ( 55 26 Fe) nucleus captures an electron in this way the nucleus changes into a manganese ( Mn) nucleus. Write a nuclear equation to summarise this.
Solution
Solution
This is a normal beta negative decay so the equation will be:
This is a case o balancing a nuclear equation with the electron being identical to a negative beta particle. The electron is present beore the interaction and so appears on the let-hand side o the equation giving:
90 38
2
Sr
0 -1
e+
90 39
_
Y + 00
Under certain circumstances a nucleus can capture an electron rom the innermost shell o electrons surrounding the nucleus.
55 26
Fe +
0 -1
e
55 25
Mn
Half-life Radioactive decay is a continuous but random process there is no way o predicting which particular nucleus in a radioactive sample will decay next. However, statistically, with a large sample o nuclei it is highly probable that in a given time interval a predictable number o nuclei will decay, even i we do not know exactly which particular ones. We say that the nuclide has a constant probability of decay and this does not depend upon the size o the sample o a substance we have. Another way to look at this is in terms o half-life. The hal-lie is the time taken or hal the total number o nuclei initially in a sample to decay or or the initial activity o a sample to all by hal. This varies signifcantly rom nuclide to nuclide: the hal-lie o uranium-238 is 4.5 1 0 9 years, while that o phosphorus-30 is 2.5 minutes, and that o the artifcially produced nuclide ununoctium-294 is 5 milliseconds.
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7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y The nucleus o an atom has a diameter o the order o 1 0 1 5 m and is essentially isolated rom its surroundings , as atoms themselves are separated by distances o about 1 0 - 1 0 m. This means that the decay o a nucleus is independent o the physical state o the nuclide ( whether it is solid, liquid, or gas or in a chemical compound) and the physical conditions such as temperature and pressure. O nly nuclear interactions such as a collision with a particle in a particle accelerator can infuence the hal-lie o a nuclide. Figure 7 shows how the parent nuclei decay during a time equal to our hal- lives. D uring this time the percentage o the parent nuclide present has allen to 6.2 5 % . This graph shows an exponential decay the shape o this decay curve is common to all radioactive isotopes. The curve approaches the time axis but never intercepts it ( it is said to be asymptotic) .
percentage of parent isotope remaining
parent nuclide (red)
100
50
25 13 6 0
daughter nuclide (grey)
1
2 number of half-lives
3
4
Figure 7 Radioactive half-life.
C onsider a sample o radioactive material with N 0 undecayed nuclei at time t = 0. N In one hal- lie this would become __ 2 0
N In two hal- lives it would become __ 4 0
N In three hal- lives it would be __ 8 0
N And in our hal-lives it would become __ ( = 6.2 5 % o N 0 ) 16 0
1 4 1 Alternatively you could take hal to the power o our to give ( __ = __ 2 ) 16 or 6.2 5 %
At S L you will only be given calculations which have integral numbers o hal- lives, those o you studying HL will see how we deal with nonintegral hal- lives in Sub- topic 1 2 .2 .
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Investigate! Modelling radioactive decay D ifferent countries have varying regulations regarding the use of radioactive sources. It may be that you have the opportunity to investigate halflife experimentally, for example, using a thorium generator or by the decay of protactinium. If this is possible there is merit, as always, in performing a real experiment. If this is not possible then you should use software or dice to simulate radioactive decay.
Repeat this until you have just a few dice left.
278
100 100 100 100 400 92 88 79 88 347 84 79 71 80 314 67 59 61 70 257 52 51 56 57 216 48 41 36 50 175 40 36 33 42 151 36 33 28 31 128 32 28 24 21 105 28 20 16 20 84 23 18 15 19 75 17 15 11 17 60 13 15 10 12 50 11 37 6 10 10 33 9 6 9 9 28 7 3 9 9 26 7 3 8 8 21 4 2 7 8 19 4 1 7 7 18 4 1 7 6 14 4 0 5 5 13 4 0 4 5 10 3 0 2 5 6 0 0 2 4 6 0 0 2 4 6 0 0 2 4
Figure 8 Dice can be used to model radioactive decay.
0 7 7 6 3 7 5 4 6 6 4 3 3 3 2 3 3 3 3 3 3 3 3 2 2 2
number of dice remaining against number of throws 450 400 350 300 250
half-life ( T 1 ) is consistently 4 throws
N
N (student A)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
uncertainty
throw (t)
N (all students)
N (student D)
Throw the dice and remove all those which show one.
Make a note of the number of dice that you have removed.
N (student C)
N (student B)
It is possible to carry out a very simple modelling experiment using 1 00 dice ( but the more the merrier, if you have lots of time) .
2
200 150 T1
100
2
50
T1
T1
2
2
0 0
Figure 9 Modelling radioactive decay using a spreadsheet.
5
10
15 t
20
25
30
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
Record the remaining number o dice on a spreadsheet.
Use the spreadsheet sotware to draw a line graph ( or scatter chart) showing the number o dice remaining against the number o throws.
Draw a trend line or the points.
You will see that, with a small sample o just 1 00 dice, there will be quite a large variation rom point to point.
I several groups o students repeat this experiment you will be able to include error bars on your graph.
From the trend line you should calculate an average value or the hal-lie (see fgure 9) .
Measuring radioactive decay
Note
For beta and gamma radiation the count rate near to a source is measured using a Geiger counter. Strictly, this should be called a GeigerMller (GM) tube and counter. The GM tube is a metal cylinder flled with a low-pressure gas. At one end o the tube is a thin mica window (that allows radiation to enter the tube) . A high voltage is connected across the casing o the tube and the central electrode, as shown in fgure 1 0. The beta or gamma radiation entering the tube ionizes the gas. The ions and electrons released are drawn to the electrodes, thus producing a pulse o current that can be measured by a counting circuit. Alpha particles will be absorbed by the window o a GM tube but can be detected using a spark counter such as that shown in fgure 1 1 . A very high voltage is connected across the gauze on the top and a flament positioned a ew millimetres under the gauze. When the alpha particles ionize the air a spark jumps between the gauze and the flament.
to calculate the hal-lie, you should make at least three calculations and take the average o these. Students oten halve a value and halve that again and again; as a consequence o doing this the changes are getting very small and will not be reliable. It is good practice to calculate three hal-lives, starting rom dierent values, beore averaging as shown in fgure 9.
-
When you use a graph, in order
+
In the example shown the hal-lie
consistently takes our throws the number o throws would be equivalent to a measurement o time in a real experiment with a decaying source.
supply casing insulator central electrode thin mica window
For clarity, the error bars have
been omitted rom this graph. gas at low pressure
Figure 10 GeigerMller tube.
to counting circuit -source leads to high voltage supply gauze
Background count When a GM tube is connected to its counter and switched on it will give a reading even when a source o radioactivity is not present. The device will be measuring the background count. I this is carried out over a a series o equal, short time intervals, the count or each interval will vary. However, when measurements are taken over periods o one minute several times, you should achieve a airly constant value or your location. Radioactive material is ound everywhere. Detectable amounts o radiation occur naturally in the air, rocks and soil, water, and vegetation. The count rate varies rom place to place but the largest single source is rom the radioactive gas radon that can accumulate in homes and in the
spark
flament
Figure 11 Spark counter with alpha source.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS background radiation
radon gas rom the ground ood and drink cosmic rays
building and the artifcial ground sources
workplace. Figure 1 2 is a pie chart that shows the contribution o dierent sources o background radiation to the total. The sievert ( S v) is a radiation unit that takes the ionizing eects o dierent radiations into account. Most people absorb between 1 . 5 and 3 . 5 millisievert per year largely rom background radiation. There are places where the background dose o radiation is in excess o 3 0 mS v yr 1 . However, there is no evidence o increased cancers or other health problems arising rom these high natural levels. The sievert will not be examined on the IB D iploma Programme physics course.
Absorption of radiation medical
other sources
nuclear power and weapons test
Figure 12 Sources o background radiation.
The dierent radioactive emissions interact with materials according to their ionizing ability. Alpha particles ionize gases very strongly, have a very short range in air and are absorbed by thin paper; this is because they are relatively massive and have a charge o + 2 e. B eta particles are poorer ionizers but have a range o several centimetres in air and require a ew millimetres o aluminium to absorb them. They are much lighter than alpha particles and have a charge o 1 e. Gamma rays, being electromagnetic waves, barely interact with matter and it takes many metres o air or several centimetres o lead to be able to absorb them. The apparatus shown in fgure 1 3 is a simple arrangement or measuring the thickness o materials needed to absorb dierent types o radiation. The source is positioned opposite the window o the GM tube so that as high a reading as possible is achieved. D ierent thicknesses o materials are then inserted between the source and the GM tube until the count rate is brought back in line with the background count. When this happens, the absorption thickness o the named material is ound. The ollowing table summarizes the properties o , , and sources. The inormation represents rule o thumb values and there are exceptions to the suggested ranges.
counter
GM tube lead absorber source
280
Figure 13 Apparatus to measure absorption o nuclear radiation.
7. 1 D I S C R E T E E N E R G Y A N D R A D I O A C T I V I T Y
Emission
Composition
Range
Ionizing ability
a helium nucleus (2 protons and 2 neutrons)
low penetration, biggest mass and charge, absorbed by a few centimetres of air, skin or thin sheet of paper
very highly ionizing
high energy electrons
moderate penetration, most are absorbed by 25 cm of air, a few centimetres of body tissue or a few millimetres of metals such as aluminium
moderately highly ionizing
very high frequency electromagnetic radiation
highly penetrating, most photons are absorbed by a few cm of lead or several metres of concrete
poorly ionizing usually secondary ionization by electrons that the photons can eject from metals
few photons will be absorbed by human bodies
Worked example
a) D efne the terms (i) nuclide and (ii) hal- lie. b) C opy and complete the ollowing reaction 227 equation. 227 89 Ac 90 Th + + c)
(i) D raw a graph to show the variation with time t ( or t = 0 to t = 72 days) o the number o nuclei in a sample o thorium-2 2 7 (ii) D etermine, rom your graph, the count rate o thorium ater 3 0 days. (iii) O utline the experimental procedure to measure the count rate o thorium- 2 2 7.
Solution a)
c)
ecay
(i) count rate/counts per second
Actinium-2 2 7 ( 227 89 Ac) , decays into thorium- 2 2 7 ( 227 90 Th) . Thorium- 2 2 7 has a hal- lie o 1 8 days and undergoes - decay to the nuclide radium- 2 2 3 . O n a particular detector a sample o thorium- 2 2 7 has an initial count rate o 3 2 counts per second.
b) The proton number has increased by one so this must be negative beta decay. 227 89
Ac
227 90
Th +
0 -1
0_ 0
e+
30 25 20 15 10 5 0 0
20
40 time/days
60
80
(ii) Marked on the graph in green the activity is 1 0 units. There is likely to be a little tolerance on this type of question in an examination. (iii) You would not be expected to give great detail in this sort o question. You should include the ollowing points:
Use o a GM tube as detector.
Measuring the average background count rate in counts per second (this can be done by timing or 1 00 seconds with no source nearby) . This should be done three times, averaged and divided by 1 00 to give the counts per second.
Measuring the count rate three times with the source close to the detector.
C orrecting the count rate by subtracting the background count rate rom the count rate with the source in position.
(i) A nuclide is a nucleus with a particular number o protons and neutrons. (ii) Hal-lie is the time or the count rate to halve in value O R the time or hal the number o nuclei to decay into nuclei o another element.
35
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7.2 Nuclear reactions Understanding
Applications and skills
The unifed atomic mass unit Mass deect and nuclear binding energy Nuclear fssion and nuclear usion
Solving problems involving mass deect and
binding energy Solving problems involving the energy released in radioactive decay, nuclear fssion, and nuclear usion Sketching and interpreting the general shape o the curve o average binding energy per nucleon against nucleon number
Equations The massenergy relationship: E = m c 2
Nature of science Predictions about nuclides In 1869, the Russian chemist Dmitri Mendeleev took the atomic masses o chemical elements and arranged them into a periodic table. By studying his patterns, Mendeleev was able to predict missing elements; he let gaps in the table to be completed when the elements were discovered. Since the advent o the periodic table, patterns o atoms have allowed scientists to make predictions about the nuclides o which the elements are composed.
For nuclear physicists the graph o proton number against neutron number tells them whether a nuclide is likely to be unstable and, i it is, by which mechanism it will decay. The graph o nuclear binding energy per nucleon against nucleon number tells them how much energy can be generated by a particular fssion or usion reaction. These patterns have taken the guesswork out o physics and have proved to be an invaluable tool.
Patterns or stability in nuclides When a graph of the variation of the neutron number with proton number is plotted for the stable nuclei, a clear pattern is formed. This is known as the zone o f stab ility. Nuclides lying within the zone are stable, while those outside it are unstable and will spontaneously decay into a nuclide tending towards the stability zone. In this way it is possible to predict the mechanism for the decay: , - , + (or electron capture) . Nuclides having low proton numbers are most stable when the neutron proton ratio is approximately one. In moving to heavier stable nuclides the neutronproton ratio gradually increases with the heaviest stable nuclide, bismuth 2 09, having a ratio of 1 .5 2 .
282
7. 2 N U C L E A R R E A C T I O N S
Figure 1 shows this plot o neutron number against proton number. neutron number against proton number for stable nuclides 140
-emitters
neutron number (N)
120
zone of stability
100 - emitters
80 60
+ emitters or electron capture
40 20 0 0
20
40 60 80 proton number (Z)
100
Figure 1 Graph of neutron number against proton number for some stable nuclides.
Unstable nuclides lying to the let o the zone o stability are neutron rich and decay by - emission. Those nuclides to the right o the zone o stability are proton rich and decay by + emission or else by electron capture ( this is, as the name suggests, when a nucleus captures an electron and changes a proton into a neutron as a result thus increasing the neutronproton ratio) . The heaviest nuclides are alpha emitters since emission o both two protons and two neutrons reduces the neutronproton ratio and brings the overall mass down. A second pattern that is seen to aect the stability o a nucleus is whether the number o protons and neutrons are even. Almost hal the known stable nuclides have both even numbers o protons and neutrons, while only fve o the stable nuclides have odd numbers o both protons and neutrons. The elements with even numbers o protons tend to be the most abundant in the universe. The third stability pattern is when either the number o protons or the number o neutrons is equal to one o the even numbers 2 , 8, 2 0, 2 8, 5 0, 82 , or 1 2 6. Nuclides with proton numbers or neutron numbers equal to one o these magic numbers are usually stable. The nuclides, where both the proton number and the neutron number are magic numbers, are highly stable and highly abundant in the universe.
The unifed atomic mass unit The unifed atomic mass unit ( u) is a convenient unit or masses measured on an atomic scale. It is defned as one-twelth o the rest mass o an unbound atom o carbon-1 2 in its nuclear and electronic ground state, having a value o 1 .661 1 0 27 kg. C arbon was chosen because it is abundant and present in many dierent compounds hence making it useul or precise measurements. With carbon- 1 2 having six protons and six neutrons, this unit is the average
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS mass o nucleons and is, thereore, approximately equal to the mass o either a proton or a neutron. In some areas o science ( most notably chemistry) the term unifed atomic mass unit has been replaced by the term dalton ( D a) . This is an alternative name or the same unit and is gaining in popularity among scientists. For the current IB D iploma Programme Physics syllabus the term unifed atomic mass unit will be used.
Binding energy As discussed in Sub- topic 7.1 the strong nuclear orce acts between neighbouring nucleons within the nucleus. It has a very short range 1 0 1 5 m or 1 m. The stability o many nuclei provides evidence or the strong nuclear orce. In order to completely dismantle a nucleus into all o its constituent nucleons work must be done to separate the nucleons and overcome the strong nuclear orce acting between them. This work is known as the nuclear binding energy. 1 a ree proton and a ree neutron collide proton
neutron
2 the proton and neutron combine to orm a deuteron with the binding energy being carried away bey a photon deuteron photon 3 a photon o energy greater than the binding energy o the deuteron is incident on the deuteron photon
deuteron
4 the proton and neutron separate with their total kinetic energy being the diference between the photon energy and the binding energy needed to separate the proton and neutron
5 the ree proton and neutron have a greater total rest mass than the deuteron
deuteron ree proton and neutron
Figure 2 Binding energy o a deuteron.
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S uppose we could reverse the process and construct a nucleus rom a group o individual nucleons. We would expect there to be energy released as the strong nuclear orce pulls them together. This would be equal to the nuclear binding energy needed to separate them. This implies that energy is needed to deconstruct a nucleus rom nucleons and is given out when we construct a nucleus.
Mass defect and nuclear binding energy Energy and mass are dierent aspects o the same quantity and are shown to be interchangeable through perhaps the most amous equation in physics ( Einsteins massenergy relationship) : E = m c2 where E is the energy, m the equivalent mass and c the speed o electromagnetic waves in a vacuum. This equation has huge implications or physics on an atomic scale. When work is done on a system so that its energy increases by an amount + E then its mass will increase by an amount + m given by: E m = _ c2 Alternatively when work is done by a system resulting in its energy decreasing by an amount - E then its mass will decrease by an amount - m given by: -E -m = _ c2 These relationships are universal but are only signifcant on an atomic scale. When energy is supplied to accelerate a rocket, there will be an increase in the mass o the rocket. In an exothermic chemical reaction there will be a decrease in the mass o the reactants. However, these are insignifcant amounts and can be ignored without j eopardizing the calculations. It is only with atomic and nuclear changes that the percentage mass change becomes signifcant.
7. 2 N U C L E A R R E A C T I O N S
It ollows rom Einsteins relationship that the total mass of the individual nucleons making up a nucleus must be greater than the mass of that nucleus since work needs to be done in order to break the nucleus into its component parts. This dierence is known as the mass defect which is the mass equivalent o the nuclear binding energy.
Mass and energy units for nuclear changes We have seen in the last sub- topic that the electronvolt ( eV) is commonly used as the unit o energy on the atomic and nuclear scale. Nuclear energy changes usually involve much more energy than that needed or electron energy changes and so MeV ( = eV 1 0 6 ) is the usual multiplier. In the massenergy relation E = m c 2 , when the energy is measured in MeV, the mass is oten quoted in the unit MeV c 2 . This is not an SI unit but is very convenient in that it avoids having to convert to kg when the energy is in electronvolts. O n this basis, the unifed atomic mass unit can also be written as 93 1 .5 MeV c 2 . To fnd the equivalent energy ( in MeV) we simply have to multiply by c2 !
Worked example The nuclide a)
24 11
Na decays into the stable nuclide
24 12
Mg.
into a proton and an electron with the electron being emitted as a negative beta particle) .
(i) Identiy this type o radioactive decay. (ii) Use the data below to determine the rest mass in unifed atomic mass units o the particle emitted in the decay o a sodium- 2 4 nucleus 24 1 1 Na. rest mass o
24 11
Na = 2 3 . 990 96u
rest mass o
24 12
Mg = 2 3 .985 04u
(ii) The energy released is equivalent to a mass o 5 . 002 1 60 MeV c 2 . 1 u is equivalent to 93 1 . 5 MeV c 2 S o the energy is equivalent to a mass 5 .002 1 60 o _______ = 0.005 3 7u 931 .5 The energy mass o the electron must thereore be
energy released in decay = 5 . 002 1 60 MeV
23.990 96u - (23.985 04u + 0.005 37u) = 0.000 5 5 u
b) The isotope sodium- 2 4 is radioactive but the isotope sodium- 2 3 is stable. Suggest which o these isotopes has the greater nuclear binding energy.
Solution a)
(i) This is an example o negative beta decay since the daughter product has an extra proton ( a neutron has decayed
This is consistent with the value or the mass o an electron ( given as 0. 000 5 49u in the IB Physics data booklet) . b) As sodium- 2 4 has 2 4 nucleons and sodium- 2 3 has 2 3 nucleons, the total binding energy or sodium 2 4 is going to be greater than that o sodium 2 3 .
Variation of nuclear binding energy per nucleon The nuclear binding energy o large nuclei tends to be larger than that or smaller nuclei. This is because, with a greater number o nucleons, there are more opportunities or the strong orce to act between nucleons. This means more energy is needed to dismantle a nucleus into its component nucleons. In order to compare nuclei it is usual to plot the average binding energy per nucleon o nuclides against the nucleon number as shown in fgure 3 .
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS The average binding energy per nucleon is found by dividing the total binding energy for a nucleus by the number of nucleons in the nucleus. Although there is a general pattern, with most nuclides having a binding energy of around 8 MeV per nucleon, there are wide differences from this. most stable nuclides (e.g. Fe, Ni)
10
average binding energy per nucleon (MeV)
9 8 heavier nuclides (e.g. U, Pu)
7 6 5 4 3 2
lighter nuclides (e.g. H, He)
1 0 0
50
100
150 200 nucleon number
250
300
Figure 3 Plot of average binding energy per nucleon against the nucleon number.
O n the left of the plot, those nuclides of low nucleon number, such as hydrogen-2 and helium, are less tightly bound than the more massive nuclides ( hydrogen-1 or normal hydrogen consists of a single proton and has no nuclear binding energy) . As we move to the central region of the plot we reach the maximum binding energy per nucleon with nuclides such as iron-5 8 and nickel-62 . These nuclides, having the highest nuclear binding energies per nucleon, are the most stable nuclei and therefore are abundant in the universe. Further to the right than these nuclei, the pattern reverses and the heavier nuclei are less tightly bound than lighter ones. Towards the extreme right we reach the heavy elements of uranium and plutonium in these the binding energy per nucleon is about 1 MeV less than those in the central region.
Nuclear fusion The j oining together ( or fusing) of small nuclei to give larger ones releases energy. This is because the total nuclear binding energy of the fused nuclei is larger than the sum of total nuclear binding energies of the component nuclei. The difference in the binding energies between the fusing nuclei and the nucleus produced is emitted as the kinetic energy of the fusion products. Although it doesnt actually happen in this way, it is useful to think of the energy being released as the difference between the energy emitted in constructing the fused nucleus and the energy required in deconstructing the two nuclei.
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7. 2 N U C L E A R R E A C T I O N S
When two nuclei o masses m 1 and m 2 use to orm a nucleus o mass m 3 the masses do not add up as we might expect and m 1 + m 2 > m 3 S ince the total number o nucleons is conserved this means that m 3 has a smaller mass than the total or m 1 + m 2 . This loss o mass is emitted as the kinetic energy o the usion products in other words: E = ( ( m 1 + m 2 ) - m 3 ) c 2 Lets look at an example: A helium- 4 nucleus is composed o two protons and two neutrons. Let us consider the mass deect o helium-4 compared with its constituent particles. The mass o helium- 4 nucleus = 4. 002 602 u The mass o 2 ( individual) protons = 2 1 . 007 2 76u = 2 .01 4 5 5 2 u The mass o 2 ( individual) neutrons = 2 1 . 008 665 u = 2 .01 7 3 3 u The total mass o all the individual nucleons = 4.03 1 882 u S o the mass deect is 4.031 882 u 4. 002 602 u = 0.02 9 2 8u This is equivalent to an increased binding energy o 0.02 9 2 8 93 1 .5 MeV or 2 7.3 MeV this energy is given out when the nucleus is ormed rom the individual nucleons. You can probably see that the potential or generating energy is immense, but being able to produce the conditions or usion anywhere except in a star is a signifcant problem. Initially, the repulsion between the protons means that energy must be supplied to the system in order to allow the strong nuclear orce to do its work. In reality, j oining together two protons and two neutrons is not a simple task to achieve on Earth. It seems that Earth- based usion reactions are more likely to occur between the nuclei o deuterium ( hydrogen- 2 ) and tritium ( hydrogen-3 ) .
Nuclear fssion The concept o nuclear fssion is not quite as straight-orward as usion although fssion has been used practically or over 70 years. I we are able to take a large nucleus and split it into two smaller ones, the binding energy per nucleon will increase as we move rom the right-hand side to the centre o fgure 3 . This means that energy must be given out in the orm o the kinetic energy o the fssion products. S o, in nuclear fssion, the energy released is equivalent to the dierence between the energy needed to deconstruct a large nucleus and that emitted when two smaller nuclei are constructed rom its components. In terms o masses, the total mass o the two smaller nuclei will be less than that o the parent nucleus and the dierence is emitted as the kinetic energy o the fssion products. S o we see that as m 1 + m 2 < m 3 then E = ( m 3 - ( m 1 + m 2 ) ) c 2 There are some nuclei that undergo fssion spontaneously but these are largely nuclides o very high nucleon number made synthetically. Uranium- 2 3 5 and uranium- 2 38 do undergo spontaneous fssion but this has a low probability o occurring. Uranium- 2 3 6 however is not a common naturally occurring isotope o uranium but it does undergo
TOK The use o fssion and usion In the last twenty years there has been a development in techniques to extract ossil uels rom locations that were not either physically or fnancially viable previously. Given the debate relating to nuclear energy, which is the more moral stance to take: to use nuclear uel with its inherent dangers or to risk damage to landscapes and habitats when extracting ossil uels?
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS spontaneous fssion. Uranium-2 3 6 can be produced rom uranium-2 3 5 when it absorbs a low energy neutron. Uranium-2 3 6 then undergoes spontaneous fssion to split into two lighter nuclides and at the same time emits two or three urther neutrons: 235 92
U + 10 n
236 92
U
1 41 56
Ba +
92 36
Kr + 3 10 n
These isotopes o barium and krypton are j ust one o several possible pairs o fssion products o uranium- 2 3 6.
Nature of science 92 36
Fission chain reaction
Kr n n
n 235 92 92 36
Kr
141 56
U
92 36
n
Ba
Kr n
neutron
n
n
n 235 92
U
n 141 56
235 92
141 56
U
Ba
Ba 92 36
Kr n
n n 235 92
141 56
U
Ba
Figure 4 Uranium chain reaction.
The use o nuclear fssion in nuclear power stations is discussed in Topic 8. O ne o the key aspects o a continuous power production plant is that the nuclear uel is able to fssion in a controlled chain reaction. The neutrons produced in the reaction above must be capable
o producing more nuclear fssion events by encountering urther uranium nuclei. When there is sufcient mass o uranium- 2 3 5 the reaction becomes sel- sustaining, producing a great deal o kinetic energy that can be transormed into electricity in the power station.
Worked example a) C ompare the process o nuclear fssion with nuclear usion. b) Helium-4 ( 42 He) and a neutron are the products o a nuclear usion reaction between deuterium ( 21 H) and tritium ( 31 H) . 2 1
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3 1
4 2
1 0
The masses o these nuclides are as ollows: 2 1
H
2.014 102u
3 1
H
3.016 050u
He
4.002 603u
4 2
H + H He + n + energy Show that the energy liberated in each reaction is approximately 2 .8 1 0 - 1 2 J.
7. 2 N U C L E A R R E A C T I O N S
Solution a) Nuclear usion involves the joining together o light nuclei while nuclear fssion involves the splitting up o a heavy nucleus. In each case the total nuclear binding energy o the product(s) is greater than that o the initial nuclei or nucleus. The dierence in binding energy is emitted as the kinetic energy o the product(s) . In relation to the plot o nuclear binding energy per nucleon against nucleon number, fssion moves nuclei rom the ar right towards the centre whereas usion moves nuclei rom the ar let towards the centre both processes involve a move up the slopes towards higher values. b) To simpliy the calculation you should break it down into several steps. Remember to include all o the steps with so many signifcant fgures, short cuts could cost you marks in an exam. Total mass on let-hand side o equation = 2 .01 4 1 02 u + 3 .01 6 05 0u = 5 .03 0 1 5 2 u
Looking up the mass o the neutron in the data booklet ( = 1 .008 665 u) Total mass on right- hand side o equation = 4.002 603 u + 1 .008 665 u = 5 .01 1 2 68u Thus there is a loss o mass on the right-hand side ( as the binding energy o the helium nucleus is higher than that o the deuterium and tritium nuclei) Mass dierence ( m) = 5 .03 0 1 5 2 u 5 .01 1 2 68u = 0.01 8 884u As u = 93 1 .5 MeV c 2 , m = 0.01 8 884 93 1 .5 MeV c 2 = 1 7.5 9 MeV c 2 m E = ___ = 1 7.5 9 MeV c2
= 1 7.5 9 1 0 6 1 .60 1 0 - 1 9 J E = 2 .81 1 0 - 1 2 J
Nature of science Alternative nuclear binding energy plot 0
average binding energy per nucleon (MeV)
-1
50
100
nucleon number 150 200
250
300
light nuclides: H, He, etc.
-2 -3 -4 -5 -6 -7 -8 -9
- 10
0
heavy nuclides: U, Pu, etc. most stable nuclides: Fe, Ni, etc.
It is quite usual to see the binding energy plot, shown in fgure 3 , drawn upside down as shown in fgure 5 . The reason or plotting this inverted graph is that when the nucleons are at infnite separation they have no mutual potential energy but, because the strong nuclear orce attracts them, they lose energy as they become closer meaning that their potential energy becomes negative. This implies that the more stable nuclei are in a potential valley. Fission and usion changes that bring about stability, will always move to lower ( more negative) energies. This convention is consistent with gravitational and electrical potential energies which are discussed in Topic 1 0. In line with most text books at IB D iploma level we will continue to use the plot shown in fgure 3 .
Figure 5 Alternative plot of binding energy per nucleon against the nucleon number.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS
7.3 The structure of matter Understanding Quarks, leptons, and their antiparticles
Exchange particles
Hadrons, baryons, and mesons
Feynman diagrams
The conservation laws o charge, baryon number,
Confnement
lepton number, and strangeness The nature and range o the strong nuclear orce, weak nuclear orce, and electromagnetic orce
The Higgs boson
Applications and skills Describing the RutherordGeigerMarsden
Describing the mediation o the undamental
experiment that led to the discovery o the nucleus Applying conservation laws in particle reactions Describing protons and neutrons in terms o quarks Comparing the interaction strengths o the undamental orces, including gravity
orces through exchange particles Sketching and interpreting simple Feynman diagrams Describing why ree quarks are not observed
Particle properties Charge
Quarks
___ 2 e
u
c
t
- ___13 e
d
s
b
3
Baryon number
Charge
___ 1
-1
e
0
e
3 ___ 1 3
All quarks have a strangeness number o 0 except the strange quark that has a strangeness number o 1 Gravitational Weak Particles All Quarks, leptons experiencing Particles Graviton W+ , W - , Z 0 mediating
Leptons
All leptons have a lepton number o 1 and antileptons have a lepton number o 1 Electromagnetic Strong Charged
Quarks, gluons
Gluons
Nature of science Symmetry and physics Symmetry has played a major part in the development o particle physics. Mathematical symmetry has been responsible or the prediction o particles. By searching the bubble chamber tracks produced by cosmic rays or generated by particle accelerators many o the predicted 290
particles have been ound. Theoretical patterns have been very useul in developing our current understanding o particle physics and, increasingly, experiments are conirming that these patterns are valid even though their proound signiication is yet to be determined.
7. 3 T H E S T R U C T U R E O F M AT T E R
Introduction At the end o the nineteenth century, physicists experimented with electrical discharges through gases at low pressure (see fgure 1 ) . to vacuum pump fuorescence
cathode rays cathode + anode high voltage
Figure 1 Discharge tube.
In 1 86 9 , the German physicist, Johann Hittor, observed a glow coming rom the end o a discharge tube, opposite the cathode. He suggested that radiation was being emitted rom the cathode and this caused the tube to uoresce. The radiation causing the eect was later called a beam o cathode rays . E ight years later the B ritish physicist, S ir Joseph Thomson, discovered that cathode rays could be deected by both electric and magnetic felds. Thomsons experiment showed that cathode rays were charged. Ater urther experiments with hydrogen gas in the tube, Thomson concluded that cathode rays were beams o particles coming rom atoms. With atoms being neutral, Thompson deduced that the atom must also carry positive charge. Figure 2 shows Thomsons plum pudding or current bun model o the atom, consisting o a number o electrons buried in a cloud o positive charge.
+ -
Although Thomsons model was short-lived, it was the frst direct evidence that atoms have structure and are not the most elementary building blocks o matter as had been previously thought.
The scattering of alpha particles In 1 909, the German, Johannes Geiger, and the E nglishNew Zealander, E rnest Marsden, were studying at Manchester University in E ngland. E rnest Rutherord ( another New Zealander) was supervising their research. The students were investigating the scattering o alpha particles by a thin gold oil. They used a microscope in a darkened room to detect ashes o light emitted when a alpha particles collided with a zinc sulfde screen surrounding the apparatus. Geiger and Marsden were expecting the alpha particles to be deected by a very small amount as a result o the electrostatic eects o the charges in the atom. At Rutherords suggestion they moved their detector to the same side o the oil as the alpha source and were astonished to fnd that approximately one in every eight thousand alpha particles appeared to be reected ( or back- scattered ) by the thin oil. With the alpha particles travelling at about 3 % o the speed o light, reection by electrons surrounded by a cloud o positive charge was unthinkable and Rutherord described the eect as:
+ -
+
+
- +
-
-
+
+
-
+
+ -
spherical cloud o positive charge
+
-
-
+
Figure 2 Thomsons plum pudding model o the atom. scattered particles
beam o particles
source o particles
electron
most particles are undefected
thin gold oil
circular fuorescent screen
Figure 3 The RutherordGeigerMarsden apparatus.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS ... quite the most incredible event that has ever happened to me in my lie. It was almost as incredible as i you fred a 1 5-inch shell at a piece o tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result o a single collision, and when I made calculations I saw that it was impossible to get anything o that order o magnitude unless you took a system in which the greater part o the mass o the atom was concentrated in a minute nucleus. It was then that I had the idea o an atom with a minute massive centre, carrying a charge. Rutherford proposed that, as the alpha particles carry a positive charge, back-scattering could only occur if the massive part of the atom was also positively charged ( for the mechanics of this see S ub-topic 2 .4) . Thus the atom would consist of a small, dense positive nucleus with electrons well outside the nucleus. Rutherfords calculations showed that the diameter of the nucleus would be of the order of 1 0 1 5 m while that of the atom as 10 a whole would be of the order of 1 0 m; this meant that the atom was almost entirely empty space.
TOK Physicists and knowing
As will be discussed in Topic 1 2, for electrons to be able to occupy the orbits required by later models of the atom, a theory that contradicts classical physics is required. The electrons that are accelerating because of their circular motion should emit electromagnetic radiation and spiral into the nucleus. The consequence of this would be that all atoms should collapse to the size of the nucleus and matter could not exist. Since it is obvious that matter does exist, it is clear that Rutherfords model is not the whole story.
Now I know what the atom looks like. a quote rom Sir Ernest Rutherord, ater publishing the results o his alpha scattering experiment. When scientists devote time to investigating a specifc area o research, they are oten keen to publish their fndings. Although peer appraisal is an important aspect o research there have been many instances in which a scientist adheres to one chosen model without being objective about the work o others. Many scientists pride themselves on their objectivity, but is it possible to be truly objective when those who are deemed to be successul are rewarded with prestigious awards and unding?
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electron cloud
alpha particles
nucleus
Figure 4 Paths of several alpha particles in the scattering experiment.
The particle explosion In 1 928, the B ritish physicist, Paul Dirac, predicted an antiparticle of the electron (the positron) . Particles and their associated antiparticles have
7. 3 T H E S T R U C T U R E O F M AT T E R
identical rest mass but have reversed charges, spins, baryon numbers, lepton numbers, and strangeness (see later or these) . The positron was discovered in cosmic rays by the American, C arl Anderson, in 1 932. C osmic rays consist mainly o high-energy protons and atomic nuclei ejected rom the supernovae o massive stars. Anderson used a strong magnetic feld to deect the particles and ound that their tracks were identical to those produced by electrons, but they curved in the opposite direction.
non-ionizing photon (leaving no track)
electron track positron track
When an electron collides with a positron the two particles annihilate and their total mass is converted into a pair o photons o identical energy emitted at right angles to each other. The inverse o this process is called p air p roduction. This is when a photon interacts with a nucleus and produces a particle and its antiparticle; or this event to occur the photon must have a minimum energy equal to the total rest mass o the particle and the antiparticle. 1 93 2 was a signifcant year in the development o our understanding o the atom with other developments in addition to the positron discovery. In 1 93 0, B othe and B ecker, working in Germany, had ound that an unknown particle was ej ected when beryllium is bombarded by alpha particles. James C hadwick at C ambridge University proved experimentally that this particle was what we now know to be the neutron.
scattered atomic electron track more energetic electron-positron pair track
Figure 5 Electron and positron tracks in a cloud chamber.
In 1 93 6, Anderson and his Ph. D . student S eth Neddermeyer at the C aliornia Institute o Technology went on to discover the muon, using cloud chamber measurements o cosmic rays. Although at the time they believed this particle to be the theoretically predicted pion, this was the start o the recognition that there were more particles than protons, neutrons, electrons and positrons. The understanding that electrons had antiparticles suggested to D irac that the same should be true or protons and, in 1 95 5 , the antiproton was discovered at the University o C aliornia, B erkeley by Emilio S egr and O wen C hamberlain. In their experiment protons were accelerated to an energy o approximately 6 MeV beore colliding with urther protons in a stationary target. The reaction is summarized as: _
p+ p p+ p+p+p
_
Here p represents a proton and p an antiproton. The kinetic energy o the colliding protons ( let- hand side o the equation) is sufcient to produce a urther proton and an antiproton using E = m c 2 . We have encountered the neutrino and antineutrino in beta decay. According to the standard B ig B ang model, these particles are thought to be the most numerous in the universe. The electron neutrino was frst proposed in 1 93 0 by the German physicist, Wolgang Pauli, as an explanation o why electrons emitted in beta decay did not have quantized energies as is the case with alpha particles. Pauli suggested that a urther particle additional to the electron should be emitted in beta decay so that energy and momentum are conserved. It was the Italian, Enrico Fermi, who named the particle little neutral one or neutrino and urther developed Paulis theory. As the neutrino has no charge and almost no mass it interacts only minimally with matter and was not discovered experimentally until 1 95 6.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS B y the end o the 1 960s over 3 00 particles had been discovered and physicists were starting to classiy them. A ull theory explaining the structure o matter and the nature o the orces that hold particles together has become a major goal o scientifc research. The current theory is that the universe is made up o a relatively small number o undamental particles.
Classifcation o particles the standard model The S tandard Model came about through a combination o experimental discoveries and theoretical developments. Although ormulated towards the end o the twentieth century, it is supported by the experimental discoveries o the bottom quark in 1 977, the top quark in 1 995 , the tau particle in 2 001 , and the Higgs boson in 2 01 2 . The S tandard Model has been described as being the theory o ( almost) everything. O n the whole the model has been very successul, but it ails to ully incorporate relativistic gravitation or to predict the accelerating expansion o the universe. The model suggests that the only undamental particles are leptons, quarks, and gauge bosons. All other particles are believed to consist o combinations o quarks and antiquarks.
Leptons The leptons are members o the electron amily and consist o the electron ( e - ) , the muon ( ) , the tau ( ) , their antiparticles plus three neutrinos associated with each o the particles and three neutrinos associated with the antiparticles. Electrons, muons, and taus are all negatively charged and their antiparticles are positively charged. Neutrinos and antineutrinos are electrically uncharged. 1 The electron is known to have a mass o about ____ th o the mass o a 1 800 proton, making it a very light particle. The muon is also light, having a mass o about 2 00 times that o the electron. The tau is heavier and has a mass similar to that o a proton. For reasons that will be shown later, each lepton is given a lepton number. Leptons have lepton number + 1 and antileptons 1 .
Particle
Leptons e _
_
_
Antiparticles Neutrinos
e e
Antineutrinos
_
_
_
e
Charge/e 1
Lepton number (L) +1
+1 0
1 +1
0
1
Quarks S peculation about the existence o quarks began ater scattering experiments were perormed with accelerated electrons at the S tanord Linear Accelerator C enter ( S LAC ) at S tanord University in the US A between 1 967 and 1 973 . The electrons were accelerated up to energies o 6 GeV beore colliding with nuclei. In a similar way to alpha scattering, some o the electrons were scattered through large angles by the nucleons which suggested that the nucleons are not o uniorm density but have discrete charges within them. These experiments
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7. 3 T H E S T R U C T U R E O F M AT T E R supported the theories put orward independently by the Russian American physicist George Zweig and the American Murray Gell-Mann in 1 964. Gell-Mann proposed that the charges within nucleons were grouped in threes and coined the term quarks ( which he pronounced qworks) in deerence to a quote rom Finnegans Wake by the Irish novelist James Joyce: Three quarks for Muster Mark! Sure he hasn't got much of a bark. And sure any he has it's all beside the mark. Zweig reerred to quarks as aces and believed ( incorrectly, as it turned out) there were our o them as the aces in a pack o cards. As with leptons there are six quarks and six antiquarks. The quarks are labelled by their avour which has no physical signifcance apart rom identiying the quark. These avours are called up (u) , down (d) , strange (s) , charm (c) , bottom (b) , and top (t) . Quarks each carry a charge o 2 1 2 either + __ e or - __ e, and antiquarks each carry a charge o either - __ e 3 3 3 1 __ or + 3 e. O ten the e is omitted rom charges and they are written as relative 1 2 charge + __ , - __ , etc. 3 3 These quarks are split into three generations o increasing mass. The frst generation contains the up and down quarks, which are the lightest quarks. The second contains the strange and charm quarks, and the third the bottom and top quarks the heaviest quarks. The up, down, and strange quarks were the frst to be discovered, with the up and down quarks combining to orm nucleons.
Quarks with charge - ___1 e
u c t
d s b
3
3
__ _ _
Antiquarks carry the opposite charge and are denoted by u , d , c , etc.
Quark confnement It is thought that quarks never exist on their own but exist in groups within hadrons. Hadrons are ormed rom a combination o two or three quarks ( called mesons and baryons) this is known as quark confnement. The theory that explains quark confnement is known as quantum chromodynamics ( QC D ) something that is not included in the IB D iploma Programme Physics syllabus. To hold the quarks in place they exchange gluons ( the exchange particle or the strong nuclear orce see later in this sub- topic) . Moving a quark away rom its neighbours in the baryon or meson stores more energy in the interaction between the quarks and thereore requires increasing amounts o energy to increase their separation. I more and more energy is ed into the system, instead o breaking the orce between quarks and separating them ( which is what classical physics might suggest) , more quarks are produced this leaves the original quarks unchanged but creates a new meson or baryon by the massenergy relationship E = m c 2 .
u
u
d proton u d u anti-proton
d
u
d neutron u
s
u d
u
+
0 c
d
c
s
d K0
lambda
J/
Figure 6 Three quark and two quark hadrons.
Baryons qqq and antibaryons qqq These are a few of the many types of baryons. Symbol
Name
p p n
proton antiproton neutron lambda omega
V
Quarks with charge + ___2 e
u
-
Quark content uud uud udd uds sss
Electric charge +1 -1 0 0 -1
Mesons qq These are a few of the many types of mesons. Symbol
Name
+ K+ B0 c
pion kaon rho B-zero eta-c
Quark content ud su ud db cc
Electric charge +1 -1 +1 0 0
Figure 7 Baryons and mesons.
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Hadrons
strangeness n
0
-
-1
strangeness
0 0
-1
K0
+1 -
+1
+
K-
K0 -1
In order to explain which particles can exist and to explain the outcome o observed interactions between particles, the quarks are assigned properties described by a numerical value. The quark is given a baryon 1 . number ( B ) o 13 and or an antiquark the baryon number is __ 3
__
K+
0
charge
+
charge
-1
Hadrons are particles composed o quarks and include baryons (which are made up o three quarks) or mesons (which comprise o quark-antiquark pairs) . The strong interaction (see later) acts on all hadrons but not on leptons while the weak interaction acts on both leptons and hadrons. Some particle physicists have hypothesized pentaquarks consisting o our quarks and one antiquarks although experiments at the start o the century looked promising, these have yet to be confrmed experimentally.
0
-2
0
p
0
+1
Figure 8 Example patterns for some baryons and mesons (The Eightfold Way) .
S trangeness ( S) is a property that was initially defned to explain the behaviour o massive particles such as kaons and hyperons. These particles are created in pairs in collisions and were thought to be strange because they have a surprisingly long lietime o 1 0 1 0 s instead o the expected 1 0 23 s. A strange quark has a strangeness o 1 , and a strange antiquark has a strangeness o + 1 . This property o strangeness is conserved when strange particles are created but it is not conserved when they subsequently decay. Figure 8 shows eight baryons each with a baryon number +1 and eight mesons each with baryon number 0. Gell-Mann used these diagrams as a means o organizing baryons and mesons. Particles on the same horizontal level have the same strangeness, while those on the same diagonal have the same charge. Using this system, Gell-Mann predicted the eta () particle in 1 961 , which was discovered experimentally a ew months later. The name Eightold Way reers to Noble Eightold Path a way towards enlightenment in B uddhism.
Nature of science Symmetry and physics In the same way that the periodic table gives the pattern of the elements based on their electron structure, the Standard Model gives the pattern for the fundamental particles in nature. It is believed that there are six leptons and six quarks and these occur in pairs. A further symmetry is seen by each particle having its own antiparticle (which will have the same rest mass as the particle but other properties such as charge and spin are reversed) . When an electron collides with its antiparticle the positron, the two particles annihilate producing pairs of photons that travel in opposite directions (thereby conserving momentum and energy) . It is not possible for annihilating particles to produce a single photon.
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Figure 9 Particle collisions shown in a bubble chamber.
7. 3 T H E S T R U C T U R E O F M AT T E R
Examples of baryons Protons and neutrons are important in atoms and you should know their quark composition and be able to work out that of their antiparticles. S ince they are both baryons, they each consist of three quarks and have a baryon number + 1 . The proton consists of two up quarks and one down quark ( uud) , which means it has charge
( + _23 + _23 - _13 ) = + 1 The neutron consists of two down quarks and one up quark ( ddu) , which gives it a charge of 1 - _ 1 + _ 2 = 0 -_ 3 3 3
(
)
An antiproton has a baryon number of 1 and consists of two __ ___ antiup quarks and one antidown quark ( u u d ) . This gives a charge 2 1 2 of ( - __ - __ + __ = -1 3 3 ) 3 You may wish to prove to yourself that the antineutron has a charge of zero ( and baryon number of 1 ) . Since none of these particles has any strange quarks their strangeness is 0.
Examples of mesons In questions you will always be given the quark composition of mesons, so there is no need to try to remember these or the composition of any baryons apart from the proton and neutron. A + meson is also called _ a positive pion and consists of an up quark and 2 1 an antidown quark ( u d ) . The positive pion has charge ( + __ + __ = +1 . 3 ) 3 As it is not a baryon, its baryon number is 0 and, as it has no strange quarks, its strangeness is 0 too. A K + meson or positive kaon is the lightest strange meson and consists of _ an up quark and an antistrange quark ( u s ) . The positive kaon has charge ( + __23 + __13 ) = + 1 . Again, a meson is not a baryon and so it has baryon number 0. However, this particle does contain an antistrange quark and so it has strangeness of + 1 .
Conservation rules When considering interactions between particles or their decay, the equivalence of mass and energy must be taken into account. Mass may become some form of energy and vice versa using the equation E = m c 2 . In addition to the expected conservation of massenergy and momentum, no interaction that disobeys the conservation of charge has ever been observed; the same is true for baryon number ( B ) and lepton number ( L) . All lep tons have a lep ton number of + 1 and antilep tons have a lep ton number of 1 . As mentioned before, the conservation rules for strangeness are different and we will consider them later. You may be asked to decide whether an interaction or decay is feasible on the basis of the conservation rules.
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Worked examples 1
a) Show that, when a proton collides with a __ negative pion ( u d) , the collision products can be a neutron and an uncharged pion.
This suggests that the neutral pion is very __ short lived since the combination u u would mutually annihilate. In act this particle has a lietime o about 8 1 0 - 1 7 s and annihilates to orm two gamma ray photons or, very occasionally, a gamma ray photon, an electron and a positron.
b) D educe the quark composition o the uncharged pion.
Solution a) The equation or the interaction is p + - n + 0 Q: + 1 1 0 + 0
2
Explain whether a collision between two protons could produce two protons and a neutron.
Solution
B: +1 + 0 +1 + 0
Writing the equation or the baryons:
L: 0 + 0 0 + 0 This interaction is possible on the basis o conservation o charge, baryon number and lepton number. b) Writing the equation in terms o quarks: __
uud + u d ddu + ?? __
?? = u u in order to balance this equation.
p+ p p+ p+ n Q: + 1 + 1 + 1 + 1 + 0 L: 0 + 0 0 + 0 + 0 B: +1 + 1 +1 + 1 + 1 S o this interaction ails on the basis o baryon number.
Fundamental forces C urrent theories suggest that there are j ust our undamental orces in nature. In the very early universe, when the temperatures were very high, it is possible that at least three o these our orces originated as a single unifed orce. The our undamental orces are:
The gravitational force is weak, has an infnite range and acts on all particles. It is always attractive and over astronomic distances it is the dominant orce on an atomic or sub- atomic scale it is negligible.
The electromagnetic force causes electric and magnetic eects such as the orces between electrical charges or bar magnets. Like gravity, the electromagnetic orce has an infnite range but it is much stronger at short distances, holding atoms and molecules together. It can be attractive or repulsive and acts between all charged particles.
The strong nuclear force or strong interaction is very strong, but has very short- range. It acts only over ranges o 1 0 1 5 m and acts between hadrons but not leptons. At this range the orce is attractive but it becomes strongly repulsive at distances any smaller than this.
The weak nuclear force or weak interaction is responsible or radioactive decay and neutrino interactions. Without the weak interaction stars could not undergo usion and heavy nuclei could not be built up. It acts only over very short ranges o 1 0 1 8 m and acts between all particles.
As these our undamental orces have dierent ranges it is impossible to generalize their relative strengths or all situations. The table below shows a
298
7. 3 T H E S T R U C T U R E O F M AT T E R comparison o the eects that the our orces have on a pair o protons in a nucleus (since all our orces will act on protons at that range) .
Force
Range
Relative strength
Gravitational
1
binding planets, solar system, sun, stars, galaxies, clusters o galaxies
Weak nuclear
10 1 8 m
10 24
(W+ , W- ) : transmutation o elements (Wo ) : breaking up o stars (supernovae)
10 3 5
binding atoms, creation o magnetic felds
10 1 5 m
10 3 7
binding atomic nuclei, usion processes in stars
Electromagnetic Strong nuclear
Roles played by these forces in the universe
Exchange particles A very successul model that has been used to explain the mechanism o the undamental orces was suggested by the Japanese physicist, Hideki Yukawa, in 1 93 5 . Yukawa proposed that the orce between a pair o particles is mediated ( or transmitted) by particles called gauge bosons. The our undamental orces have dierent ranges and a dierent boson is responsible or each orce. The mass o the boson establishes the range o the orce. The bosons carry the orce between particles. Figure 1 0 shows how an electron can exchange a photon with a neighbouring electron, leading to electromagnetic repulsion. The exchange particle is said to be a virtual particle because it is not detected during the exchange. The exchange particle cannot be detected during its transer between the particles because detection would mean that it would no longer be acting as the mediator o the orce between the particles. The larger the rest mass o the exchange particle is, the lower the time it can be in fight without it being detected and, thereore, the lower the range o the orce. A
(a)
(b)
A
B
B
-
-
A
B
-
-
A
B
-
-
A
B
virtual photon exchanged
electrons separate
Figure 10 Exchange of virtual photon between two electrons.
A
A
electrons A and B approach
B
B
Figure 11 Analogy of exchange particles.
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Nature of science Feynman diagrams and cloud chamber tracks Dont be confused by thinking that Feynman diagrams are like the pictures of cloud or bubble chamber interactions they are not and they really should not be thought of as spacetime diagrams! The diagrams help theoretical physicists work out the probability of interactions occurring, but for the IB course we treat them as illustrations of the interactions between complex particle events.
electrons repel
time
e
-
e-
e-
efrst incident electron
virtual photon
position
second incident electron
Figure 12 Feynman diagram o the electromagnetic orce between two electrons.
time
p
Exchange particle
Acts on
Gravitational
gravitons (undiscovered)
all particles
Weak nuclear
W + , W and Z 0 bosons
quarks and leptons
Electromagnetic
photons
electrically charged particles
Strong nuclear
gluons (and mesons)
quarks and gluons (and hadrons)
Feynman diagrams These are graphical visualizations, developed by the American physicist Richard Feynman, that represent interactions between particles. These diagrams are sometimes known as spacetime diagrams. They have the time axis going upwards and the space or position axis to the right ( although many particle physicists draw these axes with space going upwards and time to the right so be careful when you are researching interactions) . S traight lines represent particles and upwards arrows show particles moving orwards in time ( downward arrows indicate an antiparticle also moving orwards in time) . Wavy or broken lines that have no arrows represent exchange particles. Points at which lines come together are called vertices ( plural o vertex) and, at each vertex, conservation o charge, lepton number and baryon number must be applied.
The electromagnetic force Figure 1 2 shows a Feynman diagram or the electromagnetic orce between two electrons. The exchange particle that gives rise to the orce is the photon. Photons have no mass and this equates to the orce having an infnite range.
n
The strong force
neutral pion ( 0 ) is exchanged between a proton (p) and a neutron (n) mediating the strong nuclear orce between these particles in the nucleus
Figure 13 Feynman diagram o the strong orce between a proton and a neutron.
300
Force
This diagram shows two electrons moving closer and interacting by the exchange o a virtual photon beore moving apart.
position
The table shows the exchange particles or the our undamental orces.
n 0
p
The simplest analogy to explain how a repulsive orce can be produced by transer o a particle is to picture what happens when a heavy ball is thrown backwards and orwards between people in two boats (see fgure 1 1 ) . The momentum changes as the ball is thrown and caught and to someone who cannot seen the ball travelling, a repulsive orce seems to exist as they move apart. In order to explain the attractive orce we need to imagine that a boomerang is being thrown between the people in the boats in this case the change o momentum brings the boats together. To understand this more ully requires the use o the uncertainty principle discussed in Topic 1 2.
This is the strongest o the orces and acts between quarks and, thereore, between nucleons. The exchange particles responsible are pions ( + , or 0 ) . Figure 1 3 shows the Feynman diagram or the strong orce between a proton and a neutron. In this case a neutral pion is exchanged between the proton and the neutron that ties them together. In hadrons, the pion carries gluons between the quarks the gluons are the exchange particles or the colour orce acting between quarks ( colour is not included on the IB D ip loma Programme Physics syllabus and so questions will be limited to the exchange of mesons between hadrons) .
7. 3 T H E S T R U C T U R E O F M AT T E R
Weak nuclear orce
ep time
This is responsible for radioactive decay by beta emission. In negative beta decay a neutron decays to a proton. In this process a W boson is exchanged as a quark changes from down to up. The W boson then immediately decays into an electron and an electron antineutrino.
e
n position
The W and Z particles responsible for weak interactions are massive. The weak interaction is the only mechanism by which a quark can change into another quark, or a lepton into another lepton.
A neutron decays into a proton by the emission of an electron and an electron antineutrino (shown with the arrow moving downwards because it is an antiparticle) . The decay is mediated by the negative W boson.
Conservation o strangeness We are now in a position to discuss conservation of strangeness. The strange quark has a strangeness of 1 and particles containing a single strange quark will also have a strangeness of 1 . The antistrange quark has a strangeness of + 1 and particles containing a single antistrange quark will also have a strangeness of + 1 . One type of K 0 (neutral kaon) has a strangeness of + 1 and so it contains an antistrange quark. A particle containing two strange quarks would have a strangeness of 2 and one containing two antistrange quarks would have a strangeness of + 2 , etc.
W-
Figure 14 Feynman diagrams for negative beta decay.
Note The shorter the range o the exchange orce the more massive the exchange particle so the exchange particles or gravitation and the electromagnetic interaction (both o infnite range) must have zero rest masses. The weak interaction will have the heaviest boson because it range is the shortest; the strong interaction has an exchange particle o intermediate mass.
S trangeness is not conserved when strange p articles decay through the weak interaction. For example, strangeness is not conserved when a strange quark decays into an up quark. S trangeness is conserved when there is a strong interaction. This is why strange particles are always produced in pairs. If two particles interact to produce a strange particle then a strange antiparticle must also appear. Figure 1 5 shows a stationary proton interacting through the strong interaction with a negative pion ( short green line) at the bottom right0 hand side of the image. These particles create a neutral kaon ( K ) and a second neutral particle called a lambda particle ( 0 ) . As these particles are neutral they produce no tracks in a bubble chamber ( a device that forms trails of bubbles along the path of a charged particle) . The K 0 track is shown as a purple broken line and the 0 as a blue broken line. The reaction is given by: p + - K0 + 0 There are no strange particles on the left-hand side of this equation. The K 0 has a strangeness of + 1 so the 0 must have a strangeness of 1 for the reaction to be viable by the strong interaction. In terms of quarks the equation for the reaction is _
__
uud + u d d s + uds Lets check this to see that charge, lepton number, baryon number, and strangeness are conserved: 2 1 1 2 2 Q: left- hand side ( + __ + __ - __ + ( - __ - __ = 0 and 3 3 ) 3 ) 3 3
right hand side
(
1 - __ 3
+
1 __ 3
)+(
2 + __ 3
-
1 __ 3
-
1 __ 3
Figure 15 Proton-pion interaction.
)=0
As 0 = 0, charge is conserved. L: as none of the particles are leptons, the lepton numbers on both sides are zero meaning lepton number is conserved.
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ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS 5 1 1 1 1 1 B : left-hand side ( + __ + __ + __ + ( + __ + __ = __ and 3 3 ) 3 3 3 3 ) 5 1 1 1 1 1 right hand side ( + __ + __ + ( + __ + __ + __ = __ 3 ) 3 3 ) 3 3 3 5 5 As __ = __ , baryon number is conserved ( as it will always be if there are 3 3 the same number of quarks on both sides of the equation.
S: there are no strange quarks on the left-hand side so the strangeness = 0; on the right- hand side there is one strange and one antistrange quark so the total strangeness = - 1 + 1 = 0. S o strangeness is also conserved. With all four quantities conserved the interaction is viable.
You should check each of these equations to show that charge, baryon number, and lepton number are conserved. Strangeness should not be conserved because each of the two interactions involves the weak nuclear force.
The kaon subsequently decays into positive and negative pions and the lambda particle decays into a negative pion and a proton. As each of these decays is a weak interaction strangeness is not conserved. These reactions are given by _K 0 + + _ __ or in terms of quarks d s u d + u d and 0 - + p __ or in terms of quarks uds u d + uud
Worked example D raw Feynman diagrams to show the following interaction:
In this case the proton and the electron collide and produce a neutron and an electron neutrino. This interaction is mediated by the W boson.
a) positive beta ( positron) decay: p n + e + + e b) protonelectron collision: p + e - n + e
c) n
c) the two types of neutronelectron neutrino collision: n + e e + n and n + e p + e -
time
e
Solution a) time
W
+
position e p
p time
eposition
b)
time
n
e We-
p position
302
W
+
n
e
Figure 16 Feynman diagram for positive beta (positron) decay. In this decay the proton decays into a neutron and emits a positron and electron neutrino. The decay is mediated by the positive W boson ( W + ) .
n
e
e+ n
Z0
Figure 17 Feynman diagram for an protonelectron collision.
position
Figure 18 Feynman diagrams for the two types of neutronelectron neutrino collision. The most likely collision between a neutron and an electron neutrino is one in which the Z 0 boson mediates the collision and the neutrino effectively bounces off the electron this is known as a neutral current interaction. The electron neutrino can occasionally also interact through the W boson by changing a neutron into a proton. These are the charged current interactions.
7. 3 T H E S T R U C T U R E O F M AT T E R
The Higgs boson The Standard Model has been very successul at tying together theory and experimental results. The original theory predicted that leptons and quarks should have zero mass this clearly did not agree with the experimental results or leptons and bosons in which fnite masses o the particles had been already measured. To solve this problem the B ritish physicist Peter Higgs introduced a theory that explained the mass o particles including the W and Z bosons. B y introducing the Higgs mechanism, the equations o the Standard Model were changed in such a way as to allow these particles to have mass. According to this theory, particles gain mass by interacting with the Higgs feld that permeates all space. The theory about the Higgs mechanism could be tested experimentally because it predicted the existence o a new particle not previously seen. This particle, called the Higgs particle, is a boson-like orce mediator; however, it does not, in this case, mediate any orce. The mass o this particle is very large and, thereore, requires a great deal o energy to be produced. With the opening o the large hadron collider (LHC ) at the European C entre or Nuclear Research (C ERN) in Geneva, a particle accelerator became available that was capable o providing sufcient energy to produce the Higgs boson. In July 2 01 2 , scientists both at C E RN and Fermilab announced that they had established the existence o a Higgs- like boson and, by March 1 4 2 01 3 , this particle was tentatively confrmed to be positively charged and to have zero spin; these are two o the properties o a Higgs boson. The discovery o the Higgs boson leads to the adaptation o the diagram o the S tandard Model as shown in fgure 1 9.
Figure 19 The Standard Model including the Higgs boson.
Nature of science International collaboration The history o particle physics is one o international collaboration and the particle research acility at C ERN is an excellent example o how the international science community can co-operate eectively. C ERN is run by 2 0 European member states with our nations waiting to j oin; in addition there are more than 5 0 countries with agreements with C ERN. The
C ERN website claims that hal the worlds particle physicists and over ten thousand scientists rom more than a hundred countries do research there. With unds at a premium it makes fnancial sense to pool resources and work internationally. In terms o international cooperation, the work undertaken by particle physicists has much in common with the ideals o the IB .
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Questions 1
D iagram 2 shows some o the principal energy levels o atomic hydrogen.
(IB) a) Light is emitted rom a gas discharge tube. O utline briey how the visible line spectrum o this light can be obtained.
a) S how, by calculation, that the energy o a photon o red light o wavelength 65 6 nm is 1 . 9 eV.
The table below gives inormation relating to three o the wavelengths in the line spectrum o atomic hydrogen.
Wavelength / 10 9 m
Photon energy / 10 19 J
1880
1.06
656
3.03
486
4.09
b) O n a copy o diagram 2 , draw arrows to represent: ( i)
the electron transition that gives rise to the red line ( label this arrow R)
( ii) a possible electron transition that gives rise to the blue line (label this arrow B ) . ( 4 marks)
b) Deduce that the photon energy or the wavelength o 486 1 0 9 m is 4.09 1 0 1 9 J. The diagram below shows two o the energy levels o the hydrogen atom, using data rom the table above. An electron transition between these levels is also shown. -2.41 10 -19 J photon emitted, wavelength = 656 nm
3
(IB) A nucleus o the isotope xenon, Xe- 1 3 1 , is produced when a nucleus o the radioactive isotope iodine I- 1 3 1 decays. a) Explain the term isotopes. b) Fill in the boxes on a copy o the equation below in order to complete the nuclear reaction equation or this decay. 1 31
-5.44 10 -19 J
c) ( i)
O n a copy o the diagram above, construct the other energy level needed to produce the energy changes shown in the table above.
( ii) D raw arrows to represent the energy changes or the two other wavelengths shown in the table above. ( 9 marks)
I 1 53 41 Xe + - +
c) The activity A o a reshly prepared sample o the iodine isotope is 3.2 1 0 5 B q. The variation o the activity A with time t is shown below. 3 .5 3 .0
2
(IB) D iagram 1 below shows part o the emission line spectrum o atomic hydrogen. The wavelengths o the principal lines in the visible region o the spectrum are shown. diagram 1 red (R)
656 nm
blue (B)
diagram 2 violet (V)
486 nm 434 nm
0 -0.54 -0.85 -1.5 -3.4
wavelength
energy/eV
A/10 5 Bq
2.5 2.0 1.5 1.0 0.5 0
0
5
10
15
20 25 t/days
30
35
40
45
On a copy o this graph, draw a best-ft line or the data points. d) Use the graph to estimate the hal-lie o I- 1 3 1 . ( 8 marks)
-13.6
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QUESTION S (IB) O ne isotope o potassium is potassium-42 ( 42 1 9 K) . Nuclei o this isotope undergo radioactive decay with a hal- lie o 1 2 .5 hours to orm nuclei o calcium. a)
C omplete a copy o the nuclear reaction equation or this decay process. 42 19
K
20
Ca +
b) The graph below shows the variation with time o the number N o potassium- 42 nuclei in a particular sample.
10 average binding energy per nucleon/MeV
4
56 Fe
9 16 O
8
138 Ba
208 Pb 235 U
7
9 Be
6
6 Li
5 4 3H
3 2
2H
1 0 0
50
N0
0
10
20
30 40 t/hours
50
60
(IB) a) Use the ollowing data to deduce that the binding energy per nucleon o the isotope 3 2 He is 2 . 2 MeV.
0 70
nuclear mass o 32 He = 3 .01 6 03 u mass o proton = 1 . 007 2 8u
The isotope o calcium ormed in this decay is stable.
mass o neutron = 1 .008 67u
O n a copy o the graph above, draw a line to show the variation with time t o the number o calcium nuclei in the sample.
b) In the nuclear reaction 2 2 3 1 1 H + 1 H 2 He + 0 n energy is released. ( i)
Use the graph in ( c) , or otherwise, to determine the time at which the ratio
is equal to 7.0. ( 7 marks)
( iii) With reerence to your graph, explain why energy is released in the nuclear reaction above.
(IB)
( 9 marks)
a) Explain what is meant by a nucleon. b) D efne what is meant by the binding energy o a nucleus. The plot below shows the variation with nucleon number o the binding energy per nucleon.
S tate the name o this type o reaction.
( ii) S ketch the general orm o the relationship between the binding energy per nucleon and the nucleon number.
number o calcium nuclei in sample ____ number o potassium-42 nuclei in sample
5
250
( 5 marks)
N0
6
c)
200
c) With reerence to the graph, explain why energy can be released in both the fssion and the usion processes.
N 1 2
100 150 nucleon number
7
a) D istinguish between nuclear fssion and radioactive decay. b) A nucleus o uranium- 2 3 5 ( 235 92 U) may absorb a neutron and then undergo fssion to produce nuclei o strontium-90 ( 90 38 Sr) 42 and xenon-1 42 ( 1 54 Xe) and some neutrons.
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7
ATO M I C , N U C L E AR , AN D PAR T I C L E P H YS I CS The strontium- 90 and the xenon- 1 42 nuclei both undergo radioactive decay with the emission o particles. ( i)
Write down the nuclear equation or this fssion reaction.
( ii) S tate the eect, i any, on the nucleon number and on the proton number o a nucleus when the nucleus undergoes decay. ( 6 marks) 8
a) A neutron collides with a nucleus o uranium- 2 3 5 and the ollowing reaction takes place. 2 35 92
U + 10 n
96 37
Rb +
1 38 55
C s + 2 10 n
S tate the name o this type o reaction. b) Using the data below, calculate the energy, in MeV, that is released in the reaction. mass o
235 92
mass o
96 37
mass o
1 38 55
U nucleus = 2 3 5 . 043 9u
Rb nucleus = 95 .93 42 u C s nucleus = 1 3 7. 91 1 2 u
mass o 10 n nucleus = 1 . 0087u c) S uggest the importance o the two neutrons released in the reaction. d) The rest mass o each neutron accounts or about 2 MeV o the energy released in the reaction. E xplain what accounts or the remainder o the energy released. ( 9 marks) 9
The diagram below illustrates a proton decaying into a neutron by beta positive ( + ) decay.
1 0 a) Possible particle reactions are given below. In each case apply the conservation laws to determine whether or not the reactions violate any o them. ( i)
e +
( ii) p + n p + 0 ( iii) p + + b) S tate the name o an exchange particle involved in the weak interaction. (1 0 marks) 1 1 (IB) When a negative kaon ( K ) collides with a proton, a neutral kaon ( K 0 ) , a positive kaon ( K + ) and a urther particle ( X) are produced. 0
K + p K + K+ + X The quark structure o kaons is shown in the table.
Particle K K+ 0 K
Quark structure _ su _ us _ ds
a) S tate the amily o particles to which kaons belong. b) S tate the quark structure o the proton. c) The quark structure o particle X is sss. Show that the reaction is consistent with the theory that hadrons are composed o quarks. (4 marks) 1 2 The Feynman diagram below represents a decay via the weak interaction process. Time is shown as upwards. The wiggly line represents a virtual exchange particle.
particle Y neutron
eparticle X
+
e
u proton d
S tate the name o: a) S tate what is meant by virtual exchange particle.
a) the orce involved in this decay b) the particle X c) the particle Y involved in the decay. ( 3 marks)
b) Determine whether the virtual particle in the process represented by the Feynman diagram is a W + , a W , or a Z 0 boson. (4 marks)
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8 E N E RGY PRO D U CTI O N Introduction In Topic 2 we looked at the principles behind the transer o energy rom one orm to another. We now look in detail at sources that provide the energy we use every day. The provision o energy is a global issue. O n the one hand, ossil uel reserves are limited and these uels can be a source o pollution and greenhouse gases, yet they are a convenient and energy- rich resource. The
development o renewable energy sources continues but they are not yet at a point where they can provide all that we require. Political rhetoric and emotion oten obscure scientifc assessments about energy resources. E veryone not j ust scientists needs a clear understanding o the issues involved in order to make sound j udgements about the uture o our energy provision.
8.1 Energy sources O B J TE XT_UND
Understanding Primary energy sources
Applications and skills Describing the basic eatures o ossil uel
Renewable and non-renewable energy sources Electricity as a secondary and versatile orm
o energy Sankey diagrams Specifc energy and energy density o uel sources
Nature of science We rely on our ability to harness energy. Our largescale production o electricity has revolutionized society. However, we increasingly recognize that such production comes at a price and that alternative sources are now required. There are elements o risk in our continued widespread use o ossil uels: risk to the planet and risk to the supplies themselves. Society has to make important decisions about the uture o energy supply on the planet.
power stations, nuclear power stations, wind generators, pumped storage hydroelectric systems, and solar power cells Describing the dierences between photovoltaic cells and solar heating panels Solving problems relevant to energy transormations in the context o these generating systems Discussing saety issues and risks associated with the production o nuclear power Sketching and interpreting Sankey diagrams Solving specifc energy and energy density problems
Equations energy power = __
time 1 Av 3 wind power equation = ___ 2
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EN ERGY PRO D U CTI O N
Primary and secondary energy We use many dierent types o energy and energy source or our heating and cooking, transport, and or the myriad other tasks we undertake in our daily lives. There is a distinction between two basic types o energy source we use: primary sources and secondary sources. A p rimary source is one that has not been transormed or converted beore use by the consumer, so a ossil uel coal, or example burnt directly in a urnace to convert chemical potential energy into the internal energy o the water and surroundings is an example o a primary source. Another example is the kinetic energy in the wind that can be used to generate electricity ( a secondary source) or to do mechanical work such as in a windmill ( a device used, or example, to grind corn or to pump up water rom underground) . The defnition o a secondary source o energy is one that results rom the transormation o a primary source. The electrical energy we use is generated rom the conversion o a primary source o energy. This makes electrical energy our most important secondary source. Another developing secondary source is hydrogen, although this is, at the moment, much less important than electricity. Hydrogen makes a useul uel because it burns with oxygen releasing relatively large amounts o energy ( you will know this i you have ever observed hydrogen exploding with oxygen in the lab) . The product o this reaction ( water) has the advantage that it is not a pollutant. However, hydrogen does not exist in large quantities in the atmosphere. So energy rom a primary source would have to be used to orm hydrogen rom hydrocarbons, or by separating water into hydrogen and oxygen. The hydrogen could then be transported to wherever it is to be used as a source o energy.
Renewable and non-renewable energy sources The primary sources can themselves be urther divided into two groups: renewable and non-renewable. Renewable sources, such as biomass, can be replenished in relatively short times ( on the scale o a human lietime) , whereas others such as wind and water sources are continually generated rom the S uns energy. Non- renewable sources, on the other hand, can be replaced but only over very long geological times. A good way to classiy renewable and non-renewable resources is by the rates at which they are being consumed and replaced. Coal and oil, both non-renewable, are produced when vegetable matter buried deep below ground is converted through the eects o pressure and high temperature. The time scale or production is hundreds o millions o years (the deposits o coal on Earth were ormed rom vegetation that lived and died during the Carbonierous geological period, roughly 300 million years ago) . There are mechanisms active today that are beginning the process o creating ossil uels in suitable wetland areas o the planet, but our present rate o usage o these ossil uels is ar greater than the rate at which they are being ormed. O n the other hand, renewable uels such as biomass use biological materials such as trees that were only recently alive. S uch sources can be grown to maturity relatively quickly and then used or energy conversion. The rate o usage o the uels can be similar to the rate at which they are being grown.
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8.1 EN ERGY SOURCES There are urther advantages in the use o biomass and other renewable sources, where the material has been produced recently. When these renewable resources are converted, they will release carbon dioxide ( one o the greenhouse gases) back into the atmosphere. B ut this is new carbon that was taken rom the atmosphere and trapped in the biomass material relatively recently. The conversion o ossil uels releases carbon dioxide into the atmosphere that was fxed in the ossil uels hundreds o millions o years ago. The carbon dioxide content o the atmosphere was more than 1 5 0% greater in the C arbonierous period than it is today, and thus the burning o ossil uels increases the overall amount o this greenhouse gas in the present- day atmosphere.
Types of energy sources In Topic 2 there was a list o some o the important energies available to us. S ome were mechanical in origin. O ther energies were related to the properties o bulk materials and atomic nuclei. O particular importance are the nuclear reactions, both fssion and usion, that you met in Topic 7 .
Primary sources The table below gives an indication o many o the primary sources that are used in the world today although not all sources can be ound in all locations. The use o geothermal energy, or example, requires that the geology o the location has hot rocks suitably placed below the surace.
Energy sources source Nonrenewable sources
Nuclear fuels Fossil fuels
Renewable sources
Energy form
uranium-235 crude oil coal natural gas
nuclear
Sun water wind biomass geothermal
radiant (solar) kinetic kinetic chemical potential internal
chemical potential
Not all the primary sources in the table are necessarily used to provide electricity as a secondary source, it depends on local circumstances. A water wheel using owing water in a river can be used to grind corn in a arming community rather than be harnessed to an electrical generator. A solar urnace may be used in an Arican village to boil water or to cook, while photovoltaic cells may produce the electrical energy required by the community. In some situations, this is oten a better solution than that o changing all the solar energy to an electrical orm that has to be reconverted subsequently. S ome energy is always degraded into an internal orm in a conversion. Nevertheless, many o the worlds primary sources are used to provide electricity using a power station where the secondary source output is in the orm o electrical energy.
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Primary energy use D ata or the present usage o various energy sources are readily available rom various sources on the Internet. Figure 1 shows two examples o data released by the US D epartment o Energy. primary energy use World energy consumption by fuel, 19902035 (quadrillion BTU)
2010 World marketed energy use 500 quadrillion Btu (12,603 Mtoe) renewables 10% nuclear 6%
liquids 35%
coal 26%
(a) Figure 1
natural gas 23%
energy/quadrillion BTU
250
projections
200 liquids 150
coal natural gas
100
renewables 50 0 1990
(b)
history
nuclear 2000
2008
2015 year
2025
2035
Total world energy usage (US Energy Information Administration, report #DOE/EIA-0484 (2010) ) . The frst example ( fgure 1 (a) ) is a chart that shows the various energy sectors that account or the total world usage o energy sold on the open market in 2 01 0. There are two units here that are common in energy data. When you carry out your own investigations o global energy usage, you will almost certainly come across these non-SI units.
The B ritish Thermal Unit ( B TU) is still used in energy resource work; it was historically very important. The B TU is defned as the energy required to raise one B ritish pound o water ( about 0.5 kg) through 1 F ( a change o roughly 0.6 K) and is equivalent to about 1 000 J.
Mtoe stands or million tonnes o oil equivalent. O ne tonne o oil equivalent is the energy released when one tonne ( 1 000 kg) o crude oil is burnt; this is roughly 42 GJ, leading to a value o 5 1 0 20 J or 2 01 0 usage total.
The US chart uses an unusual multiplier the energy total or 2 01 0 is given as 5 00 quadrillion. The quadrillion is either 1 0 1 5 or 1 0 24 depending on the defnition used. In this particular case it is 1 0 1 5 and the total world energy usage in 2 01 0 was about 5 1 0 1 7 B TU, which also leads to about 5 1 0 20 J. The second chart ( fgure 1 ( b) ) gives a proj ection rom 1 990 up to the year 2 03 5 or world energy usage. Adding the various contributions to the total indicates that the US D epartment o Energy predicts a world total energy usage in 2 03 5 o about 7.5 1 0 20 J; a 5 0% increase on the 2 01 0 fgure. O ne eature o this graph ( in the light o the enhanced greenhouse eect that we shall discuss later) is that the relative proportions o the various energy sources do not appear to be changing greatly over the timescale o the prediction.
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8.1 EN ERGY SOURCES
Investigate! Where are we today?
Textbooks are o necessity out-o-date! They represent the position on the day that part o the book was completed. So the examples here are to give you some insight into the type and quantity o data that are available to you. But they are not to be regarded either as the last word or inormation that you must memorize or the examination. All the resources shown here were accessed on the Internet without difculty. You should search or the latest tables and graphs using the source inormation printed with the graphs. ( This is known as a bibliographic reerence and it is essential to quote this when you use other peoples data.)
S earch or the latest data and discuss it in class. D ivide up the j obs so that students bring dierent pieces o data to the discussion. Ask yourselves: What are the trends now? Has the position changed signifcantly since this book was written?
Science does not stand still, and this area o environmental science is moving as quickly as any other discipline. You need to have accurate data i you are to make inormed judgments.
Predictions are based on assumptions. They cannot predict critical events that might alter the situation e.g. a disaster triggered by a tsunami.
Specifc energy and energy density Much o the extraction o ossil uels involves hard and dangerous work in mines or on oilrigs whether at sea or on land. O n the ace o it, the eort and risk o mining ossil uels does not seem to be j ustifed when there are other sources o energy available. So why are ossil uels still extracted? The answer becomes more obvious when we look at the energy available rom the ossil uel itsel. There are two ways to measure this: specifc energy and energy density. The word specifc has the clear scientifc meaning o per unit mass, or ( in S I) per kg. S o, sp ecifc energy indicates the number o j oules that can be released by each kilogram o the uel. Typical values or a particular uel can vary widely, coal, or example, has a dierent composition and density depending on where it comes rom, and even depending on the location o the sample in the mining seam. D ensity is a amiliar concept; it is the amount o quantity possessed by one cubic metre o a substance. E nergy density is the number o j oules that can be released rom 1 m 3 o a uel.
Fuel Wood Coal Gasoline (petrol) Natural gas at atmospheric pressure Uranium (nuclear fssion) Deuterium/tritium (nuclear usion) Water alling through 100 m in a hydroelectric plant
Specifc energy/ Energy density/ MJ kg1 MJ m 3 16 1 10 4 2060 (20-60) 10 6 45 35 10 6 55 3.5 10 4 8 10 7 1.5 10 1 5 3 10 8 6 10 1 5 10 3
10 3
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EN ERGY PRO D U CTI O N The table shows comparisons between some common uels ( and in the case o usion, the possible energy yields i usion should become commercially viable) . You should look at these and other values in detail or yoursel. Notice the wide range o values that appear in this table. Explore data or other common uels. As you learn about dierent uels, fnd out data or their specifc energy and energy density and add these values to your own table.
Worked examples 1
A ossil-uel power station has an efciency o 25 % and generates 1 2 00 MW o useul electrical power. The specifc energy o the ossil uel is 5 2 MJ kg 1 . C alculate the mass o uel consumed each second.
Solution I 1 2 00 MW o power is developed then, 1 200 1 00 including the efciency fgure, _________ = 4800 25 MW o energy needs to be supplied by the ossil uel. The specifc energy is 5 2 MJ kg 1 , so the mass o 48 00 uel required is ____ = 92 kg s 1 . ( That is roughly 52 1 tonne every 1 0 s, or one railcar ull o coal every 2 minutes. ) 2
When a camping stove that burns gasoline ( petrol) is used, 70% o the energy rom the uel reaches the cooking pot. The energy density o the gasoline is 3 5 GJ m 3 .
a) C alculate the volume o gasoline needed to raise the temperature o 1 litre o water rom 1 0 C to 1 00 C . Assume that the heat capacity o the pot is negligible. The specifc heat capacity o the water is 4.2 kJ kg 1 K 1 . b) Estimate the volume o uel that a student should purchase or a weekend camping expedition.
Solution a) 1 litre o water has a mass o 1 kg so the energy required to heat the water is 42 00 1 90 which is 0. 3 8 MJ. Allowing or the inefciency, 0.5 4 MJ o energy is required and this is a volume o uel 0.5 4 1 0 o ________ = 1 .6 1 0 4 m 3 or about 2 00 ml. 6
35 1 09
b) Assume that 2 litres o water are required or each meal, and that there will be 5 cooked meals during the weekend. S o 2 litres o uel should be more than enough.
Thermal power stations A thermal powe r station is one in which a primary source o energy is converted frst into internal ene rgy and then to electrical energy. The primary uel can include nuclear uel, ossil uel, biomass or other uel that can produce inte rnal e nergy. ( In principle , a se condary source could be used to provide the initial inte rnal e nergy, but this would not be sensible as it incurs substantial extra losses. ) We will discuss the die re nt types o primary conversion later. Howe ver, once the primary energy has been converte d to internal energy, all thermal power stations use a common approach to the conversion o internal into ele ctrical energy: the energy is used to heat water producing steam at high temperatures and high pressures. Figure 2 shows what happe ns. Energy rom the primary uel heats water in a pressure vessel to create steam. This steam is superheated. This means that its temperature is well above the amiliar 1 00 C boiling point that we are used to at atmospheric pressure. To attain such high temperatures the steam has to be at high pressure, hundreds o times more than atmospheric pressure. At these high pressures the water in the vessel does not boil in the
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8.1 EN ERGY SOURCES electricity generator
turbines
kinetic energy (rotation)
superheated steam (internal energy)
electrical energy
boiler condenser water
internal energy rom primary source transerred because o the temperature diference between the source and the boiler Figure 2
pylon
Energy conversions inside a power station.
way that is amiliar to us, it goes straight into the steam phase without orming the bubbles in the liquid that you see in a cooking pot on a stove. However, we will continue to use the term boiler or the vessel where the water is converted to steam even though the idea o boiling is technically incorrect. The high- pressure steam is then directed to a turbine. Turbines can be thought o as reverse ans where steam blows the blades around ( whereas in a an the blades turn to move the gas) . There is usually more than one set o blades and each set is mounted on a common axle connected to an alternating current ( ac) electricity generator. In the generator, the electrical energy is produced when coils o wire, turned by the turbine, rotate in a magnetic feld. The energy that is generated is sent, via a network o electrical cables, to the consumers. You can fnd the details o the physics o electrical energy generation and its subsequent transmission in Topic 1 1 . There are really three energy transers going on in this process: primary energy to the internal energy o water, this internal energy to the kinetic energy o the turbine, and kinetic energy o the turbine to electrical energy in the generator. It is easy to orget the kinetic energy phase and to say that the internal energy goes straight to electrical. O course, the original internal energy is produced in dierent ways in dierent types o thermal power station. In ossil and biomass stations, there is a straightorward combustion process where material is set alight and burnt. In nuclear stations, the process has to be more complicated. We shall look at the dierences between stations in the initial conversion o the energy later.
Tip Do not conuse the roles o turbine and dynamo in a power station. The energy conversions they carry out are quite diferent.
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Sankey diagrams D ierent types o thermal power station have dierent energy losses in their processes and dierent overall efciencies. The S ankey diagram is a visual representation o the ow o the energy in a device or in a process ( although in other subj ects outside the sciences, the S ankey diagram is also used to show the ow o material) . There are some rules to remember about the S ankey diagram:
Each energy source and loss in the process is represented by an arrow.
The diagram is drawn to scale with the width o an arrow being proportional to the amount o energy it represents.
The energy ow is drawn rom let to right.
When energy is lost rom the system it moves to the top or bottom o the diagram.
Power transers as well as energy ows can be represented.
Here is an example to demonstrate the use o a S ankey diagram. coal (primary electricity energyn source) (secondary energy)
electricity (fnal energy)
light radiation (useul energy) 1.0
2
65 Figure 3
2
2
1
28
lampshade
lamp internal energy
domestic wiring
distribution
100
power generation losses
30
31
33 transmission
35
1
Sankey diagram or a lamp.
It shows the ows o energy that begin with the conversion o chemical energy rom ossil uels and end with light energy emitted rom a flament lamp. The red arrows represent energy that is transerred rom the system in the orm o energy as a result o temperature dierences. It is important to recognize that in any process where there is an energy transormation, this energy is lost and is no longer available to perorm a useul j ob. This is degraded energy and there is always a loss o energy like this in all energy transers. O the original primary energy ( 1 00% shown in blue at the bottom o the diagram) , only 3 5 % appears as useul secondary energy. The remaining 65 % ( shown in red) is lost to the surroundings in the process, shown by an arrow that points downwards o the chart. The secondary energy is then shown with the losses involved in the transmission and distribution o the electricity, and to losses in the house wiring. In the lamp itsel, most o the energy ( 2 8% o the original) is transerred to the internal energy o the surroundings. Finally only 1 % o the original primary energy is let as light energy or illumination.
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8.1 EN ERGY SOURCES A S ankey diagram is a useul way to visualize the energy consumption o nations. There are many examples o this available on the Internet. Figure 4 shows the energy ows associated with the C anadian economy. Look on the Internet and you will probably fnd the S ankey diagram or your own national energy demand. Energy equivalents In terms o energy equivalencies, 1 exajoule ( EJ) is equal to: e xp o rt 7 .6 1 5 u ra n iu m 8 .4 2 3 im p o rt 0 .0 9 4 0 .8 1 3 h y d ro
e l e ctric p o we r 3 .7 4 1
1 .2 0 1 0 .3 4 1
160 million barrels o oil
energy consumed annually by 15 million average Canadian single detached homes
energy produced annually by 14 Pickering-sized nuclear stations operating at nominal capacity.
e xp o rt 0 .1 2 4
d is trib u te d e l e ct ricit y 2 .0 1 3 tra n s m is s io n l o s s e s 0 .1 7 3
1 .0 2 1 e l e ct rica l s y s t e m e n e rgy l o s s e s 1 .8 2 3 0 .0 0 3 1
re s id e n t ia l/ co m m e rcia l 2 .6 4 1
1 .2 5 1 3 .9 2 3
0 .0 8 1 ,2
e xp o rt 4.1 5 4 1 .1 6 1
im p o rt s 0 .3 8 4
0 .1 4 1
0 .43 4 0 .0 1 1
- 0 .7 0 3
1 .4 7 3 ,4
co a l
0 .5 0 1
useul energy 4.87 3
in d u s t ria l 4.1 4 1 ,4
2 .5 2 3
0 .1 8 1
1 .3 5 3
im p o rts 0 .6 1 4
lost energy 4.27 3
1 .8 7 3
1 .6 2 3
0 .8 6 1
0 .1 1 1 0 .1 8 4
b io m a s s & o th e r
0 .7 7 3
0 .2 6 1
1 .7 2 4
n a tu ra l ga s 7.69 3 ,4
the energy produced annually by over 1400 square kilometres o state-o-the-art solar cells operating under nominal conditions, enough to cover the entire Toronto urbanized area.
0 .0 1 4
e xp o rt 0 .7 4 4
p ip e l in e s
0 .0 0 1 4 0 .1 9 4
1 .8 9 3 0 .4 7 3
0 .8 8 1 4.1 0 3
p e tro l e u m 5 .9 6 3 ,4
2 .3 4 1
Version 1August 2006
0 .0 1 1 im p o rts 2 .5 9 4
Figure 4 The energy
tra n s p o rt a tio n 2.36 1
e xp o rt 4.4 5 4
0 .4 7 4 n o n -u e l 0 .9 0 3
Sources: 1) NRCan Energy Handbook2005 2) Canadian wind energy association 3) Calculated value 4) Statistics Canada 5) Natural Resources Canada
fow in Canada in 2003. The values are in exajoule. (http://ww2.nrcan.ge.ca/es/oerd) .
Nature of science Sankey diagram in another context Here is a completely dierent use o the S ankey diagram constructed in a historical context. It shows the progress o Napoleons army to and rom his Russian campaign 1 81 2 1 81 3 . The width o the strips gives the number o men in the army: brown in the j ourney to Moscow, black on the return. The graph below the diagram gives the
average temperature experienced by the army on its return j ourney in Raumur degrees. The Raumur temperature scale is named ater the French scientist who suggested a scale that had fxed points o 0 at the reezing point o water and 80 at the boiling point o water.
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EN ERGY PRO D U CTI O N
Figure 5 A Sankey
diagram showing the change in size of Napoleons army during his Russian campaign.
Worked examples 1
An electric kettle of rating 2 .0 kW is switched on for 90 s. D uring this time 2 0 kJ of energy is lost to the surroundings from the kettle. D raw a Sankey diagram for this energy transfer.
remainder of the energy is wasted in the engine, the gearbox and the wheels. Use these data to sketch a S ankey diagram for the car.
Solution Solution The energy supplied in 90 s is 2 1 000 90 = 1 80 kJ. 2 0 000 J is lost to the surroundings; this is 1 1 % of the total.
180 kJ supplied to kettle
160 kJ to heating water
Of the 1 00% of the original fuel energy, 1 2% is lost in the exhaust, and 34% is useful energy. This leaves 54% energy lost in the engine and the transmission. A convenient way to draw the diagram is on squared paper. Use a convenient scale: 1 0% 1 large square is a reasonable scale here.
1000 J as chemical energy 880 J as kinetic energy in the engine
20 kJ to surroundings
2
316
In a petrol- powered car 3 4% of the energy in the fuel is converted into the kinetic energy of the car. Heating the exhaust gases accounts for 1 2 % of the energy lost from the fuel. The
120 J as exhaust
340 J useful energy to move the car 540 J wasted in moving all the engine parts
8.1 EN ERGY SOURCES
Primary sources used in power stations This section compares the dierent ways in which the initial energy required by a thermal power station can be generated.
Fossil fuels Modern ossil-uel power stations can be very large and can convert signifcant amounts o power. The largest in the world at the time o writing has a maximum power output o about 6 GW. The exact process required beore the uel is burnt diers slightly depending on the uel used, whether coal, oil, or natural gas. The gas and oil can be burnt readily in a combustion chamber that is thermally connected to the boiler. In the case o coal, some pre-treatment is normally required. O ten the coal is crushed into a fne powder beore being blown into the urnace where it is burnt. There are obvious disadvantages to the burning o ossil uels. S ome o these are environmental, but other disadvantages can be seen as a misuse o these special materials:
The materials have taken a very large time to accumulate and will not be replaced or equally long times.
The burning o the uels releases into the atmosphere large quantities o carbon dioxide that have been locked in the coal, oil, and gas or millions o years. This has a maj or impact on the response o the atmospheric system to the radiation incident on it rom the S un ( the greenhouse and enhanced greenhouse eects) .
Fossil uels have signifcant uses in the chemical industry or the production o plastics, medicines, and other important products.
It makes sense to locate power stations as close as possible to places where ossil uels are recovered; however, this is not always possible and, in some locations, large- scale transportation o the uels is still required. A need or transport leads to an overall reduction in the efciency o the process because energy has to be expended in moving the uels to the power stations.
Nuclear fuel Sub-topic 7.2 dealt with the principles that lie behind nuclear fssion. It explained the origin o the energy released rom the nucleus when fssion occurs and showed you how to calculate the energy available per fssion. In this course we will only consider so- called thermal fssion reactors, but there are other types in requent use. A particularly common variety o the thermal reactors is the pressurized water reactor ( PWR) . Uranium- 2 3 5 is the nuclide used in these reactors. As with all our other examples o power stations, the aim is to take the energy released in the nuclear fssion and use this to create high-pressure steam to turn turbines connected to an electrical generator. However, the energy is not gained quite so easily as in the case o the ossil uels. Figure 6 shows a schematic o a PWR with the fnal output o steam to, and the return pipe rom, the turbines on the right-hand side o the diagram. The remainder o the power station is as in fgure 2 on p3 1 3 .
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EN ERGY PRO D U CTI O N
control rods
steam to drive turbines
containment building graphite moderator
uranium fuel rods steel reactor vessel
cold water from turbines
Figure 6 Basic features of a pressurised water reactor ( PWR) . Uranium is mined as an ore in various parts o the world, including Australia, C anada, and Kazakhstan ( which together produce about 6 0 % o the worlds ore every year) . The US , Russia, and parts o Arica also produce smaller amounts o uranium ore. About 9 9 % o the ore as it comes directly rom the ground is made up o uranium- 2 3 8 , with the remainder being U- 2 3 5 ; it is the U- 2 3 5 not U- 2 3 8 that is required or the fssion process. This means that an initial extraction process is required to boost the ratio o U- 2 3 5 : U- 2 3 8 . The uel needs to contain about 3 % U- 2 3 5 beore it can be used in a reactor. This is because U- 2 3 8 is a good absorber o neutrons and too much U- 2 3 8 in the uel will prevent the fssion reaction becoming sel- sustaining. The uel with its boosted proportions o U- 2 3 5 is said to be enriched.
Fission product fssion ragments decay o fssion ragments emitted gamma rays emitted neutrons
energy/MeV 160 21 7 5
The enriched material is then ormed into uel rods long cylinders o uranium that are inserted into the core o the reactor. Most o the energy ( about 2 00 MeV or 3 1 0 1 1 J per fssion) is released in the orm o kinetic energy o the fssion ragments and neutrons emitted during the fssion. Immediately ater emission, the neutrons are moving at very high speeds o the order o 1 0 4 km s 1 . However, in order or them to be as eective as possible in causing urther fssions to sustain the reaction they need to be moving with kinetic energies much lower than this, o the order o 0. 02 5 eV ( with a speed o about 2 km s 1 ) . This slower speed is typical o the speeds that neutrons have when they are in equilibrium with matter at about room temperatures. Neutrons with these typical speeds are known as thermal neutrons. The requirement o removing the kinetic energy rom the neutrons is not only that the neutrons can stimulate urther fssions eectively, but that their energy can be efciently transerred to the later stages o the power station. The removal o energy is achieved through the use o a moderator, so-called because it moderates (slows down) the speeds o the neutrons. Typical moderators or the PWR type include water and carbon in the orm o graphite. The transer o energy is achieved when a ast- moving neutron strikes a moderator atom inelastically,
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8.1 EN ERGY SOURCES
transerring energy to the atom and losing energy itsel. Ater a series o such collisions the neutron will have lost enough kinetic energy or it to be moving at thermal speeds and to have a high chance o causing urther fssion. A urther problem is that U-2 3 8 is very eective at absorbing high- speed neutrons, so i the slowing down is carried out in the presence o the U- 2 3 8 then ew ree neutrons will remain at the end o it. S o reactor designers have moderators close to, but not part o, the uel rods; the neutrons slow down in the presence o moderator but away rom the U- 2 3 8. The uel rods and the moderating material are kept separate and neutrons move rom one to the other at random. The reactor vessel and its contents are designed to acilitate this. The criteria or a material to be a good moderator include not being a good absorber o neutrons ( absorption would lower the reaction rate and possibly stop the reaction altogether) and being inert in the extreme conditions o the reactor. S ome reactor types, or example, use deuterium ( 21 H) in the orm o deuterium oxide ( D 2 O , called heavy water) rather than graphite. You should be able to recognize that the best moderator o all ought to be a hydrogen atom ( a single proton in the nucleus) because the maximum energy can be transerred when a neutron strikes a proton. However, hydrogen is a very good absorber o neutrons and it cannot be used as a moderator in this way. There is a need to regulate the power output rom the reactor and to shut down its operation i necessary. This is achieved through the use o control rods. These are rods, oten made o boron or some other element that absorbs neutrons very well, that can be lowered into the reactor. When the control rods are inserted a long way into the reactor, many neutrons are absorbed in the rods and ewer neutrons will be available or subsequent fssions; the rate o the reaction will drop. B y raising and lowering the rods, the reactor operators can keep the energy output o the reactor ( and thereore the power station) under control. The last part o the nuclear power station that needs consideration is the mechanism or conveying the internal energy rom inside the reactor to the turbines. This is known as the heat exchanger. The energy exchange cannot be carried out directly as in the ossil- uel stations. There needs to be a closed- system heat exchanger that collects energy rom the moderator and other hot regions o the reactor, and delivers it to the water. The turbine steam cannot be piped directly through the reactor vessel because there is a chance that radioactive material could be transerred outside the reactor vessel. Using a closed system prevents this. The pressurized water reactor is given its name because it transers the energy rom moderator and uel rods to the boiler using a closed water system under pressure. Water is not the only substance available or this. In the Advanced Gas-cooled Reactors ( AGR) used in the UK, carbon dioxide gas is used rather than water, but the principle o transerring energy saely through a closed system is the same.
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Nature of science Fast breeder reactors A remarkable design or nuclear reactors comes with the development o the ast breeder reactor. Plutonium-239 (Pu-239) is the fssionable material in this case just as uranium-235 is used in PWRs. However, a blanket o uranium-238 surrounds the plutonium core. Uranium-238 does not easily fssion and is a good absorber o neutrons (remember that its presence is undesirable in a PWR) . This U-238 absorbs neutrons lost rom the reactor core and is transmuted into Pu-239 the uel o the ast breeder reactor! The reactor is making its own uel
and generating energy at the same time. It has been reported that, under the right conditions, a ast breeder reactor can produce 5 kg o fssionable plutonium or every 4 kg used in fssion. This is a good way to convert the large stockpiles o virtually useless uranium-238 into something o value. There are drawbacks o course: large amounts o high- level radioactive waste rom the ast breeder reactor need to be dealt with and, in the wrong hands, the plutonium can be used or nuclear weapons.
Investigate! Running your own reactor
No schools are likely to have their own nuclear reactor or students to use! However you can still investigate the operation o a nuclear power station.
There are a number o simulations available on the Internet that will give you virtual control o the station. A suitable starting point is to enter nuclear reactor applet into a search engine.
Worked examples
Solution
1
The role o the moderator is to remove kinetic energy rom neutrons so that there is a high probability that urther fssions will occur. When neutrons are moving at high speeds, there is a very high probability that uranium-2 3 8 nuclei will absorb them without fssion occurring. So the removal o the moderator will mean that neutrons are no longer slowed down, and will be absorbed by U- 2 3 8. The fssion reaction will either stop or its rate will be reduced.
When one uranium-2 3 5 nucleus undergoes fssion, 3 .2 1 0 1 1 J o energy is released. The density o uranium-2 3 5 is 1 . 9 1 0 4 kg m 3 . C alculate, or uranium- 2 3 5 : a) the specifc energy b) the energy density.
Solution a) The mass o the atom is 2 3 5 u ( ignoring the mass o the electrons in the atom) .
3
This is equivalent to a mass o 1 .7 1 0 - 27 2 3 5 = 4.0 1 0 - 25 kg. So the specifc energy o uranium- 2 3 5 3.2 1 0 is ________ = 8.0 1 0 1 3 J kg - 1 -11
When a moving neutron strikes a stationary carbon-1 2 atom head-on, the neutron loses about 3 0% o its kinetic energy. Estimate the number o collisions that would be required or a 1 MeV neutron to be slowed down to 0.1 eV.
4. 0 1 0 - 2 5
1 b) 1 kg o uranium-235 has a volume o _______ 1 .9 1 0 4
5 10
-5
3
m . Thereore the energy density
o pure uranium-2 3 5 = 8. 0 1 0 = 1 .6 1 0 1 8 J m 3 2
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5 10
-5
Explain what will happen in a pressured water reactor i the moderator is removed.
Solution Ater one collision the remaining kinetic energy o the neutron will be 0.7 1 MeV. Ater two collisions the energy o the neutron will be 0.7 0.7 1 MeV = 0. 7 2 1 MeV
8.1 EN ERGY SOURCES
Ater n collisions the energy o the neutron will be 0.7 n 1 MeV.
S o 46 collisions are required to reduce the neutron speed by this actor o 1 0 7 .
S o 0.7 n = 0.1 1 0 6 ,
( In practice, about 1 00 are required; you might want to consider why the actual number is larger than the estimate.)
-7 or n = ______ = 45 . 2 lo g 0 . 7 10
Saety issues in nuclear power
TOK
There needs to be a range o saety measures provided at the site o a nuclear reactor to protect the work orce, the community beyond the power station, and the environment.
The reactor vessel is made o thick steel to withstand the high temperatures and pressures present in the reactor. This has the beneft o also absorbing alpha and beta radiations together with some o the gamma rays and stray neutrons.
The vessel itsel is encased in layers o very thick reinorced concrete that also absorb neutrons and gamma rays.
There are emergency saety mechanisms that operate quickly to shut the reactor down in the event o an accident.
The uel rods are inserted into and removed rom the core using robots so that human operators do not come into contact with the spent uel rods, which become highly radioactive during their time in the reactor.
The issue o the disposal o the waste produced in the nuclear industry is much debated in many parts o the world. S ome o the waste will remain radioactive or a very long time but, as you know rom Topic 7, this implies that its activity will be quite low during this long period. There are however other problems involving the chemical toxicity o this waste material which mean that it is vital to keep it separate rom biological material and thus the ood chain. The technology required to achieve this is still developing.
O all the scientifc issues o our time, perhaps nuclear energy invokes the greatest emotional response in both scientists and nonscientists alike. It is vital to carry out accurate risk assessments or all energy sources, not just nuclear. Is it possible we could orget the location o the waste sites in 50, 100, or 1000 years? Are human errors part o the equation? How can such assessments be carried out in an emotionally charged debate?
At the end o the lie o a reactor ( o the order o 2 5 5 0 years at the moment) , the reactor plant has to be decommissioned. This involves removing all the uel rods and other high- activity waste products and enclosing the reactor vessel and its concrete shield in a larger shell o concrete. It is then necessary to leave the structure alone or up to a century to allow the activity o the structure to drop to a level similar to that o the local background. S uch long-term treatment is expensive and it is important to actor these maj or costs into the price o the electricity as it is being produced during the lietime o the power station.
Nature of science Society and nuclear power The use o nuclear power has been growing throughout the world since the late 1 940s. However, society has never been truly comortable
with its presence and use. There are many reasons or this, including a lack o public understanding o the fssion process ( both advantages and
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disadvantages) , public reaction to accidents that have occurred periodically over the years, and a ear o radioactive materials. Maj or accidents have included the C hernobyl incident, the event at Three Mile Island, and the Fukushima accident o 2 01 1 that was triggered by a tsunami and possibly also an associated earthquake.
understand the scientifc acts and statistics that surround the issue o nuclear power. It is important also to understand the meaning o risk, not j ust in the context o the nuclear power industry, but also in relation to things we do every day. D o some research or yoursel. Find out the risks to your health o:
At the time o writing there has been a withdrawal o approval or nuclear plans by some governments as a result o public opinion changes ater the Fukushima accident. C ontinuing decisions about how we generate energy will be required by society as time goes on, and as our energy- generation technology improves.
living within 2 0 km o a nuclear power station
ying once rom E urope to a country on the Pacifc Rim
smoking 2 0 cigarettes every day
driving 1 000 km in a car.
Try to fnd numerical estimates o the risk, not j ust written statements.
When you debate these issues, ensure that you
Wind generators Wind generators can be used successully in many parts o the world. Even though the wind blows erratically at most locations, this can be countered by the provision o separate wind arms, each with large numbers o individual turbines all connected to the electrical power grid. There are two principal designs o generator: horizontal- axis and vertical- axis. In both cases a rotor is mounted on an axle that is either horizontal or vertical, hence the names. The rotor is rotated by the wind and, through a gearbox, this turns an electrical generator. The electrical energy is ed either to a storage system ( but this increases the expense) or to the electrical grid.
radius o rotor rotor blade radius o rotor
gearbox generator nacelle rotor height
fxed pitch rotor blade
tower
gearbox horizontal axis
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Figure 7 Wind turbine confgurations.
generator vertical axis
8.1 EN ERGY SOURCES
The horizontal-axis machine can be steered into the wind. The verticalaxis type does not have to be steered into the wind and thereore its generator can be placed o-axis.
blade radius r
air density,
It is possible to estimate the maximum power available rom a horizontal- axis wind turbine o blade area A. In one second, the volume o air moving through the turbine is vA when the speed o the wind is v ( fgure 8) . The mass o air moving through the turbine every second is vA where is the density o the air, and the kinetic energy o the air arriving at the turbine in one second is
wind speed v v
Figure 8 Volume of air entering a wind turbine in one second.
1 1 1 __ (mass) (speed) 2 = __ ( vA) v 2 = __ Av 3 2 2 2
I a wind turbine has a blade radius r then the area A swept out by the blades is r 2 and the maximum theoretical kinetic energy arriving at the turbine every second ( and hence the maximum theoretical power) is 1 __ r2 v3 2
This is a maximum theoretical value of the available power as there are a number o assumptions in the proo. In particular, it is assumed that all the kinetic energy o the wind can be used. O course, the wind has to move out o the back o the turbine and so must have some kinetic energy remaining ater being slowed down. Also, i the turbine is part o a wind arm then the presence o other nearby turbines disturbs the ow o the air and leads to a reduction in the energy rom turbines at the rear o the array. The turbine power equation suggests that a high wind speed and a long blade ( large A or r) will give the best energy yields. However, increasing the radius o the blade also increases its mass and this means that the rotors will not rotate at low wind speeds. The blade radius has to be a compromise that depends on the exact location o the wind arm. Many wind arms are placed o- shore; the wind speeds are higher than over land. This is because the sea is relatively smooth compared with the buildings, hills and so on that are ound on land. However, installation and subsequent maintenance or o- shore arrays are more expensive because o the access issues. Another ideal place or wind arms is at the top o a hill. The eect o the land shape is to constrict the ow o the wind into a more confned volume so that in consequence the wind speed rises as the air moves up the hill. Wind speeds thereore tend to be greater at the top o hills and, because the power output o the turbine is proportional to v3 , even a small increase in average wind speed is advantageous. S ome people obj ect to both on- and o- shore arrays on the grounds o visual pollution. There are also suggestions that wind arms compromise animal habitats in some places and that the turbines are noisy or those who live close by.
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EN ERGY PRO D U CTI O N The actors or and against wind being used as an energy source are summarized in this table:
Advantages
Disadvantages
No energy costs
Variable output on a daily or seasonal basis
No chemical pollution
Site availability can be limited in some countries
Capital costs can be high but reduce with economies o scale Easy to maintain on land; not so easy of-shore
Noise pollution Visual pollution Ecological impact
Worked examples 1
a) C alculate the maximum possible output o the wind turbine i the density o air is 1 .3 kg m - 3 .
A wind turbine produces a power P at a particular wind speed. I the efciency o the wind turbine remains constant, estimate the power produced by the turbine:
b) O utline why your estimate will be the maximum possible output o the turbine.
a) when the wind speed doubles b) when the radius o the blade length halves.
Solution a) Using the wind turbine equation:
Solution The equation or the kinetic energy arriving at the 1 wind turbine every second is __ r2 v3 . 2
the kinetic energy arriving at the wind turbine 1 every second is __ r2 v3 , 2
a) When the wind speed v doubles, v3 increases by a actor o 8, so the power output will be 8P.
this will be the maximum power output and 1 is __ 1 .3 2 5 2 1 1 3 = 1 .7 MW. 2
2
b) When the radius o the blade halves, r will go down by a actor o 4 and ( i nothing else P changes) the output will be __ . 4 2
A wind turbine with blades o length 2 5 m is situated in a region where the average wind speed is 1 1 m s 1 .
b) Mechanical and electrical inefciencies in the wind turbine have not been considered. The calculation assumes that all the kinetic energy o the wind can be utilized; this is not possible as some kinetic energy o the air will remain as it leaves the wind turbine.
Pumped storage There are a number o ways in which water can be used as a primary energy resource. These include:
pumped storage plants
hydroelectric plants
tidal barrage
tidal ow systems
wave energy.
All these sources use one o two methods:
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The gravitational potential energy o water held at a level above a reservoir is converted to electrical energy as the water is allowed to all to the lower level ( used in hydroelectric ( fgure 9( a) ) , pumped storage, and tidal barrage) .
8.1 EN ERGY SOURCES
The kinetic energy o moving water is transerred to electrical energy as the water ows or as waves move ( river or tidal ow or wave systems) . Figure 9( b) shows a picture o the C anadian B eauharnois run-o- the- river power station on the S t-Laurent river that can generate 1 .9 GW o power.
(a)
In this topic we ocus on the p ump ed storage system. The wind arms and nuclear power stations we have discussed so ar are known as baseload stations. They run 2 4 hours a day, 7 days a week converting energy all the time. However, the demand that consumers make or energy is variable and cannot always be predicted. From time to time the demand exceeds the output o the baseload stations. Pumped storage is one way to make up or this defcit.
intake
(b)
reservoir
power cable access main access tunnel
surge chamber
discharge
Figure 9 Water as a primary energy resource. ( a) A hydroelectric plant in Thailand. (b) A run-of-the-river plant in Canada.
powerplant chamber transformer
Figure 10 A pumped-storage generating station.
A pumped storage system ( Figure 1 0) involves the use o two water reservoirs sometimes a natural eature such as a lake, sometimes a man-made lake or an excavated cavern inside a mountain. These reservoirs are connected by pipes. When demand is high, water is allowed to run through the pipes rom the upper reservoir to the lower via water turbines. When demand is low, and electrical energy is cheap, the turbines operate in reverse to pump water back rom the lower to the upper reservoir. S ome pumped storage systems can go rom zero to ull output in tens o seconds. The larger systems take longer to come up to ull power, however, substantial outputs are usually achieved in only a ew minutes rom switch on. For a pumped storage system that operates through a height dierence o h, the gravitational potential energy available = mg h where m is the mass o water that moves through the generator and g is the gravitational feld strength. S o the maximum power P available rom the water is equal to the rate at which energy is converted in the machine and is m V P = __ g h = ( __ ) g h t t
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EN ERGY PRO D U CTI O N where t is the time or mass m to move through the generator, V is the volume o water moving through the generator in time t and is the density o the water.
Worked examples 1
Solution
Water rom a pumped storage system alls through a vertical distance o 2 60 m to a turbine at a rate o 600 kg s 1 . The density o water is 1 000 kg m 3 . The overall efciency o the system is 65 %. C alculate the power output o the system.
Solution In one second the gravitational potential energy lost by the system is mg h = 600 g 2 60 = 1 .5 MJ. The efciency is 65 % . 65 O utput power = 1 .5 1 0 6 _ = 0.99 MW 1 00 2 In a tidal barrage system water is retained behind a dam o height h. Show that the gravitational potential energy available rom the water stored behind the dam is proportional to h 2 .
Assume that the cross- sectional area o the dam is A and that the cross- section is rectangular. The volume o water held by the dam is Ah. The mass o the water held by the dam is Ah where is the density o water. When the dam empties completely the centre o mass o the water alls through a distance h __ ( because the centre o mass is hal way up the 2 height o the dam) . The gravitational potential energy o the water is h 1 mgh = ( Ah) g __ = __ Ah 2 . 2 2 Thus, gpe h 2
Solar energy Although most energy is ultimately derived rom the Sun, two systems use photons emitted by the Sun directly in order to provide energy in both large- and small-scale installations.
Solar heating panels S olar heating panels is a technique or heating water using the S uns energy. Sun
hot water, underfoor heating, and central heating
solar panel
pump and solar controller
bo ile r
mains cold water eed hot water storage cylinder
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Figure 11 Solar thermal-domestic hot water system.
8.1 EN ERGY SOURCES
A solar heating p anel contains a pipe, embedded in a black plate, through which a glycolwater mixture is circulated by a pump ( glycol has a low reezing point, necessary in cold countries) . The liquid heats up as inra-red radiation alls on the panel. The pump circulates the liquid to the hot- water storage cylinder in the building. A heatexchanger system transers the energy to the water in the storage cylinder. A pump is needed because the glycolwater mixture becomes less dense as it heats up and would thereore move to the top o the panel and not heat the water in the cylinder. A controller unit is required to prevent the system pumping hot water rom the cylinder to the panel during the winter when the panel is cold.
Solar photovoltaic panels The frst solar photocells were developed around the middle o the nineteenth century by Alexandre- E dmond B ecquerel ( the ather o Henri the discoverer o radioactivity) . For a long time, the use o solar cells, based on the element selenium, was restricted to photography. With the advent o semiconductor technology, it was possible to produce p hotovoltaic cells ( as they are properly called, abbreviated to PV) to power everything rom calculators to satellites. In many parts o the world, solar panels are mounted on the roos o houses. These panels not only supply energy to the house, but excess energy converted during sunny days is oten sold to the local electricity supply company.
electrons to external circuit
light
p-type layer
The photovoltaic materials in the panel convert electromagnetic radiation rom the S un into electrical energy. A ull explanation o the way in which this happens goes beyond the IB syllabus, but a simplifed explanation is as ollows. The photovoltaic cell consists o a single crystal o semiconductor that has been doped so that one ace is p- type semiconductor and the opposite ace is n- type. These terms n- type and p-type indicate the most signifcant charge carriers in the substance ( electrons in n- type, positive holes an absence o electrons in p-type) . Normally there is equilibrium between the charge carriers in both halves o the cell. However, when energy in the orm o photons alls on the photovoltaic cell, then the equilibrium is disturbed, electrons are released and gain energy to move rom the n- region to the p- region and hence around the external circuit. The electrons transer this energy to the external circuit in the usual way and do useul work.
Figure 12 A domestic photovoltaic panel system.
n-type layer
electron
Figure 13 Cross-section of a photovoltaic cell.
Figure 14 Connecting photovoltaic cells together.
O ne single cell has a small em o about 1 V ( this is determined by the nature o the semiconductor) and so banks o cells are manuactured in order to produce usable currents on both a domestic and commercial scale. Many cells connected in series would give large ems but also large internal resistances; the compromise usually adopted is to connect the cells in a combination o both series and parallel. The efciencies o present-day solar cells are about 2 0% or a little higher. However, extensive research and development is being carried out in many countries and it is likely that these efciencies will rise signifcantly over the next ew years.
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EN ERGY PRO D U CTI O N The advantages o both solar heating panels and photovoltaic cells are that maintenance costs are low, and that there are no uel costs. Individual households can use them. D isadvantages include high initial cost, and the relatively large inefciencies ( so large areas o cells are needed) . It goes without saying that the outputs o both types o cell are variable, and depend on both season and weather. We will look at the reasons or these variations in S ub- topic 8.2 . The mathematics o both photovoltaic cells and solar heating panels is straightorward. At a particular location and time o day, the S un radiates a certain power per square metre I ( also known as the intensity o the radiation) to the panels. Panels have an area A, so the power arriving at the surace o the panel will be IA. Panels have an efciency which is the raction o the energy arriving that is converted into internal energy ( o the liquid in the heating panels) or electrical energy ( photovoltaics) . S o the total power converted by the panel is IA.
Worked examples 1
A house requires an average power o 4.0 kW in order to heat water. The average solar intensity at the Earths surace at the house is 65 0 W m 2 . C alculate the minimum surace area o solar heating panels required to heat the water i the efciency o conversion o the panel is 2 2 % .
Solution 4000 W are required, each 1 m 2 o panel can produce 22 65 0 ___ = 1 40 W 1 00 40 0 0 Area required = ____ 1 40
= 2 8 m2 2
Identiy the energy changes in photovoltaic cells and in solar heating panels.
Solution A solar heating cell absorbs radiant energy and converts it to the internal energy o the working uid. A photovoltaic cell absorbs photons and converts their energy to electrical energy.
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R
8.2 Thermal energy transfer Understanding Conduction, convection, and thermal radiation Black-body radiation Albedo and emissivity Solar constant Greenhouse efect Energy balance in the suraceatmosphere
system
Nature of science The study o the Earths climate illustrates the importance o modelling in science. The kinetic theory or an ideal gas is a good model or the way that real gases actually behave. Scientists model the Earths climate in an attempt to understand the implications o the release o greenhouse gases or global warming. The climate is a much more complex system than a simple gas. Issues or scientists include: the availability o data or the planet as a whole, and greater computing power means that more sophisticated models can be tested. Collaboration between research groups means that debate about the accuracy o the models can take place.
Applications and skills Sketching and interpreting graphs showing
the variation o intensity with wavelength or bodies emitting thermal radiation at diferent temperatures Solving problems involving SteanBoltzmann and Wiens laws Describing the efects o the Earths atmosphere on the mean surace temperature Solving albedo, emissivity, solar constant, and the average temperature o the Earth problems
Equations Stean-Boltzmann equation: P = eAT4
2.90 10 - 3 Wiens Law: m a x (metres) = ___ T(kelvin) power intensity equation: I = __ A total scattered power albedo = _____ total incident power
Introduction We considered some of the energy resources in use today in Sub-topic 8.1 . In the course of that sub-topic some of the pollution and atmospheric effects of the resources were mentioned. In this sub-topic we look in more detail at the physics of the Earths atmosphere and the demands that our present need for energy are making on it. After a review of the ways in which energy is transferred due to differences in temperature, we discuss the Suns radiation and its effect on the atmosphere. Finally, we look at how changes in the atmosphere modify the climate.
Thermal energy transfer Any object with a temperature above absolute zero possesses internal energy due to the motion of its atoms and molecules. The higher the temperature of the object, the greater the internal energy associated with
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EN ERGY PRO D U CTI O N the molecules. Topic 3 showed that in a gas the macroscopic quantity that we call absolute temperature is equivalent to the average o the kinetic energy o the molecules. Given the opportunity, energy will spontaneously transer rom a region at a high temperature to a region at a low temperature. Heat, we say rather loosely, fows rom hot to cold. In this sub-topic we look at the ways in which energy can fow due to dierences in temperature. There are three principal methods: conduction, convection, and thermal radiation. All are important to us both on an individual level and in global terms.
Thermal conduction There are similarities between electrical conduction and thermal conduction to the extent that this section is labelled thermal conduction or correctness. However, or the rest o our discussion we will use the term conduction, taking the word to mean thermal conduction. We all know something o practical conduction rom everyday experience. B urning a hand on a camping stove, plunging a very hot metal into cold water which then boils, or melting ice in the hand all give experience o energy moving by conduction rom a hot source to a cold sink. cool water ice good conductor warm water
Figure 1 Laboratory conduction demonstrations.
Metals are good thermal conductors, j ust as they are also good electrical conductors. Poor thermal conductors such as glass or some plastics also conduct electricity poorly. This suggests that there are similarities between the mechanisms at work in both types o conduction. However, it should be noted that there are still considerable dierences in scale between the very best metal conductors ( copper, gold) and the worst metals ( brass, aluminium) . There are many lab experiments that you may have seen designed to help students recognize the dierent thermal properties o good and poor conductors. faster vibrating ions collide with slower vibrating ions
free electrons transfer energy through the metal
hot
cold
atom
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electron
Figure 2 How conduction occurs at an atomic level.
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
In conduction processes, energy ows through the bulk o the material without any large- scale relative movement o the atoms that make up the solid. C onduction ( electrical and thermal) is known as a transport phenomenon. Two mechanisms contribute to thermal conduction:
Atomic vibration occurs in all solids, metal and non-metal alike. At all temperatures above 0 K, the ions in the solid have an internal energy. So they are vibrating about their average fxed position in the solid. The higher the temperature, the greater is the average energy, and thereore the higher their mean speed. Imagine a bar heated at one end and cooled at the other (see fgure 2 ) . At the hot end, the ions vibrate at a large amplitude and with a large average speed. At the cold end the amplitude is lower and the speed is smaller. At the position where the bar is heated the ions vibrate with increasing amplitude and collide with their nearest neighbours. This transers internal energy and the amplitude o the neighbours will increase; this process continues until the bar reaches thermal equilibrium. In this case the energy supplied to each ion is equal to that transerred by the ion to its neighbours in the bar or the surroundings. Each region o the bar will now be at the same uniorm temperature.
C onduction can occur in gases and liquids as well as solids, but, because the inter- atomic connections are weaker and the atoms ( particularly in the gases) are arther apart in uids, conduction is much less important in many gases and liquids than is convection.
Although thermal conduction by atomic vibration is universal in solids, there are other conduction processes that vary in importance depending on the type o solid under discussion. As we saw in Topic 5 , electrical conductors have a covalent (or metallic) bonding that releases free electrons into what is essentially an electron gas flling the whole o the interior o the solid. These ree electrons are in thermal equilibrium with the positive ions that make up the atomic lattice o the solid. The electrons can interact with each other and the energy rom the hightemperature end o the solid diuses along the solid by interactions between these electrons. When an electron interacts with an atom, then energy is transerred back into the atomic lattice to change the vibrational state o the atom. This ree-electron mechanism or conduction depends critically on the numbers o ree electrons available to the solid. Good electrical conductors, where there are many charge carriers (ree electrons) available per unit volume, are likely also to be good thermal conductors. For example, in copper there is one ree electron per atom. You should be able to use Avogadros number, the density o copper and its relative atomic mass, to show that there are 8.4 1 0 28 electrons in every cubic metre o copper.
Nature of science Thermal and electrical resistivity Analogies are oten used in science to aid our understanding o phenomena. Electrical and thermal conduction are closely linked and so it ought to be possible to transer some ideas rom
one to the other. There is an analogy between thermal and electrical eects when thermal conduction is compared with electrical conduction or a wire o cross- section area A and length l:
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Thermal
where is the electrical resistivity o the material.
The rate o transer o internal energy Q is proportional to temperature gradient:
The thermal resistivity k o a material corresponds to the electrical resistivity and the temperature gradient corresponds to the electric potential gradient. The rate at which both internal energy and charge are carried through the wire is related to the presence o ree electrons in the metal. It is more than a coincidence o equations, physics at the atomic scale is involved. Are good electrical conductors also good thermal conductors? C ompare the values o and k or dierent metals.
( )
Q A T _ = -_ _ k l t
where k is the thermal resistivity o the material.
Electrical The rate o transer o charge q ( the current) is proportional to potential gradient:
( )
Q A _ V _ = -_ l t
Convection Convection is the movement o groups o atoms or molecules within uids (liquids and gases) that arises through variations in density. Unlike conduction, which involves the microscopic transer o energy, convection is a bulk property. Convection cannot take place in solids. An understanding o convection is important in many areas o physics, astrophysics and geology. In some hot countries, houses are designed to take advantage o natural convection to cool down the house in hot weather.
(a)
chimney
(b)
potassium permanganate crystal
(c)
potassium permanganate water
candle
glass fronted box Figure 3
Convection currents.
Figure 3 shows three lab experiments that involve convection. In all three cases, energy is supplied to a uid. Look at the glass-ronted box (a) frst. A candle heats the air underneath a tube (a chimney) that leads out o the box. The air molecules immediately above the ame move urther apart decreasing the air density in this region. With a smaller density these molecules experience an upthrust and move up through the chimney. This movement o air reduces the pressure slightly which pulls cooler air down the other chimney. Further heating o the air above the ame leads to a continuous current o cold air down the right- hand chimney and hot air up the let-hand tube. This is a convection current.
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R Similar currents can be demonstrated in liquids. Figure 3(b) shows a small crystal o a soluble dye (potassium permanganate) placed at the bottom o a beaker o water. When the base o the beaker is heated gently near to the crystal, water at the base heats, expands, becomes less dense, and rises. This also leads to a convection current as in fgure 3(c) where a glass tube, in the shape o a rectangle, again with a small soluble coloured crystal, can sustain a convection current that moves all around the tube.
day time warm air
cool air
This is the mechanism by which all the water heated in a saucepan on a stove eventually reaches a uniorm temperature.
night time
Examples of convection
warm air
cool air
There are many examples o convection in action. Figures 4 and 5 show examples rom the natural world; there are many others.
Sea breezes I you live by an ocean you will have noticed that the direction o the breeze changes during a 2 4- hour period. D uring the day, breezes blow on- shore rom the ocean, at night the direction is reversed and the breeze blows o-shore rom the land to the sea.
Figure 4 Sea
breezes.
C onvection eects explain this. D uring the day the land is warmer than the sea and warm air rises over the land mass, pulling in cooler air rom above the ocean. At night the land cools down much more quickly than the sea ( which has a temperature that varies much less) and now the warmer air rises rom the sea so the wind blows o- shore. ( You might like to use your knowledge o specifc heat capacity to explain why the sea temperature varies much less than that o the land. ) A similar eect occurs in the ront range o the Rocky Mountains in the USA. The east-acing hills warm up frst and the high-pressure region on the plains means that the wind blows towards the mountains. Later in the day, the east-acing slopes cool down frst and the eect is reversed.
Convection in the Earth
subduction
subduction mid-ocean ridge
convection
Figure 5 Convection
upwelling
At the bottom o the Atlantic O cean, and elsewhere on the planet, new crust is being created. This is due to convection eects that are occurring below the surace. The Earths core is at a high temperature and this drives convection eects in the part o the planet known as the upper mantle. Two convection currents operate and drive material in the same direction. Material is upwelling at the top o these currents to reach the surace o the Earth at the bottom o the ocean. This creates new land that is orcing the Americas, Europe, and Arica apart at the rate o a ew centimetres every year. In other parts o the world convection currents are pulling material back down below the surace ( subduction) . These convection currents have, over time, given rise to the continental drit that has shaped the continents that we know today.
convection
currents in the Earths
mantle.
Why the winds blow The complete theory o why and how the winds blow would occupy a large part o this book, but essentially the winds are driven by uneven heating o the Earths surace by the Sun. This dierential heating can be due to many causes including geographical actors and the presence o cloud. However,
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EN ERGY PRO D U CTI O N where the land or the sea heat up, the air just above them rises and creates an area o low pressure. Conversely, where the air is alling a high-pressure zone is set up. The air moves rom the high- to the low-pressure area and this is what we call a wind. There is a urther interaction o the wind velocity with the rotation o the Earth (through an eect known as the Coriolis orce) . This leads to rotation o the air masses such that air circulates clockwise around a high-pressure region in the northern hemisphere but anti-clockwise around a high-pressure area in the southern hemisphere.
Nature of science Modelling convection using a scientifc analogy Faced with a hot cup o morning coee and little time to drink it, most o us blow across the liquid surace to cool it more quickly. This causes a more rapid loss o energy than doing nothing and thereore the temperature o the liquid drops more quickly too. This is an example o forced convection when the convection cooling is aided by a draught o air. D oing nothing and allowing the convection currents to set up by themselves is natural convection. Newton stated an empirical law or cooling under conditions o orced convection. He suggested that the rate o change o the temperature o d the cooling body __ was proportional to the dt temperature dierence between the temperature o the cooling body and the temperature o the surroundings s .
The key to understanding this equation is to realize that it is about the temperature difference between the hot obj ect and its surroundings; we call this the temperature excess. Newtons law o cooling leads to a hal- lie behaviour in j ust the same way that radioactive hal- lie ollows rom the radioactive equation dN _ N dt where N is the number o radioactive atoms in a sample. Using radioactivity as an analogy, there is a cooling hal- lie so that the time or the temperature excess over the surroundings to halve is always the same or a particular situation o hot obj ect and surroundings. This is another analogy that helps us to understand science by linking apparently dierent phenomena.
In symbols, d _ ( - s) dt
Worked examples 1
E xplain the role played by convection in the fight o a hot- air balloon.
Solution The air in the gas canopy is heate d rom below and as a result its te mperature increases. The hot air in the balloon expands and its de nsity decreases below that o the cold air outside the gas e nvelope. There is thereore an upward orce on the balloon. I this e xceeds the weight o the balloon ( plus basket and occupants) then the balloon will accelerate upwards. 2
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S uggest two reasons why covering the liquid surace o a cup o hot chocolate with marshmallows will slow down the loss o energy rom the chocolate.
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Solution The marshmallows, having air trapped in them, are poor conductors so they allow only a small fow o energy through them. The upper surace o marshmallow will be at a lower temperature than the lower surace. This reduces the amount o convection occurring at the surace as the convection currents that are set up will not be so strongly driven as the dierential densities will not be so great.
Thermal radiation Basics Thermal radiation is the transer o energy by means o electromagnetic radiation. E lectromagnetic radiation is unique as a wave in that it does not need a medium in order to move ( propagate) . We receive energy rom the S un even though it has passed through about 1 5 0 million km o vacuum in order to reach us. Radiation thereore diers rom conduction and convection, both o which require a bulk material to carry the energy rom place to place. Thermal radiation has its origins in the thermal motion o particles o matter. All atoms and molecules at a temperature greater than absolute zero are in motion. Atoms contain charged particles and when these charges are accelerated they emit photons. It is these photons that are the thermal radiation.
Investigate! Black and white surfaces
Take two identical tin cans and cut out a lid or each one rom polystyrene. Have a hole in each lid or a thermometer. Paint one can completely with matt black paint, paint the other shiny white.
thermometer
black ca n
white ca n
wood block
wood block
Figure 6 Comparing emission of radiation from two surfaces.
Fill both cans with the same volume o hot water at the same temperature, replace the lids and place the thermometers in the water. Keep the cans apart so that radiation rom one is not incident on the other.
C ollect data to enable you to plot a graph to show how the temperature o the water in each can varies with time. This is called a cooling curve.
You could also consider doing the experiment in reverse, beginning with cold water and using a radiant heater to provide energy or the cans. In this case, you must make sure that the heater is the same distance rom the surace o each can and that the shiny can is unable to refect radiation to the black one.
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EN ERGY PRO D U CTI O N Experiments such as the one in Investigate! suggest that black suraces are very good at radiating and absorbing energy. The opposite is true or white or shiny suraces; they reect energy rather than absorb it and are poor at radiating energy. This is why dispensers o hot drinks are oten shiny it helps them to retain the energy.
Nature of science Radiator or not?
Making a saucepan
In many parts o the world, houses need to be heated during part or all o the year. O ne way to achieve this is to circulate hot water rom a boiler through a thin hollow panel oten known as a radiator. B ut is this the appropriate term?
We all need pans to cook our ood. What is the best strategy or designing a saucepan?
The outside metal surace o the panel becomes hot because energy is conducted rom the hot water through the metal. The air near the surace o the panel becomes hotter and less dense; it rises, setting up a convection current in the room. There is some thermal radiation rom the surace but as its temperature is not very dierent rom that o the room, the net radiation is quite low certainly lower than the contributions rom convection. Should the radiator be called a radiator?
The pan will be placed on a at hot surace heated either by ame or radiant energy rom an electrically heated flament or plate. The energy conducts through the base and heats the contents o the saucepan. The base o the pan needs to be a good conductor to allow a large energy ux into the pan. The walls o the saucepan need to withstand the maximum temperature at which the pan will be used but should not lose energy i possible. Giving them a shiny silver fnish helps this. The handle o the pan needs to be a poor conductor so that the pan can be lited easily and harmlessly. D ont make it solid, make it strong and easy to hold but as thin as possible ( think how electrical resistance varies with the shape and thickness o a conductor) . C onclusion: a good pan will have a thick copper base ( a good conductor) , sides, and handle o stainless steel ( a poor conductor) and the overall fnish will be polished and silvery.
Black-body radiation
dull black cylinder incident ray
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Figure 7 A black-body enclosure.
The simple experiments showed that black suraces are good radiators and absorbers but poor reectors o thermal energy. These poor reectors lead to a concept that is important in the theory o thermal radiation: the black-body radiator. A black body is one that absorbs all the wavelengths o electromagnetic radiation that all on it. Like the ideal gas that we use in gas theory, the black body is an idealization that cannot be realized in practice although there are obj ects that are very close approximations to it. O ne way to produce a very good approximation to a black body is to make a small hole in the wall o an enclosed container ( a cavity) and to paint the interior o the container matt black. The container viewed through the hole will look very black inside.
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S ome o the frst experiments into the physics o the black body were made by Lummer and Pringsheim in 1 8 9 9 using a porcelain enclosure made rom fred clay. When such enclosures are heated to high temperatures, radiation emerges rom the cavity. The radiation appears coloured depending on the temperature o the enclosure. At low temperatures the radiation is in the inra- red region, but as the temperature rises, the colour emitted is frst red, then yellow, eventually becoming white i the temperature is high enough. The intensity o the radiation coming rom the hole or cavity is higher when the cavity is at a higher temperature. The emission rom the hole is not dependent on the material rom which the cavity is made unlike the emission rom the surace o a container. This can be seen in the picture o the interior o a steel urnace ( see fgure 8) . In the centre o the urnace at its very hottest point, the colour appears white, at the edges the colour is yellow. At the entrance to the urnace where the temperature is very much lower, the colour is a dull red.
colour and temperature/K 1000 2000 2500 3200 3300 3400 3500 4500 4000 5000
Figure 8 Interior of furnace.
The emission spectrum from a black body Although there is a predominant colour to the radiation emitted rom a black-body radiator, this does not mean that only one wavelength emerges. To study the whole o the radiation that the black body emits, an instrument called a spectrometer is used. It measures the intensity o the radiation at a particular wavelength. Intensity is the power emitted per square metre. As an equation this is written: P I= _ A where I is the intensity, P is the power emitted, and A is the area on which the power is incident. The units o intensity are W m 2 or J s 1 m 2 . A typical intensitywavelength graph is shown in fgure 9 or a black body at the temperature o the visible surace o the S un, about 6000 K. The S un can be considered as a near- perect black- body radiator. The
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EN ERGY PRO D U CTI O N
A potter needs to know the temperature o the inside o a kiln while the clay is being fred to transorm it into porcelain. S ome potters simply look into the kiln through a small hole. They can tell by experience what the temperature is by seeing the radiating colour o the pots inside. O ther potters use an instrument called a pyrometer. A tungsten flament is placed at the entrance to the kiln between the kiln interior and the potters eye. An electric current is supplied to the flament and this is increased until the flament disappears by merging into the background. At this point it is at the same temperature as the interior o the kiln. The flament system will have previously been calibrated so that the current required or the flament to disappear can be equated to the flament temperature.
relative intensity
The potters kiln
483 nm
max T = 2.898 10 -3 m K The wavelength of the peak of the black-body radiation curve gives a measure of temperature
6000 K
visible
0
wavelength
Figure 9 Intensity against wavelength for a black body at the temperature of the Sun.
graph shows how the relative intensity o the radiation varies with the wavelength o the radiation at which the intensity is measured. There are a number o important eatures to this graph:
There is a peak value at about 5 00 nm ( somewhere between green and blue light to our eyes) . ( Is it a coincidence that the human eye has a maximum sensitivity in this region?)
There are signifcant radiations at all visible wavelengths.
There is a steep rise rom zero intensitynotice that the line does not go through the origin.
At large wavelengths, beyond the peak o the curve, the intensity alls to low levels and approaches zero asymptotically.
Figure 1 0 shows the graph when curves at other temperatures are added and this gives urther perspectives on the emission curves. 483 nm
relative intensity (arbitrary units)
Nature of science
0
visible 580 nm
6000 K 5000 K 724 nm
0
966 nm (IR)
4000 K 3000 K
wavelength
Figure 10 Intensity against wavelength for black bodies at four temperatures.
This diagram shows our curves all at dierent temperatures. As beore the units are arbitrary, meaning that the graph shows relative and not absolute changes between the curves at the our temperatures.
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R This amily o curves tells us that, as temperature increases:
the overall intensity at each wavelength increases ( because the curve is higher)
the total power emitted per square metre increases ( because the total area under the curve is greater)
the curves skew towards shorter wavelengths ( higher requencies)
the peak o the curve moves to shorter wavelengths.
The next step is to ocus on the exact changes between these curves.
Wiens displacement law In 1 893 Wilhelm Wien was able to deduce the way in which the shape o the graph depends on temperature. He showed that the height o the curve and the overall width depends on temperature alone. His ull law allows predictions about the height o any point on the curve but we will only use it to predict the peak o the intensity curve. Wiens disp lacement law states that the wavelength at which the intensity is a maximum max in metres is related to the absolute temperature o the black body T by
Tip Notice that the unit or b is metre kelvin and must be written with a space between the symbols, take care not to write it as mK which means millikelvin and is something quite diferent!
max = bT where b is known as Wiens displacement constant. It has the value 2 .9 1 0 3 m K.
SteanBoltzmann law The scientists S tean and B oltzmann independently derived an equation that predicts the total power radiated rom a black body at a particular temperature. The law applies across all the wavelengths that are radiated by the body. S tean derived the law empirically in 1 879 and B oltzmann produced the same law theoretically fve years later. The S tefanB oltzmann law states that the total power P radiated by a black body is given by P = AT 4 where A is the total surace area o the black body and T is its absolute temperature. The constant, , is known as the SteanB oltzmann constant and has the value 5 .7 1 0 8 W m 2 K 4. The law reers to the power radiated by the obj ect, but this is the same as the energy radiated per second. It is easy to show that the energy radiated each second by one square metre o a black body is T 4. This variant o the ull law is known as S teans law.
Grey bodies and emissivity In practice obj ects can be very close to a black body in behaviour but not quite 1 00% perect in the way they behave. They are oten called grey obj ects to account or this. A grey obj ect at a particular temperature will emit less energy per second than a perect black body o the
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EN ERGY PRO D U CTI O N same dimensions at the same temperature. The quantity known as emissivity, e, is the measure o the ratio between these two powers: power emitted by a radiating object e = ________ power emitted by a black body o the same dimensions at the same temperature B eing a ratio, emissivity has no units. For a real material, the power emitted can be written as P = eAT 4
Material water snow ice soil coal
Emissivity 0.60.7 0.9 0.98 0.40.95 0.95
using the same symbols as beore. A perect black body has an emissivity value o 1 . An object that completely reects radiation without any absorption at all has an emissivity o 0. All real objects have an emissivity somewhere these two values. Typical values o emissivity or some substances are shown in the table. Emissivity values are a unction o the wavelength o the radiation. It is surprising that snow and ice, although apparently white and reective, are such eective emitters ( and absorbers) in the inra- red.
Nature of science Building a theory B y the end o the 1 9 th century, the graph o radiation intensity emitted by a black body as a unction o wavelength was well known. Wiens equation ftted the experiments but only at short wavelengths. Rayleigh attempted to develop a new theory on the basis o classical physics. He suggested that charges oscillating inside the cavity produce standing electromagnetic waves as they bounce backwards and orwards between the cavity walls. S tanding waves that escape rom the cavity produce the observed black- body spectrum. Rayleighs model fts the observations at long wavelengths but predicts an ultraviolet catastrophe o an infnitely large intensity at short wavelengths. Max Planck varied Rayleighs theory slightly. He proposed that the standing waves could
not carry all possible energies but only certain quantities o energy E given by nhf where n is an integer, h is a constant ( Plancks constant) and f is the requency o the allowed energy. Plancks model ftted the experimental results at all wavelengths and thus, in 1 9 00, a new branch o physics was born: quantum physics. Planck limited his theory to the space inside the cavity, he believed that the radiation was continuous outside. Later, E instein proposed that the photons outside the cavity also had discrete amounts o energy. Planck was the scientifc reeree or E insteins paper and it is to Plancks credit that he recognised the value o Einsteins work and accepted the paper or publication even though it overturned some o this own ideas.
Worked examples 1
The S un has a surace temperature o 5 800 K and a radius o 7. 0 1 0 8 m. C alculate the total energy radiated rom the S un in one hour.
Solution P = AT 4 S urace area o Sun = 4r2 = 4 3 .1 4 ( 7 1 0 8 ) 2 = 6.2 1 0 1 8
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8 . 2 TH E RM AL E N E R G Y TRAN S FE R ___________________
S o power = 5 .7 1 0
-8
6.2 1 0
18
5 800
= 4.0 1 0 26 W In one hour there are 3600 s, so the energy radiated in one hour is 1 .4 1 0 30 J. 2
A metal flament used as a pyrometer in a kiln has a length o 0. 05 0 m and a radius o 1 . 2 1 0 3 m. D etermine the temperature o the flament at which it radiates a power o 48 W.
Solution The surace area o the flament is 2 rh = ( 2 1 .2 1 0 - 3 ) 0.05 0 = 3 . 8 1 0 4 m 2 S o the power determines the temperature as 48 = 5 .7 1 0 - 8 3 .8 1 0 4 T 4
T=
4
3
48 = 1 2 00 K. ___ 5 .7 1 0 3 .8 1 0 4
-8
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A spherical black body has an absolute temperature T1 and surace area A. Its surroundings are kept at a lower temperature T2 . D etermine the net power lost by the body.
Solution The power emitted by the body is; AT1 4 the power absorbed rom the surroundings is AT 2 4. S o the net power lost is A( T1 4 - T2 4) . Note that this is not the same as A( T1 - T2 ) 4.
Sun and the solar constant The Sun emits very large amounts o energy as a result o its nuclear usion reactions. B ecause the Earth is small and a long way rom the Sun, only a small raction o this arrives at the top o the Earths atmosphere. A black body at the temperature o the Sun has just under hal o its radiation in our visible region, roughly the same amount in the inrared, and 1 0% in the ultraviolet. It is the overall dierence between this incoming radiation and the radiation that is subsequently emitted rom the Earth that determines the energy gained by the Earth rom the Sun. This energy is used by plants in photosynthesis and it drives the changes in the worlds oceans and atmospheres; it is crucial to lie on this planet. The amount o energy that arrives at the top o the atmosphere is known as the solar constant. A precise defnition is that the solar constant is the amount of solar radiation across all wavelengths that is incident in one second on one square metre at the mean distance of the Earth from the Sun on a plane perpendicular to the line joining the centre of the Sun and the centre of the Earth. The energy rom the Sun is spread over an imaginary sphere that has a radius equal to the EarthSun distance. The Earth is roughly 1 .5 1 0 1 1 m rom the Sun and so the surace area o this sphere is 2 .8 1 0 23 m 2 . The S un emits 4 1 0 26 J in one second. The energy incident in one second on one square metre at the distance o the E arth rom the 4. 0 1 0 S un is ________ = 1 400 J. The answer is quoted to 2 s.., a reasonable 2 .8 1 0 precision or this estimate and represents about 5 1 0 - 1 0 o the entire output o the Sun.
radiation rom Sun
26 23
The value o the solar constant varies periodically or a number o reasons:
The output o the Sun varies by about 0.1 % during its principal 1 1 -year sunspot cycle. The Earths orbit is elliptical with the E arth slightly closer to the S un in January compared to July; this accounts or a dierence o about
one square metre at top o atmosphere
Figure 11 Defning the solar constant.
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EN ERGY PRO D U CTI O N 7% in the solar constant. ( Note that this dierence is not the reason or summer in the Southern Hemisphere the seasons occur because the axis o rotation o the Earth is not perpendicular to the plane o its orbit around the Sun.)
O ther longer- period cycles are believed to occur in the S un with periods ranging rom roughly hundreds to thousands o years.
Energy balance in the Earth surface atmosphere system The solar constant is the power incident on the top o the atmosphere. It is not the power that arrives at ground level. As the radiation rom the Sun enters and travels through the atmosphere, it is subj ect to losses that reduce the energy arriving at the Earths surace. Radiation is absorbed and also scattered by the atmosphere. The degree to which this absorption and scattering occurs depends on the position o the S un in the sky at a particular place. When the S un is lower in the sky ( as at dawn and sunset) , its radiation has to pass through a greater thickness o atmosphere and thus more scattering and absorption takes place. This gives rise to the colours in the sky at dawn and dusk. Even when the energy arrives at ground level, it is not necessarily going to remain there. The surace o the Earth is not a black body and thereore it will reect some o the energy back up towards the atmosphere. The extent to which a particular surace can reect energy is known as its albedo (rom the Latin word or whiteness) . It is given the symbol a: energy reected by a given surace in a given time a = _____ total energy incident on the surace in the same time Like emissivity, albedo has no units, it varies rom 0 or a surace that does not reect any energy ( a black body) to 1 or a surace that absorbs no radiation at all. Unless stated otherwise, the albedo in the Earth system is normally quoted or visible light ( which as we saw earlier accounts or nearly a hal o all radiation at the surace) . The average annual albedo or the whole Earth is about 0.3 5 , meaning that on average about 3 5 % o the Suns rays that reach the ground are reected back into the atmosphere. This fgure o 0. 3 5 is, however, very much an average because albedo varies depending on a number o actors:
Surface Ocean Fresh snow Sea ice Ice Urban areas Desert soils Pine forest Deciduous forest
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Albedo 0.06 0.85 0.60 0.90 0.15 0.40 0.15 0.25
It varies daily and with the seasons, depending on the amount and type o cloud cover ( thin clouds have albedo values o 0. 3 0.4, thick cumulo- nimbus cloud can approach values o 0.9) .
It depends on the terrain and the material o the surace. The table gives typical albedo values or some common land and water suraces.
The importance o albedo will be amiliar to anyone who lives where snow is common in winter. Fresh snow has a high albedo and reects most o the radiation that is incident on it the snow will stay in place or a long time without melting i the temperature remains cold. However, sprinkle some earth or soot on the snow and, when the sun shines, the snow will soon disappear because the dark material on its
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
surace absorbs energy. The radiation provides the latent heat needed to melt the snow. Albedo eects help to explain why satellites ( including the International S pace S tation) in orbit around the Earth can take pictures o the Earths cloud cover and surace in the visible spectrum.
Worked examples 1
Four habitats on the E arth are: orest, grassland ( savannah) , the sea, an ice cap. D iscuss which o these have the greatest and least albedo.
D ata Incident intensity rom the S un = 3 40 W m 2 Reected intensity at surace
= 1 00 W m 2
Radiated intensity rom surace = 2 40 W m 2
Solution A material with a high albedo reects the incident visible radiation. Ice is a good reector and consequently has a high albedo. On the other hand, the sea is a good absorber and has a low albedo. 2
The data give details o a model o the energy balance o the Earth. Use the data to calculate the albedo o the Earth that is predicted by this model.
Re- radiated intensity rom atmosphere back to surace
= 2 W m 2
Solution The defnition o albedo is clear. p o w e r re e cte d b y a give n su race It is ________________________ to tal p o w e r incide nt o n the su race 1 00 S o in this case the value is ___ = 0.2 9 3 40
The greenhouse efect and temperature balance The Earth and the Moon are the same average distance rom the S un, yet the average temperature o the Moon is 2 5 5 K, while that o the E arth is about 2 90 K. The discrepancy is due to the Earth having an atmosphere while the Moon has none. The dierence is due to a phenomenon known as the greenhouse effect in which certain gases in the E arths atmosphere trap energy within the Earth system and produce a consequent rise in the average temperature o the E arth. The most important gases that cause the eect include carbon dioxide ( C O 2 ) , water vapour ( H 2 O ) , methane ( C H 4) , and nitrous oxide ( dinitrogen monoxide; N 2 O ) , all o which occur naturally in the atmosphere. O zone ( O 3 ) , which has natural and man- made sources, makes a contribution to the greenhouse eect. It is important to distinguish between:
the natural greenhouse eect that is due to the naturally occurring levels o the gases responsible, and
the enhanced greenhouse eect in which increased concentrations o the gases, possibly occurring as a result o human- derived processes, lead to urther increases in the E arths average temperature and thereore to climate change.
The principal gases in the atmosphere are nitrogen, N 2 , and oxygen, O 2 , ( respectively, 70% and 2 0% by weight) . B oth o these gases are made up o tightly bound molecules and, because o this, do not absorb energy rom sunlight. They make little contribution to the natural greenhouse eect. The 1 % o the atmosphere that is made up o the C O 2 , H 2 O , C H 4 and N 2 O has a much greater eect.
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EN ERGY PRO D U CTI O N The molecular structure o greenhouse- gas molecules means that they absorb ultraviolet and inra- red radiation rom the S un as it travels through the atmosphere. Visible light on the other hand is not so readily absorbed by these gases and passes through the atmosphere to be absorbed by the land and water at the surace. As a result the temperature o the surace rises. The E arth then re- radiates j ust like any other hot obj ect. The temperature o the E arths surace is ar lower than that o the S un, so the wavelengths radiated rom the E arth will peak in the long- wavelength inra- red. The absorbed radiation had, o course, mostly been in the visible region o the electromagnetic spectrum. S o, j ust as gases in the atmosphere absorbed the S uns inrared on the way in, now they absorb energy in the inra- red rom the E arth on the way out. The atmosphere then re- radiates the energy yet again, this time in all directions meaning that some returns to E arth. This energy has been trapped in the system that consists o the surace o the E arth and the atmosphere.
Nature of science Other worlds, other atmospheres The dynamic equilibrium in our climate has been very important or the evolution o lie on Earth. Venus and Mars evolved very dierently rom Earth. Venus has similar dimensions to the Earth but is closer to the Sun with a very high albedo at about 0.76. Its atmosphere is almost entirely carbon dioxide. In consequence, the surace temperature reaches a 730 K and a runaway greenhouse eect acts. Venus and Mars are clear reminders to us o the ragility o a planetary climate.
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The whole system is a dynamic equilibrium reaching a state where the total energy incident on the system rom the S un equals the energy total being radiated away by the Earth. In order to reach this state, the temperature o the Earth has to rise and, as it does so, the amount o energy it radiates must also rise by the SteanB oltzmann law. Eventually, the Earths temperature will be such that the balance o incoming and outgoing energies is attained. O course, this balance was established over billions o years and was steadily changing as the composition o the atmosphere and the albedo changed with changes in vegetation, continental drit, and geological processes.
Why greenhouse gases absorb energy Ultraviolet and long- wavelength inra- red radiations are absorbed by the atmosphere. Photons in the ultraviolet region o the electromagnetic spectrum are energetic and have enough energy to break the bonds within the gas molecules. This leads to the production o ionic materials in the atmosphere. A good example is the reaction that leads to the production o ozone rom the oxygen atoms ormed when oxygen molecules are split apart by ultraviolet photons. The energies o inra-red photons are much smaller than those o ultraviolet and are not sufcient to break molecules apart. When the requency o a photon matches a vibrational state in a greenhouse gas molecule then an eect called resonance occurs. We will look in detail at the vibrational states and resonance in carbon dioxide, but similar eects occur in all the greenhouse gas molecules. In a carbon dioxide molecule, the oxygen atoms at each end are attached by double bonds to the carbon in a linear arrangement. The bonds resemble springs in their behaviour. The molecule has our vibrational modes as shown in fgure 1 2. The frst o these modes a linear symmetric stretching does not cause inrared absorption, but the remaining three motions do. Each one has
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O
C
O
equilibrium position
symmetric stretching bending modes anti-symmetric stretching
Figure 12 Vibrational states in the carbon dioxide molecule.
a characteristic requency. I the requency o the radiation matches this, then the molecule will be stimulated into vibrating with the appropriate mode and the energy o the vibration will come rom the incident radiation. This leads to vibrational absorption at wavelengths o 2.7 m, 4.3 m and 1 5 m. These eects o these absorptions can be clearly seen in fgure 1 3 which shows part o the absorption spectrum o carbon dioxide. In this diagram a peak indicates a wavelength at which signifcant absorption occurs.
Modelling the climate balance
relative absorption
2
3 4 5 wavelength/m
6
Figure 13 Part of the absorption spectrum for carbon dioxide.
We said earlier that about 1 400 J alls on each square metre o the upper atmosphere each second: the solar constant. We use the physics introduced in this topic to see what the consequences o this are or the E arths suraceatmospheric system. The ull 1 400 J does not o course reach the surace. O the total, about 2 5 % o the incident energy is reected by the clouds and by particles in the atmosphere, about 2 5 % is absorbed by the atmosphere, and about 6% is reected at the surace. The incoming radiation alls on the portion o the Earths surace which is normal to the Suns radiation i. e. a circle o area equal to ( radius of Earth) 2 , as only one side o the E arth aces the S un at any one time. However this radiation has to be averaged over the whole o the surace which is 4 ( radius of Earth) 2 . S o the mean power arriving at each 1 40 0 square metre is ____ = 3 5 0 W. 4 The albedo now has to be taken into account to give an eective mean power at one square metre o the surace o ( 1 a) 3 5 0 For the average E arth value or a o about 0. 3 , the mean power absorbed by the surace per square metre is 2 45 W.
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energy intensity (arbitrary units)
EN ERGY PRO D U CTI O N
2 45 = T4
0
transmittance
Figure 14 Intensity and transmittance for a completely transparent atmosphere.
energy transmitted wavelength energy not transmitted
transmittance energy intensity (arbitrary units)
0
transmittance 0
energy not transmitted
wavelength
290 K
256 K
0
wavelength
100%
wavelength
Figure 15 Intensity and transmittance for an atmosphere opaque to infra-red and ultraviolet radiation.
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We made the assumption that the Earth emits 2 45 W m - 2 and that this energy leaves the surace and the atmosphere completely. This would be true or an atmosphere that is completely transparent at all wavelengths, but Earths atmosphere is not transparent in this way. Figure 1 4 shows the relative intensitywavelength graph or a black body at 2 5 6 K. As expected, the area under this curve is 2 45 W m 2 and represents the predicted emission rom the E arth. It shows the response o an atmosphere modelled as perectly transparent at all wavelengths. ( Technically, this graph shows the transmittance o the atmosphere as a unction o wavelength, a value o 1 00% means that the particular wavelength is completely transmitted, 0 means that no energy is transmitted at this wavelength.) Not surprisingly all the blackbody radiation leaves the Earth because the transmittance is 1 or all wavelengths in this model.
256 K
100%
2 45 = 256 K _ 5 .67 1 0 4
This is very close to the value or the Moon, which has no atmosphere. We need to investigate why the mean temperature o the E arth is about 3 5 K higher than this.
wavelength
0
(b)
So T =
100%
energy intensity (arbitrary units)
(a)
_________
wavelength
0
The knowledge o this emitted power allows a prediction o the temperature o a black body T that will emit 2 45 W m 2 . Using the S teanB oltzmann law:
256 K
In act the atmosphere absorbs energy in the inra- red and ultraviolet regions. A simple, but slightly more realistic model or this absorption will leave the transmittance at 1 00% or the visible wavelengths and change the transmittance to 0 or the absorbed wavelengths. Figure 1 5 shows how this leads to an increased surace temperature. When the transmittance graph is merged with the graph or blackbody radiation to give the overall emission rom the E arth into space, the area under the overall emission curve will be less than 2 45 W m 2 because the inra- red and ultraviolet wavelengths are now absorbed in the atmosphere and these energies are not lost ( Figure 1 5 ( a) ) . This defcit will be re- radiated in all directions; so some returns to the surace. In order to get the energy balance correct again, the temperature o the emission curve must be raised so that the area under the curve returns to 245 W m 2 . As the curve changes with the increase in temperature, the area under the curve increases too. The calculation o the temperature change required is difcult and not given here. However, or the emission rom the surace to equal the incoming energy rom the Sun, allowing or the absorption, the surace temperature must rise to about 2 90 K. The net eect is shown in Figure 1 5 (b) with a shited and raised emission curve compensating or the energy that cannot be transmitted through the atmosphere. The suggestion that the atmosphere completely removes wavelengths above and below certain wavelengths is an over- simplifcation.
8 . 2 TH E RM AL E N E R G Y TRAN S FE R Figure 1 6 shows the complicated transmittance pattern in the inrared and indicates which absorbing molecules are responsible or which regions o absorption.
transmittance (percent)
100 80 60 40 20 0 0
1
O2
2
H 2O
3
4
5
CO 2 H 2 O CO 2 O 3
6
7 8 9 wavelength/m
10
11
H 2O CO 2 O 3 absorbing molecule
12
13
14
H 2 O CO 2
15
CO 2
Figure 16 Transmittance o the atmosphere in the inra-red.
The energy balance of the Earth The suraceatmosphere energy balance system is very complex; fgure 1 7 is a recent diagram showing the basic interactions and you should study it careully.
global energy fows W m -2 102
refected solar radiation 101.9 W m -2 refected by clouds and atmosphere
341
79 emitted by 169 atmosphere
79
outgoing longwave radiation 238.5 W m -2
239
incoming solar radiation 341.3 W m -2
40 30
atmospheric window greenhouse gases
absorbed by 78 atmosphere latent 80 heat 17 refected by surace 23
161 absorbed by surace
356 396 80 17 surace thermals evaporadiation transpiration net absorbed 0.9 W m -2
40
333 back radiation 333
absorbed by surace
Figure 17 Factors that make up the energy balance o the Earth (ater Stephens and others. 2012. An update on earths energy balance in light of the latest global observations. Nature Geoscience.) .
347
8
EN ERGY PRO D U CTI O N
Global warming There is little doubt that climate change is occurring on the planet. We are seeing a signifcant warming that may lead to many changes to the sea level and in the climate in many parts o the world. The act that there is change should not surprise us. We have recently ( in geological terms) been through several Ice Ages and we are thought to be in an interstadial phase at the moment ( interstadial means between Ice Ages ) . In the seventeenth century a Little Ice Age covered much o northern E urope and North America. The River Thames regularly roze and the citizens held airs on the ice. In 1 6 08, the D utch painter Hendrick Avercamp painted a winter landscape showing the typical extent and thickness o the ice in Holland ( fgure 1 8) .
Figure 18 Winter landscape with skaters (1608) , Hendrick Avercamp.
Many models have been suggested to explain global warming, they include:
changes in the composition o the atmosphere ( and specifcally the greenhouse gases) leading to an enhanced greenhouse eect
increased solar are activity
cyclic changes in the Earths orbit
volcanic activity.
Most scientists now accept that this warming is due to the burning o ossil uels, which has gone on at increasing levels since the industrial revolution in the eighteenth century. There is evidence or this. The table below shows some o the changes in the principal greenhouse gases over the past 2 5 0 years.
Gas Carbon dioxide Methane Nitrous oxide Ozone
Pre-1750 concentration 280 ppm 700 ppb 270 ppb 25 ppb
Recent concentration 390 ppm 1800 ppb 320 ppb 34 ppb
ppm = parts per million; ppb = parts per billion
348
% increase since 1750 40 160 20 40
8 . 2 TH E RM AL E N E R G Y TRAN S FE R
The recent values in this table have been collected directly in a number o parts o the world ( there is a well- respected long- term study o the variation o carbon dioxide in Hawaii where recently the carbon dioxide levels exceeded 400 ppm or the frst time or many thousands o years) . The values quoted or pre- 1 75 0 are determined rom a number o sources:
Analysis o Antarctic ice cores. C ores are extracted rom the ice in the Antarctic and these yield data or the composition o the atmosphere during the era when the snow originally ell on the continent. C ores can give data or times up to 400 000 years ago.
Analysis o tree rings. Tree rings yield data or the temperature and length o the seasons and the rainall going back sometimes hundreds o years.
Analysis o water levels in sedimentary records rom lake beds can be used to identiy historical changes in water levels.
An enhanced greenhouse eect results rom changes to the concentration o the greenhouse gases: as the amounts o these gases increase, more absorption occurs both when energy enters the system and also when the surace re- radiates. For example, in the transmittancewavelength graph or a particular gas, when the concentration o the gas rises, the absorption peaks will increase too. The surace will need to increase its temperature in order to emit sufcient energy at sea level so that emission o energy by Earth rom the top o the atmosphere will equal the incoming energy rom the Sun. Global warming is likely to lead to other mechanisms that will themselves make global warming increase at a greater rate:
the ice and snow cover at the poles will melt, this will decrease the average albedo o Earth and increase the rate at which heat is absorbed by the surace. a higher water temperature in the oceans will reduce the extent to which C O 2 is dissolved in seawater leading to a urther increase in atmospheric concentration o the gas.
Nature of science An international perspective There have been a number o international attempts to reach agreements over the ways orward or the planet. These have included:
The Kyoto Protocol was originally adopted by many ( but not all) countries in 1 997 and later extended in 2 01 2 .
The Intergovernmental Panel on Climate Change.
AsiaPacifc Partnership on C lean D evelopment and C limate.
The various other United Nations C onventions on C limate C hange, e.g. C ancn, 2 01 0.
Do some research on the Internet to fnd what is presently agreed between governments.
O ther human- related mechanisms such as deorestation can also drive global warming as the amount o carbon fxed in the plants is reduced. This is a problem that has to be addressed at both an international and an individual level. The world needs greater efciency in power production and a maj or review o uel usage. We should encourage the use o non- ossil-uel methods. As individuals we need to be aware o our personal impact on the planet, we should be conscious o our carbon ootprint. Nations can capture and store carbon dioxide, and agree to increase the use o combined heating and power systems. What everyone agrees is that doing nothing is not an option.
349
8
EN ERGY PRO D U CTI O N
Questions 1
b) The ow rate o water in the pipe is 400 m 3 s 1 . C alculate the power delivered by the alling water.
(IB) a) A reactor produces 2 4 MW o power. The efciency o the reactor is 3 2 % . In the fssion o one uranium- 2 3 5 nucleus 3 .2 1 0 1 1 J o energy is released.
3
D etermine the mass o uranium- 2 3 5 that undergoes fssion in one year in this reactor.
The energy losses in a pumped storage power station are shown in the ollowing table.
b) D uring its normal operation, the ollowing set o reactions takes place in the reactor. 1 0 239 92 239 93
n+
238 92
U
U
Np
239 93 239 94
239 92
Source of energy loss
U
Np + Pu +
0 -1 0 -1
_
e+ v _
e+ v
C omment on the international implications o the product o these reactions. 2
(IB)
Percentage loss of energy
friction and turbulence of water in pipe
27
friction in turbine and ac generator
15
electrical heating losses
5
a) C alculate the overall efciency o the conversion o the gravitational potential energy o water in the tank into electrical energy.
(IB) The diagram shows a pumped storage power station used or the generation o electrical energy.
b) S ketch a Sankey diagram to represent the energy conversion in the power station. tank
4
A nuclear power station uses uranium-2 3 5 ( U- 2 3 5 ) as uel.
pipe
310 m
(IB)
a) O utline: turbine
( i)
( ii) the role o the heat exchanger o the reactor and the turbine in the generation o electrical energy.
lake
Water stored in the tank alls through a pipe to a lake through a turbine that is connected to an electricity generator. The tank is 5 0 m deep and has a uniorm area o 5 .0 1 0 4 m 2 . The height rom the bottom o the tank to the turbine is 3 1 0 m. 3
3
The density o water is 1 .0 1 0 kg m . a) S how that the maximum energy that can be delivered to the turbine by the alling water is about 8 1 0 1 2 J.
350
the processes and energy changes that occur through which the internal energy o the working uid is increased
b) Identiy one process in the power station where energy is degraded. 5
(IB) The intensity o the Suns radiation at the position o the Earth is approximately 1 400 W m 2 . Suggest why the average power received per unit area o the Earth is 3 5 0 W m 2 .
QUESTION S 6
a) State the region o the electromagnetic spectrum to which the resonant requency o nitrogen oxide belongs.
(IB) The diagram shows a radiation entering or leaving the Earths surace or a simplifed model o the energy balance at the Earths surace.
b) Using your answer to ( a) , explain why nitrogen oxide is classifed as a greenhouse gas.
atmosphere TA = 242 K
9 transmitted through radiated by atmosphere 245 W m -2 atmosphere radiated by Earths 0.700 T4A surface = T4E Earths surface TE
a) State the emissivity o the atmosphere. b) D etermine the intensity o the radiation radiated by the atmosphere towards the Earths surace.
( IB) The diagram shows a simple energy balance climate model in which the atmosphere and the surace o Earth are treated as two bodies each at constant temperature. The surace o the E arth receives both solar radiation and radiation emitted rom the atmosphere. Assume that the Earths surace and the atmosphere behave as black bodies. 344 W m -2
c) C alculate TE . e = 0.720
8
a)
O utline a mechanism by which part o the radiation radiated by the Earths surace is absorbed by greenhouse gases in the atmosphere. Go on to suggest why the incoming solar radiation is not aected by the mechanism you outlined.
a = 0.280
atmospheric radiation
242 K
solar radiation
Earths surface
288 K
b) C arbon dioxide ( C O 2 ) is a greenhouse gas. S tate one source and one sink ( that removes C O 2 ) o this gas.
D ata or this model:
(IB)
average temperature o the atmosphere o E arth = 2 42 K
The graph shows part o the absorption spectrum o nitrogen oxide ( N 2 O ) in which the intensity o absorbed radiation A is plotted against requency f.
emissivity e o the atmosphere o Earth = 0.72 0 average albedo a o the atmosphere o E arth = 0.2 80 solar intensity at top o atmosphere = 344 W m 2 average temperature o the surace o E arth = 2 88 K a) Use the data to determine:
A/arbitrary units
7
atmosphere
( i) the power radiated per unit area o the atmosphere ( ii) the solar power absorbed per unit area at the surace o the Earth.
2
3
4
5
f/ 10 13 Hz
351
8
EN ERGY PRO D U CTI O N b) It is suggested that, i the production o greenhouse gases were to stay at its present level, then the temperature o the Earths atmosphere would eventually rise by 6 K. C alculate the power per unit area that would then be ( i) radiated by the atmosphere ( ii) absorbed by the Earths surace. c) Estimate the increase in temperature o the Earths surace.
352
1 0 ( IB) It has been estimated that doubling the amount o carbon dioxide in the Earths atmosphere changes the albedo o the Earth by 0.01 . Estimate the change in the intensity being refected by the Earth into space that will result rom this doubling. S tate why your answer is an estimate. Average intensity received at E arth rom the Sun = 3 40 W m 2 Average albedo = 0.3 0
9 WAVE PH E N O M E N A ( AH L) Introduction In this topic we develop many of the concepts introduced in Topic 4. In general, a more mathematical approach is taken and we consider
the usefulness of modelling using a spreadsheet both for graphing and developing relationships through iteration.
9.1 Simple harmonic motion Understanding The defning equation o SHM Energy changes
Nature of science The importance o SHM The equation or simple harmonic motion (SHM) can be solved analytically and numerically. Physicists use such solutions to help them to visualize the behaviour o the oscillator. The use o the equations is very powerul as any oscillation can be described in terms o a combination o harmonic oscillators using Fourier synthesis. The modelling o oscillators has applications in virtually all areas o physics including mechanics, electricity, waves and quantum physics. In this sub-topic we will model SHM using a simple spreadsheet and see how powerul this interpretation can be.
Applications and skills Solving problems involving acceleration,
velocity and displacement during simple harmonic motion, both graphically and algebraicallyO B J TE XT_UND Describing the interchange o kinetic and potential energy during simple harmonic motion Solving problems involving energy transer during simple harmonic motion, both graphically and algebraically
Equations
2 angular velocityperiod equation: = _ 2
defning equation or shm: a = - x
T
displacementtime equations: x = x 0 sin t;
x = x 0 cos t velocitytime equations: v = x 0 cos t; v = - x 0 sin t velocity-displacement _______ equation:
v = ( x 0 2 - x 2 )
1 m 2 ( x0 2 - x2 ) kinetic energy equation: E K = ___ 2 1 total energy equation: E = ___ m 2 x 2 T
2
0
__
period o simple pendulum: T = 2 ___gl __
m period o massspring: T = 2 ____ k
353
9
WAVE P H E N O M E N A ( AH L )
Introduction In this sub-topic we treat S HM more mathematically but restrict ourselves to two systems the simple pendulum and o a mass oscillating on a spring. Each o these systems is isochronous and is usually lightly damped; this means that a large number o oscillations occur beore the energy in the system is transerred to the internal energy o the system and the surrounding air.
Angular speed or frequency ()
screen metal sphere turntable light source drive belt
Figure 1 Comparison of SHM and circular motion.
In Sub-topic 6.1 we considered the angular speed in relation to circular motion; it is the rate o change o angle with time and is also called angular frequency. It is measured in radians per second (rad s - 1 ) . This quantity is important when we deal with simple harmonic motion because there is a very close relationship between circular motion and SHM. This relationship can be demonstrated using the apparatus shown in fgure 1 . A metal sphere is mounted on a turntable that rotates at a constant angular speed. A simple pendulum is arranged so that it is in line with the sphere and oscillates with the same periodic time as that o the turntable a little trial and error should give a good result here. The pendulum and the turntable are illuminated by a light that is projected onto a screen. The shadows projected, onto the screen, o the circular motion o the sphere and oscillatory motion o the pendulum show these motions to be identical.
Circular motion and SHM In mathematical terms the demonstration in fgure 1 is equivalent to proj ecting the two- dimensional motion o a point onto the single dimension o a line. Imagine a point P rotating around the perimeter o a circle with a constant angular speed . The radius o the circle r j oins P with the centre o the circle O . At time t = 0 the radius is horizontal and at time t it has moved through an angle radians. For constant angular speed = _t , rearranging this gives = t. Proj ecting P onto the y- axis gives the vertical component o r as r sin . Proj ecting P onto the x- axis gives the horizontal component o r as r cos . The variation o the vertical component with time or angle is shown on the right o fgure 2 and takes the orm o a sine curve. B ecause the rate o rotation is constant, the angle or t is proportional to time. I we drew a graph o y against t the quantities T 2 and would be replaced by T and __ respectively. 2 y
y
y = r sin
r P r = t
0 x = r cos
x
0
2
t
-r
Figure 2 Projection of circular motion on to a vertical line. The equations o the proj ections are y = y 0 sin t and x = x 0 cos t.
354
9 .1 S I M PLE H ARM O N I C M O TI O N
Here y 0 and x 0 are the maximum values o y and x, which in this case are identical to r. These are the amplitudes o the motion. You may remember rom S ub- topic 4.1 that in S HM the displacement, velocity and acceleration all vary sinusoidally with time. Thus, the proj ection o circular motion on the vertical or horizontal takes the same shape as SHM. This is very useul in analysing S HM.
The relationship between displacement, velocity, and acceleration In S ub-topic 4.1 we saw that, starting rom the displacementtime curve, we could derive the velocitytime and accelerationtime curves rom the gradients. This can be done graphically, but another technique is to dierentiate the equations with respect to time ( dierentiation is equivalent to fnding the gradient) . We have seen rom the comparison with circular motion that the displacement takes the orm: x = x0 sin t or x = x0 cos t depending on when we start timing. It really doesnt matter whether the projection is onto the x-axis or the y-axis thereore we can use x and y interchangeably. Lets start with x = x 0 sin t
Note
You will meet diferential
calculus in your Mathematics or Mathematics Studies course. This is not the place to teach you to diferentiate you will not be expected use calculus on your IB Physics course. However, or students studying physics, engineering and allied subjects at a higher level, calculus will orm a major aspect o your course. In this case, we will diferentiate the equations, but it is the results that are important not the method o obtaining them.
dx = x 0 cos t, where x 0 This means that the velocity is given by v ( = __ dt ) is the amplitude and the angular requency.
The maximum value that cosine can take is 1 so the maximum velocity is v 0 = x 0 thus making the equation or the velocity at time t become v = v 0 cos t We know that the acceleration will be the gradient o a velocitytime graph so we have dv a = __ = - v 0 sin t = - x 0 2 sin t dt
As or cosine, the maximum value that sine can take is 1 so a 0 = v 0 = x 0 2 giving a = - a 0 sin t C omparing the equations x = x 0 sin t and a = - x 0 2 sin t we can see a common actor o x 0 sin t meaning that a = - 2 x This may remind you, in S ub- topic 4.1 , we saw that a = - kx or S HM. S o the constant k must actually be 2 ( the angular requency squared) . We will look at the signifcance o 2 very soon. In examinations, you can be asked to fnd maximum values or velocity and acceleration this makes the calculations easier because the maximum o sine and cosine are each 1 and, thereore, we dont need to include the sine or cosine term in our calculations. Thus the maximum velocity will be v0 = x0 and the magnitude o the maximum acceleration will be a 0 = x0 2 (the direction o a 0 will be opposite that o x0) .
Modelling SHM with a spreadsheet Much can be learned about SHM by using a spreadsheet to graph it. Figure 3 is a screen shot o part o a spreadsheet set up or this purpose
355
9
WAVE P H E N O M E N A ( AH L ) B C D A time/s displacement/m velocity/m s1 acceleration/m s2
6.30 6.10 5.52 4.58 3.36 1.93 0.37 1.21 2.71 4.05 5.12 5.87 6.26 6.24 5.84 5.06 3.96 2.62 1.11 0.48 2.03 3.45 4.66 5.57 6.13 6.30 6.07 5.46 4.51 3.27 1.83 0.26 1.31 2.81 4.13 5.18 5.91 6.27 6.23 5.79 5.00 3.88 2.52 1.00 0.58 2.13 3.54 4.73 5.62
0.00 1.98 3.83 5.45 6.71 7.56 7.92 7.79 7.16 6.09 4.62 2.87 0.93 1.06 2.99 4.73 6.17 7.22 7.81 7.92 7.52 6.64 5.35 3.72 1.85 0.13 2.11 3.95 5.54 6.78 7.60 7.93 7.76 7.11 6.00 4.51 2.74 0.84 1.20 3.11 4.84 6.25 7.28 7.84 7.90 7.47 6.57 5.25 3.60
G
H
I
J
K
L
M
N
O
P
Q
x0 = 5 0 0.2 = 1.26
SHM displacement-time graph 6.00 4.00 2.00 x/m
0.00 1.25 2.41 3.43 4.23 4.76 4.99 4.91 4.51 3.83 2.91 1.81 0.59 0.67 1.88 2.98 3.89 4.55 4.92 4.99 4.73 4.18 3.37 2.34 1.17 0.08 1.33 2.49 3.49 4.27 4.79 5.0 4.89 4.48 3.78 2.24 1.73 0.50 0.75 1.96 3.05 3.94 4.58 4.94 4.98 4.71 4.14 3.31 2.27
F
t1 = t =
t/s 0.00 0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
2.00 4.00 6.00
SHM velocity-time graph 8.00 6.00 4.00 v/m s 1
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 20. 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0 6.2 6.4 6.6 6.8 7.0 7.2 7.4 7.6 7.8 8.0 8.2 8.4 8.6 8.8 9.0 9.2 9.4 9.6
E
2.00 t/s
0.00 2.00
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
4.00 6.00 8.00
SHM acceleration-time graph 10.00 8.00 6.00 4.00 2.00 0.00 2.00 4.00 6.00 8.00 10.00
a/m s 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
t/s 0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
Figure 3 Spreadsheet for SHM. ( as before this uses Microsoft E xcel but other spreadsheets will have similar functions) . C olumn A contains incremental times starting from zero ( cell H1 is copied into A2 this means that the starting time can be changed) . C ell H2 determines the time increments ( in this case 0.2 s) by adding the contents of H2 to each previous cell; the times can be increased down the column. The formula in cell A3 , in this case, is = A2 + $ H$ 2 and the formula in cell A4 is = A3 + $ H$ 2 , etc.
356
9 .1 S I M PLE H ARM O N I C M O TI O N
The times generated, together with chosen values o x0 and , are used to generate the displacement, velocity and time curves. The values or x0 and are inserted into cells J1 and J2 respectively these can then be changed at will. The equation or the displacement, x = x0 sin t is converted into the ormula =$J$1 *SIN(A2*$J$2) , which is copied into column B. The equation or the velocity, v = x0 cos t is converted into the ormula =$J$1 *$J$2*COS(A2*$J$2) which is copied into column C. Finally, the equation or the acceleration, a = - x0 2 sin t, is converted into the ormula =(- 1 ) *$J$1 *($J$2) ^ 2*SIN(A2*$J$2) and is copied into column D.
Note We have used a spreadsheet to show the solutions to the SHM equation. Later we will discuss how we can actually solve the SHM equation using iteration.
To produce the curves shown in fgure 3 , you now need to:
insert a scatter chart
choose the data series
do a little ormatting to improve the size and position o the chart and label the axes, etc.
The SHM equation and 2 Re- visiting the equation a = - 2 x discussed earlier in this sub- topic, we see that it fts the defnition o S HM ( motion in which the acceleration is p rop ortional to the disp lacement rom a fxed p oint and is always directed towards that fxed p oint) . The equation is the defning equation or SHM. The sinusoidal graphs provide the solutions to this equation with respect to time. This may seem strange because there is no apparent time actor in a = - 2 x. This is where 2 comes in. We saw with circular motion that represents the angular speed. Thereore 2 is simply the square o 2 1 this measured in rad 2 s 2 . 2 has dimensions equivalent to ( ____ . tim e ) In circular motion we defned as being __ or simply __t when it is t constant. For a complete revolution the angle will be 2 radians and the 2 time will be the periodic time T. This means that = ___ or, alternatively, T 1 __ = 2 f; comparing this equation with f = T explains why we can call the angular requency.
Worked example An obj ect perorms S HM with a period o 0.40 s and has amplitude o 0.2 0 m. The displacement is zero at time zero. C alculate:
We are only asked to fnd the magnitude o the velocity so it doesnt matter which o the cosine curves the motion really takes.
a) the maximum velocity
Using v = x 0 cos t ( 2 0.1 0) 2 t rad t = _ = __ _ T 0.40 2 2 _ _ S o v = 0. 2 0 = 0 cos 0.40 2 This could have been done without the ull calculation once we had decided that it was a cosine. 0. 1 0 s represents the time or a quarter o a period ( 0. 40 s) ; the value o any cosine at a quarter o a period ( or __ radian) is zero. 2
b) the magnitude o the velocity ater 0.1 0 s c) the maximum acceleration o the obj ect.
Solution 2 a) v 0 = x 0 = x 0 _ T 2 _ v 0 = 0.2 0 = 3 .1 m s 1 0.40 b) As the displacement is zero at time zero this must be a sine or negative sine wave. Thus the velocity will be a cosine or negative cosine.
(
2 c) a 0 = x 0 2 = 0.2 0 _ 0.40
)
2
49 m s 1
357
9
WAVE P H E N O M E N A ( AH L )
The velocity equation The derivation o the SHM equation is or reerence and you dont need to reproduce it ollowing it through, however, will help you to understand what is going on. We have already seen that v = x 0 cos t and x = x 0 sin t 2 and you may know that sin + cos 2 = 1 . ________
This means that cos = 1 - sin ( dont orget that squaring either the positive or negative cosine will give ( + ) cos 2 . This means that must be included in the square root o this. 2
________
S o v = x 0 1 - sin 2 t
x x 2 2 __ B ut sin t = __ x so sin t = ( x ) 0
v = x0
0
_______
( ) x 1- _ x0
2
=
___________
x
2 0
( )
x - x0 2 _ x0
2
______
= x 0 2 - x 2
______
The equation v = x 0 2 - x 2 is useul or fnding the velocity at a particular position when you know the amplitude and period (or requency or angular requency) you dont need to know the time being considered.
Worked example
Note
An obj ect oscillates simple harmonically with requency 60 Hz and amplitude 2 5 mm. C alculate the velocity at a displacement o 8 mm.
Solution = 2 f = 1 2 0 ( there is no need to calculate the value here but do not leave in the answer in an examination.
tells us that the obj ect could be going in either o the two opposite directions. D ont orget x 0 2 - x 2 ( x 0 - x ) 2 ; it is a common mistake or students to equate these two expressions.
______
v = x 0 2 - x 2
____________________
= 1 2 0 ( 2 5 1 0 - 3 ) 2 - ( 8 1 0 - 3 ) 2 = 8.9 m s 1
Simple harmonic systems 1. The simple pendulum
The simple pendulum represents a straightorward system that oscillates with SHM when its amplitude is small. When the pendulum bob (the mass suspended on the string) is displaced rom the rest position there is a component o the bobs weight that tends to restore the bob to its normal rest or equilibrium position. A condition o a system oscillating simple harmonically is that there is a restoring orce that is proportional to the displacement rom the equilibrium position this is, in eect, tying in with the equation defning SHM because F = ma and a = - 2 x this means that F = - m 2 x.
l Ft
m
mg sin mg cos
Figure 4 Restoring force for simple pendulum.
358
Figure 4 shows the orces acting on a pendulum bob. The bob is in equilibrium along the radius when the tension in the string F t equals the component o the weight in line with the string ( = mgcos). The component o the weight perpendicular to this is not in equilibrium and provides the restoring orce.
9 .1 S I M PLE H ARM O N I C M O TI O N
S o the restoring orce must be equal to the mass multiplied by the acceleration according to Newtons second law o motion:
Note
mg sin = ma
The period of a simple
pendulum is independent of the mass of the pendulum.
x sin _ l g _ x = ma rearranging gives - m l
For a small angle
( )
pendulum is independent of the amplitude of the pendulum.
( The minus sign is because the displacement ( to the right) is in the opposite direction to the acceleration ( to the let) in fgure 4. C ancelling m
( )
g a= - _ x l
g 2 = _ l
l is the length of the
pendulum from the point of suspension to the centre of mass of the bob.
this compares with the defning equation or S HM a = - 2 x leading to
The equation only applies to
small oscillations (swings making angle of less than 10 with the rest position) .
__
2 As the periodic time or SHM is given by T = _ , this shows that T = 2
The period of a simple
_gl .
Investigate! 1
Experimenting with the simple pendulum:
attach a piece o thread o length about 2 metres to a pendulum bob ( a lump o modelling clay would do or this)
split cork
clamp
suspend the thread through a split cork to provide a stable point o suspension align a pin mounted in another piece o modelling clay with the rest position o the bob to act as a reerence point ( sometimes called a fducial marker)
ruler l timer
set the bob oscillating through a small angle and start timing as the bob passes the fducial marker (you should consider why your measurements are likely to be more reliable when the bob is moving at its astest)
G-clamp bob
time thirty oscillations (remember that each oscillation is rom A to C and back to A)
measure the length o the thread rom the point o suspension to the centre o mass o the bob ( take this to be the centre o the bob)
repeat the procedure and fnd an average period or the pendulum
< 10
A
C B
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WAVE P H E N O M E N A ( AH L )
2
repeat your measurements or a range o lengths
frst plot a graph o periodic time T against length l
reerring to the simple pendulum equation you may see that this graph should not be linear what should you plot in order to linearize your results?
how can you use the graph to calculate a value or the acceleration o reeall g?
dont orget to do an error analysis and compare your result with the accepted value o g.
increment the change in velocity will be the acceleration multiplied by the time increment when you add this to your previous velocity you will get the new velocity ( actually its the average velocity through the time increment)
because velocity is the change in displacement divided by the time increment, you can fnd the change in displacement by multiplying your current velocity value by your chosen time increment
the new value o the displacement will be the previous value added to the change in displacement
This now eeds back into fnding the next value o the acceleration and the cycle repeats and you generate your data or which you can plot displacement- time, velocity- time and acceleration-time graphs ( a example o the spreadsheet is included on the website) .
Use o iteration Making use o the powerul iterative unctionality o a spreadsheet can be a valuable learning aid. With iteration a graphical investigation is possible rom basic principles and basic equations without knowing the solution and without advanced mathematics.
you will need to set up a worksheet with the headings time, displacement, acceleration, change in velocity, velocity and change in displacement you now need to set up the initial conditions: 1 . choose a time increment or the iteration ( make it airly large until your spreadsheet is working well) 2 . start your time column at zero and decide how long you want to run the iteration or ( make this, say, 1 0 time increments initially) 3 . choose an amplitude value or your oscillation 4. set the initial displacement value equal to the amplitude
The ollowing ow chart illustrates the iteration:
n =0 choose t, tn , xn , k n = 0 a n+1 = -k xn = a n+1 t n+1 = n +
n
n+1
x = n+1 t xn+1 = xn + x
enough increments ?
no
yes
2
5 . choose a constant ( k or ) 6. set the initial velocity to be zero
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the acceleration will always be - k multiplied by the displacement
because acceleration is the change in velocity divided by your time
fnish
S ubscripts n represent the current value o a variable and n + 1 the next value; at each loop the next value always replaces the previous one.
9 .1 S I M PLE H ARM O N I C M O TI O N
2. Massspring system We have ocused on a simple pendulum as being a very good approximation to S HM. A second system which also behaves well and gives largely undamped oscillations is a massspring system. We will consider a mass being oscillated horizontally by a spring ( see fgure 5 ) ; this is more straightorward than taking account o including the eects o gravity experienced in vertical motion. We will assume that the riction between the mass and the base is negligible. The mass, thereore, exchanges elastic potential energy ( when it is ully extended and compressed) with kinetic energy ( as it passes through the equilibrium position) .
Note
The period of the
mass-spring system is independent of amplitude (for small oscillations) .
The period of the
mass-spring system is independent of the acceleration of gravity.
restoring force relaxed spring mass
extended spring
mass
base
base initial position of left edge
initial position of left edge
x
position of left edge when spring extended
Figure 5 Restoring force for mass-spring system. When a spring ( having spring constant k) is extended by x rom its equilibrium position there will be a restoring orce acting on the mass given by F = - kx ( the orce is in the opposite direction to the extension) . Using Newtons second law ma = - kx which can be written as
( )
k a= - _ m x this compares with the defning equation or S HM a = - 2x
( )
k 2 = _ m
As the periodic time or S HM is given by 2 T= _ this shows that __
T = 2
m
_k
When the spring is compressed the quantity x represents the compression o the spring. When the mass is to the let o the equilibrium position the compression is positive but the restoring orce will be negative. This, thereore, leads to the same outcomes as or extensions.
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Energy in SHM systems
Worked example The massspring system is used in many common accelerometer designs. A mass is suspended by a pair of springs which displaces when acceleration occurs. An accelerometer contains a mass of 0.080 kg coupled to a spring with spring constant of 4.0 kN m -1 . The amplitude of the mass is 20 mm. Calculate:
We have seen that, in a simple pendulum, there is energy interchange between gravitational potential and kinetic; in a horizontal massspring system the interchange is between elastic potential and kinetic energy. B ecause each system involves kinetic energy we will focus on this form of energy and bear in mind that the potential energy will always be the difference between the total energy and the kinetic energy at a particular time. The total energy will be equal to the maximum kinetic energy. From Topic 2 we know that the kinetic energy of an obj ect of mass m, moving at velocity v, is given by 1 m v2 EK = _ 2
a) the maximum acceleration b) the natural frequency of the mass.
Solution
( )
We also know that the equation for the velocity at a particular position is
k a) a = - _ m x
(
______
v = x 0 2 - x 2
)
4.0 1 0 3 a = - _ 2 0 1 0-3 0.08 = 1 000 m s 2 __
m 1 b) T = 2 _ so f = _ T k
__
1 = _ 2
k _ m
________
1 f= _ 2
C ombining these equations gives the kinetic energy at displacement x: 1 E K = _ m 2 ( x0 2 - x2 ) 2 This tells us that the maximum kinetic energy will be given by 1 m 2 x 2 E Kmax = _ 0 2 and this must be the total energy ( when the potential energy is zero) so we can say
4.0 1 0 = 3 6 Hz _ 0.08 3
1 m 2 x 2 ET = _ 0 2 The potential energy at any position will be the difference between the total energy and the kinetic energy so 1 m 2 x 2 - _ 1 m 2 x 2 - x2 = _ EP = ET - EK = _ ( 0 ) 1 m 2 x2 0 2 2 2 Figure 6 illustrates the variation with displacement of energy: the green line shows the total energy, the red the potential energy and the blue the kinetic energy. At any position the total energy is the sum of the kinetic energy and the potential energy as we would expect. 1.200 1.000
energy/J
0.800 0.600 0.400 0.200 0.000 0.30
0.20
0.10
0.00 0.10 displacement/m
Figure 6 Variation of energy with displacement.
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0.20
0.30
9 .1 S I M PLE H ARM O N I C M O TI O N
The variation o the potential energy and kinetic energy with displacement are both parabolas. With all the quantities in the total energy equation being constant, the total energy is, o course, constant or an undamped system. In addition to looking at the variation o energy with displacement we should consider the variation o energy with time. Again, let us start with the kinetic energy. The velocity varies with time according to the equation v = x 0 cos t so the kinetic energy will be 1 m v2 = _ 1 m( x cos t) 2 EK = _ 0 2 2 or 1 m x 2 2 cos 2 t EK = _ 0 2 When the cosine term equals 1 , this gives the maximum kinetic energy. The maximum kinetic energy occurs when the potential energy is zero and so is numerically equal to the total energy at that instant. 1 E T = _ m x0 2 2 2 The potential energy will be the dierence between the total energy and the kinetic energy so EP = ET - EK 1 1 m x 2 2 co s 2 t = _ m x0 2 2 - _ 0 2 2 1 = _ m x 0 2 2 si n 2 t 2 This relationships are shown on fgure 7. The green line represents the total energy, the red curve the potential energy and the blue curve the kinetic energy.
Note
1.20
the sum of the kinetic and potential energies.
1.00
0.80
energy/J
The total energy is always
The graphs (unlike sine
and cosine) never become negative.
0.60
The period of the energy
change is half that of the variation with time of displacement, velocity, or acceleration.
0.40
0.20
0.00 0.0
2.0
4.0
6.0
8.0 time/s
10.0
12.0
The frequency of the energy
change is twice that of the variation with time of displacement, velocity, or acceleration.
Figure 7 Graph showing variation of energy with time.
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9.2 Single-slit difraction Understanding
Applications and skills
The nature o single-slit diraction
Describing the eect o slit width on the
diraction pattern Determining the position o frst intererence minimum Qualitatively describing single-slit diraction patterns produced rom white light and rom a range o monochromatic light requencies
Nature of science Development o theories That rays travel in straight lines is one o the frst theories o optics that students encounter. It comes as a surprise when this theory cannot explain the diraction seen at the edges o shadows cast by small objects illuminated using point sources. Although partial shadows can be explained by considering light sources to be extended, this cannot account or the diraction rom a point source. The wave theory o diraction and how diraction can be explained in terms o waveront propagation rom secondary sources is a good example o how theories have been developed in order to explain a wider variety o phenomena.
angle between frst minimum and central maximum = ____ a
Introduction
intensity
Io
Equations
In S ub- topic 4. 4 we introduced diffraction and saw that when a wave passes through an aperture it spreads into the geometric shadow region. We also saw that the diffraction pattern consists of a series of bright and dark fringes. We will now consider the diffraction pattern in more detail.
5% Io angle 3
2
1
0
1
2
3
central maximum
Figure 1 Variation o intensity with angle or a difraction pattern.
Graph o intensity against angle Figure 1 shows a single-slit diffraction pattern together with a graph of the variation of the intensity of the diffraction pattern with the angle measured from the straight- through position. 1 , 2 , and 3 are the angles with the straight- through position made by the minima.
Note
The central maximum has twice the angular width o
the secondary maxima (each o these have the same angular width) .
No light reaches the centre o the minima but, in going
towards the maxima, the intensity gradually increases it is very dicult to decide the exact positions o the maxima and minima.
The intensity alls o quite signifcantly rom the
principal maximum to the secondary maxima the intensity o the frst secondary maximum is approximately 5% o that o the principal maximum,
364
the second is about 2% and the third is about 1% (fgure 1 is not drawn to scale the secondary maxima are all larger than a scale-diagram would show) .
The intensity is proportional to the square o the amplitude.
9 . 2 S I N G LE- S LI T D I FFR ACTI O N
The single-slit equation
TOK
Returning to the work o S ub- topic 4. 4 we saw how waves can superpose to give constructive intererence ( when they meet in phase) or destructive intererence ( when they meet anti- phase) .
Small angle approximation
(a)
incident plane waves a
(b)
single slit
top o slit
towards frst minimum a
waveront
bottom o slit
path dierence = a sin
Figure 2 Deriving the single slit equation. Figure 2 (a) shows a single-slit o width a. Each point on the waveront within the slit behaves as a source o waves. These waves interere when they meet beyond the slit. For the two waves coming rom the edges o the slit making an angle with the straight through, there is a path dierence o a sin . Waves rom a point halway along the slit will have a a path dierence o __ sin rom the waves coming rom each o the 2 edges. When this path dierence equals hal a wavelength, the waves rom halway along the slit will interere destructively with the waves coming rom the bottom edge o the slit. It ollows that or each point in the bottom hal o the slit there will be a point in the top hal o the slit a a at a distance __ sin rom it. This means that when __ sin = __ , there is 2 2 2 destructive intererence between a wave coming rom a point in the upper hal o the slit and an equivalent wave rom the lower hal o the slit.
We say that diraction is most eective when the aperature is o the same order o magnitude as the width o the slit; we can demonstrate this eectively using a ripple tank. In the equation = ____ a i we make a, then = 1 radian (or 57.3) or, i you avoid the small angle approximation, sin = 1 and = 90. Examine how closely diraction in a ripple tanks agrees with small angle approximation prediction. With poor agreement (as this should show) why do we continue to use the equation = ____ a?
B ecause we are dealing with small angles we can approximate sin to and, by cancelling the actor o two, we arrive at the equation = _ a n The position o other minima ( shown in fgure 1 ) will be given by = ___ a ( where n = 2 , 3 , 4, ...) but you will not be examined on this relationship.
Note
This is the equation or the angle o the frst minimum. n that the minima are not actually separated It ollows rom the equation = _____ a
by equal distances. However, or values o that are less than about 10, it is a good approximation to consider the minima to be equally spaced.
The principal or central maximum occurs because the pairs o waves rom the
top and bottom halves o the slit will travel the same distance and have no path dierence.
The angular width o the principal maximum (rom frst minimum on one side
to frst minimum on the other) is 2.
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Worked example
Solution
a) Explain, by reerence to waves, the diraction o light at a single slit.
a) The waveront within the slit behaves as a series o point sources which spread circular wave ronts rom them. These secondary waves superpose in ront o the slit and the superposition o the waves produces the diraction pattern.
b) Light rom a heliumneon laser passes through a narrow slit and is incident on a screen 3 .5 m rom the slit. The graph below shows the variation with distance x along the screen o intensity I o the light on the screen. I
b) (i) In this case the graph is drawn or distance not angle, so we need to calculate the angle in order to be able to use = __ a. -3
s 3.0 1 0 = __ = ________ D 3.5
= 0.86 1 0 3 rad with s being the distance rom the centre o the principal maximum to the frst minimum. It is more reliable to measure 2 rom the frst minimum on one side o the principal maximum to the frst minimum on the other side o the principal maximum. D is the distance rom the slit to the screen. -9
630 1 0 a = __ = _________ 0.86 1 0 -3
10
5
0 x/mm
5
(i) The wavelength o the light emitted by the laser is 63 0 nm. D etermine the width o the slit. (ii) S tate two changes to the intensity distribution o the central maximum when the single slit is replaced by one o greater width.
10
= 0.73 1 0 3 m or 0. 73 mm (ii) With a wider slit more light is able to pass through. This will result in an increase in the intensity o the beam and so the peaks will all be higher. a in the equation = __ a increases but remains constant; the angle will decrease which means the principal maximum will become narrower and the minima will move closer together.
Single slit with monochromatic and white light
Figure 3 Single slit with monochromatic and white light.
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Figure 3 shows two images o the light emerging rom a singleslit. The upper image is obtaine d using monochromatic ( one requency) green light and the lower image is obtained using white light. B oth the angular width o the ce ntral maximum and the angular se paration o successive secondary maxima depend on the wavelength o the light this is the re ason why the secondary maxima produced by the diraction o white light are coloured. You will se e that, or the secondary maxima, the blue light is le ss deviated than the other colours as it has the shortest wavele ngth. The e dges o the principal maximum are coloured rather than pure white . This is because the principal maxima or the colours at the blue end o the visible spectrum are le ss spread than the colours at the red end; the e dges are thereore a combination o red, orange, and ye llow and have an orange hue.
9.3 IN TERFEREN CE
Investigate! Diraction at a single slit
Using a laser pointer, shine the light on the gap between the j aws o a pair o digital or vernier callipers ( orming a slit) and proj ect it onto a white screen. The screen needs to be at least 3 m rom the callipers.
Measure the distance between the callipers and the screen, D, using a tape- measure.
C alculate the wavelength o the light using s a as we know that 2 _ D B oth the dependence o the width o the central maximum and the separation o the maxima on the wavelength o light can be investigated by using laser pointers with dierent colours.
Adj ust the callipers so that the image is clear.
Measure the separation, s, o the frst minima on the screen using a metre ruler. D
Research the wavelength ranges o the laser light and then suggest how you could modiy the experiment to check the calibration o the callipers at small j aw separation.
s
2 laser callipers
We will return to diraction when we look at resolution in S ub-topic 9.4.
screen
Figure 4 Investigating laser light passing through slit.
9.3 Interference Understanding Youngs double-slit experiment
Applications and skills Qualitatively describing two-slit intererence
Modulation o two-slit intererence pattern by
one-slit diraction eect Multiple slit and diraction grating intererence patterns Thin flm intererence
Nature of science Thin flm intererence The observation o colour is not simply a question o colour pigmentation. Certain mollusc and beetle shells, buttery wings, and the eathers o hummingbirds and kingfshers can produce beautiul colours as a result o thin flm intererence known as iridescence. The observed colour changes depend on the angle o illumination or viewing, and are caused by multiple reections rom the suraces o flms with thicknesses similar to the wavelength o light. These natural aesthetics require the analysis o the physics o intererence.
patterns, including modulation by one-slit diraction eect Investigating Youngs double-slit experimentally Sketching and interpreting intensity graphs o double-slit intererence patterns Solving problems involving the diraction grating equation Describing conditions necessary or constructive and destructive intererence rom thin flms, including phase change at interace and eect o reractive index Solving problems involving intererence rom thin flms
Equations D ringe separation or double slit: s = ______ d diraction grating equation: n = d sin light reection by parallel-sided thin flm: constructive intererence 2dn = (m + ___12 ) destructive intererence 2dn = m
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Introduction In Sub-topic 4.4 we considered intererence o the waves emitted by two coherent sources. We saw that intererence is a property o all waves, and we considered the equal ringe spacing in the Young double-slit experiment. In this sub-topic we ocus on the intensity variation in a double-slit experiment and see how intererence is achieved using multiple slits. We then look at a second way o achieving intererence, using division o amplitude instead o division o waveront as with double-slit intererence.
Intensity variation with the double-slit Figure 1 shows the image o the light rom a heliumneon laser that has passed through a double slit. The alternate red and dark ringes are equally spaced as we saw in Sub-topic 4. 4. However, looking at the image closely we see that there are extra dark regions.
Figure 1 Double-slit difraction pattern or light rom HeNe laser. We know rom Sub-topic 9.2 that a single slit produces a diraction pattern with a very intense principal maximum and much less intense secondary maxima. A double slit is, o course, two single slits so each o the slits produces a diraction pattern and the waves rom the two slits interere. The two eects mean that the intensity o the intererence pattern is not constant, but is modifed by the diraction pattern to produce the intensity. Figure 2(a) below shows how the relative intensity would vary or a doubleslit intererence pattern without any modifcation due to diraction. By using relative intensity we avoid the need to think about the actual intensity values and units. Figure 2(b) shows the variation o relative intensity with angle or a single slit. Figure 2(c) shows the superposition o the two eects so that the single-slit diraction pattern behaves as the envelope o the intererence pattern. Shaping a pattern in this way is called modulation and is important in the theory o AM (amplitude modulation) radio. You will note that the ringe spacing does not change between fgures 2(a) and (c) but that the bright ringes occurring between 1 1 and 1 4 in fgure 2(a) are reduced to a much lower intensity. An intererence maximum coinciding with a diraction minimum is suppressed and does not appear in the overall pattern. We saw in S ub- topic 4.4 that the ringe separation s or light o wavelength is given by D s=_ d where D is the distance rom the double slit to the screen and d is the separation o the slits. The value o s used in this equation is or the intererence pattern not the modulated pattern caused by diraction.
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9.3 IN TERFEREN CE
relative intensity (a)
20
15
10
5 0 10 5 angle rom straight through position/
15
20
15
20
15
20
relative intensity (b)
20
15
10
5 0 5 10 angle rom straight through position/
relative intensity (c)
20
15
10
5 0 10 5 angle rom straight through position/
Figure 2 Combination o difraction and intererence.
Multiple-slit interference We have now seen the interference patterns produced both by a single slit and a double slit. What happens if there are more than two slits? The answer is that the bright fringes, which come from constructive interference of the light waves from different slits, remain in the same positions as for a double slit but the pattern becomes sharper. The bright fringes are narrower and their intensity is proportional to the square of the number of slits. Why is this? At the centre of the principal maximum the waves reaching the screen from all of the slits are in phase and so it is very bright here. B y moving to a position close to the centre of this maximum the path difference between the light from two adj acent slits has changed by such a small amount that it hardly affects the interference and so it is still bright. Slits further away from each other will have more likelihood of a greater path difference and therefore meeting out of phase.
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9
WAVE P H E N O M E N A ( AH L ) Imagine having three slits: at the principal maximum the path dierence between the waves will be small as their paths are nearly parallel. Each o the three wave trains will meet nearly in phase at positions around the principal maximum peak, which makes it airly wide. Now imagine having 1 0 0 slits: or every slit there will a second slit somewhere that transmits a wave with a hal wavelength path dierence rom the frst. When the waves interere at a position close to the centre o the maximum, destructive intererence occurs and reduces the overall intensity. This will be true or the waves coming rom the two slits adj acent to this pair, the two next to those and so on. Increasing the number o slits will give destructive intererence close to the centre o the maxima this reduces the width o the central maximum ( and the other maxima too) . The extra energy that is needed to increase the intensity o the maxima must come rom the regions that are now darker, so the maxima are more intense. The mathematics o modulation o waves is quite complex and or the purpose o the IB Physics course you will simply need to recognize that the modulation is happening and remember the eect o increasing the number o slits. Figure 3 shows the intererence patterns obtained when red laser light passes through various numbers o slits o identical width. Figures 3 and 4 show that the single-slit diraction pattern always acts as an envelope or the multiple-slit intererence patterns.
Figure 3 The efect o increasing the number o slits on an intererence pattern. In fgure 4 the relative intensities have been drawn the same size in order to show the increased sharpness; however, or two, three, and fve slits the actual intensities are in the ratio 1 : 9 : 2 5 .
370
9.3 IN TERFEREN CE
single slit
double slit
relative intensity
20
15 10 5 0 5 10 15 angle rom straight through position/
relative intensity
20
20
three slits
15 10 5 0 5 10 15 angle rom straight through position/ fve slits
relative intensity
20
15 10 5 0 5 10 15 angle rom straight through position/
20
relative intensity
20
20
15 10 5 0 5 10 15 angle rom straight through position/
20
Figure 4 The eect o increasing the number o slits on variation o intensity.
white light source
The difraction grating A diraction grating is a natural consequence o the eect on the intererence pattern when the number o slits is increased. D iraction gratings are used to produce optical spectra. A grating contains a large number o parallel, equally spaced slits or lines ( normally etched in glass or plastic) typically there are 600 lines per millimetre ( see fgure 5 ) . When light is incident on a grating it produces intererence maxima at angles given by
diraction grating
Figure 5 Diraction grating.
n = d sin The spacing between the slits is small, which makes the angle large or a fxed wavelength o light and n. This means that we cannot use the small angle approximation or relating the wavelength to the position o the maxima as we did or a double-slit. Figure 6 shows a section o a diraction grating in which three consecutive slits deviate the incident waves towards a maximum. The slits are so narrow and the screen is so ar away rom the grating that the angles made by the diracted waves are virtually identical. Providing the path dierence between waves coming rom the same part o successive slits is an integral number o wavelengths, the waves will reach the screen in phase and give a maximum. In the case o a diraction grating the distance d is taken as the length o both the transparent and opaque sections o the slit (they are taken to be o equal width) . From fgure 6 we see that the path dierence between those waves coming rom A and B and the path between waves coming B and C will each be d sin . This means
A d
B C
portion o grating
light diracted
Figure 6 Deriving the diraction grating equation.
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9
WAVE P H E N O M E N A ( AH L ) that n = d sin (where n is the order o the maximum and is zero or the central maximum, 1 or the frst maximum on each side o the centre, etc.) . We can see that the angular positions o the intererence maxima depend on the grating spacing, d. The shape o the diraction envelope, however, is determined by the width o the clear spaces ( as or a single slit) . The maxima must lie within the envelope o the single slit diraction pattern i they are to be relatively intense; this sets an upper limit on how wide the transparent portions o the grating can be.
Figure 7 Dispersing white light with a difraction grating.
O ne o the uses o a diraction grating is to disperse white light into its component colours: this is because dierent wavelengths produce maxima at dierent angles. Figure 7 shows that light o greater wavelength ( or any given order) is deviated by a larger angle. This is in line with what we would predict rom using n = d sin . Each successive visible spectrum repeats the order o the colours o the previous one but becomes less intense and more spread out.
Grating spacing and number o lines per mm It is usual or the number o lines per mm or ( N) to be quoted or a diraction grating, rather than the spacing. N must be converted into d in order to use the diraction grating equation n = d sin . 1 This is straightorward because d = __ but N must frst be converted to N the number o lines per metre by multiplying by 1 000 beore taking the reciprocal.
Nature of science
For a diraction grating to produce an observable pattern, the grating spacing must be comparable to the wavelength o the waves. The wavelength o visible light is between approximately 400700 nm. A grating with 600 lines per mm has a spacing o approximately 2 000 nm or our times the wavelength o green light. This spacing produces very clear images.
The wavelengths o low- energy X-rays is around 1 0 - 1 0 m or 0.1 nm. The 600 lines per millimetre optical grating will not produce any observable maximum . .. thereore, what will diract X- rays? The spacing o ions in crystals is o the same order o magnitude as X- rays and the regular lattice shape o a crystal can perorm the same task with X- rays as a diraction grating does with visible light. The diracted patterns are now commonly detected with charge-coupled device ( C C D ) detectors.
Worked example
Solution
A diraction grating having 6 0 0 lines per millimetre is illuminated with a parallel beam o monochromatic light, which is normal to the grating. This produces a se cond- order maximum which is observed at 42 . 5 to the straightthrough direction. C alculate the wave length o the light.
N = 6.00 1 0 5 lines per metre.
Appropriate wavelengths or efective dispersion
372
1 1 d = __ = ________ = 1 .67 1 0 - 6 m. N 6.00 1 0 5
-6
dsin 1 . 6 7 1 0 sin 42 . 5 n = d sin so = ____ = ________________ ( dont n 2
orget to have your calculator in degrees) = 5 .64 1 0 - 7 = 5 64 nm ( all data is given to three signifcant fgures so the answer should also be to this precision) .
9.3 IN TERFEREN CE
Nature o science Using a spectrometer A spectrometer is a useul, i expensive, piece o laboratory equipment that allows the wavelengths o light emitted by sources to be analysed. The instrument consists o a collimator, turntable and telescope. A diraction grating or other light disperser is placed on the turntable, which is careully levelled. The collimator uses lenses to produce a parallel beam o light rom a source this light beam is then incident on the grating that disperses it. The end o the collimator urthest rom the grating has a vertical adj ustable slit that serves as the source when we talk about spectral lines we iner that a spectrometer is being
used because the lines are the dispersed images o the collimator slit. B eore measurements are made the telescope is ocused on the collimator slit this means that the dispersed light will also be in ocus. It is usual or sources that are said to be monochromatic to actually emit a variety o dierent wavelengths. B y using the spectrometer, the angle in the diraction grating equation can be measured. This is usually done by measuring the angle between the two frst-order images on either side o the straight- through position i.e. 2 . With a knowledge o the grating spacing d, the wavelength can be determined.
adjustable slit collimator difraction grating turntable
light source angular scale
te l e
sco
pe
Figure 8 Key eatures o a spectrometer.
Interference by division of amplitude It was mentioned in the introduction to this sub-topic that there are two ways o providing coherent sources that are able to interere. Youngs double slit and multiple slits all derive their interering waves by taking waves rom dierent parts o the same waveront. Because the interering waves have all come rom the same waveront they will be in phase with each other. Wherever the waves meet they will interere and a ringe pattern can be obtained anywhere in ront o the sources (the slits) . Since this intererence can be ound anywhere the ringes are said to be non-localized. D ivision o amplitude is a method o achieving intererence using two waves that have come rom the same point on a waveront. Each wave has a portion o the amplitude o the original wave. In order to achieve intererence by division o amplitude, the source o light must come rom a much bigger source than the slit used or division o waveront intererence. The image produced will, however, be localized to one place instead o being ound anywhere in ront o the sources.
Thin flm intererence Figure 9 shows a wave incident at an angle to the normal to the surace o a flm o transparent material ( such as low- density oil or detergent) having reractive index n. This diagram is not drawn to scale
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incident wave
B
C
Waves reected by the flm (you may be tested on this in IB Physics examinations) In this case A has been reected rom the top surace o the flm and, because the reection is at an optically denser medium, there is a phase change o radian (equivalent to hal a wavelength) . The wave B travels an optical distance o 2tn beore it reracts back into the air. Thus the optical path dierence between A and B will be 2tn. I there had been no phase change then this optical distance would equal m or constructive intererence. However, because o As phase change at the top surace the overall eect will be destructive intererence. Thus or the light reecting rom the flm when 2tn = m there will be destructive intererence and when 2tn = (m + ___12 ) there will be constructive intererence.
air transparent medium o reractive index n
t
air A
B
C
Figure 9 Intererence at parallel-sided thin flm.
Waves transmitted through the flm (you will not be tested on this in IB Physics examinations) I the flm is thin and is small the (geometrical) path dierence between waves that have passed through the flm (i.e. between A' and B' or between B' and C') will be very nearly 2t, in other words B' travels an extra 2t compared with A'. Because the waves are travelling in a material o reractive index n, the waves will slow down and the wavelength becomes shorter. This means that, compared with travelling in air, they will take longer to pass through the flm. This is equivalent to them travelling through a thicker flm at their normal speed. The optical path dierence will, thereore, be 2tn. I this distance is equal to m where m = 0, 1, 2 etc. (we are using m to avoid conusion with the reractive index n) then there will be constructive intererence. I the optical path dierence is equal to an odd number o hal wavelengths, (m + ___12 ), then there will be destructive intererence.
Nature of science Coating o lenses When light is incident on a lens some o it will be reected and some transmitted. The reected light is eectively wasted, reducing the intensity o any image ormed ( by the eye using corrective lenses, in a camera or in a telescope) . B y coating the lens with a transparent material o quarter o a wavelength thickness, the light reected by the coating and the light reected by the lens can be made to interere destructively and thus eliminate the reection altogether. Magnesium uoride is oten used or the coating and is optically denser than air but less dense than
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glass ( having a reractive index o 1 . 3 8) . When it is used to coat a lens the waves reecting rom both the magnesium uoride and the glass will undergo a phase change o radian . . . so the phase changes eectively cancel each other out. The optical path dierence between the waves will be equal to the reractive index o magnesium uoride multiplied by twice the thickness o the coating. For destructive intererence the optical path dierence is 2 nt = __ 2 so the thickness o the coating is given by t = __ . 4n With white light there will be a range o
9.3 IN TERFEREN CE
wavelengths and so it is impossible to match the thickness o the coating to all o these wavelengths. I the thickness is matched to green light, then red and blue will still be reected giving the lens the appearance o being magenta
( = red + blue) . B y using multiple layers o a material o low reractive index and one o higher reractive index, it is possible to reduce the amount o light reected to as little as 0. 1 % or a chosen wavelength.
Worked example
Solution
a) Name the wave phenomenon that is responsible or the ormation o regions o dierent colour when white light is reected rom a thin flm o oil oating on water.
a) Although there is reection involved, the colours come about because o intererence.
b) A flm o oil o reractive index 1 .45 oats on a layer o water o reractive index 1 .3 3 and is illuminated by white light at normal incidence. illumination air oil water
When viewed at near normal incidence a particular region o the flm looks red, with an average wavelength o about 65 0 nm. (i) Explain the signifcance o the reractive indices o oil and water with regard to observing the red colour. (ii) C alculate the minimum flm thickness.
b) (i) n is the reractive index o the oil. Because the waves are travelling in oil, they move more slowly than they would do in air and so the eective path dierence between the waves reected at the airoil interace and the waves at the oilwater interace is longer by a actor o 1 .45. The wave reected at the airoil interace undergoes a phase change equivalent to __ (oil is denser 2 than air) . The waves reected at the oil water interace undergo no phase change on reection at the less dense medium. So or the bright constructive red intererence 1 2tn = ( m + __ ) 2 i m = 0, 2 tn = _ or = 4tn 2 65 0 (ii) t = _ = 1 1 0 nm ( 4 1 .45 )
Nature of science Vertical soap flms Figure 1 0 shows the colours o light transmitted by a vertical soap flm. Over a ew seconds the flm drains and becomes thinner at the top and thicker at the bottom. When the flm is illuminated with white light, the reected light appears as a series o horizontal coloured bands. The bands move downwards as the flm drains and the top becomes thinner. The top o the flm appears black just beore the flm breaks. It has now become too thin or there to be a path dierence between the waves coming rom the two suraces o the flm. The phase change that occurs or the light reected by the surace o the flm closest to the source means that, or all colours, there is cancellation and so no light can be seen.
Figure 10 Thin flm intererence in a vertical soap flm.
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TOK The aesthetics of physics The colour in the soap flm makes a beautiul image. Does physics need to rely upon the arts to be aesthetic? Herman Bondi the Anglo-Austrian mathematician and cosmologist implied that Einstein believed otherwise when he wrote: What I remember most clearly was that when I put down a suggestion that seemed to me cogent and reasonable, Einstein did not in the least contest this, but he only said, Oh, how ugly. As soon as an equation seemed to him to be ugly, he really rather lost interest in it and could not understand why somebody else was willing to spend much time on it. He was quite convinced that beauty was a guiding principle in the search for important results in theoretical physics. H. Bondi How does a mathematical equation convey beauty? Is the beauty in the mathematics itsel or what the mathematics represents?
9.4 Resolution Understanding The size o a diracting aperture The resolution o simple monochromatic
two-source systems
Applications and skills Solving problems involving the Rayleigh
criterion or light emitted by two sources diracted at a single slit Describing diraction grating resolution
Equations Rayleighs criterion: = 1.22 ____ b resolvance o a diraction grating: R = _______ = mN
Nature of science How ar apart are atoms? When we view objects, we are limited by the wavelength o the light used to make an observation shorter wavelengths such as X-rays, gamma rays and ast-moving electrons will improve the resolution that is achievable. However, even using the shortest wavelengths obtainable does not allow us to locate the exact position o objects on the
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atomic and sub-atomic scale. The process o making an observation using these waves disturbs the system and increases the uncertainty with which we can locate an object. Heisenbergs uncertainty principle places a limit on how close we can be to fnding the exact position o objects on the quantum scale.
9.4 RESOLU TI ON
Introduction Resolution is the ability o an imaging system to be able to p roduce two sep arate distinguishable images o two sep arate obj ects. The imaging system could be an observers eye, a camera, a radio telescope, etc. Whether obj ects can be resolved will depend on the wavelength coming rom the obj ects, how close they are to each other and how ar away they are rom the observer.
Difraction and resolution
images ully resolved
We have seen that when light passes through an aperture a diraction pattern is ormed. For an optical system the aperture could be the pupil o the observers eye or the obj ective lens o a telescope. When there are two sources o light two diraction patterns will be ormed by the system. How close can these patterns be or us to still recognize that there are two sources? Two obj ects observed through an aperture will produce two diracted images which may or may not overlap. In the late nineteenth century the E nglish physicist, John William Strutt, 3 rd B aron Rayleigh, proposed what is now known as the Rayleigh criterion or resolution o images. This states that two sources are resolved i the p rincip al maximum rom one diraction p attern is no closer than the frst minimum o the other p attern.
images just resolved
The limit to resolution is when the principal maximum o the diraction pattern rom one source lies on the frst minimum diraction pattern rom the second source ( and vice versa) . D iracted images urther apart than this limit will be resolved and those closer will be unresolved. Figure 1 shows the diraction intensity patterns produced by two obj ects the same distance apart but viewed through a circular aperture rom dierent distances. The variation o intensity with angle or each o the diraction patterns is shown below the image. According to the Rayleigh criterion, the uppermost pair o images are ully resolved because the principal maximum o each diraction pattern lies urther rom the other than the frst minimum. The central images are j ust resolved since the principal maximum o one diraction pattern is at the same position as the frst minimum o the second diraction pattern. The bottom images are unresolved as the principal maximum o one diraction pattern lies closer to the second pattern than its frst minimum.
images unresolved
Figure 1 Difraction intensity patterns o two objects viewed through a circular aperture.
Nature of science Other criteria or resolution? Rayleighs criterion is not the only one used in optics. Many astronomers believe that they can resolve better than Rayleigh predicts. C M S parrow developed another criterion or telescopes that leads to an angular separation
at resolution about hal that o Rayleigh. His criterion is that the two diraction patterns when added together give a constant amplitude in the regions o the two central maxima.
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Resolution equation For single slits we saw in S ub-topic 9.2 that the frst minimum occurs when the angle with the straight- through position is given by = __ a where is the wavelength o the waves and b is the slit width. With a circular aperture the equation is modifed by a actor o 1 .22, but derivation o this actor is beyond the scope o the IB Physics course. So we have = 1 .2 2 __ a
in this case a is the diameter o the circular aperture (or very commonly the diameter o the lens or mirror orming the image) . The pupil o the eye has a diameter o about 3 mm and, taking visible light to have a wavelength in the order o 6 1 0 7 m, the minimum 1 .22 6 1 0 2 1 0 - 4 rad. angle o resolution or the eye is = ___________ 3 10 O ptical deects in the eye mean that this limit is probably a little small. -7
-3
The primary mirror o the Hubble Space Telescope has a diameter o 2.4 m. 1 .2 2 6 1 0 For this telescope, the minimum angle o resolution is = ___________ 2.4 3 1 0 - 7 rad. This is a actor o about a thousand smaller than that achieved by the unaided eye which means that the Hubble Space Telescope is much better at resolving images. As this telescope is in orbit above the atmosphere, it avoids the atmospheric distortion which degrades images achieved by Earth- based telescopes. These concepts are covered in more detail in O ption C ( Imaging) . -7
Worked example A student observes two distant point sources o light. The wavelength o each source is 5 5 0 nm. The angular separation between these two sources is 2 . 5 1 0 4 radians subtended at the pupil o a students eye.
Note
You may wish to support
your answer by drawing the two intensityangle curves i you think your answer may not be clear.
It is common or
students to miss out the word difraction in their answers this is crucial to score all marks.
a) S tate the Rayleigh criterion or the two images on the retina to be j ust resolved. b) E stimate the diameter o the circular aperture o the eye i the two images are j ust to be resolved.
Solution a) The images will be j ust resolved when the diraction pattern rom one o the point sources has its central maximum at the same position as the frst minimum o the diraction pattern o the other point source. 1 .2 2 5 5 0 1 0 b) = 1 .2 2 __ = > b = 1 .2 2 __ = _____________ b 2 .5 1 0 -9
-4
= 2 .7 1 0 3 m 3 mm
Nature of science Difraction and the satellite dish When radiation is emitte d rom a transmitting satellite dish the waves diract rom the dish. The diameter o the dish behaves as an aperture. The angle made by the frst
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minimum with the straight- through is related to diameter b by the equation = 1 .22 _ b
9.4 RESOLU TI ON
When the wavelength increases so does the angle through which the waves become diracted. This means that the diracted beam now covers a larger area. However, increasing the diameter o a dish narrows the beam and it covers a smaller area. In satellite communications the footp rint is the portion o the E arths surace over which the satellite dish delivers a specifed amount o signal power. The ootprint will be less than the region covered by the principal maximum because the signal will be too weak to be useul at the edges o the principal maximum. Figure 2 shows the ootprint o a communications satellite.
Small values o the dish diameter will give a large ootprint but the intensity may be quite low since the energy is spread over a large area.
The ootprint o a satellite has social and political implications ranging rom unwarranted observation to sharing television programmes. satellite
ootprint
Figure 2 Satellite ootprint.
Resolvance o difraction gratings We have seen that diraction gratings are used to disperse light o dierent colours. Such gratings are usually used with spectrometers to allow the angular dispersion or each o the colours to be measured. When using a spectrometer, knowing the number o lines per millimetre on the diraction grating and measuring the angles that the wavelengths o light are deviated through allows the wavelengths o the colours to be determined. We have seen that there is a limit to how close two objects can be beore their diraction images are indistinguishable the same is true o the dierent wavelengths that can be resolved using a particular diraction grating.
double-slit angular dispersion o principal maximum
When we previously looked at intererence patterns or multiple slits in S ub- topic 9.3 , we saw that increasing the number o slits improves the sharpness o the maxima ormed. Figure 3 shows that, when light o the same wavelength is viewed rom the same distance x, the angular dispersion D or the principal maximum with the double slit is larger than angular dispersion G or the diraction grating. A sharper principal maximum is one with less angular dispersion. With wider maxima there is more overlap o images rom dierent sources and lower resolution.
D X
Using this argument we see that, when beams o light are incident on a diraction grating, a wider beam covers more lines ( a greater number o slits) and will produce sharper images and better resolution.
difraction grating angular dispersion o principal maximum
The resolvance R or a diraction grating ( or other device used to separate the wavelengths o light such a multiple slits) is defned as the ratio o the wavelength o the light to the smallest dierence in wavelength that can be resolved by the grating .
G X
The resolvance is also equal to Nm where N is the total number o slits illuminated by the incident beam and m is the order o the diraction. R = _ = Nm The larger the resolvance, the better a device can resolve.
Figure 3 Angular dispersion o principal maximum or difraction grating compared with a double-slit.
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Nature of science Resolution in a CCD C harge- coupled devices ( C C D s) were originally developed or use in computer memory devices but, today, appear in all digital cameras and smartphones. When you buy a camera or phone you will no doubt be interested in the number o pixels that it has. The pixel is a picture element and, or example, a 2 0 megapixel camera will have 2 1 0 7 pixels on its C C D . The resolution o a C C D depends on both the number o pixels and their size when compared to the proj ected image. The smaller the camera, the more convenient it is to carry, and ( at present) cameras with pixels o dimensions as small as 2 .7 m 2 . 7 m are mass- produced. In general C C D images are resolved better with larger numbers o smaller pixels. Figure 4 shows two images o the words
IB Physics the top uses a small number o large pixels, while the lower one is ar better resolved by using a large number o small pixels. The upper image is said to be pixellated.
B Ph sics
Figure 4 Images with a small number of large pixels and a large number of small pixels.
Worked example Two lines in the emission spectrum o sodium have wavelengths o 5 89. 0 nm and 5 89. 6 nm respectively. C alculate the number o lines per millimetre needed in a diraction grating i the lines are to be resolved in the second- order spectrum with a beam o width 0. 1 0 mm.
Solution 5 89.0 R = ___ = _____ ( both values are in nanometres so 0.6
this actor cancels) You would be equally j ustifed in using 5 89.6 ( or 5 89.3 the mean value) in the numerator here.
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R = 981 .7 ( no units) Thus 981 .7 = Nm = N 2 N = 498.8 lines This is in a beam o width 0.1 0 1 0 3 m so, in 1 mm, there needs to be 4988 5 000 lines. S ince diraction gratings are not normally made with 4988 lines mm 1 , the sensible choice is to use one with 5 000 lines mm 1 !
9 . 5 TH E D O PPLE R E FFE CT
9.5 The Doppler efect Understanding The Doppler eect or sound waves and
light waves
Nature o science From water waves to the expansion o the universe In his 1842 paper, ber das farbige Licht der Doppelsterne (Concerning the coloured light o the double stars) , Doppler used the analogy o the measurement o the requency o water waves to reason that the eect that bears his name should apply to all waves. Three years later, Buys Ballot verifed Dopplers hypothesis or sound using stationary and moving groups o trumpeters. During his lietime Dopplers hypothesis had no practical application; one hundred years on it has arreaching implications or cosmology, meteorology and medicine.
Applications and skills Sketching and interpreting the Doppler eect
when there is relative motion between source and observer Describing situations where the Doppler eect can be utilized Solving problems involving the change in requency or wavelength observed due to the Doppler eect to determine the velocity o the source/observer
Equations Doppler equation or a moving source: v f = f __ v us or a moving observer: v uo f = f __ v
(
)
(
)
or electromagnetic radiation:
_ = _ v _ c f
Introduction When there is relative motion between a source of waves and an observer, the observed frequency of the waves is different to the frequency of the source of waves. The apparent change in pitch of an approaching vehicle engine and a sounding siren are common examples of this effect. The D oppler effect has wide- ranging implications in both atomic physics and astronomy.
The Doppler efect with sound waves Although we use general equations for this effect, we build up the equations under different conditions before combining them. In the following derivations ( that need not be learned but will help you to understand the equations and how to answer questions on this topic) the letter s refers to the source of the waves ( the obj ect giving out the sound) and the letter o refers to the observer; f will always be the frequency of the source and f the apparent frequency as measured by the observer.
Note It is only the component o the wave in the source observer direction that is used; perpendicular components o the motion do not alter the apparent requency o the wave.
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1. Moving source and stationary observer v f
v f us
A
B
S
S' us f
Figure 1 The Doppler efect or a moving source and stationary observer.
At time t = 0 the source is at position S and it emits a wave that travels outwards in all directions, with a velocity v as shown in fgure 1 . At time t = T ( i.e. one period later) the wave will have moved a distance equivalent to one wavelength or ( using v = f) a distance = _vf to reach positions A and B . When the source is moving to the right with a u velocity u s , in time T it will have travelled to S , a distance u s T( = __ ) . To f a stationary observer positioned at B it will appear that the previously emitted crest has reached B but the source that emitted it has moved orwards and now is at S . Thereore, to the observer at B , the apparent u wavelength ( ) is the distance SB = _vf - __ f s
s
v - us = _ f
Thus, the wavelength appears to be squashed to a smaller value. This means that the observer at B will hear a sound o a higher requency than would be heard rom a stationary source the sound waves travel at speed v ( which is unchanged by the motion o the source) so v v f = _ = f _ v - us
(
)
To an observer at A the wavelength would have appeared to be stretched to a longer value given by S A meaning that v + us = _ f and the observed requency would be lower than that o a stationary source, meaning that v v f = _ = f _ v + us
(
)
The two equations or f can be combined into a single equation v f = f _ v us
(
)
in which the sign is changed to - or a source moving towards a stationary observer and to + or a source moving away rom a stationary observer.
2. Moving observer and stationary source In this case the source remains stationary but the observer at B moves towards the source with a velocity u o . The source emits crests at a
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9 . 5 TH E D O PPLE R E FFE CT
frequency f but the observer, moving towards the source, encounters the crests more often, in other words at a higher frequency f. Relative to the observer, the waves are travelling with a velocity v + u o and the frequency is f. The wavelength of the crests does not appear to have changed and will be = _vf . The wave equation ( v = f) applied by the observer becomes v + u o = f _vf meaning that the frequency measured by the observer will be v + uo f = f _ v
(
)
When the observer moves away from the source the wave speed appears to be v - u o and so f = f
v- u (_ v ) o
Again the two equations for f can be combined into a single equation
(
v uo f = f _ v
)
in which the sign is changed to - for an observer moving away from a stationary source and to + for an observer moving towards a stationary source.
Worked example A stationary loudspeaker emits sound of frequency of 2 .00 kHz. A student attaches the loudspeaker to a string and swings the loudspeaker in a horizontal circle at a speed of 1 5 m s 1 . The speed of sound in air is 3 3 0 m s 1 . An observer listens to the sound at a close, but safe, distance from the student. a) Explain why the sound heard by the observer changes regularly. b) D etermine the maximum frequency of the sound heard by the observer.
Solution a) As the loudspeaker approaches the observer, the frequency appears higher than the stationary frequency because the apparent wavelength is shorter meaning that the wavefronts are compressed. As it moves away from the observer, the frequency appears lower than the stationary frequency because the apparent wavelength is stretched. The overall effect is, therefore, a continuous rise and fall of pitch heard by the observer. b) The maximum observed frequency occurs when the speaker is approaching the observer so: v 330 _______ f = f( _____ v - u ) = 2 000 ( 3 3 0 - 1 5 ) = 2 1 00 Hz ( this is 2 s.f. precision in s
line with 1 5 m s 1 )
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The Doppler efect with light The D oppler eect occurs not only with sound but also with light ( and other electromagnetic waves) ; in which case the requency and colour o the light diers rom that emitted by the source. There is a signifcant dierence in the application o D oppler eect or sound and light waves. S ound is a mechanical wave and requires a medium through which to travel; electromagnetic waves need no medium. Additionally, one o the assumptions or postulates o special relativity is that the velocity o light waves is constant in all inertial reerence rames this means that, when measured by an observer who is not accelerating, the observer will measure the speed o light to be 3 . 00 1 0 8 m s - 1 irrespective o whether the observer moves towards the source or away rom it and, thereore, it is impossible to distinguish between the motion o a source and an observer. This is not true or sound waves, as we have seen. Although the D opple r eect e quations or light and sound are derived on complete ly dierent principles, providing the speed o the source ( or observer) is much less than the speed o light, the equations give approximately the corre ct results or light or sound. The quantities u s and u o in the D oppler e quations or sound have no signifcance or light, and so we need to de al with the re lative velocity v between the source and observer. Thus, in either o the equations v f = f _ v us
(
)
(
)
or v uo f = f _ v
the wave speed is that o electromagnetic waves ( c) and one o the velocities u s or u o is made zero while the other is replaced by the relative velocity v.
Note
Substituting values into
the second equation will give the same resulting relationship you may like to try this but remember you do not need to know any of these derivations.
The equation is only
valid when c v and so cannot usually be used with sound (where the wave speed is 300 m s 1 unless the source or observer is moving much more slowly than this speed) .
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The frst o these equations becomes
(
)
c v -1 1 _ f = f _ = f _ v = f(1 + c ) _ c+ v 1 + c
(
)
This can be expanded using the binomial theorem to approximate to v v _ f = f ( 1 - _ c ) f- f c ( ignoring all the terms ater the second in the expansion) . This can be written as v f- f f _ c or v f f _ c This equation is equivalent to v _ c
9 . 5 TH E D O PPLE R E FFE CT
Worked example As the S un rotates, light waves received on E arth rom opposite ends o a diameter show equal but opposite D oppler shits. The speed o the edge o the S un relative to the E arth is 1 . 9 0 km s - 1 . What wavelength shit should be expected in the helium line having wavelength 5 87 . 5 6 1 8 nm?
3
1 .90 1 0 = 5 87.5 61 8 ________ = 0.003 7 nm 3.00 1 0 8
S ince one edge will approach the E arth the shit rom this edge will be a decrease in wavelength ( blue shit) and that o the other edge ( receding) will be an increase in wavelength ( red shit) .
Solution v Using f = f _vc this is equivalent to _ c. 1 .90 km s 1 = 1 .90 1 0 3 m s 1
Nature of science Applications o the Doppler efect 1. Astronomy The D oppler eect is o particular interest in astronomy it has been used to provide evidence about the motion o the obj ects throughout the universe. The Doppler eect was originally studied in the visible part o the electromagnetic spectrum. Today, it is applied to the entire electromagnetic spectrum. Astronomers use Doppler shits to calculate the speeds o stars and galaxies with respect to the Earth. When an astronomical body emits light there is a characteristic spectrum that corresponds to emissions rom the elements in the body. B y comparing the position o the spectral lines or these elements with those emitted by the same elements on the Earth, it can be seen that the lines remain in the same position relative to each other but shited either to longer or shorter wavelengths (corresponding to lower or higher requencies) . Figure 2 shows an unshited absorption spectrum imaged rom an Earth-bound source together with the same spectral lines rom distant astronomical objects. The middle image is red-shited indicating that the source is moving away rom the Earth. The lower image is blue-shited showing the source to be local to the Earth and moving towards us. Frequency or wavelength shits can occur or reasons other than relative motion. Electromagnetic waves moving close to an obj ect with a very strong gravitational feld can be red- shited this, unsurprisingly, is known as
unshifted
redshifted
blueshifted
Figure 2 The Doppler shifted absorption spectra.
gravitational red-shift and is discussed urther in O ption A. The cosmological red-shit arises rom the expansion o space ollowing the B ig B ang and is what we currently detect as the cosmic microwave background radiation ( this is urther discussed in O ption D ) .
2. Radar Radar is an acronym or radio detection and ranging. Although it was developed or tracking aircrat during the S econd World War, the technique has wide- ranging uses today, including:
weather orecasting
ground- penetrating radar or locating geological and archaeological arteacts
providing bearings
radar astronomy
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use in salvaging
collision avoidance at sea and in the air.
transmitterreceiver ultrasound beam blood cells v
direction o blood ow
blood vessel
Figure 3 Radar screen used in weather orecasting.
Radar astronomy diers rom radio astronomy as it can only be used or Moon and planets close to the Earth. This depends on microwaves being transmitted to the object which then reects them back to the Earth or detection. In order to use the Doppler equations, we must recognize that the moving object frst o all behaves as a moving observer and then, when it reects the microwaves, behaves as a moving source. This means, or microwave sensing, the equation f f _vc is adapted to become f 2 f _vc .
3. Measuring the rate o blood fow The D oppler eect can be used to measure the speed o blood ow in blood vessels in the body. In this case ultrasound ( i.e. longitudinal mechanical waves o requency above
Figure 4 The Doppler efect used to measure the speed o blood cells.
approximately 2 0 kHz) is transmitted towards a blood vessel. The change in requency o the beam reected by a blood cell is detected by the receiver. The speed o sound and ultrasound is around 1 5 00 m s 1 in body tissue so the equation f f _vc is appropriate or blood cells moving at speeds o little more than 1 m s 1 . As the blood does not ow in the direction o the transmitter receiver, there needs to be a actor that will give the necessary component o the blood velocity in a direction parallel to the transmitter receiver. As with radar, there will need to be a actor o two included in the equation the shit is being caused by the echo rom a moving reector. As can be seen rom fgure 4, the equation or the v co s rate o blood ow will be f 2 f _____ c . The D oppler eect owmeter may be used in preerence to in- line owmeters because it is non- invasive and its presence does not aect the rate o ow o uids. The act that it will remain outside the vessel that is carrying the uid means it will not suer corrosion rom contact with the uid.
Worked example Microwaves o wavelength 1 5 0 mm are transmitted rom a source to an aircrat approaching the source. The shit in requency o the reected microwaves is 5 .00 kHz. C alculate the speed o the aircrat relative to the source.
Using 8
c 3.00 1 0 = ________ = 2 .00 1 0 9 Hz. c = f gives f = __ 150 10 -3
Using the D oppler shit equation or radar fc 5.00 1 0 3.00 1 0 f 2 f _vc = > v ___ = _________________ 2f 2 2.00 1 0 1 = 3 75 m s 3
8
9
Solution The data gives a wavelength and a change o requency so we need to fnd the requency o the microwaves.
386
QUESTION S
Questions 1
b) Explain why the magnitude of the tension in the string at the midpoint of the oscillation is greater than the weight of the pendulum bob.
(IB) The variation with displacement x of the acceleration a of a vibrating obj ect is shown below. 3000
c) The pendulum bob is moved to one side until its centre is 2 5 mm above its rest position and then released.
a/m s 2
point of suspension rigid support
2000 1000 x/mm 0.6 0.4 0.2
0
0.2
0.4
0.6
0.80 m
1000 2000 25 mm 3000 pendulum bob
a) State and explain two reasons why the graph indicates that the obj ect is executing simple harmonic motion.
(i) S how that the speed of the pendulum bob at the midpoint of the oscillation is 0.70 m s 1 .
b) Use data from the graph to show that the frequency of oscillation is 3 5 0 Hz.
(ii) The mass of the pendulum bob is 0.057 kg. The centre of the pendulum bob is 0.80 m below the support. Calculate the magnitude of the tension in the string when the pendulum bob is vertically below the point of suspension.
c) State the amplitude of the vibrations. ( 9 marks) (IB) a) A p e ndu lum co nsists o f a b o b that is susp e nde d fro m a rigid sup p o rt b y a light ine xte nsib le string. The p e ndulu m b o b is mo ve d to o ne side and the n re le ase d. The ske tch grap h sho ws ho w the disp lace me nt o f the p e ndulu m b o b unde rgo ing simp le harmo nic mo tio n varie s with time o ve r o ne time p e rio d. displacement
0
3
( 1 0 m arks ) (IB) a) A particle of mass m attached to a light spring is executing simple harmonic motion in a horizontal direction. State the condition (relating to the net force acting on the particle) that is necessary for it to execute simple harmonic motion. b) The graph shows how the kinetic energy E K of the particle in (a) varies with the displacement x of the particle from equilibrium.
0
0.07 EK /J
time
0.06 0.05 0.04 0.03
C opy the sketch graph and on it clearly label
0.02
(i) a point at which the acceleration of the pendulum bob is a maximum.
x/m
05 0.
03
02
01
04 0.
0.
0.
0.
0.
0
0.
01
02
0.
03
04 0.
05 0.
(ii) a point at which the speed of the pendulum bob is a maximum.
0.01
2
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WAVE P H E N O M E N A ( AH L ) (i) On a copy o the axes above, sketch a graph to show how the potential energy o the particle varies with the displacement x.
Q
Z W
(ii) The mass o the particle is 0.30 kg. Use data rom the graph to show that the requency f o oscillation o the particle is 2.0 Hz.
b
Y
X
P
( 8 m arks ) 4
(IB) a) D escribe what is meant by the diffraction of light.
The angle is small.
b) A parallel beam o monochromatic light rom a laser is incident on a narrow slit. The diracted light emerging rom the slit is incident on a screen.
a) O n a copy o the diagram, label the hal angular width o the central maximum o the diraction pattern. b) S tate and explain an expression, in terms o , or the path dierence ZW between the rays ZP and XP.
screen slit parallel light wavelength 620 nm
0.40 mm
c) D educe that the hal angular width is given by the expression = _ b d) In a certain demonstration o single slit diraction, = 45 0 nm, b = 0.1 5 mm and the screen is a long way rom the slits.
C
1.9 m
The centre o the diraction pattern produced on the screen is at C . Sketch a graph to show how the intensity I o the light on the screen varies with the distance d rom C . c) The slit width is 0.40 mm and it is 1 .9 m rom the screen. The wavelength o the light is 62 0 nm. D etermine the width o the central maximum on the screen. ( 8 marks) 5
(IB) Plane waveronts o monochromatic light are incident on a narrow, rectangular slit whose width b is comparable to the wavelength o the light. Ater passing through the slit, the light is brought to a ocus on a screen. The line XY, normal to the plane o the slit, is drawn rom the centre o the slit to the screen. The points P and Q are the frst points o minimum intensity as measured rom point Y. The diagram also shows two rays o light incident on the screen at point P. Ray ZP leaves one edge o the slit and ray XP leaves the centre o the slit.
388
screen
slit
C alculate the angular width o the central maximum o the diraction pattern on the screen. ( 8 marks) 6
(IB) Monochromatic parallel light is incident on two slits o equal width and close together. Ater passing through the slits, the light is brought to a ocus on a screen. The diagram below shows the intensity distribution o the light on the screen. I
A
B distance along the screen
a) Light rom the same source is incident on many slits o the same width as the widths o the slits above. O n a copy o the diagram, draw a possible new intensity distribution o the light between the points A and B on the screen.
QUESTION S a) S tate the phase change that occurs when light is reected rom
A parallel beam o light o wavelength 45 0 nm is incident at right angles on a diraction grating. The slit spacing o the diraction grating is 1 .2 5 1 0 6 m.
( i) surace A ( ii) surace B .
b) D etermine the angle between the central maximum and frst order principal maximum ormed by the grating.
The light incident on the plastic has a wavelength o 62 0 nm. The reractive index o the plastic is 1 .4.
( 4 marks) 7
b) C alculate the minimum thickness o the flm needed or the light reected rom surace A and surace B to undergo destructive intererence.
(IB) Light o wavelength 5 90 nm is incident normally on a diraction grating, as shown below.
( 5 marks)
grating 6.0 10 5 lines per metre
9 frst order
light wavelength 590 nm
(IB) The two point sources A and B emit light o the same requency. The light is incident on a rectangular narrow slit and, ater passing through the slit, is brought to a ocus on the screen.
zero order frst order
The grating has 6.0 1 0 5 lines per metre. a) D etermine the total number o orders o diracted light, including the zero order, that can be observed.
A
b) The incident light is replaced by a beam o light consisting o two wavelengths, 5 90 nm and 5 89 nm. State two observable dierences between a frst-order spectrum and a second-order spectrum o the diracted light. ( 6 marks) 8
(IB) Monochromatic light is incident on a thin flm o transparent plastic as shown below.
C
monochromatic light
A
B flm
The plastic flm is in air. Light is partially reected at both surace A and surace B o the flm.
B point sources slit screen
a) B is covered. S ketch a graph to show how the intensity I o the light rom A varies with distance along the screen. Label the curve you have drawn A. b) B is now uncovered. The images o A and B on the screen are j ust resolved. Using your axes, sketch a graph to show how the intensity I o the light rom B varies with distance along the screen. Label this curve B . c) The bright star Sirius A is accompanied by a much ainter star, Sirius B. The mean distance o the stars rom Earth is 8.1 1 0 1 6 m. Under ideal atmospheric conditions, a telescope with an objective lens o diameter 25 cm can just resolve the stars as two separate images. Assuming that the average wavelength emitted by the stars is 5 00 nm, estimate the apparent, linear separation o the two stars. ( 6 marks)
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WAVE P H E N O M E N A ( AH L ) 1 0 (IB) a) Explain what is meant by the term resolvance with regards to a diraction grating. b) A grating with a resolvance o 2 000 is used in an attempt to separate the red lines in the spectra o hydrogen and deuterium. ( i)
The incident beam has a width o 0.2 mm. For the frst order spectrum, how many lines per mm must the grating have?
( ii) E xplain whether or not the grating is capable o resolving the hydrogen lines which have wavelengths 65 6.3 nm and 65 6. 1 nm.
a) Explain, using a diagram, any dierence between f and f. b) The requency f is 3 .00 1 0 2 Hz. An observer moves towards the stationary car at a constant speed o 1 5 .0 m s 1 . C alculate the observed requency f o the sound. The speed o sound in air is 3.30 1 0 2 m s -1 . (5 marks) 1 3 (IB) The wavelength diagram shown below represents three lines in the emission spectrum sample o calcium in a laboratory. A
B
C
( 6 marks) 1 1 A source o sound approaches a stationary observer. The speed o the emitted sound and its wavelength, measured at the source, are v and respectively. C ompare the wave speed and wavelength, as measured by the observer, with v and . Explain your answers. ( 4 marks) 1 2 The sound emitted by a cars horn has requency f, as measured by the driver. An observer moves towards the stationary car at constant speed and measures the requency o the sound to be f.
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wavelength
A distant star is known to be moving directly away rom the Earth at a speed o 0.1 c. The light emitted rom the star contains the emission spectra o calcium. C opy the diagram and sketch the emission spectrum o the star as observed in the laboratory. Label the lines that correspond to A, B , and C with the letters A*, B *, and C *. Numerical values o the wavelengths are not required. ( 3 marks)
10 F I E L D S ( A H L ) Introduction E arlier in the book you were introduced to gravitation and electrostatics. At that stage, these were treated as separate sets of ideas, but there are strong similarities between the
concepts in both topics. In this topic we are going to discuss these similarities and develop the concepts of gravitation and electrostatics in parallel.
10.1 Describing felds Understanding Gravitational felds Electrostatic felds Electric potential and gravitational potential Field lines Equipotential suraces
Applications and skills Representing sources o mass and charge,
lines o electric and gravitational orce, and feld patterns using an appropriate symbolism Mapping felds using potential Describing the connection between equipotential suraces and feld lines
Equations workelectric potential equation: W = QVe workgravitational potential equation: W = mVg
Nature of science Our everyday experience o orces is that they are direct and observable. They act directly on objects and the response o the object is calculable. Acceptance o the feld concept means acceptance o action at a distance. One object can inuence
another without the need or contact between them. The paradigm shit rom one world view to another is dicult. It took a signifcant eort in the history o science and demands a leap in conceptual understanding rom physicists.
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F I E L D S ( AH L )
Nature of science Scalar and vector felds S trictly, the felds here are vector felds as the orces have magnitude and direction. O ther felds are dierent in that the quantities they represent only have magnitude they are scalar felds. The temperature associated with each point around a camp fre is an example o a scalar feld.
The parts o this chapter devoted to electric felds have a blue background or a blue line in the margin; the sections dealing with gravity have a green background or a green line in the margin. You can concentrate on one type o feld by sticking to one colour. B eore we introduce new ideas, here is a review o those ideas that we considered in the earlier topics.
Fields A feld is said to exist when one obj ect can exert a orce on another obj ect at a distance.
Electric felds An electrostatic orce exists between two charged objects.
A gravitational orce exists between two objects that both have mass.
There are two types o charge:
A gravitational feld is associated with each mass. Any other mass in this feld has a gravitational feld acting on it.
negative, corresponding to a surplus o electrons in the object
positive, corresponding to a defcit o electrons in the object.
Nature of science Action-at-a-distance The action-at-a-distance envisaged here is imagined to act simultaneously even though it does not. Topic 1 2 teaches you that contemporary models o physics explain interactions using the concept o exchange particles that have a fnite ( and calculable) lietime.
Gravity felds
When the two objects have the same sign o charge, then the orce between them is repulsive.
Gravity is always an attractive orce. Repulsion between masses is never observed.
When the objects have opposite signs, then the orce between them is attractive. All electric charges give rise to an electric feld. An electrostatic orce acts on a charge that is in the feld o another charge.
Field strength Electric felds
Gravity felds
The defnition o feld strength is similar in both types o feld. It arises rom a thought experiment involving the measurement o orce acting on a test object. Near the surace o the Earth, the gravitational feld strength is 9.8 N kg- 1 and, on a clear day, the electric feld strength is about 100 N C- 1 . Both felds point downwards. In both gravity and electrostatics there is a problem with carrying out the measurement o feld strength practically. The presence o a test object will distort and alter the feld in which it is placed since the test object carries its own feld.
392
orce acting on positive test charge electric feld strength = ___________________________________________________ magnitude o test charge
orce acting on test mass gravitational feld strength = _____________________________________ magnitude o test mass
The direction o the feld is the same as the direction o the orce acting on a positive charge. We have to pay particular attention to direction in electrostatics because o the presence o two signs o charge.
Gravitational orce is always attractive. This means that the direction o feld strength is always towards the mass that gives rise to the feld.
The unit o electric feld strength is N C1 .
The unit o gravitational feld strength is N kg1 .
10.1 DE SCRI BI N G FI ELD S
Energy ideas so ar Electric felds
Potential diference
Gravity felds
Potential energy
In Topic 5 electric potential dierence In Topic 2 gravitational potential energy was used in electrostatics and current was used as the measure o energy electricity as a measure o energy transer. transer when a mass m is moved in a gravitational feld. W V= _ Q EP = mgh V is the electric potential dierence and W is the work done on a positive test charge o size Q.
Where EP is the change in gravitational potential energy, m is the mass, g is the gravitational feld strength, and h is the change in vertical height. EP is measured in joules.
V is measured in volts, which is equivalent to joule coulomb 1 .
This equation applies when g is eectively constant over the height change being considered. Remember that g has to be the local value o the feld strength near the Earths surace 9.8 N kg1 is the value to use, but urther out a smaller value would be required.
The electric potential dierence reers to work done per unit charge, gravitational potential energy reers to work done = ( force (weight) distance moved) .
Now read on ... The recap o earlier work at the start o this topic should have reminded you o the similarities and dierences between gravitational and electric felds. The dierences include:
Gravity is most noticeable when masses are very large the size o planets, stars, and galaxies.
Electric orces largely control the behaviour o atoms and molecules but, because there is usually a complete balance between numbers o positive and negative charges, the eects are not easily noticed over large distances.
Gravitational orces on the other hand are perceived to act over astronomical distances.
D espite these dierences, the two felds share an approach that allows the study o one to inorm the other. S tudying the two types o feld in parallel will help you understand the links. I you need to concentrate on either gravity or electric felds, look at the relevant background colour or line in the margin.
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Field lines Field lines help us to visualize the shapes o electric felds that arise rom static charges. They also aid the study o the magnetic felds associated with moving charges. Magnetic felds and their consequences are discussed urther in Topic 1 1 . I an obj ect in a feld has the relevant feld property ( mass or gravity, charge or electricity, etc.) , then it will be inuenced by the feld. In a Topic 5 Investigate!, small particles ( semolina grains) were suspended in a uid. This experiment gives an impression o the feld shape between charged parallel plates and other arrangements o charge. However the experiments are tricky to carry out and give no quantitative data. Here is another qualitative experiment that gives more insight into the behaviour o electrostatic felds.
Investigate! Field between parallel plates
A small piece o oil that has been charged can be used to detect the presence o an electric feld.
The detector is made rom a rod o insulator a plastic ruler or strip o polythene are ideal. Attached to the rod is a small strip o oil: thin aluminium or gold oil or D utch metal are suitable. The dimensions o the oil need to be about 4 cm 1 cm and the oil can be attached to the rod using adhesive tape.
Touch the oil briey to one o the plates. This will charge the oil. You should now see the oil bend away rom the plate it touched.
The angle o bend in the oil indicates the strength o the electric feld. E xplore the space between the plates and outside them too. Notice where the feld starts to become weaker as the detector moves outside the plate region. D oes the orce indicated by the detector vary inside the plate region or is it constant?
394
Set up two vertical parallel metal plates connected to the terminals o a power supply that can deliver about 1 kV to the plates. Take care when carrying out this experiment ( use the protective resistor in series with the supply i necessary) . B egin with the plates separated by a distance about one- third o the length o their smaller side.
Turn o the supply and change the spacing between the plates. D oes having a larger
separation produce a larger or a smaller feld?
C hange the pd between the plates. D oes this aect the strength o the feld?
An alternative way to carry out the experiment is to use a candle ame in the space between the plates. What do you notice about the shape o the ame when the feld is turned on? C an you explain this in terms o the charged ions in the ame?
The oil detector itsel can also be used to explore the feld around a charged metal sphere such as the dome o a van der Graa generator.
oil detector -
+ + + + charged plate ++ + + + + + +
- + high voltage power supply
Figure 1 Electric feld detector.
10.1 DE SCRI BI N G FI ELD S
E xperiments such as this suggest that the feld between two parallel plates:
is uniorm in the region between the plates
becomes weaker at the edges ( these are known as edge effects) as the feld changes rom the between- the- plates value to the outside- the- plates value ( oten zero) . You should be able to use the properties o the feld lines rom Topic 5 to explain why there can be no abrupt change in feld strength.
+
+
+
+
+
+
+
-
-
-
-
-
-
-
Figure 2 Electric feld line pattern or parallel plates.
A close study o the feld shows feld lines like those in fgure 2 . At the edge, the feld lines curve outwards as the feld gradually weakens rom the large value between the plates to the much weaker feld well away rom them. For the purposes o this course, you should assume that this curving begins at the end o the plate ( although in reality it begins a little way in as you may have seen in your experiment with the feld detector) . Try to predict the way in which the shape o the feld lines might change i a small conductive sphere is introduced in the middle o the space between the two plates. Look back at some o the feld pattern results rom Topic 5 and decide how the feld detector used here would respond to the felds there. S ometimes you will see or hear the feld lines called lines o orce. It is easy to see why. The feld detector shows that orces act on charges in a feld. The lines o orce originate at positive charge ( by defnition) and end at negative charge as indicated by the arrow attached to the line. I the line is curved, the tangent at a given point gives the direction o the electric orce on a positive charge. The density o the lines ( how close they are) shows the strength o the orce.
Linking feld lines with electric potential The Investigate! indicates that the strength o the electric feld depends on the potential dierence between them. Another experiment will allow you to investigate this in more detail.
Investigate! Measuring potentials in two dimensions
This experiment gives a valuable insight into the way potential varies between two charged parallel plates.
You need a sheet o a paper that has a uniorm graphite coating on one side, one type is called Teledeltos paper. In addition, you will need a 6 V power supply ( domestic batteries are suitable) , two strips o copper oil, and two bulldog clips. Finally, you will need a high- resistance voltmeter ( a digital meter or an oscilloscope will be suitable) and connecting leads.
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F I E L D S ( AH L )
C onnect the circuit shown in the diagram. Take particular care with the connections between the copper strips and the paper. O ne way to improve these is to paint the connection between copper and paper using a liquid consisting o a colloidal suspension o graphite in water.
Press the voltage probe onto the paper. There should be a potential dierence between the lead and the 0 V strip.
C hoose a suitable value or the voltage ( say 3 V) and explore the region between the copper strips. Mark a number o points where the voltage is 3 V and draw on the paper j oining the points up. Repeat or other values o voltage. Is there a consistent pattern to a line that represents a particular voltage?
Try other confgurations o plates. O ne important arrangement is a point charge, which can be simulated with a single point at 6 V and a circle o copper oil outside it at 0 V. O ne way to create the point is to use a sewing needle or drawing pin ( thumb tack) . You will need the colloidal graphite to make a good connection between the point and the paper.
An alternative way to carry out the experiment, this time in threedimensions, is to use the same circuit with a tank o copper( II) sulate solution and copper plates. voltage probe at end o fying lead
0V 0V +
-1.5 V
+
+
+
+6.0 V +4.5 V + +
- 3.0 V -4.5 V
6V cells
voltmeter
+3.0 V - -6.0 V
-
-
-
0V
+1.5 V
Figure 3 Equipotentials in two-dimensions.
Figure 4 Lines o equal potential.
Tip Think about why the feld line and the equipotential must be at 90 to each other. As a hint, remember the defnition o work done in terms o orce and distance moved rom Topic 2.
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The lines o equal potential dierence between two parallel plates drawn in this experiment resemble those shown in fgure 4. These lines are called equip otentials because they represent points on the two-dimensional paper where the voltage is always the same. When a charge moves rom one point to another along an equipotential line, no work is done. Figure 4 also shows the electric feld lines between the plates. You will see a simple relationship between the feld lines and the equipotentials. The angle between them is always 90. I you know either the shape o the equipotential lines or the shape o the feld lines then you can deduce the shape o the other. The same applies to the relationship between rays and waveronts in optics.
10.1 DE SCRI BI N G FI ELD S
You will have noticed that the word dierence has disappeared and that we are using the plain word potential. As long as we always reer to some agreed reerence point ( in this case the 0 V copper strip on the paper) , the absolute values o potential measured on the voltmeter are identical to the values o potential dierence measured across the terminals o the meter. To illustrate this, the parallel- plate equipotentials in fgure 4 are re- labelled on dierent sides o the diagram taking a dierent point in the circuit to be zero. You will see that the potential dierences are unchanged, but the change o reerence position shits all the potential values by a fxed amount. This defnition o a reerence position or potential will be considered in the next sub- topic.
Tip Calculate the change in potential going rom -4.5 V to -3.0 V as well as rom +1.5 V to +3.0 V. In both cases, we go rom a low potential point to a higher potential point so the potential dierence should be positive and the same in both cases (+1.5 V) . Remember dierences are calculated rom fnal state minus initial state.
+ -
+ -
+
+
-
+
+
-
-
+ +
-
-
+ -
Figure 5 Equipotentials in three dimensions.
The conducting-paper experiment is two-dimensional. The oil detector allowed you to explore a three-dimensional feld. The equipotentials between the charged plates in three dimensions are in the orm o sheets parallel to the plates themselves always at 90 to the feld lines. So another consequence o extending to three dimensions is that it is possible to have equipotential suraces and even volumes. An example o this is a solid conductor: i there is a potential dierence between any part o the conductor and another, then charge will ow until the potential dierence has become zero. All parts o the conductor must thereore be at the same potential as each other the whole conductor is an equipotential volume. C onsequently the feld lines must emerge rom the volume at 90 whatever shape it is. In summary, equipotential suraces or volumes:
link points having the same potential
are regions where charges can move without work being done on or by the charge
are cut by feld lines at 90
should be reerred to a zero o potential
do not have direction (potential, like any energy, is a scalar quantity)
can never cross or meet another equipotential that has a dierent value.
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Equipotentials and feld lines in other situations Figure 6 shows the equipotentials and electric feld lines or a number o common situations.
+
+
-
-
-
(b)
(a)
-80 V outer conductor -90 V -100 V equipotential
equipotentials
+ + + + + + +
inner conductor (c)
-
(d)
(e)
Figure 6 Equipotentials around charge arrangements.
(a) Field due to a single point charge Earlier we looked at the feld due to a single point charge. In the case o a positive charge, the feld lines radiate out rom the point and the feld in this case is called a radial feld. For a negative charge the only change is that the arrows now point inwards to the charge. What are the equipotentials or this case? The key lies in the 90 relationship between feld line and equipotential. Figure 6( a) shows the arrangement o equipotentials around a single point charge. They are a series o concentric shells centred on the point charge. This arrangement is shown as a two-dimensional arrangement, but it is important always to think in three dimensions. Two- dimensional diagrams are however easier to draw and perectly acceptable in an examination.
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10.1 DE SCRI BI N G FI ELD S
(b) Field due to two point charges of the same and opposite sign These arrangements bear some similarities to the magnetic feld patterns in the space between two bar magnets; they demonstrate the use o the rules or the electric feld lines.
(c) Field due to a charged sphere A hollow or solid conducting sphere is at a single potential.The feld outside the sphere is identical to that o a point charge o the same magnitude placed at the centre o the sphere. This time however, there are no feld lines inside the sphere.
(d) Field due to a co-axial conductor C o- axial conductors are commonly used in the leads that connect a domestic satellite dish to a TV. There is a central conductor with an earthed cylinder outside it, separated and spaced by an insulator. The symmetry o the arrangement is dierent rom that o a sphere and so the symmetry o feld pattern changes too.
(e) Field between a point charge and a charged plate The key to this diagram is to remember that the feld lines are radial to the point charge when very close to it and the lines must be at 90 to the surace o the plate.
Field and equipotential in gravitation We did not use the concept o gravitational feld lines in Topic 6. However, feld lines can be used to describe gravitational felds. The concepts o feld lines and equipotentials transer over to gravity without difculty. Although there is no easy experiment that can make the gravitational feld lines visible ( unlike magnetic or electric feld lines) , the direction can be determined simply by hanging a weight on a piece o string! The direction o the string gives the local direction o the feld pointing towards the centre o the planet. The arrangement o gravitational feld lines that surround a point mass or a spherical planet is similar to the pattern or electric feld lines around a negative point charge. Again, the feld is radial or both cases with the feld lines directed towards the mass because gravitational orce can only be attractive. All that has been written about equipotentials in electric felds also applies to gravitational equipotentials. Imagine a mass positioned at a vertical height above the E arths surace, say 1 0 m. I the mass moves to another location which is also 1 0 m above the surace, then no work has to be done ( other than against the non- conservative orces o riction) to move it. The mass begins and ends its journey on the same equipotential and no work has been transerred. These two points lie on the surace o a sphere whose centre is at the centre o the planet.
Figure 7 Field lines and equipotentials around a planet.
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F I E L D S ( AH L ) There is another way to interpret gravitational equipotentials and feld lines. It is possible that you regularly use equipotentials without even realizing it!
line of steepest descent
Figure 8 Contour map (contour lines have been removed for clarity) .
The contour lines on this map are equipotentials. I you walk along a contour line then your vertical height does not change, and you will neither gain nor lose gravitational potential energy. Mountain walkers have to be experts in using contour maps, not j ust to know where they are, but to choose where it is sae to go. Looking across the contours where they are closest together gives the line o steepest descent. This gives the direction o the gravitational feld down the slope at that point. Just as or electric felds, a gravitational feld is uniorm and directed downwards at 90 normal to the surace when close to the surace o the obj ect.
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10.1 DE SCRI BI N G FI ELD S
Worked examples 1
An electrostatic device has electric feld lines as shown in the diagram.
2
The diagram shows electric equipotential lines or an electric feld. The values o the equipotential are shown. Explain where the electric feld strength has its greatest magnitude.
10 V
20 V
D raw the diagram and add to it fve possible equipotential lines or the arrangement.
30 V
Solution
40 V
The equipotential lines must always meet the feld lines at 90. O ne possible line is a vertical line in the centre o the diagram. O n each side o this the lines must bend to meet the lines appropriately.
50 V
Solution The work done in moving between equipotentials is the same between each equipotential. The work done is equal to force distance. So the orce on a test charge is greatest where the distance between lines is least. This is in the region around the base o the 2 0 V equipotential. Providing that potential change is the same between neighbouring equipotential lines, then the closer the equipotential lines are to each other, the stronger the electric feld strength will be.
Potential at a point Gravitational potential S o ar the terms potential and potential difference have been used almost interchangeably by saying that potential needs a reerence point. B ut this needs clarifcation. We need to defne potential careully or both gravitation and electrostatics and, in particular, we need to decide on the reerence point we will use or our zero. It is easier to begin with gravity: Gravitational p otential dierence is the work done in moving a unit mass (1 kg in our unit system) between two p oints. To calculate the change in gravitational potential o an obj ect when it leaves one point and arrives at another, we need to know both the
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F I E L D S ( AH L ) gravitational energy needed to achieve the move and also the mass o the obj ect. Then the change in the gravitational potential o the w o rk do ne to m o ve th e o b j e ct arrival point relative to the departure point is _____________________ . m ass o the o b j e ct
In symbols W Vg = _ m where Vg is the change in gravitational potential, W is the work done, and m is the mass o the obj ect. The unit o gravitational potential is j oule kilogram 1 ( J kg 1 ) . You can expect to use MJ kg 1 or changes in potentials on a planetary scale. What makes a good zero reerence or potential? O ne obvious reerence point could be the surace o the oceans. When travel is based only on E arth this is a reasonable reerence to choose. We reer to heights as being above or below sea level so that Mount E verest is 8848 m asl ( above sea level) . D espite being an appropriate local zero o potential or E arth- based activities, sea level is not good enough as soon as astronauts leave the E arth. It makes no sense to reer planet Mars to the E arth sea level we need another measure. The reerence point that makes most sense although it might seem implausible on frst meeting is infnity. Infnity is not a real place, it is in our imagination, but that does not prevent it rom having some interesting and useul properties! Gravitational p otential is defned to be zero at infnity. Newtons law o gravitation tells us that the orce o attraction F between two point masses is inversely proportional to the square o their separation r 1 F _2 r The defnition o gravitational potential reects the act that there is no interaction between two masses when they are separated by this imaginary infnite distance. Newtons law predicts that i the separation between two masses increases by one thousand times, then the attractive orce goes down by a actor o one million. S o i r is very large, then the orce becomes very small indeed. At infnity we know that the orce is zero even though we cannot go there to check. Now we can study what happens as the two masses begin to approach each other rom this infnite separation. Imagine that you are one o the masses and that the second mass ( originally at infnite distance) is moved a little closer to you. This mass is now gravitationally attracted to you ( and vice versa) . Work would now need to be done on the system to push the mass away rom you against the orce o attraction back to its original infnitely distant starting point. Energy has to be added to the system to return the mass to infnity. O n arrival at infnity the mass will have returned to zero potential
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10.1 DE SCRI BI N G FI ELD S
( because it is at infnity) and thereore whenever the masses are closer than infnity, the system o the two masses has a negative potential. The more closely they approach each other, the more negative the stored energy becomes because we have to put increasing amounts o energy back into the system to return the masses to an infnite separation ( fgure 9) . Gravitational p otential at a p oint is defned to be equal to the work done p er unit mass (kilogram) in moving a test mass rom infnity to the p oint in question. 3m
+30J kg-1
2m
+20J kg-1
1m
+10J kg-1
Tip At infnity, the gravitational potential is maximum and equal to zero, by the convention. At separations less than this the masses are bound to each other gravitationally and so have negative gravitational potential.
0J kg-1
-6.3 10 7 J kg-1 -4 -1 10 7 J kg-1 0
10 7 J
-5 kg-1
10 7 J
kg-1 8.0 Mm
10 Mm
40 Mm
Figure 9 School lab and the Earth.
O n the Earth, sea level corresponds to a gravitational potential o 6.2 5 1 0 7 J kg 1 . ( We will calculate this in the next section.) Near the surace, raising one kilogram through one metre rom sea level requires 9.8 ( N) 1 ( m) o energy, in other words, about 1 0 J o energy or each kilogram raised. S o the potential becomes more positive by 1 0 J kg 1 or each metre raised. You could label the wall o your school laboratory in potential values! C omparing the gravitational potential at sea level with the gravitational potential at the top o Mount Everest shows that the potential at the top o the mountain is 86 71 0 J kg 1 greater than at sea level. This makes the gravitational potential at the top o Everest equal to 6.2 4 1 0 7 J kg 1 . Not apparently a large change, but then infnity is a good deal urther away! I we know the change in potential, then we can use it to calculate the total gravitational potential energy required or a change in height. For a mountaineer and kit o total mass 1 2 0 kg, the gravitational potential energy required to scale Mount Everest is about 1 0 7 J ( that is 86 71 0 J kg 1 1 2 0 kg) starting at sea level.
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F I E L D S ( AH L ) Returning to the idea o our contour maps, we could re-label all the contours changing the values rom heights in metres to potentials in joule kg 1 . The shape would be the same. Instead o knowing the dierence in height, we now know the amount o energy that will be used or released when each kilogram o the mass moves rom one place on the map to another.
Electrical potential The situation needs to be modifed or electric potential. Again, the zero o potential is defned to be at infnity and the change in electric potential Ve is defned to be
Tip These charges are not bound to each other. They have a high electric potential energy. When returned to infnity they will have zero electric potential energy and this value is a minimum.
work done in moving charge between two points W Ve = ____ = _ Q magnitude of charge The energy stored in the system per unit charge is what we call the potential o the system. I we have two charges o opposite sign then they attract each other, and the situation is exactly the same as in gravitation. In other words, when the charges are some fnite distance apart, energy is required to move them to infnite separation. As the charges separate, the electric potential will gradually climb to zero. The situation changes when the two charges have the same sign: now both charges repel each other.
Tip Dont orget there are three separate points in this defnition:
404
work done in moving unit charge
the test charge is positive
the test charge moves rom infnity to the point
Imagine two negative charges held at some fnite distance apart. Energy was stored in the system when the charges were brought together rom infnity. I they are released, the two charges will y apart ( to infnity where the repulsive orce becomes zero) . The system does work on the charges. There must have been a positive amount o energy stored in the system beore the charges were allowed to separate. The potential is still defned to be zero at infnity, but this time as one charge is moved rom infnity towards the other charge, the overall value o the potential rises above the zero level to become positive. Work has been done on the charge. The electric p otential at a p oint is the work done in bringing a unit p ositive test charge rom infnity to the point.
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10.2 Fields at work Understanding Potential and potential energy
Applications and skills Determining the potential energy o a point
Potential gradient Potential dierence
Escape speed
Orbital motion, orbital speed, and orbital energy
Forces and inverse-square law behaviour
Nature of science Electric charge is invisible to our senses, yet we accept its existence. We are aware o it through the orce laws that it obeys and via the ield properties that we assign to it. In the same way, the motion o planets and satellites on a larger scale also requires that scientists develop ways to visualize and then report their theories to non-scientists.
mass and the potential energy o a point charge Solving problems involving potential energy Determining the potential inside a charged sphere Solving problems involving the speed required or an object to go into orbit around a planet and or an object to escape the gravitational feld o a planet Solving problems involving orbital energy o charged particles in circular orbital motion and masses in circular orbital motion Solving problems involving orces on charges and masses in radial and uniorm felds
Equations potential energy
equations:
kQ Ve = _____ r V
feld strength
equations: relation between
feld strength and potential:
GM Vg = ______ r Vg
e E = _______ r
g = _______ r
Ep = qVe
Ep = mVg
kQq = _______ r q q
GMm = ________ r m m
1 2 1 2 FE = k ________ FG = G __________ r2 r2
orce laws:
___
2GM escape speed: ve s c = ________ r __ ______ GM orbital speed: v = o rb it
r
Introduction In the second part o this topic we look at the underlying mathematics o felds and how it applies to gravitational and electric felds. Parts o this chapter devoted to electric felds have a blue background or margin line; the sections dealing with gravity have a green background or margin line. You can concentrate on one type o feld by sticking to one colour.
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Forces and inverse-square law behaviour Electric felds
Gravity felds
Both felds obey an inverse-square law in which the orce between two objects is inversely proportional to the distance between them squared. In a vacuum: kq 1 q 2 FE = + __ r2 where FE is the orce between two point charges q 1 and q 2 and r is the distance between them. This is known as Coulombs law. The constant k in the equation is _1_ 4 0 where 0 is known as the permittivity o a vacuum or the permittivity o ree space.
Gm 1 m 2 FG = __ r2 where FG is the orce between two point masses m 1 and m 2 and r is the distance between them. This is known as Newtons law o gravitation. The constant in the equation is G known as the universal gravitational constant. The value o G is a universal constant.
I the feld is in a medium other than a vacuum then the 0 in the equation is replaced by , the permittivity o the medium. The sign in the equation is positive. The sign o the overall result o a calculation indicates the direction in which the orce acts. Negative indicates attraction between the charges; positive indicates repulsion.
Nature of science Inverse-square laws B oth felds obey the inverse-square law in which 1 the orce depends on _______ . Sub-topic 4.4 distance showed how the inverse-square law also arises in the context o radiation rom the geometry o space. 2
1 The value o n in __ has been tested a number o r times since the inverse- square law behaviour o electric and gravitational felds was suggested, and n is known to be very close to 2 ( or C oulombs law, within 1 0 - 1 6 o 2 ) n
Tip Charges q1 and q2 can be positive or negative. A positive orce FE indicates a repulsive orce. A negative orce FE indicates an attractive orce. The masses m 1 and m 2 are always positive, and so in this notation a positive orce FG always indicates an attractive orce.
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One consequence o a orce law being inversesquare is that the orce between the charges or masses becomes weaker as the distance increases, but never becomes zero at any real distance. We said earlier that the orce was zero when the objects were separated by an infnite distance. There is no actual infnity point in space, but there is one in our imagination. We use infnity as a useul concept or our energy ideas. In one sense, infnity can be pushed urther away as the instruments that measure feld strength become more precise!
We saw in Topic 5 that, close to any surace, the feld is uniorm. The argument was that, locally, the surace appears at and that there is cancellation o the components o feld parallel to the surace due to each charge. This leaves only components perpendicular to the surace to contribute to the feld. The electric felds between parallel plates or the gravitational or electric felds close to a curved surace are uniorm and are useul in developing ideas about felds.
1 0 . 2 F I E L D S AT W O R K
Investigate! Charged parallel plates
This experiment illustrates how the feld strength and other actors are connected or parallel plates and will help you to understand these actors.
Repeat the measurements o distance apart and charge stored. D ont orget to switch o the power supply beore you measure the separation o the plates.
S et up a pair o parallel plates, a 5 kV power supply, a well-insulated ying lead, and a coulombmeter ( a meter that will measure charge directly in coulombs) as shown in the circuit. Later you will need to replace the parallel plates with ones that have dierent areas. You will also need to record the distance between the plates.
C arry out another experiment where the power supply em and the plate separation are unchanged but plates o dierent areas are used. Record the plate areas.
C arry out another experiment where the em V o the power supply is changed but the plate area and distance are not changed.
Plot your results as Q versus V, Q versus A, and Q versus d. where Q is the charge on the plates, V is the potential o the plate, A is the area o the plates, and d is the separation o the plates.
S et and record a suitable voltage V on the power supply. Your teacher will tell you the value to use. Take great care at all times with the high voltages used in this experiment.
Measure and record the distance d between the parallel plates.
Touch the ying lead to the right- hand plate briey and then remove it rom the plate. This charges the plates.
d a rea A cha rge Q
Zero the coulombmeter and then touch its probe to the right-hand plate that you j ust charged. Record the reading Q on the meter. You may wish to repeat this measurement as a check. Change the distance between the plates without changing the setting on the power supply.
-
+ + + + + + + + +
0V
pla te
cou lom b m e ter
V f y in g lea d
0V
-
+
power su pply
Figure 1 How the charge stored on parallel plates varies. -
The results o the experiment show that
+ V
QV
QA 1 Q_ . d These results can be expressed as
-
+
-
+ F
+Q
VA Q_ d The constant in the equation turns out to be 0 ; in principle careul measurements in this experiment could provide this answer too.
-
+
-
+
-
+
-
+
So VA Q = 0 _ d where 0 = 8.85 4 1 0 - 1 2 m - 3 kg - 1 s 4 A 2 .
d
Figure 2 Field strength between parallel plates.
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Electric feld strength and potential gradient The uniorm electric feld between two charged plates provides us with another way to think about electric feld strength.
In Topic 5 (page 1 77) we mentioned that electric feld strength could be measured as the change in potential divided by the change in distance (we call this the potential gradient) . We can now show that this result is correct. In fgure 2, a positive charge that has a size Q is in a feld between two charged plates. This feld has a strength E and the orce F that acts on the charge is thereore (rom the defnition o electric feld strength given in Sub-topic 5.1 ) F = EQ It is easy to determine the work done by the feld on the charge when the charge is moved at constant speed because the feld is uniorm and the orce is constant. As usual, the work done is force distance moved and the lines o orce are in the same direction as the distance moved so there is no component o distance to worry about Work done = EQ x where x is the distance moved. When the charge goes rom one plate to the other the distance moved is d. The work done on the charge is EQd. The potential dierence V between the plates is (see Topic 5.1 ) work done in moving a charge V = ___ magnitude of charge So EQd work done V= _ =_ Q Q leading to V E= _ d This shows that electric feld strength can be written and calculated in two ways:
force actin g on ch arge ______________ , this gives the unit N C 1 m agn itu de of ch arge
poten tial ch an ge ___________ , this gives the unit V m1 . distan ce m oved
So as well as two equations, there are two possible units or electric feld strength and both are correct. Writing the equations in ull or the uniorm feld between parallel plates V _ E=_ = F Q d
In practice, it is easier to measure a potential dierence with a voltmeter in the laboratory than to measure a orce. This is why an electric feld strength is commonly expressed in volt metre -1 .
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Worked examples 1
A pair o parallel plates with a potential dierence between them o 5 .0 kV are separated by 1 2 0 mm. C alculate:
a) C alculate the potential dierence required to produce a orce o 3 .6 1 0 1 4 N on the droplet.
a) the electric feld strength between the plates
b) The plates are now moved closer to each other with no change to the potential dierence. The orce on the droplet changes to 1 .4 1 0 1 3 N. C alculate the new separation o the plates.
b) the electric orce acting on a doubly ionized oxygen ion between the plates.
Solution V 5 000 a) E = __ = ____ = 4. 2 1 0 4 V m 1 0.1 2 d
Solution
F = QE = 3.2 1 0 1 9 4.2 1 0 4 = 1 .34 1 0 1 4 N
3 .6 1 0 1 4 F a) E = _ = 3 .2 1 0 4 N C 1 = __ Q 1 1 .2 1 0 1 9 V = Ed = 3 .2 1 0 4 0.080 = 2 600 V
The orce is directed towards the negative plate.
b) The orce changes by a actor o
b) The charge o the ion is 2 e = + 3 .2 1 0 1 9 C .
2
A pair o parallel plates are separated by 80 mm. A droplet with a charge o 1 1 .2 1 0 1 9 C is in the feld.
1 . 4 1 0 1 3 ________ = 3 .9 3 . 6 1 0 1 4
The separation decreases by this actor, to 80 ___ = 2 1 mm. 3.9
Electric feld strength and surace charge density The expression VA Q = 0 _ d rearranges to Q V _ = 0 _ A d Q __ is , the surace charge density on the plate. We say that or the A
uniorm feld between the plates, the electric feld strength E can be V written as __ . d S o between the plates E=_ 0 Remember, this equation is or the feld between two parallel plates. Each plate contributes hal the feld, so the electric feld E close to the surace o any conductor is equal to E=_ 2 0
Tip
where is the charge per unit area on the surace and 0 has its usual meaning. This is dimensionally correct because the units o are C m 2 and the C m _______ 1 units o 0 are C 2 N 1 m 2 , __ is C N m which simplifes to N C , the 2
0
2
1
2
same units as or E.
Generally, the term gradient refers to a change per unit of distance. The term rate refers to a change per unit to time. (Sometimes gradient is used a word for slope)
You will be given this equation in an examination i you need to use it.
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F I E L D S ( AH L ) 0V
(a)
2V
4V
6V
+
-
+ +
-
Figure 3(a) is a reminder o the feld pattern between charged parallel plates.
+
(b)
Graphical interpretations o electric feld strength and potential
8 V 10 V 12 V
The equipotentials are equally spaced. Figure 3 ( b) shows the data rom measurements o the potential at a series o points between the plates plotted as a graph o potential against distance rom the zero potential plate. It is a straight line because the electric feld is uniorm. A graph o electric feld strength against distance rom the 0 V plate is a straight line parallel to the distance axis ( Figure 3 ( c) ) .
+ 12
potential/V
10 8 6
ch an ge in V The gradient o the potentialdistance graph is ________ and because it is ch an ge in x V __ a straight line, this is equal to d , in other words, E.
4 2
This leads to:
0 0
1 2 3 4 5 distance rom 0 V plate/cm
E, electric feld strength = p otential gradient =
6
V _ e
x
( , as usual, stands or change in) .
electric feld strength/NC - 1
(c)
distance rom 0 V plate/cm 0 1 2 3 4 5
6
0 -100 -200 -300
Figure 3 Electric feld strength and potential.
(a) orce
F work done = area under graph = F r r
distance
electric orce
(b) work done = area under graph
V The equation E = ___ can be re- written as E x = V, and in this x orm has a graphical interpretation.
electric feld strength
work done per unit charge = potential change between r1 & r2 potential at r3 A r1
B
to
r2 r3 distance
Figure 4 Work done and work done per unit charge.
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You may be wondering whether the argument changes i the moving charge is an electron. I an electron moves rom 0 V towards the 1 2 V plate (a position o higher potential) its potential energy is reduced because the potential energy change Ep is given by eV as usual but here e is negative and V is positive. The electron will accelerate in this direction i ree to do so gaining kinetic energy rom the feld at the expense o the electrical potential energy. V In numerical terms (fgures 3( b) and ( c) ) , the gradient ___ is + 2 00 V m - 1 x -1 and so the electric feld strength is - 2 00 N C .
distance (c)
The minus sign that appears in the equation needs an explanation. It means that the direction o the vector electric feld is always opposite to the variation o the potential ( Ve ) o a positive charge. S o, travel in the opposite direction to that o the feld means moving to a position o higher potential and thus a positive gain in potential. In Figure 3 ( a) the motion o the positive charge is to the let, Ve is negative ( fnal state minus initial state: 0 V - 1 2 V) ; so, according to the equation, E will be positive and this tells us that the motion is in the direction o the feld. In other words, going upstream in the feld ( against the feld) means going to higher potential so a gain in potential ( positive change in Ve ) . Going downstream, Ve is negative and E is positive ( product o two minus signs) indicating the motion is in the direction o feld.
A graph o orce against distance or a constant orce is shown in fgure 4( a) . The work done when this orce moves an obj ect through a distance r is equal to the area under the graph. In symbols F r = W When an electric orce varies with distance then the work done is still the area under the graph ( Figure 4( b) ) . Electric feld strength ( Figure 4( c) ) is the orce per unit charge, and so the area under a graph o electric feld strength against distance is equal to the work done per unit charge in other words the change in potential.
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kQ E=_ r2 Q here is the charge creating the feld ( not a charge that would react to the feld) and show that Ve is given by Q Ve = _ or 4 0 r
kQ _ r 1 using the constant k = _. 4 0 This is the expression or the potential V at a distance r rom a point charge. It predicts that the closer we are to a (positive) charge, the greater (more positive) is the potential. C onversely, the closer we are to a negative charge, the more negative is the potential. These predictions correspond to the conclusions we reached when we considered fgure 3.
1 Notice the relative shapes of the curves: ____ r ___ 1 is more sharply curving than the r of the potential. Notice also the links between the two graphs. 2
(a) electric feld
The equation Ve E = _ r can be written in calculus orm as dVe E = _ dr This new orm can be used to take the equation or the feld strength
Tip
V = area = E r (r small) E r
distance
(b) electric potential
Figure 4( c) shows the variation o electric feld strength against distance or a single point positive charge this is an inverse- square variation. Two areas are shown on this graph, one area ( A) that shows the potential change between r1 and r2 , the other ( B ) shows the potential at r3 in this case the area included goes all the way to infnity.
feld E = V
r
V (r small) r
distance
Figure 5 Electric feld strength and potential against distance or a positive point charge.
Potential is the work done in moving a positive unit charge ( rom infnity) to a particular point, so the work required to move a charge o size q rom infnity to the point will be: Ve q. This is the potential energy that charge q possesses due to its position in the feld that is giving rise to the potential. I a charge q is in a feld that arises rom another single point charge Q then its potential energy Ep = qVe , or kqQ qQ _ Ep = _ r = 4 r 0 Graphs o feld strength and potential against distance or the feld due to a single positive point charge are shown in fgure 5 .
Worked examples 1
Two parallel metal plates have a potential dierence between them o 1 5 00 V and are separated by 0.2 5 m. C alculate the electric feld strength between the plates.
Solution
potential dierence Electric feld strength = __ separation 1 500 = _ = 6 kV m1 (or kN C 1 ) 0.25
2
A point charge has a magnitude o 0.48 nC . C alculate the potential 1 .5 m rom this charge.
Solution Q Ve = _ 4 0 r S ubstituting 0. 48 1 0 9 Ve = __ = 2 .9 V 4 0 1 .5
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Nature of science Linking electrostatics and gravity The conclusion reached here arises because o the inversesquare law. It also applies to a hollow planet. We can show, in a similar way, that the gravitational feld strength is zero. This helps us later to show how the gravitational feld strength varies inside a solid planet o uniorm density.
Potential inside a hollow conducting charged sphere S o ar we have considered point charges and the parallelplate arrangement in detail. Now we turn to another important charge confguration: the hollow conducting charged sphere. O utside a charged conducting sphere, the feld is indistinguishable rom that o the point charge. The argument is that because the feld lines leave the surace o the sphere at 90 these lines must be radial, so an observer outside the sphere cannot distinguish between a charged sphere and a point charge with the same magnitude o charge placed at the centre o the sphere.
The feld inside the sphere B ecause the sphere is a conductor, all the surplus charge must reside on the outside o the sphere. This ollows because:
the charges will move until they are as ar apart as possible
the charges will move until they are all in equilibrium ( which in practice means that they have to be equidistant on the surace) .
To fnd the feld strength inside the conductor we need to compute the orce that acts on a positive test charge Q placed anywhere inside the conductor. We choose a point inside the conductor at random and place our test charge there. This is shown in fgure 6. The next step is to fnd the orce acting on this test charge. To make this easier we consider two cones that meet at the test charge. These cones have the same solid angle at the apex. O ne o these cones is large and the other is small. The test charge is close to the conductor surace on the side o the small cone but not so close on the side o the large cone.
surface of charge conducting sphere 2xl rl
+Q rs
cm -2 , surface charge density
Figure 6 Field inside a charged conductor.
2xs
C all the distance rom the test charge to the sphere rs or the small cone and rl or the large cone. The cone radii are xs and xl or the small and large cones respectively. The surace charge density on the cone is and is the same over the whole sphere. Thereore
For the small cone
For the large cone
Area of the end of the cone
xs 2
xl 2
Charge on the end of the cone
xs 2
xl 2
rs
rl
Distance of area from test charge Force on test charge due to area
kQxs 2 __ rs 2
2 kQx l _ _ rl 2
Directed away from the surface
Directed away from the surface
The geometry o the cones is such that xs xl = _ _ tan _ = rs rl 2
( )
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and thereore kQxs2 kQxl2 _ _ is equal to rs2 rl 2
E 1 r2
The orces acting on the test charge due to the two charge areas are equal in size and opposite in direction. Remarkably, they completely cancel out leaving no net orce due to these two areas o charge. We chose the angle o the cones arbitrarily so this result applies or any angle we could have chosen. S imilarly, we chose the position or the test charge at random. Thereore this proo must apply or any pair o cones and any test point inside the sphere. There is thereore no net orce on the test charge anywhere inside the sphere and consequently there is no electric feld either. This proo relies on the inverse- square law and the act that the area o the ends o the cones also depends on the ( distance of the test charge from the area) 2 . B ecause the electric feld is zero the potential inside the sphere must be a V is zero inside the sphere. V being constant constant since the gradient ___ r must mean that V is zero so no work is required to move a charge at constant speed inside the charged sphere. The potential inside is equal to the potential at the surace o the sphere.
r
0
V 1 r
0
r
Figure 7 Field and potential inside and outside the conductor.
S o we can now plot both the electric feld strength and the electric potential or a charged conducting sphere and these graphs are shown in fgure 7.
Nature of science Potential and potential energy The connection between feld strength and potential gradient is universal and applies to all felds based on an inverse-square law. It tells us about the undamental relationship between orce and the work the orce does and it tells us this in a way that is independent o the mass (or the charge) o the test object that is moved into the feld. Thus, feld strength is a concept that represents orce but with the mass or charge term o the test object
removed. O course, the mass or charge element that gives rise to the feld in the frst place is still important and remains in E and g. So Mass/charge independent Field strength Potential
Mass/charge dependent force p otential energy
Gravitational potential Gravitation As with electric felds Vg g = _ r where g is the gravitational feld strength, Vg is the gravitational potential, and r is distance. For the case o the feld due to a single point mass M with a point mass m a distance r rom it, the potential Vg at r is GM Vg = _ r
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F I E L D S ( AH L ) The potential energy o m at r rom mass M is
distance
(a)
GMm Ep = mVg = _ r
1 r2 gravitational feld strength distance 1 r gravitational potential (b)
These equations lead to graphs similar to those or electric felds, which are illustrated in fgure 8( a) . The orces are always attractive or gravity and we saw that this has the consequence that the potential is always negative with a maximum o zero at infnity. The graphs in fgure 8 are plotted to reect this. The graphs should be shown in three-dimensions, but are plotted in two dimensions or clarity. When thinking o the plotted surace o this 1 negative __ r potential in 3 -D , treat it as a potential well ( fgure 8( b) ) . This well will trap any particle that has insufcient kinetic energy to escape the mass. Later when we discuss how spacecrat leave the Earth you can imagine that a spaceship without enough energy can only climb up so ar. From this point the spaceship can circle around the mass ( that is orbit the mass) , but climb no higher.
Figure 8 Gravitational feld strength and potential variation with distance.
Worked examples 1
At a point X the gravitational potential is 8 J kg 1 . At a point Y the gravitational potential is 3 J kg 1 . C alculate the change in gravitational potential energy when a mass o size 4 kg moves rom X to Y.
Solution A change in potential is calculated rom fnal state minus initial state so change here is ( 3 ) ( 8) = + 5 J kg 1 . S o energy required is + 2 0 J, moving rom a lower point in the potential well to a higher point ( climbing toward infnity where the value is maximum and equal to zero) . 2
Which diagram shows the gravitational equipotential suraces due to two spheres o equal mass? A.
B.
C.
D.
Solution The spheres attract and so the gravitational feld lines j oining them resemble the pattern between two unlike charges ( this pattern is similar to that in A) . The equipotential suraces must cut these feld lines at 90 so the best ft to the appropriate pattern is C . 3
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C alculate the potential at the surace o the Earth. The radius o the Earth is 6.4 Mm and the mass o the Earth is 6.0 1 0 24 kg.
1 0 . 2 F I E L D S AT W O R K
Solution GM Vg = _ r 6.7 1 0 1 1 6.0 1 0 24 ___ Vg = 6.4 1 0 6 = 6.3 1 0 7 J kg 1
distance rom centre o Earth
rE
Potential inside a planet In a charged conducting sphere the mobile charges move to the outside o the sphere and this results in zero feld inside. This is not the same or the gravitational feld o a solid massive sphere (massive in this context means having mass not having a large mass) . This case was mentioned in Topic 6 in the context o a tunnel drilled rom the surace to the centre o the Earth. The essential physics is that i we are at some intermediate depth between surace and centre we can think o the E arth as having two parts: the solid sphere beneath our eet and the spherical shell above our head. The inner solid sphere, a miniature Earth with a smaller radius, behaves as a normal Earth but with a dierent gravitational feld strength given by G redu ced m ass of th e Earth g = ___________________ radiu s of sm all Earth
outside Earth
outside Earth
g inside Earth
Figure 9 Gravitational feld inside the Earth.
Nature of science Delivering the mail!
2
The outer spherical shell does not give rise to a gravitational feld. The argument is similar to that which we used to prove that the electric feld inside a charged conductor was zero. 4 The reduced mass of the Earth is equal to __ r 3 where is the density o 3 the Earth ( assumed constant) , so
4G r g = _ 3 where r is the reduced radius o the small E arth The gravitational feld is proportional to the distance rom the centre o the Earth until we reach the surace. This is an unexpected result. The inverse- square law actually gives rise to a linear relationship because 1 the __ behaviour o the orce cancels with the r3 variation o the mass o r the reduced Earth. 2
Leaving the Earth Picture a spacecrat stationary on its launch pad. A space launch rom Earth may put a satellite into Earth orbit. Or it may be a mission to explore a region ar away rom Earth requiring the crat to escape Earths gravity and eventually come under the inuence o other planets and stars. What minimum energy is needed to allow the crat to escape the Earth?
Orbiting In Topic 6 there are some basic ideas about circular orbits o a satellite about a planet. We showed that the linear orbital speed v o a satellite about a planet is ____
vorbit =
GM _ r
The acceleration g is always directed toward the centre o Earth and is proportional to r. So a vehicle inside a tunnel drilled right through the Earth will perorm simple harmonic motion i riction is neglected (see Topic 9) . The vehicle will stop momentarily when arriving at the antipode 84 minutes ater release. This is also the period o a satellite that goes round the Earth in low-earth orbit (technically, just skimming the surace) . Get the timing right and mail bags could be exchanged at the surace! A chordal tunnel dug between Vancouver and Montreal or between New York and Rio de Janeiro or between London and Johannesburg ( not passing through the center o the Earth) also allows the mail to be delivered. Use some o the ideas and equations in Topic 9 to assess the engineering issues with this plan!
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F I E L D S ( AH L ) and that the orbital time T or this orbit is _____
Torbit =
4 r _ GM 2 3
here M is the mass o the planet and r is the radius o the orbit.
(b)
(a)
rotation of Earth
geostationary orbit
polar orbit polar axis
Figure 10 Polar and geostationary orbits (not to scale) and the track of a geosynchronous orbit.
Worked example C alculate the orbital time or a satellite in polar orbit.
Solution The orbital time or these satellites is obtained by using the appropriate value or r in the T equation. The radius o the orbit is the orbital height (1 00 km) plus the radius o the Earth (6.4 Mm) , so r = 6.5 1 0 6 m. _______________
T=
4 (6.5 1 0 ) __________________ 6.67 1 0 6.0 1 0 2
The p olar orbit is used or satellites close to the E arths surace ( close in this context means about 1 00 km above the surace) . The orbit is called polar because the satellites in this orbit are oten put into orbit over the poles. Imagine you are viewing the E arth and the orbiting satellites rom some way away in space. You will see the satellite orbiting in one plane ( intersecting the centre o the E arth) with E arth rotating beneath it. In the course o 2 4 hours, the satellite will view every point on the E arth. For the orbital radius o a polar orbit, the linear speed o the satellite is 7800 m s 1 or 2 8 000 km per hour.
6 3
1 1
24
= 5 2 00 s ( about 87 min)
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Although satellites can be put into orbits o any radius ( provided that the radius is not so great that a nearby astronomical body can disturb it) , there are two types o orbit that are o particular importance.
Geosynchronous satellites orbit at much greater distances rom the E arth and have orbital times equal to one sidereal day which is roughly 2 4 hours. This means that the geosynchronous satellite can be made to stay in the same area o sky and typically ollows a fgureo- eight orbit above a region o the planet. A typical track in the southern hemisphere is shown in fgure 1 0( b) . This track shows where the satellite is overhead during the 2 4 hour cycle. B ecause the plane o the orbit coincides with the centre o the E arth but is not through
1 0 . 2 F I E L D S AT W O R K
the E quator, the overhead position wanders. To see this, use a globe imagining the satellite going around its orbit as the E arth rotates below it at the same rate. A geostationary orbit on the other hand is a special case o the geosynchronous orbit. In this case the satellite is placed in orbit above the plane o the Equator and will not appear to move i viewed rom the surace. This has the advantage that a receiving antenna ( aerial) or satellite dish on E arth also does not have to track the transmitting antenna on the satellite. A TV satellite dish pointing towards the south ( equator) in the northern hemisphere will be always aligned with the geostationary satellite. These satellites are generally used or communication purposes and or collecting whole- disk images o the E arth or weather orecasting purposes.
Worked examples
____________________________
r=
a) A geostationary satellite has a orbital time o 2 4 hours. C alculate the distance o the orbit rom the surace o the Earth.
86 400 6.67 1 0 6.0 1 0 ____ 4 1 1
2
24
3
2
= 42 1 0 6 m = 42 000 km
b) C alculate the gravitational feld strength at the orbital radius o a geostationary satellite.
This is ( 42 000 6400) = 3 6 000 km above the surace ( the subtraction is to fnd the distance rom the surace) .
Solution
b)
a) 2 4 hours is 86 400 s.
GME g = _2 ( rE + r) 6. 7 1 0 1 1 6.0 1 0 24 = ___ ( 4. 2 1 0 7 ) 2 = 0. 2 3 N kg 1
Rearranging the orbital time equation ______
r=
T GM _ 4 2
3
2
Escaping the Earth The total energy o a satellite is made up o the gravitational potential energy and the kinetic energy ( ignoring any energy transerred to the internal energy o the atmosphere by the satellite) . To escape rom the surace o the Earth, work must be done on the satellite to take it to infnity. For an unpowered proj ectile this is simply equal to the kinetic energy meaning that the total o the ( negative) gravitational potential energy and the ( positive) kinetic energy must add up to zero. Thus to reach infnity gravitational potential energy + kinetic energy = 0 and thereore GME m S 1 m v2 = 0 _ + _ RE 2 S E where RE is the distance o the satellite rom the centre o the Earth, vE is the escape speed, and ME and m s are the masses o the Earth and the satellite respectively. The gravitational potential energy is negative because this is a bound system. Kinetic energy is always positive. ( It is important to keep track o the signs in this proo.)
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F I E L D S ( AH L ) To escape the Earths gravitational feld completely the total energy o the satellite must be ( at least) zero. For the case where it is exactly zero ( or the satellite to j ust reach infnity) vE =
_____
2 GME _ RE ____
= 2 gR E
where g is the gravitational feld strength at the surace. This speed must be about 1 1 2 00 m s 1 ( 400 00 km hour 1 ) or something to escape. It is important to be clear about the true meaning o escape speed. This is the speed at which an unpowered obj ect, something like a bullet, would have to be travelling to leave the Earth rom the surace. In theory, a rocket with enough uel can leave the E arth at any speed. All that is required is to supply the 63 MJ or each kilogram o the mass o the rocket ( the gravitational potential o the Earths feld at the surace is 62 . 5 MJ kg 1 ) . However, in practice it is best to reach the escape speed as soon as possible. Earth
Moon
distance
0 L
gravitational potential
- 0.77 mJ kg-1
-63 mJ kg-1
Figure 11 Gravitational potential between the Earth and the Moon (not to scale) .
S imilarly, i a spacecrat begins its j ourney rom a parking orbit, then less uel will be required rom there because part o the energy has already been supplied to reach the orbit. The graph o potential against distance helps here. The graph in fgure 1 1 shows the variation in gravitational potential rom the Earth to the Moon. B eore takeo, a spacecrat sits in a potential well on the Earths surace. When the rockets are fred the spacecrat gains speed and moves away rom the Earth. It needs enough energy to reach the maximum o the potential at point L (this is known as the Lagrangian point and it is also the point where the gravitational feld strengths o the Earth and Moon are equal and opposite) . Once at L the spacecrat can all down the potential hill to arrive at the Moon.
Orbit shapes We have now identifed two speeds: orbital speed close to the surace ( 7. 8 km s 1 ) and the escape speed ( 1 1 . 2 km s 1 ) . At speeds between these values the spacecrat will achieve dierent orbits. These are shown in fgure 1 2 .
v > 11.2 km s -1 hyperbola v = 7.8 km s -1 circle
v = 11.2 km s -1 parabola 7.8 km s -1 < v < 11.2 km s -1 ellipse
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Figure 12 Orbital shapes of satellites.
1 0 . 2 F I E L D S AT W O R K
At speeds less than 7.8 km s 1 the crat will return to Earth.
At a speed o 7.8 km s 1 the crat will have a circular path.
At speeds between 7.8 km s 1 and 1 1 . 2 km s 1 the orbit will be an ellipse. At a speed o 1 1 .2 km s 1 the crat will escape ollowing a parabolic path. At speeds greater than 1 1 .2 km s 1 the crat will escape the Earth along a hyperbolic path. ( It will not however necessarily escape the S un and leave the S olar S ystem. )
You will only have to consider circular orbits or the examination.
Nature of science Using the Earths rotation and the planets Rotation Many launch sites or rockets are close to the Equator. This is because the Equator is where the linear speed o the E arths surace is greatest. This equatorial velocity is about 5 00 m s 1 rom west to east meaning that a rocket taking o to the east has to attain a relative speed o 1 1 .2 0. 5 = 1 0.7 km s 1 , a signifcant saving in uel.
Slingshots Another technique much used in missions to distant planets is to use slingshot techniques in which a spacecrat is attracted towards a planet and then swings around it to shoot o in a dierent direction possibly at a greater speed. The interaction is eectively a collision between the crat and the planet in which momentum is conserved but the crat gains some energy rom the orbital energy o the planet. The spacecrat Voyager 2 was launched in 1 977 and in the course o its j ourney has passed Jupiter, S aturn, Uranus, and Neptune, using each planet to change its traj ectory achieving a Grand Tour o the planets. The spacecrat is still travelling and is now a considerable distance rom Earth ( it has its own Twittereed i you wish to know where
it is today) . It is expected to lose all its electrical energy sometime in 2 02 5 and will then continue, powered down, through space carrying messages rom the people o Earth to any other lie orm that may encounter it.
Gases in the atmosphere The Earths atmosphere only contains small amounts o the lighter gases such as hydrogen and helium. All gases in the atmosphere are at the same average temperature __ T and thus have the same mean kinetic energy EK or the molecules __ ( because, rom Topic 3 , EK T) . 1 Kinetic energy is __ mv2 , thereore the mean speed 2 o the molecules o a particular element with mass 1 __ . C alculations show that m is proportional to ___ m the lightest gases in the atmosphere have mean speeds that are close to ( but less than) escape speed. B ut even so, the very astest o these molecules can escape the atmosphere so that eventually almost all the light gases disappear. The Moon has no atmosphere, because its escape speed is lower than that o the E arth and ultimately almost all its gas molecules leave ater release. The low-density atmosphere o Mars is slowly leaking away into space.
Charges moving in magnetic and electric felds Charges moving in magnetic felds We saw in Topic 5 that a orce acts on a charge moving in a magnetic feld. In Figure 1 3 an electron moves to the right into a uniorm magnetic feld in which the feld lines are oriented at 90 to the direction in which
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F I E L D S ( AH L ) the electron moves ( the feld lines are out o the plane o the paper in the fgure) .
magnetic feld out o plane o peper current orce magnetic feld out orce
o plane o peper current
conventional
orce
current -
magnetic feld out o plane o peper
Figure 13 Force acting on an electron in a uniorm magnetic feld.
Flemings let-hand rule predicts the eect on the electron. The orce will be at right angles to both the velocity and the direction o the magnetic feld. In using Flemings rule, remember that it applies to conventional current, and that here the electron is moving to the right so the conventional current is initially to the let. The electron accelerates in response to the orce, and its direction o motion must change. The direction o this change is such that the electron will still travel at right angles to the feld and the magnetic orce will continue to be at right angles to the electrons new direction. This is exactly the condition required or the electron to move in a circle. The magnetic orce acting on the electron is providing the centripetal orce or the electron. As the electron continues in a circle so the magnetic orce direction alters as shown in fgure 1 3 . The orce acting on the electron depends on its charge q, its speed v, and the magnetic feld strength B. The centripetal orce will lead to a circular orbit o radius r and m ev2 _ r = Bqv S o the radius o the circle is ( written in a number o ways) 1 _
m ev ( 2 m e Ek ) 2 p r= _= _= _ Bq Bq Bq
Figure 14 Electrons moving in a circle demonstrated in a fne-beam tube.
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where p is the momentum and m e is the mass o the electron. It is possible to use observations o this electron motion to measure the specifc charge on the electron. This is the charge per unit mass measured in C kg 1 . A fne- beam tube is used in which a beam o
1 0 . 2 F I E L D S AT W O R K
electrons is fred through a gas at very low pressure ( which means that the electrons do not collide with too many gas atoms) . When a uniorm magnetic feld is applied at right angles to the beam direction, the electrons move in a circle and their path is shown by the emission o visible light rom atoms excited by collisions with electrons along the path. The electrons o mass m e are accelerated using a potential dierence V beore entering the feld. Their energy is 1 _ m v2 = qV 2 e So
y helical path
____
v=
_ m 2 qV e
which with
B
Bqr v= _ m e gives q 2V _ _ m e = B 2 r2
+ +q z
In a case where the electron beam is moving into a magnetic feld that is not at 90 to the beam direction, then it is necessary to take components o the electron velocity both perpendicular to the feld and parallel to it. The perpendicular component leads to a circular motion exactly as beore. The only dierence will be that m e v sin r= _ Bq
x
Figure 15 A positively-charged particle moving in a magnetic feld not at 90 to its direction o motion.
where is the angle between the feld direction and the beam direction. S o the radius o the circle will be smaller than the perpendicular case. The parallel component o velocity does not lead to a circular motion. The electrons will continue to move at the component speed ( v cos ) in this direction. The overall result is that as the beam enters the feld it will begin to move in a helical path. Large particle accelerators can use magnetic felds to steer the charged particles as they are accelerated to higher and higher speeds. The acceleration is carried out using electric felds.
Charges moving in electric felds The situation is dierent or electrons moving into the uniorm electric feld produced by a pair o charged parallel plates. +V
parabolic path d
-
-
E=
V d
X 0V
Figure 16 Motion o an electron in a uniorm electric feld.
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10
F I E L D S ( AH L ) The orce on the electron now acts parallel to the feld lines. In the case shown in the fgure 1 6 the electron is accelerated vertically. B ecause the orce is constant in both magnitude and direction ( uniorm feld) the acceleration will also be constant. Mechanics equations rom Topic 2 are used to analyse the situation: V E = __ as usual or a uniorm feld, and d
qE qV F _ _ a vertical = _ me = me = m d e where m e is the mass o the electron. There is no horizontal acceleration and the horizontal component o the velocity does not change. The time t taken to travel between the plates o length x is thereore x t= _ vhorizontal S o the vertical component o the velocity as the electron leaves the plates is ( using v = u + at) qV x vvertical = a vertical t = _ _ vhorizontal m ed To obtain the fnal speed with which the electron leaves the feld, it is necessary to combine the two components in the usual way as ____________
v 2vertical + v 2horizonta l The deection s o the electron while in the feld is proportional to 1 t2 ( s = __ at2 ) . This orm is equivalent to the equation or a parabola, 2 so the traj ectory o the electron is parabolic not circular unlike the case with a magnetic feld. This derivation ignores the eects o gravity on the electron. Is this j ustifed?
Charge moving in magnetic and electric felds D espite the act that magnetic felds lead to circular orbits and electric felds lead to parabolic traj ectories, it is possible to use them in a particular confguration to do a useul j ob. The trick is to combine the magnetic and electric felds at right angles to each other.
+
E-
FE B +
FB -
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Figure 17 Crossed electric and magnetic felds can cancel out.
1 0 . 2 F I E L D S AT W O R K For the arrangement shown in the diagram, the electric orce is vertically upwards and the magnetic orce is downwards. When these orces are equal then there will be no net orce acting on the charged particle. For this to be the case FE = FB ;
qE = Bqv
leading to E v= _ B This means that or a particular ratio o E : B there is one speed at which the orces will be balanced. C harged particles travelling more slowly than this speed will have a larger electric orce than magnetic orce, and the net orce will be upwards on the diagram. For aster electrons the magnetic orce will be larger and the electron will be deected downwards. This provides a way to flter the speeds o charged particles and the arrangement is known as a velocity selector.
Nature of science Bainbridge mass spectrometer The B ainbridge mass spectrometer is a good example o a number o aspects o electric and magnetic felds being brought together to perorm a useul j ob. There are two parts to the instrument: a velocity selector with crossed electric and magnetic felds and a deection chamber with j ust a magnetic feld ( in this diagram, into the paper) . It is let as an exercise or the reader to see that the motion o equally charged ions o the same speed but dierent mass travel in circles o dierent radii in the deection chamber. The ions arrive at the photographic plate or electronic sensor at dierent positions and, by measuring these positions, their mass ( and relative abundance) can be determined.
photographic plate or electronic sensor path o ion with smaller m q vacuum + cathode anode + ion source
gas
positive ions velocity selector
magnetic feld into paper defection chamber
Figure 18 Bainbridge mass spectrometer.
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F I E L D S ( AH L )
Concept map or feld theory Figure 1 9 gives a visual summary o all the relationships presented in this topic. It applies to both gravitational and electric felds. B oth sets o equations are represented on it. The diagram shows the cycle o feld strength potential potential energy orce and orce the relationships between them.
Comparison o gravitational and electric felds The table below sets out some o the important similarities between electric and gravitational orces and summarizes the contents o Topics 5, 6, and 1 0. E
orce = - gradient - dE = dr
F
r
r
F=
kQq r2
F=
g=
GM r2
V
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g =
Potential energy = potential quantity
FIELD STRENGTH
- dV dr
GM Eg = r Ve =
r1
g
kQ r
r
V = area under graph
Quantity = charge or mass g
Figure 19 Relationships between feld quantities.
kQq r
Potential = potential energy quantity
Potential = area under feld strength-distance graph
r
GMm r
Ee =
POTENTIAL
Field strength = - gradient o potential-distance graph
kQ E= 2 r
r1
Force = feld strength quantity
r
Eg =
POTENTIAL ENERGY
FORCE
Field strength = orce quantity
r
Potential energy = area under orce- distance graph
Force = - gradient o potential energy- distance graph
GMm r2
energy = area under graph
1 0 . 2 F I E L D S AT W O R K
Electric Qq F = __ (Coulombs law) 4 0 r2
GMm _(Newtons law) F= _ r2
Modifcation when not in a vacuum
Replace 0 with
No change
Defnition
F E= _ q
F g=_ m
Unit
N kg1 or m s 2
Distance r rom a point object
N C1 or V m 1 Q E = __ 4 0 r2
At r rom centre o sphere o radius R, r R
Q E = __ 4 0 r2
GM g = _ r2
At r rom centre o sphere o radius R, r < R
E= 0
4Gr g = __ 3
Defnition
Electric potential energy per unit charge
Gravitational potential energy per unit mass
Unit
V J C1
J kg1
For two point charges or masses
Q Vp = __ 4 0 r
GM Vg= _ r
Force law Field strength
Potential
Dierences
value at infnity: Attractive: zero and maximum Repulsive: zero and minimum
Gravitational
Constant in orce law
Between charges Attractive and repulsive depending on charge sign
_1_(= Coulomb constant k) 4 0
GM g=_ r2
Between masses Only attractive
G
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F I E L D S ( AH L )
Questions 1
b) The graph shows the variation o the gravitational potential V with distance r rom the centre o Mars. R is the radius o Mars which is 3 .3 Mm. ( Values o V inside the planet are not shown. )
( IB) The mass o Earth is ME and the radius o Earth is RE . At the surace o Earth the gravitational feld strength is g. A spherical planet o uniorm density has a mass o 3 ME and a radius 2 R E . C alculate the gravitational feld strength at the surace o the planet. ( 1 mark)
( IB) A spacecrat travels away rom a planet in a straight line with its rockets switched o. At one instant the speed o the spacecrat is 5 400 m s 1 when the time t = 0. When t = 600 s, the speed is 5 1 00 m s 1 . C alculate the average gravitational feld strength acting on the spacecrat during this time interval. ( 1 mark)
3
(IB) a)
0.0 - 1.0 - 2.0 - 3.0 - 4.0 - 5.0 - 6.0 - 7.0 - 8.0 - 9.0 - 10.0 - 11.0 - 12.0 - 13.0
1
2
3
4
5
6
7
V/MJ kg- 1
2
r/R 0
State, in terms o electrons, the dierence between a conductor and an insulator.
b) Suggest why there must be an electric feld inside a current- carrying conductor.
A rocket o mass 1 2 Mg lits o rom the surace o Mars.
c) The magnitude o the electric feld strength inside a conductor is 55 N C 1 . Calculate the orce on a ree electron in the conductor.
( i) C alculate the change in gravitational potential energy o the rocket at a distance 4R rom the centre o Mars.
d) The electric orce between two point charges is a undamental orce that applies to charges whereas gravity is the gravitational orce that applies to two masses. State one similarity between these two orces and two other dierences.
( ii) D etermine the magnitude o the gravitational feld strength at a distance 4R rom the centre o Mars. c) D etermine the magnitude o the gravitational feld strength at the surace o Mars.
e) The orce on a mass o 1 . 0 kg alling reely near the surace o Jupiter is 2 5 N. The radius o Jupiter is 7.0 1 0 7 m.
d) The gravitational potential at the surace o Earth is 63 MJ kg 1 . Without any urther calculation, compare the escape speed required to leave the surace o Earth with that o the escape speed required to leave the surace o Mars.
( i) State the value o the magnitude o the gravitational feld strength at the surace o Jupiter. ( ii) C alculate the mass o Jupiter. ( 1 3 marks)
( 1 0 marks) 4
5
An astronaut in orbit around Earth is said to be weightless. Explain why this is.
( IB) This question is about the gravitational feld o Mars. a) D efne the gravitational potential energy o a mass at a point.
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6
( IB) A small sphere X o mass M is placed a distance d rom a point mass. The gravitational orce on sphere X is 90 N. S phere X is removed and a second sphere Y o mass 4M is placed a distance 3 d rom the same point mass. C alculate the gravitational orce on sphere Y.
11
ELECTRO M AGN ETI C I N D U CTI O N ( AH L)
Introduction The physics of electromagnetic induction has profound implications for the way we generate electrical energy and therefore for the way we live. E very year throughout the world about 1 0 2 0 J of energy are converted into an electrical form using electromagnetism. This vast
co nversion of energy enables us to feed our ever- growing appetite for technologies that require electricity. It also raises considerab le issues about the sustainability of the energy sources used for the conversion.
11.1 Electromagnetic induction Understandings Electromotive orce (em) Magnetic ux and magnetic ux linkage Faradays law o induction Lenzs law
Nature of science Much o the physics in this sub-topic was discovered through painstaking experimentation by a handul o scientists. Pre-eminent among them was Michael Faraday who observed currents induced in a coil when magnetic felds were varied nearby. These observations led him to the laws o electromagnetic induction that we use in our largescale generation o electrical energy.
Applications and skills Describing the production o an induced em by
a changing magnetic ux and within a uniorm magnetic feld Solving problems involving magnetic ux, magnetic ux linkage, and Faradays law Explaining Lenzs law through the conservation o energy
Equations Flux: = BA cos
Faraday's / Neumann's equation: = - N _ t
em induced in moving rod: = Bvl in side o coil with N turns: = BvlN
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Inducing an emf In Topic 5 we saw that when an electric charge moves in a magnetic feld, then a orce acts on the charge ( and thereore on the conductor in which it is moving) . In a reverse sense, a movement or change in a magnetic feld relative to a stationary charge gives rise to an electric current. This phenomenon is called electromagnetic induction.
Investigate! Making a current B egin with a magnet and a coil or a solenoid o wire. Arrange the coil horizontally and connect it to a galvanometer ( a orm o sensitive ammeter, with the zero in the middle o the scale) . You can use a laboratory coil, or you can wind your own rom suitable metal wire using a cylinder as a ormer.
Move the bar magnet so that its north- seeking pole approaches one end o the coil and observe the eect on the meter. Record the direction o the current as indicated by the meter and the peak value shown.
Repeat the movement, moving the bar magnet with its south- seeking pole towards the coil.
Move the bar magnet away rom the coil.
C hange the speed with which you move the magnet.
C ompare the current directions or each case and also the size o the current produced.
Now try moving the coil and keeping the magnet still. Does this change your observation?
Now try moving the coil and the magnet at the same speed and in the same direction. What is the size o the current now? I your coil allows it, you might also try making the magnet enter the coil at an angle rather than along its axis. You could also try moving the magnet completely through the coil and out o the other side. Try to interpret this complex situation when you have understood the simpler cases.
Relate the direction o current ow in the coil to the magnetic poles produced at the ends o the coil using the ideas in Topic 5 . ( Figure 1 ( b) reminds you o the rule. )
(a)
(b)
(c) S
N
S
N
N
S
N
S
N
S
N
Figure 1 (a) the experiment, (b) the N and S rule, and (c) typical results.
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S
11 .1 E LE CTR O M AG N E TI C I N D U CTI O N
The results or this simple experiment are shown in fgure 1 ( c) . You should ocus both on the direction o the conventional current ow and on the magnitude o the current. You are observing results similar to those made by Faraday in the 1 9th century. A number o conclusions emerge rom these simple experiments:
The current only appears when there is relative motion between the coil and the magnet; either or both can move to produce the eect. However, i both coil and magnet move with no relative motion between them then no current occurs. O nly movement o the wire in the coil relative to the feld is important.
When the north- seeking pole o the magnet is inserted into the coil, the current in the coil tends to reduce the magnets motion by producing another north-seeking pole at the magnet end o the coil. Push a south- seeking pole in and another south pole appears at the magnet end o the coil. It is as though the system acts to repel the bar magnet and to reduce its movement. The system appears to oppose any change in the magnetic ux; the greater the rate o change, the greater is this opposition. We will look at this again later.
In the same way, when a magnetic pole is moved away rom the coil, the opposite pole is ormed by the current in the coil in an attempt to attract the magnet and reduce its speed o motion.
Moving the coil at greater speeds relative to the magnet increases the sizes o the currents. The eect is at a maximum when the axis o the magnet between its poles is perpendicular to the area o cross-section o the coil.
The keys to understanding these eects lie in what we said earlier in Topic 5 about the motion o charges in a magnetic feld and what we know about the internal structure o the conducting wire that makes up the coil. Figure 2 ( a) shows electrons in a metal rod that is moving through a uniorm, unchanging magnetic feld. Free electrons in the wire are moving upwards with the rod in an external magnetic feld that acts into the page. A orce acts on each electron to the right. This orce is equivalent in direction to that which would act on a positive charge moving down the page ( a conventional current downwards) . This direction is perpendicular to both the feld and the direction o motion o the rod and is determined using Fleming' s let- hand rule. magnetic feld into page
L
(a)
R
L
- - - R
+ + +
(b)
L
conventional current
R
(c)
electron ow
Figure 2 Electrons orced to move in a magnetic feld experience a orce. In fgure 2 ( b) , with no external connection between L and R the electrons accumulate at the right- hand end ( R) o the rod making it negatively charged, and a lack o electrons at the let-hand end ( L)
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) makes it positive. A potential dierence exists between L and R, L being at the higher potential. When there is no external connection between L and R, no current will circulate. Charges will accumulate at the end o the rod, that is electrons will move to one end leaving the other end positive. Without a current in a closed circuit no work is required, no transormation o energy takes place. I the circuit is closed externally between L and R ( fgure 2 ( c) ) , a ow o electrons will occur as shown. Inside the rod the conventional current ows rom R to L. The electrons ow out o the right-hand end o the rod and this is a conventional current in the external circuit rom L to R. A current has been generated, or induced, in the circuit. The system is acting to move the electrons through the resistor and, because this is a transormation o energy into an electric orm, the source o the energy is termed an electromotive orce (em) . As usual, we can identiy the amount o energy transerred or each coulomb o charge that moves around the circuit and we use the term induced emf to show that the em arises rom an induction eect. (The word induction is another term used in the early days o the study o electromagnetism.)
Nature of science Electromagnetic force Perhaps you can now see why the term electromagnetic orce arose in the early days o electromagnetic induction and why it still persists. S ome physicists obj ect that no orce acts when the term em is applied to electric cells, batteries, piezoelectric devices, and so on thereore, they say, em is not a good expression. It is true that it is difcult to see how the word orce can apply in the case o a cell. B ut in the case o electromagnetic induction, it is clear that a orce is acting on the electrons in the conductor that is being moved and so the term em continues to be used in physics. The act that we oten use the abbreviation em rather than the ull expression is a reminder that we should not ocus on the term orce but rather on the units o em: J C - 1 .
Lenzs law An important aspect o electromagnetic induction is that the induced em exists whether the charge fows in a complete circuit or not. In the case shown in fgure 2 ( b) the circuit is incomplete. E lectrons ow along the rod until an excess o them sets up an electric feld which repels urther electrons, stopping urther ow. It is only when the circuit is complete that charge ows. An induced em is always generated by the system and the induced current will exist only i there is a complete circuit. We can look at the system in terms o a possible direction rule. Flemings let-hand rule was able to predict the orce on, and thereore a ow o, the electrons. This ow o electrons is equivalent to a conventional current
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motion o the conductor
orce F
Nature of science Another rule
magnetic feld B
magnetic feld
right hand left hand Flemings let-hand rule or motor eects
current I
induced current Flemings right hand rule or induction eects
Figure 3 Flemings rules. acting in the opposite direction. You have two choices: either use Flemings let-hand rule to work out the orce direction rom frst principles and let this lead you to the conventional current direction (the argument is given above) , or you can use another rule (Flemings right-hand rule) , which uses the symmetry between our let and right hands to give the relationship between the motion o the conductor, the direction o the feld and the direction o the induced conventional current. It' s your choice!
Another possibility is to use the right hand with its thumb in the direction o the current and fngers in the direction o the magnetic feld. B y pushing you get the direction o the orce on a positive charge or a conventional current. I an electron is the moving charge, point the thumb in the opposite direction to get the orce in opposite direction This convention can also be used to get the direction o the magnetic feld due a conventional current e.g. by curling the fngers around a wire with current.
The observations you made in the Investigate! can be interpreted by looking closely at the directions o current in the coil relative to the movement o feld and coil. A rule that describes this was summed up by the German scientist Heinrich Lenz in 1 833. He stated that the direction of the induced current is such as to op p ose the change that created the current. C heck your experimental notes ( or the diagrams in fgure 1 that sum them up) and see i your results confrm this law. In act, Lenzs law is little more than the conservation o energy. Suppose that, rather than opposing the induced eect, the change were to enhance it. This would imply an attraction instead o a repulsion between magnet and coil; the magnet would be pulled into the coil, accelerating as it goes. This would increase the speed and lead to an even greater acceleration. The magnet would move aster and aster into the coil, gaining kinetic energy rom nowhere. C onservation o energy tells us this cannot happen. Another way to look at the consequences o Lenz' s law is to realize that you cannot do work without having some opposition. The induced current in the coil is such that the induced feld produced by this current opposes the motion o the magnet you are holding. I the circuit is open, there is no current, no opposition, and no electric energy produced. I you move the magnet very ast you will clearly eel the opposite orce acting on you! E lectrons had not been discovered in Lenzs time and we can see how his law arises rom frst principles. In Topic 5 we saw that when a current ows within a magnetic feld the current produces a magnetic feld which distorts the original feld pattern. The result was that a orce acted on ( and could accelerate) the current- carrying conductor. This was the eect that led us to the basis o an electric motor.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) B X W
B -
constant speed
motor eect
conventional charge ow
Z Y rod rolling to right
motor equal i v external eect and is to be orce orce opposite constant to
Figure 4 Conducting rod rolling in a magnetic feld. In electromagnetic induction, there is a current in the conductor as a result o the motion o the conductor. Figure 4 extends the example o the moving rod that is now rolling to the right at a constant speed through the magnetic feld on a pair o rails. The rails conduct and orm part o a complete electrical circuit WXYZ. ( Rolling means that we do not have to worry about riction between the rod and the rails. ) C harges, driven by the induced em, ow around the circuit giving rise to an induced current in the direction shown. This induced current interacts with the uniorm magnetic feld to give rise to a orce - the motor eect orce that we discussed in Topic 5 ( page 2 3 4) . I you now use Fleming' s let- hand rule you will see that the induced current leads to a motor eect ( a orce) acting to the let in fgure 4, that is, opposite to the direction in which the conductor is moving. I the rod is to move at a constant speed then ( by Newtons frst law o motion) an external orce must be exerted on it. This is where the energy conversion comes in. The work done by the external orce to keep the conductor moving at a constant speed appears as electrical energy in the conductor.
Magnetic fux and fux density We can use the physics rom Topic 5 to extend these qualitative ideas. The magnitude o this magnetic orce is BIl where B is the magnetic feld strength, I is the induced current in the rod, and l is the length o the rod. Flemings let- hand rule shows that the magnetic orce arising rom the induced current opposes the original orce. In other words, the opposing magnetic orce is to the let i the original applied orce is to the right. The net orce is zero and the rod moves at constant speed. It is possible or work to be done because o the opposition to the motion presented by the external magnetic feld that produces a magnetic orce on the induced current. No opposition, no work possible ( or needed! ) . From Newtons frst law, to keep the rod in fgure 4 moving at constant velocity, a constant orce equal to BIl must act on the rod to the right.
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The energy we have to supply thereore in a time t is orce distance moved which is BIl x where x is the distance moved to the right by the rod. The induced em is equal to the energy per coulomb supplied to the system. In other words energy supplied __ = charge moved
BIl x _ BIl x =_ = Q I t
and thereore
= Blv where v is the speed o the rod. C ancelling and rearranging gives B A =_ = B t
rate o change o area
This is because the area that is swept through by the rod in a time t is lx = A ( because it is the amount by which the area changes in time t) S o, in words, induced em = magnetic fux density rate o change o area This introduces you to an alternative term in magnetism or what was earlier called the magnetic feld strength magnetic fux density. The magnetic eld strength is numerically equivalent to magnetic fux density. In Topic 5 we used the term magnetic feld strength because in that topic we were concerned with basic ideas o feld. In both electrostatics orce orce _____ and gravity, the term feld strength has a meaning o ____ m ass or ch arge
depending on the context. We defned magnetic feld strength in Topic 5 as orce __ current length However, this defnition does not take account o the old but helpul view o magnetic feld lines as lines directed rom a north- seeking to a south- seeking pole with a line density that is a measure o the strength o the feld. It is this visualization o a feld in terms o lines ( close together when the feld is strong, and well- separated when the feld is weak) that links magnetic feld strength to magnetic ux density. When lines o orce (feld lines) are close together, then the magnetic feld strength is large. There will be many lines through a given area. We say also that the magnetic ux density is large. The total number o lines per square metre is a measure o the magnetic ux density and thereore the total number o lines in a given area is a measure o the magnetic ux. Flux is an old E nglish word that has the meaning o ow and one way to think about ux is to imagine a windsock used to show the direction and speed o wind at an airfeld. I the wind is strong then the ux
normal to area B, magnetic fux density
= BA cos area, A
Figure 5 Flux and fux density.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) density is high. The ux is the number o streamlines going through the sock. I the wind has the same speed or two windsocks o dierent sizes then the windsock with the larger opening will have a larger ux even though the ux density is the same. I magnetic ux density is defned as the number o ux lines per unit area, then ux itsel must be equivalent to fux density area so that, in symbols, =B A This equation assumes that B and A are at right angles. Figure 5 shows the relationship between area and ux density when this is not the case. In this case, we need to consider the component o the feld normal to the plane. A normal to the area is constructed and this normal makes an angle with the lines o ux. S o, = BA cos so that when = 0 ( area normal to feld lines) = BA, and when = 90, = 0. To sum up:
Magnetic ux density B is related to the number o feld lines per unit area. It is a vector quantity.
Magnetic ux is equal to BA ( also written as ) . It is a scalar quantity.
The equation = BA cos is used i the area is not normal to the lines.
The unit o ux is the weber ( Wb) and is defned in terms o the em A induced when a magnetic feld changes. The equation = B ___ can be t
ch an ge in fu x ( BA) or ______________ and so re-written as = _____ tim e taken or ch an ge t
a rate o change o fux o one weber p er second induces an em o one volt across a conductor For a particular conductor, knowledge o a magnetic feld in terms o the magnetic ux and the rate at which it changes allows a direct calculation o the magnitude o the induced em that will appear. We can now make a direct link between ux density and feld strength. The magnetic fux density
) o rc e _____________ strength ( ).
fux , measured in ( __________________ a re a o ve r w h ich it a c ts
weber metre 2 is numerically equal to the magnetic eld cu rre n t le n g th
O ne tesla (T) one weber p er square metre (Wb m 2 ) We thereore also have a link between changes in the magnetic feld strength and the induced em.
Magnetic fux linkage A Finally, one more quantity appears. O ur derivation o = B ___ above t used a single rod rolling along two rails. This is equivalent to a single rectangular coil o wire that is gradually increasing in area. Imagine this single turn o coil gradually increasing its area to include more and more feld lines. As beore the em across the ends o the coil will be
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equal to the rate o change o area multiplied by the magnetic ux density. I there are N turns o wire in the coil then the induced em ( N ) A will be N times greater so that = NB ___ = _____ . N is known as the t t magnetic fux linkage. The unit o ux linkage is oten written as weber turns, although this is entirely equivalent to weber because the number o turns is simply a number. The relationships between these interlinked quantities can be shown as: magnetic ux linkage, measured in Wb-turns
N
magnetic ux measured in Wb
= BA
magnetic ux density measured in Wb m -2
B
magnetic feld strength measured in T
Magnetic induction is summed up in a law devised by Faraday himsel that is known as Faradays law, it states that the induced em in a circuit is equal to the rate o change o magnetic fux linkage through the circuit. In our usual notation this is written algebraically as N =- _ t The negative sign is added to include Lenzs law. In its ull mathematical orm, the equation is also known as Neumanns equation. This equation combines Faradays and Lenzs ideas. It reminds us that, ( as we shall see in the next section) magnetic ux can be changed in three dierent ways: by changing the area o
[
]
( co s ) A cross- section with time ( ___ , by changing with time ______ or by t ) t B changing magnetic ux density with time ( ___ ). t
Nature of science Cutting lines of force Faraday frst introduced the fe ld- line model though his model is not quite the same as our modern inte rpre tation o a magnetic feld. He considered the lines o orce to be at the edges o tubes o orce , like elongated elastic bands. At the time, an invisible ae ther , thought to have elastic properties, was conside re d to fll space. Later on, Faraday and others took the concept o the feld line urther by suggesting that it was the action o the conductor
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cutting the tubes o orce that led to an induced em. This is a helpul way to think o the process, although it conceals the link between a charge being moved in a magnetic feld and the magnetic orce that acts on the charge as a result. B ut we need always to remember that Faraday and the others did not know o the existence o the electron, and that they were very amiliar with the ideas o feld lines. In the nineteenth century, Maxwell refned these models by including both electrostatic and electromagnetic orces in one set o equations. This illustrates two things about the nature o science: the way in which scientists allow a discovery to illuminate prior knowledge in a dierent way, and the power o the visual image to help us to comprehend a phenomenon.
Changing felds and moving coils An em can be electromagnetically induced in a number o ways that, on the ace o it, appear dierent rom each other:
A wire or coil can move in an unchanging magnetic feld ( the example o the rolling rod above) .
The magnetic feld can change in strength but the conductor does not move or change.
A coil can change its size or orientation in an unchanging magnetic feld.
C ombinations o these changes can occur.
We will look at these cases in the context o a rectangular coil interacting with a uniorm ( but possibly changing) magnetic feld.
X
Y
B, feld out o page
area, A
X
Y X zero feld
X
= 2BA
= BA (a)
(b)
G (c)
Figure 6 Moving coils and changing felds.
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ipped though 180
Y
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Case 1 : Straight wire moving in a uniorm feld This is the case o the rolling rod above. The change in area per second is lv, the length l o the rod multiplied by v, the speed o the rod. The induced em is thereore = Bvl when the wire moves at 90 to the feld lines. I the wire motion is not at 90 to the feld then, as usual, the component o feld at 90 to the direction should be used.
Case 2: Coil moving The coil can move as shown in fgure 6(a) rom one position in a magnetic feld to another position where the feld may be dierent. I the coil begins and ends in positions where the feld is identical, then there is no change in the ux linkage and there is no induced em. Although the coil is cutting lines, the same number is being cut on opposite sides o the coil. Two ems are induced but in opposite senses and they thereore cancel out. I the coil moves rom a position where the ux is to a position where the ux is zero, the change in ux linkage is N and the induced em is N = __ time taken for change
An interesting variant o this occurs where a coil in a feld is ipped through 1 80 ( fgure 6( b) ) . To visualize this, look at things rom the point o view o the coil. The feld lines appear to reverse their direction through the coil, and so the change in ux is ( ) , in other words, 2 N 2 . S o the em induced will be equal to _______________ . tim e taken to rotate coil When a coil rotates in a feld the em produced instantaneously depends on the rate o change o the ux linkage, and this in turn depends on the angle the coil makes instantaneously with the feld. I the coil rotates at a constant angular speed, then the em output varies in a sinusoidal way. This is the basis o an alternating current ( ac) generator as we shall see later.
Case 3: Magnetic feld changes S ometimes the feld changes rom one value to another it gets stronger or weaker but the coil does not move. The act o cutting feld lines is not so obvious here. S uppose the feld is being turned on rom zero. B eore the feld begins to change there are no feld lines inside the coil. You can think o the lines as moving rom the outside into the area bounded by the coil. The change stops when the ux density is at its fnal value. In so doing the lines must have cut through the stationary coil. N B Again, the solution is not difcult. Now = - ____ becomes = - NA ___ t t because only B is changing. You need to know the rate at which the feld is changing with time ( or the total change and the total time over which it happens) . Another example is the case o two coils ace- to-ace as in fgure 6( c) . O ne coil is connected to a galvanometer alone, the other coil is connected to a circuit with a cell, a variable resistor and a switch. In what way will you expect the galvanometer reading to change when the switch is closed and remains closed? When the switch is opened? When the switch is closed and the resistance in the circuit is varied?
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Worked examples 1
The graph shows the variation o magnetic ux with time through a coil o 5 00 turns. 2.5
the gradient o this graph ( as always) as the ux is proportional to the time we can use any corresponding values o and t. 2 mWb = _ _ = 0. 5 V 4 ms t thus the induced em = 5 00 0.5 V = 2 5 0 V
2.0
2
1.5
A small cylindrical magnet and an aluminum cylinder (which is non-magnetic) o similar shape and mass are dropped rom rest down a vertical copper tube o length 1 .5 m.
/10 -3 Wb
a) Show that the aluminum cylinder will take about 0.5 s to reach the bottom o the tube. b) The magnet takes 5 s to reach the bottom o the tube. E xplain why the obj ects take dierent times to reach the bottom.
1.0
Solution 1 a) Use a kinematic equation, e.g. s = ut + __ a t2 . 2 1 __ 2 Then 1 .5 = 2 9.8 t
0.5
t = 0. 5 5 s. 0.0 0
1
2 t/10
3 -3
4
5
s
C alculate the magnitude o the em induced in the coil.
Solution The change in ux is 2 1 0 3 Wb and this occurs in a time o 4.0 ms. the rate o change o ux is
b) As the magnet alls the copper tube experiences a changing magnetic ux, and as a result an em is induced in the walls o the tube. This em results in a current in the tube. The current leads to another magnetic feld that opposes the motion o the magnet by Lenzs law. There is an upward orce on the magnet so that its acceleration is less than the value or ree all. In the case o the aluminium cylinder no current arises and it alls with the usual acceleration.
Nature of science Applications of electromagnetic induction There are many applications o electromagnetic induction over and above the generation o electrical energy. They include electromagnetic braking, which is used in large commercial road vehicles; the use o an induction coil to generate the large pds required to provide the spark that ignites the petrolair mixture in a car engine; and the generation o the signal in geophones and metal detectors. In each o these examples, a changing magnetic feld leads to the generation o an em and demonstrates the physics developed in this sub-topic.
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11.2 Power generation and transmission Understanding Alternating current (ac) generators Average power and root mean square (rms)
Applications and skills Explaining the operation o a basic ac generator,
values o current and voltage Transormers Diode bridges Hal-wave and ull-wave rectifcation
Nature of science
The provision o abundant, cheap electrical energy has been one o the ways in which an abstract science and its technological development have directly aected the lives o many people on the planet. Who could have imagined the enormous impact that Faraday's discoveries would make? ...certainly not Faraday himsel! This is an example o research in pure science leading to great practical applications.
including the eect o changing the generator requency Solving problems involving the average power in an ac circuit Solving problems involving step-up and stepdown transormers Describing the use o transormers in ac electrical power distribution Investigating a diode bridge rectifcation circuit experimentally Qualitatively describing the eect o adding a capacitor to a diode bridge rectifcation circuit
Equations rms and peak values: I0 __ I = _______ r.m .s .
2
0 __ potential dierence: Vr.m .s . = _______
V
2
0 r.m.s. = ________ resistance: R = ____ I I
V
0
V
r.m.s.
maximum power: Pm a x = I 0 V0 1 I V average power: Pa ve ra ge = ___ 2 0 0
p _____ Np Is ___ transormer equation: _ s = Ns = Ip
Introduction We have seen that a rod rolling along two parallel rails generates induced em and induced current; this is hardly a sensible way to generate electrical energy on a large scale. The practicalities o generation were solved by scientists rom the 1 83 0s onwards, frst or direct current, and later or alternating current.
Alternating current (ac) generators In this IB course we ocus on the ac generator because it is commonly used or the generation o energy. S uch a generator consists o a coil with a large number o turns; the coil rotates relative to a magnetic feld.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
(a)
G
fxed coil N
magnet
S
turn table
(b) ux linkage
or time
rate o change o ux linkage and induced em (c)
time
Magnetic feld lines cut the coil as the turntable rotates and an induced em is generated. However the em varies as the turntable rotates. Figure 1 ( b) shows how the ux linkage varies with , the angle between the normal to the coil and feld lines. When is equal to 90, the feld lines lie in the plane o the coil and so the ux through the coil is zero ( cos 90 = 0 in equation = BA cos ) . When is equal to 0 the feld lines are at 90 to the plane o the coil and the ux through the coil is a maximum. When the turntable rotation speed is constant, this graph also shows how the ux varies with time. The graph is a sine curve.
ac generator
N
S
slip rings V brush (d) N
For the moment we will imagine a fxed coil placed between the poles o a U- shaped magnet that stands at the centre o a rotating turntable ( fgure 1 ( a) ) . The turntable can turn at dierent angular speeds and the coil can have dierent numbers o turns and cross-sectional areas. The coil is connected to a galvanometer or to a data logger that registers the em across the coil. When the magnetic ux in the coil is maximum, the em induced ( current) is minimum ( 0) and vice- versa. C hanging the speed o the turntable changes the requency o the em as well as its amplitude. Increasing the number o turns or increasing the area o the coil will increase the amplitude o the em but leave the requency unchanged i the turntable speed does not change.
S N
S N
Figure 1 Basic ac generator.
The em induced in the coil is equal to - ___ , in other words the negative t o the gradient o the ux-time graph. Figure 1 ( b) also shows how the induced em varies with time; this graph can be obtained either by dierentiation or by a consideration o the gradient o the ux-time graph. When the normal to the coil and the feld lines are parallel ( is 90) then the em is zero because the coil does not instantaneously cut the feld lines at all. However, while some ac generators have a rotating magnetic feld, others have fxed magnets and a rotating coil ( Figure 1 ( c) ) . The principle is however the same. The direction o charge ow in the coils varies with the position o the coil. The use o a direction rule shows this. When the let- hand wire is moving upwards as shown in Figure 1 ( d) , the conventional current direction in this wire is to the back o the S coil. The right- hand wire is moving downwards at the same instant and the current in this wire is towards the ront o the coil. C harge ows clockwise ( looking rom above) in the coil and out into the external circuit. Hal a cycle later the sides o the coil will have exchanged position. C harge is again owing clockwise, but because the coil has rotated, the current ( so ar as the meter is concerned) is in the opposite direction. Figure 1 ( d) shows this too. I there were wires permanently connecting the coil to the meter they would quickly become twisted. Energy needs to be extracted rom the generator without this happening. S lip rings are used or this. The ends o the coil terminate in two rings o metal that rotate with the coil about the same axis. Two stationary brushes, connected to the external part o
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the circuit, press onto the rotating rings and charge ows out into the circuit through these connections ( Figure 1 ( c) ) . The essential requirements or an ac generator are thereore:
a rotating coil
a magnetic feld
relative movement between the coil and the magnetic feld
a suitable connection to the outside world.
Real-lie generators are more sophisticated than our simple models and an Internet search will allow you to see many dierent types o ac generator. For real generators, there is another issue that arises because an induced current is generated in the coils. As we saw in S ub- topic 1 1 . 1 , any moving conductor carrying an induced current in a magnetic feld has two orces acting on it: the orce that moves it, and an opposite orce that arises because o the induced current. This also applies to the rotating current- carrying coil in the ac generator. Flemings lethand rule and Lenzs law show that this orce opposes whatever is turning the coil. I a generator coil is being rotated clockwise by an external agent, then the magnetic orces that arise rom the induced current interacting with the magnetic feld will exert a turning orce anticlockwise on the coil. S ome o the energy provided by the external agent turning the coil has to be used to overcome this anticlockwise magnetic eect. This reduces the induced current that can be made available to the external circuit. However, remember that with no opposition, no work is done and no energy will be transerred rom the agent ( doing the turning) to the coil ( and its associated electrical circuit) . This is easily demonstrated using a bicycle dynamo ( a device similar to our frst rotating-magnet generator) ( see fgure 2 ) . In this type o dynamo a permanent magnet is rotated in the gap inside a coil. When the lamp is switched o so that no induced current is produced, the dynamo is relatively easy to turn ( remember, there will still be an induced em across the terminals) . When the dynamo supplies current and lights the lamp, more eort is required to rotate the dynamo at the same speed since an opposing magnetic orce will act on the current ( coil) . This is Lenzs law in action.
S
N
m o t i o n o f wh e el
iron core S
N
rotating magnet coil to bicycle lamp
Figure 2 Bicycle dynamo.
Nature of science Modelling an ac generator Although the ollowing derivation will not be required in the examination, it shows you how Faradays law can be used to model the behaviour o a simple ac generator.
The coil has an average length l and an average width o w with N turns ( these dimensions are shown in fgure 3 ( a) ) . The area o the coil A is thereore l w.
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Thereore the ux linkage varies with t as NB cos t and a graph showing how ux linkage varies with time has a cosine shape with maximum and minimum values o NBA at t = 0 and NBA when the coil is halway through one cycle.
(a)
w B l
(b) fux linkage N
At the instant when the normal to the plane o the coil is at an angle to the magnetic feld, the ux linkage through the coil is N , which is N BAcos . The coil spins at a constant angular speed In time t the angle swept is ( = t) .
em
A supply in which the current and voltage vary as a sine wave is an alternating supply. Although is used or convenience in the mathematics o the em induced in the generator, or everyday purposes we use the requency. This has the same meaning as elsewhere in physics: the number o cycles that occur each second. A generator that rotates 50 times in one second has an alternating current output at a requency o 5 0 Hz.
em
time
time
Figure 3
The model above shows that the output em o a generator is sinusoidal and that one complete rotation o the coil through 3 60 gives one cycle o the alternating current. time
when the rotation speed is doubled:
time
Figure 4 Increasing the rotation speed o the coil.
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rate o change o fux linkage and induced em
The value o the induced em at any instant is equal to the ( - ) rate o change o the ux linkage ( N) and this is the the ( - ) gradient o the ux linkagetime graph. S o the graph o induced em is a ( negative) sine curve with an equation o which has maximum and minimum values or the em o 0 = BAN.
Figure 4 shows the eect on the em o changing the angular speed o the coil ( without changing any other eature o the coil or feld) . I the angular speed o the coil is increased then:
the coil will take a shorter time to complete one cycle and so there will be more cycles every second hence the requency increases.
the time between maximum and minimum ux linkage will decrease and thereore ( as the ux linkage is constant) the rate o change increases and hence the peak em increases.
O ther ways to increase the output o the em, but without changing requency, include increasing:
the magnetic feld strength ( B)
the number o turns on the rotating coil ( N)
the larger coil area ( A) .
1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
Worked example A coil rotates at a constant rate in a uniorm magnetic feld. The variation o the em E with angle between the coil and the feld direction is shown. C opy the graph below and, on the same axes, add the em that will be produced when the rate o rotation o the coil is doubled. Explain your answer.
The rate o change o the ux linkage will double, so the magnitude o the peak em will also double. This is because Faradays law states that the induced em is proportional to the rate o change o ux linkage. However, the coil now rotates in hal the time, so the time or one cycle will be halved. O n the same scales, the new graph is: E 0
0
180
360
/degree
0
180
Measuring alternating currents and voltages The current and voltage o an alternating supply change constantly throughout one cycle. Measuring these quantities is not straightorward because, with positive and negative hal- cycles, the average value or current or voltage over one cycle is zero.
360
/degree
current
E 0
Solution
0
time
O ne way around this problem is to consider the power supplied to a resistance R connected to the generator. The instantaneous power dissipated in the resistance is I2 R where I is the instantaneous current.
The powertime graph is always positive ( which we would expect because the power is I2 R and a number squared is always positive) .
The power graph cycles at twice the requency o the current.
To see this, suppose that the time period o the ac generator is very large taking 1 0 s to turn once through one cycle. I you watch a flament lamp supplied with an ac supply o such a low requency you will see the lamp ash on and o twice in each cycle. The lamp is on when the em is near its maximum (positive) and minimum (negative) values.When we look at a flament lamp powered by the AC mains, persistence o vision prevents us seeing its ashing like this because it is switched on and o at 1 00 or 1 20 times per second (twice the normal mains requency o 50 or 60 Hz) . Alternating values are measured using the equivalent direct current that delivers the same power as the ac over one cycle. A lamp supplied rom a dc supply would have the same brightness as the average brightness o our ac lamp ashing on and o twice a cycle. This equivalent dc current is that which gives the average value o the power in the powertime graph. B ecause the average value o a sin 2 graph is halway between the peak and zero values and the curve is symmetrical about this line (shown on fgure 5 (b) ) , the areas above and below the average line are the same.
(a)
power dissipated in resistor
Figure 5 shows both the currenttime graph and the power dissipatedtime graph with the same time axes. Notice the dierence between the two:
(b) 0
mean value of power
time
Figure 5 Currenttime and power time graphs for ac.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) 1 2 So the mean power that an ac circuit supplies is __ I R where Imax is the 2 max peak_____ value of the current. The dc current required to give this power I 1 I 2 which is ___ __ . is _ 2 2 max This value is known as the root mean square (rms) current I V ___ __ . In a similar way V __ Irms = ___ rms = . The power P dissipated in a
m ax
m ax
m ax
2
2
resistance = Irms Vrms
I m a x ___ Vm ax I m a x Vm a x __ __ = ______ = ___ 2 2
2
with the usual equivalents: V2rms 1 2 1 _ __ P = IrmsVrms = __ I R = rms 2 2 R You will not need to know how to prove the relationship between peak values and rms values for the examination. Many countries use alternating current for their electrical supply to homes and industry. We will look at some of the reasons for this later, but different countries have made differing decisions about the potential differences and frequencies at which they transmit and use electrical energy. Thus, in some parts of the world the supply voltage is about 1 00 V; in others it is roughly 2 5 0 V. Likewise, frequencies are usually either 5 0 Hz or 60 Hz.
Worked examples 1
Solution
The diagram below shows the variation with time t of the emf E generated in a rotating coil.
a) the peak value of the emf is 3 60 V, so the rms 360 __ = 2 5 5 V value is ___ 2
1 1 b) f = _ = ____ = 5 0 Hz T 0.02
E/V 360
2 0 0
10
30
20
t/ms
-360
C alculate: a) the rms value of the emf b) the frequency of rotation of the coil.
A resistor is connected in series with an alternating current supply of negligible internal resistance. The peak value of the supply voltage is 1 40 V and the peak value of the current in the resistor is 9.5 A. C alculate the average power dissipation in the resistor.
Solution
peak current peak pd The average power = __ 2 1 = _ 1 40 9.5 = 670 W 2
Transformers One of the reasons why ac supplies are so common is because a device called a transformer can be used to change alternating supplies from one pd to another. Transformers come in many shapes and sizes ranging from devices that convert voltages at powers of many megawatts down to the small devices used to power domestic devices that need a low-voltage supply. A transformer consists of three parts:
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an input ( or primary) coil
an output ( or secondary) coil
an iron core on which both coils are wound.
1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
(a)
primary winding Np turns primary current
laminations
m agn e ti c f u
secondary winding Ns turns
(b)
x,
Ip
secondary l current s
primary voltage Vp secondary voltage Vs
(c) transormer core
Figure 6 Transormers in theory and practice. Figure 6 shows a schematic diagram o a transormer ( a) , a real-lie transormer ( b) , and also the transormer symbol used in IB examinations ( c) . The operation o a transormer, like the ac generator, relies on electromagnetic induction.
Alternating current is supplied to the primary coil.
A magnetic feld is produced by the current in the primary coil and this feld links around a core made rom a magnetic material, usually sot iron. (The basic ideas behind the production o this feld were covered in Topic 5 .)
B ecause the primary current is alternating, the magnetic feld in the core alternates at the same requency also. The feld goes frst in one direction around the core and then reverses its direction.
The transormer is designed so that the changing ux also links the secondary coil.
The secondary coil has a changing feld inside it and an induced alternating em appears at its terminals. When the coil is connected to an external load, charge will ow in the circuit o the secondary coil and its load.
Energy has been transerred rom the primary to the secondary circuit through the core.
Suppose that an alternating pd with a peak value o Vp is applied to the primary coil and that this results in a ux o in the core. The ux linked to the secondary coil o Ns turns is thereore Ns and the induced em in the secondary coil is Vs = Ns ___ . There is also a ux linkage to the t primary coil even though the primary current was originally responsible or setting up the feld in the frst place. This gives rise to p = N p ___ t
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) where p and N p are the induced em and the number o turns in the primary coil. This induced primary em will oppose the applied pd. I we assume that the resistance o the primary coil is negligible, then p is the same or both will be equal in magnitude to Vp . Thus, because ___ t V V coils __ = __ = __ and N N N p
p
s
p
p
s
p Vp Np _ _ _ s = Vs = N s
This is known as the transormer rule. It relates the number o turns on the coils to the input and output voltages:
When N s > N p the output voltage is greater than the input voltage. This is known as a step -up transormer.
When N s < N p the output voltage is less than the input voltage. This is known as a step -down transormer.
The terms step- up and step-down reer to changes in the alternating voltages not to the currents.
You may wonder how zero current in the primary coil can lead to any energy transer to the secondary. The answer is: it cannot. The equation applies to the case where there is no current in the secondary ( in other words it has not yet been connected to a load) . O nce again Lenzs law has a part to play. When the secondary coil supplies current ( because a resistor is now connected across its terminals) , then because charge ows through this secondary coil, another magnetic feld is set up in the coil and through the core. This magnetic feld tries to oppose the changes occurring in the system and so tends to reduce the ux in the core. This in turn reduces the opposing em in the primary ( oten called a back em or this reason) and so now there will be an overall current in the primary that allows energy to be transerred. Electrical engineers use a more complex theory o the transormer than the one presented here to take account o this and other actors. B ut or our simple theory, which assumes that the transormer loses no energy, the energy entering the primary is equal to the energy leaving the secondary so Ip Vp = Is Vs where Ip and Is are the currents in the primary and secondary circuits respectively. This is a second transormer equation that you should know be able to use. In act, many transormers have an efciency that is close to 1 00% as energy losses can be reduced by good design. The efciency o a transormer is energy supplied by secondary coil ____ 1 00% energy supplied to primary coil Ways to improve the efciency include:
Laminating the core
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Iron is a good conductor and the changing ux in the transormer produces currents owing inside the iron core itsel. These are known as eddy currents. To prevent these, transormer designers use thin
1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N layers o insulating material that separate layers (sheets) o iron in the core (shown in Figure 6(a) ) . This has the result that although the magnetic properties o the iron are largely unaltered, the electrical resistance o the core is signifcantly increased. The currents are orced to travel along longer paths within the layers hence increasing electrical resistance and reducing the current. Laminations reduce the energy losses that result rom a reduction in the amount o ux and rom a rise in temperature o the iron that would occur i the eddy currents were large.
Note Hard and soft magnetic materials We sometimes talk about magnetic materials being "sot" or "hard". A sot material, such as iron, can be easily magnetised by another magnet or a current-carrying coil. When the magnetic inuence is removed, however, the iron loses all or most o its magnetism easily. Such materials are excellent or the cores o transormers because they respond well to the variations in magnetic feld. Hard materials such as some iron alloys (a steel) do not magnetise easily but are good at retaining the magnetism. These materials are best or the manuacture o permanent magnets such as the ones you use in the laboratory.
Choosing the core material The magnetic material o the core is a sot magnetic material. It can be magnetized and demagnetized very readily and also maintain high uxes too. These are all desirable properties or the core.
Choosing the wire in the coils Low- resistance wires are used in the primary and secondary coils as high resistances would lead to heating losses ( called j oule heating) in the coils.
Core design It is important that ux is not allowed to leak out o the core. As much ux as possible should link both coils so that the maximum rate o change o ux linkage occurs.
Worked example A transormer steps down a mains voltage o 1 2 0 V to 5 V or a tablet computer. The computer requires 1 .0 W o power to operate correctly. There are 2 300 turns on the primary coil o the transormer. a) C alculate the number o turns on the secondary coil. b) C alculate the current in the secondary coil when it is operating correctly.
c) The input current to the primary is 0.0090 A. C alculate the efciency o the transormer.
Solution V N V N 5 2300 a) __ = __ so N s = ______ = _______ = 96 turns V N V 120 p
p
s
s
s
p
p
tab le t p o w e r 1 .0 b) C urrent in the secondary Is = _________ = ___ 5 V = 0. 2 0A s
p o w e r su p p lie d b y se co ndary co il c) Efciency = _______________________ p o w e r su p p lie d to p rim ary co il 5 0.2 = __________ = 0.93 ( or 93 % ) 1 20 0.0090
Investigate! Transormer action
In this experiment you are asked to investigate the basic ideas behind the transormer. You will require an ac lowvoltage power supply to provide an input, an oscilloscope to view the output o your coils, and the coils themselves. O ne coil should have signifcantly less turns than the other,
typical values might be 2 40 turns on one coil and 1 1 00 turns on the other. You will also need additional apparatus or part o the experiment. ( S ome coil kits have special laminated iron cores that clip together to give a continuous magnetic circuit, )
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
C onnect the coil with ewer turns to the supply (this is the primary circuit) and the other coil to the oscilloscope (this is the secondary) . Set the coils close so that the secondary coil is linked to the fux produced by the primary. Look closely at the output rom the secondary coil and compare the relative sizes o the input and output and also their phases. Try the ollowing changes to see how they aect the output:
Alter the separation and orientation o the coils ( reverse one coil relative to the other and observe the dierence) . Alter the requency o the primary current. Link the
coils by inserting a piece o iron or several iron nails through the centre o both coils.
With the coils linked with iron compare the output voltages when the 2 40 turn coil is the primary and, later, when the 1 1 0 turn coil is the primary. I you have access to other coils with dierent numbers o turns collect data to complete the table.
Number Primary Number of Secondary of primary pd, secondary pd, V / V s turms Ns turns Np Vp / V
Np Vp _ _ Vs Ns
oscilloscope low-voltage power supply
3 2 1 0
4
5 6 7 8
9 10 11 12
primary coil. 240 T
secondary coil, 1100 T
V
Figure 7
Transformers in action Many countries have an electrical grid system so that each separate community within the country does not have to provide its own energy. In the event o power ailures in one part o the grid system, energy can be diverted where needed. When energy has to be sent over long distances it is advantageous to send it at very high voltages. This is because transmitting at high voltages and low currents helps to reduce the energy lost in the power transmission lines that are used in the grid. A laboratory example will help here: Figure 8 shows two alternative ways to transmit electrical energy rom one point in a school laboratory to another. In both circuits a power supply acts as the generator providing energy at an alternating potential dierence o 1 2 V. This energy is transmitted through two transmission lines ( leads strung between two retort stands) . In one case ( Figure 8( a) ) , the energy is tramsitted at 1 2 V, but in the other ( Figure 8( b) ) a step- up transormer with a turns ratio o 1 : 1 0 is used to transmit the energy at 1 2 0 V. At the other end o this transmission line, a step-down transormer returns the pd supply to the voltage level required by the lamps in the circuit.
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1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
dimly lit lamps
(a)
(b)
1:10 step-up
12 V ac power supply
120 V pd used in transmission line
10:1 step-down
12 V ac power supply
brightly lit lamps
Figure 8 Transmission at high voltages. A numerical calculation will help you to compare the two cases: In each case, there are three lamps each rated at 0.3 A, 1 2V (a total power requirement o 3.6 W) and a transmission line o total resistance 1 .5 ohms. Without the transormers, the total current required by lamps will be 0.9 A and power loss in transmission line = I2 R = 0.81 1 .5 = 1 .2 W. With the voltage stepped up to 1 2 0 V along the transmission line, the total current will be stepped down to 0.09 A during transmission. This means the power loss in the transmission line = I2 R = 0.081 1 .5 = 0.01 2 W. S o, stepping up the voltage by a actor o 1 0 reduces the current by a actor o 1 0 and thereore reduces the power lost during transmission by a actor o 1 00 ( i.e. 1 0 2 ) . This is an important saving or electricity supply companies ( and their customers! ) . Transormers play a crucial role in increasing the efciency o transmission. In practice, grid systems are usually like the one shown in fgure 9. Numbers are not given on this diagram as they vary rom country to country. Find out the details o the grid voltages used where you live. high voltage transmission
power generation
generator transformer
transformer lower voltage distribution
transformer small commercial and residential
light industry medium factories
heavy industry large factories
Figure 9 A grid system.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Nature of science HVDC and international collaboration The argument that high transmission voltages lead to better efciency does not apply simply to ac. The argument is still valid or dc but historically the transormation between voltages was easier using ac. However, things are now changing. As the physics and engineering o electrical transmission improve, so it is advantageous to use high-voltage direct- current transmission ( HVD C ) . Although the cost o the equipment to convert between two dc voltages is greater than the cost o a transormer, there are other actors in the equation. Countries use dierent supply requencies and this is a major problem when eeding electricity rom one country into the grid o another. Using undersea cables over long distances with ac also involves larger currents than might be expected as the cables have capacitance and induction eects. Additional currents are required to move charges every cycle.
S ome o the longest HVD C transmission paths in the world include the 2 400 km long connection between the Amazon B asin and south- eastern B razil that carries about 3 GW o electrical power, and the Xiangj iabaS hanghai system in C hina that carries 6.4 GW over a distance o 2 000 km. Governments work together to maintain electricity supplies. There are many examples o electrical links between countries provided so that one nation can supply energy to another nation during times o shortage. These are not necessarily times o crisis. There are short- term uctuations in the demand or electricity. S ometimes energy is ed rom country to country at one time o the day and then back again later. E xamples o such links include the electrical links rom the Netherlands to the UK and the HVD C link between Italy and Greece.
Rectifying ac The convenience o ac or domestic distribution is clear, but it raises the question o how devices that can only operate on dc, including computers and electronic equipment, can be made to unction when connected to an ac supply. The process o converting an ac supply into dc is called rectifcation. A device that carries this out is known as a rectifer. We consider two varieties o rectifer in this course: hal wave and ull wave. In hal wave, only hal o each cycle o the current is used whereas in ull wave a more complex circuit leads to the use o both halves o the ac cycle.
Hal-wave rectifcation Figure 1 0( a) shows the basic circuit. A single diode is connected in series with the secondary terminals o a transormer and the load resistor. The load resistor represents the part o the circuit that is being supplied with the rectifed current. D iodes are devices that only allow charge to ow through them in one direction ( the symbol or the diode has an arrowhead that points in the direction o conventional current allowed by the device) . We say that the diode is forward biased when it is conducting. When no charge can move through it ( because the pd across the device has the wrong polarity) we say that the diode is reverse biased. The load resistor will only have a pd across it or about hal a cycle ( fgure 1 0( b) ) . The waveorm is not quite hal a cycle wide because the diode does not conduct rom exactly 0 V; it requires a small orward pd or conduction to begin.
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1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
The way in which the circuit operates is straightorward. When terminal X on the transormer is positive with respect to terminal Y, then there will be a current through the diode in its orward conducting direction. When the ac switches so that X is negative with respect to Y, then the diode is reverse biased and there is no conduction or current. An obvious problem is that although the charge now ows in only one direction, the current is not constant. We need a way to smooth the waveorm so that it more closely resembles the constant value o a dc supply. O ne way to achieve this is to use a resistor and capacitor connected in parallel between the diode and the load ( fgure 1 0( c) ) . D uring the frst hal-cycle, the capacitor charges up and the potential dierence across it approaches the peak o the transormer output em. When the current is zero in the second ( negative) hal- cycle, the capacitor discharges through the resistor at a rate determined by the time constant o the circuit ( Figure 1 0( d) ) . I the time constant is much larger than the time or hal a cycle, the amount o charge released by the capacitor ( and thereore the pd across it) will be small and the pd will not change very much. When the diode conducts again in the next hal- cycle, the charge stored on the capacitor will be topped up. From now on there will be a dischargecharge cycle with the pd across the capacitor varying much less than in the basic circuit ( Figure 1 0( e) ) . When a capacitor is used in this way, it is oten reerred to as a smoothing or reservoir capacitor. The small variation in the output pd is known as the ripple voltage. The choice o capacitor and resistor values depends on the application in use. A large time constant provides good smoothing, but at the cost o a more expensive capacitor and a trade- o in the shape o the waveorm due to the large charging currents that are drawn rom the transormer.
X
Y (a)
output pd
time
(b)
input
+ ++ ++ -- -(c) charging
++ -+ (d) discharging
Full-wave rectifcation For some applications, the amount o ripple in a hal- wave rectifer cannot be tolerated. In such cases, ull-wave rectifers are used. Figure 1 1 ( a) below shows one way to achieve ull- wave rectifcation. This arrangement uses two diodes and requires a centre- tap on the transormer. This means that there is a connection hal way along the length o the wire that has been wound to make the secondary coil. This can be clearly seen at Y on the transormer symbol in the diagram. Also in the diagram is the resistorcapacitor pair that will smooth the rectifed wave.
load resistor
10 V ac
ripple
pd
time
(e)
Figure 10 Hal-wave rectifer.
The best way to understand how the circuit works is to imagine that terminal Y is always at zero potential. Then, or hal the time X will be positive and Z will be negative relative to point Y. For the other hal o the cycle, these polarities reverse. When X is positive relative to Y, diode D 1 will conduct. Notice that only hal o the secondary coil connected to D 1 takes part in conduction at any time. One hal cycle later, D 1 will not conduct because X will be negative and Z will be positive relative to Y. D 2 conducts during this hal cycle. Current is again supplied to a capacitorresistor combination during both halves o the cycle leading to ull-wave rectifcation.
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11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) D1 X
(a)
C R
pd
time
pd
time
Y
D2 Z X (b) Y
D4 D3
secondary o transormer
X
Y
Z
D1 D2
+
A B
Figure 11 Full-wave rectifcation. In order to achieve a particular value o peak pd, twice as many turns are required on the secondary compared to the hal- wave arrangement. This disadvantage can be overcome using a diode bridge ( fgure 1 1 ( b) ) . The ull secondary coil is now used and the diode arrangement allows the whole o the coil to supply current throughout the cycle. The disadvantage is the need or our diodes and a more complex circuit arrangement. When X is positive relative to Z then the j unction between D 1 and D 4 is positive relative to the j unction between D 2 and D 3 . O D 1 and D 4, D 1 is the one that conducts so that point A o the capacitor will be positive too. S imilarly, D 3 conducts and point B o the capacitor becomes negative. The capacitor charges and current is supplied to the rest o the circuit. When the polarity o the secondary coil switches, X becomes negative relative to Z and the conducting diodes are now D 2 and D 4. The connections are arranged so that the direction o the conventional current in D 4 is away rom B and the direction in D 2 is towards A. This is the same as in the previous hal cycle and the polarity o charge delivered to the capacitor is unchanged. The best way to learn this arrangement is not necessarily to memorize the orientation o the diodes but by understanding how they unction in order to provide a consistent pattern o polarity at the capacitor during the ull cycle. These examples o rectiying circuits are said to be passive, meaning that none o the devices ampliy or modiy the waveorms. Some modern power supplies or computers are active devices. Switched- mode power supplies change the requency o the supplied ac to a much higher value to allow the signal to be modifed by an electronic circuit. The advantage is that less energy is wasted in resistance but this is achieved only at the expense o much greater complexity in the electronic circuits.
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1 1 . 2 P O W E R G E N E R AT I O N A N D T R A N S M I S S I O N
Nature of science The war of the currents In the 1 880s a commercial battle broke out between the direct-current distribution system developed by Edison and the alternating-current system o Westinghouse. Both companies were attempting to corner the US market. In those days, electricity was principally used to provide energy or flament lamps. Edisons system had the advantage that batteries could be used as a backup i the power ailed (which was a requent event when electrical distribution began) . On the other hand, Westinghouses ac system could be transmitted with lower energy losses using larger and ewer distribution stations. Westinghouse made ull use o scientists rom around the world in developing ac
distribution, including Nikola Tesla who had made much progress in transormer design. The ull story o the battle is complex, but more and more companies ( principally the newly ormed General E lectric C ompany) began to ollow Westinghouses lead and eventually dc supplies largely disappeared. However, as late as the 1 980s, direct current was generated in the UK to supply the dc powered printing presses in London. In the US , the hotel where Tesla lived used dc into the 1 960s. C onsolidated E dison ( as Edisons company became known) shut down its last dc supply on 1 4 November 2 007.
Wheatstone and Wien bridge circuits The our- diode rectifer arrangement is one o a class o circuits known as bridge circuits. The Wheatstone bridge ( fgure 1 2 ) was popularized by S ir C harles Wheatstone in 1 83 3 and is a completely resistive arrangement that can be used to estimate the resistance o an unknown resistor. It is normally used in a dc context. The working o the circuit is straightorward ( fgure 1 2 ) . Three known resistors R 1 , R 2 and R 3 , are connected in the circuit with a ourth unknown resistor RX. O ne o the known resistors is variable and is adj usted until the current in a galvanometer that " bridges" the pairs o resistors is zero. This galvanometer is usually a centre-zero meter ( that shows negative and positive values with the 0 marking in the centre o the scale) with high sensitivity. When the current in the galvanometer is 0, the bridge is said to be balanced. No charge is owing through the galvanometer at the balance point because the potentials at B and D are identical so there is no potential dierence across B D . This means that the potential dierences across R1 and R X are the same as each other, and the potential dierences across R2 and R 3 are also identical. V1 = I1 R 1 = I2 R X
A RX
R1 + B
D
G
R2
R3 C
Figure 12 The Wheatstone bridge.
V2 = I1 R 2 = I2 R 3
R I R S o __ = __ = __ I R R 1
X
3
2
1
2
thereore R1 R3 RX = _ R2 Knowledge o three o the resistors means that that R X can be calculated. For accurate work, the bridge requires that the known resistors have accurately measured values and that the ammeter is sufciently sensitive to give precise results.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Investigate! Using the Wheatstone bridge
The details o this experiment will depend on the equipment that you have in your school. Your teacher will advise you on this.
2
RX
1 m wire
R The way to determine the ratio __ is to fnd the R 3 2
G
R3
O ne common way to carry out the experiment is to use a long ( 1 m) straight wire o uniorm cross- section perhaps taped along a metre ruler. The resistance per unit length o such a wire should be constant along its length. Figure 1 3 shows the way the wire is connected in the circuit.
point on the metre wire at which the current in the ammeter is zero. Then the ratio o the lengths o the two parts o the wire is equal to the ratio o their resistances.
tapping key
l2
3
ammeter ( the galvanometer) is zero.
Set up the circuit you are to use. C hoose R 1 so that it is similar in value to the unknown resistor.
l1
R S et the ratio __ so that the current in the R
The equation or the bridge becomes l R 3 = R 1 __ l 1
2
where l1 and l1 are the relevant lengths on the metre wire.
Figure 13
Nature of science Unbalanced Wheatstone bridges It is possible to make use o the Wheatstone bridge when it is out o balance. Suppose the unknown resistor RX is replaced by a thermistor. At a particular temperature the bridge can be made to balance with a particular combination o three fxed resistances and the thermistor itsel. When the temperature changes rom this starting point, the thermistor resistance will also change, increasing i the temperature drops, and becoming smaller i the temperature rises. These changes will take the bridge out o balance with a
CX R3 RX
C2 R4
Figure 14 Wien bridge.
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R2
potential dierence appearing across the meter that bridges the resistor pairs. The whole circuit can be calibrated so that a known temperature change at the thermistor gives rise to a known out-o-balance voltage across the meter. Such a circuit can be made to be extremely sensitive i the sensitivity o the meter and the resistor values are chosen careully. Other sensors can be chosen too: or example, a light-dependent resistor or changes in light intensity, and a length o resistance wire under tension (a strain gauge) or changes in length.
The Wien bridge circuit is a modifcation o the Wheatstone arrangement to allow the identifcation o resistance and capacitance values or an unknown component. This bridge operates with an alternating power supply. The bridge is modifed by the addition o a capacitor in series with R 2 . Again the current in the centre arm o the bridge is minimized. R2 , C2 and the requency o the supply need to be adj usted or the minimum current. Knowing the values o the components allows the operator to calculate the value o RX and CX. The theory or this bridge will not be tested in the examination.
1 1 . 3 C APACI TAN CE
11.3 Capacitance Understanding Capacitance Dielectric materials Capacitors in series and parallel
Applications and skills Describing the eect o dierent dielectric
Resistorcapacitor (RC) series circuits Time constant
Nature of science This sub-topic contains many important links and analogies. Some are straightorward, the link between the energy stored in a spring and the energy stored in a capacitor, or example. Others, however, reect the importance o the over-arching ideas o exponential growth and decay that underpin many areas o sciences and social sciences. Rates o reaction in chemistry, changes in living populations in biology, and radioactive decay in physics are all related by the important idea that the rate o change depends on the total instantaneous number and a constant probability o change. This is one o the many ways in which scientists use mathematics to model reality.
materials on capacitance Solving problems involving parallel-plate capacitors Investigating combinations o capacitors in series or parallel circuits Determining the energy stored in a charged capacitor Describing the nature o the exponential discharge o a capacitor Solving problems involving the discharge o a capacitor through a fxed resistor Solving problems involving the time constant o an RC circuit or charge, voltage, and current
Equations defnition o capacitance: C = ___V Q
combining capacitors in parallel:
C p a ra l l e l = C 1 + C 2 + ... 1 1 1 series: ________ = ____ + ____ + ... C series C1 C2 capacitance o a parallel-plate capacitor: C = ___A d
1 2 energy stored in a capacitor: E = ___ CV 2
time constant: = RC ___t
exponential discharge charge: Q = Q 0 e - ___t
current: I = I 0 e -
___t potential dierence: V = V0 e -
Introduction In Topic 1 0, a pair o charged parallel plates was used to produce a uniorm feld. A charge was transerred to the plates using a power supply. In this topic we look in detail at this transer o charge onto, and rom, the plates.
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Capacitors in theory
beore charging
-
+
-
+
-
electron + movement +
-
-
C harge is being separated by the system and stored. This requires energy and, not surprisingly, this is provided by the cell, the only source o em in the circuit. Energy is being stored on the plates as the electrons arrive there.
-
V< E
during charging
-
E
-
+ + + + + + + + + E
charging fnished
E
Figure 1 shows an arrangement o two parallel plates ( in a vacuum or simplicity) connected to a cell. The plates are initially uncharged. When the switch is closed, electrons begin to ow. There is no current between the plates because o the insulation between them. E lectrons are removed rom the plate connected to the positive terminal o the cell and are transerred to the plate connected to the negative.
E
+
The arrangement o parallel plates in which the two plates are separated by an insulator is called a capacitor. The insulator might be a vacuum, it could also be air or another gas providing that a spark (and thereore charge) cannot jump between the plates. It could also be a non-conducting material such as a plastic.
Figure 1 Charging a capacitor.
To understand this, imagine the frst electron that moves through the cell rom the positive plate in the diagram to the negative one. The cell will not need much energy to do this as the plates were initially uncharged. However, the second electron to move will fnd it slightly more difcult ( fgure 1 ) . This is because the frst electron is now on the right- hand plate and repels this second electron. The cell has to use more energy to move the second electron. The energy required rom the cell increases with each subsequent electron until fnally there is insufcient potential energy available in the cell to move any more electrons. At this point, i we were to disconnect the cell and measure the potential dierence across the capacitor then we would fnd that the capacitor pd is equal to the em o the cell. Had this been done earlier in the charging, then the unconnected capacitor pd would be less than the cell em. The capacitor pd cannot exceed the em o the power supply. The charge Q on the capacitor is stored at a potential dierence V. The ratio o the charge stored to the potential dierence between the plates is defned to be the cap acitance o the capacitor. charge stored on one plate Q capacitance, C = ___ = _ V potential difference between plates The charge stored on one plate is the same as the charge transerred through the cell and moved rom one plate to the other.
Tip Notice that, even though one plate has charge +Q and the other -Q, the charge stored is Q not 2Q. This is because it is only one group of charges (electrons) that are being moved between the plates.
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The unit of capacitance is the coulomb per volt, or farad (symbol: F) and is named ater Michael Faraday, the English physicist, who developed so much o our understanding about current electricity and magnetism. In undamental units the arad is equivalent to A2 S 4 kg 1 m 2 . O ne arad is a very large unit o capacitance since 1 C o charge is a very large amount o charge and you will usually use capacitances measured in milliarads ( mF) , microarads ( F) and picoarads ( pF) . Try not to conuse mF and F.
1 1 . 3 C APACI TAN CE
Worked examples 1
A pair o parallel plates store 2 .5 1 0 6 C o charge at a potential dierence o 1 5 V. C alculate the capacitance o this capacitor.
Solution
2
C alculate the potential dierence across a capacitor o capacitance 0.1 5 F when it stores a charge o 7.8 1 0 8 C .
Solution
Q 2 .5 1 0 C = _ = _ = 1 . 7 1 0 - 7 F = 1 70 nF. V 15 -6
Q 7.8 1 0 - 8 V= _ = _ = 0.5 2 V C 1 .5 1 0 - 7
Nature of science Changing names The name capacitance has not always been used. In the late nineteenth century and even later it was called capacity in English ( in French capacit dun condensateur and in Spanish la capacidad de un condensador) These were in many ways obvious names, and the idea o the capacity o a container can provide a model to help you understand capacitance ideas. Imagine two containers o water the same height, one narrow, the other wide. B oth containers, initially empty, are flled with water at the same rate ( the same volume o water every second) . O bviously, the narrower container overows frst. Thinking in terms o the top o both containers having the same gravitational potential, the liquid in the narrow container reaches this potential frst. Put another way, when both containers are ull, or the same potential ( height o the water surace
table
Figure 2 Water container analogy.
above the table is equivalent to the potential dierence o the capacitor) the wider container will hold more liquid ( equivalent to more charge) than the other. This is another link within physics that leads to analogous relationships.
Investigate! Charging a capacitor with a constant current
It is unusual or a capacitor to charge or discharge with a constant current. B ut this exercise will help you to understand the charging process or any capacitor.
S et up the circuit. A value o 470 F is suitable or this experiment.
You also require a clock with a second hand. You may wish to use a data logger with a
voltage sensor in place o the voltmeter and the clock ( the data logger can make both measurements or you) .
B egin by shorting out the capacitor with the ying lead so that it is completely discharged. The voltmeter should indicate zero.
In the experiment you need to alter the resistance o the variable resistor to keep the
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(a)
charging current at a constant value. I the resistance remains at a constant value, the current will all, reaching zero when the capacitor is ully charged so you will need to decrease the resistance as the experiment proceeds.
A
470 F
S et the variable resistor to its maximum value. C lose the switch with the ying lead connected across the capacitor. Remove the ying lead, start the clock, and record measurements o time elapsed t and potential dierence V across the capacitor. Record the value o the constant current I. Use your measurements to plot a graph to show how the charge Q stored on the capacitor varies with potential dierence V. To obtain the charge stored, remember that Q = It.
V flying lead
(b) 3.0 2.5 charge/mC
100 k
20 1.5 1.0 0.5 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 pd/V
Figure 3(a) and (b) Charging a capacitor with constant current.
The results o this experiment are shown in fgure 3(b) . The graph is a straight line with a gradient that is equal to C. The potential dierence across the plates o a capacitor is directly proportional to the charge it carries.
Energy stored in a capacitor The graph shown in fgure 3(b) can give more inormation than the gradient. The potential dierence V is the energy per unit o charge stored on the capacitor and Q is the charge, so the product o these two quantities is the energy stored and this quantity is equal to the area under the graph. Algebraically, this area is 1 1 Q V _ ( base height) = _ 2 2 There are two additional ways that this expression or the energy can be written: Q2 1 1_ 1 Energy stored on capacitor = _ QV = _ = _ CV2 2 2 C 2 This is an idea that is analogous to the potential energy stored in a stretched 1 string where there was also a actor o __ . (We looked at this in Topic 2.) 2
Worked examples 1
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C alculate the energy that can be stored on a 5 000 F capacitor that is charged to a potential dierence o 2 5 V.
2
A capacitor o value 1 00 mF stores an energy o 2 5 0 J. C alculate the pd across the capacitor.
Solution
Solution
1 CV2 = 0.5 5 1 0 - 3 2 5 2 Energy stored = _ 2 = 1 .6 J
Energy stored V =
_________
2 energy __ = C
____
5 00 = 71 V _ 0. 1
1 1 . 3 C APACI TAN CE
Capacitance of a parallel-plate capacitor In Topic 1 0 we saw that or a pair o parallel plates
Worked example 1
Q V _ = 0 _ A d where Q is the charge stored, A is the area o overlap o the plates, V is the potential dierence between the plates, and d is the separation o the plates. Rearranging this expression gives
Two parallel plates both have an area 0.01 5 m 2 and are placed 2 . 0 1 0 3 m apart. C alculate the capacitance o this arrangement.
Solution
Q A _ = 0 _ V d Q and __ is the capacitance C o the two- plate system. V
Finally,
A C = 0 _ d 0.01 5 = 8.9 1 0 - 1 2 _ 2 1 0 -3 = 6. 7 1 0 - 1 1 F
A C = 0 _ d This enables us to calculate the capacitance o a capacitor given the area o overlap o the plates and their separation. However, in real capacitors, edge eects will reduce this value. E arlier we mentioned the edge eects, the feld distortions at the edge o the plates that lead to dierences between the theoretical results here and what happens in practice with real capacitors. You saw some o the results o edge eects when you considered electric feld- line shapes at the edges o plates in Topic 1 0. I the gap between the plates is flled with an insulator o permittivity , this becomes A C= _ d
Capacitors in practice C apacitors are used extensively in electronic circuits. They store charge and can be used to reduce the eects o uctuations in a circuit. They are also the basis o timing circuits. This means that electric component designers need to be able to design capacitors o various sizes and capacitance values.
__
The equation C = 0 Ad gives the designers a basis or this. It is evident that there are three ways to increase the capacitance o the plates:
Make the plate overlap area larger because C A
1 Set the plates closer together because C __ d
C hange the value o the constant in the equation ( in other words change the permittivity rom that o a vacuum to some other value) .
Increasing the area o plate overlap increases capacitance because more charge can be stored or a given potential dierence between the plates Q and thereore __ increases. V The second way to increase C is to move the plates closer together. To understand the mechanism here, imagine the plates charged and
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) Ecap Edielectric Enet -
plate
++++++-
++++++-
++++++-
++++++-
+ + + + + + + +
dielectric
Figure 4 How a dielectric increases capacitance.
Dielectric strength of dielectrics I the potential dierence across a capacitor is increased, a maximum feld strength will be reached ater which the dielectric becomes a conductor. A spark will then jump between the plates. This maximum feld is called the dielectric strength o the dielectric. For air this is about 1 kV mm - 1 (the exact value depends on temperature, pressure and how dry the air is) ; or teon the dielectric strength is 60 kV mm - 1 . I during thunderstorms the electric feld strength in air between charged clouds or between a charged cloud and the ground exceeds this critical value, then a lightning strike may occur.
isolated, i.e. not connected to a power supply. The charge on the plates cannot change as there is no route along which the charges can move between them. The plates attract because they are positive and negative. I they are allowed to move closer together at a constant speed ( without touching) then they will do work on whoever is moving them. This energy must come rom somewhere, the only possible source is the capacitor itsel and so the potential dierence must drop and once Q increases. again __ V The third way to change capacitance is to insert a material between the plates that replaces the air ( or vacuum) that we have so ar imagined to fll the space. This means that the new material can do two things or the capacitor designer: it can be used to separate the plates by a fxed amount, and also change the properties o the capacitor at the same time. In Topic 5 we mentioned that materials have their own permittivity which is greater than that o a vacuum. To treat this mathematically, when a material o permittivity is present we replace 0 in the equations where it occurs with alone. Some typical values or were also given in Topic 5 or a number o dierent materials. A particularly useul thing happens when a dielectric material is inserted between capacitor plates to fll the whole space between them. A dielectric is an electrical insulator that is polarized when placed in an electric feld. The origin o the polarization is in the molecules o the dielectric substance. Each molecule is slightly more positive at one end than the other. This means that when a molecule is in an electric feld, it responds either by moving slightly or by rotating so that the more positive end o the molecule moves in the direction o the electric feld. I the molecules are in a solid and the solid is packed between the plates o a capacitor then this reduces the electric feld strength between the plates. When the dielectric is in place ( fgure 4) , the molecules inside it respond to the feld produced by the capacitor. Notice the feld directions careully. In the diagram the original feld Ecap o the capacitor is rom right to let, but the dielectric feld Edielectric ( indicated by the charges at the surace o the dielectric) is rom let to right. Thereore the net feld between the plates is equal to the capacitor feld minus the dielectric feld. The overall feld Enet in the dielectric V ( V is the potential is reduced and because Enet = E cap - E dielectric = __ d dierence) then V decreases too as d is fxed. The insertion o the dielectric reduces the potential dierence between the plates because some o the stored energy o the capacitor has been used to align the dielectric molecules. The overall charge stored is unchanged Q so __ is increased and hence so is the capacitance o the capacitor. V D ielectrics increase the capacitance o a capacitor. Another way to describe the action o the dielectric is by saying that the presence o the dielectric raises the potential o the negative plate and lowers the potential o the positive plate; this reduces the potential difference between the plates.
This explanation o the dielectric eect is a simplifed one. There are other reasons or the increase in capacitance, but in many cases the explanation can be given in terms o potential change as here.
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1 1 . 3 C APACI TAN CE Figure 5 ( a) shows some typical designs used or practical capacitors. A very common type o capacitor is the electrolytic capacitor shown as a cutaway in fgure 5 ( b) . The designer here uses the dielectric advantageously by using it to space the two metal oils apart. The capacitor layers are then rolled up together ( rather like rolling up three carpets initially on top o each other) . This type o capacitor must be connected into a circuit correctly. The dielectric material is a chemical ( the electrolyte in the diagram) that is a good dielectric when the electric feld direction is correct. In this design the layers are very thin ( good because d is small) and the plate area is large ( even better) giving some o the largest values o capacitance possible in a given volume.
(a)
Materials used or the dielectric include; paper, mica ( a mineral that can be cut into thin layers) , Teon, plastics, ceramics, and the oxides o various metals such as aluminium.
(b)
negative charge connection positive charge connection
The table shows how good some o these materials can be at improving the ability o a capacitor to store charge at a given voltage. For each material the number given is the ratio o the permittivity o the material to that o a vacuum, in symbols __ . ( This ratio is called the relative permittivity or sometimes the dielectric constant, but you will not be asked about this in the examination.)
dielectric
0
Material vacuum air paper mica polystyrene ceramic aluminium oxide tefon paran water (pure)
metal plate aluminum
_
plastic insulation
0 1 1.000 54 1 4 5 3 10015 000 911 2.1 2.3 80
paper separator
Tip I you are using electrolytic capacitors in the laboratory, take care that the polarity o the capacitor in the circuit is correct. I the polarity is reversed, the dielectric can become hot and cause the capacitor to explode.
+ oil oxide layer oil
tissue soaked in electrolyte
Figure 5 Practical capacitors.
Worked example 1
In a laboratory experiment, two parallel plates, each o area 1 00 cm 2 , are separated by 1 .5 cm. C alculate the capacitance o the arrangement i the gap between the plates is flled with:
b) Polystyrene has a dielectric constant o 3 , which means that = 3 8. 9 1 0 - 1 2 . C apacitance is now 1 .8 1 0 1 1 F.
a) air b) polystyrene.
Solution a) C onvert the centimetre units to metres: area = 1 00 1 0 4 = 1 0 2 m 2 ; separation = 0.01 5 m 1 0 -2 A = 8.9 1 0 - 1 2 _ C = 0 _ = 5.9 1 0 - 1 2 F 0.01 5 d
2
C alculate the area o overlap o two capacitor plates separated by a thickness o 0.01 0 m o air. The capacitance is 1 nF.
Solution Cd 1 1 0 - 9 0.01 __ A= _ = 8.9 1 0 - 1 2 = 1 .1 m
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Combining capacitors in parallel and series A urther way to modiy capacitance values is to combine two or more capacitors together in much the same way that resistors were combined in Topic 5 . Like resistors, capacitors can be connected in parallel and in series. C1 (a)
+Q 1
V
-Q 1 V +Q
+Q 2
-Q
-Q 2 C C2
V1
C = C1 + C2
V2
(b) +Q
-Q +Q
-Q
+Q
-Q
V C1
C2
C 1
C
=
1
C1
+
1
C2
Figure 6 Capacitors in parallel and series.
Parallel Parallel capacitors have the same potential dierence across them when connected as in fgure 6(a) . The total charge stored is Q 1 + Q 2 (as shown on the diagram) and these charges are Q 1 = VC1 and Q 2 = VC2 . The single capacitor that is equivalent to the two parallel ones has a charge o Q = VC.
Worked example C alculate the capacitance o the network below.
So Q = Q 1 + Q 2 ( conservation o charge) and thereore
C
VC = VC1 + VC 2 3C
C ancelling the V terms gives C = C1 + C2
2C
Solution
When two cap acitors are connected in p arallel, the total cap acitance is equal to the sum of the cap acitances.
Series
C ombining the parallel capacitors and 3 C gives:
C apacitors in series ( fgure 6( b) ) store the same amount o charge Q as each other because the same current charges both capacitors ( an example o Kirchhos frst law in action) . Kirchhos second law tells us that the potential dierences across the capacitors V1 and V2 must add up to give the em o the cell V.
1 1 1 2 ___ = __ + __ = __ so Ctotal= 1 . 5 C. C 3C 3C 3C
Thus
The capacitors in parallel have capacitance o C + 2 C = 3 C
to ta l
V = V1 + V2 ( conservation o energy)
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1 1 . 3 C APACI TAN CE
Using the defnition o capacitance Q Q Q _ = _+ _ C C1 C2 Q cancels to give
_1 = _1 + _1 C
C1
C2
S o this time, the recip rocal of the total cap acitance is equal to the sum of the reciprocals of each cap acitance. This is a reversal o the equations or combining resistors and can be a convenient way to remember the equations.
Discharging and charging a capacitor In an earlier Investigate! a capacitor was charged with a constant current. This is an unusual situation that required continuous changes to the total resistance in the circuit to achieve a constant ow o charge. When the total resistance is constant then the charging current varies with time. This section examines the nature o this variation.
Investigate! Discharging a capacitor
S et up the circuit shown in fgure 7( a) . This circuit has two unctions: to charge the capacitor and then to discharge it with the power supply disconnected rom the circuit. S uitable values or the components are: capacitance, 1 00 F; resistance 470 k. C harge the capacitor by connecting the ying lead to point X. Then begin the discharge by disconnecting the ying lead.
R
X fying lead (b)
Record the variation o the potential dierence across the capacitor with time. The capacitance and resistance values are designed to allow you to carry out the experiment by yoursel. However, you may fnd it even easier with two people, one recording the data. Alternatively use a voltage sensor together with a data logger to collect and display the data.
7 6 5 4 3 2 1
Plot the graph o potential dierence against time or the discharge. Your results will probably resemble those shown in fgure 7( b) .
V
C
pd/V
(a)
0 0
20
40
60 80 time/s
100 120 140
Figure 7 Circuit or capacitor discharge and results.
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Discharging You can think o the capacitor as taking the place o a power source and, at any instant, the current I and the pd VC across the capacitor are related by
R
VC = IR C Vc I
Here both VC and I change with time. This equation is obtained by applying Kirchhos second law to the circuit loop in Figure 8. Thereore
Figure 8 Discharging circuit.
VC Q _ = _ R t where Q is the charge on the capacitor and t is the time that has elapsed since discharging began. S o, because Q VC = _ C Q Q _ = -_ RC t A negative sign has appeared in the right- hand side o the expression. As the capacitor discharges, the charge on the capacitor alls and in each t the change in the charge is a negative value. Rearranging gives Qt Q = -_ RC and we replace RC by the constant where = R C so that Q t Q = -_
7
This new equation allows us to analyse how the charge on the capacitor varies with time during the discharge. The frst step is to recognize what the equation says. It predicts that the loss of charge 1 rom the capacitor in a time interval t will be equal to __ o the total charge that was stored on the capacitor at the beginning o the time interval.
6
charge/mC
5 4 3
Figure 9 shows what this means in terms o a graph o Q against t. It has Q the same shape as fgure 7( b) because C = __ and so Q V. V
2
Q t
1 0 0
20
40
60 80 time/s
100 120 140
Figure 9 Charge versus time for a discharging capacitor.
The next step is to model the discharge using Q = ____ . We will use a spreadsheet to do this and begin by frst concentrating on the part o the Qt graph j ust ater the discharge begins ( at time t = 0) at which time the original charge on the capacitor is Q 0 . I the charge on the capacitor did not change, then ater a time interval t the charge would still be Q 0 and the graph would be parallel to the time axis. B ut this is not what happens, the equation tells us that the Q t charge goes down (remember the minus sign! ) by ____ . This change is shown on the graph. 0
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Q t The charge remaining on the capacitor ater t will be Q 0 - ____ .
The graph line has a gradient o - __ during the irst time interval t.
0
Q0
1 1 . 3 C APACI TAN CE
What happens next? A second time interval begins and charge continues to move off the capacitor plates. B ut at the start of this interval the charge and, therefore, the pd across the capacitor are less than when t was 0. C all the new charge stored Q 1 . The change in charge this time Q t because the charge is no longer Q 0 . B ecause Q 1 is less is Q = - ____ than at the start, Q will also be less over this second time interval. The Q gradient of the graph becomes less and has a new value of __ . 1
1
B
C
D
t/s
delta Q/C
Q/C
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160
2.13E- 07 1.67E-07 1.32E-07 1.04E-07 8.17E-08 6.43E-08 5.06E-08 3.99E-08 3.14E-08 2.47E-08 1.95E-08 1.53E-07 1.21E-08 9.49E-09 7.47E-09 5.88E-09 4.63E-09
1.00E-06 7.87E-07 6.20E-07 4.88E-07 3.84E-07 3.02E-07 2.38E-07 1.87E-07 1.48E-07 1.16E-07 9.14E-08 7.20E-08 5.67E-08 4.46E-08 3.51E-08 2.76E-08 2.18E-08
E
F
G
H
I
4.70E+05 1.00E-04 10 1.00E-06
R C delta t Q0
J
K
L
M
N
O
how charge stored on a capacitor varies with time during discharge of the capacitor 1.20E-06 1.00E-06 8.00E-07 Q/C
A
6.00E-07 4.00E-07 2.00E-07 0.00E+00 0
10
20
-13.6
30
40 t/s
50
60
70
80
40 discharge time/s
60
-13.8 - 14 -14.2 -14.4 -14.6 -14.8
0.25
-15
0.20 current/A
In (Q/C)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
-15.2 -15.4
0.15 0.10 0.05
-15.6 -15.8
0.00 0
20
40 60 discharge time/s
80
100
0
20
80
Figure 10 Spreadsheet model for capacitor discharge.
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) In each successive t the change in the charge will be less than in the previous time interval and as time goes on the graph line will curve.
C olumn A shows the time increasing in steps o 1 0 s ( this is a deliberately large time increment, 1 s or 0.1 s would be better) . Q t
C olumn B shows the calculation o Q using Q = - ___ . RC
C olumn C shows the new value o Q at the end o the time interval.
The value o Q in column C is then applied as the initial charge in the next time interval on the next row o the sheet ollowing a loop with initial conditions changing at each cycle.
The program calculates successive values or the charge that remains on the capacitor or every 1 0 s o the discharge. The program then plots a graph o charge o the capacitor against time. ( I you want to produce your own version o this spreadsheet, the ormulae are:
in cell B 3 and below = C 3 * $G$3 /($G$1 * $G$2 )
in cell C 9 and below = C 3 -B 3
the symbol $ is required to ensure that the spreadsheet uses the same row and column or C, R and t when the columns are copied downwards rom row to row i these were not there then G3 would become G4 then G5 then G6 and so on as the rows are copied down the sheet.
At the top let-hand corner o the spreadsheet are the constants required in the solution: R, C, t and Q 0 . R and C have the same value as in the Investigate! so this solution should match up with the results you obtained then. Look closely at this graph ( which only has the frst 80 s o the discharge plotted) and compare it to your own results or the discharging experiment. The graphs have interesting properties. Examine the time it takes the charge on the capacitor to halve. The initial charge at t = 0 is 1 1 0 6 C (written as 1 .00E-6 in Excel notation) and hal this value (5 .00E-7) occurs at about t = 2 9 s. C ompare this with the time taken or the charge to halve rom 5 .00E-7 to 2 .5 0E-7. This is rom t = 2 9 s to t = 5 8 s another time o 2 9 s. The next halving rom 2 .5 0E-7 to 1 .2 5 E-7 takes another 29 s. For this combination o R and C it always takes 2 9 s or the charge to halve. I you used values or 1 00 F and 470 k in your experiment check that your results give a similar result. This behaviour is similar to that shown by decaying radioactive nuclei. The activity o the radioactive substance halves in one hal-lie. For a capacitor, hal the charge leaves the capacitor during a hal- lie. S uch behaviour is characteristic o exp onential decay. The equation Q 1 t _ = -_ Q has a similar orm to the radioactivity decay equation
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1 1 . 3 C APACI TAN CE N _ = - t N (N is the number o nuclei present at a given instant and is the probability o decay) and thereore we expect them to share the same solution N = N 0 e - t and _t
Q = Q0 e -
The capacitor discharge equation can also be written ( by taking natural logs o both sides o the expression) t log e Q = log e Q 0 _ 1. A graph o log e Q against t should be a straight line with gradient - _ The log e Q t graph or the spreadsheet results is shown in fgure 1 0. It has an intercept on the y-axis o 1 3 .8 and a gradient o 0.02 1 s 1 . This tells us that log e ( Q 0 /C ) = 1 3 .8 ( remember that the units are written inside the brackets to make the whole log number unitless) . Thereore Q 0 = e - 1 3.8 = 1 .0 C which was the original charge the model gave the capacitor. R was 470 000 and C was 1 00 F in the model, 1 so RC = ___ = 47 s and the model predicts Gk to be 0.02 1 s 1 . This is G k also confrmed by the gradient. The computer spreadsheet allows us to model one other aspect o capacitor discharge. As the charge ows, discharging the capacitor, there is a current in the circuit. The current is equal to the change in charge per second over each 1 0 s time interval. This can be modelled by assigning the average current during a time interval to the value o the current at a time halway through the time interval ( i.e. at 5 s or the frst 01 0 s interval, at 1 5 s or the second 1 02 0 s interval and so on) . Another graph in fgure 1 0 shows that the current j ust like charge and potential dierence decays exponentially with a hal- lie behaviour. The product R C = is known as the time constant or the circuit. The name is ully j ustifed because the units o RC are ampere second volt coulomb coulomb _ __ ohm arad = _ = _ ampere ampere = ampere volt = second. The capacitor time constant is not the same as a hal-lie. Ater a time o RC since the start, Q will be Q = Q 0 e -1 because t = RC = So Q 1 = 0.3 7 _ = _ e Q0 D uring each time interval the charge drops to 3 7% o its value at the start o the interval. Ater 2 this will be ( 0.3 7) 2 or 1 3 .7% . Ater 5 the charge remaining on the capacitor will be less than 1 % o its initial value.
Tip To obtain the instantaneous current in the resistor rom the graph o charge on the capacitor against time, remember that the rate at which charge leaves the capacitor is also the rate at which charge ows through the resistor. In other words, the gradient o the tangent o the Q-t graph will give you the value o the current in the resistor.
Tip There are three equations that give the variation o charge, current and potential diference with time or a discharge circuit: t _
Q = Q0 e- t _
I = I0 e -
t _
V = V0 e -
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11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L )
Worked examples 1
A 2 2 0 F capacitor charged to 3 0 V discharges through a 3 3 0 k resistor. a) Calculate the time taken or the capacitor to discharge to 1 0 V. b) C alculate the charge moved rom the capacitor in this time.
2
C alculate the number o multiples o the time constant that are required or a capacitor to lose 90% o its charge.
Solution Q = Q 0e
t -_ RC t ___
t For this case 0.1 = e - , thereore - ___ = ln 0.1 RC = 2 . 3 , making t = 2 .3 RC. RC
Solution t -_
t - ___ 3 -6
a) V = V0 e RC so 1 0 = 3 0 e 330 1 0 220 1 0 . t and t = 80 s. Thereore ln 0.3 3 = - ____ 72.6
2 .3 time constants are required to lose 90% o charge.
b) Q = CV. So Q = 220 1 0 - 6 20 = 4.4 mC.
Charging Again we approach the solution o these equations through an experiment ollowed by making a spreadsheet model.
Investigate! Charging a capacitor
This is very similar to the experiment during which you varied the total resistance in a circuit to achieve a constant current. This time, however, the resistance does not change. As the potential dierence at which the capacitor is storing charge increases, the current in the circuit will decrease.
the capacitor every 5 s until the pd becomes constant.
Plot a graph o pd against time and examine it closely. To what extent does the behaviour resemble the discharge case? R
Suitable values or the fxed resistor and capacitor are, as beore, 1 00 F and 470 k.
Make sure the capacitor is discharged, by short- circuiting it with the ying lead, and then disconnect the ying lead to begin the experiment. Wait or a data logger to do its j ob or take manual readings o the voltage across
Vem
fying lead
C
V
Figure 11
A model for charging The capacitor is in series with a resistor o resistance R and a cell o em Vem. When the circuit is switched on with the capacitor uncharged, charge begins to ow. Using Kirchhos second law Vem = VC + VR and thereore VR = Vem - VC = IR
468
1 1 . 3 C APACI TAN CE
so ( Vemf - VC ) I = __ R In the time interval t, the change in the charge on the capacitor Q is related to the change in the potential difference between the plates V by Q CV I= _= _ t t therefore ( rearranging) (Vemf - VC ) t V = __ RC This expression can be integrated mathematically, but a numerical solution illustrates more of the physics of the charging.
B
C
delta V/C
Vc/C
0 1.28E+00 1.00E+00 7.91E-01 6.23E-01 4.90E-01 3.86E-01 3.04E-01 2.39E-01 1.88E-01 1.48E-01 1.17E-01 9.19E-02 7.23E-02 5.69E-02 4.48E-02 3.53E-02 2.78E-02 2.19E-02 1.72E-02 1.36E-02 1.07E-02 8.40E-03 6.61E-03 5.21E-03 4.10E-03 3.23E-03 2.54E-03 2.00E-03 1.57E-03 1.24E-03
0 1.28E+00 2.28E+00 3.07E+00 3.70E+00 4.19E+00 4.57E+00 4.88E+00 5.11E+00 5.30E+00 5.45E+00 5.57E+00 5.66E+00 5.73E+00 5.79E+00 5.83E+00 5.87E+00 5.90E+00 5.92E+00 5.94E+00 5.95E+00 5.96E+00 5.97E+00 5.98E+00 5.98E+00 5.98E+00 5.99E+00 5.99E+00 5.99E+00 5.99E+00 6.00E+00
D
E
F
G
H
I
J
K
L
R 4.70E+05 C 1.00E-04 delta t 10 Vemf 6
7 6 5 Vc/V
A 1 2 time/s 3 0 4 10 5 20 6 30 7 40 8 50 9 60 10 70 11 80 12 90 13 100 14 110 15 120 16 130 17 140 18 150 19 160 20 170 21 180 22 190 23 200 24 210 25 220 26 230 27 240 28 250 29 260 30 270 31 280 32 290 33 300 34
4 3 2 1 0 0
50
100 150 200 250 300 350 400 t/s
Figure 12 Spreadsheet model for capacitor charging.
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11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) The three columns of the spreadsheet model are:
time is increased row by row by an amount t; the formula for A4 is =A3 + $F$3 ( as before the $ signs force the spreadsheet to use the same row and column each time) .
deltaVc is the change in Vc since the last time; it is given by the equation above and B4 translates into = ($F$4-B 3 ) * $F$3 / ($F$1 * $F$2 ) in the spreadsheet language. Look carefully at the contents of each cell and you should see why this is correct.
Vc is the new value of Vc that incorporates the deltaVc from the previous line; it is a simple addition = C 3 + B 4
The spreadsheet is copied for about 5 0 rows ( not all shown here) and then the values of t and Vc are used to plot a graph. D oes this graph have similarities to the one you plotted from experimental data? What are the similarities between charge and discharge?
TOK Exponential decay The similarity between radioactivity and capacitor discharge is shared with many other phenomena throughout the whole o science and extends into economic theory. In radioactivity the reason or the behaviour is that each nucleus o a particular isotope has an identical probability o decay per second. Thereore the number decaying in one second is directly proportional to the number o nuclei that remain undecayed. In capacitor discharge, the potential diference (which is proportional to the charge) between the plates is related to the discharge current Q through V = IR so ___QC = R _______ . We can use our water analogy again, this time or t water owing rom the bottom o a large tank. When the tank is ull the pressure (potential) rom the weight o water at the bottom is large and the ow rate is large too. When the tank is nearly empty the pressure is lower and the ow rate smaller. There are other examples o exponential changes in the natural world. Sea animals with shells grow at a rate that depends on the mass o ood they can eat and this mass depends on their mouth size. So the shell grows exponentially and the shape o the shell can be in the orm o an exponential curve. Population growth is an exponential change i there are no predators to remove a species. Why do so many academic areas have this linking relationship?
470
QUESTION S
Questions 1
S tate the magnitude o the magnetic ux linking the coil.
( IB) A bar magnet is suspended above a coil o wire by means o a spring, as shown below.
B
spring
area S
magnet
3
( IB)
coil electromagnet
The ends o the coil are connected to a sensitive high-resistance voltmeter. The bar magnet is pulled down so that its north pole is level with the top o the coil. The magnet is released and the variation with time t o the velocity v o the magnet is shown below.
The current in the circuit is switched on. a) S tate Faradays law o electromagnetic induction and use the law to explain why an em is induced in the coil o the electromagnet. b) S tate Lenzs law and use the law to predict the direction o the induced em in part a) .
v
0
c) Magnetic energy is stored in the electromagnet. State and explain, with reerence to the induced em, the origin o this energy. ( 8 marks)
t
0
a) C opy the diagram and on it: (i) mark with the letter M, one point in the motion where the reading o the voltmeter is a maximum (ii) mark with the letter Z, one point where the reading on the voltmeter is zero.
4
( IB) A small coil is placed with its plane parallel to a long straight current- carrying wire, as shown below.
current-carrying wire
b) Explain, in terms o changes in ux linkage, why the reading on the voltmeter is alternating. ( 4 marks) 2
A uniorm magnetic feld o strength B completely links a coil o area S. The feld makes an angle to the plane o the coil.
small coil
471
11
E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) a) Use Faradays law o electromagnetic induction to explain why, when the current in the wire changes, an em is induced in the coil.
b) O utline why the core is laminated. c) The primary coil o an ideal transormer is connected to an alternating supply rated at 2 3 0 V. The transormer is designed to provide power or a lamp rated as 1 2 V, 42 W and has 45 0 turns o wire on its secondary coil. D etermine the number o turns o wire on the primary coil and the current rom the supply or the lamp to operate at normal brightness. ( 7 marks)
current
0
magnetic fux
0
em
The diagram below shows the variation with time t o the current in the wire.
0
0
t
0
t
0
t
6
(IB) The graph shows the variation o potential dierence V with time t across a 2 2 0 F capacitor discharging through a resistor. C alculate the resistance o the resistor. ( 3 marks) 14 12 10
(i) C opy the diagrams and sketch, on the axes, graphs to show the variation with time t o the magnetic fux in the coil.
V/V
b)
c) Such a coil may be used to measure large alternating currents in a high-voltage cable. Identiy one advantage and one disadvantage o this method. ( 8 marks) 5
6 4
(ii) S ketch, on your axes, a graph to show the variation with time t o the em induced in the coil. (iii) S tate and explain the eect on the maximum em induced in the coil when the coil is urther away rom the wire.
8
2 0
7
a)
0
10
20
30 t/s
40
50
60
A circuit is used to charge a previously uncharged capacitor. The supply has an em o 1 2 . 0 V and negligible internal resistance and is in series with resistance R. The graph shows how the potential dierence across the capacitor varies with time ater the switch is closed. S
(IB) The diagram below shows an ideal transormer. laminated core 12.0 V secondary coil primary coil
a) Use Faradays law to explain why, or normal operation o the transormer, the current in the primary coil must vary continuously.
472
V 200 F
QUESTION S
C alculate:
15
(i) the initial energy stored by the capacitor (ii) the efciency o the system.
10 pd/V
( 1 0 marks) 5
9
0 0
50
100 time/s
150
200
(i) D etermine the time taken or the potential dierence across the capacitor to reach hal the maximum value. (ii) C alculate R. (iii) C alculate the initial charging current. b) A 1 00 F capacitor is added in series with the 2 00 F capacitor. (i)
( 1 0 marks) a)
1 0 In a timer, an alarm sounds ater a time controlled by a discharge circuit powered by a 9.0 V cell o negligible internal resistance. The alarm time is varied using resistor R. The capacitor is charged by moving the twoway switch to position S 1 . The timing starts when the switch is moved to S 2 . S1
X
S2
C alculate the eective capacitance o the combination.
(ii) D raw a graph showing how the potential dierence across the combination varies with time when the combination is charged.
8
A capacitor stores a charge o 3 0 C when the pd across it is 1 5 V. C alculate the energy stored by the capacitor when the pd across it is 1 0 V. ( 2 marks)
(i) A capacitor has a capacitance o 1 F. O utline what this tells you about the capacitor. (ii) S ketch a graph to show how the charge on the 1 F capacitor varies with the potential dierence across it over a range o 6 V. (iii) E xplain how you could determine the energy stored by the capacitor or a potential dierence o 6 V.
b) A 0.047 F capacitor is charged to a pd o 2 0 V, disconnected rom the supply and connected to a small motor without discharging. The motor can then lit an obj ect o mass 0.2 5 kg through a height o 0.90 m beore the capacitor is ully discharged.
R
6.0 V
to alarm
2200 F
Y
An alarm rings when the potential dierence across R reaches 3 .0 V. a) In one setting the time constant o the circuit when the capacitor is discharging is 3 . 0 minutes. Sketch a graph to show how the potential dierence across R varies with time or two time constants o the discharge. b) Use your graph to state the time at which the alarm will sound. c) C alculate the resistance o the variable resistor when the time constant is 3 .0 minutes. d) D etermine the maximum value o the resistance R that is needed or the timer to operate or up to 5 minutes. e) S tate how a capacitor could be connected to the circuit to increase the range o the timer. ( 1 1 marks)
473
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E L E C T R O M AG N E T I C I N D U C T I O N ( AH L ) 11
1 2 Three capacitors, one of unknown value, are connected in a circuit. The supply has a negligible internal resistance. The total charge stored on the capacitors is 400 C when the potential difference between A and B is 1 2 .0 V.
switch A
R
V
200 F
8.0 V
S 20 F
A
12 V
X
switch
60 F
150 k
B
B X
a) R
V
8.0 V
(i) C alculate the total capacitance of the circuit. (ii) C alculate X. (iii) C alculate the potential difference across the 2 0 F capacitor.
200 F
The diagram shows two graphs of the variation of p.d. with time for the discharge through resistor of value R of ( i) a 2 2 0 F capacitor and ( ii) the same 2 2 0 F capacitor in series with a capacitor of unknown capacitance. 12
b) The supply is disconnected and the capacitors discharge through the 1 5 0 resistor. C alculate the: (i) time taken for the potential difference between A and B to fall to 6. 0 V (ii) potential difference between A and B 8.0 s after opening the switch.
10
( 1 0 marks) V/V
8 6
1 3 A parallel plate capacitor is made from two circular metal plates with an air gap of 1 .8 mm between them. The capacitance was found to be 2 .3 1 0 1 1 F.
A 4 B 2
C alculate: 0 0
10
20 time/s
30
40
a) O utline why the p.d. in experiment B falls more rapidly than in experiment A. b)
(i) D etermine R. (ii) D etermine the capacitance of the unknown capacitor. ( 8 marks)
474
a) the diameter of the plates b) the energy stored when the potential difference between the plates is 6.0 V. ( 5 marks)
12
Q U AN TU M AN D N U CLE AR P H YS I C S ( A H L )
Introduction This topic develops material we frst met in Topic 7. The physics included here was ground-breaking when it was frst proposed (mostly in the early to mid 20th century) ; there were many highly respected physicists who totally rejected much o
the theory. Today these ideas are seen as being airly uncontroversial but there are still those who believe that reality should be modelled by theories that are simpler or more universal. Time will, no doubt, tell whether we have currently got it right!
12.1 The interaction of matter with radiation Understandings Photons The photoelectric efect Matter waves Pair production and pair annihilation Quantization o angular momentum in the Bohr
model or hydrogen The wave unction The uncertainty principle or energy and time, and position and momentum Tunnelling, potential barrier, and actors afecting tunnelling probability
Applications and skills Discussing the photoelectric efect
experiment and explaining which eatures o the experiment cannot be explained by the classical wave theory o light Solving photoelectric problems both graphically and algebraically Discussing experimental evidence or matter waves, including an experiment in which the wave nature o electrons is evident Stating order o magnitude estimates rom the uncertainty principle
Equations
Nature of science Scientists increasing dependence on quantum phenomena Much o what is described by quantum mechanics depends upon probability and seems in many ways counter-intuitive. Einstein, who remained unconvinced about quantum mechanical interpretations o matter, said that he did not believe that God plays dice.
Planck relationship: E = hf Einstein photoelectric equation: Em a x = hf -
13.6 n nh quantization o angular momentum: mvr = _ 2 Probability density: P(r) = || 2 V Heisenberg relationships: positionmomentum h xp _ 4 h energytime: Et _ 4
Bohr orbit energies: E = - _ eV 2
475
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L )
Introduction In Topic 4 we treated light as a wave, but in Topic 7 we needed to modiy our views o this in order to describe energy changes that happen in atoms and to explain atomic spectra. In S ub- topic 7.3 we looked at the Rutherord model o the atom with a nucleus surrounded by orbiting electrons. We said that this model did not obey the laws o classical physics but had much to commend it in terms o agreement with experiment. We now consider the urther leap o aith that needed to be made in order to reconcile the Rutherords experimental results with theory; this leap o aith meant abandoning aspects o classical physics and our everyday experiences and to accept many outlooks which go against our intuition. We will now consider the photoelectric eect and go on to look at B ohrs old quantum theory and its development into the modern interpretation o the quantum theory; acets o physics that would mean that physicists could never be truly certain o anything again!
Nature o science In the later part o the nineteenth century physicists were attempting to explain the radiation emitted by a black body a perect emitter and absorber o radiation. The radiation emitted by a black body was modelled using classical physics by the RayleighJeans law, whereby the intensity is proportional to the square o the requency. This theory worked well or the visible and inrared parts o the spectrum, where it matched the practical curve, but it ailed in the ultraviolet region by implying that ultraviolet radiation would be emitted with an infnite intensity something that clearly was not possible ( see fgure 1 ) . In 1 900, the German physicist Max Planck suggested that the ultraviolet catastrophe would be rectifed i electrons oscillating in the atoms o hot bodies were to have energies that were quantized in integral values o hf where f is the requency o the electrons and h is Plancks constant. The theory that he developed worked well at interpreting black- body radiation, but it
intensity of radiation/arbitrary units
The ultraviolet catastrophe the ultraviolet catastrophe 5000K
Rayleigh-jeans law
Planck radiation formula 1000 2000 wavelength of radiation/nm
3000
Figure 1 The ultraviolet catastrophe the experimental curve and one based on classical physics. was not well- received by the physics community because he was unable to j ustiy why the energies o electrons should be quantized. It was not until Einstein used similar assumptions to those o Planck in order to explain the photoelectric eect that physicists began to take Plancks ideas seriously.
The photoelectric efect Demonstration o the photoelectric efect The photoelectric eect can be demonstrated using a gold-lea electroscope (or a coulombmeter) . A reshly cleaned sheet o zinc should be mounted on the electroscope plate and the sheet charged negatively by connecting it to a high negative potential (o around 3 kV) . When a range o
476
1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
electromagnetic radiation rom inra-red to ultraviolet is incident on the sheet the divergence o the lea only alls when ultraviolet radiation is used; the lea then immediately collapses. This is because the zinc sheet and electroscope lea are discharging by the emission o electrons.
negatively charged zinc plate
ultraviolet radiation
Explanation o the photoelectric efect E instein explained the photoelectric eect in the ollowing way:
Light can be considered to consist o photons, each o energy = hf
Each photon can only interact with a single electron.
There is a minimum photon requency called the threshold frequency ( f0 ) below which no electron can be emitted.
Energy is needed to do the work to overcome the attractive orces that act on the electron within the metal this energy is called the work function ( ) .
Any urther energy supplied by a photon becomes the kinetic energy o the emitted electron ( oten called a photoelectron) .
Increasing the intensity o light simply increases the number o photons incident per second.
gold lea alls immediately that the zinc plate is illuminated with ultraviolet radiation
Figure 2 Gold-lea electroscope demonstration o the photoelectric efect.
Explaining observations from the gold leaf experiment The zinc sheet has a certain work unction and photons must have a greater energy than this to be able to emit electrons. The ultraviolet radiation has the highest requency o the radiation used and so it is these photons that are able to ree the electrons rom the metal. a) With very intense visible or infra-red light incident on the sheet the leaf remains diverged. Increasing the intensity only increases the number o photons incident per second; with long wavelength ( or low requency) none o the photons have enough energy to liberate electrons so the lea remains diverged. b) If low intensity ultraviolet is used, the leaf still falls immediately. O ne photon interacts with one electron so no time is needed to build up the energy to release an electron; as soon as the electron absorbs a sufciently energetic photon it will be ej ected rom the metal. c) Placing a sheet of glass between the ultraviolet source and the zinc p revents the leaf from falling. Glass will only transmit low energy visible photons; it absorbs the higher energy ultraviolet ones so none reach the zinc.
hf < no electrons emitted
hf = electrons brought to surace but stay there
d) If the zinc sheet is charged p ositively the leaf remains diverged for all wavelengths of radiation. Raising the potential o the metal presents a much deeper potential well or the electrons to escape rom, meaning that the photons no longer have sufcient energy to liberate the electrons. This is similar
hf > electrons ejected
metal o work unction
Figure 3 Photoelectric emission and the work unction.
477
12
Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) to replacing the zinc with a metal o greater work unction. This is analogous to someone being trapped in a well with vertical sides and only being able to j ump out in one leap.
Einsteins photoelectric equation We are now in a position to look at how E instein expressed the photoelectric eect in a single equation. His equation takes the orm: Emax = hf - In this equation Emax represents the maximum kinetic energy o the emitted electron, hf is the energy o the incident photon, and is the work unction o the metal. Each o the terms in the equation represents a quantity o energy that could be measured in j oule. However, it is quite common to express the energies in electronvolts simply because this avoids using very small powers o ten. The kinetic energy is expressed as a maximum value because the work unction is defned as the minimum energy required to liberate an electron those embedded urther inside the metal will take more energy than this to be liberated.
Worked examples 1
The work unction or aluminium is 6.6 1 0 1 9 J. a) C alculate the photoelectric threshold requency or aluminium. b) C alculate the maximum kinetic energy o the electrons emitted when photons o requency 1 . 2 1 0 1 5 Hz are incident on the aluminium surace.
Solution
b) C alculate the maximum speed with which a photoelectron may be emitted rom a potassium surace by photons o this energy. The work unction or potassium is 1 .5 eV.
Solution a) First the energy needs to be converted into joules: 2 .2 eV = 2 . 2 1 .6 1 0 - 1 9 = 3 .5 1 0 - 1 9 J
The work unction or aluminium is 6.6 1 0
1 9
J.
3 .5 1 0 E Then E = hf f = __ = _________ h 6.63 1 0 -19
-34
a) Using Einsteins equation Emax = hf - at the threshold value Emax = 0 so = hf0 where f0 is the threshold requency. 6. 6 1 0 - 1 9 = __ f0 = _ h 6.63 1 0 - 34 = 1 .0 1 0 1 5 Hz This is to two signifcant fgures in line with the work unction value. b) Using Emax = hf - Emax = 6.63 1 0 - 34 1 .2 1 0 1 5 - 6.6 1 0 - 1 9 = 1 .4 1 0 - 1 9 J 2
Photons incident on a metal surace have energy 2 .2 eV. a) C alculate the requency o this radiation.
478
= 5.3 1 0
14
Hz
b) Again using Emax = hf - but sticking to electronvolts to start with: Emax = 2 . 2 - 1 .5 = 0. 7 eV S o the maximum kinetic energy = 0.7 eV = 0.7 1 .6 1 0 - 1 9 J = 1 .1 2 1 0 - 1 9 J 1 This means that __ m v2 = 1 .1 2 1 0 - 1 9 J 2 e
__________
So v =
2 1 .1 2 1 0 ____________ m -19
e
Looking up the value or the mass o electron in the data booklet gives m e = 9.1______________ 1 1 0 - 31 kg 2 1 .1 2 1 0 - 1 9 v = __ 9.1 1 1 0 - 31 = 5 .0 1 0 5 m s - 1
1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
Investigate! Millikans photoelectric experiment In 1 91 6, the American physicist Robert Millikan designed an elegant experiment with which to test Einsteins photoelectric equation. We can use a photocell or essentially the same experiment with an arrangement as shown in fgure 4.
potential and e is the elementary charge on an e lectron) .
Einsteins equation Emax = hf - can now be rearranged to give eVs = hf - hf0 where f0 is the threshold requency.
The flters are usually provided with a range o transmitted wavelengths which gives a value or their uncertainties. This means E insteins equation needs urther rearrangement into
anode Vs +
- electrons -
pA
vacuum
hc hc eVs = _ - _ o
-
where o is the threshold wavelength and c is the speed o electromagnetic waves in a vacuum.
pA cathode
D ividing throughout by e gives
potential of anode is made negative so electrons cannot quite reach it and the picoammeter reading becomes zero
(
hc hc hc _ 1 1 _ Vs = _ - _ = _ e - e e o o This is o the orm:
Figure 4 Photocell for measuring the Planck constant. A variety o coloured flters are used with a white light source to allow incident photons o dierent requencies to all on the cathode o the photocell. The electrons are emitted rom the photocell and travel across the vacuum towards the anode.
y = mx + c
1 A graph o Vs against __ :
hc will be o gradient __ e and
1 1 have an intercept on the __ axis equal to __ o
have an intercept on the Vs axis equal to hc - ___ ( as shown in fgure 5 ) e o
In this way the electrons complete the circuit and a small current o a ew picoamps registers on the picoammeter.
Vs gradient =
The maximum kinetic energy is obtained by adj usting the voltage o the anode with respect to the cathode using a potential divider ( not shown on the diagram) the anode is actually made negative! When the pote ntial die rence across the tube is j ust sufcie nt to preve nt electrons rom crossing the tube , the maximum kinetic energy o the emitte d electrons is equal to eVs ( whe re Vs is the stopping
)
0
0
1 0
hc e
1
-
hc e 0
Figure 5 Plancks constant graph.
479
12
Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L )
Note Dierent metal cathodes will have dierent work unctions so they will always give graphs o the same gradient as fgure 5 but they will produce parallel lines with dierent intercepts.
The wave theory and the photoelectric eect The reason that the wave theory ails to account or the photoelectric eect is explained by the instantaneous nature o the emission that occurs when light alls onto a metal surace. Waves provide a continuous supply o energy, the intensity o which is proportional to the square o the wave amplitude. According to classical wave theory, when low intensity electromagnetic radiation o any requency is incident on a metal surace, given sufcient time, enough energy should eventually accumulate to allow an electron to escape rom its potential well. However we fnd that actually a small number o photons with requencies above the threshold will always ej ect electrons rom a metal surace immediately; below this requency, no electrons are ej ected. This is completely contradictory to wave theory. The outcome o the photoelectric eect leaves us in a slightly uncomortable position o needing waves to describe some properties o light, such as intererence and diraction, but needing particle theory to explain the photoelectric eect. This does not mean that either theory is wrong rather that both are incomplete. Light appears to have characteristics that can be attributed to either a wave or a particle we call this waveparticle duality and is another example o the outcomes o experiments ailing to be in accord with our everyday perception.
Worked example
Solution 15
Photons o requency o 1 . 2 1 0 Hz are incident on a metal o work unction o 1 . 8 eV. a) C alculate the energy transerred to the metal by each photon. b) C alculate the maximum kinetic energy o the emitted electrons in electronvolts. c) D etermine the stopping potential or the electrons.
a) Using E = hf to calculate the photon energy in j oules ( we are not asked or this to be in eV so it is fne to leave it in j oules) . E = 6.63 1 0 - 34 1 . 2 1 0 1 5 = 8.0 1 0 - 1 9 J b) We now need to work in electronvolts so the 8.0 1 0 photon energy = ________ = 5 .0 eV 1 .6 1 0 -19 -19
With = 1 .8 eV this means the maximum kinetic energy is ( 5 . 0 - 1 .8) = 3 .2 eV c) Working in electronvolts now really comes into its own since the stopping potential will be 3.2 V.
Matter waves In his 1 924 PhD thesis, the French physicist, Louis de Broglie (pronounced de broy) , used the ideas o symmetry to suggest that i something classically considered to be a wave had particle-like properties, the opposite would also be true. Matter could, thereore, also have wave-like properties. He suggested that the wavelength associated with a particle is given by h = _ p Here h is the Planck constant and p is the momentum o the particle (= mv) . This wavelength is known as the de Broglie wavelength. D e B roglie used ideas rom the special theory o relativity and the photoelectric eect in order to derive this equation or light ( you dont need to learn this derivation) .
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The total energy o an obj ect ( rom the special theory o relativity) is the _________ total o the rest energy and kinetic energy, given by E = p 2 c 2 + m 0 2 c 4 ( the frst term in the equation is a kinetic energy term and the second term is the rest mass energy term) . For___ a photon the second term is zero 2 2 ( photons have no rest mass) so E = p c or E = pc. As we have seen rom the photoelectric eect E = hc h pc = __ or = __ p
__h c equating these
With this equation derived or light, de B roglie simply speculated that the same thing would be true or matter, and very soon he was shown to be correct!
Electron difraction In 1 92 5 , two American physicists, C linton D avisson and Lester Germer, demonstrated de B roglies hypothesis experimentally by observing intererence maxima when a beam o electrons was reected by a nickel crystal. In 1 92 8, the B ritish physicist George Thomson independently repeated D avisson and Germers work at the University o Aberdeen. Figure 6( a) and ( b) shows a laboratory arrangement or demonstrating the eect using the transmission o electrons through a thin slice o crystal. E lectrons rom a heated cathode pass through a thin flm o carbon atoms ( the crystal) . I the electrons behaved like particles they would be only slightly deviated by collisions with the carbon atoms and would orm a bright region in the centre o the screen. evacuated tube crystal
power supply and connections not shown
series o bright and dark rings heated cathode difracted electron beams
(a)
(b)
Figure 6 Electron difraction tube.
The bright rings indicate where the electrons land on the screen. Where there is a bright glow there is a high probability o electrons reaching that point where there is darkness there is a low probability o the electrons reaching that point. The same pattern builds up slowly, even i there are only a ew electrons travelling in the tube at any one time. This pattern is very similar to the intererence pattern that is obtained with light using a diraction grating and can be explained by assuming that electrons behave in a similar way to waves: When electrons are accelerated through a potential dierence they gain kinetic energy 1 mv2 eV = _ 2
(
)
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) Assuming that the accelerated electrons do not travel close to the speed of light then the momentum of the electrons is given by p = mv, meaning that p 2 = ( mv) 2 2
p 1 so __ mv 2 = ___ = eV 2 2m
_____
and p = 2 meV using the de B roglie relationship h = _ p h _____ this gives = _ 2 meV
Worked example C alculate the de B roglie wavelength of electrons accelerated through a potential difference of 3 . 00 kV.
Solution In an IB question on this you would be asked to develop this equation step by step and then do this calculation. h 6.63 1 0 - 34 _____ = ____ ________________________________ = _ 2 meV 2 9.1 1 0 1 0 - 31 1 .60 1 0 - 1 9 3 000 = 2 .2 4 1 0 1 1 m
This is similar to the wavelength of the X- rays used to form diffraction patterns when they are incident on crystals. Increasing the accelerating voltage increases the energy and momentum of the electrons. The wavelength, therefore, decreases and so produces smaller diameter rings with smaller spacing between them. This is analogous to light passing through a diffraction grating: the diffraction angle ( ) in the equation n = d sin is reduced when light of a shorter wavelength is used.
Nature of science Waveparticle duality From our discussion of the photoelectric effect and electron diffraction we have seen that both electrons and photons sometimes behave as waves and sometimes as particles. This gives rise to questions such as is matter a wave or a particle? and is light a wave or a particle?. The answer to each of these questions is neither: matter is matter and light is light, therefore they are neither waves nor particles. However, in each case we need the wave model to explain some properties of each and we need the particle model to explain other properties. Nobody knows the mechanisms
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by which electrons and photons behave, because they do not enter into the realms of everyday experience. We interpret their behaviour by using mathematics ( of increasing complexity) , but it is impossible to unravel the properties of waves or electrons by relating these to everyday occurrences. That individual photons can pass through individual slits and produce a pattern consisting of regions of high photon densities and regions of low photon densities makes no sense when we think of the interference of particles yet this is what happens in the quantum world.
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The Bohr model In order to interpret the scattering o alpha particles as discussed in S ubtopic 7.3 , Niels B ohr proposed a model o an atom in which electrons could only occupy orbits o certain radii. His model was based on the ollowing three assumptions: 1
E lectrons in an atom exist in stationary states.
2
3
C ontradicting classical physics, electrons could remain in these orbits without emitting any electromagnetic radiation.
Electrons may move from one stationary state to another by absorbing or emitting a quantum of electromagnetic radiation
I an electron absorbs a quantum o radiation it can move rom one stationary state to another o greater energy; when an electron moves rom a stationary state o higher energy to one o lower energy it emits a quantum o radiation.
The dierence in energy between the stationary states is given by E = hf.
The angular momentum of an electron in a stationary state is quantized in integral values of 2h
___
This can be represented mathematically by: nh mvr = _ 2 Angular momentum is the ( vector) product o the momentum o a particle and the radius o its orbit so, or a particle in a circular orbit, the angular momentum will be constant. This assumption is equivalent to suggesting that an integral number o de B roglie-type wavelengths fts the electrons orbital: h h _ = _ p p( = mv) = The circumerence o an orbit ( o radius r) = 2 r When the number o complete waves ftting this orbit is n then each
visualization o electron waves or frst three Bohr orbits electron wave resonance n = 1, 1 = 2r1
2 r = _ n
n = 2, 2 2 = 2r2 n = 3, 3 3 = 2r3
Equating the two values or gives h 2 r _ _ p = n
n=3
or, rearranging and substituting mv or p, gives nh mvr = _ 2
n=2 n=1
This pattern is shown in fgure 7 and corresponds to standing waves.
Energies in the Bohr orbits O ne o the triumphs o the B ohr atom was that it produced an equation that agreed with the experimental equation or the spectrum o the hydrogen atom. B y measuring the total kinetic and potential energy o
Figure 7 Standing waves in an atom or n = 2 and n = 3.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) the hydrogen atom we fnd that, or an electron in the nth energy level ( where n = 1 represents the ground state, n = 2 the frst excited state, etc. and is called the p rincip al quantum number) , the total energy E in electronvolts at each level is given by:
n= n=4 n=3
E =0 E = -0.85 eV E = -1.51 eV
n=2
E = -3.40 eV
n=1
E = -13.6 eV
Figure 8 Energy levels in a hydrogen atom.
1 3.6 E = -_ n2 S ee fgure 8. This total energy is negative because the electron is bound to the nucleus and energy must be supplied to the system in order to completely separate the electron rom the proton. B ohr went on to modiy this equation so that it could accommodate other hydrogen- like ( i.e. one electron) systems such as singly ionized helium and doubly ionized lithium, etc. The model ailed, however, to be extended to more complicated systems o atoms and it could not explain why certain allowed transitions were more likely to occur than others. D espite its aws, the B ohr model proved to instigate a more undamental approach to the atom; this is now called quantum mechanics.
Worked example In his theory o the hydrogen atom, B ohr reers to stable electron orbits. a) S tate the B ohr postulate that determines which stable orbits are allowed.
When in a stable orbit an electron does not emit radiation.
b) D escribe how the existence o such orbits accounts or the emission line spectrum o atomic hydrogen.
When an electron drops to a lower energy level or orbit it emits a photon.
The requency o the emitted photon is proportional to the dierence in energy between the two levels.
A transition between two levels results in a spectral line o a single wavelength being emitted.
The B ohr model o the hydrogen atom can be extended to singly ionized helium atoms. The model leads to the ollowing expression or the energy En o the electron in an orbit specifed by the integer n. k En = _2 n where k is a constant. In the spectrum o singly ionized helium, the line corresponding to a wavelength o 3 6 2 nm arises rom ele ctron transitions betwe en the orbit n = 3 to the orbit n = 2 . c) D educe the value o k.
Solution a) The angular momentum ( mvr) o an electron in a stationary state is quantized in integral h values o ___ 2
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b) There are many things that can be said here but some of the key points are:
hc k c) E = __ = __ so the dierence in energy n between the two levels will be 2
(
)
1 1 = k _ - _ = 0.1 3 9 k 4 9 hc 6.63 1 0 - 34 3 1 0 8 k = _ = ___ 0.1 3 9 3 62 1 0 - 9 0.1 3 9 = 3 . 95 1 0 - 1 8 J
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Schrdingers equation
intensity/arbitrary units
22
Waveparticle duality explains a bright intererence ringe as being the place where there is a high probability o fnding a particle. The position o particles is described mathematically by probability waves. As with classical waves, probability waves superpose with one another to produce the expected intererence pattern. A low-intensity beam o photons incident on a single slit one at a time will build up a distribution that is identical to the expected diraction pattern, provided that we wait a sufciently long time. A similar pattern is obtained by fring electrons at a slit o suitably small width (see fgure 1 0) . From this distribution we can predict the places where electrons will not reach but we are unable to predict where an individual electron will be detected although, statistically, there will be approximately 2 2 times the number o electrons arriving at the area around the principal diraction maximum compared to the number around the secondary maximum.
1 (a)
(c)
0
(b)
1
2 3 4 5 6 difraction angle/degrees
7
Figure 9 Difraction pattern intensity distribution.
(d)
Figure 10 Difraction pattern being built up by individual electrons.
The concept o probability was developed into the quantum mechanical model o the atom in 1 92 6 by the Austrian ( and later AustrianIrish) physicist, E rwin Schrdinger. S chrdingers wave function describes the quantum state o particles. His wave equation has many similarities to a classical wave equation but it is not a derived equation. It has, however, been thoroughly verifed experimentally and its solution or the single electron hydrogen atom agrees with the B ohr relationship 1 3.6 E = - ____ . n 2
The wave unction is not a directly observable quantity but its amplitude is very signifcant. With light waves we observe the intensity, not the amplitude, and we have seen that the intensity is proportional to the square o the amplitude. For the wave unction, where the square o the amplitude is a maximum there is the greatest probability o fnding a photon. When the square o the amplitude is zero there is zero probability o fnding the photon. The quantity may be thought o as the amplitude o the de B roglie wave corresponding to a particle ( although it does not have any physical signifcance) ; however, the square o the amplitude o the wave unction 2 is proportional to the
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) probability per unit volume o fnding the particle this is known as the probability density. Mathematically we write this as: P( r) = 2 V Here P( r) is the probability o fnding a particle a distance r rom a chosen origin and V is the volume being considered. For double slit intererence, in terms o the probability wave, the wave unction is considered to be such that a single photon or electron passes through both slits and be everywhere on the screen until it is observed or measured. When this happens, the wave unction collapses to the classical case and the particle is detected. This is known as the Copenhagen interpretation, so named by the German physicist, Werner Heisenberg. It relates to the interpretation o quantum mechanics used by Heisenberg, B ohr and their co-workers between 1 92 4 and 1 92 7. It can be summarized as nothing is real unless it is observed. So matter or light can be considered to be a wave or a particle. I it behaves like a particle then it is a particle. I it behaves like a wave, then it is a wave.
energy/eV 0 n=3
n=2 n=1 -13.6
Figure 12 Electron standing waves in a potential well.
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Figure 11 The Copenhagen interpretation.
In the simplifed ( one-dimensional) version o the hydrogen atom, as shown in fgure 1 2 , an electron would be detected somewhere between the nucleus and the outside edge o the atom these are shown by the edges o a potential well. A more realistic model would show the potential 1 varying as the inverse o the distance rom the nucleus ( V - __ r ). Within the well, the electron energy must be such that the wave unction has nodes at the sides. In the electron wave model the probabilities o fnding an electron within the nucleus or outside
1 2 . 1 T H E I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N
the atom are both zero so the wave amplitude is zero at these points. The electron is most likely to be ound ( highest probability) where the amplitude is maximum; this is midway between the nodes.
Worked example The graph below shows the variation with distance r rom the nucleus o the square o the wave unction, 2 , o an electron in the hydrogen atom according to the S chrdinger theory. The nucleus is assumed to be a single point.
S tate where the electron is most likely to be ound. Explain whether the electron could be ound either in the nucleus or at positions corresponding to the largest value o r shown on the graph.
Solution
2
0 0
a
r
The position around a has the highest value o the square o the wave unction. This is the position where there is the highest probability o fnding the electron. B ecause the square o the wave unction is zero at the position o the nucleus, the electron defnitely cannot be ound there but there is a fnite chance o fnding the electron at large values o r since the graph has not allen to zero. The electrons probability cloud would be densest at a, but it will still have a little density at large values o r.
The Heisenberg uncertainty principle We have seen that when a quantum is diracted it is only possible to predict its subsequent path in terms o the probability o the wave unction. The outcome o this experiment is in line with Heisenbergs uncertainty principle. h This is usually written as x p ___ and places a limit on how precisely 4 we are able to know the position and momentum o something in the quantum mechanical realm.
I we wish to know where an electron is positioned at a given time, the principle tells us that there is an uncertainty ( given by x) with which we can know that position. I x is very small, then the uncertainty ( p) in knowing the momentum o the electron is very large. Thereore, it is not possible to precisely determine the position o the electron and its momentum at the same time; the product o the two uncertainties h will always be greater than or equal to ___ . 4 I we imagine the wave unction o a ree electron (one that is not in any feld that would change its motion) to be a sine wave then we can measure its wavelength perectly. Since we know its wavelength perectly, we also h know its momentum perectly (p = __ ) . This implies that the position o the electron has an infnite uncertainty and is spread out over all o space. In order to detect a quantum particle it would be necessary to use something that has a comparable size to that particle. Using radiation with which to detect a nucleus would need a wavelength o 1 0 1 5 m. h We know rom the de B roglie relationship ( = __ p ) that the shorter the
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) wavelength is, the greater the momentum so, with a wavelength o 6.63 1 0 1 0 1 5 m, the momentum will be _________ 1 0 - 1 8 N s. 10 - 34
Note
-15
In the limit when one o the quantities is known perectly the other quantity has an infnite uncertainty.
O n a nuclear level this is a very large momentum and would mean that any radiation o this wavelength would impart energy to the nucleus which would then make its position eectively immeasurable. In dealing with electrons diracting through a narrow gap, perhaps o size 1 0 1 8 m, the uncertainty principle also applies. In passing through the gap the uncertainty o the electrons position in the gap will be hal the gap width. This then puts a limit on the precision with which we can know the component o the momentum o the electron parallel to the gap (and h 1 1 0 - 1 6 N s and thereore its wavelength) . Thus p ____________ 4 0 . 5 1 0 h ___ -17 p 1 0 m ( or approximately ten times the gap size. That is a large uncertainty in relation to the gap size! ) -18
Worked example The diagrams show the variation, with distance x, o the wave unction o our dierent electrons. The scale on the horizontal axis in all our A
0
C
B
0
0
D
0
x
0
diagrams is the same. For which electron is the uncertainty in the momentum the largest?
x
0
x
0 x
0
Solution The square o the wave unction is proportional to the probability o fnding the electron. Each o B , C and D have positions where the square o the wave unction is greatest ( close to zero in B and centrally in C and D ) . This means that there is a high probability o fnding the electron at one position and, thereore, there is low uncertainty in ( the Heisenberg) position but, as a consequence,
there will be large uncertainty in momentum. For A there are many positions where the electron could be ound with equal probability at each o the maxima or minima there is large uncertainty in position but low uncertainty in momentum. B ecause C has the position where the electron is most well- defned it must have the largest uncertainty in its momentum.
Pair production and annihilation C lose to an atomic nucleus, where the electric feld is very strong, a photon o the right energy can turn into a particle along with its antiparticle. This could be an electron and a positron or a proton and
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an antiproton. The outcome will always be a particle and an antiparticle in order to conserve charge, lepton number, baryon number and strangeness. The particle and antiparticle are said to be a pair and the eect is known as p air p roduction. The antiparticle will have a mass equal to that o the particle meaning that the photon must have enough energy to create the masses o the two particles. The minimum energy needed to do this is given by:
time e e
-
+
E = 2 mc2 where m is the ( rest) mass o the particle/antiparticle and c is the speed o electromagnetic waves in a vacuum. Figure 1 3 shows a Feynman diagram or the production o an electronpositron pair. The gamma ray photon ( ) must have an energy o at least 1 .02 MeV ( which is twice the rest energy o an electron) . Any photon energy in excess o this amount is converted into the kinetic energy o the electronpositron pair and the original electron. Pair production can also occur in the vicinity o an orbital electron but in this case more energy will be needed as the orbital electron itsel gains considerable momentum and kinetic energy. Figure 1 4 shows pair production taking place near an atomic electron. The photon is non- ionizing and leaves no track but the newly ormed electron and positron can be seen spiralling in opposite directions in the applied magnetic feld. The recoiling electron gains a great deal o kinetic energy and this is why it hardly bends in the magnetic feld although it can be seen to bend in the same direction as the other electron. Theory shows that the threshold energy needed or this type o pair production is 4mc2 ( = 2 .04 MeV) .
Figure 13 Feynman diagram o electron positron pair production.
The equation or this interaction is: + e- e - + e - + e + When a particle meets its antiparticle they annihilate, orming two photons. The total energy o the photons is equal to the total mass-energy o the annihilating particles. S ometimes a pair o particles annihilate but then one o the photons produces another pair o particles. The positron that is ormed in the interaction, as shown in fgure 1 4, quickly disappears as it is re- converted into photons in the process o annihilation with another electron in matter.
position o original electron
newly ormed positron original electron gains signifcant kinetic energy
path o incident gamma ray photon (no track actually seen)
newly ormed electron
Figure 14 Bubble chamber tracks o electronpositron pair production.
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Pair production and the Heisenberg uncertainty principle There are pairs o variables, other than x and p, to which the uncertainty principle also applies o these energy and time are two important conj ugate variables. This version o the uncertainty principle is written as: h E t _ 4 S ome interesting things can happen in the quantum world. It is ound experimentally that the threshold energy required or the production o an electronpositron pair can be much less than the expected 1 .02 MeV in the presence o a heavy nucleus. Imagine a 1 0 eV photon in the vicinity o a heavy nucleus: the low energy photon produces an electronpositron pair. A very short while later the electron and the positron collide to produce two 5 eV photons. This looks like a huge violation o the conservation o mass- energy but it is allowable under the uncertainty principle. D uring the lietime o the composite electronpositron pair there is uncertainty regarding the total energy. I the uncertainty was equal to 1 .02 MeV, what would be the limit on the uncertainty o the lietime o the pair? The time t can be calculated by the energytime ormulation o the uncertainty principle: h 6.63 1 0 - 34 t = _ = ___ = 3 .2 1 0 - 2 2 s 1 .02 1 0 6 1 .6 1 0 - 1 9 4 E 4 Here we have converted the energy in MeV into J by multiplying by the electronic charge. This lietime is so short that a measurement o the energy o the pair would have an uncertainty o at least 1 . 02 MeV and the experiment would not be able to detect the violation o the law o conservation o energy. I we cannot perorm an experiment to detect a violation o the conservation law, then quantum mechanics says there is some probability o the process occurring. Another way o explaining this example is to say that nature will cheat i it can get away with it! This example has been verifed experimentally but the theory behind it is beyond that covered in the IB D iploma physics course.
Quantum Tunnelling According to quantum mechanics, a particles wave unction has a fnite probability o being everywhere in the universe at the same time. The probability may be infnitesimally small away rom the eect that we would expect rom classical physics but it is, nevertheless, fnite. This means that, or example, an electron in the ground state o a hydrogen atom could escape the attraction o the nucleus with less than the expected 1 3 . 6 eV. In agreement with the uncertainty principle a particle can eectively borrow energy rom its surroundings, pass through a barrier and then pay the energy back, providing it does not take too long. Figure 1 5 shows a situation in which the wave unction o the electron extends across a physical barrier giving the electron a fnite possibility o
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it existing there ( ignoring any modifcation o the probability curve by the presence o the barrier) .
probability
Quantum tunnelling is responsible or the relatively low temperature usion that occurs in main sequence stars such as the S un. The repulsive orces between a pair o protons that are to use means that they require kinetic energies o j ust over 1 MeV this requires them to be at a temperature o around 1 0 1 0 K. This is much higher than the temperature o the core o the S un ( which is around 3 1 0 7 K) . As the result o the high pressures and quantum tunnelling there is a small chance that hydrogen atoms can use at a temperature below that expected. B ecause o the immense numbers o atoms in the S un, even with a very low probability o quantum tunnelling usion occurring, there is still a great deal going on. In the case o the S un there is estimated to be in excess o our million tonnes o hydrogen using in this way in every second.
physical barrier
fnite probability displacement
Figure 15 Quantum tunnelling to pass through a physical barrier.
The scanning tunnelling microscope ( STM) , invented in 1 981 by IB M Zurich, has revolutionized the study o material suraces and has been used to manipulate individual strands o D NA thus oering the potential to repair genetic damage. STMs use the currents generated when electrons tunnel into a surace in order to map out the structure o the surace. Quantum tunnelling is currently used in quantum tunnelling composites ( QTC s) . QTC s are the basis o touch screen technology. This technology has applications in smartphones, computer tablets, cameras and monitors. The entanglement o quantum particles is starting to bear ruit in photon teleportation, quantum cryptography and computing.
Worked example The graph shows the variation with distance x o the wave unction o an electron at a particular instant o time. The electron is confned within a region o length 2 .0 1 0 1 0 m.
Solution a) The wave unction is a property o the electron everywhere in space. The square o the wave unction is proportional to the probability o fnding the electron somewhere. b) ( i) the ( de B roglie) wavelength o the electron is
0
0.0
0.5
1.0
1.5
x/10 - 10 m 2.0
2 1 0-10 _______ = 3 .3 1 0 1 1 m 6
and so h 6 .6 3 1 0 p = ( __ = _________ = ) 2.0 1 0 23 N s 3 .3 1 0 - 34
1 1
( ii) The electron is confned to 2 .0 1 0 1 0 m, this means the uncertainty in its position ( x) = 2 .0 1 0 1 0 m. a) S tate what is meant by the wave function o an electron. b) Using data rom the graph estimate, or this electron: ( i) its momentum ( ii) the uncertainty in its momentum.
Using the uncertainty principle this means that the uncertainty in momentum will be at least: 6.63 1 0 - 34 h p = _ = __ = 4 2 .0 1 0 - 1 0 4 x
(
)
2 .6 1 0 25 N s
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12.2 Nuclear physics Understanding Rutherord scattering and nuclear radius Nuclear energy levels
Applications and skills Describing a scattering experiment including
The neutrino The law o radioactive decay and the decay
constant
Nature of science Why we need particle accelerators When Rutherord, Geiger and Marsden perormed their scattering experiments they needed to use naturally occurring sources o alpha particles with relatively low energies o between 3 and 7 MeV. This is not enough energy or them to penetrate the electrostatic potential energy barrier around the nucleus. Although there are cosmic rays reaching the Earth with energies a million times greater than can be produced in a particle accelerator (10 2 1 eV compared with 10 1 5 eV on Earth) , these are unpredictable and cannot be used. In CERNs large hadron collider (LHC) the energy available has been raised to 7 TeV. By using particle accelerators, detectors and sophisticated computers the number o known particles has increased rom the proton, neutron and electron to the enormous number possible in the standard model. I our knowledge o the atom is to progress still urther, particle accelerators are the most likely way orward. Plans are already in place to construct a very large hadron collider (VLHC) with a 240 km circumerence chamber and beam energy o at least 50 TeV.
location o minimum intensity or the diracted particles based on their de Broglie wavelength Explaining deviations rom Rutherord scattering in high energy experiments Describing experimental evidence or nuclear energy levels Solving problems involving the radioactive decay law or arbitrary time intervals Explaining the methods or measuring short and long hal-lives
Equations relationship between radius o nucleus and ___ 1
nucleon number: R = R0 A 3 decay equation or number o nuclei at time t: N = N0 e - t decay equation or activity at time t: A = N0 e - t angle o electron diraction frst minimum: sin ____ D
Introduction In S ub- topic 7.3 we looked at the Rutherford model of the atom with a nucleus surrounded by orbiting electrons. We will now look more closely at the implications of the alpha scattering experiment and see how similar experiments have provided a much better understanding of the nucleus. We will then consider a more mathematical approach to the radioactive decay of nuclei.
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Rutherford scattering and the nuclear radius In S ub- topic 7.3 we saw that the main results o the alpha scattering experiment were that:
most o the alpha particles passed through the gold lea undefected
some alpha particles were defected through very wide angles
some alpha particles rebounded in the opposite direction.
The interpretations o these results were:
most o the atom is empty space
the atom contains small dense regions o electric charge
these small dense regions are positively charged.
We will now look at the analysis o this ground- breaking experiment in some detail.
The method of closest approach Alpha particles that backscatter are those colliding head- on with a gold nucleus. As only about 1 in 8000 alpha particles are scattered through large angles it must mean that the probability o a head- on collision is very small and that the nucleus occupies a very small portion o the total atomic volume ... but how much? maximum electrical potential energy
-particle
Au (79e)
path of alpha particle
rc position of closest approach
Figure 1 Method of closest approach.
Figure 1 shows an alpha particle that is incident head- on with a gold nucleus. As the alpha particle becomes closer to the nucleus its kinetic energy alls and its electrical potential energy increases. When the alpha particle is at its closest to the nucleus, its kinetic energy has allen to zero and it has momentarily stopped moving. Taking an alpha particle o kinetic energy E at the position o closest approach ( = distance rc rom the nucleus) we can equate its kinetic energy to the electrical potential energy giving: kZe 2 e E = _ rc where k is the C oulomb constant, Z is the proton number o gold ( making Ze the charge on the gold nucleus) and 2 e represents the alpha particle charge.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) S o for an alpha particle k2 Ze2 rc = _ E In Rutherfords experiment E = 7.68 MeV and Z = 79, this gives a value for rc of 8.99 1 0 9 2 79 ( 1 .60 1 0 - 1 9 ) 2 rc = ____ = 2 .96 1 0 1 4 m ( 7.68 1 0 6 1 .60 1 0 - 1 9
Note You should not conuse the Fermi radius (which is related to a nucleus) with the Bohr radius (which is related to an atom) : there is a actor o 1 0 5 diference in these values.
This derivation is an approximation because we have treated the gold nucleus as being a point mass. If the alpha particle were to have penetrated the nucleus, then the C oulomb force would not have been applicable as we now know that the strong nuclear force is dominant within the nucleus. Rutherford went on to obtain an expression for the number of alpha particles scattered through a variety of angles and he also arrived at values that were in agreement with the upper limit of the gold nucleus being approximately 3 1 0 1 4 m. However, further experiments using more energetic alpha particles have approached the nucleus closer than this value. At this separation the strong nuclear force becomes dominant and the alpha particle must have penetrated the nucleus. As the volume V of a nucleus must be proportional to the number of its nucleons, we would expect V A ( where A is the nucleon number) and __ so the nuclear radius R would be expected to be A . This gives 1 3
1 _
R = R0 A 3 Here R0 is called the Fermi radius and has a value of 1 .2 1 0 1 5 m and is measured experimentally.
Nuclear density If we imagine the nucleus to be spherical, it follows that its volume can be calculated using the following equation: 4 4 AR 3 V = _ R 3 = _ 0 3 3 The density of nuclear material will be given by M Au 3u = _= _ = _3 4 __ 3 V 4R 0 AR 0 3 where u is the uniform atomic mass and Au is the total mass of a nucleus of nucleon number A. As each of the quantities in the equation is a constant, it implies that the density of any nucleus is independent of the number of nucleons in the nucleus. S ubstituting values into this equation gives: 3 1 .66 1 0 - 27 = __ = 2 . 3 1 0 1 7 kg m 3 4 ( 1 .2 1 0 - 1 5 ) 3 This is an incredibly dense material a volume of 1 cm 3 would have a mass of 2 00 000 tonne! In nature the only obj ect that has a nuclear density of this value is a neutron star a type of stellar remnant resulting from the gravitational collapse of a massive star, which is composed almost entirely of neutrons.
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1 2 . 2 N U C L E A R P H YS I C S
Deviations rom Rutherord scattering
The method o closest approach gives an approximation o the size o a nucleus. More reliable values or the size o a nucleus can be ound using electron diraction.
repulsive 0 attractive
The scattering experiments perormed by Rutherord, Geiger, and Marsden were limited by the energies o the alpha parties emitted by the radioactive sources available to them. When their experiments are repeated using more energetic, accelerated alpha particles it is ound that, at these higher energies, the Rutherord scattering relationship does not agree with experimental results. At higher energies the alpha particles were able to approach the target nucleus so closely that the strong nuclear attractive orce overcomes the electrostatic repulsion. Figure 2 shows how the strong nuclear orce and the repulsive coulomb orce ( between protons) vary with distance.
orce between nucleons/10 - 4 N
+3.0
coulomb repulsion between two protons 1.0
2.0
3.0
nucleon separation/m
strong nuclear orce between two nucleons
-3.0 equilibrium position or protons sum o two orces = zero
Figure 2 Variation o the strong nuclear orce and coulomb orce with distance.
Electron difraction As electrons are leptons ( and not hadrons) they are not aected by the strong nuclear orce but are aected by the charge distribution o the nucleus. High-energy electrons have a short de B roglie wavelength o the order o 1 0 1 5 m. As this is also the order o magnitude o the size o a nucleus, it means that diraction analogous to that observed with light incident on a narrow slit or small obj ect can be observed. For light incident on a small circular obj ect o diameter D, the angle that the frst diraction minimum makes with the straight- through position ( = 0) is given by sin _ D
where is the wavelength o the light
The elastic scattering o high energy electrons by a nucleus produces a similar eect. With the arrangement shown in fgure 3 ( a) , the intensity o the diracted beam is seen to be a maximum in the straight-through position, alling to a minimum beore slightly increasing again. The minimum diers rom light because it never reaches zero or scattered electrons. Again, the relationship can be approximated by
Note With angles greater than 10 the small angle approximation that sin cannot be applied to the electron scattering.
sin _ D Here D is the nuclear diameter and is the de B roglie wavelength o the electrons. electron intensity
electron beam detector thin metal sample in a vacuum amplifer and meter (a) outline o experiment
min angle o diraction (b) typical results
Figure 3 Experimental arrangement and results.
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) To achieve an appropriate de B roglie wavelength the electrons used in this scattering experiment need energies in the region o 400 MeV. At these energies, electrons are travelling close enough to the speed o light to mean that relativistic corrections should be applied to both the electron momentum and the associated wavelengths. With the electron rest energy o approximately 0.5 MeV, the total energy can be taken as 400 MeV without any serious error consideration. The wavelength o an electron is given by hc = _ E For an electron with energy 400 MeV 6.63 1 0 - 34 3 .00 1 0 8 = ___ = 3 .1 1 0 - 1 5 m 400 1 0 6 1 .60 1 0 - 1 9 This is the same order o magnitude as the size o a nucleus.
Worked example The nuclear radius o calcium- 40 has an accepted value o 4.5 4 m. It is being investigated using a beam o electrons o energies 42 0 MeV. a) How closely does the value o the radius o calcium-40, obtained using the relationship __ = R 0 A , agree with the accepted value? 1 3
b) C alculate the de B roglie wavelength o an electron having energy 42 0 MeV. c) Determine the angle that the frst minimum in the diraction pattern makes with the straightthrough direction. d) Explain why 5 0 MeV electrons would be unlikely to provide reliable results in this experiment.
Solution a) A is the nucleon number that, or calcium- 40, ___ __ 3 is 40. Thus A = 40 = 3 . 42 which gives a value or R o 3 . 42 R 0 = 3 .42 1 .2 m = 4.1 0 m. 1 3
There is an error o 0. 44 m between the accepted value and that obtained using
the relationship given ( 4.5 4 4. 1 0) . As a 0 . 44 = 9.7% 1 0% . percentage this is = ____ 4. 5 4 hc 6. 63 1 0 - 34 3 . 00 1 0 8 b) = _ = ___ E 42 0 1 0 6 1 .60 1 0 - 1 9 = 2 . 9 1 0-15 m c) Using sin __ sin - 1 ( __ ) D D
Thus, as the radius is 4.5 4 m, the diameter is 9.08 m giving 2 . 9 1 0-15 sin - 1 __ = 1 8. 6 9. 08 1 0 - 1 5 d) As the de B roglie wavelength is given by hc = __ , 5 0 MeV electrons would have a de E 0 B roglie wavelength o 42 = 8.4 times greater 50 than that o 420 MeV electrons. This means that
___
(
2 . 4 1 0-14 sin - 1 __ 9.08 1 0 - 1 5
)
This is not calculable as sin cannot be greater than one. The de B roglie wavelength is too long to be diracted by a calcium nucleus.
Using electrons of higher energies When electrons o much greater energies than 42 0 MeV are used in scattering experiments, something very dierent happens. The collisions are no longer elastic ( the bombarding electrons lose kinetic energy) . This energy is converted into mass as several mesons are emitted rom the nucleus. At still higher energies, the electrons penetrate deeper into the nucleus and scatter o the quarks within protons and neutrons this is
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known as deep inelastic scattering and provides direct evidence for the quark model of nucleons.
Energy levels in the nucleus Much of our evidence for the nucleus having energy levels comes from the radioactive decay of nuclides. The emission of gamma radiation is analogous to the emission of photons by electrons undergoing energy level transitions. The emission of alpha or beta particles by radioactive parent nuclei often leaves the daughter nucleus in an excited state. The daughter nucleus then emits one or more gamma ray photons as it reaches the ground state. Figure 4 shows some of the decay routes of americium-2 41 . E ach nucleus emits an alpha particle having one of a number of possible energies ( three of which are shown) to become a nucleus of neptunium- 2 3 7. D epending on the energy of the emitted alpha particle, the neptunium nucleus can be in the ground state or an excited state. From this it will decay into the ground state by emitting a single gamma photon or, when it decays in two steps, two photons. From the differences in the energies of the alpha particles we can see that the energy level E1 will be (5 .5 45 5 .486) = 0.05 9 MeV above the ground state (E0 ) and energy level E2 will be (5 .5 45 - 5 .443) = 0.1 02 MeV above the ground state. From the differences in the energy levels we can calculate the energies of each of the three gamma ray photons that could be emitted they will be 0.1 02 MeV, 0.05 9 MeV, and 0.043 MeV. 241 Am 95
5.443 MeV 5.486 MeV
E2
5.545 MeV
E1
237 Np 93
E0 ground state
Figure 4 Decay of americium-241.
As we see from the decay of americium, the energies of the alpha particles are also quantized and provide evidence for the nucleus having energy levels. The mechanism by which the alpha particle leaves the nucleus is more complex than that producing the emission of the gamma ray photons. Alpha particles form as clusters of two protons and two neutrons inside the nucleus well before they are emitted as alpha particles. The nucleons are in random motion within the nucleus but their kinetic energies are much smaller than those needed to escape from the nucleus. This is because the strong nuclear force provides a potential energy barrier which the alpha particle needs to overcome before it can escape from the nucleus ( when the electrostatic repulsion will ensure that it accelerates away from the nucleus) . From a classical
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) mechanics point o view the alpha particle simply should not leave the nucleus ( see fgure 5 ) . From the quantum mechanical standpoint, the wave unction or the alpha particle is not localized to the nucleus and allows an overlap with the potential energy barrier provided by the strong nuclear orce. This means that there is a fnite but very small probability o observing the alpha particle outside the nucleus. Although the probability is extremely small, some alpha particles will tunnel out o the nucleus. Experimentally it is ound that, with a higher potential barrier and greater thickness to cross, a nucleus will have a longer lietime. This explains the very long hal- lives o uranium and polonium. The Russian- born American physicist, George Gamow, was the frst person to describe alpha decay in terms o quantum tunnelling ( discussed in S ub- topic 1 2 . 1 ) . When the wave unction is at its maximum the probability o tunnelling is greatest, meaning that alpha particles with specifc energies are most likely to be emitted. energy energy
kinetic energy o uppermost cluster
wave unction outside nucleus has less momentum and longer wavelength
alpha particle cluster with insufcient kinetic energy to penetrate the potential energy barrier
distance
distance
wave unction in potential energy well has large kinetic energy (and momentum) and short wavelength
edge o nucleus where nucleons are tightly bound by the strong nuclear orce - inside the nucleus there are nucleons with a variety o kinetic energy
Figure 5 Classical mechanics view o alpha decay.
Figure 6 Quantum tunnelling view o alpha decay.
Negative beta decay We looked at beta decay in S ub- topic 7.1 and returned to it in S ubtopic 7.3 when it was explained that an anti- neutrino accompanies the electron emitted in negative beta decay. The frst theory o beta decay was proposed in 1 93 4 by Fermi; at this time the existence o quarks was unknown and neutrinos were hypothetical. Experiments show that beta particles emitted by a source have a continuous energy spectrum and are not o discrete single energy as are alpha particles and gamma photons. Figure 7 shows a typical negative beta- energy spectrum.
intensity
energy spectrum o beta decay electrons rom 210 Bi
0
0.2
0.4 0.6 0.8 1.0 kinetic energy, MeV
Figure 7 Negative beta-energy spectrum.
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1.2
Possible explanations or this spectrum were that massenergy and momentum were not conserved in beta decay. These were very unlikely solutions since both o these principles are considered to be undamental to physics. Pauli suggested that, i a third particle was to be emitted in the decay, not only would this solve the massenergy and momentum problems but it would also allow spin angular momentum to be conserved in the emission. The emission (o what has proved to be) an electron antineutrino meant that or a particular nucleus the energy would be shared between the electron (the beta particle) and the antineutrino.
1 2 . 2 N U C L E A R P H YS I C S
TOK Awe and wonder or turn-of? Rumour has it that Murray Gell-Mann, when searching or a name to call what is now known as the quark, had a epiphany upon seeing the word quark in the novel Finnegans Wake by the Irish novelist James Joyce. The poet, John Updike, on reading about neutrinos chose to write the ollowing:
Cosmic Gall Neutrinos, they are very small. They have no charge and have no mass And do not interact at all. The earth is just a silly ball To them, through which they simply pass, Like dustmaids down a draughty hall Or photons through a sheet o glass. They snub the most exquisite gas, Ignore the most substantial wall, Cold shoulder steel and sounding brass, Insult the stallion in his stall, And scorning barriers o class, Infltrate you and me! Like tall And painless guillotines, they all Down through our heads into the grass. At night they enter at Nepal And pierce the lover and his lass From underneath the bed you call It wonderul; I call it crass. Telephone Poles and Other Poems, John Updike, (Knop, 1960)
Figure 8 Super-Kamioka Neutrino Detection Experiment Mount Kamioka near Hida, Japan.
Updike clearly was not a believer in the neutrino. Does he have a strong case or his disbelie? Is the concept o something as challenging as the neutrino inspirational or is it too anciul to be credible?
The law of radioactive decay As we saw in S ub-topic 7.1 , radioactive decay is a random and unp redictable process. There is no way of telling which nucleus in a sample of material will decay next. What we do know is that the more radioactive nuclei present, the greater the probability of some decaying. With a sample of many millions of nuclei the rules of statistics can be applied with a virtual certainty. The probability that an individual nucleus will decay in a given time interval (of one second, one minute, one hour, etc.) is known as the decay constant, . The units for are time 1 ( s 1 , minute 1 , h 1 , etc. ) . The activity of a sample A is the number of nuclei decaying in a second it is measured in becquerel (B q) . In a sample of N undecayed
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L ) nuclei, the activity will be equal to the number of nuclei present multiplied by the probability that one will decay in a second. In equation form: A = N N will decrease with time; this relationship is often written with a minus sign ( although this is not the case on the IB D iploma Programme physics syllabus)
Note Again, we can make use iteration to avoid using calculus when solving the decay equation. The ollowing fow chart shows an algorithm or doing this:
Using calculus notation, this is can be written as: dN _ = - N dt This is the relationship that gives an exponential decay and is analogous to capacitor discharge as seen in S ub- topic 1 1 .3 with the constant being analogous to - 1 .
n=0 choose t, tn , Nn ,
In general, when the rate of change of a quantity is proportional to the amount of the quantity left to change, an exponential relationship will always be obtained. We will now show this for radioactive decay the process will involve integral calculus and so you will not be tested on this in examinations.
Nn = - Nn t N n+1 = N n + Nn
A = - N
n
n+ 1
Rearranging the equation dN _ = - N dt
t n+1 = t n + t
gives enough increments? yes fnish
no
dN _ = - dt N this can be integrated from time t = 0 to time t = t when the number of undecayed nuclei will fall from N = N0 to N = N N
N0
t
dN _ = N
dt 0
Integrating this gives [ln N] NN = - t 0
So ln N - ln N0 = - t or
( )
N ln _ = - t N0 Raising both sides to the power of e gives N _ = e - t or N = N0 e - t N0 As A is proportional to N, this equation can be written in terms of the activity to give A = A 0 e - t Here A 0 is the activity of sample of radioactive material at time t = 0 and this can also be written as A = N0 e - t
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Worked example 1 mol of a substance contains 6.02 1 0 23 particles so this quantity amounts to
Radium-2 2 6 emits alpha particles. The decay constant is 1 .3 5 1 0 1 1 s 1 What mass of radium 2 2 6 is needed to give an activity of 2 2 00 B q?
1 .63 1 0 1 4 __ = 2 . 7 1 0 - 1 0 mol 6.02 1 0 23 As a mol of radium 2 2 6 has an approximate mass of 2 2 6 g the sample has a mass of
Solution 2 2 00 A A = N so N = _ = 1 . 63 1 0 1 4 = __ 1 .3 5 1 0 - 1 1
2 .7 1 0 - 1 0 2 2 6 1 0 - 3 kg = 6.1 1 0 - 1 1 kg
Decay constant and half-life We dealt with integral numbers of half-lives in S ub- topic 7.1 . Now we consider cases where the number of half-lives is not a whole number. The half-life is the time that it takes for the number of radioactive nuclei N to halve. So, in this time, N falls from N0 to __ . 2 0
- t N S ubstituting these values into N = N0 e - t gives __ = N0 e where t1 /2 is 2 the half-life. 0
1 __ 2
When we rearrange this and take logs to base e we get
( ) N0 __
( )
2 1 ln _ = ln _ = - t1 /2 N0 2 1 C alculating log to base e of __ gives 2
- 0.693 0.693 t1 /2 = _ = _ -
Worked example A laboratory prepares a 1 0 g sample of caesium- 1 3 4. The half- life of caesium- 1 3 4 is approximately 2 . 1 years. a) D etermine, in s 1 , the decay constant for this isotope of caesium. b) C alculate the initial activity of the sample. c) C alculate the activity of the sample after 1 0.0 years.
Solution 0.693 0.693 a) = _ = ___ t 1 /2 2 .1 3 65 2 4 3 600 = 1 .05 1 0 - 8 s - 1 ( = 0.3 3 y- 1 ) 1 34
b) 1 mol of C s has a mass of approximately 1 3 4 g so 1 0 g 10 10 comprises of _______ mol or = 7. 5 1 0 - 8 mol. 1 34 -6
This means that there are 7.5 1 0 - 8 6. 02 1 0 23 = 4. 49 1 0 1 6 atoms of C s A = N = 1 .05 1 0 - 8 4. 49 1 0 1 6 = 4. 7 1 0 8 B q c) A = A 0 e - t = 4.7 1 0 8 e - 0.33 1 0 ( working in years) A = 1 .7 1 0 7 B q
Note A second way to tackle this is to calculate the number of half10 lives that have elapsed = ______ 2.1 = 4.76 The amount that will be left is then 0.5 4.7 6 = 0.037, the activity will then be 0.037 4.7 10 8 = 1.7 1 0 7 Bq
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L )
Investigate! Measuring short half-lives The GM tube was discussed in Sub-topic 7.1 . This is a very important instrument used to measure hal-lives o radioactive materials. For those with airly short hal-lives (rom a ew seconds to a ew hours) it is a straightorward process to measure the count rate using a GM tube and counter or a data logger.
Measure the background count rate with your apparatus with no radioactive sources in the vicinity.
Position the source close to the window o the G- M tube so that almost none o the radiation is absorbed by the air.
Take readings o the count rate at appropriate time intervals until the count rate is the same as the background count.
S ubtract the background count rate rom your readings to give the corrected count rate ( R) ; assuming that the radiation is emitted equally in all directions, the count rate shown on the counter will be proportional to the activity o the source.
Plot a graph o the natural log o the corrected count rate, ln( R) , against time.
As R A we can write R = R 0 e - t Taking natural logs o this gives ln R = ln R0 - t So a graph o ln R ( on the y- axis) against t will be o gradient - and intercept on the ln R axis o ln R0 as shown in fgure 9.
Note the notation or the unit o a log quantity logs have no units but R does so the unit is bracketed to R. The corrected count rate is in counts per second, as count has no unit this is equivalent to s 1 .
What is the advantage o plotting this log graph when compared with plotting count rate against time?
How do you use the gradient to calculate the hal- lie o the radioactive nuclide?
C ompare your value with the accepted value to give you an idea o the uncertainty.
What is the dierence between the activity o source and the count rate?
In (R/s -1 ) ln(R0 )
t/s
Figure 9 Graph of natural log of the corrected count rate against time.
Measuring long half-lives S ome nuclides have very long hal- lives, or example uranium- 2 3 8 has one o j ust under 4.5 billion years. When a radioactive nuclide has a hal-lie that is long compared to the time interval over which radioactive decay observations are possible, there is no apparent rate o decay and it is not possible to measure the hal- lie in the manner suggested using a GM tube.
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In these cases a pure sample o the nuclide in a known chemical orm needs to be separated, its mass measured and then a count rate taken. From this reading the activity can be calculated by multiplying the count rate by the ratio: area o sphere o radius equal to the position o the G- M tube window _______ area o G- M tube window The decay constant is then determined rom the mass o the specimen using the method shown in the frst worked example in this section.
Investigate! Decay in height of water column setting so, i the clip is not ully undone, you will need to count how many turns you make rom the clip being ully tight.
Perspex tube with open top
As the water level in the long tube reaches a previously chosen point near the top o the tube, start measuring the height at regular time intervals you will need to perorm a trial experiment to allow you to j udge what a sensible time interval should be one that will give you a minimum o six points on your graph.
Record your results in a table.
Repeat the readings twice more always starting the clock when the water level is at the same mark on the scale.
Find the average o the readings or each time interval.
Plot graphs o height against time and the natural log o height against time.
From each o these graphs calculate the hallie and decay constant o the water.
D etermine which o the two values is the most reliable.
Consider how this experiment models radioactive decay and in which ways it is dierent.
height metre ruler
adjustable clip
capillary tube rubber tube
Figure 10 Decay in height of a water column.
This is an analogue o radioactive decay using a long vertical tube o water, which is connected to a capillary tube.
Make sure the clip is closed, then fll the long tube almost to the brim with water. Undo the clip completely to allow the water to ow through the capillary tube you will need to repeat the experiment with the same
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Q U AN T U M AN D N U C L E AR P H YS I C S ( AH L )
Questions 1
When light is incident on a metal surace, electrons may be ej ected. The ollowing graph shows the variation with requency f o the maximum kinetic energy Emax o the ej ected electrons.
( i) S tate the connection between photon energy and the energy o the emitted electron. ( ii) Using your answer to ( i) calculate the work unction o the surace o the metal plate.
Emax
0
( 8 marks)
3 0
f
a) Sketch the graph obtained or the variation with requency o the maximum kinetic energy o the emitted electrons when a dierent metal with a lower threshold requency is used. b) Explain why you have drawn the graph in this way. ( 4 marks)
2
(IB) In order to demonstrate the photoelectric eect, the apparatus shown below is used.
a) S tate one aspect o the spectrum o atomic hydrogen that B ohrs model did not explain. B ohr proposed that the electron could only have certain stable orbits. These orbits are specifed by the relation nh mvr = _ 2
with n = 1 , 2 , 3
where m is the mass o the electron, v its speed, r the radius o the orbit and h the Planck constant. This is sometimes known as B ohrs frst assumption.
B y using Newtons second law and C oulombs law in combination with the frst assumption, it can be shown that
glass tube vacuum metal plate V
Monochromatic light is incident on the metal plate. The potentiometer is adj usted to give the minimum voltage at which there is zero reading on the microammeter. a) State and explain what change, i any, will occur in the reading o the microammeter when: ( i) the intensity o the incident light is increased but the requency remains unchanged ( ii) the requency o the light is increased at constant intensity. b) For light o wavelength 5 40 nm, the minimum reading on the voltmeter or zero current is 1 .9 V.
504
In 1 91 3 Niels Bohr developed a model o the hydrogen atom which successully explained many aspects o the spectrum o atomic hydrogen.
b) S tate a second assumption proposed by B ohr.
A
monochromatic light
(IB)
n2h2 r= _ 4 2 mke2
1 . where k = _ 4 0
It can also be shown that the total energy En o the electron in a stable orbit is given by ke2 En = _. 2r c) Using these two expressions, deduce that the total energy En may be given as K where K is a constant. En = _2 n d) S tate and explain what physical quantity is represented by the constant K. e) O utline how the S chrdinger model o the hydrogen atom leads to the concept o energy levels. ( 1 0 marks)
QUESTION S 4
d) The radius ( in metre) R o a nucleus with nucleon number A is given by
(IB) A beam o electrons is incident normally to the plane o a narrow slit as shown below.
1 _
R = 1 .2 1 0 - 1 5 A 3 . ( i)
slit
beam of electrons
x
State in terms o the unifed atomic mass unit u, the approximate mass o a nucleus o mass number A.
( ii) The volume o a sphere o radius R 4 R . D educe that the is given by v = ____ 3 density o all nuclei is approximately 2 1 0 1 7 kg m 3 . 3
( 8 marks)
The slit has width x equal to 0. 01 mm. As an electron passes through the slit, there is an uncertainty x in its position. a) C alculate the minimum uncertainty p in the momentum o the electron. b) Suggest, by reerence to the original direction o the electron beam, the direction o the component o the momentum that has the uncertainty p. ( 3 marks)
6
(IB) A nucleus o the nuclide xenon, Xe- 1 3 1 , is produced when a nucleus o the radioactive nuclide iodine, I-1 3 1 decays. a) Explain the term nuclide. b) C omplete the nuclear reaction equation or this decay. 1 3 1 I 1 3 1 Xe + - + 54
5
(IB) An - particle approaches a nucleus o palladium. The initial kinetic energy o the - particle is 3 .8 MeV. The particle is brought to rest at point P, a distance d rom the centre o the palladium nucleus. It then reverses its incident path. palladium nucleus
-particle P d
a) C alculate the value, in j oules, o the electric potential energy o the - particle at point P. Explain your working. b) The proton number o palladium is 46. C alculate the distance d.
c) The activity A o a reshly prepared sample o I- 1 3 1 is 6.4 1 0 5 B q and its hal-lie is 8.0 days. ( i)
Sketch a graph to show the variation o the activity o this sample over a time o 25 days.
( ii) D etermine the decay constant o the isotope I- 1 3 1 ( in day - 1 ) . The sample is to be used to treat a growth in the thyroid o a patient. The isotope should not be used until its activity is equal to 0.5 1 0 5 B q. ( iii) C alculate the time it takes or the activity o a reshly prepared sample to be reduced to an activity o 0.5 1 0 5 B q ( 1 1 marks)
c) Gold has a proton number o 79. Explain whether the distance o closest approach o this - particle to a gold nucleus would be greater or smaller than your answer in ( b) .
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(IB)
( iii) Use the following data to determine the maximum energy, in J, of the
a) A stable isotope of argon has a nucleon number of 3 6 and a radioactive isotope of argon has a nucleon number of 3 9.
particle in the decay of a sample of argon- 3 9.
( i)
S tate what is meant by a nucleon.
Mass of argon-39 nucleus = 38.96431 u
( ii)
O utline the quark structure of nucleons.
Mass of K nucleus = 3 8. 963 70 u
( iii)
c) The half- life of argon- 3 9 is 2 70 years. ( i)
S uggest, in terms of the number of nucleons and the forces between them, why argon- 3 6 is stable and argon- 3 9 is radioactive.
( ii) Explain how you would calculate the half-life using the quantities you have stated in ( i) .
b) Argon- 3 9 undergoes decay to an isotope of potassium ( K) . The nuclear reaction equation for this decay is
39 18
State what quantities you would measure to determine the half-life of argon- 3 9.
( 2 1 marks)
Ar K + - + x
( i)
S tate the proton number and the nucleon number of the potassium nucleus and identify the particle x.
( ii)
The existence of the particle x was postulated some years before it was actually detected. Explain the reason, based on the nature of energy spectra, for postulating its existence.
8
(IB) a) O utline a method for the measurement of the half-life of a radioactive isotope having a half- life of approximately 1 0 9 years. b) A radioactive isotope has a half-life T1 /2 . D etermine the fraction of this isotope that remains in a particular sample of the isotope after a time of 1 .6 T1 /2 . ( 5 marks)
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A RELATI VI TY Introduction Earlier we described the rules of classical mechanics as developed by Newton and others. These rules provided the basis for physics for 3 00 years. Ultimately, however, Newtonian and Galilean mechanics struggles: it cannot deal with things that move very fast or with obj ects that are
very small. The insights of Einstein and others around the beginning of the twentieth century enabled physicists to change their understanding of time and space this type of change is called a paradigm shift. This topic describes our present view of what we now called spacetime.
A.1 The beginnings of relativity Understandings Reerence rames Galilean relativity and Newtons postulates
concerning time and space Maxwell and the constancy o the speed o light Forces on a charge or current
Nature of science Einsteins great insight was to realize that the speed o light is constant or all inertial observers. This has enormous consequences or our understanding o space and time. A paradigm shit occurred and the Newtonian view o time was overturned. There are many other examples o paradigm shit in science but perhaps none quite as proound as this.
Applications and skills Using the Galilean transormation equations Determining whether a orce on a charge or
current is electric or magnetic in a given rame o reerence Determining the nature o the felds observed by dierent observers
Equations Galilean transormation equations:
x' = x - vt u' = u - v
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Reference frames A reerence rame allows us to reer to the position o a particle. Reerence rames consist o an origin together with a set o axes. In this course we generally use the Cartesian reerence rame, in which position is defned using three distances measured along axes that are at 90 to each other. The axes o a three-dimensional graph make up a Cartesian reerence rame. O ther rames are, o course, available. Sailors use latitude and longitude; these together with the distance o an obj ect rom the centre o the Earth constitute a dierent reerence rame. Astronomers usually use angles when defning the position o a star that is being observed; they only need two angles because the distance to the star ( or observational purposes) is irrelevant. For some o the rames o reerence in use, one or more o Newtons laws o motion do not hold. Our own rame on the surace o a rotating planet shows this straight away. Everything o-planet appears to be spinning around us. An object that is at rest relative to us on planet Earth is not moving at a constant velocity at all. This has consequences that we have already seen in earlier topics; the fctitious centriugal orce and the Coriolis orce used to explain the movement o weather systems are two cases in point. For Newtons frst law to be valid we need to be careul about the nature o the reerence rame in which we use the law. We defne an inertial frame of reference as a rame in which an object obeys Newtons frst law: so that it travels at a constant velocity because no external orce acts on it. Do inertial rames exist? The best way to fnd one is to take a spaceship out into deep space, well away rom the gravitational eects o planets and stars, and then turn o the engines. No orces act rom outside or inside the spacecrat and this will be a true inertial rame o reerence.
Galilean relativity and Newtons postulates E ven though we have to take some trouble to reach one, there are an infnite number o inertial rames o reerence in the universe and there are a number o ways in which we can move between them. y
y'
y
The frst obvious way is to step sideways rom one rame to another. This is known as a translation ( fgure 1 ( a) ) .
The set o axes can be rotated to orm another set. This is known as a rotation (fgure 1 (b) ) . We shall not consider these in detail in this course.
O ne rame can move relative to another rame with a constant relative velocity. This is known as a boost. ( fgure 1 ( c) )
y'
x'
X
x'
x'
x (a) y
x (b)
x
later time t
y'
t=0 vt
v x'
x x x = vt + x'
x'
(c)
Figure 1 Translations, rotations, and boosts.
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It is easy to see that i an object moves with constant velocity in one reerence rame then under any o these three conditions it will be measured as having a constant (but dierent) velocity in the other reerence rame. The principle o relativity is oten associated with Albert Einstein. In act, Galileo was probably the frst person to discuss the principle. He describes how, in a large sailing ship, butteries in a cabin with no windows would be observed to y at random whether the ship were moving at constant velocity or not. An observer in the cabin could not deduce by observing
A . 1 T H E B E G I N N I N G S O F R E L AT I V I T Y
the butteries whether the ship was moving. And the butteries would certainly not be pinned against the back wall o the cabin. This principle o relativity as described by Galileo tells us about the nature o our universe.
The translation rule is equivalent to saying that there is no special place in the universe; position is relative.
The rotation rule says that there is no special direction; direction is relative.
The boost rule says that there is no special velocity. Stationary is not an absolute condition and any obj ect can be described as stationary with respect to another obj ect. ( As we shall see, the disagreement between Galileos ideas and those o Einstein is crucial here.)
Thinking, or a translation, in terms only o the x-direction (that is, in one dimension) , i the distance between the origins o two inertial rames S' and S is X then a position x in S is related to position x' in rame S' by x' = x - X For a boost, i one inertial rame moves relative to the other by a constant relative velocity v along the x-axis, then the distance between the origins o the reerence rames must be changing by v every second and this distance is vt where t is the time since the rames coincided. Imagine that the origins o the two reerence rames S and S' ( Figure 1 (c) ) were at the same position ( that is, they were coincident) at time t = 0. Technically, we say that clocks in the rames were adjusted so that x = x = 0 when t = t = 0. At a later time t, the origins o S and S' will be separated by vt where v is the velocity o rame S' relative to rame S. Thereore a position x in rame S will be related to position x in S' by x = x' + vt and x' = x - vt The velocities also transorm in an obvious way. I the velocity in S is u and the velocity in S' is u' , then u' = u - v This set o equations that link two reerence rames by their relative velocity are known as the Galilean transformations. Newton developed Galileos ideas urther in his Principa Mathematica by suggesting two important postulates ( a postulate is an assertion or assumption that is not proved and acts as the starting point or a proo) :
Newton treated space and time as fxed and absolute. This is implied in our use o t in both equations above (t' does not appear, only t) . A time interval between two events described in rame S is identical to the time interval between the same two events as described in rame S' . The evidence o our senses seems to confrm this (but remember that we do not travel close to the speed o light in everyday lie) . Newton recognized that two observers in separate inertial rames must make the same observations o the world. In other words, they will both arrive at the same physical laws that describe the universe.
Nature of science The way Newton put it. These postulates were expressed somewhat dierently by Newton in the Principa Mathematica the book ( in Latin) that he wrote to publish some o his discoveries. In translation his postulates were: I. Absolute, true, and mathematical time, o itsel, and rom its own nature, ows equably without relation to anything external . II. Absolute space, in its own nature, without any relation to anything external, remains always similar and immovable. ( Translation by Mottes ( 1 971 ) , revised by C aj orio, University o C aliornia Press.)
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Worked example 1
Solution
In a laboratory an electron travels at a speed o 2 1 0 4 m s 1 relative to the laboratory and another electron travels at 4 1 0 4 m s 1 relative to the laboratory. Use a Galilean transormation to calculate the speed o one electron in the rame o the other when they are travelling a) in opposite directions b) in the same direction.
a) The closing speed o the electrons is 2 1 0 4 ( 4 1 0 4) = 6 1 0 4 m s 1 b) The relative speed o one electron relative to the other is 2 1 0 4 ( 4 1 0 4) = 2 1 0 4 m s 1
TOK
Charges and currents a puzzle
Changing perspectives
The choice o inertial rame can make a radical dierence to the perception o a situation.
In Topic 2 we began by describing Newtons second law o motion in a simple way as: orce = mass acceleration Later we showed that this was better expressed as
C onsider a positively- charged particle moving initially at velocity v some distance away rom and parallel to a wire carrying a current. In the wire the electrons are also moving with speed v. The positive charges in the wire are stationary. We will use two inertial reerence rames in this example. O ne rame is at rest relative to the positive charges in the wire. The other rame is at rest relative to the moving charge q. (a)
(b)
orce = rate o change o momentum A similar change o expression is possible here: Newtons frst law is usually given as a variant o Every object continues in its state o rest or uniorm motion unless net external orces act on it (and you should continue to use this or a similar wording in your own work) . But this is not the only possibility. A succinct and interesting way to express Newtons frst law is as Inertial rames exist. To what extent do concise orms o scientifc laws help or hinder our understanding?
v
v
+q +
v
+q
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Figure 2 Moving charges in reference frames. To an observer at rest ( fgure 2 ( a) ) with respect to the stationary positive charges the situation in the wire seems clear. The numbers o negative and positive charges in the wire are equal and thereore the lone moving charge + q does not experience an electrostatic ( electric) orce. The movement o the electrons in the wire to the right ( a conventional current to the let) gives rise to circular magnetic feld lines centred on the wire and going into the page above the wire. The single moving charge is thereore moving perpendicularly with respect to this feld. Flemings let-hand rule indicates that there is a magnetic orce acting perpendicularly outwards on the charge and thereore the charge will be accelerated in this direction. The stationary observer detects a magnetic repulsive orce acting between the charge and the current-carrying wire. What is the situation rom the point o view o the moving charge q? An observer moving with the charge ( fgure 2 ( b) ) sees the single charge + q as stationary; electrons in the wire that also appear to be stationary, but the positive charges in the wire appear to be moving to the let at speed v. We could imagine that the moving positive charges lead to a
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A . 1 T H E B E G I N N I N G S O F R E L AT I V I T Y
magnetic feld around the wire j ust as beore but this does not help! To the moving observer the charge + q is stationary relative to the resulting magnetic feld and thereore no magnetic orce should arise. What happens is that ( as we shall see later) the relative movement o the positive charges in the wire leads to a contraction in the spacing o these charges as perceived by the observer moving with + q. There are more positive charges than negative charges per unit length as detected by the moving observer and so there is eectively a net repulsive orce acting on + q ( we shall see how this change arises rom length contraction in a later sub- topic) . Now the observer moving with q explains the orce acting on q as electrostatic in origin, not magnetic as beore. However, both observers report a orce and in both cases the orce acts outwards rom the wire. The physical result is the same even though the explanations dier. A mathematical analysis rom the standpoint o the two inertial rames also confrms that the magnitude o the orce is identical in both cases. Another situation is that o two point electric charges moving in parallel directions at the same speed as each other. This case (fgure 3) is not identical to that o the charge moving near a current-carrying wire. There is no balance o positive and negative charges to complicate matters this time. Again, we consider the situation rom the standpoint o two separate inertial rames: that o an observer stationary relative to the point charges, and that o an observer moving at a dierent velocity relative to the charges. Observations made in the rame o the point charges are easier to understand (fgure 3(a) ) : the observer moving with the charges sees two positive charges that repel and no magnetic attraction between the two. This observer describes the repulsion as purely electrostatic.
v
+q
+q
(a)
stationary, observes a magnetic feld due to both charges moving
+q
+q
(b)
Figure 3 Two point charges moving parallel.
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TOK Validating a paradigm shift The development of Einsteins relativity ideas required a shift in the view that scientists took of the physical rules that govern the universe. How do scientists ensure that the need to shift perspectives is valid?
TOK Maxwells advances Einstein wrote: The precise ormulation o the timespace laws was the work o Maxwell. Imagine his eelings when ...equations he had ormulated proved to him that electromagnetic felds spread in the orm o polarised waves, and at the speed o light!... it took physicists some decades to grasp the ull signifcance o Maxwell's discovery, so bold was the leap that his genius orced upon the conceptions o his ellow workers. (Albert Einstein, Science, May 24, 1940) To what extent do you think this is true?
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For an observer no longer stationary relative to the point charges ( fgure 3 ( b) ) , the situation is changed. From this point o view, the repulsive electric feld is increased ( through relativistic length contraction) . There is also a magnetic feld that was not apparent when the observer was stationary relative to the charges. This is because a moving charge gives rise to a magnetic feld. E ach charge now appears to be moving within the magnetic feld due to the other charge and consequently there is an attraction that the observer describes as magnetic in origin. There is both an increased ( electrostatic) repulsion and a new ( magnetic) attraction compared with the stationary observer rame. Again, a mathematical analysis shows that the orce between the charges is identical or all observer inertial rames o reerence. This is what we expect given that all physical systems observed in inertial rames must obey the same laws. As Einstein himsel said: What led me ... to the special theory o relativity was the conviction that the electromagnetic orce acting on a body in motion in a magnetic feld was nothing else but an electric feld.
Maxwell and electromagnetism In 1 861 James Maxwell established the connection between electrostatics, electromagnetic induction, and the speed o light. He developed our equations that between them describe the whole o electrical and magnetic theory and lead to the recognition that light is a orm o electromagnetic radiation. The our equations incorporate the value o the speed o light travelling in a vacuum ( ree space) in a undamental way. The conclusion that must be drawn rom Maxwells equations is that, i observers in dierent inertial rames make observations o the speed o light then, i they are to agree about physical laws, they must observe identical values or the speed o light. B ut the Galilean transormations predict a dierent result rom this. The Galileo predicts that the speed o light diers in dierent rames by the magnitude o the relative velocity between the rames. S o the inescapable conclusion that ollows rom Maxwell ( same speed o light or all observers) is directly contrary to the assumption o absolute time and absolute space as postulated by Newton, and as embodied in the Galilean transormations. Physics had reached an impasse; it required the genius o Maxwell to recognize the problem. It required another genius, Einstein, to move the subj ect orward again hal a century later.
A . 2 L O R E N T Z T R A N S F O R M AT I O N S
A.2 Lorentz transformations Understanding The two postulates of special relativity Clock synchronization The Lorentz transformations Velocity addition Invariant quantities (spacetime interval, proper
time, proper length, and rest mass) Time dilation Length contraction The muon decay experiment
Nature of science Einsteins theory of relativity stems from two postulates. He deduced the rest of the theory mathematically. This is an example of pure deductive science at work.
Applications and skills Solving problems involving velocity addition Solving problems involving time dilation and
length contraction Solving problems involving the muon decay experiment
Equations Lorentz transformation equations:
1 _____ = ___ v2 1- _ c2 x' = (x - vt) ; x' = (x - vt)
(
)
(
vx ; t' = t - _ vx t' = t - _ 2 2 c uv __ u' = uv 1- _ c2 t = t 0
c
)
The two postulates of special relativity Newtons two postulates o space and time rom Sub- topic A.1 require an absolute time that is the same or all observers in inertial rames. This implies, through the Galilean transormation, that a moving observer will observe a dierent value or the speed o light in ree space rom that o a stationary observer. Maxwells discoveries began to prompt serious questions about the peculiar behaviour o light and the validity o Galilean transormations. In 1 887, two US scientists, Michelson and Morley, using very precise apparatus showed experimentally that any such dierence between moving and stationary observers was below the measurement limits o their experiment. Scientists such as Lorentz, Fitzgerald, and Poincar attempted to explain both Maxwells conclusions and the results o the Michelson-Morley experiment. Einsteins great leap orward was to recognize that i Maxwells our electromagnetic equations were to be true in all inertial rames (which had to be the case) then some modifcations o Newtons postulates were required. He was able to show that some equations that had already been developed by Lorentz as a way o avoiding the problems o the null result o Michelson and Morley could, alternatively and more properly, be derived assuming only Einsteins own modifcations o Newtons postulates.
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R E L AT I VI T Y Einsteins two postulates are:
The laws o physics are the same in all inertial rames o reerence ( Newtons frst postulate too) .
The speed o light in ree space ( a vacuum) is the same in all inertial rames o reerence ( replacing the concept o absolute time and space, Newtons second postulate) .
The Lorentz transformation Earlier we saw that, or boosts, the Galilean transormations lead to the equations x = x X, x= x vt, and u= u v. There is an additional new transormation that arises rom Newtons postulate that time is absolute. This is represented as t'= t. To make the Maxwells equations consistent or all inertial reerence rames, Lorentz introduced a actor given by 1 _____ = _ v2 1- _ c2
v is ( as usual) the speed o one inertial rame relative to the other and c is the speed o light in ree space, is called the Lorentz factor. As v tends to c, tends to infnity. Lorentz then used this actor to modiy the Galilean expressions so that x' = ( x - vt) and t' = t 8 7 6 5 4 3 2 1 0 0
0.2
0.4
0.6 v c
0.8
1
1.2
Figure 1 How the Lorentz factor varies with speed. Figure 1 shows how the Lorentz actor changes with speed. The scale on the x- axis is in the ratio o _vc and shows us that, or speeds up to 2 0% o that o light, the Lorentz actor remains close to 1 . I the Galilean transormations were true or all speeds then the graph would show a horizontal line at value 1 or all values o _vc . S trictly, the equation involving x and x is used to translate rom a position in one rame to the position in the other rame at the same instant in time. In the Galilean transormation, lengths measured within one rame transorm without change into the same length in any other rame because length x is the dierence between two positions, and x = x1 - x2 = ( x1' - vt) - ( x2' - vt) = x1' - x2' = x' .
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In the Lorentz transformation this equality of x and x' is no longer true and x' = ( x - v t) This is a crucial result because it tells us that if one inertial reference frame is moving at a constant velocity relative to another, then an observer in one frame making a length measurement of an obj ect in the other frame will not agree with the measurement made by an observer in the other frame. S pace is no longer absolute. The expression x' = ( x - vt') gives the position x of the obj ect as observed in the moving reference frame. Sometimes we observe the position in the moving frame x' and we need the value of x in the stationary frame. This is given by x = ( x' + vt) and is known as the inverse Lorentz transformation. The Lorentz transformation for time is v t' = t - _2 x c -v with an inverse transformation of t = ( t + ___ x ) c
(
)
2
Like space, time has also lost the property of being absolute. Time measured in different frames differs when there is relative velocity between the frames. Also, terms in x now appear in the time equations and terms in t in the expressions for x' and x. We have assume d so far that there is no relative motion betwee n the frames in the y or z directions. If this is true the n the re will be no relativistic changes in these directions e ither ( this can be proved formally) . The assumption of no motion in directions y and z and the pre vious expressions lead to the complete set of Lore ntz transformations which are here compared with their Galilean e quivale nts.
Galilean
Lorentz
Inverse Lorentz
x' = x - vt
v ct x' = ( x - _ c )
v ct' x = ( x' + _ c )
y' = y
y' = y
y = y'
z' = z
z' = z
z = z'
t' = t
-v x t= t- _ c2 vx ct' = ( ct - _ c )
(
)
-v x t= t+ _ c2 v x' ct = ( ct' + _ c )
(
)
In each case, if v c then 1 and the Lorentz equations reduce to the Galilean equations with which we are already familiar. An additional change in the Lorentz equations in this table is the expression of time using ct rather than t alone ( this gives the time equation the dimensions of distance) . A second change is to include the speed of light, c, twice in the distance equations. These changes make the equations appear more symmetric and help to explain why later in this topic we use axes of ct against x to draw spacetime diagrams.
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Worked examples 1
a) The distance between x1 and x2
C alculate the Lorentz factor for an obj ect travelling at 2 .7 1 0 8 m s 1 .
b) The time interval between t1 and t2 .
Solution
Solution 5 a) = _ = 1 .67 3 From the Lorentz transformation
This speed is 0.9c. 1 1 _____ = __ ________ = 2 .3 = _ 2 ( 1 0.9 2 ) v 1- _ 2 c
2
x2 - x1 = ( ( x2 - x1 ) - v( t2 - t1 ) ) S ubstituting gives
C locks in two frames S and S are adj usted so that when x = x = 0, t = t = 0. Frame S as a speed of 0. 8c relative to S. Event 1 occurs at x1 = 5 0 m, y1 = 0, z1 = 0, and t1 = 0.3 s. Event 1 occurs at x2 = 80 m, y2 = 0, z2 = 0, and t2 = 0.4 s. C alculate, as measured in S,
x2 - x1 = 1 .67 ( ( 80 - 5 0) - 0.8 3 1 0 8 (4 1 0 -7) - (3 1 0 -7) ) x2 - x1 = 1 0 m
(
)
( x2 - x1 ) v_ b) t2 - t1 = ( t2 - t1 ) - _ c c v ( t2 - t1 ) = 1 . 67 ( t2 - t1 ) - _c _ c
(
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t2 - t1 = 3 0 ns ( 2 s.f. )
Velocity addition An obj ect is moving in frame A with a constant velocity u A. Frame A itself is moving with a constant velocity v with respect to frame B . What do the Lorentz equations have to say about the velocity u B of the obj ect when it is viewed by an observer in frame B ?
v uA
frame B
frame A
Figure 2 Relativistic relative velocities. Galilean relativity has no problem with this, the answer is simple: u B = u A + v. B ut this cannot be correct from a relativistic view. S uppose the rocket in frame A is moving at the speed of light. Then if v is positive u B will exceed c; this is not allowed by Einsteins second postulate. x We need to use the Lorentz equations. The speed u A is equal to __ when t viewed in frame A ( here xA and tA correspond to x' and t' in the earlier transformation equations) . A
A
x S imilarly u B = __ ( x and t from before) t B
B
so ( x A + vt A ) xB u B = _ = __ vx A tB t A + _ c2
(
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using the Lorentz transormations. The positive signs are there because we are expressing xB and tB in terms o xA and tA rather than the other way round as we did beore.
Worked example 1
You should satisy yoursel that substituting xA = u AtA into this expression gives uA + v uB = _ u Av 1 + _ c2 This is the case where an observer at rest in frame B is measuring the speed o the obj ect. The inverse case o an observer in rame A ( sitting in the rame A rocket in fgure 2 ) measuring the speed o the obj ect in rame B gives a similar expression with a change in sign. uB - v uA = _ u Bv 1 - _ c2 A way to get the signs correct or a particular combination o rame velocities is to begin with the Galilean transormation ( that is, when uv c) and see what signs you would expect in this case. Then remember that the sign matches top and bottom o the equation.
Jean and Phillipe are in separate rames o reerence, neither o which is accelerating. Jean observes a spacecrat moving to his right at 0.8c. Phillippe observes a spacecrat moving to his let at 0.9c. C alculate the velocity o Phllippes rame o reerence relative to Jeans.
Solution S uppose Jean is the observer at rest and Phillippe is moving relative to Jean with speed v. u - v u A = ______ u v can be re___ B
Invariant quantities
1 -
Spacetime interval
B
c2
u - u arranged to give v = ______ u u ____ B
In developing his theory o special relativity, Einstein realized that absolute time and absolute space are not invariant (unchanging) properties when moving rom one inertial reerence rame to another. However, not all quantities change when moving between inertial rames. In the previous section we chose to express time in the Lorentz transormations not as plain t but as the product o the speed o light and time, ct. This quantity ct has the dimensions o length and this leads us to an invariant quantity known as the spacetime interval s that is defned or motion in the x direction as s 2 = ( c t) 2 - x2 ( sometimes called the invariant interval) . ( You may see this defned in some books as x2 ( c t) 2 , in other words, as the negative o our defnition.)
1 -
A
B
A
c2
This gives the relative velocity o the reerence rame in terms o the known individual speeds in the reerence rames. So
( 0 . 8 c ( 0 . 9 c) ) v = ___________ 0.7 2 ____ 1 +
c2
1 .7c = 0.989c. = ____ 1 .72
The spacetime interval is invariant because in another rame o reerence
(
v v v x' _ _ ct = c( t2 - t1 ) = ( ct2 ' + _ c x2 ' ) - ( ct1 ' + c x1 ' ) = c t' + c
)
and v v _ x = x2 - x1 = ( x2 ' + _ c ct2 ' ) - ( x1 ' + c ct1 ' ) = ( x' + v t' ) thereore
(
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2
v x' s2 = 2 c t' + _ - 2 ( x' + v t' ) 2 c
(
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v2 = 2 ( c 2 - v 2 ) t' 2 - 2 1 - _ x' 2 c2 = ( c t' ) 2 - x' 2 This is obviously identical to the original defnition using the same quantities ( and no others) measured in the new rame.
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R E L AT I VI T Y In three dimensions the spacetime interval becomes s2 = ( ct) 2 - x2 - y2 - z2 .
TOK So what about time travel? The spacetime interval has a bearing on the cause and eect relationship between two objects or events. Intervals can be classifed as space-like, time-like, or light-like depending on the value o s 2 or the two events. Space-like intervals I s2 is (ct) 2 . This means that the distance between the events is too great or light (or anything travelling slower) rom one event to have any eect on the other. They do not occur in each others past or uture and although there is a reerence rame in which they occur at the same time, there is no reerence rame in which the events can occur at the same place. Time-like intervals In this case, s2 is > 0 (ie positive) and (ct) 2 > (x2 + y 2 + z2 ) . Now there is sucient time or there to be a cause and eect relationship between the events because the time part o the spacetime interval is greater than the spatial separation. Light-like intervals The intermediate case is where (ct) 2 = (x2 + y 2 + z2 ) and thereore s2 = 0. The spatial distance and the time interval are exactly the same. Such events are linked by, or example, a photon travelling at the speed o light. Time travel has always been a ascination o science-fction authors. The spacetime interval can tell us the extent to which two events in space and time can aect each other. To what extent do fctional works that you know mirror scientifc truth?
Rest mass A second invariant quantity is the rest mass m 0 o a particle. This is an important quantity and is defned as the mass o a particle in the rame in which the particle is at rest. This will be discussed at more length in a later sub- topic, as the concept o mass is bound up with what we mean by energy.
Proper time t = 2L c
L
D
L
mirror position 1
D
mirror position 2
S ome o the most dramatic dierences between our everyday perceptions o space and time and the predictions made by special relativity concern the time and length dierences that arise between rames o reerence moving relative to each other.
vt'
Figure 3 A light clock from stationary and moving frames of reference.
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Figure 3 shows a simple light clock that consists o two mirrors acing each other across a room. An observer sits at rest in the room and watches light reect between the mirrors. The distance across the room is L and so the time taken or the light to return to a mirror is t = 2cL .
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A . 2 L O R E N T Z T R A N S F O R M AT I O N S
Anothe r observer moves to the le t parallel to the mirrors at constant ve locity v relative to the mirrors, watching the reections. The righthand diagram shows the bottom mirror at two positions as se en by the moving observer: when the light leaves the bottom mirror and when it returns. In the rame o this observer, the light appears to travel to the right at an angle to the direction o motion ( but, o course, at the same spee d o light) . The distance travelle d by the light is now 2 D and the time observe d between reections at the same 2D mirror is now t' = ___ c . The distance travelled horizontally by the moving observer in time t is vt and by an application o Pythagoras theorem: v 2 t 2 D2 = L 2 + _ 42 ct' D=_ 2 Rearranging or t gives 2L _ c _ t' = ______ v2 1 - _2 c and so t ______ t' = _ v2 1 - _ c2
which reduces to t' = t. This result also ollows directly rom the Lorentz transormations: Two events ( the light leaving rom, and returning to, the bottom mirror) occur at t1 and t2 . These events occur at the same place ( the mirror) so we do not need to include the x terms in our proo ( because x1 = x2 = x) . The time interval between these events is t = t2 t1 . In the observer rame thereore vx vx t' = t2 - t1 = t2 - _ - t1 - _ c2 c2
(
(
) ) = t
The same result as beore and time interval in the observer rame = time in the mirror rame t = t0 The quantity t0 , the time interval in the stationary rame, is known as the p rop er time. Its defnition is the time interval between two events measured in the reerence rame where both events occur at the same position. Proper time is our third invariant quantity. Alternatively, it is the shortest possible measured time interval between two events. The result shows that time measured in a moving rame is always longer than the time measured in a rame that is stationary relative to a clock. The eect is known as time dilation ( dilated means expanded) . I the moving observer in our example here also has a clock then an observer stationary with respect to the mirror rame observes the moving clock running slower than the mirror clock. The situation is symmetric. We will discuss this point urther in the next section.
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Worked example 1
Tom is ying a plane at 0.9c. The landing lights o the plane ash every 2 s as measured in the reerence rame o the plane. S am watches the plane go by. C alculate the time between ashes as observed by Sam.
Proper length The length o an obj ect also changes when observed in rames moving relative to the obj ect. When discussing proper time we had to be careul to speciy that the positions at which time was measured were the same or both measurements. This time, the length L o an obj ect ( where L = x2 x1 ) must have x1 and x2 measured at the same time so that t1 and t2 in the Lorentz equations are both equal to t. In reerence rame S, x1 and x2 represent the ends o an obj ect o length L 0 . This is the frame in which the object is at rest. In S' these ends become x1 ' and x2 ' with a length L' . The reerence rame o S moves at speed v relative to S' . In S
Solution When v = 0.5 c, = 2 .3 ( this was calculated in an earlier example) . The proper time is 2 s. The time between ashes or Sam = t = 2 .3 2 = 4.6 s. The time is dilated in S ams rame o reerence.
L 0 = x2 - x1 which in S' using the Lorentz transormations x2 + vt2 x1 + vt1 _____ - _ _____ = _ 2 v v2 1- _ 1- _ 2 c c2
The ends o the rod are measured at the same time and so t1 and t2 are equal. So L' ______ L0 = _ v2 1 - _ c2
and L0 L = _ This leads to the ourth invariant quantity ( L 0 ) called the p rop er length defned as the length o an obj ect measured by an observer at rest relative to the obj ect. B y implication the two measurement events have to be made at the same time. The proper length can also be regarded as the longest measured length that can be determined or an obj ect. There is direct experimental evidence or time dilation and length contraction ( which are two sides o the same coin) . Muons are particles that can be created either in high- energy accelerators or ( more cheaply! ) in the upper atmosphere when cosmic rays strike air molecules. These muons have very short mean lietimes o about 2 .2 s. When travelling at 0.98c, the distance the muon will travel in one mean lietime is roughly 660 m ar less than the height o 1 0 km above the Earths surace where they are created. O n a Newtonian basis very ew muons would be expected to reach the surace as the time to reach it is about 1 5 mean lietimes. Yet a considerable number o muons are detected at the surace. This is due to time dilation ( or length contraction, whichever 1 _______ = 5 . S o, in the viewpoint you choose) . At a speed o 0.98c, = ________ 1 - 0 . 9 8 2
reerence rame o the Earth, the mean lietime becomes 1 1 s. The time to travel 1 0 km at 0.98c is 3 3 s so that a signifcant number o muons will remain undecayed at the surace.
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In the rame o reerence o the muon, the 1 0 km ( as measured by an 1 0 km observer on the E arth) rom atmosphere to Earths surace is only _____ ( to the muon) . This is 2 km in the muons rest rame corresponding to a travel time o about 3 mean lietimes allowing many more muons to reach the surace than Galilean relativity would suggest. D epending on the observers viewpoint, either time dilation or length contraction can be used to explain the observed large number o muons at the surace.
Clock synchronization The problem o synchronizing the clocks we have used in our discussions is an important one. S uppose an observer is standing close to a clock and simultaneously viewing another clock 1 km away. B ecause the speed o light is invariant, the distant clock even i originally synchronized 10 will appear to register a time _____ = 3 0 s later than the nearby clock. 3 10 However, i the two clocks are in the same inertial rame, they will be synchronized ( tick at the same rate) .
Worked example 1
C lare and Phillippe y identical spacecrat that are 1 6 m long in their own rame o reerence. C lares spacecrat is travelling at a speed o 0.5 c relative to Phillippes. Calculate the length o a) C lares aircrat according to Phillippe b) Phillippes aircrat according to C lare.
4
8
O ne way to achieve this synchronization is to synchronize both clocks when they are close together and then move one to its fnal position, but to do it very slowly so that the two clocks continue ( approximately) to share a reerence rame with each other. Another method would be to have both clocks at their fnal positions and to use a third clock moving slowly between them to transer the times rom one to another.
Solution = 1 .1 5 or this relative speed. a) The length o C lares 16 aircrat is ____ m according 1 .1 5 to Phillippe, this is 1 3 .9 m. b) B ecause the situation is symmetrical C lare will also think that Phillippes spacecrat is 1 3.9 m long.
Nature of science GPS a study in relativity S atellite navigation units ( satnavs) in cars and other domestic devices that use the global positioning system are now very common. They can pinpoint their position to within a ew metres. At the time o writing there is a network o 2 4 satellites orbiting at about 2 1 0 7 m above the Earth with orbital periods o about 1 2 hours. The orbits are such that at least our satellites are above the horizon at any point on the surace at all times. Inside the satellite is an atomic clock that is accurate to about 1 0 9 s and transmits a signal to the receivers on or above the Earth.
The receivers triangulate the signals rom satellites above the horizon to arrive at a positional fx to within metres within a ew seconds. Wait a little longer with some special receivers and this precision can rise to orders o millimetres. It is now routine or a satnav in a moving vehicle to show its speed and heading in real time. The design o both the satellite transmitters and the GPS receivers need to take account o relativity. The atomic clocks are adj usted so that once in orbit they run at the same rate as Earth- bound clocks. The GPS receivers have microcomputers that carry out the required calculations to make the relativistic corrections.
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A.3 Spacetime diagrams Understanding
Applications and skills
Spacetime diagrams
Representing events on a spacetime diagram
Worldlines The twin paradox
as points Representing the positions o a moving particle on a spacetime diagram by a curve (the worldline) Representing more than one inertial reerence rame on the same spacetime diagram Determining the angle between a worldline and the time axis or specifc speed on a spacetime diagram Solving problems on simultaneity and kinematics using spacetime diagrams Representing time dilation and length contraction on spacetime diagrams Describing the twin paradox Resolving o the twin paradox through spacetime diagrams
Nature of science The spacetime (or Minkowski) diagram is a perect example o a picture adding value to conceptual understanding. At frst, Einstein did not realize the ull potential o these diagrams but came to see that they had considerable value in making some o the dicult ideas o relativity including simultaneity more obvious.
Equations v angle o worldline to ct axis = tan - 1 ( _ ) c
Spacetime diagrams and worldlines In 1 908, Minkowski introduced a way to visualize the concept of spacetime. This will help you to understand many of the ideas and concepts that arise in special relativity. Physicists are well used to graphs as ways to visualise data. Minkowski attempted to represent the four- dimensional nature of spacetime using a graphical picture known as a sp acetime diagram or sometimes as a Minkowski diagram. t/s t t
P'
worldline of Q
3
R'
speed = c
2 1 x P (a)
Figure 1 Spacetime diagrams.
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x/m 0 (b)
4
8
R
12 (c)
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A. 3 S PACE TI M E D I AG R AM S
S pacetime diagrams show the position o an obj ect in one dimension ( x) at a time ( t) in an inertial rame. The axes themselves constitute the inertial rame. The diagram resembles ( but should not be conused with) the ordinary distancetime graphs with which you are amiliar in mechanics, except that time is plotted on the y- axis and position on the x- axis. Figure 1 ( a) shows the spacetime diagram or a particle that is stationary with respect to the inertial rame that is represented by the diagram. At t = 0 the particle P is on the x-axis. As time goes on, because the object is stationary, it does not change its position (x) in the reerence rame. Time is, o course, increasing. Line PP shows the traj ectory o the particle through spacetime and is known as the worldline o the particle. Figure 1 ( b) shows a dierent particle Q moving at a constant velocity in the reerence rame o this spacetime diagram. At t = 0, Q is at the origin o the diagram ( x = 0) and it is moving at 4 m s - 1 to the right. Each second ater the origin time, Q is 4 m urther to the right and so its worldline in the spacetime diagram is a line at an angle to the axes. It should be easy or you to see that i a urther particle R were to be accelerating relative to the reerence rame o the diagram, then the worldline o R would be a curve ( fgure 1 ( c) ) . There must be a limit to the R worldline because nothing can exceed the speed o light in ree space. So the gradient o the dotted line on fgure 1 ( c) shows the maximum limiting speed o R. This dotted line also represents the world line o a photon in the diagram ( the minimum gradient o RR) . Returning to particle Q which is moving with constant velocity in the reerence rame o P, we could think o Q as being stationary in its own reerence rame and in this event the spacetime diagram or Q in its own rame would be identical in shape to fgure 1 ( a) . We can combine these two separate spacetime diagrams or dierent inertial rames moving at constant speed relative to each other and this combination o axes will be o particular help later in this sub- topic when we resolve some o the paradoxes that special relativity appears to create.
t
t' worldline o Q x'
Figure 2 shows the combined spacetime diagram or Q and P drawn or the reerence rame o P. Notice what has happened to the Q axes (t' x' ) , they have swung away rom t and x and become closed up together. The Q x'-axis is now at the same angle as the previous Q worldline in the P reerence rame. Q is, o course, moving along a worldline parallel to the time axis in the Q reerence rame. We have so ar used t and x or our axes, but this is not the only convention used. S ometimes you will see time plotted in units o ct. This is convenient because it means that the limiting line that represents the speed o light will have a gradient o 1 on the spacetime diagram. You can expect to see both conventions used in IB examinations. An additional convention is that sometimes physicists defne c to be equal to 1 so that, in calculations, large values or the answers do not trouble them. Equally, expect to see speeds quoted as, or example, 0.95 c meaning 95 % o the speed o light in ree space ( 2 .85 1 0 8 m s 1 ) .
x
Figure 2 Two inertial rames one spacetime diagram.
ct
particle X worldline c
cT x
Figure 3 Defnition o .
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R E L AT I VI T Y S ome simple geometry ( Figure 3 ) shows that when we are using ct-x axes, then the angle between the worldline o a particle and the ct axis is given by opposite v X 1 _ X=_ tan = _ = _ =_ c c cT T adjacent or v = tan - 1 ( _ c) When v = c, then = 45 ( tan 45 = 1 ) and the worldline or a photon starting at the origin o the spacetime diagram is a line at 45 to both ct and x axes.
Simultaneity There are signifcant changes to our ideas about the order in which things happen or whether two events happen simultaneously under special relativity. This is because the speed o light is always observed to have the same value by observers in dierent rames. The classic thought experiment to illustrate this is the example o a train carriage moving at constant velocity past an observer standing on a station platorm ( Figure 4) . A person in the carriage ( Jack) switches on a lamp that hangs rom the centre o the ceiling. Jack observes that the light rom the lamp reaches the two end walls o the carriage ( R and L) at the same moment. railway carriage lamp R
L Jack
Jill
platform
simultaneous
ct'
not simultaneous
ct L
R
L
R
x' x (a)
Jacks frame
x (b)
Jills frame (ct' - x')
Figure 4 Simultaneity at work in a spacetime diagram. Jill who is on the platorm, does not agree with this observation. The light rom the lamp moves to both ends o the carriage at the same speed c. However, in the time it takes the light to get to the ends, the let-hand end o the carriage has moved towards the light and the right-hand end has
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A. 3 S PACE TI M E D I AG R AM S
moved away. The consequence is that the light (according to Jill) hits the let-hand end frst (event L) beore the right (event R) . This result becomes clear in a spacetime diagram. In the reerence rame o Jack (ct-x) the events R and L occur at the same instant because they are on a line parallel to the x axis and are at the same ct coordinate (fgure 4(a) ) . In Jills rame plotted on fgure 4(b) (ct-x) it is clear that L occurs beore R when you look these events in terms o the ct axis. There is a danger o conusion here because it is possible to misunderstand and think that the loss o simultaneity is to do with the transmission o the inormation. In other words, that this dierence o opinion between Jack and Jill arises because the light travels through dierent distances rom the ends o the carriage to their eyes. That is not the explanation o what is happening. The lack o simultaneity arises because the speed o light is always constant even i a particular observer is moving relative to the light source. As ar as Jack is concerned, he is always midway between the carriage endwalls. As ar as Jill is concerned, once the photons have let the lamp, then they travel at c and the carriage will continue to move while the photons themselves are in transit. O ne o the reasons or this conusion is the use o the term observer, which is a very common one in books and articles about relativity. We oten think o an observer as being located at one point in the inertial reerence rame. This is not the true meaning. It is better to think o the observer as being in overall charge o an ( infnitely) large number o clocks and rulers that are located throughout the observers rame. Jack ( the stationary observer in this case) can take a reading at the instant when the light hits the end wall o the carriage without having to worry about the time taken or this inormation to travel rom the carriage to his position. Another way is to think about the observer as being a whole team o observers with each one able to make measurements o his or her immediate region o the reerence rame. This can be explained in terms o the Lorentz transormations. Imagine that two events are simultaneous in one rame o reerence. This will mean that or both events the time coordinate will be ct or any value o x. However, in a rame moving at v relative to the frst rame, the Lorentz transormation shows that: vx t- _ c2 _ t' = ______ v2 1 - _2 c
Thereore unless x is the same or both events then t cannot be the same or both events they will not be simultaneous. Conversely, this also tells us that i two events are simultaneous in the second, moving rame then they must be occurring at the same position (x) .
Time dilation and length contraction re-visited Figure 5 ( a) shows the spacetime diagram or an observer in rame B who is viewing an automobile moving at a constant velocity v relative to B . The diagram shows the stationary rame A or the automobile with the B spacetime axes included. We need to be quite clear about what we mean by the term time interval here. It is the time between two events
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R E L AT I VI T Y measured at the same place in the reerence rame ( in other words, the proper time) . The spacetime diagram should show that a measurement in any other rame leads to a time interval greater than the proper time. ct
proper time time in A
ct ct'
ct'
time in B
time that A thinks B measures
proper time
x' frame B x
x' x
frame A (a)
(b)
Figure 5 Measuring times in two reference frames. At t = 0 the origins o both reerence rames coincide and the automobile is at the origin in both rames. The automobile is stationary in rame A and its worldline lies along the time axis as usual. Frame B is moving at constant velocity and its axes are modifed as usual in rame A. The event that marks the end o the proper time interval is transormed along the x axis to meet the ct axis in order to obtain the time that B will measure at what B thinks is the same instant in the moving rame ( fgure 5 ( a) ) . The question o what A (in the stationary rame) thinks is the same instant is dierent (fgure 5 (b) ) . However, whichever view we take o the measurement in rame B (whether rom the A or B standpoint) , the time measured in B is always greater than the proper time because the length o the line in the spacetime diagram is always along the hypotenuse o the triangle, whereas the proper time is one o the other sides o the triangle. A spacetime diagram also helps understanding o length contraction. To see why distance measurements change in a moving rame, again we must understand what is being measured. We need to consider the distance between two points at the same instant in time as j udged by the two observers in dierent rames. ct frame B
ct' frame A
worldline of 0 cm in B
worldline of 30 cm in B worldline of 30 cm in A
worldline of 0 cm in A
p ro p e r le
n gth
29 30 26 27 28 23 24 25 20 21 22 17 18 19 14 15 16 13 12 9 10 11 7 8 4 5 6 1 2 3
x2 '
CM
simultaneous in A simultaneous in B
x1 ' x' frame A
x1
x2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 CM
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Figure 6 Measuring lengths in two reference planes.
x frame B
A. 3 S PACE TI M E D I AG R AM S
Figure 6 shows a ruler stationary in rame A. The ruler lies along the x- axis because this is where we will determine the proper length. The rules or measuring proper length indicate that the two position determinations must be made simultaneously. The other rame B is considered to be the stationary rame or the purposes o the spacetime diagram. Thus, A must be moving relative to B and as a result has the axes ct- x. The diagram shows what happens. The proper length o the ruler is the distance measured between the worldlines and parallel to the x axis. The equivalent simultaneous measurement in B will always be shorter than that in A the length is contracted.
The twin paradox Many o the ideas we have discussed so ar in the study o special relativity come together in a series o paradoxes. The twin paradox is the most amous o these and can be very simply stated: ct
Mark and Maria are twins. Mark decides to go on a journey to a distant star at a high speed with Lorentz actor equal to . Ater reaching the star taking a time T in Marias rame o reerence. Mark returns with his journey also taking a time T (to Maria) to do so. At the end o his journey 2T Maria has aged 2T but she is amazed to fnd that Mark has only aged by __ . This is nothing more than time dilation so where does the paradox arise? Think about Marks experience. He sits in his spacecrat his rame and watches Maria and the E arth move away at high speed ( with the same as beore) so why is Maria not younger than him on his return? We would expect some symmetry between the two rames. The answer to the paradox is that there is no symmetry at all between the two cases. Maria has remained in an inertial rame throughout Marks j ourney. Mark has not. He needs to accelerate our times during the journey: at the start o the trip, when he slows down at the star, when he accelerates back up to top speed and fnally when he decelerates to arrive at the Earth. Moving out o an inertial rame o reerence even once breaks the symmetry and as a consequence Mark and Maria age at dierent rates relative to each other over the whole journey. The spacetime diagram (fgure 7(a) ) shows what happens. The rame or Maria is ct- x, the rame or Mark is ct- x. Mark leaves Maria at the origin o her reerence rame and she remains here throughout. O course, she moves along the ct axis as time increases. Meanwhile, Mark moves along his worldline which is at his origin x = 0 or (in Marias rame) at x = vt where v is Marks speed relative to Maria. Mark reaches the star at event P. We can draw two lines o simultaneity; one or Mark and one or Maria. In Marias rame she thinks Mark arrives at the star at time Q. R is the time that Mark thinks Maria observes when he arrives at the star. They disagree about the simultaneity o events Q and R as we should expect. At this stage, both Mark and Maria think that the other is younger by a actor o as predicted by the time dilation result rom earlier. There is a problem in thinking about the return j ourney because i it is to happen at all, Mark has to change speed. It is not necessary or him to do this, though. We can imagine that as soon as he reaches the star, he synchronizes his clock with another clock on a spacecrat that belongs
Mark worldline
P ct'
Q
Mark arrives at star
R
x' x
Maria worldline
(a)
ct Jay worldline Mark worldline
S Q
Mark arrives at star
ct'
R
x' x
Maria worldline
(b)
Figure 7
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R E L AT I VI T Y to Jay. Jay is already on his way to Earth ( and thereore Maria) at the same speed that Mark had when approaching the star. Figure 7( b) shows the added worldline or Jay. As Jay leaves the star, he thinks that Maria is at S. I Mark were to slow down at the star, turn round, and go back to the Earth, the acceleration procedure would make Maria appear to rapidly age rom Q to S.
Worked example In the distant uture a network o our warning beacons W, X, Y and Z is set up to warn spaceship commanders o the approach lanes or planet Earth. The beacons ash in sequence. The spacetime diagram shows the reerence rame in which the beacons are at rest and one cycle o the sequence. The worldline or a spaceship is also shown.
ct ct'
Y
X
ct ct' worldline of spaceship
W
Z x
( ii) The order o the ashes in the spaceship rame has to be obtained rom constructing lines parallel to the x axis. In this rame, Z is observed to ash frst, W and X then ash simultaneously, fnally Y ashes. x W
X
beacon frame
Y
Z
a) D etermine the order in which the our beacons ash according to:
b) To decide on the arrival o the light rom the beacons it is necessary to add the photon worldlines to the diagram. These are lines that begin at the beacon ash and travel at 45 to the axes. ct
( i) an observer stationary in the rame o the beacons ( ii) an observer on the spaceship b) D etermine the order in which the observer on the spaceship sees the beacons ash. c) C alculate the speed o the spaceship.
light from Y
Solution a) ( i) T h e s p a ce tim e dia g ra m in th e ra m e o th e b e a co n s in dica te s th e ch ro n o lo g ica l o rde r in w h ich the b e a co n s la s h : W a n d Z s im u lta n e o u s ly, th e n X, a n d th e n Y.
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light from W
light from X
light from Z x
The intersection o the photon worldline with the ct axis gives the arrival time at the spacecrat. The order is Z, Y, X, W. v c) tan = _c = 0.5 8 v = 0.5 8c
A . 4 R E L AT I V I S T I C M E C H A N I C S ( A H L )
A.4 Relativistic mechanics (AHL) Understanding Total energy and rest energy Relativistic momentum Particle acceleration Electric charge as an invariant quantity Photons MeV c 2 as the unit o mass and MeV c 1 as the
unit o momentum
Applications and skills Describing the laws o conservation o
momentum and conservation o energy within special relativity Determining the potential dierence necessary to accelerate a particle to a given speed or energy Solving problems involving relativistic energy and momentum conservation in collisions and particle decays
Nature of science A urther paradigm shit occurred in physics when Einstein realized that some conservation laws (momentum and energy) broke down as inviolate laws o physics unless modifcations were made to them. This led him (amongst other things) to ormulate one o the most amous equations in the whole o physics.
Equations total relativistic energy E = m 0 c 2 rest-mass energy E0 = m 0 c 2 relativistic kinetic energy EK = ( - 1) m 0 c 2 relativistic momentum p = m 0 v energy-momentum relation E2 = p 2 c 2 + m 0 2 c 4 acceleration through pd V qV = EK
So far we have dealt almost exclusively with the basic concepts of space and time and we have shown that they change their nature when we have regard to the relative motion of inertial frames. Now we have to identify and explain the changes that need to take place in the other laws of physics for them to be invariant for all inertial observers.
Total energy and rest energy In a previous sub- topic we mentioned that the rest mass m 0 of a particle was an invariant quantity. E instein, in one of his famous 1 905 papers, proved that when an obj ect loses energy its mass changes and by so doing he suggested that energy and mass are related. This led him to possibly the most well- known physics equation in the world and possibly the most misunderstood, E = mc2 . When a particle is viewed from its rest frame then its mass will be observed as the rest mass m 0 . This means that the rest energy of the particle is E0 = m 0 c2 . Like the rest mass, this is an invariant quantity for a particular particle. Some physicists also use the concept of relativistic mass, meaning the mass observed when the particle is in a reference frame moving relative
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R E L AT I VI T Y to the observer. However, the consequences of energy and mass having an equivalence means that it is not necessary to use both relativistic mass and energy; we will only use the term total energy to signify both these ideas. The total energy can easily be converted into an equivalent mass if required using Einsteins equation ( but we will not do so in this course) . The total energy E of a particle is equal to the sum of the rest energy E0 and the kinetic energy EK: E = EK + E 0 ignoring potential energy and its changes, and energy dissipation. This leads to a set of equations relating the energies and masses: E = m 0 c2 and therefore EK = mc2 - m 0 c2 = m 0 ( - 1 ) c2 10 special relativity newton
kinetic energy EK
8 6 4 2 0 0
0.5
1 speed/v/c
1.5
2
Figure 1 Relativistic kinetic energy against speed. Figure 1 shows the shape of the graph of EK against speed ( expressed as a fraction of c) . When the speed is zero, the energy is the rest energy. At high speeds approaching the speed of light, the curve becomes asymptotic to the line _vc = 1 .
Relativistic momentum Momentum must still be conserved within the special theory. However in order to achieve this, the expression for momentum must incorporate to change from the Newtonian mu to m 0 u. In fact, both momentum and energy are j ointly conserved within the theory with the proviso that energy has to include the whole bundle of energies associated with a particle including its rest energy. The momentum p of the particle is expressed as p = mu and ( as before) the total energy is E = m 0 c2 . C ombining these equations and eliminating u leads to an expression for the total energy: E2 = p 2 c2 + m 0 2 c4 = ( pc) 2 + ( m 0 c2 ) 2
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A . 4 R E L AT I V I S T I C M E C H A N I C S ( A H L ) This is known as the energymomentum relation. It has an important property. I the equation is rearranged as
Worked example 1
( m 0 c2 ) 2 = E2 - ( pc) 2 then the quantity on the let- hand side is the invariant mass it does not change between inertial rames. Thereore the right- hand side must also be invariant, so we can write ( m 0 c2 ) 2 = E2 - ( pc) 2 = E' 2 - ( p' c) 2 The knowledge that ( E2 p 2 c2 ) in the rame is invariant allows us to, or example, move easily between the particle rame in a particle accelerator and the lab rame. Many tests o these equations were made rom the time o E insteins frst suggestions o the relationships. The equations were always verifed but some o the determinations were indirect and had many sources o error. Perhaps the most direct and convincing experiment was that developed by B ertozzi in 1 964. He accelerated electrons to a high speed and measured their speed as they travelled through a vacuum. Immediately ater passing through the speed- measuring apparatus, the electrons were absorbed by an aluminium disc, thus transerring their energy to the internal energy o the disc. The temperature o the disc increased as a result and the energy o the incident electrons could be measured directly. B ertozzis results are shown in fgure 2 . This was a direct and convincing verifcation o Einsteins theory.
C alculate the speed at which a particle must travel or its total energy to equal fve times its rest mass energy.
Solution E = mc2 = 5 mc2 1 _____ As = 5 = _ v2 1- _ c2 2 v v2 1 24 _ _ 1- _ = and = __ 25 25 c2 c2
Thereore, v = 0.98c.
1.1 1.0
v2 /c 2
0.9 0.8 special relativity
0.7
bertozzi 0.6 0
5
10
15
20
25
30
kinetic energy/mc 2
Figure 2 Results of Bertozzis experiment.
Nature of science Operating a particle accelerator Although one o the most direct verifcations o special theory was only carried out in the 1 960s, it was clear that the theory was the appropriate one to use. Cyclotrons had already been designed and built, and in these particle accelerators the circulating particles gain considerable energy. The energy that they gain (in our rame o reerence) is ound not to
1 mv2 and a relativistic correction or this vary with __ 2 is required. Essentially, the particles become more massive than would be expected rom a Newtonian consideration. As cyclotrons were developed that could supply more and more energy there came a point where a synchronization mechanism was required to allow or the relativistic changes. This led to the development o synchrocyclotrons and other high-energy accelerators.
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Particle acceleration The obvious way to accelerate a charged particle is to place it in an electric feld. As we saw in Topic 5 this leads to an electric orce on the particle and to a gain in energy that can be expressed in terms o the charge on the particle and the potential dierence through which it moves. The key here is that charge q is our fth invariant quantity so the term does not enter into our specifcation o charge in a moving rame. Thus the charge o the particle during the acceleration does not change and we can see directly ( as in earlier parts o this book) that qV = EK where EK is the change in kinetic energy and V is the potential dierence. This leads to a new set o units that are extensively used in particle and relativistic physics. Rather than use kg or mass and kg m s 1 or momentum it is much easier to think and work in terms o the energy equivalent o these units. Thus, or mass we use eV c 2 ( or, more commonly, multiples o this, MeV c2 and GeV c 2 ) and or momentum MeV c 1 and GeV c 1 . What has eectively happened is that c in the energymomentum relation has been made equal to 1 . It is as though the equation has been written: E2 = p 2 + m 0 2 One o the worked examples below is a repeat o an earlier example to show you how this change in units works.
Photons Photons have zero mass. What does this mean or their properties within the special theory? The starting point is the energymomentum relation once again, but this time m 0 is zero and so the right-most term disappears leaving E2 = p 2 c2 hf _ h E= _ S o the momentum o a photon is p = _ c c =
TOK
Worked examples 1
Conservation laws This discussion o conservation rules that have needed to be modifed in the light o the special theory calls into question the nature o all conservation rules. Do the laws o natural science dier rom the laws that exist in other branches o knowledge, or example, in economic theory?
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A particle o charge + e has a rest mass o 9. 1 1 0 31 kg. It is accelerated rom rest to a speed o 2 . 4 1 0 8 m s 1 . C alculate, or this particle: a) the rest energy in MeV b) the total energy ater acceleration c) the kinetic energy ater acceleration d) the potential dierence through which it must be accelerated to reach this speed.
Solution 8.2 1 0 - 1 4 a) The rest energy = m 0c2 = 8.2 1 0 1 4 J; this is __ = 0.5 1 MeV. 1 .6 1 0 - 1 9
A . 4 R E L AT I V I S T I C M E C H A N I C S ( A H L )
b) The speed is equivalent to = 1 .7. The total energy = m0 c2 = 1 .7 8.2 1 0 14 = 1 .4 1 0 13 J = 0.87 MeV c) The kinetic energy = ( 1 .7 1 ) 8.2 1 0 1 4 = 5 .5 1 0 1 4 J = ( 0.87 0.5 1 ) MeV = 0.3 6 MeV d) To attain a kinetic energy o 3 60 keV must require a pd o 3 60 kV. 2
C alculate the momentum o a photon o visible light o wavelength 5 60 nm.
Solution h p= _ 6.6 1 0 - 34 SO p = _ = 1 . 2 1 0 - 27 kg m s 1 . 5 .6 1 0 - 7 3
A small insect with a mass o 1 .5 1 0 3 kg fies at 0.48 m s 1 . C alculate the momentum and kinetic energy o the insect.
Solution This is a non- relativistic solution as the speed is much less than that o light. 1 eV 1 .6 1 0 - 1 9 J and 1 kg 9 1 0 1 6 J = 9 1 0 4 TJ. 1 Kinetic energy = _ 1 .5 1 0 3 0. 48 2 = 1 .7 1 0 4 J 2 -4 1 .7 1 0 = __ = 1 .1 1 0 1 5 eV = 1 . 1 1 0 3 TeV 1 .6 1 0 - 1 9 Momentum = 1 .5 1 0 3 0.48 = 7.2 1 0 4 kg m s - 1 = 7.2 1 0 4 3 1 0 8 1 .6 1 0 1 9 = 1 .3 5 1 0 24 eV c 1 = 1 . 4 1 0 1 2 TeV c 1 .
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A.5 General relativity (AHL) Understanding
Applications and skills
The equivalence principle
Using the equivalence principle to deduce and
The bending o light Gravitational redshit and the Pound-Rebka
Snider experiment Schwarzschild black holes Event horizons Time dilation near a black hole Applications o general relativity to the universe as a whole
Nature of science Ater publishing his special theory, which applied to non-accelerated reerence rames, Einstein tackled the general theory incorporating the efects o acceleration and gravity. This required intuition, imagination and creative thinking on his part. He needed to modiy his ideas o spacetime in the light o the efects o mass and the curvature that it produces. In this way he was responsible or yet another paradigm shit in physics.
explain light bending near massive objects Using the equivalence principle to deduce and explain gravitational time dilation Calculating gravitational requency shits Describing an experiment in which gravitational redshit is observed and measured Calculating the Schwarzschild radius o a black hole Applying the ormula or gravitational time dilation near the event horizon o a black hole
Equations
f gh Gravitational requency shit: _ = __ 2 f c 2GM _ Schwarzchild radius: Rs = c2 t0 _____ gravitational time dilation t = __
1 -
Rs ____ r
The equivalence principle In Topic 2 we introduced the idea that there are two types o mass: inertial and gravitational and we suggested that these are equivalent. Mass that is gravitationally attracted is taken to be the same as the mass that responds to a orce by accelerating. This equivalence had been recognized since the time o Galileo but was frst discussed in detail by the German physicist Mach at the end o the nineteenth century ( having been touched on by various philosophers beore him) . E instein and others named the central idea: Machs principle. Mach had rej ected Newtons view o absolute time and space, taking a relational view o the universe in which any motion can only be seen with respect to other obj ects in the universe. Thus we cannot say merely that an obj ect is rotating but must reer to the axis about which it rotates. Machs writings, although controversial amongst scientists, partly led to E instein developing the general theory o relativity that he published in 1 91 6. In this he proposed a principle o equivalence: E insteins p rincip le of equivalence states that gravitational effects cannot be distinguished from inertial effects.
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A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) A thought experiment helps to explain the equivalence principle; the experiment involves two observers and an elevator ( lit) . The elevator is a long way rom any other mass so that it is essentially a gravity- ree zone. O ne o the observers ( X) is in the elevator and carries an obj ect that has mass; X cannot see out o the elevator. Another observer ( Y) is outside the elevator and not connected to it in any way but can view what happens inside.
X
Y
Y
X
(a) b
Figure 1 The elevator "thought" experiment.
Initially the elevator is moving at constant velocity. X releases the obj ect and it stays exactly where it is placed. ( Remember that there are no nearby masses or planets to attract the mass.) X repeats the experiment ( Figure 1 ( a) ) but this time, as X releases the obj ect, the elevator begins to accelerate at 9.8 m s 2 in the direction o the elevator roo ( we will call this " upwards" or brevity) . X will think that the obj ect accelerates downwards at 9.8 m s 2 . Y observes rom outside that the obj ect stays where it is in space and that the lit is accelerating upwards around it. C onsider the same experiment repeated ( Figure 1 (b) ) with the elevator stationary relative to and close to the surace o the Earth. When X releases the obj ect it will be accelerated downwards at 9.8 m s 2 . Y will explain this acceleration as due to the attraction o the Earth. The important point here is that in both experiments the obj ect accelerated downwards, in one case due to an inertial ( acceleration) eect and in the other due to the eects o gravity. B ut to observer X in the elevator these two cases are identical and cannot be distinguished. It is this inability to decide what causes the acceleration that lies at the heart o the principle o equivalence. There is in principle no experiment that the observer in the lit can carry out to decide which eect is which. In fgure 1 ( b) we can either regard the lit as accelerating within a universe ( i. e. everything else) that is stationary or we can view the lit as stationary in the rame o observer X with the rest o the universe accelerating around it. According to E instein there is no absolute motion, only relative motion exists. This means that the special status held by observers in inertial rames o reerence in the special theory no longer exists in the general theory and accelerated or not all observers have the same status and obey the same laws. However, it may be that some rames allow the laws to be stated in a more straightorward way.
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Gravitational redshift The general theory predicts that gravity can aect the motion o light itsel. The equivalence principle helps us understand why.
h
F F
R
(a)
R
(b)
Figure 2 Equivalence in action.
Imagine a spaceship that is stationary ( or moving at constant velocity) relative to the rest o the universe. Inside the ship are an observer and two identical light sources, one light source at the rear o the ship ( lamp R) and one close to the observer at the ront ( lamp F) . The spaceship begins to accelerate Figure 2(a) ) . At the instant that the acceleration begins, both light sources start to emit light o the same requency. (It helps to imagine that the light emission begins as the spaceship starts to accelerate, but this is not an essential part o the argument.) h The light rom R take s a inite time t = ___ c to re ach the obse rve r ( whe re h is the distance b e twe e n the light source s) . D uring this time the ve locity o the space ship and ob se rve r will have change d by an amount v = g t, whe re g is the accele ration. Howe ve r, the light in transit rom R to the ob se rve r will not change its spe e d be cause ( as usual) the spe e d o light is inde pe nde nt o the ob se rve r. The obse rve r will obse rve a longe r wave le ngth or the light rom R compare d to the light rom F as lamp R will appe ar to have b e e n D opple r re dshite d.
This can be made quantitative. Merging the two equations above gives g h v = g t = _ c v is the change in speed o the observer over the time period, t is relative to the speed o light source R when it emitted the observed light, and so the requency shit f observed is f _ g h v _ = c = _ f c2 where f is the requency o the emitted light. The equivalence principle predicts that the eects in a gravitational feld should be indistinguishable rom an inertial system, so we do not need necessarily to say more. However, it is instructive to consider the problem rom another dierent ( but ultimately identical) standpoint. Figure 2(b) shows the spaceship at rest relative to, and sitting on, the surace o the Earth. Again lamp R emits light to the observer. Each emitted light
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A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) photon has energy hf when it leaves the source. To reach the observer some o this energy has to be traded o against the gravitational feld as gravitational potential energy. This loss o photon energy (reduction in f) is equivalent to an increase in the wavelength o the light (f = c) and so again, the light rom R is redshited compared to the light rom lamp F. These discussions o requency and wavelength shits lead to the idea o gravitational time dilation. We can regard the arrival o the wave as a series o ticks o a clock. A redshit means that the clock is observed to tick more slowly than the original. S o the observer at the ront o the rocket thinks that lamp R is ticking more slowly than lamp F and the observer at the top o a mountain on Earth thinks that time is running more slowly or an observer at sea level. This eect is gravitational time dilation. It is dierent rom the time dilation eect observed when inertial rames are moving relative to each other. We mentioned the GPS system earlier ( p 5 2 1 ) in the context o special relativity, which predicts that the clocks on the satellites all behind ground clocks by about 7 s every 2 4 hours due to their relative motion with respect to the surace. General relativity, on the other hand, predicts that the satellite clocks should advance compared with the surace clocks by about 45 s in the same time period. The net result is that the clocks in the GPS satellite gain on clocks back on E arth by about 3 8 s every day. This actor swamps the 2 0 ns accuracy required o the E arth-bound GPS receivers. I relativistic eects are not taken into account, then the errors in a position become serious ater about 1 00 s and accumulate at a rate o tens o kilometres every day. This would be completely unacceptable or any navigational systems. The GPS receivers in our cars and on our mobile phones are constantly carrying out relativistic corrections to adj ust or the unavoidable time changes due to relativistic eects.
Worked examples 1
The requency o a line in the emission spectrum o sodium is measured in a rame o reerence in which the sodium source is stationary and well away rom gravitational inuences. C alculate the ractional requency shit that will be measured by an observer also stationary with respect to the sodium source but placed 1 km above the source close to the surace o Earth.
This is a redshit so the wavelength is increased by this ractional amount. 2
C alculate the dierence in time per day due to gravitational redshit o a clock at the top o E verest compared to a clock at sea level. Mount Everest is 8800 m above sea level. Assume that the acceleration due to gravity is constant over this height dierence.
Solution
Solution
f T ___ = __ where T is the time or one day. T f g h gT h T This means that ___ = ____ and T = ____ = T c2 c2 9 . 8 8 6 40 0 8 8 0 0 _______________ this is 80 ns every day. The
The ractional requency shit is given by f _ g h _ = f c2
( 3 1 0 8) 2
g h 9.8 1 000 so the wavelength shit is ___ = ____ = ________ c 9 10 -13 1 .1 1 0 2
16
observer on the mountain thinks that the sea-level clock is running more slowly so the mountain clock appears to gain time.
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Nature of science Tidal orces, and a more precise defnition The earlier explanation o the elevator thought experiment was too simplistic. The equivalence o gravity and acceleration is true only or uniorm gravitational felds. Real gravitational felds will almost always be non-uniorm and inhomogeneous. At the surace o the E arth with human-sized experiments, the feld is very close to uniorm and nearby obj ects appear to all in parallel directions with the same acceleration. B ut two obj ects in a very large elevator will converge towards the centre o the Earth as they all. An observer in the elevator sees the two obj ects apparently moving closer together and will wonder why. This is known as a tidal eect ( fgure 3 ) and means that a precise statement o the principle o equivalence should include a clause stating that it applies when tidal eects can be
Figure 3 Tidal efects.
neglected. Alternatively, we can say that over an infnitesimally small spacetime region the laws o physics in general relativity are equivalent to those under special relativity.
Although the need or relativistic corrections in the GPS satellite navigation system indicates the truth o Einsteins general theory, there have been ormal scientifc tests made over the past century to veriy that it is correct. They include experiments some o which were suggested by Einstein himsel:
gamma detector
gamma source
gamma source
gamma detector
Figure 4 PoundRebkaSnider experiment.
The precession o the perihelion o Mercury ( the point in its orbit where Mercury is closest to the S un)
D eection o light by the S un
Gravitational redshit o light
O ne o the most recent o these is the experiment devised by Pound, Rebka and S nider in 1 95 9 to measure gravitational redshit. This experiment was the last o Einsteins suggestions to be attempted using the ( then) recently discovered technique o Mossbauer spectroscopy. This technique allows small shits in the requencies o gamma rays to be measured. In a very precise experime nt, Pound and his co- workers fred a beam o gamma rays ve rtically upwards towards a detector place d about 2 2 m above the gamma source. The y repeated the experiment with the gamma source fring downwards to the de tector. I the upward and downward ractional changes in energy o the gammas are compared then E (_ E )
upwards
( )
E - _ E
downwards
2 g h = _ c2
For the PoundRebkaSnider experiment, h = 2 2 .6 m and g = 9.81 m s 2 which leads to a theoretical dierence in the ractional energies o 4.9 1 0 1 5 . They measured the dierence to be ( 5 .1 0.5 ) 1 0 1 5 which compares well with the theoretical value. This was a convincing verifcation o the general theory.
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A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) I a single planet orbits a star, then Newtonian gravitation and mechanics predict that the orbit o the planet will ollow the same elliptical path orever. The presence o other nearby planets and moons, however, disturbs this motion. In practice, planets in the solar system orbit the S un in an ellipse that rotates gradually around the S un; this is known as precession. The rate at which this rotation occurs can be predicted very accurately on the basis o Newtonian mechanics. Observations made in 1 85 9 showed that the precession o Mercury is aster than predicted by Newtonian mechanics. The change to the precession rate is small but much larger than the error in its measurement; there is no doubt that the precession prediction is incorrect. One o the frst successes o Einsteins general theory was that it correctly predicted the value o the observed precession rate o Mercury. E insteins third suggestion was that a massive star would deect light rom its straight- line path. S hortly ater Einstein published his theory, the English physicist E ddington travelled to the west coast o Arica with colleagues to observe stars during the 1 91 9 total eclipse o the S un. This enabled them to confrm that the mass o the S un deects light according to the general theory. This will be discussed in more detail in the next section. A ourth eect was teste d by Irwin S hapiro in 1 9 6 4. E instein had p redicted that the time or a radar ( microwave) pulse o radiation to go past the S un and return to E arth ater relection rom Venus and Mercury would take longe r than expected b ecause o the eect o the S uns gravity. S hapiro measured this time and made a similar measure ment when the signal did not travel close to the S un ( because the planets had moved on in their orbits) . He ound that the delay existed as E instein had predicted and that its magnitude was as expected. S imilar experiments have been repeated in various ways since the 1 9 6 0 s, and always indicate a time delay as predicted b y E instein.
The bending of light S o ar we have discussed the general theory in terms o the equivalence it suggests between gravitational and inertial eects and the changes it predicts or astronomical observations. The general theory, o course, oers much more than this. For Newtonian gravitation, the gravity feld is a model o reality that can be analysed using the law o gravitation. In the general theory, Einstein constructed a set o ten equations known as the Einstein feld equations. These equations indicate that gravity is the eect observed when spacetime is curved ( distorted) by the presence o mass and energy. An analogy or spacetime curvature is that o a rubber sheet. The sheet represents a two-dimensional space in a three-dimensional spacetime continuum. ( This means that it is only an analogy to the real world because one dimension has been suppressed.) In the absence o energy or mass, the sheet is at, horizontal and undistorted. However, i a mass is placed on the sheet, then the sheet deorms under the inuence o the mass as shown in fgure 5 .
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R E L AT I VI T Y light path
star apparent position o star apparent light path sun
earth
Figure 6 A primary star and lensing efect.
Figure 5 Rubber sheet analogy or gravity.
The eect o this deormation on the passage o light rom a distant star can be seen in fgure 5 . Light comes rom the real location o the star but, as it passes near the Sun, the direction o the light is altered by the warping o spacetime. Ater leaving the vicinity o the S un the light appears to have come rom a dierent direction. In 1 91 9 E ddington was able to measure this apparent shit in the star position when the S un moves close to the line between the Earth and the star. Measurements were made in B razil and Arica ( both places where the eclipse was total) and they confrmed the predictions made by the general theory. However the data were hard to collect and it is only recently that reworkings o the data have confrmed that Eddingtons conclusion was not aected by observational errors and confrmation bias. O course, the defection eect is happening in the our dimensions o spacetime, not the three o our analogy. This means that a threedimensional ( to us) gravitational lensing can be observed. Figure 6 is a striking image taken by the Faint O bj ect C amera o a European S pace Agency satellite showing the primary image o a star and our additional images o it ormed by this lensing eect.
event horizon
rs
R photon emitted
The Einstein feld equations do not always have exact solutions (in the sense that two simultaneous equations with two unknowns have an exact solution) . For example, at present, the equations cannot provide an exact solution or the spacetime o two binary stars one o the commonest star arrangements. Physicists usually make simpliying approximations when studying the implications o the equations. I, or example, the assumptions o low speeds and very weak gravity are made, then the feld equations can be manipulated to give Newtons law o gravitation which, thereore, proves to be a special case o general relativity.
Schwarzschild black holes mass M
Figure 7 The Schwarzschild radius.
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Within a ew weeks o the publication o the general theory, Karl S chwarzchild was able to produce one o the frst exact solutions o the Einstein feld equations. He assumed the presence in spacetime
A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) o a spherical, non-rotating, uncharged mass and was able to derive equations or the gravitational feld that surrounds such an obj ect. Using S chwarzchilds equations, the photon redshit that results rom the eect o the gravitational feld o the mass can be derived and is given ( approximately) by Rs gM _ = _= _ 2r rc2
Worked example C alculate the S chwarzchild radius or: a) the Earth ( mass = 6 1 0 24 kg) b) the Sun ( mass = 2 .0 1 0 30 kg) .
where is the emitted wavelength shited by and r is the distance rom the centre o the mass to the point at which the photon is emitted 2 GM . This approximate equation assumes that r is and the constant Rs = ____ c much greater than Rs . Notice the similarity between this equation and the earlier expression or the ractional change in requency 2
f _ g h v _ = c = _ f c2 The constant in the expression, R s , has the dimensions o length and is 2 GM . known as the S chwarzschild radius; it is equal to ____ c
Solution 2 GM a) The S chwarzchild radius = ____ c 2
2 6 .6 7 1 0- 11 6 1 0 24 = ___________________ 9 1 016
= 8.9 mm b) A similar substitution or the mass o the S un gives: 3. 0 km.
2
O utside the S chwarzchild radius ( r > R s) gravity obeys the usual rules, but inside the spherical region defned Rs where r < R s the normal structure o spacetime does not apply. The change between the two regimes where r = R s is known as an event horizon. Events happening inside the event horizon cannot aect an observer outside the horizon. Although the event horizon is not a true boundary ( observers are able to pass through it rom outside to inside) , the event horizon represents the surace where the gravitational pull is so large that nothing can escape not even light. A dierent interpretation o the event horizon is that it is the surace at which the speed needed to escape rom the mass becomes equal to the speed o light. Light emitted rom inside the event horizon cannot escape. The strong gravitational feld ( extreme warping o spacetime) associated with the black hole, attracts nearby mass to it. Mass will appear to collapse towards the centre o the black hole. C locks in a strong gravitational feld run more slowly than in the absence o gravity. The same phenomenon is observed near an event horizon. As a clock moves towards the event horizon rom the outside, an external observer will see the clock slow down with the clock never quite passing through the horizon itsel. The light emitted rom the clock will also be increasingly gravitationally redshited as the clock approaches the S chwarzschild radius ( we already know that this is equivalent to time dilation) . We might expect the situation in the rame o the clock to be dierent and, indeed, the clock ( in its rame) will pass through the horizon in a fnite amount o proper time. The proper time interval to or an observer at distance R o within the gravitational feld o the mass that has a Schwarzchild radius o R s is related to the time interval td measured by a distant observer by ______
to = td
Rs 1 - _ R0
This equation assumes that the obj ect giving rise to the event horizon does not rotate.
Worked example Ticks separated by intervals o 1 .0 s are emitted by a clock that is 2 1 0 5 m rom the event horizon o a black hole o mass 3 1 0 31 kg. C alculate the number o ticks detected by an observer in a distant rocket in a period o 1 0 minutes in the rame o reerence o the rocket.
Solution The S chwarzchild radius 2 6 .6 7 1 0 3 10 is ___________________ = 9 10 -11
31
16
4.5 1 0 4 m_____
Rs
tp = to 1 - __ and when the R measured time interval on the rocket ship is 600 s the proper time at the horizon is 0
__________
600
1 -
4. 5 1 0 4 ________ 2 .0 1 0 31
= 5 2 9 s. There will be 5 92 ticks.
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R E L AT I VI T Y This applies to all bodies that have mass and the worked examples show typical S chwarzchild radii or planetary and solar bodies. In theory, any object that can be compressed sufciently or all its mass to be inside the event horizon will demonstrate these eects. The crucial point is whether this compression is possible or not. For almost all obj ects it is not, because the gravitational sel-attraction o a planets mass is insufcient to overcome the repulsion o the electrons in the atomic shells o atoms. However, or very dense objects, the mass can ft inside the Schwarzchild radius and this object then becomes a black hole. A star needs, typically, to be about three times the mass o the Sun or this to occur. B lack holes are massive, extremely dense obj ects rom which matter and radiation cannot escape. S pacetime inside the event horizon is so warped that any path taken by light inside the S chwarzchild radius will curve arther into the black hole.
Nature of science Links between classical and quantum physics For many years ater Einsteins discovery it was thought that nothing could escape a black hole. It is now recognized that this is only true or black holes under a classical theory. Under quantum theory a black hole can be shown to radiate in a similar way to a black body. This result was unexpected and led to other connections between black holes and the classical study o thermodynamics. The work o the physicist S tephen Hawking and others has led to an understanding o the thermodynamics and mechanics o black holes,
so that it is now recognized that black holes ( or rather the strong gravity feld near them) can lead to the emission o Hawking radiation. This would in principle allow a black hole to be observed and studied. In practice, the radiation emitted by mass being accelerated towards the event horizon will swamp the Hawking radiation. This emission o this radiation implies that black holes can evaporate over time leading to a dynamic process o hole creation and disappearance,
Applications of general relativity to the universe as a whole There are many topics in cosmology that rely on general relativity. S ome o these have been solved; some remain the subj ect o active research. O thers remain as tantalizing theoretical predictions.
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S tudies o the E instein feld equations lead to a number o dierent solutions ( o which S chwarzchilds was one o the frst) in particular the solutions o Friedmann, Lemaitre, Robertson and Walker ( FLRW) . These allow the behaviour o the universe over its lietime to be modelled and the equations have been highly successul in their application. Many aspects o the early universe are illustrated in the FLRW solutions including the large- scale structure o the universe, the way in which chemical elements were created, and the presence o the cosmic background radiation. Additionally, knowledge o the rate at which the universe is expanding allows the total mass o the universe to be estimated.
A . 5 G E N E R A L R E L AT I V I T Y ( A H L ) The answer the equations produce does not agree with the amount o matter that we can actually see around us in the universe. Physicists now suggest the presence o an ( at present) unobservable dark matter and dark energy.
There is a suggestion that single super- massive black holes can be ound at the centre o galaxies. The mass o these obj ects can range up to a mass o several billion times that o our S un. Such black holes will have been inuential in the ormation o galaxies and other larger structures in the universe. As interstellar dust and other materials all into the galactic black holes, a number o arteacts appear, many o which are predicted by the general theory. These include the emission o j ets o very energetic particles at speeds close to that o light that can in principle be observed. Astronomers look careully or evidence o black holes at the centre o galaxies. There is a strong candidate in the Milky Way with an obj ect ( Sagittarius A*) that has a diameter similar to the radius o Uranus but a mass about 4 million times greater than that o the S un.
B inary black holes, normally in orbit around each other, can merge. The general theory suggests that the resulting event should lead to the emission o gravitational waves. O ther events in the universe may also cause gravitational-wave emission. There are experiments in operation and others being devised that will, it is hoped, detect the eects o these gravitational waves and provide urther verifcation o some o E insteins predictions.
TOK The unnecessary constant One orm o the Einstein feld equations is 8G T 1 g R+ g = _ R - _ 2 c4 (this is not going to be tested in the examination!) . R and g tell us about the curvature o spacetime, and T is concerned with the matter and energy in the universe. The constants G and c have their usual meaning. The reason or including this equation is to draw your attention to the third term on the let-hand side o the equation and the constant (capital lambda) that it contains. Einstein called the cosmological constant. In the early part o the twentieth century when Einstein was working on the general theory, the scientifc view was that the universe was stationary and unchanging. Einstein realized that his theory predicted a universe that was neither static nor unchanging according to the equations it should be growing larger. Einstein added the term to adjust the theory to predict a stationary state. A ew years later the astronomer Edwin Hubble ound strong evidence that the universe is expanding. At this point, Einstein realized that his inclusion o had been unnecessary. He later described putting the constant into the equation as my greatest blunder. What other examples are there o scientists whose work was initially rejected (either by themselves or others) but which was later accepted?
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Questions 1
E xplain what is meant by an inertial frame of reference. ( 1 mark)
2
An electron is travelling parallel to a metal wire that carries an electric current. D iscuss the nature o the orce on the electron in terms o the rame o reerence o:
3
4
5
(IB) The radioactive decay o a particular nuclide involves the release o a - particle. A betaparticle detector is placed 0.3 7 m rom the actinium source, as measured in the laboratory reerence rame. The Lorentz actor o the beta particle ater release is 4. 9. a) C alculate, or the laboratory reerence rame:
a) a proton stationary with respect to the wire
(i)
b) the moving electron.
(ii) the time taken or the - particle to reach the detector.
Two electrons are travelling directly towards one another. Each has a speed o 0.002 c relative to a stationary observer. C alculate the relative velocity o approach, as measured in the rame o reerence o one o the electrons according to the Galilean transormation. ( 2 mark)
(IB) a) D efne: (i) proper length
b) The events described in ( a) can be described in the - particles rame o reerence. For the rame o the beta particle: (i)
(ii) proper time
(IB) a) E xplain what is meant by an inertial frame of reference. b) An observer in reerence rame A measures the relativistic mass and the length o an obj ect that is at rest in their reerence rame. The observer also measures the time interval between two events that take place at one point in the reerence rame. The relativistic mass and length o the obj ect and the time interval are also measured by a second observer in reerence rame B . B is moving at constant velocity relative to A.
Muons are accelerated to a speed o 0.95 c as measured in the reerence rame o the laboratory. They are counted by detector 1 and any muons that do not decay are counted by detector 2 . The distance between detector 1 and 2 is 1 3 70 m. v = 0.95 c muons detector 2 1370 m
c) (i) Hal the number o muons pass through detector 2 as pass through detector 1 in the same given time. in the laboratory rame
(ii) in the rest rame o the muons. (iii) D etemine the separation o the counters in the muon rest rame. c) Use your answers in ( b) to explain what is meant by the terms time dilation and length contraction. ( 1 1 marks)
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State whether the observer in rame B measures the quantities as being larger, the same size or smaller than when measured in rame A.
(ii) C ompare the density o the obj ect measured in rame B with the same measurement made in A. ( 7 marks)
b) D etermine, the hal- lie o the muons (i)
state the speed o the detector
(ii) calculate the distance travelled by the detector. ( 7 marks)
6
detector 1
the speed o the -particle
7
Some students are marooned on a planet carrying out feld work when it becomes clear that the star o the planet system is about to become a supernova. A spaceship is despatched to rescue them and the students can be beamed
QUESTION S
aboard the spaceship without the need or it to change speed. The spacetime diagram shows the rame o the star and planet and the rame o the spaceship. The star and the planet do not move relative to each other. The star becomes a supernova at spacetime point S .
ct
ct
ip ra m
e
x
a ce
time axis o planet rame
sh
S fnishing line
xi s
o
sp
a) D iscuss who wins the race in the reerence rame o:
ti m
ea
(i)
S
ce sp a
a xi
s
p o s
ac
ip esh
ra
me
(iii) the reeree. b) Discuss whether there is agreement about the result o the race. ( 1 0 marks)
x planet
space axis o planet rame
a) C opy the diagram to scale and on it identiy: (i)
the spacetime point at which the spaceship arrives at the planet
(ii) the spacetime point o the star when the spaceship arrives at the planet. b) D iscuss whether the spaceship arrives at the planet beore or ater the supernova: (i)
in the rame o the spaceship
(ii) in the rame o the planet. c) D iscuss when the IB students see the supernova. ( 9 marks)
8
F
(ii) S
9
star
F fnishing line
Two athletes compete in a race. Athlete S is slow and is awarded a handicap so that he has less ar to run than athlete F. The spacetime diagram or the race is shown or the rame o reerence o the reeree who is stationary relative to the fnishing lines.
(IB) a) D efne rest mass. b) An electron o rest mass m 0 is accelerated through a potential dierence V. Explain why, or large values o V, the equation 1 __ m 0 v2 = eV cannot be used to determine the 2 speed v o the accelerated electron. c) D etermine the mass equivalence o the change in kinetic energy o the electron when V is 5 MV. ( 7 marks)
1 0 (IB) a) A charged particle o rest mass m 0 and carrying charge e, is accelerated rom rest through a potential dierence V. Deduce that eV = 1 + _2 m 0c where is the Lorentz actor and c is the speed o light in ree space. b) C alculate the speed attained by a proton accelerated rom rest through a potential dierence o 5 00 MV. ( 5 marks)
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A
R E L AT I VI T Y 1 1 A proton is accelerated rom rest through a potential dierence o 2 . 0 1 0 9 V. C alculate, in MeV c1 the fnal momentum o the proton. ( 3 marks) 1 2 (IB) a) D istinguish between the rest massenergy o a particle and its total energy.
acceleration
initial direction of light beam
initial direction of light beam
b) The rest mass o a proton is 93 8 MeV c2 . State the value o its rest massenergy. c) A proton is accelerated rom rest. D etermine the potential dierence through which it must be accelerated to reach a speed o 0.98c, as measured by a stationary observer in the laboratory reerence rame. ( 9 marks)
acceleration
base
(i)
base
Describe the path taken by the light beam as observed by the inertial observer.
(ii) Describe the path taken by the light beam as observed rom within the spaceship. (iii) Explain the dierence between the paths.
1 3 (IB) a) In both the special and general theories o relativity, Einstein introduced the idea o spacetime. D escribe, with a diagram, what is meant by spacetime with reerence to a particle that is ar rom other masses and is moving with a constant velocity. b) Explain how the general theory o relativity accounts or the gravitational attraction between the E arth and an orbiting satellite.
b) Explain how this dierence relates to the equivalence principle. ( 7 marks)
1 5 (IB) a) S tate the principle o equivalence. b) A spacecrat is initially at rest on the surace o the Earth. Ater accelerating away rom Earth into deep space, it then moves with a constant velocity. A spring balance supports a mass rom the ceiling.
c) D escribe what is meant by a black hole. d) Estimate the radius o the S un necessary or it to become a black hole. ( 1 2 marks)
1 4 (IB) a) A spaceship in a gravity- ree region o space accelerates uniormly with respect to an inertial observer, in a direction perpendicular to its base. A narrow beam o light is initially directed parallel to the base. D iagram 1: View with D iagram 2: View with respect to respect to inertial observer observer in ship
0
0
0
(a)
(b)
(c)
The diagrams show the readings on the spring balance at dierent stages o the motion. Identiy, with an explanation, the reading that will be obtained when the spaceship is: (i)
at rest on the Earths surace
(ii)
moving away rom Earth with acceleration
(iii) moving at constant velocity in deep space.
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QUESTION S c) The spacecrat now accelerates in deep space such that the acceleration equals that o ree all at the E arths surace. State and explain which reading would be observed on the spring balance. ( 1 0 marks)
1 8 (IB) On 2 9 March 1 91 9, an experiment by Eddington carried out during a total eclipse provided evidence to support Einsteins general theory. The diagram below (not to scale) shows the relative position o the Sun, Earth and a star S on this date.
1 6 (IB) The gravitational feld o a black hole warps spacetime.
S
sun
earth
orbit path of earth about sun
Eddington measured the apparent position o the star and six months later, he again measured the position o the star rom Earth. a) (i) (ii)
D escribe what is meant by the centre and the surface o a black hole. D efne the Schwarzchild radius.
(iii) C alculate the Schwarzchild radius or a star that has a mass 1 0 times that o the S un ( solar mass = 2 .0 1 0 30 kg) . b) In 1 979, astronomers discovered two very distant quasars separated by a small angle. Examination o the images showed that they were identical. O utline how these observations give support to the theory o general relativity. ( 9 marks)
1 7 One prediction rom the principle o equivalence is the eect known as gravitational lensing. a) S tate the principle of equivalence. b) Use the principle to explain gravitational lensing. ( 5 marks)
a) S tate why the experiment was carried out during a total solar eclipse. b) Explain why the position o the star was measured again six months later. c) C opy the diagram and draw the path o a ray o light rom S to the E arth as suggested by the general theory. d) Explain how Einsteins theory accounts or the path o the ray. e) Label the apparent position o the star as seen rom Earth. ( 6 marks)
1 9 (IB) a) (i)
D escribe, with reerence to spacetime, the nature o a black hole.
(ii) D efne, with reerence to ( a) ( i) , the S chwarzchild radius. (iii) A star that has a mass o 4.0 1 0 31 kg evolves into a black hole. C alculate the S chwarzchild radius o the black hole, stating any assumption that you make.
547
A
R E L AT I VI T Y b) A spacecrat approaches the black hole in ( a) ( iii) . I it were to continue to travel in a straight line it would pass within 1 000 km o the black hole. (i) S uggest the eect the black hole would have on the motion o the spacecrat. (ii) Explain gravitational attraction in terms o the warping o space-time by matter. ( 1 0 marks) 2 0 Alan and B renda are in a spaceship that is moving with constant speed. C lose to Alan is a light source fxed to the oor o the spaceship. B oth Alan and B renda measure the same value or the requency o the light emitted by the source.
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Alan
Brenda
The spaceship begins to accelerate. a) Explain why B renda observes the source close to Alan to emit light o a lower requency during the acceleration. b) O utline how the situation described in ( a) leads to the idea o gravitational redshit. ( 5 marks)
B E N G I N E E RI N G PH YS I CS Introduction Engineering physics covers some o the key topics that you would expect to meet on an undergraduate engineering course. These are quite diverse topics covering rotational motion, thermodynamics, uid mechanics, and orced vibrations and
resonance. The mathematical content o these topics is naturally restricted to be in line with a preuniversity course; however, those students with a strong mathematical inclination will fnd much in these sub-topics to stimulate urther research.
B.1 Rigid bodies and rotational dynamics Understanding Torque Moment o inertia
Applications and skills Calculating torque or single orces and couples Solving problems involving moment o inertia,
Rotational and translational equilibrium Angular acceleration
Equations o rotational motion or uniorm
angular acceleration Newtons second law applied to angular motion Conservation o angular momentum
Nature of science Extended objects The work covered until now has almost invariably used the model o an object being represented by a single particle o zero dimensions. This model works well or many situations but alls short when orces are not applied along a line passing through the centre o mass o the object. In such cases an extended object can still be modelled; however, objects o dierent shape will not always move in the same way when orces are applied to them. Using simpliying models, engineers were able to design and manuacture sophisticated machines that had signifcant impact during the industrial revolution o the late eighteenth to early nineteenth century.
torque, and angular acceleration Solving problems in which objects are in both rotational and translational equilibrium Solving problems using rotational quantities analogous to linear quantities Sketching and interpreting graphs o rotational motion Solving problems involving rolling without slipping
Equations Torque equation: = Fr sin Moment o inertia: I = mr2 Newtons second law for rotational motion:
= I Relationship between angular requency and requency: = 2f Equations o motion or constant angular acceleration: = i + t 2 = 2i + 2 = i t + _1 t2 2 Angular momentum: L = I Rotational kinetic energy: E K = _1 I 2 2 rot
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E N G I N E E R I N G P H YS I C S
Introduction Note
The angular acceleration is diferent rom both centripetal acceleration ( a c) and tangential acceleration ( a t) this is the rate o change o the linear tangential velocity with time.
As an object moving in a circle undergoes angular acceleration, its tangential acceleration increases as does its centripetal acceleration.
The tangential acceleration o the body is related to the angular acceleration and the radius o the circle (r) by the relationship: at = r
This is because vt = r at
v This means that = _______ = ____r rt where vt is the tangential velocity o the object. Thus the linear quantity (velocity or acceleration) is the angular quantity multiplied by the radius.
This sub- topic is very closely related to the linear mechanics and circular motion that you studied in Topics 2 and 6. There are quantities that are directly analogous to the linear quantities o displacement, velocity, acceleration, orce, etc. With the knowledge that you have already absorbed and an understanding that a rigid body is an extension o a point obj ect you will be able to solve most rotational dynamics problems.
Uniorm motion in a circle You will remember that or a body moving around a circle with a constant linear speed ( or angular speed) there needs to be a means o providing a centripetal orce. This can be a contact orce, such as tension or riction ( or a component o these orces) , or a orce provided by a feld, such as Newtons law o gravitation or C oulombs law. In all cases the orce ( or component) providing the centripetal orce will be given by: mv2 2 F= _ r = m r These terms were defned in Topic 6. Remember that is known as the angular velocity or angular requency and is related to requency f by the equation: = 2 f
Angular acceleration Let us consider an obj ect moving in the circle so that it is given an angular acceleration and its angular speed increases. We defne angular acceleration () as the rate of change of angular sp eed with time. = _ t is measured in radian per second squared ( rad s 2 )
Worked example The diagram shows a hula- hoop ( a large plastic ring) rolling with constant angular speed along a horizontal surace prior to rolling down a uniorm inclined plane. When it reaches a second horizontal surace it, again, moves with a constant angular speed.
hula-hoop
A
S ketch graphs to show the variation with time o a) the angular velocity and b) the angular acceleration o the hula- hoop.
B
Figure 1
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C
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Solution a) The graph shows the hula- hoop travelling with constant angular velocity along A and C . It has a greater value along C since it has now undergone angular acceleration. As B is o constant gradient, the angular acceleration is constant here. b) The second graph shows zero angular acceleration throughout A and C and a constant angular acceleration along B . You should compare these graphs with those for a point object moving along a frictionless surface.
When the hula-hoop is travelling at the higher angular velocity it covers the same distance in a shorter time
/arbitrary units
C B A t/arbitrary units /arbitrary units
B
A
C
t/arbitrary units
Figure 2
Equations of motion In Topic 2 we met the our equations o motion under constant linear acceleration: v = u + at 1 s = ut + __ at 2 2
v2 = u 2 + 2as (v + u)t s=_ 2 In rotational dynamics there are our analogous equations that apply to a body moving with constant angular acceleration: = i + t 1 t2 = it + __ 2
2 = 2i + 2 ( i + )t =_ 2 Here i = initial velocity (in rad s 1 ) , = fnal velocity (in rad s 1 ) , = angular displacement (in rad) or sometimes just " angle" , = angular acceleration (in rad s2 ) , t = time taken or the change o angular speed (in s) .
Note The IB Physics Data Booklet does not include the last of these rotational equations it is probably the most straightforward of the equations because it simply equates two ways of expressing the average angular speed.
Worked example
Solution
A wheel is rotated rom rest with an angular acceleration o 8. 0 rad s 2 . It accelerates or 5 .0 s.
a) = i + t = > = 0 + 8.0 5 . 0 = 40.0 rad s 1 ( i + )t ( 0 + 40.0) 5 . 0 b) = _ = __ = 1 00 rad 2 2 Each revolution makes an angle o 2 radian 1 00 so the number o revolutions = ___ 2
D etermine: a) the angular speed b) the number o revolutions that the wheel has rotated through.
= 1 5 .9 revolutions.
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Moment of inertia The moment o inertia ,I, o a body is the rotational equivalent o the role played by mass in linear dynamics. In a similar way to the inertial mass o an object being a measure o its opposition to a change in its linear motion, the moment o inertia o an object is its resistance to a change in its rotational motion. Objects such as ywheels that need to retain their rotational kinetic energy are designed to have large moments o inertia as shown in fgure 3. The moment o inertia o an obj ect depends on the axis about which it is rotated. For a particle ( a single point) o mass m rotating at a distance r about an axis, the moment o inertia is given by: Figure 3
A pair o fywheels with much o the mass distributed to be around the perimeter.
I = mr2 Moments o inertia are scalar quantities and are measured in units o kilogram metres squared ( kg m 2 ) . For an obj ect consisting o more than one point mass, the moment o inertia about a given axis can be calculated by adding the moments o inertia or each point mass.
m
r
I = mr2
axis o rotation
This summation is a mathematical abbreviation or m 1 r1 2 + m 2 r2 2 + m 3 r1 3 + . . . + m nrn2 when the obj ect is made up o n point masses. The moment o inertia o a simple pendulum o length l and mass m is given by Ipendulum = ml2 while that o a simple dumb-bell consisting o two masses m connected by a light rod o length l, when rotating about the centre o the rod, will be
Figure 4 Moment o inertia
o a
point mass.
( )
l I dumb-bell = m _ 2
2
( )
l + m _ 2
2
( )
l = 2m _ 2
2
1 ml2 =_ 2
axis o rotation into plane o the page light string
light connecting rod l/2
l
point masses o value m
bob o mass m l/2
Note With linear motion there is just one single mass value or an object. In rotational motion, the moment o inertia is also a unction o the position o the axis o rotation. This means that there are an infnite range o possible moments o inertia or any one object.
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simple pendulum
Figure 5 Moment o inertias o a
m axis o rotation simple dumb-bell into plane o page
simple pendulum and dumb-bell.
In each o these cases we assume that the mass is a point mass and that the string and the connecting rod are so light that their masses can be ignored. For a more complicated structure we need to use integral calculus in order to derive ormulae or moments o inertia. In IB D iploma Programme physics questions, any equations that need to be used or moments o inertia will be provided in the question.
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Torque Torque , , can be defned as shown in fgure 6. C onsider orce F acting at point P on an obj ect. The direction o F is such that it makes an angle to the radius o the circle in which the obj ect rotates. The torque will be given by: = Frsin r sin
This is sometime called the orce multiplied by the " arm o the lever" . Torques are also called " moments" , but to avoid conusion between " moment" and " momentum" we advise sticking to the term torque. They are ( pseudo) vector quantities with the direction perpendicular to the plane o the circle in which the obj ect rotates. Imagine your right hand gripping the axis about which the obj ect rotates so that the fngers curl round the axis, when your thumb is " up" it will be pointing in the direction o the torque ( as shown in fgure 7) . This rule also applies to the directions o angular displacement , , angular velocity , , and angular acceleration , , which, like torque, are all vector quantities. The maximum torque that a orce can apply to a body is when the orce is perpendicular to the arm o the lever. In this case = 90 and so sin = 1 . This means that = Fr
F radius (r)
P
Figure 6 Defnition
o torque.
F applied orce radius rom axis
r
torque direction
Couples A coup le consists of a p air of equal and op p osite forces that do not act in the same straight line ( see fgure 7) . This combination o orces produces a torque that causes an obj ect to undergo angular acceleration without having any translational acceleration.
d
F P d
Figure 7
Direction o torque.
-F
P -F
(a)
F (b)
Figure 8 Two examples o couples.
The torque o the couple about the pivot P in both fgure 8( a) and ( b) is equal to Fd ( that is the product o one o the orces and the perpendicular distance between them) .
A single resultant orce acting through the centre o mass o an object will produce translational acceleration. This means that the object will move in a straight line. An object in ree-all is an example o this as (ignoring air resistance) the pull o gravity is the only orce acting on it.
The same orce, but acting through a point displaced rom the centre o mass o the object, will produce a combination o both linear and angular acceleration. This means that the object will move in a helical path.
Note Torque is measured in units of newton metre (N m) but dont confuse this with the unit of work or energy (the joule). In the case of torque the force is perpendicular to the direction in which the object moves but in the case of work the force and direction moved are the same. Torque is a vector quantity while work is a scalar.
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E N G I N E E R I N G P H YS I C S Figure 9 shows a orce acting at the top o an obj ect the eect o this orce can be split into the ( i) same orce acting at the centre o mass o the obj ect ( producing translation) and ( ii) a couple ( producing rotation) . 1 1 The couple consists o __ F at the top and - __ F at the bottom o the obj ect. 2 2 This is a general p rincip le that can be app lied to any force not acting through the centre of mass of an obj ect. A couple acting on an obj ect will produce angular acceleration with no linear motion at all. This means that the object will move in a circular path ( as shown in fgure 8) . 1 2F
F
=
+ F
centre of mass
-1F 2 translation Figure 9
rotation
Force acting through point displaced from centre of mass.
Newtons frst law or angular motion rotational equilibrium We saw in Topic 2 that a body in (translational) equilibrium does not accelerate but remains in a state o rest or travels with uniorm velocity. This means that there can be no resultant orce acting on the body in line with Newtons frst law o motion. For an object to be in rotational equilibrium there can be no external resultant torque acting on it it will then remain in a state o rest or continue to rotate with constant angular velocity. For rotational motion Newtons frst law may be stated as: An object continues to remain stationary or to move at a constant angular velocity unless an external torque acts on it. When the obj ect is in rotational equilibrium: total clockwise torque = total anticlockwise torque. This statement is oten called the " principle o moments" .
Newtons second law or angular motion angular acceleration Again, in Topic 2 we considered the simpler statement o Newtons second law o ( translational) motion equating orce to the product o mass and acceleration. The rotational equivalent o Newtons second law relates the angular acceleration and torque on a body o moment o inertia. = I
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In IB D ip loma Programme Physics questions, the axis to which the torque is ap p lied and the axis about which the moment of inertia is taken will always be the same ( although there are quite simple ways o dealing with situations when this is not the case) .
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Newtons third law for rotational motion You can probably guess that this version o the law says that action torque and reaction torque are equal and opposite this pair o torques, like action and reaction orces or linear motion, act on dierent bodies. I body A applies a torque to body B , then body B applies an equal and opposite torque to body A.
Angular momentum Linear momentum ( p) is a vector quantity defned as being the product o mass and velocity. Angular momentum ( L) is the rotational equivalent to this and is defned as being the product o a bodys moment o inertia and its angular velocity; this, too, is a vector. L = I Angular momentum is measured in units o kg m 2 rad s 1 .
Worked example Solution
A couple, consisting o two 4.0 N orces, acts tangentially on a wheel o diameter 0.60 m. The wheel starts rom rest and makes one complete rotation in 2.0 s. C alculate:
1 t2 = > 2 = 0 + _ 1 2 .0 2 a) = it + _ 2 2 4 = _ = = 3 .1 4 rad s 2 4. 0 b) torque o couple = Fd = 4.0 0.60 = 2.4 N m 2 .4 _ 2 as = I = > I = _ = 3 .1 4 = 0.76 kg m
a) the angular acceleration b) the moment o inertia o the wheel.
The conservation of angular momentum In linear dynamics we fnd the total ( linear) momentum o a system remains constant providing no external orces act on it. In rotational dynamics, the total angular momentum o a system remains constant providing no external torque acts on it. Figure 1 0 shows a small mass being gently dropped onto a reely spinning disc. The addition o the mass increases the combined moment o inertia o the disc and the mass and so the angular velocity o the system now alls in order to conserve the angular momentum. small mass
r
spinning disc Figure 10 An
example of the conservation of angular momentum.
The conservation o angular momentum has many applications in physics, such as the speeding up o an ice skater when pirouetting with decreasing angular momentum, or a gymnast putting in turns or twists by changing the distribution o mass.
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Rotational kinetic energy You probably eel confdent about the analogies between linear and rotational dynamics, so it should not surprise you that the relationship or angular kinetic energy is 1 E K = __ I 2 2 ro t
This, like all energies, is a scalar quantity and could, in principle, be ound by adding the ( translational) kinetic energies o all the particles making up a rotating obj ect. Two more useul rotational analogies are given in the table below.
Quantity
Linear equation
Angular equation
work
W = Fs
W =
power
P = Fv
P =
Rolling and sliding
R + v
v R
R - v Figure 11
Disc rolling on a fat
surace.
I an obj ect makes a perectly rictionless contact with a surace it is impossible or the obj ect to roll it simply slides. When there is riction the obj ect can roll; as the point o contact between the rolling body and the surace along which it rolls is instanteously stationary, the coefcient o static riction should be used in calculations involving rolling. The point o contact must be stationary because it does not slide. Figure 1 1 shows a disc o radius R rolling along a at surace such that its centre o mass has a velocity v. Each point on the perimeter o the disc will have a tangential velocity = R. This means that the top o the disc will have total velocity o R + v and the point o contact with the surace will have a velocity o R - v. As the disc is not slipping, the bottom o the disc has zero instantaneous velocity and so R = v. This means that the top o the disc will have an instantaneous velocity = 2 v. The total kinetic energy o a body that is rolling without slipping will 1 1 be = __ I 2 + __ mv2 . When an obj ect rolls down a slope so that it loses a 2 2 vertical height h, the loss o gravitational potential energy will become the total kinetic energy. This gives us 1 1 mgh = __ I 2 + __ mv 2 2 2
Worked example A solid ball, o radius o 45 mm, rolls down an inclined plane o length 2 .5 m. The sphere takes a time o 6.0 s to roll down the plane. Assume that the ball does not slip. The moment o inertia, I, o a solid ball o 2 mass M and radius R is given by I = __ MR 2 5 a) C alculate the velocity o the sphere as it reaches the end o the plane.
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b) C alculate the angular velocity o the sphere as it reaches the end o the plane. c) D etermine the angle o inclination o the plane. d) C omment on whether the assumption that the ball does not slip was appropriate in this instance.
B . 1 R I G I D B O D I E S A N D R O T AT I O N A L D Y N A M I C S
Solution a) For the ball starting at rest rolling along the slope or time t the ( translational) equations o motion give: 1 s = __ at2 and v = at 2
Substituting the values or s and t 2s _ 2 2 .5 a=_ = = 0.1 4 m s 2 t2 6.0 2 v = at = 0.1 4 6.0 = 0.84 m s 1
substituting or the moment o inertia equation we have: 2 MR2 _ 5 1 __ 2 _ + M Mgh = 2 v R2
(
)
C ancelling M and R2 gives
(
)
2 + 1 =_ 7 2 1 gh = __ v2 _ v 2 5 10 v 7 So h = __ __ g where h is the height o the end 10 o the inclined plane. 2
0.84 b) The angular speed = _vr = _______ = 1 8.7 45 1 0 1 1 9 rad s 3
__1 mv
1 c) rom the equations mgh = __ I 2 + 2 and v = R
(
1 2 _ mgh = __ v I2 + m 2 R
7 0.84 2 h = _ _ = 0.05 m 10 9.81
2
2
0.05 sin = _ 2.5
)
(
)
0.05 = sin 1 _ = 1 .1 2 .5
d) This angle is very small and so there is no problem with assuming that the ball rolls without slipping.
Investigate! Measuring the moment o inertia o a fywheel A fywheel is mounted so that it can rotate on a horizontal axle (the axle orming part o the fywheel) . A mass, suspended by a string, is wound round the axle o a fywheel ensuring that it does not overlap. When the mass is released, the gravitational potential energy o mass is converted into the linear kinetic energy o the mass + the rotational kinetic energy o the fywheel + work done against the rictional orces (all these values are taken when the string loses contact with the axle) . 1 1 I 2 + n W mgh = _ mv2 + _ 1 2 2 The symbols here have their usual meaning with n 1 being the number o turns o string and W the work done against rictional orces during each revolution. When the string disengages, the rotational kinetic energy makes a urther n 2 rotations beore it comes to rest so the rotational kinetic energy o the fywheel must be equal to n 2 W or 1 I 2 _ I 2 = n 2 W making W = _ 2 n2 2 2
n I 1 1 This means mgh = __ mv 2 + __ I 2 + ____ 2 2 2n 1
or
n 1 1 mgh = __ mv 2 + __ I 2 1 + __ n 2 2
(
1
2
)
2
v h In addition to this = vr and __ = __ t 2
Measure the radius r o the axle o the system using vernier or digital callipers.
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Attach a mass hanger to one end o the piece o string and make a loop at the other end so that, as the mass hanger j ust touches the oor, the loop slips o the peg in the axle.
Loop the string over the peg, wind up the string or a whole number o revolutions n 1 and measure the height h o hanger above the oor. Make sure the windings do not overlap. view rom ront
view rom side
axle
peg r axle
fywheel axis o rotation
slotted masses mass hanger
h
Figure 12
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Measuring the moment o inertia o a fywheel.
Adj ust the number o slotted masses on the hanger.
Measure the time t rom the moment o release until the hanger reaches the oor.
C ount the number o revolutions n 2 the ywheel makes ater the string comes o the peg until it comes to rest.
Repeat several times and calculate mean values rom which a value or the moment o inertia o the ywheel and axle I can be deduced.
C onsider how the data could be used graphically to fnd a value or I.
B . 2 TH E RM O D YN AM I CS
B.2 Thermodynamics Understandings The frst law o thermodynamics The second law o thermodynamics Entropy
Applications and skills Describing the frst law o thermodynamics as a
Cyclic processes and pV diagrams Isovolumetric, isobaric, isothermal, and
adiabatic processes Carnot cycle Thermal eciency
Nature of science Dierent viewpoints o the second law o thermodynamics Thermodynamics is an area o physics which, through dierent scientifc eras, has been shaped by a combination o practice and theory. When Sadi Carnot wrote his treatise about heat engines, he was a believer in the caloric standpoint yet his ideas were suciently developed to inuence the development o machines in the industrial revolution. Clausius, Boltzmann, Kelvin, and Gibbs were all responsible or dierent statements o the second law o thermodynamics a law that has a undamental impact on whether or not a process, allowed by the frst law o thermodynamics, can actually occur.
statement o conservation o energy Explaining sign convention used when stating the frst law o thermodynamics as Q = U + W Solving problems involving the frst law o thermodynamics Describing the second law o thermodynamics in Clausius orm, Kelvin orm and as a consequence o entropy Describing examples o processes in terms o entropy change Solving problems involving entropy changes Sketching and interpreting cyclic processes Solving problems or adiabatic processes or 5 _
monatomic gases using p V 3 = constant Solving problems involving thermal eciency
Equations First law o thermodynamics: Q = U + W
3 2 Q Entropy change: S = _ T Equation o state or adiabatic change: Internal energy: U = _ nRT
5 _
p V 3 = constant (or monatomic gases) Work done when volume changes at constant pressure: W = pV useul work done Thermal eciency: = __________________________ energy input Tc o l d Carnot eciency: Ca rn o t = 1- _ Th o t
Introduction When using thermodynamics, there is a convention that we talk about the body that we are interested in as being the system. Everything else that may have an impact on the system is known as the surroundings. The system is separated from the surroundings by a boundary or wall. Everything including the system and the surroundings is called the universe.
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surroundings universe
system
the fask represents the boundary in this case Figure 1
Thermodynamic system and surroundings.
As we saw in Topic 3 , James Joule showed that work done on a system or energy transerred to the system because o temperature dierences result in the same outcome: the internal energy o that system increases. The nature o the way the system changes and how that energy reveals itsel depends on whether or not the phase o the substance changes. When there is no change o state the most apparent eect is the increase in the mean random kinetic energy o the particles; when there is a change o state the increase in the potential energy is the most signifcant eect. Remember that the internal energy of a system is the total of the potential energy and the random kinetic energy of all the particles making up the system. We oten simpliy discussion by considering the system to be an ideal gas. In this case the internal energy is entirely kinetic and we can use the relationship derived in Sub-topic 3 .2 . This showed that, or n moles o an ideal gas, the internal energy U is related to the absolute temperature T 3 by the equation U = __ nRT, where R is the universal molar gas constant. 2
The frst law o thermodynamics The internal energy o a system may change as any combination o ( i) doing work on the system ( or allowing the system to do work on the surroundings) and (ii) transerring energy to or rom the system as a result o a dierence in temperature. Saying this amounts to stating the conservation o energy. There are a variety o versions o the equation or the frst law o thermodynamics and each has its merits they will all be sel-consistent and understandable but you will need to make sure that you read the defnition o each o the terms in the equation. The preerred version o the equation or the IB Physics syllabus is written as: Q = U + W When each o these quantities is p ositive:
Q represents the energy transerred from the surroundings to the system because the surroundings are at a higher temperature than the system
U represents the increase in the internal energy o the system ( this is not simply a change, it is an increase)
W represents the work done by the system as it expands and pushes back the surroundings.
When each o these quantities is negative:
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Q represents the energy transerred from the system to the surroundings because the system is at a higher temperature than the surroundings
B . 2 TH E RM O D YN AM I CS
U represents the decrease in the internal energy o the system
W represents the work done on the system as the surroundings compresses it.
Worked example A systems internal energy alls by 2 00 J as a result o energy transer and work being done. The system does 5 00 J o work on the surroundings.
Solution
a) S tate and explain whether the system is at a higher or lower temperature than its surroundings.
b) Using the frst law o thermodynamics Q = U + W, U must be negative and W must be positive.
b) C alculate the amount o energy transerred causing the reduction in internal energy.
a) The system must gain energy in order to be able to do this amount o work and so its temperature must be below that o the surroundings.
Q = - 2 00 + 5 00 = 3 00J S o 3 00 J are transerred to the system rom the surroundings.
Nature of science Human metabolism and the frst law Let us consider a human body as a thermodynamic system. When we eat, our internal energy increases because we are taking in ood in the orm o chemical potential energy. When there is no energy transer because o temperature dierences this must be a process which is related to work although we will not discuss the biochemistry here. S o, using the frst law with no energy transer: 0 = U + W As U is positive this means that W is negative. The chemical energy in the ood we eat does three main things: it allows us to do work on
our surroundings, it allows us to transer energy to the ( usually) cooler surroundings and the remainder is stored in our bodies ( as at) . With a well- adj usted, balanced diet the build up o at and change in internal energy is zero and the ood we eat allows us to stay warm and do work and be active. S o the ideal system is to make the net change in the internal energy o our bodies zero eat too much and we will build up at ( U is positive) , eat too little and we will lose weight ( U is negative) . In reality, human metabolism is much more complicated than this, but the frst law o thermodynamics does represent a model that has many more applications than the gases we will now ocus on.
Using the frst law or ideal gases When discussing the changes that can be made to the state o the gas, it is usual to consider an ideal gas enclosed in a cylinder by a moveable piston. The gas represents the system, the cylinder and the piston represent the boundaries or walls. Everything else becomes the surroundings.
Calculating the work done in an isobaric change An isobaric change is one which occurs at constant pressure. Consider an ideal gas at a pressure p enclosed in a cylinder o cross-sectional area A. When the gas expands it pushes the piston a distance x so that the volume
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E N G I N E E R I N G P H YS I C S o the gas increases by V (= Ax) . When energy is supplied to the gas rom the surroundings the pressure can remain constant at value at p. The work done by the gas on the surroundings during this expansion will be W. The orce F o the gas on piston = pA. This means that the work done during expansion = Fx = pAx = pV.
piston
W = p V V= A X
X
area A
Figure 3 shows a p- V graph or an isobaric change. The area under the graph is equal to the work done. The arrow on the line connecting the end points AB shows that the gas is expanding and doing work on the surroundings an arrow in the opposite direction would show a compression.
volume V
ideal gas = system
cylinder Figure 2
Applying the frst law o thermodynamics to this, we get Q = U + pV. This could result in any number o possibilities but they must be consistent with this equation; Q must be positive or an expansion and negative or a compression. For example, supplying 2 0 J to the gas could increase the internal energy by 1 9 J and allow the gas to do 1 J or work or increase the internal energy by 1 9.1 and do 0.9 J o work etc.
Work done by an ideal gas.
p area under line = work done by gas on surroundings = p V A
p
V The equation o state or an isobaric change is __ = constant. T
B
Work done for non-isobaric changes
p
V Figure 3
This will be a positive quantity because the system is doing work on the surroundings. When the gas is compressed W will be negative.
When changes do not occur at constant pressure we can still calculate the work done rom the area under a p- V graph. We can make the assumption that the pressure will be unchanged over a small change in volume and, thereore, we approximate the overall change to a series o small isobaric changes as shown in fgure 4.
V
Work done in an isobaric change.
The area o the frst constant pressure rectangle = p 1 V1 = W1 p
A
The area o the second rectangle would be p 2 V2 = W2 , etc.
area of isobaric rectangle = p 1 V1 = work done
p1 p2
Thereore, the total work done = n Wn = n p n Vn or n rectangles this is the area under the curve. S o or any p- V graph the area under the curve will give the work done this depends on the path taken and not j ust on the end points.
B
Isothermal changes V1
V
V2
Figure 4 Work done in
a non-isobaric change.
Note Isothermal changes normally take place very slowly and the boundary between the system and the surrounding must be a good conductor of energy.
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Isothermal changes are those that occur resulting in the internal energy of the system staying constant. The internal energy o an ideal gas consists o the sum o the mean random kinetic energies o the particles o gas. The mean kinetic energy is proportional to the temperature o the gas. An isothermal change, thereore, means that there is no change in the temp erature of the system. Using the frst law o thermodynamics Q = U + W with U = 0, this leaves Q = W. As there is no change in internal energy, the energy transerred to the system because o a temperature dierence between the system and the surroundings Q will allow the system to do work W on the surroundings. As W = pV this can only mean that the gas is expanding the direction arrow on the graph should go rom A to B .
B . 2 TH E RM O D YN AM I CS
The other possibility is that - Q = - W so the energy transerred rom the system to the surroundings is equal to the work done on the system by the surroundings. In this case the gas is being compressed the direction arrow on the graph should go rom B to A.
p
We saw in Topic 3 that when the temperature does not change pV = constant. This is the equation o the line on a p- V graph or an isothermal change. The lines are known as isotherms. As shown in fgure 6, changes at higher temperatures will always produce isotherms that are urther rom the origin than those at lower temperatures. For a given volume the pressure will always increase in moving rom a low temperature isotherm to one at higher temperature.
A
B area = work done V
Adiabatic changes
Figure 5 Work done in
This is the name given to a change in which no energy is transferred between the system and the surroundings. This does not mean that the system and the surroundings are always at the same temperature, although that could be so, but it is more likely that there is a wellinsulated barrier between them. Adiabatic changes usually happen very quickly, which means that there is no time or the energy to transer.
an isothermal change.
p T1 > T2
Applying the frst law o thermodynamics to a system undergoing an adiabatic change gives:
T1
Q = U + W but, as Q = 0, this can mean either U = - W ( an increase in internal energy occurs because o work being done on the system) or - U = W ( a decrease in internal energy occurs because the system is doing work on the surroundings) . For an ideal monatomic gas the equation or an adiabatic change takes the orm:
T2 V Figure 6 Isotherms at diferent temperatures.
Note
5 _
p V 3 = constant 5 _
You dont need to derive the equations or adiabatic changes. The exponent or V does vary i the gas molecules are diatomic (two atom molecules) or polyatomic (three or more molecules in the atom) . The value is actually the ratio o the principal molar specifc heats but, as this is not going to be examined, we leave you to do urther research on this.
5 _
This can be written as p 1 V1 3 = p 2 V2 3 when there is a constant temperature. As the gas is an ideal gas the equation p 1 V1 p 2 V2 _ = _ T1 T2 also applies to an adiabatic change. 5 5 _ _ p 1 V1 p 2 V2 D ividing p 1 V1 3 = p 2 V2 3 by _ = _ T1 T2 gives 2 _
2 _
p isotherms
T1 V1 3 = T2 V2 3 In the p- V equation or an adiabatic change V is raised to a power o greater than one. This means that or an adiabatic change the line will be steeper than that or an isothermal change as shown in fgure 7. The area under the adiabatic change, as or all other changes, will be the work done. The direction o the arrow will determine whether work is done on or by the gas.
adiabatic process
Isovolumetric changes These changes occur at constant volume and, thereore, mean that no work can be done by or on the system. Q = U + W = U + p V
work done V Figure 7
Adiabatic and isothermal changes.
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With no change in the volume, the frst law o thermodynamics becomes Q = U ( when the energy transerred to the system increases the internal energy) or - Q = - U ( when the energy transerred rom the system decreases the internal energy) .
B
Figure 8 shows an isovolumetric change on a p- V graph. In this case the change shows an increase in temperature ( moving to a higher isotherm on an isothermal graph) .
A V Figure 8 Isovolumetric change.
p The equation o state or an isovolumetric change is __ = constant. T
Worked example a) D istinguish between an isothermal process and an adiabatic process as applied to an ideal gas. b) An ideal gas is held in a cylinder by a moveable piston and energy is supplied to the gas such that the gas expands at a constant pressure o 1 .5 1 0 5 Pa. The initial volume o the cylinder is 0. 040 m 3 and its fnal volume is 0.1 2 m 3 . The total energy supplied to the gas during the process is 7.5 1 0 3 J. ( i) S tate and explain the type o change that the gas undergoes. ( ii) D etermine the work done by the gas. ( iii) C alculate the change in internal energy o the gas.
Solution
transerred because o the temperature dierence between the gas and the surroundings and work done one on the other. An adiabatic process is one in which there is no energy exchanged between the system and the surroundings. This means that changes in the internal energy (and hence temperature) occurs because o work done by or on the ideal gas. b)
( i) As the pressure does not change the gas undergoes an isobaric expansion. ( ii) Work done = area under the p- V graph or the change ( = p V) = 1 .5 1 0 5 ( 0.1 2 - 0.04) = 1 .2 1 0 4 J ( iii) Using the frst law o thermodynamics Q = U + W
a) An isothermal process is one that takes place at constant temperature (and constant internal energy) so there is an interchange o energy
U = Q - W because only 7.5 1 0 3 J is supplied to the gas and 1 2 .0 1 0 3 J o work is done, the internal energy must all by 4.5 1 0 3 J
Cycles and engines hot reservoir T1 Q1
heat engine
work = Q 1 - Q 2
Q2 cold reservoir T2
Figure 9
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The principle of a heat engine.
Work can be converted into internal energy eectively through rictional orces. Work done by riction is usually undesirable because it increases the temperature o the system ( good) and the surroundings ( bad) . The reverse process o continuously converting energy into work is more difcult to achieve, but it can be done using a heat engine that operates through a cycle o changes. In 1 82 4 the French physicist and engineer, S adi C arnot, published the frst description o a heat engine; in this he described what has come to be known as the " C arnot cycle" . The principle o a heat engine is to take in energy at a high temperature, rej ect energy at a low temperature and use the remainder o the energy to do work on the system as illustrated in fgure 9. We can think o the system as being a gas enclosed in a cylinder with a rictionless moveable piston. When energy Q 1 is supplied to the gas rom a hot reservoir at temperature T1 , the gas expands and moves the piston, doing work. This will stop as soon as the gas pressure is equal to that o the surroundings. The gas now needs to be returned to its original state beore it can do urther work. This can only happen i some o the energy (Q 2 ) that was initially absorbed is rejected to a cold reservoir at lower temperature T2 .
B . 2 TH E RM O D YN AM I CS
The thermal efciency o the heat engine will be given by: Q1 - Q2 useul work done W = __ = _ = _ Q1 Q1 energy input
p A
In the C arnot cycle we imagine a completely riction- ree engine that is able to take a gas through a cycle o two isothermal and two adiabatic changes as shown in fgure 1 0.
B T1
S tarting at point A where the gas is at its highest temperature T1 , it expands isothermally to B by absorbing energy Q 1 . The internal energy o the gas does not change so all the energy absorbed is doing work.
D Q2
T2
C
V
The gas now expands adiabatically to C where the temperature alls to T2 . No energy is now being absorbed but the gas still does work on the surroundings by losing some internal energy. D uring the expansion AB C the area under the two expansion curves gives the work done on the surroundings.
T1 > T2
Q1
Figure 10
The Carnot cycle.
Note
At C the gas now needs work being done on it so it is compressed isothermally to D and rej ects energy Q 2 . The internal energy does not change so the work done on the gas is all rej ected as energy. Finally, the gas is urther compressed adiabatically rom D back to A. The work done on the gas is all used to increase the internal energy in returning the gas to temperature T1 . D uring the compression C D A the area under the two compression curves gives the work done by the surroundings on the gas. The area enclosed by the curve is the net work done by the gas on the surroundings in one cycle. The C arnot heat engine is said to be reversible. This is a theoretical concept in which, at any part in the cycle, the system can be returned to a previous state without any energy transerence this must be done infnitely slowly and means that the system returns exactly to its initial state at the end o the cycle. It can be shown that, or the C arnot cycle ( and all reversible heat engines) , that the thermal efciency is given by T1 - T2 T = _ = 1 - _2 T1 T1
Temperatures used here must be in kelvin.
This is the equation or the maximum eciency o a heat engine.
The maximum eciency is increased by raising the temperature o the hot reservoir and/or by lowering the temperature o the cold reservoir.
The maximum eciency can never equal 100% as this would mean that the cold reservoir was at absolute zero or else the hot reservoir was at an infnitely high temperature neither o these requirements is possible.
In practice thermodynamic cycles can be achieved but none will be more ecient than the Carnot cycle.
This is written in the IB Physics S yllabus as Tcold Carnot = 1 - _ Thot
Worked example A quantity o an ideal gas is used as the working substance o a heat engine. The cycle o operation o the engine is shown in the p- V graph opposite. C hange C A is isothermal.
p I10 5 Pa A
12.0
B
8.0
The temperature o the gas at A is 3 00 K. a) D uring the change AB the change in internal energy o the gas is 7.2 kJ.
C
4.0
( i) C alculate the temperature, at B , o the gas. ( ii) D etermine the amount o energy transerred during change AB .
0
0
2
4
6
V I10 -3 m 3
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The increase in internal energy = 7.2 kJ so the energy transerred to the gas must equal 4.8 + 7.2 = 1 2 . 0 kJ.
b) S tate why, or the change B C , the change in the internal energy o the gas is numerically the same as that in AB . c) C alculate: ( i) the net work done in one cycle ( ii) the efciency.
Solution a)
b) The gas undergoes the same change in temperature and, as the gas is ideal, this means that the change in internal energy depends solely on the temperature. c)
pV ( i) For an ideal gas __ = constant T V V As pressure is constant __ = __ = >TB T T A
B
A
B
V T 6.0 1 0 300 = ______ = _____________ = 900 K V 2.0 1 0 B
-3
A
-3
A
( ii) W = p V = 1 2 .0 1 0 5 ( 6.0 1 0 - 3 - 2 .0 1 0 - 3 ) = 4. 8 kJ
hot reservoir T1
( i) We need to fnd the area enclosed by the cycle. Each large square is equivalent to 1 .0 1 0 - 3 2 .0 1 0 5 J = 2 00 J Estimate that there are 1 4 large squares, making a total 2 800 J or 2 .8 kJ. W 2 .8 ( ii) = _ = _ = 0.2 3 or 2 3 % 12 Q1
Heat pumps and refrigerators Heat pumps and rerigerators act in a similar manner to a heat engine working in reverse. They take in energy Q 2 at a low temperature T2 , do work W on the working substance and rej ect energy Q 1 at the high temperature T1 . Although they are very similar in their working the heat pump is designed to add energy to the high temperature reservoir ( or example, the room being heated by extracting energy rom the ground) whilst the rerigerator is designed to remove energy rom the low temperature reservoir ( the cool box) . Air conditioning units are another example o heat pumps.
Q1 refrigerator or heat pump W = Q1 - Q2 Q2 cold reservoir T2 Figure 11
The principle of a heat pump and refrigerator.
Worked example The diagram shows the relationship between the pressure p and the volume V o the working substance o a rerigerator or one cycle o its operation. The working substance is a volatile liquid which is made to vaporize and condense.
S uggest the reason why both the changes rom C D and AB are isothermal, isobaric changes. b) S tate during which process o the cycle energy is absorbed rom the cold reservoir and during which process energy is transerred to the hot reservoir.
p B
a) The working substance at point C o the cycle is entirely in the liquid phase.
A
c) S tate how the value o the work done during one cycle may be determined rom the pV diagram. C
D V
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Solution a) B oth changes are isobaric and isothermal because there is no pressure or temperature change. Each o the changes occurs because o a change o phase o the working substance. From C to D the liquid vaporizes and rom A to B the vapour condenses.
b) Energy is absorbed during C to D as the liquid needs energy to vaporize and it is ejected during A to B in order or the vapour to condense. c) The area enclosed by the cycle will always indicate the net amount o work in this case it is work done on the system (working substance) .
The second law of thermodynamics Although this is a undamental law that has its origins in practical experiences, the second law o thermodynamics can be stated in a number o dierent ways. The frst law o thermodynamics equates work to energy, the second law deals with the circumstances in which energy can be converted to work. Each o the statements is equivalent to the others and communicates an expression o the impracticability o reversing real thermodynamic processes. The C lausius version o the second law can be stated as: It is imp ossible to transfer energy from a body at a lower temp erature to one at higher temp erature without doing work on the system. The Kelvin ( or Kelvin-Planck) version states: It is imp ossible to extract energy from a hot reservoir and transfer this entirely into work. I the second law was not true it would be possible to power ships by energy extracted rom the sea this cannot be done because there needs to be a cold reservoir into which the dierence between the energy extracted and the work done would be rej ected. When, ater a time, we return to ull cup o coee we do not expect to fnd it at a higher temperature than when it was made. Energy passes rom the hot coee to the cooler room until the two bodies are at the same temperature i we wanted the coee to heat up we would have to transer energy to it using a heating coil or else do work on it by stirring it rapidly! The internal energy o an obj ect is related to the random motion o the molecules o the obj ect. B y trying to convert this internal energy into work we are trying to convert random motion into something more ordered. It is impossible to do this because we cannot take control over the individual motion o a colossal number o molecules.
Entropy A third version o the second law o thermodynamics involves the concept o entrop y S. This quantity can be defned ( or a reversible change) in terms o the equation Q S = _ T
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Note
Q is the energy absorbed by the system
T is always positive so when energy is absorbed by a system and Q is positive there will be a positive change in entropy an increase. When energy is rejected by the system the entropy will decrease.
For an adiabatic change Q = 0 and so S = 0
A substance taken through a complete reversible cycle will undergo no change in T2
= entropy as = 1- ____ T Q2
Q1
1
1
1
Entropy is a scalar quantity and has units o j oule per kelvin ( J K 1 )
Worked example 0.2 0 kg o ice at 0 C melts. The specifc latent heat o usion o water is 3 .3 1 0 5 J kg 1 . C alculate the change in entropy o the ice as it melts.
Solution E nergy needed to melt the ice = mL = 0.2 0 3 .3 1 0 5 = 6.6 1 0 4 J Q 6. 6 1 0 4 S = _ = _ = 2 .4 1 0 2 J K 1 T 2 73
Q2
1- ____ this means ____ = ____ T T Q
T is the temperature in kelvin at which this occurs.
2
All heat engines reject energy to the surroundings and generate an overall increase in entropy of the universe.
Entropy as a measure of disorder We have now seen that or real processes the entropy o the universe increases in act this is another way o stating the second law o thermodynamics frst suggested by B oltzmann. Real processes always degrade the energy, i.e. change the energy rom being localized to being more spread out. I some hot water is mixed with cold water in a completely insulated container there is no loss o energy, however, the opportunity to use the energy to do work is now restricted by having cold water. It is not a sensible proposition to separate the most energetic molecules to produce some hot water and some cold water rom the mixture the energy has changed rom the localized situation in the hot water molecules to a situation where it is spread- out amongst all o the molecules. Imagine having 1 0 coins all placed heads up on the table. The coins are picked up and shaken beore being returned to the table, without looking to see where they are placed; some coins will, thereore, have heads up but others will have heads down. The coins have moved rom an ordered 1 10 1 state to a disordered state. There is a small likelihood ( __ = ____ that 2 ) 1 024 the coins would be replaced with all 1 0 heads up. B y increasing the number o coins rom 1 0 to 1 00 it decreases the likelihood o them all being heads up to 1 in 1 .3 1 0 30, and, this really isnt likely to happen! The coin experiment mirrors nature in that a system does not naturally become more ordered. We have now seen that the entropy o a system naturally increases with the disorder o the system. This is not coincidental since it can be shown that entropy is a measure of the disorder of a system.
(
Figure 12
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Ten coins all heads up.
)
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TOK The arrow of time Many scientists have discussed increasing entropy as representing the "arrow o time". Because the disorder o a large system will increase with time, fnding such a system with increased order would be equivalent to time going backwards. Perpetually increasing entropy is
contrary to Newton's laws o motion in which the change o state o a system is equally predictable whether we go orwards or backwards in time. Does this statement o the Second law prevent the possibility o time travel?
Worked example
Solution
a) S tate what is meant by an increase in entropy of a system.
a) When the entropy increases, there is an increase in the degree of disorder in the system.
b) S tate, in terms of entropy, the second law of thermodynamics.
b) The total entropy of the universe increases.
c) When a chicken develops inside an egg, the entropy of the egg and its contents decreases. E xplain how this observation is consistent with the second law of thermodynamics.
c) E ntropy of the surroundings must increase more than the decrease of entropy in the developing egg. The energy generated by the biochemical processes within the egg becomes more spread out as a consequence of some passing into the surroundings.
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B.3 Fluids and fuid dynamics (AHL) Understandings Density and pressure
Applications and skills Determining buoyancy orces using
Buoyancy and Archimedes principle Pascals principle
Hydrostatic equilibrium The ideal uid
Streamlines The continuity equation
The Bernoulli equation and the Bernoulli efect Stokes law and viscosity Laminar and turbulent ow and the Reynolds
number
Archimedes principle Solving problems involving pressure, density and Pascals principle Solving problems using the Bernoulli equation and the continuity equation Explaining situations involving the Bernoulli efect Describing the rictional drag orce exerted on small spherical objects in laminar uid ow Solving problems involving Stokes law Determining the Reynolds number in simple situations
Equations Buoyancy orce: B = V g Pressure in a uid: p = p 0 + gd Continuity equation: Av =constant The Bernoulli equation: ___ 1 v2 + gz + p = constant 2
Stokes law: FD = 6rv
vr Reynolds number: R = _
Nature of science Fluids in motion The study o the transportation o mass is very important in medicine and engineering. Knowledge o how the velocity and pressure changes throughout a moving uid is vital in our understanding o many physical processes, including how blood ows around the body, how aircrat y, and how jet engines operate. The visualization o uid ow can be extremely appealing and has been used extensively in the visual arts. 570
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L)
Introduction The gaseous and liquid states o matter are j ointly known as fuids substances that can ow and take the shape o their containers. There are important dierences between gases and liquids:
It is easy to compress a gas but liquids, like solids, are almost incompressible.
Gases are not restricted by a surace but liquids are you can have hal a cup o coee but not hal a cup o air! The most energetic molecules may, however, be able to break through the surace to orm a vapour above the liquid.
Although this topic is called uids and uid dynamics, it is almost entirely restricted to idealized uids those that cannot be compressed, are non-viscous, and ow in a steady manner. These properties are best represented by a low viscosity liquid, such as water, and not by a ' sticky' one, such as oil ( which has much internal riction) .
Static fuids density and pressure The density o a substance is given by the ratio o the mass m o the substance to the volume V o the substance: m = _ V Pressure is the ratio o the perpendicular contact orce acting on a surace to the area o the surace: F p=_ A Although the direction o orce is at right angles to the surace on which it acts, pressure is a scalar quantity and acts in all directions; with the orce in newton and the area in metres squared, pressure is measured in pascal ( Pa) . In a uid, pressure is ound to increase with depth and, at a given depth, the pressure is ound to be equal in all directions. This results in a perpendicular orce acting on any surace at a given depth. I this was not the case, any pressure dierences would cause the liquid to ow until the pressure was constant. C onsider a cylinder o height h and base area A in a uid o density as shown in fgure 2 the top o the cylinder is at the surace o the uid. The orces acting on the bottom o the cylinder are the weight W o the column o liquid above it acting downwards and the pressure orce pA rom the liquid acting upwards. These are in equilibrium so
orce F p= F A area A Figure 1
Defnition o pressure.
cross-sectional area A
h
W
uid o density
W = pA W = mass o cylinder gravitational feld strength = mg, this means that W = Ahg = pA and thereore pA
p = hg In a liquid that is not in a sealed container the atmospheric pressure p 0 will also act on the liquid surace. This makes the total pressure p at depth h become hg + p 0 p = hg + p 0
Figure 2
Pressure in a uid.
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Nature of science Measuring density of immiscible liquids using a U-tube When two liquids that do not mix are added to the limbs o a U-tube, their levels settle in a way that is dependent on their relative densities. At the level XY in fgure 3 the pressure will be constant, so
liquid A o density A
p 0 + h A A g = p 0 + h B B g C ancelling p 0 and g gives
atmospheric pressure = p 0
liquid B o density B
h A A = h B B When we know the density o one liquid we are then able to calculate that o the second liquid by taking a ratio o the heights. This experiment is an example o hydrostatic equilibrium.
X
Y
Figure 3 Hydrostatic equilibrium to measure density.
Pascals principle This principle says that the pressure applied at one point in an enclosed fuid under equilibrium conditions is transmitted equally to all parts o the fuid. The principle allows hydraulic systems to operate. For example, the hydraulic jack is used to raise heavy objects such as cars.
Fx Fy Ax
Ay
pressure throughout fuid = p x Figure 4 Hydraulic jack.
A small orce Fx applied to a piston o small cross-sectional area A x produces a pressure Px in a liquid such as oil. This pressure is communicated through the liquid so that it acts on a second, larger piston o cross-sectional area A y. This will then produce a larger orce Fy so that Fy Fx px = _ = _ Ax Ay The device cannot ampliy the energy, so the distance moved by Fx must be ar more than that moved by Fy in order to conserve energy ( work being orce distance) .
pxA fuid o density
Archimedes principle cylinder o cross-sectional area A
h
(p x + p x) A Figure 5 Archimedes principle.
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When you dive into a swimming pool you quickly eel the buoyancy orce o the water acting on you. The magnitude o this upward orce is expressed by Archimedes' principle; this says that or an obj ect wholly or p artially immersed in a fuid there will be an up ward buoyancy orce acting on the obj ect which is equal to the weight o fuid that the obj ect disp laces. The buoyancy orce is oten known as the up thrust. We can deduce this relationship by considering a cylinder o height h and cross- sectional area A ully immersed in a liquid o density . The pressure is p x at the top and ( p x + p x) at the bottom. The latter is higher because o the extra depth o uid. The orces on the top and bottom o the cylinder will be p xA and ( p x + p x) A. We ignore the atmospheric pressure because it acts on both the top and the bottom o the cylinder.
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L) The buoyancy orce B will, thereore, be the dierence between these orces B = ( p x + p x ) A - p xA = p xA p x = hg making B = hgA
Note When an object is foating:
The buoyancy orce acting on the object must equal the weight o the object.
The buoyancy orce will equal the weight o fuid that is displaced by the submerged portion o the foating object.
V = hA And so B = Vg, which is the weight o the uid displaced by the cylinder. To emphasize that this applies to a uid, the IB Physics data booklet shows this equation as B = V g
Worked example A block o wood o mass 5 0 kg is completely immersed in water. The density o the wood is 1 .1 1 0 3 kg m 3 and that o the water 1 .0 1 0 3 kg m 3 a) C alculate the resultant orce on the block. b) When the block is cut into planks and made into a boat it oats, displacing 0.1 5 m 3 o water when empty. C alculate the buoyancy orce acting on the boat.
Solution a) The weight o the block = mg = 5 0 9.81 = 4.90 1 0 2 N B uoyancy orce = weight o water displaced = B = V g m 50 _ 3 V = volume o block = _ = 1 .1 1 0 3 = 0. 045 m B = V g = 1 .1 1 0 3 0.045 9. 81 = 4.86 1 0 2 N The overall orce is, thereore, 4.90 1 0 2 - 4.86 1 0 2 = 4 N downwards b) The weight o water displaced by the boat = 0.1 5 1 .0 1 0 3 9.81 = 1 .5 1 0 3 N S o B = 1 .5 1 0 3 N ( around three times the original value)
Fluid dynamics There are many instances o uid ow. Examples include the smoke rom a fre, hot water in a central heating system and wind spinning around in a tornado. Fluid dynamics also deals with solid objects moving through stationary uids. An ideal uid oers no resistance either to a solid moving through it or it moving through or around a solid object. An ideal uid is said to be non-viscous. We will start our discussion o uid dynamics by considering a non-viscous, incompressible uid in a stream tube.
Streamline (or laminar) fow Ideal uids ow in a very predictable way. We can represent the paths taken by particles within a uid using the concept o streamlines. In
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E N G I N E E R I N G P H YS I C S streamline fow, the motion o a particle passing a particular point is identical to the motion o all the particles that preceded it at that point. Streamlines will be close together when the particles move quickly and urther apart when they move more slowly. A group o streamlines is known as a stream tube. Fluid never crosses the surace o a stream tube. Figure 6 shows the streamlines in a tube that narrows and widens again.
Figure 6 Streamlines in
a tube that narrows.
The continuity equation With steady fow the mass o fuid entering one end o a stream tube must be equal to that leaving the other end. The stream tube may be a physical tube or pipe or it may be a series o streamlines within the fuid.
Ay vy vx
Ax
Y X Figure 7
Streamlines at the boundary of a stream tube.
For a length o stream tube the fuid enters at X through an area o cross- section A x at velocity vx and leaves at Y through an area o crosssection A y with velocity vy. In a short time, t, the fuid leaving X will travel a distance vx t thus meaning that a volume A xvx t and a mass A xvx t enter the tube in this time. In the same time a mass A yvy t will leave the stream tube through Y. As there cannot be any discontinuities in an ideal liquid these masses must be equal so A xvx t = A yvy t meaning that when t is cancelled rom both sides A xvx = A yvy We see rom this that Av = constant This is known as the continuity equation and the product Av is known as the volume fow rate ( or, sometimes, j ust fow rate) . Volume fow rate is measured in units o m 3 s 1 .
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B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L)
The Bernoulli equation This equation is a generalized equation which deals with how a uid is able to both speed up and rise to a higher level as it passes through a stream tube. The equation is derived rom the conservation o energy and, although the derivation is shown below, you will not need to repeat this in an IB Physics examination. In your data booklet the B ernoulli equation is written as 1 __ v 2 + gz + p = constant 2
Here is the uid density, v is its speed, g the gravitational feld strength, z the height above a chosen level, and p the pressure at that height. You will notice in the equation that, i was replaced by m in each o the frst two terms, we would have kinetic energy and potential energy. m As density is equal to __ , multiplying the B ernoulli equation by V V would give 1 __ mv 2 + mgz + pV = dierent constant 2
This has now turned the B ernoulli equation into the conservation o energy because it implies that the kinetic energy + gravitational potential energy ( z is height here) + work done ( remember the frst law o thermodynamics) remains the same. The equation in the B ernoulli orm expresses each o the terms as an energy density ( energy per unit volume) and is measured in J m 3 . This equation also tells us that, or any point in a continuous steady ow, the total o all the quantities will remain the same: i the pressure changes then one or more o the other terms must also change.
Derivation of the Bernoulli equation y Fy = p y A y Y Y' vy
x
hy
Fx = p x A x hx
X
X' vx
Figure 8 The Bernoulli equation.
C onsider a uid entering and leaving a pipe that becomes wider and higher. It is obvious that the pressure at the inlet must be greater than the pressure at the outlet. When the inlet pressure is p x the orce Fx at the inlet is p xA x and that at the outlet is p yA y. In a short time, t, the uid entering the pipe at X moves a short distance x rom X to X at velocity vx. This means that the work done on the uid is Fx x = p xA x x. In the same time, the same mass o uid moves a distance y in moving rom Y to Y and the work done will be Fy y = p yA y y.
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E N G I N E E R I N G P H YS I C S The net work done will be p xA x x - p yA y y As this happens in the same time, t, and because equal masses must mean equal volumes or an ideal fuid o constant density, this gives A x x = A yy = V thus net work done = V( p x- p y ) 1 the change o kinetic energy or this mass = __ m ( vy 2 - vx 2 ) 2
and the change in potential energy = mg( h y - h x ) O verall, or the conservation o energy: work done = gain in kinetic energy + gain in gravitational potential energy so 1 V( p x - p y ) = __ m v 2 - vx2 ) + mg ( h y - h x ) 2 ( y
D ividing by V and collecting terms or x and y we get 1 1 vx2 + gh x = p y + __ vy2 + gh y p x + __ 2 2
This is the B ernoulli equation. In the IB Physics data booklet the symbol z is used or heights.
Applications of the Bernoulli equation From the B ernoulli equation it ollows that, when a fuid speeds up, there must be a decrease in either the pressure or the gravitational potential energy or both o these. When the fow is horizontal there can be no change in the gravitational potential energy, so there must be a reduction in pressure. Aerooils, used on aircrat wings and mounted on racing cars, are shaped so that the air fows aster over the more curved surace than over the fatter surace in most cases it is the aerooil that moves, but the relative eect is identical. The aster moving air produces a lower pressure and less orce acts on that section o the aerooil this causes the aerooil to be orced upwards ( a lit) or downwards ( a down thrust) depending on which way it is positioned. The eect o the shape o the aerooil causes the streamlines to be closer together on the curved side and to be urther apart on the fatter side. longest path, high velocity streamlines
direction o air fow
shorter path, low velocity Figure 9
The aerooil.
Venturi tubes When a tube narrows, the speed o the fuid increases at the narrow part o the tube beore returning to the original value at the wider part. This means that the pressure is higher at the edges o the tube and lower
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B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L) in the centre. This is the principle o the Venturi gauge, which is used or measuring uid speeds as demonstrated in fgure 1 0. The higher air pressure has a greater eect on the right limb o the manometer. This pressure dierence can be calibrated to measure relative uid speeds and ow rates. narrow high speed wide low speed
lower pressure
higher pressure
height proportional to pressure diference
Figure 10 Venturi gauge.
Pitot static tubes Pitot static tubes are used or measuring the velocity o a uid o density . The tubes X and Y must be on the same streamline. The opening o tube X is perpendicular to the direction o uid ow but the opening o tube Y is parallel to the ow. The uid can then ow into the opening o tube Y. The two tubes must be ar enough apart or tube X not to aect the velocity at Y. When there is steady ow, no uid particles can move rom one streamline to another and so none enter tube X. Tube X measures the static pressure p x. Particles will enter tube Y where their kinetic energy will be brought to zero. Pitot tube
static tube
hy
hx ow direction velocity = v X Figure 11
Y
owing liquid
Pitot static tubes.
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E N G I N E E R I N G P H YS I C S Applying the B ernoulli equation to the streamline along XY gives 1 vx2 = p y p x + __ 2
which gives ________
vx =
_2 ( p
y
- px )
As p x = h xg + p 0 and p y = h yg + p 0 the velocity equation becomes vx =
X
_________
2 g( h y -
hx )
For an idealized uid this will be the velocity along the streamline XY. This arrangement is the one used or measuring the velocity o liquids, but it can be adapted to measure the ow o gases. A Pitot tube is installed on aircrat wing to measure the speed o the aircrat relative to the air.
hx
Y
Flow out of a container You may be amiliar with the leaking can demonstration. Figure 1 2 shows liquid pressure increasing with depth. This is another application o the B ernoulli equation. The line XY represents a streamline j oining water at the surace to a hole in the side o the can. Each o these two points will be open to the atmosphere and so they will both be at atmospheric pressure. This means that p x = p y = p 0 . Figure 12
The B ernoulli equation becomes:
Leaking can demonstration.
1 1 __ vx2 + gh x = __ vy2 + gh y 2 2
Here h x is the height above Y and h y = 0. I we assume the container is wide enough and the holes small enough or the velocity o the water surace to approximate to zero, the equation reduces to 1 gh x = __ vy2 2
C ancelling and re- arranging the equation gives ____
vy = 2 gh x
This confrms that the velocity is related to the height o water above opening. Viscosity and turbulence have been neglected.
Worked example X
tank 12.0 m
The diagram shows a large water tank, open to the atmosphere. The tank eeds a pipe o crosssectional area 4.0 1 0 3 m 2 which, in turn, eeds a narrow tube o cross- sectional areas 5 .0 1 0 4 m 2 . The narrow tube is sealed with a closed valve. The water surace in the tank is 1 2 .0 m above the centre o the two tubes. density o water = 1 .00 1 0 3 kg m 3
Y
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Z
valve
atmospheric pressure = 1 . 0 1 0 5 Pa
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L)
a) C alculate the pressure at Y. b) C alculate the pressure at Z. c) The valve is now opened so that water leaves the system. The pressure at the open valve is now atmospheric.
Note That this is not an equation given in the IB Physics data booklet. I you were tested on it, the question would need to be structured in order to lead you through the development o this equation.
( i) D etermine the velocity o the water at Z. ( ii) C alculate the velocity o the water at Y. ( iii) C alculate the pressure at Y.
( ii) Using the continuity equation A yvy = A zvz
Solution
4.0 1 0 3 vy = 5 . 0 1 0 4 1 5 .3
a) and b) The pressure at Y and Z will be equal because they are at the same level. The pressure will be p = p 0 + gd = 1 .0 1 0 5 + ( 1 . 00 1 0 3 9. 81 1 2 . 0) = 2 .1 8 1 0 5 Pa 2 .2 1 0 5 Pa
vy = 1 .9 m s 1
c) ( i) Assuming the surace area is large enough or the kinetic energy at X to be zero we have shown that ___ _____________ v = 2 gh = 2 9. 81 1 2 . 0 = 1 5 .3 m s 1
( iii) Using the B ernoulli equation between Y and Z ( they are on the same level so there will be no dierence in gravitational potential energy) . 1 1 vz2 = p y + __ vy2 p z + __ 2 2 1 and as p z = p 0 then p y = p 0 + __ v 2 - vy 2 ) 2 ( z 1 py = 1 .0 1 05 + __ 1.00 1 03 ( 1 5.3 2 - 1.9 2 ) 2
= 2 .2 1 0 5 Pa
Viscosity o fuids We have seen that ideal uids are non-viscous. Real uids do, o course, have viscosity with some uids being more viscous than others. For a viscous uid in laminar ow, each layer impedes the motion o its neighbouring layers. In a pipe, the layers adjacent to the inner walls o the tube are stationary while the layers in the centre o it travel at the highest speed. The viscosity o uids is highly temperature dependent, with most liquids becoming less viscous at higher temperatures and gases becoming more viscous at higher temperatures. The viscosity o a uid is measured in terms o its coefcient o viscosity ( at a given temperature) . This is a quantity measured in units o pascal second ( Pa s) .
Stokes law When a sphere, o radius r, alls slowly through a viscous uid it pulls cylindrical layers with it. Under these conditions o streamline ow, the sphere experiences a viscous drag orce FD . This is given by FD = 6rv where v is the velocity o the sphere and is the coefcient o viscosity o the uid. I the sphere is contained in a tube, the tube would need to be very wide or S tokes equation to apply. When a small metal ball is released in oil, the ball will initially accelerate until the drag orce ( plus the buoyancy orce) is equal to the weight o the ball the ball will then travel at its terminal speed so 4 6rvt + __ r3 g = mg 3
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E N G I N E E R I N G P H YS I C S or 4 4 r3 g = __ r3 g 6rvt + __ 3 3
and 2 r 2 g( - ) vt = __ 9 Here vt is the balls terminal speed, is the density o the uid and is the density o the ball.
Investigate!
metal ball
Measuring the coecient o viscosity o oil
0
Using the apparatus shown in fgure 1 3 you can investigate Stokes equation.
Use six or more balls o dierent radii ( but made o the same material) small balls should reach the terminal velocity in a ew centimetres i you use a viscous liquid such as engine oil or glycerol.
light gate 10 20 metre ruler 30 40
You will need to know the densities o the liquid and the material o the balls.
50 seconds
You will need to make sure that the balls reach their terminal velocity use a long, wide transparent cylinder.
60
oil
I you use steel balls, a magnet can be used to remove them rom the oil.
80
Think about ways o releasing the balls consistently. Think how you are going to measure the diameter o balls accurately.
timer
70
90 light gate 100 Figure 13
Stokes law investigation.
Turbulent fow laminar
turbulent
Figure 14 Laminar and
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turbulent fow.
At low velocities uids ow steadily and in layers that do not mix because the uid is viscous the layers travelling closest to the walls will move slowest as shown in fgure 1 4. When the uid velocity is increased or obstacles proj ect into the uid the ow becomes turbulent. The ow is no longer laminar and the particles in the dierent layers mix with each other. This causes the smooth streamlines, seen during laminar ow, to break up and orm eddies and vortices. It is not easy to predict when the rate o ow is sufciently high to cause the onset o turbulence. A quantity, known as the Reynolds number R, gives a practical " rule o thumb" that can be used to predict whether the ow is great enough to become turbulent. This is a dimensionless quantity which is calculated rom the equation: vr R=_
B . 3 FLU I D S AN D FLU I D D YN AM I CS ( AH L) For a uid v is velocity, r the radius o the pipe ( or other dimension or a river, length o a plate etc.) , the density and the coefcient o viscosity. Using this defnition a Reynolds number that is less than 1 000 is taken to represent laminar ow. I the Reynolds number is greater than 1 000, it does not mean that the ow will be turbulent because there is a transition stage between the uid being laminar and becoming turbulent. It is generally accepted that, using this defnition o the Reynolds number, a value o above 2 000 will be turbulent.
Worked example O il ows through a pipe o diameter 3 0 mm with a velocity o 2 .5 m s 1 . The oil has viscosity 0. 3 0 Pa s and density o 890 kg m 3 a) D etermine whether or not this ow is laminar. b) C alculate the maximum velocity at which the oil will ow through the pipe and still remain laminar.
Solution C oefcient o viscosity is oten j ust called " viscosity" . vr 2 .5 1 5 1 0 3 890 ___ a) R = _ = 111 = 0.3 0 This value is below 1 000 so the ow is laminar. R 1 000 0. 3 0 __ 1 b) v = _ r = 1 5 1 0 3 890 = 2 2 m s
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B.4 Forced vibrations and resonance (AHL) Understandings Natural requency o vibration
Applications and skills Qualitatively and quantitatively describing
Q actor and damping Periodic stimulus and the driving requency Resonance
Nature of science Risk assessment and resonance Resonance can be a useul phenomenon and, as discussed in this sub-topic, it is utilized in many walks o lie. Despite this, the possibility o one system interacting with another has its drawbacks loud music with heavy bass notes or the noise during construction work can carry over long distances and can, thereore, impinge upon peoples lives. Most nations promote the inclusion o resonance hazards during risk assessment applicable when carrying out building or maintenance work. For example, devising saety measures such as incorporating sound insulation in wall cavities or installing double or triple glazing to shield rom the efects o unwanted sounds rom outside.
examples o under-, over-, and criticallydamped oscillations Graphically describing the variation o the amplitude o vibration with driving requency o an object close to its natural requency o vibration Describing the phase relationship between driving requency and orced oscillations Solving problems involving Q actor Describing the useul and destructive efects o resonance
Equations Quality actor equations energy stored Q = 2 ______ energy dissipated per cycle energy stored Q = 2 resonant requency ___ power loss
Introduction All mechanical systems, and some electrical systems, will vibrate when they are set in motion. We have seen this in S ub- topics 4.1 and 9.1 when studying simple harmonic motion. More intricate examples o oscillations, that are not simple harmonic, include the motion o a tall building in the wind, the shaking o an unbalanced washing machine as its drum spins, or the vibrations caused by a car engine misfring. In some instances, the vibrations are useul but, oten, the unction o the system will beneft rom being damped when the amplitude o oscillation is restricted. D ampers are increasingly being incorporated in tall buildings to restrict potential earthquake damage.
Free vibrations When a mechanical system is displaced rom its rest position and allowed to vibrate, without any external orces being applied, it will oscillate at its natural requency f0 . S uch vibrations are called free vibrations.
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B . 4 F O R C E D V I B R AT I O N S A N D R E S O N A N C E ( A H L )
Oscillations and damping At a given time the amplitude o an oscillation depends on how much the system is damped as well as the size o the driving orce. This happens when an oscillating system experiences a resistive orce which causes the amplitude o the object oscillating to decay. The resistive orce acts in the opposite direction to the motion o the system and it increases with the speed o the oscillation. This means that the damping orce is a maximum when the system passes through its equilibrium position as it will have a maximum speed at this point ( and it will be zero at the maximum displacement at which point the system momentarily stops) . In opposing the damping orce, the system must do work and this reduces the energy that it stores causing the amplitude to decay. The resistive nature o the orce slows the oscillator down which increases the time period ( although this eect is negligible when a system is lightly damped) . Figure 1 shows how the displacement o an oscillator varies with time when there is light damping. This is also known as under-damp ing. The envelope o the curve takes an exponential shape when the damping or resistive orce is proportional to the speed; this means the ratio o the amplitudes at hal- period intervals is a constant value 0.25 0.2 1 0.1 8 ____ ____ etc. ) . ( ____ 0.21 0.1 8 0.1 5
Investigate! Iteratation with damped SHM
( the symbols having their usual meaning and b being the damping actor . .. a number between 0 and 1 ) .
The term bv should now be subtracted rom each acceleration value in the spreadsheet iteration discussed in the SHM spreadsheet investigate! in sub-topic 9.1 (the second box in the ow chart is modifed to a n + 1 = - kxn - bvn) .
You should try out dierent values or b in order to j udge the impact that it has on the damping o the shm.
An example o this spreadsheet is provided on the webpage.
displacement/m 0.25 0.20 0.15
The equation or a velocity dependent S HM can be written as ma = - kx - bv
0.10 0.05 0
2
4
6
8
10
12
14
16
18
20 t/s
-0.05 -0.10 -0.15 -0.20
Figure 1
Under-damped vibration.
Although the amplitude decreases with time, the requency and time period o the oscillator are approximately constant. Heavier damping may completely stop a system rom oscillating. When the system is displaced and it returns to its equilibrium position without overshooting it, the system is " heavily damped" . A system returning to its equilibrium position in the shortest possible time is said to be critically damped, while a system taking longer than this to return to the equilibrium position is over-damped. Critical damping is necessary in many mechanical systems using this type o damping in car suspension systems avoids oscillations on uneven suraces; such oscillations may lead to a loss o control and a very uncomortable ride. For a similar reason, fre doors in buildings are oten ftted with automatic closers that are critically damped.
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E N G I N E E R I N G P H YS I C S amplitude
over-damped critically damped time
under-damped Figure 2
Variation o amplitude with time or diferent degrees o damping.
These curves can be modelled by adding a velocity- dependent damping term to the simple harmonic motion equation. Figure3
Child being pushed on
amplitude o the orced vibration
swing.
zero damping
light damping heavy damping natural requency driving requency
Figure 4 Efect o damping on
resonance.
Note
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For zero damping the amplitude would be infnite, however, this could not happen in practice because infnite amplitude would require an infnite amount o energy to be supplied! As the damping increases the peak moves slightly to the let o the natural requency. Increased damping reduces the sharpness o the resonance peak.
Forced vibrations When an external orce acts on a mechanical system, the orce may have its own requency o vibration, which may aect the motion o the mechanical system. A simple example o this is a child on a swing. When the child is let to his own devices he will only be able to swing at a particular requency which is a property o the childswing system. Pushing the child provides an external orce that causes the motion o the swing to change. The amplitude o the swing will only increase i the pushes are applied at a rate matching the swing requency. When the applied requency does not match the swing requency the amplitude might decrease. Forced vibrations are those that occur when a regularly changing external orce is applied to a system resulting in the system vibrating at the same requency as the orce.
Resonance When a mechanical system is orced to oscillate by a driving orce that has the same requency as the natural requency o the mechanical system, it will vibrate with maximum amplitude. This is called resonance. When the requency o the driving orce becomes closer to the natural requency o the system the amplitude o the oscillation will be greater. This can be seen in fgure 4, which shows the variation with the driving requency o the systems amplitude. The degree o damping alters the systems amplitude response. Resonance occurs in many physical systems ranging rom tuning radios to the tidal eects o the Moon and the unction o lasers.
Q factor The Q or " quality" actor is a criterion by which the sharpness o resonance can be assessed. It is defned by the relationships: energy stored Q = 2 ___ energy dissipated per cycle energy stored = 2 resonant requency __ power loss This is an arbitrary defnition with the 2 being included so that using the equation with a real system becomes simplifed! In the equations or many rotating or oscillating systems the 2 actor will cancel ...making estimation o Q actors straightorward. The two defnitions are equivalent relationships:
B . 4 F O R C E D V I B R AT I O N S A N D R E S O N A N C E ( A H L ) writing Es or energy stored, Ed or the energy dissipated per cycle, f0 or the resonant requency, T0 or the resonant period, P or power loss ( = power dissipated) and E or the energy lost in time t. The second relationship becomes Es E 1 _ _ Q = 2 f0 __ = 2 P T0 _ E t E t s = 2 _ _ T0 E s
tim e co n side re d t The actor __ = ________________ = n ( number o oscillations) T tim e o r o n e o scillatio n 0
Es
S o Q = 2 n __ E E E = 2 __ = 2 __ E __ E s
n
s
d
e ne rgy sto re d or in words Q = 2 ___________________ e ne rgy dissip ate d p e r cycle
This is the frst relationship. The Q actor is a numerical quantity and has no unit. A system with a high Q actor is lightly damped and will continue vibrating or many oscillations as the energy dissipated per cycle will be small. As a rule o thumb, the Q actor is approximately the number o oscillations that the system will make beore its amplitude decays to zero (without urther energy input) . The larger the Q actor, the sharper the resonance peak on the graph o amplitude against driving requency. A high quality actor means a low loss o energy. For light damping Q = m/b where b is the damping actor relating the resistive orce Fr to the speed v: Fr = - bv S ome typical values o Q actors are:
Oscillator
Q factor
critically damped door
0.5
loaded test tube oscillating in water
10
mass on spring
50
simple pendulum
200
oscillating quartz crystal
30 000
Worked example
Solution
An electrical pendulum clock has a period o 1 . 0 s. An electrical power supply o 2 5 mW maintains its constant amplitude. As the pendulum passes its equilibrium position it has kinetic energy o 40 mJ.
a) The pendulum has a requency o 1 .0 Hz.
a) E xplain how these quantities apply to the Q actor relationship. b) C alculate the Q actor or the pendulum clock.
As it is storing 40 mJ, the rate o energy supplied must equal the rate at which energy is lost i.e., the power is supplied at a rate o 25 mW. energy stored b) Q = 2 resonant requency __ power loss 40 1 0 3 = 2 1 _ 2 5 1 0 3 =1 0
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E N G I N E E R I N G P H YS I C S The Q factor is an especially important quantity for electrical oscillations transmitting radio waves. When selecting radio and television stations it is essential that the transmitting and receiving circuits are tuned to the same resonance frequency.
Barton's pendulums Figure 5 shows an arrangement of pendulums that can be used to investigate resonance. The apparatus is known as B artons pendulums named after the B ritish physicist, E dwin B arton.
2
1
0
3 5
4 paper cones
Figure 5 Bartons pendulums.
brass bob
As the diagram shows, the set up consists of a number of paper cone pendulums of varying lengths. All are suspended from the same string as the brass bob that acts as the " driver" pendulum. When the driver pendulum is displaced from its rest position and released, it forces all the paper cone pendulums to oscillate with the same frequency, but with different amplitudes. This is an example of forced oscillations. The " cone" that has the same length as the driver pendulum has the greatest amplitude because it has the same natural frequency as the driver pendulum. In calculating the period using the equation for the period of a simple pendulum, the " equivalent length" of each pendulum should be taken from the centre of the bob to the horizontal broken line shown on the diagram. C areful observation shows that:
cone 3 ( which has the same length as the driver pendulum) always lags behind the driver pendulum by one quarter of period ( equivalent to 90 or __ radian) 2
the shorter cones ( 1 and 2 ) are almost in phase with the driver pendulum
the longer pendulums ( 4 and 5 ) are almost in anti- phase ( 1 80 or radian out of phase) with the driver pendulum.
Figure 6 shows the variation, with forcing frequency, of the phase lag for a pair of pendulums. The length of the driver pendulum string is adjusted when it is shorter than the driven pendulum the forcing frequency is higher and the driven pendulum will lag behind the driver by radian, etc. phase lag/rad driver leads by half a period lighter damping heavier damping /2 driver leads by quarter of a period
driver and driven in phase 0 f0 Figure 6 Phase relationship for the displacement of a
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driver frequency
forced vibration.
B . 4 F O R C E D V I B R AT I O N S A N D R E S O N A N C E ( A H L )
Nature of science Examples of resonance In nature, resonance is a very common phenomenon having examples in virtually all areas o physics. Further examples include:
The human voice uses resonance to produce loud sounds rom a relatively weak source the vocal cord.
The sound box o musical instruments amplifes the energy, causing air in the box to resonate at the same requency.
O zone in the stratosphere has natural requencies o vibration that match the requency o ultraviolet and absorb this radiation.
Microwave cookers emit electromagnetic waves that match the natural requency o water molecules in ood.
The optical cavities in lasers set up standing waves or light in order to produce coherent beams.
MRI ( magnetic resonance imaging) scans use resonating protons in atoms to provide vital inormation about body cells.
Resonance can have drawbacks as well as benefts, or example:
Vibration set up by soldiers marching across bridges has meant that they are told to " break step" . In June 2 000, Londons Millennium Footbridge ( fgure 7) was ound to sway alarmingly rom the resonances set by pedestrians crossing it. Fitting dampers solved the problem but ailed to prevent the bridge being known as " the wobbly bridge" .
The Tacoma Narrows Bridge in Washington State, USA, collapsed in November 1 940 as a result o cross winds matching its natural requency.
Figure 7 The London Millennium Footbridge.
Feedback can be heard at rock concerts a loud howling sound is produced when microphones or pickups are too close to loudspeakers, giving an uncontrolled amplifcation o the sound.
The clamped hacksaw blade will have a natural requency that depends on the length projecting.
The amplitude o the end o the blade can be determined using a vertically clamped millimetre scale ( not shown) .
This apparatus can be used to measure the variation o amplitude with ( a) supply requency f ( b) proj ecting length L.
Vibrations in machinery can cause vibrations in nearby obj ects overtaking mirrors on lorries that are idling can be seen to vibrate with large amplitude.
Investigate! Resonance of a hacksaw blade Apparatus similar to that shown in fgure 8 can be used to investigate the resonance o a hacksaw blade.
An electromagnet is powered by a signal generator that behaves as a variable requency ac supply.
The requency can be varied by adj usting the signal generator.
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The general orm o the relationship may be x0 = kfn or x0 = kL n
This suggests a loglog graph should generate a straightline graph.
at both ends. The magnet is positioned somewhere near one end o the wire.
Melde' s string oers a similar investigation when a long metallic wire carries a current in a magnetic feld ( see S ub- topic 4.5 ) :
An alternating small current is set up in the wire.
B y changing the tension in the wire the natural requency o the wire will approach the requency o the AC current.
When resonance is reached, standing waves will appear along the wire, the harmonic depending on the conditions.
Take care because the wire can become red-hot!
The wire is placed between the poles o a strong U- shaped magnet and is clamped
signal generator wooden blocks
hacksaw blade
G-clamp
electromagnet
laboratory jack
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Figure 8 Resonance of a hacksaw blade.
QUESTION S
Questions 1
An obj ect o mass 3 .0 kg is attached to a string which is wrapped round a thin uniorm disc o radius 0.2 5 m and mass 7. 0 kg. The disc rotates about a horizontal axis passing through its centre. Assume the rotation o the disc is rictionless.
7 kg
(b)
(a)
3
The angular speed o a rotating disc is increased rom 2 0 rad s - 1 to 85 rad s - 1 in a time o 6.0 s by a constant torque. The disc has mass o 8.0 kg and radius is 0.3 5 m. C alculate: a) the work done by the torque in this time b) the average power applied to the disc during this time. ( The moment o inertia o a disc is given by the equation in question 1 . ) ( 7 marks)
C alculate: a) the angular acceleration o the disc when the mass is released rom rest b) the angular acceleration i the orce were applied by pulling the cord with a constant orce o 3 0 N. The moment o inertia I o a thin uniorm disc o mass M and radius R is given by 1 I = __ MR 2 ( 5 marks) 2 2
A skater has moment o inertia 2 .85 kg m 2 with her arms outstretched. She rotates with an initial angular speed o 2 .0 rad s - 1 as shown in the diagram. B y bringing in her arms the skaters moment o inertia reduces to 1 .5 kg m 2 . For the skater, calculate:
4
( IB) The graph shows the variation with volume o the pressure o a fxed mass o gas when it is compressed adiabatically and also when the same sample o gas is compressed isothermally. 7.0
C
6.0 pressure/ 10 5 Pa
3.0 kg
5.0
B
4.0 3.0
a) her fnal angular speed b) the change in her rotational kinetic energy.
A
2.0
( 5 marks)
1.0
2.0
3.0
4.0
5.0
6.0
Volume /10 -3 m 3
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B
E N G I N E E R I N G P H YS I C S a) S tate and explain whether line AB or AC represents the isothermal compression.
O n a copy o the diagram above: ( i) draw arrows to show the direction o the changes
b) O n a copy o the graph, shade the area that represents the dierence in work done in the adiabatic change and in the isothermal change.
(ii) label with the letter A an isobaric change ( iii) label with the letter B the change during which energy is transerred to the working substance because o a temperature dierence.
c) D etermine the dierence in work done, as identifed in ( b) . d) Use the frst law o thermodynamics to explain the change in temperature during the adiabatic compression. ( 9 marks)
5
c) Use data rom the diagram in ( b) to estimate the work done during one cycle o the working substance. d) ( i) B y reerence to entropy change, state the second law o thermodynamics.
(IB) An ideal gas at an initial pressure o 4.0 1 0 5 Pa is expanded isothermally rom a volume o 3.0 m3 to a volume o 5.0 m 3 .
( ii) The cycle o the working substance in ( b) reduces the temperature inside the rerigerator. E xplain how your statement in ( d) ( i) is consistent with the operation o a rerigerator. ( 1 1 marks)
a) C alculate the fnal pressure o the gas. b) O n graph paper, sketch a graph to show the variation with volume V o the pressure p during this expansion. 7
c) Use the sketch graph in ( b) to: ( i) estimate the work done by the gas during this process ( ii) explain why less work would be done i the gas were to expand adiabatically rom the same initial state to the same fnal volume. ( 7 marks) 6
( i) the volume rate o ow in m 3 s 1 through the tube ( ii) the speed o the water in the narrower part o the tube.
p I10 5 Pa
b) The diagram below shows the pressure volume ( p- V) changes or one cycle o the working substance o a rerigerator.
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a) C alculate:
(IB) a) S tate what is meant by isobaric change.
13 12 11 10 9 8 7 6 5 4 3 2 1 0
A horizontal tube o cross-sectional area 3 .0 1 0 4 m 2 narrows to 1 .6 1 0 4 m 2 . Water ows through the tube at a speed o 0. 40 m s 1 at the wider part.
b) Explain how the ow o water in the pipe is accelerated. ( 6 marks) 8
The diagram shows a venturi meter which is used to measure the ow rate o low-density liquid through a horizontal pipe. When the liquid is owing, the mercury manometer is used to measure the dierence between the pressure o the liquid in the pipe at the entrance to the venturi meter and that in the throat o the meter: entrance pipe
throat
exit pipe liquid fow
0
0.2
0.4
0.6 0.8 1.0 V I10 -3 m 3
1.2
1.4
1.6
mercury manometer
80 mm
QUESTION S a) In one measureme nt the die rence in levels o the me rcury in the manome ter arms is 8 0 mm. C alculate the pressure diere nce betwe en the liquid in the e ntrance to the ve nturi meter and that in the thro at, e xpre ssing your answer in pascals. density o me rcury = 1 . 4 1 0 4 kg m - 3 b) the corss- sectional areas o the pipe and venturi meter throat are 4.0 1 0 - 2 m 2 and 1 .0 1 0 - 2 m 2 respectively. D etermine the ratio o the speed o the liquid owing in the throat o the venturi meter to its speed in the entrance pipe. c) The liquid has a density o 8. 0 1 0 2 kg m - 3. ( i)
Use the B ernoulli equation to estimate the speed o the liquid in the horizontal pipe.
( ii) Hence estimate the ow rate o the liquid in kg s - 1 . ( 1 0 marks) 9
a) ( i) Use the continuity equation to show that an incompressible liquid moving rom a wider pipe to a narrower pipe must increase its velocity. ( ii) S tate B ernoullis relation or the ow o an incompressible inviscid uid along a horizontal stream line. b) Figure 1 shows a cross-section through a simple laboratory flter pump. B
( i)
Explain how a ow o water rom A to C through the pump produces a partial vacuum at D .
( ii) C alculate the velocity o the water emerging rom the nozzle B . ( iii) C alculate the rate, in m 3 s - 1 , at which water ows through the pump. ( density o water = 1 000 kg m - 3 ) ( 1 3 marks) 1 0 A spring or which the extension is directly proportional to the weight hung on it is suspended vertically rom a fxed support. When a weight o 2 . 0 N is attached to the end o the spring the spring extends by 5 0 mm. A mass o 0 . 5 0 kg is attached to the lower end o the unloaded spring. The mass is pulled down a distance o 2 0 mm rom the equilibrium position and then released. a) ( i) S how that the time period o the simple harmonic vibrations is 0. 70 s. ( ii) O n a sheet o graph paper sketch the displacement o the mass against time, starting rom the moment o release and continuing or two oscillations. S how appropriate time and distance scales on the axes. b) The massspring system described in part (a) is attached to a support that can be made to vibrate vertically with a small amplitude. Describe the motion o the massspring system with reerence to requency and amplitude when the support is driven at a requency o ( i) 0.5 Hz
( ii) 1 .4 Hz ( 8 marks)
A
C D
1 1 The Millennium Footbridge in London was discovered to oscillate when large numbers o pedestrians were walking across it.
Figure 1
The Pressure at C is at atmospheric pressure ( 1 00 kPa) and the water at C is moving very slowely. The Pressure at D is 45 kPa. Nozzle B has a diameter o 2 . 0 mm. C alculate
a) What name is given to this kind o physical phenomenon? b) E xplain the conditions that would cause this phenomenon become particularly hazardous?
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B
E N G I N E E R I N G P H YS I C S c) Suggest two measures which engineers might adopt in order to reduce the size of the oscillations of a bridge. ( 7 marks) 1 2 a)
A forced vibration could show resonance. E xplain what is meant by: ( i) forced vibrations ( ii) resonance.
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b) ( i) E xplain what is meant by damping. ( ii) What effect does damping have on resonance? ( 5 marks)
C I M AG I N G Introduction D ata collection in science often depends on using our senses to make an observation. S ometimes the image we perceive needs to be enhanced in some way. For example, doctors
may need to see inside the human body in order to make a diagnosis. This topic deals with the physics of imaging in both visual and non- visual contexts.
C.1 Introduction to imaging Understanding Thin lenses
Applications and skills Describing how a curved transparent interace
Converging and diverging lenses Converging and diverging mirrors
Ray diagrams Real and virtual images Linear and angular magnifcation
Spherical and chromatic aberrations
Nature of science Virtual images cannot be ormed directly on a
screen. The technique o ray tracing allows the position and size o a virtual image or a virtual object to be inerred. This is an example o deductive logic in action.
modifes the shape o an incident waveront Identiying the principal axis, ocal point, and ocal length o a simple converging or diverging lens on a scaled diagram Solving problems involving not more than two lenses by constructing scaled ray diagrams Solving problems involving not more than two curved mirrors by constructing scaled ray diagrams Solving problems involving the thin lens equation, linear magnifcation, and angular magnifcation Explaining spherical and chromatic aberrations and describing ways to reduce their eects on images
Equations 1 =_ 1 +_ 1 Thin lens equation: _ v u f 1 Power o a lens: P = _ f
h ho Angular magnifcation: M = _i o
v
i Linear magnifcation: m = _ = - _ u
Magnifcation o a magniying glass:
D + 1; M D M n e a rp o i n t = _ =_ in f n ity f f
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Converging and diverging mirrors Topic 4 introduced the rules o reection at plane suraces. In this section we look at what happens when suraces are no longer at. A curved reecting surace allows the rays to be manipulated, leading to images o various types ormed by mirrors. The shapes o convex and concave suraces are shown in fgure 2 . Mirrors made with these shapes are commonly ound in a number o situations: as mirrors used or makeup or shaving where a magnifed image is needed, or as a way or seeing a wide angle o view.
Figure 1 Concave and convex mirrors and their refections.
We will mainly consider mirror suraces that have a spherical profle. Figure 2 shows what happens when light is incident on both convex and concave reectors that have been ormed rom the surace o a hollow sphere. The line that goes through the centre o the mirror surace at 90 to the surace is the p rincip al axis. concave mirror
convex mirror
Q
principal axis
C
F
P
C
F
Q f ocal length R radius o curvature (a)
ocal length
(b)
Figure 2 When rays that are parallel and close to the principal axis are incident on a concave mirror then they all reect through a point known as the principal focus F. F is also known as the focal p oint o the mirror. It is easy to see why this has to be the case. The second law o reection tells us that the angle o incidence at the mirror is equal to the angle o reection. The surace o the mirror is part o a sphere (drawn as a circle in the two-dimensional diagram) and thereore the line joining the centre o the sphere C to the mirror at the point where the light ray is incident is the normal to the surace. This defnes the angle o incidence i. The angle o reection r must equal i. I the incident ray is close to the principal axis then, ater reection, the ray will cross the principal axis at a point F that is hal way rom the pole P o the mirror to the centre o the sphere. The distance rom F to the pole is known as the focal length f. When the distance between the principal axis and the incident ray is small, f is hal the radius o curvature o the mirror. We shall look at what happens when incident rays lie well away rom the principal axis later. For now, our assumption is always that the distance between rays and the axis is small. When a convex mirror is used ( fgure 2 ( b) ) the position is very similar, except that this time the reected ray appears to have come from the ocus. There is the same relationship between f and the radius o the mirror as beore.
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Investigate! Images from a mirror These experiments are intended to introduce you to the images ormed by a mirror. The mirrors that will work best or these experiments have a ocal length o about 1 5 2 5 cm. You need to know the approximate ocal length o each mirror beore you begin.
(1) Concave B egin by taking the concave mirror and looking into it with an approximate distance between your eye and the mirror that is ( a) about double the ocal length, ( b) less than the ocal length. What do you notice about the image you see? Is the image the right way up or upside down? C an you describe what happens when the eye is exactly at the ocal point? Now set the mirror on the bench using suitable stands so that the light rom an illuminated object is incident on the mirror and then refected to orm
an image on a screen. Make sure that the object mirror distance u is greater than the ocal length. Find out the range o u or which the image is larger, smaller, and the same size as the object.
(2) Convex This time simply look into the mirror. Where is the image ormed? Is it similar to the image ormed by a plane mirror, what are the dierences? lamp
concave mirror
wooden block
white paper ruler
Figure 3.
To explain the positions o images ormed by curved mirrors we use the technique o drawing a ray diagram. A ray diagram establishes the relationship between an obj ect and its image that results when rays rom the obj ect are refected in the mirror. Using the idea that light is reversible we can identiy p redictable rays that we will use to construct a scaled ray diagram. Reversibility means that when light rays are refected back along their original path they will trace out the incident path but in the reverse direction.
X
Y Z
The predictable rays we will use are:
X: initially parallel to the principal axis and then going, ater refection, through the ocal point.
Y: initially through the ocal point and then going, ater refection, parallel to the principal axis.
Z: a ray through the centre o curvature that goes ater refection back along its original path.
Figure 4 Predictable rays.
Using these rays we can determine the position and the nature ( size and type) o the image ormed by a curved mirror. Remember that X, Y, and Z are close to the principal axis and the mirror is almost fat at the centre. The drawing technique uses several conventions:
The mirror is drawn in two dimensions as a straight line at 90 to the principal axis. O nly the top and bottom o the mirror are curved to indicate whether the mirror is convex or concave.
The vertical scale and the horizontal scale need not be the same. O bj ects and mirrors are sometimes only centimetres high but can
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The obj ect is represented by a vertical arrow drawn to show the largest dimension o the obj ect to scale.
At least two predictable rays are drawn rom the top o the obj ect arrow. The position o the top o the image o the arrowhead will be shown by the direction o the rays ater reection.
Arrows are drawn to show the direction in which the rays travel.
The step-by- step construction o one ray diagram each or a concave and a convex mirror is shown in sequence in fgure 5 . Concave
object C
C
C
C
Convex 1 Draw the principal axis and the mirror. Add the centre o curvature and the ocal point. Add the object to scale (vertical and horizontal scales can be diferent) .
F
F
2 Draw one ray rom the top o the object parallel to the principal axis. Ater reection the ray goes through/appears to have come rom the ocal point. Note that a line or the convex mirror is dotted because it is a construction line, not a ray.
3 Draw another ray either (a) through the ocal point and then reecting parallel to the principal axis or (b) to and rom the centre o curvature.
F
F image
object
4 Where the rays/construction lines cross is the position o the image. The nature and position o the image give inormation about it.
This image is smaller than the object (diminished) , upside down (inverted) , and real (because the rays cross) .
F
C
F
C
F
C
image F
C
This image is smaller than the object (diminished) , the right way up (erect) , and virtual (because the rays appear tohave come rom the image and do not cross) .
Figure 5 Step-by-step instructions or drawing concave mirror and convex mirror ray drawing. The concave mirror has a number o separate cases each o which depends on the position o the object. I the rays rom the top o the object converge to a single point then this marks the top o the arrowhead. This is now below the principal axis (whereas the object was above) because the image is an inversion o the object (turned upside down) . This point where the rays converge is still called the top o the image.
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The rays in these cases orm a real image. Rays that intersect ( converge) in this way can be used to orm an image on a screen. This is what was happening when you saw a picture orming on the screen in the Investigate!
C .1 I N TRO D U CTI O N TO I M AG I N G
However, when the obj ect is closer than the ocal point, the fnal ray directions diverge ( move apart) and do not cross. When these diverging rays are constructed back ( notice that construction lines are always drawn as broken lines in the worked examples as they are not rays) , they orm a dierent type o image known as a virtual image. This type o image cannot be ormed on a screen but it can be seen or manipulated using another mirror or a lens. Virtual images in the Investigate! were ormed by the lens in your eye. A convex mirror by itsel can never orm a real image rom a real obj ect. As the rays always diverge, the convex mirror is called a diverging mirror. For the concave mirror, unless the obj ect is closer to the lens than the ocal point, the rays converge. C oncave mirrors are usually called converging mirrors in some books they are reerred to as converging ( concave) mirrors.
Magnifcation B ecause the diagrams or both types o mirrors are scaled, the fnal scale sizes o the obj ect and image can be used to measure the linear magnifcation m o the mirror system. hi height o image m = __ = _ height o obj ect ho
object v __ u,
Geometry shows that m is also equal to where u and v are the distances rom the mirror pole to the obj ect and the image respectively.
ho
o i
C
hi F image
In some mirror examples that we shall see later, the obj ect and image heights cannot be easily identifed. In these cases angular magnifcation M is used. M = __ where i and o are the angles i
o
subtended at the mirror by the rays rom the image o the top o the image and the rays rom the top o the obj ect, respectively.
Figure 6 Magnifcation by a mirror, linear and angular.
Aberrations in a mirror S o ar we have assumed that the rays rom the obj ect are travelling close and approximately parallel to the principal axis. When this is not the case then the spherical mirror introduces distortions ( aberrations) into the images. The principal reason or this is that at large distances rom the pole o the mirror the rays no longer reect through the ocal point. When rays parallel to the principal axis are incident over the whole mirror surace then they orm a pattern known as a caustic curve. You may have noticed this pattern orming when strong sunlight strikes the reecting surace o a circular cup o coee or another opaque liquid. The inner surace o the cup reects the light onto the liquid surace. The position o ocal point and the shape o the caustic can be clearly seen. To reduce this problem in cases where a large mirror surace is needed, a parabolic surace can be used. Rays parallel to the principal axis o a parabolic mirror always pass through the ocus and no caustic orms in these circumstances ( fgure 7( b) ) . An example is the large mirror used or an astronomical telescope.
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(a) spherical mirror
(b) parabolic mirror
Figure 7 Spherical mirror distortions and the parabolic mirror. The reversibility of light tells us that a parabolic mirror shape can also be used to generate a parallel beam of light if a light source is placed at the focus. This is often used to produce a searchlight beam that diverges very little even at large distances from the mirror. In some situations, two or more mirrors can be combined in optical instruments such as telescopes. We will give examples of these in a later section.
Worked example 1
c)
C onstruct a ray diagram for a concave mirror when: a) the obj ect is between F and C
object C
b) the obj ect is at C
F
c) the obj ect is at F d) the obj ect is between the pole of the mirror and F.
d) object
Solution a) C
2 image C
object
F
image
F
C onstruct a ray diagram for a convex mirror when the obj ect is between the pole of the mirror and the focal distance.
Solution b)
C
image
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image
object
F
object
F
C
C .1 I N TRO D U CTI O N TO I M AG I N G
Converging and diverging thin lenses A mirror has only one surace and the light always travels in the air. When light enters a transparent medium it is reracted and the speed and wavelength o the light change. This change o speed makes the analysis o light passing through a curved interace into a dierent medium more complex than the reection case. Lenses usually have two interaces through which the light enters and leaves. B eore considering this double reraction in detail we will look at the eect o a single interace between two dierent media. In this example, the light travels slower in the second medium than the frst. Figure 8(a) shows parallel rays o light as they travel through a convex surace rom one medium to another o greater optical density. Parallel plane wave ronts are associated with these rays and are also shown. The centre o these waveronts meets the curved convex surace earlier than the outer edges o the wave ront. As the wave enters the medium it slows down and the wavelength o the light becomes smaller. This means that the centre o the wave is travelling at a slower speed in the second medium than the wave edges that have not yet reached the interace.
curved interace (a)
waveronts
ray ray
(b)
O
I
Figure 8 Wave ronts through a medium.
The overall eect on the wave when it has ully entered the medium is that what was a parallel wave has become curved. The rays are no longer parallel either. They meet at a ocal point. O ne interpretation o the action o a curved surace is that it adds curvature to waveronts ( in the convex case) or removes curvature ( in the concave case) . The curvature o a surace can be defned in general 1 terms as __ where R is the radius o the surace. The smaller the radius R o the surace, the more curved it is. Now we can see what happens when a real lens is used to modiy rays ( fgure 8( b) ) . There are two curved interaces and it is the cumulative eect o both interaces that determine the properties o the lens ( fgure 9) . We use a similar model to that o the mirrors to defne the parts o the lens ( fgure 1 0) . The op tical centre o the lens is the point in the lens that does not deviate rays o light passing through it. As with the mirror we defne a p rincip al axis, focal p oint and focal length. Rays parallel to the principal axis pass through the ocal point ater reraction or a convex lens. For a concave lens, the parallel rays appear to have come rom the ocal point. We shall assume that the convex or concave lenses have lens thicknesses that can be ignored compared to the distances between obj ect and image. The thin-lens theory that uses this assumption is more straightorward than when the thickness o the lens is included. We will also consider only spherical lenses, that is, those with suraces shaped so as to orm part o the surace o a sphere. Many modern lenses are designed to have an asp herical shape ( non- spherical) , such lenses, or example, spectacle lenses, can be made thinner and lighter than their spherical counterparts.
Figure 9 Rays through convex and concave lenses. rays rom infnity
optical centre
ocal point F
O
F principal axis
F
O
F principal axis
Figure 10 Defnitions o the parts o a lens.
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Investigate! Finding an approximate value for the focal length of a convex lens
We have used the idea of rays that travel parallel to the principal axis. These can be obtained (approximately) by using rays that have come from a distant object. The diagram shows how these arise. Two rays that leave the same place on an object and then travel a large distance compared with the diameter of the lens have to be travelling very close to parallel if they are both to enter the lens.
Hold your lens near a window so that rays from a distant obj ect outside ( more than about 2 0 m away will do) form an image on a sheet of card ( called a screen) held behind the lens.
The distance from the card to the lens is, approximately, the focal length. Get someone to measure this distance while you hold the lens and the card. Notice that the image is upside- down.
long distance C screen 2F
F
O
O
Figure 11
Investigate! The relationship between object and image for a thin convex lens
For this experiment you will need a convex lens of approximate focal length 0.1 5 m, an illuminated obj ect, a screen, and a ruler to measure the distance between the obj ect and the lens ( the obj ect distance) and the lens and the screen ( the image distance) . Set the obj ect about 0.4 m from the lens. Move the screen until there is a clearly focused picture of the obj ect on the screen you should expect it to be upsidedown. Measure the obj ect distance u and the image distance v and record them.
Repeat the procedure for a range of obj ect distances ( do not attempt to have the obj ect distance closer than the focal length) .
Plot your data on a graph of: 1 1 __ __ v against u
lamp (object)
lens u
screen (image) v optical bench
Figure 12
The results of this experiment can be summed up using scaled ray diagrams. Ray diagrams for both convex and concave lenses are constructed on a similar basis to those for mirrors. Additional rules are that:
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The lens is drawn as a vertical line to emphasize that it is thin. S ymbols at the top and bottom of the lens indicate whether it is convex or concave.
C .1 I N TRO D U CTI O N TO I M AG I N G
The ocal length should be shown on both sides o the lens.
There are three predictable rays:
A ray rom the top o the obj ect parallel to the principal axis goes through the ocal point ater reraction by the convex lens or, in the case o a concave lens, deviates so that it appears to have passed through the virtual ocus.
A ray rom the top o the obj ect through the optical centre o the lens does not deviate.
A ray rom the top o the obj ect through the ocal point travels parallel to the principal axis ater reraction.
O nly two o these rays are needed to complete the diagram and show the position size and nature o the image. The third can be used as a check.
Rays should have an arrow to show the direction in which they travel.
(a) object
(b) image image 2F
O
F
F
2F
object object F
2F
O
F
F
F
2F image
object
2F
F
O
F
2F
image
Figure 13 Ray diagrams for a convex (converging) lens. Figure 1 3 ( a) shows the ray diagrams that match some o the cases you tested in the Investigate! When the obj ect is closer to the lens that the ocal point, a real image cannot be ormed and a virtual image results ( fgure 1 3 ( b) ) . This is the ray diagram or a magnifying glass. D iverging lenses have ewer ray diagrams than the converging lenses, the fnal image is always virtual. Such images cannot be ormed on a screen and an additional converging lens ( such as the lens in the eye) is required to orm a real image that can be proj ected ( fgure 1 4) .
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object
image
F
F
Figure 14 Ray diagram for a concave ( diverging) lens.
The lens equation The Investigate! shows that the position o the image depends both on the position o the obj ect and the ocal length o the lens. The lens equation quantifes this relationship. In curvature terms, a particular lens o ocal length f adds the same amount o curvature to any waveront passing through it. The amount 1 added is __ . This is because a lens with a small f gives more curvature to f incident parallel rays than does a lens where f is large ( the small f lens 1 1 __ bends the rays more) . The quantities __ u and v determine the curvature o the obj ect and image waveronts. Figure 1 5 shows how the symbols are defned and relate to each other. Thereore we can write: curvature o waveronts leaving lens = curvature o waveronts entering lens + curvature added by lens so 1 1 1 _ _ _ v = u + f
u
f v
Figure 15 The lens equation. However, the waveronts rom the obj ect are spreading out when they reach the lens whereas the waveronts that have let the lens to orm the image are converging. The curvatures o the obj ect waveronts are in the opposite sense to those o the image. To allow or this we should consider the u value to be negative with respect to v. This leads to the lens equation: 1 1 1 _ _ _ u + v = f
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S ome care needs to be taken with the signs given to u, v and f. The rule that applies when using the lens equation in this orm is:
Note In fact, although this will not be tested in the examination, the lens equation also applies to mirrors. In this case f is equal to 1/R.
real obj ects and images are treated as positive
virtual obj ects and images are treated as negative
the ocal length o a converging lens is positive
the ocal length o a diverging lens is negative
this is known as the real-is-p ositive sign convention; you may see texts that use other conventions.
The ability or power o the lens to add (or subtract) curvature to a waveront is: 1 power, P ( in dioptre) = __. f( in metre) A strong lens has a small ocal length and adds a large curvature to any waveront passing through it. The power o a lens is measured in diop tres ( not an S I unit) . A converging lens with a ocal length o 5 0 cm ( 0.5 m) has a power o +2 D ; the lens is said to be a positive lens. A diverging lens o ocal length 1 2 .5 cm has a power o 8D ; the lens is said to be a negative lens. I you wear eye glasses or contact lenses you may have seen these units used on your lens prescription. The defnitions o linear and angular magnifcation are identical to those or curved mirrors. height o image v m = __ = _ u height o obj ect i M=_ o where the o and i are the angles subtended by the obj ect and the image at the lens.
Worked examples 1
An obj ect o height 5 cm is placed 1 2 cm rom a converging lens o power + 1 0 D . C alculate:
c) The positive sign tells us that the image is real, 60 = 5 times and is thereore magnifed by __ 12 2 5 cm high. The image is upsidedown and on the opposite side o the lens to the obj ect.
a) the ocal length o the lens b) the position o the image c) The nature o the image.
Solution 1 a) f = _ = 0.1 0 m = 1 0 cm D 1 1 1 1 1 1 _ _ _ _ _ b) _ u + v = f so v = f - u 1 1 2 1 _ _ _ _ v = 1 0 - 1 2 = 1 20 ; v = +60 cm rom the lens
2
An obj ect o height 5 cm is placed 8. 0 cm rom a converging lens o ocal length + 1 0 cm. C alculate the position and nature o the image.
Solution 1 = _ 1 - _ 1 = _ 1 1 -2 _ - _ = _; v = - 40 cm v u 8 10 80 f The image is ormed 40 cm rom the lens; it is virtual, magnifed 4 times, and is the right way up. It orms on the same side as the obj ect. This is the magniying glass arrangement.
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3
C alculate the position and nature o an image placed 1 2 cm rom a diverging lens o ocal length 2 4 cm.
Although not asked or in the question, the ray diagram is given as an example o a more complicated arrangement. frst lens
Solution
second lens
36 1 =_ 1 - _ 1 =_ 1 1 -1 _ - _ = -_ = _ v u -24 +1 2 24 1 2 8 f The image is ormed 8 cm rom the lens on the same side as the obj ect. It is diminished with 8 a magnifcation o ( ) __ = 0.67. The image is 12 upright and virtual. 4
a)
fnal image f1
f2
A converging lens o ocal length + 1 5 cm is 2 5 cm rom an obj ect. C alculate the position o the image.
b) Another converging lens also o ocal length + 2 5 cm is placed 1 8 cm rom the lens on the image side. C alculate the new position o the image.
Solution 1 = _ 1 - _ 1 =_ 1 1 a) _ - _; v = 3 7. 5 cm. v u 15 25 f b) The new u is 3 7.5 1 8 = 1 9. 5 cm rom the lens. This is a virtual obj ect or the second lens and so will have a negative sign in the lens equation. 1 1 1 _ _ _ v = 2 5 - - 1 9.5 ; v = + 1 1 cm A real image orms 1 1 cm rom the second ( f = 2 5 cm) lens.
f1
f2 intermediate (virtual) object
The frst lens and its image is drawn frst ( green lines) the rays beyond the position o the second will not orm so are drawn dashed as construction lines. Then the second lens is added together with its ocal points (red lines) . There are a number o possible predictable rays that can be drawn. One shown here is the ray that is aimed at the optical centre o the second lens this ray will not deviate as it goes through the lens and must also go through the top o the virtual image. The other ray or the second lens comes rom the top o the original object, through f1 and then must go parallel to the principal axis ater being reracted by the lens. Ater passing through the second lens, it must go through f2 and defnes the top o the fnal real image.
The simple magnifying glass A single convex lens used as a magniying glass is the oldest optical instrument recorded. Around 2 400 years ago, Aristophanes magnifed small obj ects using spherical asks flled with water like those used by present- day chemists. We have already drawn the ray diagram or the magniying glass, but because o its importance, both alone and in conj unction with other lenses, it merits a section o its own. The human eye has a near p oint distance D that is the closest distance at which the eye can ocus on an obj ect without strain. For a normal eye this distance is taken to be 2 5 cm ( although it varies greatly rom individual to individual) . Thereore, the best we can do to see the fne detail in an obj ect is to hold it 2 5 cm away rom the eye. The magniying glass helps us to improve on this.
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h
D without lens image at near point
image at infnity
h f
object at f
image f
object
F
f with lens
Figure 16 Image ormed by a magniying glass at near point and infnity O ne way ( and the least tiring method or prolonged viewing) is to position the obj ect at the ocus o the converging lens and thereore orm the image at infnity where it can be viewed with relaxed eye muscles ( fgure 1 6) . I f or the lens is smaller than 2 5 cm the obj ect will need to be closer than the near point with the eye muscles relaxed. In such circumstances, the eye should be as close to the lens as possible. In this case, the angle subtended by the unaided eye is the size o the h h image __ whereas the angle subtended by the aided eye is __ so the D f angular magnifcation M is h _ f D _ = _. f h _ D S o a magniying glass with ocal length 0. 1 m will give a magnifcation o 2 .5 . However this is not the best that the magniying glass can do. Another way to use the lens, but more tiring because the muscles are not relaxed, is to place the eye close to the lens as beore and to adj ust the position o the obj ect so that the image orms at the near point. The obj ect can now be much closer to the eye than would be comortable without the lens.
u
h' _
u The angular magnifcation now changes to M = __ where u, h' and h are h _ defned in fgure 1 7. D
The lens equation gives 1 1 1 _ =_ + _ u -D f D - u D D D __ S o _____ = __ or __ u u - 1= f f
S o or this setting o the magniying glass D M= _+ 1 f
h' G
h image f
object D
Figure 17 Angular magnifcation or a magniying glass.
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Note You may see that some books give the equation as M = ___D - 1. This results when the sign o D is not introduced during the proo but is used in the fnal calculation. Using the numerical examples above together with the equation with the negative sign, when the image is at infnity is M 2.5 and when the image is at the near point it is 3.5. So the magnitude o the magnifcation still remains larger or the second case.
For a magniying glass with ocal length 0.1 0 m and a near point o 0.2 5 m, the magnifcation is 3 . 5 .
Spherical and chromatic aberrations Chromatic aberrations The absolute reractive indices o the materials used or making lenses vary with wavelength. Dierent colours travel through the lenses at dierent speeds and thus orm ocal points at dierent distances rom the lens. crown int
chromatic aberration
Worked example A magniying glass has a ocal length o 7.5 cm. It is used by a person with a near point o 2 5 cm to view an obj ect. C alculate the angular magnifcation o the obj ect when the image is at: a) infnity b) the near point.
Solution 25 D; M = _ a) M = _ = 3.3 7. 5 f b) M = infnity angular magnifcation + 1 = 4.3
achromatic doublet
Figure 18 Chromatic aberration and a way to correct it. This problem is known as chromatic aberration and it leads to the appearance o colour ringes around the image. The best place to view the image is at the intersection o the green rays. This is the point known as the circle of least confusion. A circle drawn around all the rays here is at its smallest. One common way to correct or chromatic aberrations is to use a doublet lens, one o the lenses is converging with positive power and the other diverging with a negative and smaller magnitude power. The lenses produce chromatic aberrations in opposite senses. Over the whole visible range, the colour perormance o the combined lens is better than with one lens alone.
Spherical aberrations Like mirrors, lenses also suer rom aberrations caused through the geometry o the spherical lens. As fgure 1 9 shows, rays that are ar rom the optical centre are brought to a ocus closer to the convex lens than those near the lens centre. This is sp herical aberration and leads to an obj ect consisting o a square grid distorting as shown. This is known as barrel distortion.
Figure 19 Rays that produce spherical aberration and the efect on a square grid.
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C .1 I N TRO D U CTI O N TO I M AG I N G The easiest way to cure spherical aberration is to reduce the aperture ( diameter) o the lens perhaps by putting an obstacle with a hole cut in the centre over the lens. O course, this will reduce the amount o light energy arriving at the image position and will make the image appear less bright.
Nature of science Same travel times edge ray O
centre ray
shorter physical path but spends more time in lens
short time in lens longer physical path I same time to go from O to I
TOK Sign conventions what is their efect? There are two common sign conventions used in optics: the real-is-positive convention used here and the New Cartesian convention. Earlier we used a conventional current o positive charges in electricity. To what extent, i any, do sign conventions afect our understanding and use o science?
Figure 20 There is another interpretation o how lenses work. We have regarded the lens as reracting light rays or as adding curvature to or subtracting curvature rom a waveront. Think about two rays going through a converging lens ( fgure 2 0) . O ne goes through the centre o the lens where it is thickest. The other goes through the edge o the lens where there is not much glass. The image orms at the place on the other side o the lens where the time taken by the light to take these two paths is the same. The ray through the centre travels arther than the other ray in the glass, and because the speed o the light is slower in the glass, this ray spends longer in the glass than the edge ray. The edge ray has a longer overall path. What is special about the image position is that it is the ( unique) place where all rays rom the obj ect take the same time to reach their respective places on the image. This has to be the case given our waveront interpretation. The lens keeps the wave together so that all parts meet at the same time at the same place on the image. Taking the same time and having the same eective path distance is what makes a lens a useul thing to manipulate rays. This is an example o Fermats principle o least time.
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C.2 Imaging instrumentation Understanding Optical compound microscopes
Applications and skills Constructing and interpreting ray diagrams
Simple optical astronomical reracting
telescopes Simple optical astronomical reecting telescopes Single-dish radio telescopes Radio intererometry telescopes Satellite-borne telescopes
Nature of science Optical instruments have been developed over the centuries to improve our ability to observe very distant objects in the sky and very small objects on Earth. Instruments to improve our vision were used in the time o the Egyptians. Authors described combinations o lenses in mediaeval Europe and it is certain that the knowledge o optics had been known in the Arab world since at least 1000 years BP. This line o improvement in instrumentation is clear in optics.
o optical compound microscopes at normal adjustment Solving problems involving the angular magnifcation and resolution o optical compound microscopes Investigating the optical compound microscope experimentally Constructing or completing ray diagrams o simple optical astronomical reracting telescopes at normal adjustment Solving problems involving the angular magnifcation o simple optical astronomical telescopes Investigating the perormance o a simple optical astronomical reracting telescope experimentally Describing the comparative perormance o Earth-based telescopes and satellite-borne telescopes
Equations f0 Magnifcation o astronomical telescope M = _ fe
Optical compound microscope Although the convex lens used as a magniying glass provides magnifcation, small ocal lengths are required to produce large D magnifcations ( M = __ ) . Lenses with small ocal length have large f surace curvatures and this can give rise to so much aberration that eatures in the image cannot be seen clearly. The compound microscope is designed to produce a virtual magnifed image o a small obj ect. The term compound means that it is composed o more than one lens: an obj ective lens, close to the obj ect, with a very short ocal length fo , and an eyep iece lens into
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which the user looks As with all optical systems, you should try to understand the unction o each component in the microscope:
eye lens O
The obj ective orms a real, highly magnifed image o the obj ect at a position that is closer to the eyepiece than fe . The eyepiece then acts as a magniying glass the obj ectives real image as its own obj ect.
The fnal image, viewed by the observers eye, is virtual and very highly magnifed.
The fnal image should not be closer to the eye than the near point D and when the image is ormed here then the microscope is said to be in normal adj ustment. Many experienced microscope users do not ocus the microscope in ull normal adj ustment, however. It is oten much more convenient or the fnal image to be ocused at the plane o the bench on which the microscope stands. A notebook next to the microscope on the bench will then also be in ocus or note keeping or drawing and can be viewed by the other open eye.
objective lens fo
fe I1 O D
Figure 1 Compound microscope.
The ray diagram ( see fgure 1 ) shows how the microscope in normal adj ustment orms its image. When learning about microscopes and telescopes, try to understand the role o the separate elements as this will make it easier or you to draw the ray diagrams in examinations. It is a good idea to devise a strategy or drawing the diagram as it is easy to end up with a fnal image that is too large or the paper or o the page. Figure 3 ( on page 61 1 ) shows one route to achieve a good sketch. Notice that the vertical size o the lenses does not necessarily reect their actual sizes in the instrument itsel. The eyepiece and obj ective apertures are usually very small in diameter, but are shown large on the diagram so that it appears that the rays actually go through them. In practice, with real microscopes the lenses are small to minimise the aberrations and so that the lenses can be ground very accurately. As usual, i the ray diagram is drawn to scale, then measurements can be taken o the obj ect and image scaled sizes to establish the overall magnifcation. L For the obj ective lens the magnifcation is equal to __ where L is the f o
length in the microscope tube between the obj ective ocal point and the eyepiece ocal point ( so that the total physical length o the microscope D tube is fo + L + fe ) . The angular magnifcation o the eyepiece is __ as f e
it is a magniying glass ( remember that D is the near point distance) . The angular magnifcation o the complete microscope is given by the product o the magnifcation o the two lenses acting separately and this is approximately DL _ fo fe
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Investigate! Make a microscope
Take two converging lenses, with ocal lengths perhaps 5 cm and 1 5 cm. D etermine their ocal lengths by using rays o light rom a distant obj ect and measuring the approximate ocal length between the lens and a piece o card.
Figure 2
Fix the short ocal length ( obj ective) lens securely to the end o a short ruler ( length 3 0 cm or so) using modelling clay or similar material. Set up the ruler so that it is a short distance rom an obj ect a well- illuminated piece o graph paper makes a good obj ect or this experiment.
Fix the other ( eyepiece) lens so that when you look into it, you see a magnifed virtual obj ect.
B y comparing the size o the grid on the obj ect with the size o the grid on the image, estimate the magnifcation o the microscope.
D oes your microscope approach the theoretical value o the magnifcation? The approximations assume that fo and fe are very small compared with D and L. Is this true in this case?
Resolution of a microscope All light that goes through an aperture such as a lens is diracted. Point obj ects become image disks as a result. I the image disks o two adj acent point obj ects overlap, it is difcult to tell them apart and the two images are not resolved. This is a signifcant problem in microscopy.
Nature of science Electron microscopes E lectron microscopes are used to image the smallest obj ects. The electrons are accelerated and as a result have extremely small wavelengths, much less that that o light. Glass lenses cannot be used to ocus the electrons but magnetic and electrostatic felds can, and these are used to bend the paths o the electrons to produce a ocusing eect. Look at some web sites that explain how the various types o electron microscope work. Try to understand how the felds replace the lenses o the compound optical microscope.
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Rayleigh suggested that two images were j ust resolved when the minimum o the diraction pattern o one image coincided with the central maximum o the pattern o the other. This means that when the images are j ust resolved, the angle subtended at the eye by rays rom each image is given by: sin = 1 .2 2 _ d where is the wavelength o the light and d is the diameter o the aperture (in this case circular) . For the microscope the aperture diameter is the eective diameter o the lenses. This is known as the Rayleigh criterion. You can fnd more discussion o the Rayleigh criterion in Topic 9. The criterion is oten modifed or use with microscopes and the eective aperture is replaced by N the numerical aperture o the lens. This is equal to n sin where n is the reractive index o the medium in which the lens is placed (n = 1 in the case o most microscopes that are in the air) and is the hal angle o the maximum cone o light that can enter the lens. The criterion becomes: sin = 1 .2 2 _ N ( You will not need to use this equation in the examination.)
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First, draw the principal axis. Add the lenses a scaled distance apart i you have been given dimensions. Then draw the intermediate image use an intermediate image height that will allow the fnal image to be on the page.
principal axis
Work backwards rom the intermediate image to orm the object to the let o the objective. Either choose the ocal lengths yoursel or use the scaled values i they have been provided. Dont draw rays to the right o the intermediate image yet.
objective lens eyepiece lens
object
fo
fe
fe
fo
Finally, draw the rays to the right o the intermediate image. These do not deviate between the lenses, but ater going through the eyepiece they must appear to have come rom the top o the virtual image. To complete the diagram add the arrows to all rays (not construction lines) and label all ocal points and both lenses.
virtual image
Figure 3 Strategy or drawing the compound microscope diagram. The equations suggest that the resolution o a microscope can be improved by:
using a short wavelength or the light ( sometimes microscopists use ultraviolet radiation or taking photographs o specimens to improve the resolution)
using as a wide an aperture or the obj ective as is consistent with reducing aberrations ( the design o the fnal eye piece lens is fxed by the average size o the eye)
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using a liquid of high refractive index between the specimen and the obj ective lens. The liquid obj ective allows a wider cone of light rays to be collected by the lens and this extra information improves the resolution of the image. Typically an oil of refractive index around 1 . 6 is used in special microscopes that have oil- immersion obj ectives.
Astronomical refracting telescopes Figure 4 An astronomical reector.
O bj ects in the night sky have always fascinated astronomers, and telescopes for viewing the sky were, like the microscope, developed early in the history of science. objective eyepiece
fo , fe
fo
fe
Begin by drawing the principal axis, the lenses and the common ocal points Add an intermediate image. Do not make it too small allow enough room or the construction later.
objective eyepiece
fo , fe
fo
fe
Draw construction lines to show you the fnal direction o the rays when they have been reracted by the eyepiece. These are construction lines so should be dashed.
objective eyepiece
fo , fe
fo
fe
The frst ray should be the ray that ends at the top o the intermediate image and goes through the optical centre. This ray is not deviated at the objective. Ater passing through the eyepiece the ray must travel parallel to the construction lines.
objective eyepiece
fo
fo , fe
o
image at
fe
e
Choose two more rays parallel to the frst ray beore they enter the objective. Ater passing through the objective they must go to the top o the intermediate image. Ater passing through the eyepiece they must go parallel to the frst ray (and the construction lines) . Finally, add urther construction lines to show the direction o the fnal image. Then label the diagram and add the ray arrow
Figure 5 Telescope ray diagram and drawing sequence.
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Also like the microscope, the astronomical reracting telescope uses two converging lenses. The obj ective lens has a long ocal length fo and an eyep iece lens that has a short ocal length fe . The telescope is designed primarily to ocus the light rom very distant obj ects and in its normal adj ustment is set up or this purpose only. Light rom infnity (that is, parallel rays) rom the object are ocused by the objective lens at its ocal point. The ocal points o the objective and eyepiece lenses are coincident (at the same place) inside the telescope and thereore the eyepiece treats the image ormed by the objective as though it is a real object placed at fe. We know rom earlier that this leads to parallel rays emerging rom the eyepiece and these then enter the eye o the observer. The angle subtended by the rays rom the top o the image and the principal axis is much greater than the angle subtended by the rays rom the top o the object. The image is magnifed. The ray directions show that the viewed image is also upsidedown, but this is not an issue or astronomers (maps o the Moon are oten inverted to take account o this) . The angular magnifcation o the telescope ollows directly by looking at the incident and emerging rays and the angles, and e, they make with the principal axis
Note In astronomy the term power is sometimes used in place o angular magnifcation. The maximum practical power or any given astronomical telescope is limited by the need to produce a high resolution image. This in itsel is determined by the atmospheric conditions.
o fo M= _= _ fe e
Investigate! (fe about 1 01 5 cm) . The combined total o fo + fe should not exceed the length o the ruler.
Making a telescope
Measure and record the ocal lengths o both lenses.
Fix the eyepiece lens at the end o the ruler with modelling clay as beore.
Holding or clamping the ruler horizontally, fx the objective lens a distance equal to fo + fe rom the eyepiece while looking through the eyepiece. You may need to move the objective orward and backwards along the ruler a short distance to get the optimum ocus or your own eye.
View a distant obj ect ( say a brick wall) and estimate how much larger one brick is when viewed through the telescope compared with using your naked eye.
D oes your telescope approach the theoretical angular magnifcation give above?
Figure 6 The basic method is similar to that o the microscope in that the instrument is constructed on a ruler or optical bench. This time a metre ruler is required together with a long ocal length objective (fo about 5 0 cm or longer) and a short ocal length eyepiece
Reracting telescopes are requently used by amateur astronomers and also by proessional astronomers or some applications. However, the largest reracting telescopes made have objective apertures up to 2 m and this represents the limit o the technology or grinding the glass lens. It is also difcult to support the lens given that it needs to be at the end o a long tube. Mirrors can be made much larger and thereore collect more light energy meaning that more distant objects can be viewed.
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Worked example 1
A small obj ect is placed 3 0 mm rom the obj ective lens o a compound microscope in normal adj ustment. A real intermediate image is ormed 1 5 0 mm rom the obj ective lens. The eyepiece lens has a ocal length o 75 mm and orms a virtual image at the near point. For the observer, the near point distance is 3 00 mm. C alculate the overall magnifcation o the telescope.
2
An astronomical telescope in normal adj ustment has an obj ective ocal length o 1 5 0 cm and an eyepiece ocal length o 5 .0 cm. C alculate: a) the angular magnifcation o the telescope b) the overall length o the telescope.
Solution f 1 50 a) M = __ = ___ = 3 0 ( angular magnifcation 5 f o
Solution 1 50 Magnifcation o the obj ective = ___ = 5 . 30 5 1 1 1 1 ____ __ ___ For the eyepiece __ = __ u + - 300 ; u = 300 ; 75 u = 60 mm. S o magnifcation o eyepiece 300 = 5 . = ___ 60
e
and linear are the same or the telescope in normal adj ustment) b) O verall length = fo + fe = 1 .5 5 m
O verall magnifcation = 5 5 = 2 5
Astronomical refecting telescopes The frst known reecting telescope was completed by Newton in 1 668. The basis o the telescope is straightorward. Rays rom the top and bottom o a distant obj ect are reected by a concave mirror and come to a ocus at the ocal point where an image o the obj ect orms. An observer positioned at the ocal point will see an image in the mirror. Rays rom the top and bottom tend to conuse the diagram. The convention in many reecting telescope ray diagrams is to show only the rays parallel to the principal axis and you may well see this in some books. B ecause the rays are parallel or at very small angles to the principal axis, either spherical mirrors or parabolic mirrors can be used in the reecting telescopes. For astronomical use, parabolic is chosen so that the spherical aberrations discussed in the earlier section are eliminated.
prime
prime focus primary mirror
Figure 7 Picture of astronomical telescope and basic ray diagram.
As with compound microscopes, the angular resolution o a reecting telescope is given by sin = 1 .2 2 _ d where d is the diameter o the aperture. So a large aperture or the mirror is an advantage as this increases the resolving power o the instrument. There are many advantages o reecting telescopes over their reracting equivalents:
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There is one surace which is usually ground to shape and then coated with a reective metallic surace. The rays o light do not have to pass through a number o layers o glass. The telescope can thereore in principle have no chromatic aberration.
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The mirror only has to be supported rom one side. This makes engineering large reecting telescopes more straightorward.
With only one surace to grind, the telescopes are cheaper or a comparable quality o image.
O nly one surace has to be made perect.
There are also some disadvantages however:
The mirror surace is vulnerable to damage and needs to be cleaned ( unless covered which introduces aberration) .
The optics can easily get out o alignment i the support or the mirror shits in some way.
In addition, this simple design means that the observers head or a camera blocks some o the light travelling along the tube thereby reducing some o the benefts o using a large mirror. For this reason, various arrangements are used to transer the image rom inside to outside the main tube o the telescope. There are a number o dierent design variants or achieving this, some o which use lenses and other devices to produce a high- quality image. In this course we will examine only two simple systems: the newtonian telescope and the cassegrain telescope. B oth are named or their inventors.
Newtonian mounting
newtonian mounting
cassegrain mounting
Figure 8 Newtonian and cassegrain mirror systems. In this system a small secondary plane mirror diverts the rays rom inside to outside the tube. Plane mirrors o high quality can be produced that introduce little or no distortion to the image. This is one o the cheapest designs o telescope and is very popular with amateur astronomers, some o whom grind their own mirrors and build their telescopes rom scratch.
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Cassegrain mounting Replacing the plane mirror with a hyperbolic secondary mirror enables the rays to be sent through a hole in the primary mirror surace. This means the observer can be in line with the telescope and can look in the observing direction. A principal advantage o this mounting is that the ocal length o the telescope can be eectively longer than the tube itsel. However, the secondary mirror inside the tube cuts o some o the incoming light, and the principal mirror has lost some o its reecting surace. In general, all telescope types have specifc observational advantages and disadvantages.
Radio telescopes Many o the principles o the reecting telescopes are carried over into the single- dish radio telescopes. These instruments are intended to collect electromagnetic signals in the radio region that originate rom astronomical obj ects. rays from distant object
F
Figure 9 Radio telescopes. A parabolic dish has the property that all rays parallel to the principal axis are brought to a ocus at the same point. I a radio antenna (aerial) is placed at this point then it will collect all the ocused radio waves that arrive simultaneously. The larger the area o the dish, the greater the power collected, so in principle the designer should aim or as large an area as possible. Another advantage o a large dish diameter is that, as with the optical reecting telescopes, the resolution o the telescope improves ( in the Rayleigh criterion equation decreases) as the diameter becomes larger. Resolution is a particularly important actor as the wavelengths used by the telescopes are those o the radio segment o the electromagnetic spectrum. These wavelengths are o the order o centimetres to tens o metres and are much larger than those o the visible light used with optical instruments ( order o 1 0 7 m) . S o a telescope working at radio wavelengths needs to have an aperture at least 1 00 000 times greater than its optical counterpart to have the same resolution. Figure 10 The Arecibo Observatory, Puerto Rico.
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This raises engineering problems or a dish that has to be steered to point at particular obj ects in the sky. There is the difculty o moving the dish and problems associated with the dish deorming rom the ideal parabolic shape under its own weight. S ome radio telescopes get over this problem by building the dish into a cavity in the ground.
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The disadvantage is that the dish can no longer be steered. Arecibo O bservatory in Puerto Rico is a good example o this ( fgure 1 0) . S ome exibility can be engineered into the system by moving the antenna, which in this telescope is suspended at the ocus and can be moved on a suspended railway track that can be seen in the photograph. The Arecibo telescope is 3 00 m across but a larger telescope o 5 00 m diameter, also in a natural crater, is being constructed in Pingtang C ounty, Guizhou Province, south-west C hina. The C hinese telescope is designed so that the panels that make up the dish are moveable to allow the principal axis o the telescope to be steered within limits. B y contrast, the largest dish telescope in the world at present is the Green B ank 1 00 m diameter telescope in West Virginia USA ( fgure 1 1 ) .
Interferometer telescopes To overcome the inherent design problems o a dish and the inability to steer a telescope built in a crater, a recent trend is to construct intererometer telescopes. from source
Figure 11 The Green Bank Telescope, West Virginia, USA.
baseline
beam combiner
Figure 12 Interferometer telescope. S uch telescopes come in various ormats with two or more radio telescopes combined, but the basic idea behind all o the arrangements is that signals rom a source are combined in the individual components o the telescope to produce a total signal. The signals can be combined in such a way that the whole telescope is equivalent to a single dish with an eective diameter equal to a baseline B. The baseline is the dimension across the individual dishes that make up the telescope. The resolution o such an array is sin = _. B O ne ormat that is requently used is a series o small steerable dishes. These are individually o low cost and do not have the engineering problems associated with large dishes. The signals are combined using computers to give a fnal signal.
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I M AG I N G Another approach is to use a very long baseline. Signals rom steerable dishes situated in dierent countries can be combined to give a network, the baseline o which can approach the size o the planet. For example the E uropean Very- long- baseline intererometry Network ( EVN) is a combination o 1 2 telescopes and can, itsel, be combined with intererometers in the UK and the US A to make a global network that achieves a very high resolution. There are plans to build a square kilometre array ( S KA) that will combine telescopes in Australia, S outh Arica and other countries in the southern hemisphere with a baseline o the order o 3 000 km. The telescope is due to be completed by 2 02 4. O ther intererometer ormats are possible and these at present include linear arrays o two long antennae or arrangements o antennae built in the shape o a plus sign.
Nature of science Discoveries in radio astromony Not everyone has to be a proessional. S ome branches o science are known or the number o discoveries made by amateurs. Astronomy is one o these. The frst radio telescope was constructed in 1 931 by a radio engineer named Karl Jansky who worked in the B ell Telephone Laboratories in the US. He studied sources o static noise in radio signals and discovered that some o the static originated outside the Earth. His telescope was a mobile collection o radio antennae (aerials) . The frst true dish radio telescope was built by a North American amateur radio enthusiast named Grote Reber (call sign W9GFZ) who was inspired by Janskys
work. He spent considerable time (rom 1 93338) constructing a series o parabolic dish reectors that were eventually sensitive enough to reproduce Janskys results. Reber and Jansky were the frst true pioneers o radio astronomy. Today, amateur astronomers have excellent telescopes coupled to digital cameras. This makes it possible or them to contribute to science in a signifcant way. Proessional astronomers cannot view all areas o the sky night ater night. For example, amateur astronomers discover deep sky supernovae in distant galaxies and alert proessional observers who then make observations using larger telescopes.
Earth and satellite-borne telescopes Recently, high-quality telescopes have been placed in satellite orbit above the Earth or on spacecrat that are aimed away rom the Earth into the S olar S ystem. O bservational astronomy carried out on the surace is subj ect to a number o limitations, as described overlea:
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When we look at the night sky with the naked eye, the stars appear to twinkle. This is because our observations are made through tens o kilometres o air. The atmosphere introduces brightness variations and position errors into the images o the stars through local short- term changes in the air density caused by heating and convection eects. A common way to reduce this problem is to build optical telescopes on mountain tops in places where the atmosphere is relatively stable.
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Some recent telescopes use up-to- date techniques to remove light variation using adaptive optics. When viewing a distant dim star the optics make reerence also to light rom a nearby bright star that is assumed to have a constant intensity and position. Parts o the telescope mirror suraces are then moved rapidly to correct or the distortion o the light, which is assumed to be the same or both reerence and observed stars. However, observations that are made rom a satellite platorm above the atmosphere do not need these corrections.
TOK What do we see?
Stars emit radiation right across the electromagnetic spectrum, and all this inormation is o value to astronomers. Many wavelengths rom the stars are absorbed by the atmosphere ( fgure 1 3 ) and cannot be detected at the Earths surace. The only way to measure these is through the use o a telescope or sensor mounted on a satellite. X- rays are a particularly important region o the spectrum to astronomers and orbiting satellites are almost the only way to study emissions at these wavelengths rom stars ( fgure 1 3 ) .
ultraviolet X-ray blocked by atmosphere
visible infra-red blocked window by atmosphere
radio window long-wavelength radio blocked
10 cm
There is an increasing problem o electromagnetic pollution rom artifcial sources o radiation associated with cities on Earth. Optical telescopes need to be constructed in increasingly remote areas, and radio telescopes also need to be placed in places where the radio spectrum is relatively uncluttered.
1 m
Any optical or radio instrument extends our senses beyond their design limits. This can be in terms of wavelengths to which we are not normally sensitive, or in terms of magnifying beyond what we would normally see. To what extent do the images we see or interpret represent true reality?
opaque
1 km
100 m
10 m
1m
1 cm
1 mm
100 m
10 m
100 nm
10 nm
1 nm
0.1 nm
transparent
wavelength/m
Figure 13 Absorption of the electromagnetic spectrum by the atmosphere.
An Earth- based telescope, whether radio or optical, can either be fxed or steerable. A fxed telescope can only view the parts o the sky that move in ront o it. A steerable instrument can view ( within the limits o the local horizon) the whole o the hemisphere on which it is centred. Similarly, some satellite telescopes are fxed so that they observe certain sections o the sky. O ther sky- survey satellites allow the entire sky to be mapped over the period o the satellites lietime.
The expense o placing an observing satellite into orbit and the costs o building surace- bound telescopes is so great that international collaboration is common in astronomy with research groups rom around the world booking observation time on the instruments.
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C.3 Fibre optics Understandings
Applications and skills
Structure o optic fbres
Solving problems involving total internal reection
Step-index fbres and graded-index fbres Total internal reection and critical angle Waveguide and material dispersion in
optic fbres Attenuation and the decibel (dB) scale
Nature o science Modern communications rely heavily on the use o fbre optics. A relatively simple piece o science has led through applied science to a transormation o all types o communications across the globe.
and critical angle in the context o fbre optics Describing how waveguide and material dispersion can lead to attenuation and how this can be accounted or Solving problems involving attenuation Describing the advantages o fbre optics over twisted pair and coaxial cables
Equations 1 critical angle equation n = _ sin c I _ attenuation (dB) = 10 log I0
Structure and use o optic fbres The concepts o total internal reection and critical angle were introduced in Sub- topic 4. 4. These ideas are used in the modern technology o fbre optics ( Figure 1 ) which began to be developed or communication purposes in the early 1 970s. However, the idea o sending light along a light pipe was frst demonstrated by C olladon and B abinet as early as 1 82 0. Nowadays, under-sea fbres link nations with international collaboration and agreement on common standards or the transmission o the inormation. cladding core
cladding
Figure 1 Total internal reection in a fbre.
Communications and the advantages o optic fbres The principal methods using physical links or communication ( as opposed to the radiation o electromagnetic waves as in radio) are:
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Twisted p air. In this technique developed by Alexander B ell, the wires connecting one operator with another were twisted together ( hence the name) . External electrical signals can induce an em in both wires and this em is to some extent cancelled out by the
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twisting. However, the noise cancellation is not necessarily complete and i two pairs o twisted- pair cables are next to each other then signals rom one pair can be induced in the other cable pair giving cross-talk and a lack o security. Twisted- pair cables are generally used or low- requency applications.
C oaxial cable. This cable is constructed with a central core conductor insulated rom a metallic shield that is earthed at zero potential relative to the changes in the signal voltage ( but which carries the return current) . The cable is completed with a tough protective outer cover. S uch cable is generally used or radiorequency signals. You may well have seen an example o it carrying the ( very weak) signals rom your tv antenna or satellite dish to the tv itsel. This type o cable rej ects outside electrical noise as the earth shield acts as the surace o a conducting shell. Little noise rom outside can distort the weak signal travelling along the cable and equally the signal itsel cannot radiate signifcantly beyond the earthed conductor. However, the cable is bulky and expensive. O p tic fbres. In this technique the signal to be transmitted down the fbre modulates an electromagnetic wave ( usually at visible or inra- red wavelengths) . This modulated wave is shone along a very thin glass fbre. The thickness o the fbre is so small that the light strikes the walls at angles ( much) greater than the critical angle and so the light propagates along the fbre. E lectrical noise does not aect the passage o the light through the fbre. There are wavelength windows in optic fbre glasses at which the loss o signal with distance ( the attenuation o the glass) is very low. The fbre is surrounded by a cladding material with a reractive index smaller than that o the core in order to ensure that the critical angle condition is maintained.
twisted pair
plastic jacket metallic shield dielectric insulator centre core co-axial cable
optic fbre
cladding core cladding
Figure 2 Physical communication links.
D igital signals are transmitted along the optic fbres. S uch signals consist o a sequence o changes between two states; these might be on/o, variations between two fxed requencies, or abrupt changes in the phase o the signal. For our discussions we will assume a simple model where the light is either on or o, so that a series o light pulses is transmitted down the cable. The advantage o optic fbres over conductors are:
Immunity to electromagnetic intererence unlike most electrical cables.
Low attenuation loss over long distances in glass compared to metal conductors.
B roader bandwidth ( total range o usable requencies) in the glass compared to wires. This means that one fbre can carry millions o telephone conversations.
Glass is an insulator and does not suer rom the inherent problems o a electrical- cable telephone system where electrical connections to earth ( shorts) are a problem.
Very small diameter so that compact bundles o fbres can be constructed thus increasing the capacity o the system even more or the same physical space as an electrical cable.
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Greater security to signal detection rom outside the fbre ( fbre tapping) . S ome optical fbres have been designed with a dual core and are said to be completely secure to outside tapping.
The main types o fbre are:
S tep -index fbre in which there is an abrupt change in the reractive index at the interace between core and cladding.
Graded-index fbre in which there is a gradual reduction in the reractive index rom the centre to the outside o the core.
S ingle-mode fbre, which has a diameter o only a ew times the wavelength o the light used in it. At such small diameters, the rules o geometrical optics that we use or our discussion o optic fbres break down and wave theories are required. We are not going to consider this type o fbre urther in this course.
input power
We will look at the behaviour o step- and graded- index fbres in more detail later.
Attenuation and dispersion time
O ptic fbres have very low loss. I sea water had the same optical properties as glass used in fbres, then the details at the bottom o the 1 1 km deep Mariana Trench, o the coast o Japan, would be clearly visible rom the ocean surace. Nevertheless, eventually the signal weakens ( attenuates) so that it needs to be amplifed beore it can continue its j ourney. The device that carries out this re- amplifcation is known as a rep eater.
output power
At the point where the signal enters the cable, the pulse will have an abrupt on/o change in intensity. time
Figure 3 Attenuation with time in an optical fbre. The time axes are to the same scale.
Note This defnition uses the intensity (or power) o the signal. The intensity I o the signal is related to its amplitude A by I A 2 . So
The repeater needs to do two things: re- shape the pulse into its original square ormat and also boost the amplitude o the signal. We will discuss the reasons or the loss o amplitude and the dispersion later, but or the moment we concentrate on how attenuation is measured.
A attenuation = 10 log10 _____
The signal amplitude can all by many orders o magnitude beore it needs to be boosted. To deal easily with large ratio changes a logarithmic scale called the B el scale is used. In this scale, the attenuation in bels ( B ) is defned to be the logarithm to base 1 0 o the ratio o the intensity ( or power) o a signal to a reerence level o the signal.
This can be re-written as
So
attenuation = 10 log10 ____II 0
translates to 2
A 20
A attenuation = 20 log10 ____ A 0
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Figure 3 shows how the power o the pulse varies with time beore and ater passing along the cable: the power is reduced ( by a very large actor) so that the total energy in the pulse ( the area under the power time curve) is reduced also. This is attenuation; energy has been lost to the cable. Separate disp ersion eects change the shape and cause the pulse to spread out as it travels along the fbre. I two pulses that were initially separate overlap through dispersion, then the receiving system cannot disentangle them rom each other. D ispersion imposes an upper limit on the rate at which a particular fbre can transmit inormation.
I attenuation in bel = log 1 0 _ I0
C. 3 FI B RE O PTI CS where I is the attenuated ( output) power level o the signal and I0 is the input intensity level ( the subscript is a zero not a letter o) . This means I that 1 0 B is equal to a power ratio __ o 1 0 1 0 , which is a large ratio. As I power ratios tend to be smaller than this, it is more usual to quote power ratios in decibels, where I attenuation in decibel ( dB ) = 1 0 log 1 0 _ I0
( ) 0
A change in power o 1 0 dB is equivalent to a power ratio o 1 0 times. A change o 3 dB is equivalent to a change by a actor o 2 in the power.
Link
Attenuation per unit length dB/100 m Coaxial cable 15300 or requencies up to about 1 GHz depending on signal requency and cable design Twisted pair 550 or requencies up to about 300 MHz Optic fbre 0. 02 at 10 1 4 Hz
The table shows some typical values or attenuations in optic fbres, twisted cable, and coaxial cable.
Worked example
Solution
A signal o power 7.5 mW is input to a fbre that has an attenuation loss o 3 .0 dB km 1 . The signal needs to be amplifed when the power has allen to 1 .5 1 0 1 8 W. C alculate the distance required between amplifers in the system.
7 .5 1 0 Power loss = 1 0 log 1 0 ( ________ ) = 1 57 dB 1 .5 1 0 -3
-18
The fbre loses 3 dB every kilometre so another 1 57 amplifer will be required in ___ = 52 km. 3
The requencies most used or fbre optics are in the range 0. 81 .5 m. Attenuations in these ranges arise rom two principal causes:
Absorp tion This loss arises rom the chemical composition o the glass and rom any impurities that remain in it ater manuacture. Glass absorbs some inra- red wavelengths strongly and these wavelengths have to be avoided or transmission.
S cattering Rayleigh scattering ( named ater the B ritish physicist Lord Rayleigh, who developed the ideas o image resolution) is the main scattering loss. It is caused by small variations in reractive index o the glass introduced during manuacture. Rayleigh scattering accounts or about 95 % o the attenuation.
inrared absorption total absorption loss/dB km -1
Grace Hopper and her nanoseconds The US computer scientist Grace Hopper was amous as one o the frst people to construct a compiling language or computers. She was also amous or her lectures on computer science. To illustrate the problem o communication at high speed she would hand out lengths o wire 30 cm long. This is the distance travelled by light in a vacuum in one nanosecond. She used this to link the speeds and sizes o computers in the minds o her audience. To what extent does the fnite speed o light put limitations on our ability to carry out computations aster and aster?
rayleigh scattering
1.5
TOK
1.0
0.5
0 1.0
1.2
1.4 wavelength/m
Figure 4 Attenuation mechanisms in optic fbres.
1.6
1.8
Figure 5
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I M AG I N G Light in the core interacts with the molecules o the glass. I the light continues in the general orward direction there is no attenuation, but i the light is scattered in directions other than rom which it came, then attenuation occurs. This depends very strongly on the size o the small variations in density, etc. in the glass and also on the wavelength o the light. The loss due to Rayleigh scattering is proportional to the 4 and is greatest at short wavelengths. The graph ( fgure 4) shows how the two eects o absorption and scattering contribute to the overall attenuation o a typical fbre or wavelengths greater than 1 um.
Nature of science Why is the sky blue? Rayleigh scattering is the reason why the sky is blue. B ecause the blue and violet light rom the S un have shorter wavelengths than the other colours, they are preerentially scattered out o the direct beam by the gas molecules in the air compared to longer wavelengths. We see the S un as having a yellow disk with a blue sky. At
cladding acceptance cone
Two other eects contribute to distortions o the light as it passes through the fbre:
core
sunset, the S uns rays pass through a thicker atmosphere and the presence o particles in the air scatters longer wavelengths too, so the S un and the clouds now appear red. In act, this is a simplistic explanation o a complex phenomenon. I you want to know more, investigate both Rayleigh and Mie scattering.
cladding
Material disp ersion This is similar to the chromatic aberration we met earlier in S ubtopic C .1 . B oth occur because the reractive index o the material used or the lens or fbre depends on wavelength. Reractive index is w ave sp e e d in vacu u n equal to the ratio o: _________________________ . w ave sp e e d in tran sm itting m e diu m
Figure 6 Dispersion.
The reractive index o glass decreases as the wavelength increases and so the wave speed in the glass also increases with increasing wavelength ( red light travels aster than violet) . Using light with a wide range o colours ( a large bandwidth) means that the long wavelength light will reach the end o the fbre ahead o the shorter wavelengths leading to a spreading o the pulse. The answer is to restrict the wavelengths but this, at the same time, means that ewer wavelengths can be used and so there may be ewer channels available or communication.
Waveguide (or modal) disp ersion Even i monochromatic light is used, a large diameter step-index optic fbre will still be aected by waveguide dispersion. Rays o light that propagate along the fbre can travel by dierent routes depending on their initial angle o incidence at the end o the fbre. C ompare a ray that travels along the central axis o the fbre with one that is at a large angle to the axis and reects many times.
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Although the scale o this drawing ( fgure 7) is unrealistic ( because, to scale, it has ar too large a diameter or any real optic fbre) , you should see that the large- angle ray takes a much longer path than the central ray to travel rom one end o the fbre to the other. The large-angle ray will arrive later and the time dierence will appear as an increase in the pulse width. fbre cross-section
fbre n
distance
fbre n
distance
Figure 7 Graded-index fbres. To make a correction or this eect, graded- index fbres are used. As described earlier the reractive index o the core is not constant. It has a high index in the centre o the core and a decreasing index towards the corecladding interace. The speed o the light is slowest in the centre with the speed increasing towards the cylinder wall. This means that large-angle rays travel more quickly than central axis rays during the periods when they are close to the wall. C ombining this with a smaller overall diameter o core reduces ( but does not eliminate) the waveguide dispersion eect.
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C.4 Imaging the body (AHL) Understanding Detection and recording o X-ray images in
medical contexts Generation and detection o ultrasound in medical contexts Medical imaging techniques (magnetic resonance imaging) involving nuclear magnetic resonance (NMR)
Nature of science Decisions made by a physician can involve an assessment o risk. The doctor will use some imaging techniques in the knowledge that the ionizing radiations involved can harm the patient. The real question is whether the techniques lead to a possible overall beneft.
Applications and skills Explaining eatures o X-ray imaging, including
attenuation coecient, hal-value thickness, linear/mass absorption coecients, and techniques or improvements o sharpness and contrast Solving X-ray attenuation problems Solving problems involving ultrasound acoustic impedance, speed o ultrasound through tissue and air, and relative intensity levels Explaining eatures o medical ultrasound techniques, including choice o requency, use o gel, and the dierence between A and B scans Explaining the use o gradient felds in NMR Explaining the origin o the relaxation o proton spin and consequent emission o signal in NMR Discussing the advantages and disadvantages o ultrasound and NMR scanning methods, including a simple assessment o risk in these medical procedures
Equations
I1 I0
Attenuation (dB) : L I = 10 log _ Intensity: I = I 0 e - x
Linear absorption: x 1 /2 = ln2 Acoustic impedence: Z = c
Introduction Medicine and physics combine in the world o medical diagnosis and treatment. D octors have come to rely on technology rom developments in physics. There has been a revolution in the methods that enable a doctor to visualize the interior o a patients body.
X-ray images in medicine Rontgen discovered X-rays in 1 895 and, within one month o the publication o his original scientifc paper, the radiation had been used or medical purposes. Nowadays, X- radiation is used both diagnostically and therapeutically. Here we concentrate on the use o X- rays to generate an image o the interior o the body that will inorm doctors in making their diagnoses.
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C. 4 I M AG I N G TH E B O D Y ( AH L) The basic principles behind the imaging are that X-rays are shone through part or all o the body. Materials such as bone absorb some o the X-rays or scatter them out o the beam, so that a reduced intensity emerges rom the patient. Other, soter materials such as muscle and tissue do not absorb or scatter the rays so well. The X-rays are then incident on a photographic plate or a sensor. The plate is developed or a computer computes the data rom the sensor, in both cases an image on flm or monitor is produced. Where the intensity o the X-rays is high (not absorbed by the tissue) , the plate is exposed (darkened) or the sensor detects the arrival o many photons. When signifcant absorption has occurred, ewer photons leave the body and the plate is not exposed to the same degree. As X-rays image are traditionally viewed as negatives, this means that the photograph is darkened where the rays are not absorbed and remains transparent (white) where the radiation has been absorbed by the body.
Figure 1 X-ray images.
The X- rays themselves are produced when electrons travelling at high speed are decelerated in a heavy- metal target such as tungsten. The electrons are frst accelerated in a vacuum through tens o thousands o volts in an electric feld and then strike the target. O n colliding with the target, electrons are slowed down rapidly and as a result lose energy to internal energy o the target ( 99% o the incident energy) and as energy in the orm o X- ray photons ( 1 % ) . Given the large amount o internal energy, the X- ray tubes need to be cooled and the anode target is oten rotated to avoid hot spots developing on it. D etails o the generation o the X- rays are not required or the examination. X- rays are attenuated as they pass through material, and this means that some o the photons are removed. O thers change direction ( so that they can no longer be considered part o the beam) . The principal mechanisms or removing photons or changing their direction are:
C oherent scattering A process similar to the scattering mechanisms described in the previous sub- topic on fbre optics. It is the predominant mechanism or low- energy X-ray photons up to energies o about 3 0 keV.
Photoelectric effect This is identical to the eect described in Topic 1 2 . The incoming electron has enough energy to remove an inner-shell electron rom an atom. When other electrons rom higher energy states lose energy to occupy the inner shell, a photon o light is emitted. This mechanism is most important in the energy range 3 01 00 keV. The photoelectric scattering is particularly important as it provides contrast on the image between tissue and bone.
C omp ton scattering In this mechanism, which occurs at energies generally greater than those used in diagnostic X-rays, the high-energy X-ray photon ejects an outer-shell electron rom an atom and, as a result, a photon o lower energy moves o in a dierent direction to the original photon. This scattering mode is o principal importance in therapeutic X-ray medicine where X-rays o high energy are involved.
Pair p roduction As explained in Topic 1 2, at high energies in excess o 1 MeV, a photon can interact with a nucleus to produce an electronpositron pair.
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I M AG I N G Attenuation also occurs when the intensity o the X- ray beam decreases with distance rom the X- ray tube as the beam diverges. For the photons in a monochromatic beam o X- rays, the chance o an individual photon being scattered or absorbed is related to the probability o this photon interacting with an atom in the material being X- rayed. Suppose a photon has a 1 0% chance o removal by a particular thickness o material ( fgure 2 ) .
81 000 66 000 53 000 43 000 90 000 73 000 60 000 48 000
100 000 photons
1
2
3
4
5
6
7
8
4 5 thickness
6
7
8
100 000
80 000
60 000 hal thickness 40 000
20 000
0 0
1
2
3
Figure 2 Absorption efects lead to a hal thickness. Then i 1 00 000 photons are incident on material o this thickness, 90 000 ( 90% o 1 00 000) will remain in the beam when it leaves the material. I the material were twice as thick, then 81 000 ( 90% o 90 000) will remain ater this urther thickness. This is an identical argument to that used in radioactive decay ( except that there the argument related the time taken or the ensemble o atoms to decay to the probability o decay or one atom) . However, this similarity means that, like radioactivity, the intensity o the X-rays obeys an exponential relationship and we can state that: I = I0 e - x l
where I is the intensity o the attenuated beam, I0 is the intensity o the incident beam, x is the thickness o the material and l is the linear absorp tion coefcient measured in units o metre 1 . Although the linear absorption coefcient ollows directly rom the way the absorption probability is defned, it is not the most convenient absorption coefcient to use as it depends on the density o the absorbing material. As an example, compare water vapour ( steam) with ice. Ice will scatter more eectively than steam because it has a greater density and thereore the X- ray photons are more likely to encounter
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C. 4 I M AG I N G TH E B O D Y ( AH L)
The relationship between the linear absorption coefcient l and the mass absorption coefcient m is l m = _ where is the density o the material. The units o mass absorption coefcient are m 2 kg 1 . As a consequence, when the mass absorption coefcient is used, the appropriate intensity equation is I = I0 e -
intensity
lead
distance
soft x-rays
lead intensity
a water molecule every centimetre o travel in ice than in steam. It is more convenient to quote a single mass absorp tion coefcient which is density independent and depends only on the element or compound that is absorbing the X- rays.
x
m
distance
soft x-rays
This equation is only strictly correct when the beam is monochromatic because values o the absorption coefcient vary with the energy o the X-ray photons. Highly penetrating radiation has a short wavelength (about 0.01 nm) and the absorption coefcients are small at these wavelengths; such radiation said to be hard. Long wavelength photons (about 1 nm) are termed sot X-rays and the absorption coefcients are much larger.
Figure 3
In the same way that radioactive materials have a defned hal- lie, so or X-rays the thickness o material required to halve the intensity is called the hal thickness x _1 . 2
Again, ollowing radioactive decay: ln 2 ln 2 _ x 1 /2 = _ l or x 1 /2 = m The value o l may be determined rom the gradient o a graph o ln I against x. Frequently the beam will pass through a number o layers o dierent materials all with dierent thicknesses and absorption coefcients. This is easy to treat mathematically i the layers have parallel plane interaces. S uppose that layer 1 has a thickness x' and a linear absorption coefcient l' and that layer 2 has a thickness x'' and a linear absorption coefcient l''. Then the intensity when the beam leaves layer 1 , I1 is I1 = I0 e - 'x' l
and the intensity when the beam leaves layer 2 , I2 is I2 = I1 e - ''x'' l
Thereore x
I2 = I0 e - ' x' e - '' x'' l
x!
l
layer 1 layer 2
or I2 = I0 e - ( ' x' + '' x'') l
l
For each layer add the product o the linear absorption coefcient and the layer thickness and then use the exponential unction to fnd the fnal intensity o the beam. When the mass absorption coefcients are known the equation becomes: I2 = I0 e m - (' '
m
x' + '' m'' x'')
I2 = I0 e -(1 x + !1 x!)
I1 = I0 e -1 x
IO
l
l !
Figure 4
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Worked examples 1
A sample o a metal o density 4000 kg m 3 has a hal thickness o 1 2 mm with a particular wavelength o radiation. C alculate, or the metal, the:
2
a) linear absorption coefcient
Solution
C alculate the thickness o muscle tissue that will reduce the intensity o a certain X-radiation by a actor o 1 0 3 , assuming that the linear absorption coefcient or muscle is 0.035 cm 1 .
b) mass absorption coefcient.
I1 = I0 e - 'x' l
Solution
S o 0.001 = e - 0.035 x
ln 2 -1 a) l = _ x1 /2 = 58 m
ln 0.001 = - 0.03 5 x 6.9 x = _ = 1 97 cm 0.03 5 About 2 m o muscle tissue is required to reduce the intensity by 1 000 .
l 58 _ -2 2 -1 b) m = _ = 4000 = 1 .4 1 0 m kg
aluminium flter
collimator
Improving the image Collimation Figure 5 shows a common arrangement used to take diagnostic X- ray images. A number o eatures are used to enhance both the sharpness and contrast o the image ormed on the photographic plate:
The X- ray beam is fltered by passing it through a thin plate o aluminium. This selectively removes low- energy photons because the absorption coefcient is much greater at low energies. These low- energy photons would simply be absorbed by the patient ( adding to the radiation dose) and would serve no useul imaging purpose. The plate also reduces the intensity o the beam somewhat, but overall the penetrating power o the beam increases and the X- rays become harder.
The beam as it leaves the tube is very divergent. It is collimated by passing through a series o channels in lead plates. O - axis rays are absorbed by the lead and do not reach the patient. This produces a narrower, more parallel beam. This is benefcial because photons scattered rom the o- axis angles tend to blur the photographic image.
As or the X- rays that reach the patient: some do not interact, some are absorbed and disappear rom the system, and others are scattered. Again, the scattering leads to blurring on the image. A grid system o lead plates below the patient, arranged parallel to the beam, is used to absorb the scattered rays but allows the on- axis rays to reach the imaging system.
The X-rays fnally arrive at the flm cassette and this contains two uorescent screens, one each side o the flm. As the X- rays interact with these screens they cause the materials in the screens to uoresce and emit light that improves the blackening o the photographic negative. Many o the X- rays would not contribute to the image without this arrangement.
lead plate uorescent material
flm flm cassette
Figure 5 Contrast techniques in X-ray imaging.
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C. 4 I M AG I N G TH E B O D Y ( AH L)
Contrast S ome tissues in the body are hard to distinguish on the photographs without enhancements to the contrast. Heavy elements with large absorption coefcients can be introduced into the body to help with this. B arium and bismuth are examples o elements used. Patients with stomach disorders may be asked to drink barium sulate ( B aS O 4 ) , which coats the lining o the stomach and makes its outline very clear on the image. Iodine can be introduced intravenously to produce clear images o the cardiovascular system. O ther ways to enhance detection include using charge- coupled sensors in place o flm together with computer systems to orm electronic images viewed on monitors. The advantage o X- ray imaging is that it is a quick and inexpensive technique, costing ar less than an MRI or C T scan. The X- ray machines can be highly portable, meaning that patients may not have to travel to a central location or a scan. However, unlike ultrasound and MRI techniques, X- rays involve ionizing radiation and this presents a risk to both the patient and the radiographer. Techniques to reduce the X- ray intensity and exposure time are constantly being improved to reduce the risks. Nevertheless this exposure needs to be kept in perspective. The average chest or dental X- ray provides a much smaller dose o radiation to the body than a commercial intercontinental ight at 1 2 km above the Earth.
Computed tomography C omputer ( computerized) tomography ( C T scan) , also known as computed axial tomography ( C AT scan) was introduced during the 1 970s. It gives a much greater range o grey scales to the image and provides an axial scan an image o a slice through the patient. X- rays are incident on the area o the patient under investigation and the scattered and direct photons are detected by a series o small detectors placed in an arc around the patient. Ater the frst exposure the X- ray tube and the detectors move around the patient taking exposure ater exposure until an entire range covering 3 60 around the body has been made ( fgure 6) . The detection method uses scintillation counters that respond to the presence o a photon by emitting a ash o light, which is then detected by a photomultiplier that produces an enhanced electric current or each ash that occurs.
X-ray tube beam of X-rays
A computer builds up the inormation rom each detector and each exposure to produce a complete image o the slice o the patient up to a ew tens o millimetres thick. I urther slices are required, the patient can be moved under computer control to a new position and the process is repeated. The sensitivity o the system is very high, and many eatures can be seen and interpreted by doctors. A disadvantage o the C T scan, however, is that the cumulative dose to the patient is high, and there is a greater risk o damage to the patient than with a normal X- ray. However, exposures rom the scanners are dropping all the time as improved detection techniques are developed.
scattered X-rays
series of rotating X-ray detectors
Figure 6 CT scanning.
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I M AG I N G
Ultrasound in medicine Generation Ultrasound is a sound wave generated at a requency above that at which humans hear. The lower requency limit o ultrasound is taken to be 2 0 kHz. The range o requencies used in medicine is rom 2 to 2 0 MHz. Like electromagnetic radiation, ultrasound waves can be absorbed by matter, and reected and reracted when they meet a boundary between two media. The usual rules or reection and reraction o light at an interace also apply to ultrasound. B rothers Pierre and Paul- Jacques C urie frst observed the piezoelectric eect that is used to generate the ultrasound. They ound that when a quartz crystal is deormed, it produces a small em between opposite aces o the crystal. C onversely, applying a potential dierence across the crystal causes it to deorm and, under the right circumstances, to vibrate at high requencies. This property o piezoelectricity was ound to be shared by other materials besides quartz including some synthetic ceramic materials. These newer materials are now used in preerence to quartz or producing ultrasound. To make the scan a piezoelectric transducer ( which can both emit and receive the ultrasound) is placed in contact with the skin. A gel is used between the transducer and skin to prevent a large loss o energy that would occur at the transducerair and airskin interaces. A single pulse o the ultrasound is transmitted into the tissues and the system then waits or the reected wave ( an echo) to return rom each interace inside the patient. There are a number o ways to display the scan but the simplest display ( known as an A- scan) is to show a graph o reected signal strength against time. Figure 7( a) shows a typical A-scan with the various echo returns labelled. A knowledge o the speed o the ultrasound and the time or return enables the size and location o an organ to be determined ( remembering that, like all echoes, the ultrasound has to travel to the interace and return) . refections at boundary
probe
transducer movement organ 3 signal strength b) B scan
4 2 3 X
time
5 scan Y 4
time
2
scan X
Figure 7 A and B scans.
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1
1 time a) A scan
6
gel
bone
signal strength
abdomen wall
5 6 Y
patient
signal strength
C. 4 I M AG I N G TH E B O D Y ( AH L) A more complex type o scan is known as the B - scan ( fgure 7( b) ) . This requires a computer to combine a series o scans produced by the transducer. The operator rocks the transducer backwards and orwards to illuminate the internal suraces and the computer builds up a slice image ( in a similar way to the C T scan) through the patient by combining the signals rom a whole series o A-scans. The transmitted ultrasound that passes through the patient is not used in ultrasound diagnostic techniques, so it is o maj or importance to minimize the absorption o the ultrasound by tissues and to maximize the energy in the returned echo. To compare the ultrasound perormance o dierent types o material and tissue we use the acoustic impedance o the medium. This is a number that indicates how easy it is to transmit ultrasound through a particular material. The acoustic impedance Z o a material is ound to depend on the speed o sound c in the material and the density o the material: Z = c The S I unit or acoustic impedance is kg m 2 s 1 ; this is sometimes abbreviated to the non- SI unit the rayl ( named or Lord Rayleigh) but we will use the SI unit here. The table gives some typical values or speeds, densities and Z values or various tissues, though it should be remembered that c and thereore Z depend on the ultrasound requency being used.
Medium
Bone Soft tissue Liver Muscle Water Air at 15C
Speed / m s 1
Density / kg m 3
4100 1500 1550 1600 1480 340
1900 1050 1070 1080 1000 1.21
Acoustic impedance / 10 6 kg m 2 s 1 7.8 1.6 1.7 1.7 1.5 4.1 10 4
The proportion o the incident wave energy that is reected at an interace depends on dierences between the acoustic impedances o the two media. It can be shown that the ratio o the initial intensity I0 to the reected intensity Ir is: Ir ( Z2 - Z1 ) 2 _ = _2 I0 ( Z2 + Z1 ) where Z1 and Z2 are the acoustic impedances o the frst material and the second material respectively. The resolution o the image is o great importance in ultrasound imaging. The obvious approach might be to aim or the highest requency possible as this will give the shortest wavelength and the best resolution. However, the attenuation o the wave also increases markedly with requency as does the resolution itsel ( typically about 2 mm) i the requency is taken beyond a certain maximum. Generally the doctor has to accept a compromise, but requencies in the range 2 5 MHz are used or most applications.
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I M AG I N G Not all medical ultrasound use leads to an image in the conventional sense. There are many other diagnostic and therapeutic uses or ultrasound. These include the detection o blood ow and measurement o its speed using Doppler shits, and recent innovations where microbubbles o gas are introduced to enhance the image o blood vessels (in a similar way to the injection o contrast enhancers in X-radiography) . The advantages o ultrasound include:
It is a non- invasive technique.
It is relatively quick and inexpensive.
There are no known harmul eects.
It is o particular value in imaging sot tissues.
However:
Image resolution can be limited.
Ultrasound does not transmit through bone.
Ultrasound cannot image the lungs and the digestive system as these contain gas which strongly reects at the interace with tissue.
Nuclear magnetic resonance (NMR) in medicine Magnetic resonance imaging ( MRI scans) are medical diagnostic tools that use the phenomenon o nuclear magnetic resonance ( NMR) . In use, a patient is placed in a strong uniorm magnetic feld produced by an electromagnet. Other magnetic felds around the patient are varied and signals emitted rom the tissues are detected, measured and transormed into an image using a computer. The images produced have good resolution and good contrast or parts o the body that contain large proportions o water (and hence hydrogen nuclei) . The resolution or NMR techniques depends on the resonance requency o the signal, so the resolution is proportional to the magnetic feld strength. At the time o writing, a typical resolution or NMR imaging is 2 mm. MRI is a technique preerred or diagnosis o brain and central nervous system disorders. An explanation o MRI comes in two parts: the basic NMR eect itsel and then a description o the way it is modifed or medical use. top precession S
S N
N
S
high energy state
S conventional current loop high energy state
B
N
proton precession
N
Figure 8 Larmor precession.
The basic NMR efect
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Protons possess the properties o charge and spin and this leads them to behave as small magnets. The situation is analogous to charges moving around a circular loop o wire producing a magnetic feld that acts through the centre o the loop.
C. 4 I M AG I N G TH E B O D Y ( AH L)
Under normal conditions, hydrogen- rich materials such as water have the proton spins arranged randomly to give no overall magnetic feld. However, when a strong magnetic feld is imposed on the material the spins o the protons, and thus their magnetic felds, line up with the imposed magnetic feld.
Normally the protons line up in the lowest energy state, but i a radio- requency ( r) feld o an appropriate requency is applied to the system, some o the protons will ip into the opposite high energy state. This is analogous to a bar magnet ipping into a state where the north-seeking pole o the bar magnet is next to the northseeking pole o the feld source ( see fgure 8) .
The appropriate requency to cause the ip is known as the Larmor requency and, crucially or MRI, is directly proportional to the strength o the magnetic feld in which the material is placed. The Larmor requency in Hz is 4. 2 6 1 0 7 B where B is the magnetic feld strength in T.
What happens to the spinning protons when the r is applied can be thought o in terms o a spinning childs top. A top precesses around its spin axis in response to the Earths gravitational feld. The protons can be thought o as similarly precessing at the Larmor requency.
As the protons precess, the changing direction o the magnetic feld that they produce induces an em in a coil o wire nearby.
When the r feld is switched o, the protons in the high- energy state relax ( return) to the low- energy state.
MRI modifcations The process is essentially the same as the basic phenomenon. The patient is placed in a strong uniorm magnetic feld. However an additional gradient feld is added to the strong feld.
i Larmor requency is 64.8 MHz then signal must be coming rom here
Larmor requency = 64.5 MHz 1.54
65.6
1.52
64.8
1.50
63.9
Larmor re uenc /MHz
uniorm feld + gradient feld/T
patient
distance
Figure 9 A specifc Larmor requency comes rom a unique slice in the patient.
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TOK Henry Thoreau said It's not what you look at that matters, it's what you see. To what extent do you agree with this comment in the context of imaging inside the body?
The gradient feld is designed so that the total feld ( uniorm + gradient) varies linearly across the patient. The Larmor requency itsel depends linearly on B and thereore the Larmor requency varies predictably with position in the body.
The r feld is switched on and proton precession occurs. The relaxation o the protons ater the r feld has been removed leads to an electromagnetic signal that contains inormation about the number o protons emitting at each Larmor requency. This inormation is recovered rom the signal allowing a computer to plot the inormation spatially on an image.
MRi produces excellent images or diagnostic purpose without exposing the patient to radiation ( whether rom X- rays or radio- isotopes) . The energy o the photons in MRI are well below the 1 eV levels that correspond to molecular bonds and also are below the energies o X- ray photons. However, MRi is not entirely without risk. Factors involved in the risk include:
Strong magnetic felds. S ome patients are unsuitable or an MRI scan i they have, or example, a knee or hip j oint replacement that would distort the magnetic feld or a heart pacemaker that could be severely aected by currents induced in it when the strong magnetic felds change. There is no known risk rom a strong magnetic feld by itsel.
Radio-requency (r) felds. These can lead to local heating in the tissues o the patient.
Noise. The large changes in magnetic feld strength within the scanner cause parts o it to attempt to change shape during the scan. This can give rise to high intensities o sound that patients can fnd disturbing.
Claustrophobia. A strong uniorm feld is difcult to produce and can only be maintained over a small volume o space. This means that, typically, the patient is scanned while in a small- diameter tunnel, which some people fnd uncomortable.
Additionally, there may be elements o discomort or some patients in that a scan can take up to 90 minutes to complete; lying still in a confned space may prove difcult or some. In particular, young children and babies need to be sedated as they cannot understand the need to remain still.
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QUESTION S
Questions 1
Four rays o light rom O are incident on a thin concave ( diverging) lens. The focal points o the lens are labelled F. The lens is represented by the straight line XY.
( i)
O n a copy o the diagram, draw rays to locate the position o the image o the obj ect ormed by the lens.
( ii) Explain whether the image is real or virtual.
X
b) A convex lens o ocal length 62 .5 cm is used to view an insect o length 8.0 mm that is crawling on a table. The lens is held 5 0 mm above the table.
F
( i)
F
C alculate the distance o the image rom the lens.
( ii) C alculate the length o the image o the insect. ( 8 marks) Y
a) D efne the term focal point.
3
b) O n a copy o the diagram: ( i)
complete the our rays to locate the position o the image ormed by the lens
(IB) a) A parallel beam o light is incident on a convex lens o ocal length 1 8 cm. The light is ocused at point X as shown below.
( ii) show where the eye must be placed in order to view the image. c) S tate and explain whether the image is real or virtual.
X
d) The ocal length o the lens is 5 0. 0 cm. C alculate the linear magnifcation o an obj ect placed 75 . 0 cm rom the lens. e) Hal o the lens is now covered such that only rays on one side o the principal axis are incident on the lens. D escribe the eects, i any, that this will have on the linear magnifcation and the appearance o the image. (1 4 marks) 2
P
S tate the value o the distance PX. b) A diverging lens o ocal length 2 4 cm is now placed 1 2 cm rom the convex lens as shown below.
(IB) a) The diagram shows a small obj ect O represented by an arrow placed in ront o a converging lens. The ocal points o the lens are labelled F.
X P
12 cm F
O
F
c) ( i) ( ii)
Explain why point X acts as a virtual obj ect or the diverging lens. C alculate the position o the image as produced by the diverging lens.
L
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I M AG I N G d) A lens combination, such as a diverging and a convex lens, is reerred to as a telephoto lens. Suggest why a telephoto lens is considered to have a longer ocal length than that o a single convex lens. ( 7 marks) 4
( ii) Using the completed ray diagram above, derive an expression in terms o O and E or the angular magnifcation o an astronomical telescope. Assume that the fnal image is at infnity. c) When speciying an astronomical telescope, the diameter o the obj ective lens is requently quoted. S uggest a reason or quoting the diameter. ( 8 marks)
(IB) The diagram below shows the image o a square grid produced by a lens that does not cause spherical aberration. 6
(IB) A compound microscope consists o two convex lenses o ocal lengths 1 . 2 0 cm ( lens A) and 1 1 . 0 cm ( lens B ) . The lenses are separated by a distance o 2 3 .0 cm as shown below. ( The diagram is not drawn to scale.) O
1.30 cm
a) O n a copy o the diagram, draw the shape o the image when produced by a lens that causes spherical aberration.
lens A f = 1.20 cm
b) D escribe one way in which spherical aberration can be reduced. ( 4 marks)
5
23.0 cm
lens B f = 11.0 cm
An obj ect O is placed 1 .3 0 cm rom lens A. An image o O is ormed 1 5 .6 cm rom A.
(IB)
a) This image orms an obj ect or lens B . C alculate the obj ect distance or lens B .
The diagram below shows two lenses arranged so as to orm an astronomical telescope. The two lenses are represented as straight lines.
b) C alculate the distance rom lens B o the image as produced by lens B .
objective lens
c) C alculate the magnifcation o the microscope. ( 5 marks)
eye lens
7
(IB) a) S tate one cause o attenuation and one cause o dispersion in an optical fbre.
focal length fO
focal length fE
The ocal lengths o the obj ective lens and o the eye lens are O and E respectively. Light rom a distant obj ect is shown ocused in the ocal plane o the obj ective lens. The fnal image is to be ormed at infnity. a) C omplete the ray diagram to show the ormation o the fnal image. b) ( i)
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S tate what is meant by angular magnifcation.
b) An optical fbre o length 5 .4 km has an attenuation per unit length o 2 .8 dB km 1 . The signal power input is 80 mW. ( i)
C alculate the output power o the signal.
( ii) In order or the power o the output signal to be equal to the input power, an amplifer is installed at the end o the fbre. State the gain, in decibels ( dB ) , o the amplifer at the end o the fbre.
QUESTION S c) The signal to noise ratio ( S NR) , in dB , is defned as Psignal
SNR = 1 0 log ____ where Psignal and Pnoise are P noise
the powers o the signal and noise respectively. The S NR o the signal in ( b) beore amplifcation was 2 0 dB . C alculate the S NR ater amplifcation. ( 7 marks) (IB)
d) Annotate your sketch- graph to show the hal-value thickness x 1 /2 . e)
The variation with time t o the output power rom the optic fbre is shown in diagram 2 .
( i)
output power
1 0 a)
State the name o one o the mechanisms responsible or the attenuation o diagnostic X-rays in matter. ( 6 marks)
S tate and explain one situation, in each case, where the ollowing diagnostic techniques would be used.
0
0
t diagram 1
0
0
( i)
attenuation o the signal
( ii)
signal noise.
t diagram 2
b) The duration ( time width) o the signal increases as it travels along the optic fbre.
( ii)
( ii) Ultrasound
b) Apart rom health hazards, explain why dierent means o diagnosis are needed. ( 6 marks)
a) State and explain the eature o the graphs that shows that there is:
( i)
S tate two reasons or this increased time duration. S uggest why this increase in the width o the pulse sets a limit on the requency o pulses that can be transmitted along an uninterrupted length o optic fbre. ( 7 marks)
1 1 B eam energies o about 3 0 keV are used or diagnostic X- rays. This results in good contrast on the radiogram because the most important attenuation mechanism is not simple scattering. a)
( i)
I IO
attenuation coefcient
( ii) hal-value thickness. c)
The attenuation coefcient at 3 0 keV varies with the atomic number Z as attenuation coefcient Z3
The data given below list average values o the atomic number Z or dierent biological materials.
Biological material fat muscle bone
State what is meant by X-ray quality.
A parallel beam o X-rays o intensity I0 is incident on a material o thickness x as shown below. The intensity o the emergent beam is I.
Outline the most important attenuation mechanism that is taking place at this energy.
b) E xplain what is meant by:
(IB) a)
X- rays
( iii) Nuclear magnetic resonance
The scales are the same on both diagrams.
9
Draw a sketch-graph to show how intensity I varies with distance x.
The variation with time t o the input power to an optic fbre is shown in diagram 1 .
input power
8
c)
( i)
Atomic number Z 5.9 7.4 13.9
C alculate the ratio: attenuation coefcient or bone ____ attenuation coefcient or muscle
( ii) Suggest why X-rays o 3 0 keV energy are useul or diagnosing a broken bone, but a dierent technique must be used or examining a at- muscle boundary.
x
( 1 3 marks) b) D efne hal-value thickness.
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I M AG I N G 1 2 (IB)
d
a) State and explain which imaging technique is normally used: ( i)
O
to detect a broken bone
ultrasound transmitter and receiver
( ii) to examine the growth o a etus. The graph below shows the variation o the intensity I o a parallel beam o X-rays ater it has been transmitted through a thickness x o lead.
b)
The pulse strength o reected pulses is plotted against time t where t is the time elapsed between the pulse being transmitted and the time that the pulse is received.
15 10 5 0
0
(i)
2
4
6 x/mm
8
10
12
Use the graph to estimate the hal-value thickness x 1 /2 or this beam in lead.
c)
b) The diagram below shows an ultrasound transmitter and receiver placed in contact with the skin.
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25 50 75 100 125 150 175 200 225 250 275 300 t/s
Identiy the origin o the reected pulses A, B and C and D .
The above scan is known as an A- scan. State one way in which a B -scan diers rom an A- scan.
d) State one advantage and one disadvantage o using ultrasound as opposed to using X-rays in medical diagnosis. ( 1 0 marks)
1 5 S tate and explain the use o: a)
S tate a typical value or the requency o ultrasound used in medical scanning.
C
( ii) The mean speed in tissue and muscle o the ultrasound used in this scan is 1 . 5 1 0 3 m s 1 . Using data rom the above graph, estimate the depth d o the organ beneath the skin and the length l o the organ O .
( 1 1 marks)
E xplain which photon energy would be most suitable or obtaining a sharp picture o a broken leg. ( 2 marks)
D
B
( i)
( iii) A second metal has a hal-value thickness x 1 /2 or this radiation o 8 mm. C alculate what thickness o this metal is required to reduce the intensity o the transmitted beam by 80% .
1 3 The attenuation o X-rays depends not only on the nature o the material through which they travel but also on the photon energy. For photons with energy o about 3 0 keV, the halfvalue thickness o muscle is about 5 0 mm and or photons o energy 5 keV, it is about 1 0 mm.
A
0
( ii) Determine the thickness o lead required to reduce the intensity transmitted to 20% o its initial value.
1 4 a)
l
The scan is to estimate the depth d o the organ labelled O and also to fnd its length, l.
pulse strength/relative units
I/arbitrary units
20
layer of skin and fat
a barium meal in X-ray diagnosis
b) a gel on the skin during ultrasound imaging c)
a non-uniorm magnetic feld superimposed on a much larger constant feld in diagnosis using nuclear magnetic resonance. ( 7 marks)
D A S T R O P H YS I C S Introduction Astrophysics probes some of the most fundamental questions that humanity has sought to answer since the dawn of civilization. It links the experimental discipline of astronomy with the theoretical understanding of everything in the universe cosmology. It offers insights into the universe and provides answers, albeit tentatively, to its size, age, and content. Astrophysicists
can theorize on the life cycles of stars and gain an appreciation of how the universe looked at the dawn of time. The weakest of the four fundamental forces, gravity, comes into its own on an astronomic scale. It provides the mechanism to attach planets to stars, stars to other stars (in galaxies) , and galaxies to other galaxies (in clusters and super clusters) .
D.1 Stellar quantities Understandings Objects in the universe The nature o stars Astronomical distances Stellar parallax and its limitations Luminosity and apparent brightness
Nature of science When we look upwards, away rom the Earth, on a dark clear night we see many points o light in the sky. Some o these, such as the planets, artifcial satellites, and aircrat, are visible because they are reecting light rom the Sun. Others, such as stars and galaxies, are visible because o the light that they emit. The light rom distant stars and galaxies has travelled truly astronomical distances to reach us and we are able to construct a historical account o space rom analysing this. However, the light rom the dierent sources will have travelled varying distances to reach us and have been emitted at a range o times. This is analogous to looking at a amily photograph containing a mixture o many generations o a amily at dierent stages in their lives. It is a credit to humanity that we are able to conjecture so much about space when we can only justiy our reasoning with circumstantial evidence. In this way astrophysics mirrors orensic science we can never obtain evidence at the actual time an event happens.
HEAD A_UND
Applications and skills O B J TE XT_UND Identiying objects in the universe Qualitatively describing the equilibrium
between pressure and gravitation in stars Using the astronomical unit (AU) , light year (ly) and parsec (pc) Describing the method to determine distance to stars through stellar parallax Solving problems involving luminosity, apparent brightness, and distance
Equations
1 parsec defnition: d (parsec) = ___ p (arc-second)
luminosity equation: L = AT
4
L apparent brightness equation: b = __ 2 4 d
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Introduction In this frst sub-topic we rapidly move outwards rom our solar system to the remainder o our galaxy and beyond. In doing this we briey examine some o the obj ects that make up the universe. We then consider the range o units that we use or astronomical measurements. We end by considering how we can estimate the distance and luminosity o relatively near stars by treating them as black-body radiators and using their apparent brightness.
Objects that make up the universe The solar system comets
Pluto asteroids
Mercury Earth
Venus Mars Sun
Saturn
Neptune
Jupiter Uranus
(not drawn to scale)
Figure 1 The solar system. The solar system is a collection o planets, moons, asteroids, comets, and other rocky obj ects travelling in elliptical orbits around the S un under the inuence o its gravity. The S un is a star ormed, we believe, rom a giant cloud o molecular hydrogen gas that gravitated together, orming clumps o matter that collapsed and heated up. A gas disc around the young, spinning S un evolved into the planets. It is thought that the planets were ormed about 4.6 1 0 9 years ago ( see fgure 1 ) .
Figure 2 Thermal emission from the young star Fomalhaut and the debris disc surrounding it.
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The high temperature close to the S un permitted only those compounds with high condensation temperatures to remain solid, gradually accreting ( sticking together) particles to orm the our terrestrial planets: Mercury, Venus, Earth, and Mars. Further away rom the S un the Gas Giants or j ovian planets comprising o Jupiter, S aturn, Uranus, and Neptune were ormed rom cores o rock and metal and an abundance o ice. The huge quantity o ice meant that these planets became very large and produced strong gravitational felds that captured the slow moving hydrogen and helium. Pluto used to be called the ninth planet but, in 2 006, it was downgraded to a dwar planet. The planets move in elliptical orbits round the Sun with only Mercury occupying a plane signifcantly dierent to that o the other planets. Further out rom the S un, beyond Neptune, is the Kuiper belt. This is similar to the asteroid belt but much larger; it is the source o short- period comets and contains dwar planets ( including Pluto) . The Kuiper belt is set to be the next rontier o exploration in our solar system.
D .1 S TE LL AR Q U AN TI TI E S
S ix o the planets have moons orbiting them: Mars has two moons, while Jupiter has at least 5 0 acknowledged moons with several provisional ones that might be asteroids captured by its gravitational feld. About 4.5 billion years ago, the Earths moon is believed to have been ormed rom material ej ected when a collision occurred between a Mars- size obj ect and the Earth. Asteroids are rocky obj ects orbiting the S un with millions o them contained in solar orbit in the asteroid belt situated between Mars and Jupiter. Those o size less than 3 00 km have irregular shape because their gravity is too weak to compress them into spheres. S ome o the asteroids, such as C eres with a diameter o about 1 0 6 m, are large enough to be considered as minor planets. C omets are irregular obj ects a ew kilometres across comprising rozen gases, rock, and dust. O bservable comets travel around the S un in sharply elliptical orbits with periods ranging rom a ew years to thousands o years. As they draw near to the S un the gases in the comet are vaporized, orming the distinctive comet tail that can be millions o kilometres long and always points away rom the S un.
Figure 3 Comet Hale-Bopp.
Stars gas and radiation pressure Like the S un, all stars initially orm when gravity causes the gas in a gravity nebula to condense. As the atoms move towards one another, they lose gravitational potential energy that is converted into kinetic energy. This raises the temperature o the atoms which then orm a protostar. When the mass o the protostar is large enough, the temperature and pressure at the centre will be sufcient or hydrogen to use into helium, with the release o very large amounts o energy the star has ignited. Ignition produces emission o radiation rom the core, producing a radiation pressure that opposes the inward gravitational orces. When this is balanced the star is in a state o hydrostatic equilibrium and will remain stable or up to billions o years because it is on the main sequence. As the hydrogen is used up the star will eventually undergo changes that will move it rom the main sequence. D uring these changes the colour o the star alters as its surace temperature rises or alls and it will change size accordingly. The original mass o material in the star determines Figure 4 Hydrostatic equilibrium. how the star will change during its lietime.
Groups of stars D espite the difculties in assessing whether stars exist singly or in groups o two or more, it is thought that around fty per cent o the stars nearest to the S un are part o a star system comprising two or more stars. B inary stars consist o two stars that rotate about a common centre o mass. They are important in astrophysics because their interactions allow us to measure properties that we have no other way o investigating. For example, careul measurement o the motion o the stars in a binary system allows their masses to be estimated. A stellar cluster is a group o stars that are positioned closely enough to be held together by gravity. S ome clusters contain only a ew dozen stars while others may contain millions. All o the stars in a star cluster were ormed at the same time rom the same nebula. The Pleiades ( fgure 5 ) is a stellar cluster o about 5 00 stars that can be seen with the naked eye;
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A S T R O P H YS I C S this is an example o an op en cluster. O pen clusters consist o up to several hundred stars that are younger than ten billion years and may still contain some gas and dust. They are located within our galaxy, the Milky Way, and so lie within a single plane. Globular clusters contain many more stars and are older than eleven billion years and, thereore, contain very little gas and dust. There are 1 5 0 known globular clusters lying j ust outside the Milky Way in its galactic halo. Globular clusters are essentially spherically shaped.
Figure 5 The Pleiades.
A constellation, unlike a stellar cluster, is a pattern ormed by stars that are in the same general direction when viewed rom the E arth. They are more signifcant historically than physically as many ancient societies attributed them with religious importance. Today, the patterns made by the constellations are helpul to astronomers in locating areas o the sky or telescopic study. Naturally, some o the stars in a constellation are much closer to the E arth than others. B ecause o proper motion they will appear very dierent in, say, ten thousand years time. S uch stars are not held together by gravity.
Nebulae
Figure 6 The Crab Nebula.
Regions o intergalactic cloud o dust and gas are called nebulae ( singular is nebula) . As all stars are born out o nebulae, these regions are known as stellar nurseries. There are two dierent origins o nebulae. The frst origin o nebulae occurred in the matter era around 3 8 0 000 years ater the B ig B ang. D ust and gas clouds were ormed when nuclei captured electrons electrostatically and produced the hydrogen atoms that gravitated together. The second origin o nebulae is rom the matter which has been ej ected rom a supernova explosion. The C rab Nebula, shown in fgure 6 , is a remnant o such a supernova. O ther nebulae can orm in the fnal, red giant, stage o a low mass star such as the S un.
Galaxies
Figure 7 The Andromeda galaxy with two smaller satellite galaxies.
A galaxy is a creation o stars, gas, and dust held together by gravity and containing billions o stars. The Milky Way contains about 3 1 0 1 1 stars and, probably, at least this number o planets. S ome galaxies exist in isolation but the maj ority o them occur in groups known as clusters that have anything rom a ew dozen to a ew thousand members. The Milky Way is part o a cluster o about 3 0 galaxies called the Local Group which includes Andromeda ( fgure 7) and Triangulum. Regular clusters consist o a concentrated core and are spherical in shape. Irregular clusters also exist, with no apparent shape and a lower concentration o galaxies within them. S ince the launch o the Hubble Space Telescope it has been observed that even larger structures, called superclusters, orm a network o sheets and flaments; approximately 90% o galaxies can be ound within these. In between the clusters there are voids that are apparently empty o galaxies. The Milky Way and Andromeda are members o the most common class o galaxies spiral galaxies. These are characterized by having a discshape with spiral arms spreading out rom a central galactic bulge that contains the greatest density o stars. It is increasingly speculated that, at the centre o the galactic bulge, there is a black hole. The spiral arms
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contain many young blue stars and a great deal o dust and gas. O ther galaxies are elliptical in shape, being ovoid or spherical these contain much less gas and dust than spiral galaxies; they are thought to have been ormed rom collisions between spiral galaxies. Irregular galaxies are shapeless and may have been stretched by the presence o other massive galaxies the Milky Way appears to be having this eect on some nearby dwar galaxies.
Astronomical distances Resulting rom the huge distances involved in astronomical measurements, some unique, non- S I units have been developed. This avoids using large powers o ten and allows astrophysicists to gain a eel or relative sizes and distances. The light year (ly) : The speed o light is one o the most undamental constants in physics; all inertial observers measure light as travelling at the same speed ( see Option A or a background to this) . This property can be used to defne a set o units based on the speed o light. For example, the distance travelled by light in one minute is called a light minute. As it takes light approximately 8 minutes to travel rom the Sun to the Earth, the distance between them is 8 light minutes. The light year (ly) is a more commonly used unit and is the distance travelled by light in one year. 15
1 ly = 9.46 1 0 m The astronomical unit (AU): This is the average distance between the Sun and the Earth. It is really only useul when dealing with the distances o planets rom the Sun. 1 AU = 1 .5 0 1 0 1 1 m 8 light minutes The parsec (pc): This is the most commonly used unit o distance in astrophysics. 1 pc = 3.26 ly = 3.09 1 0 1 6 m D istances between nearby stars are measured in pc, while distances between distant stars within a galaxy will be in kiloparsecs (kpc) , and those between galaxies in megaparsecs (Mpc) or gigaparsecs (Gpc) . The distances used in astronomy are truly enormous, and this has meant that a variety o indirect methods have been developed or their measurement. The method used to measure the distance o an astronomical object rom the Earth is dependent on its proximity.
Stellar parallax O n Earth, surveyors measure distances by using the method o triangulation. A known or measured length is taken as a baseline. The angles that a distant obj ect makes at either end o the baseline are then measured ( using a theodolite) . From these angles the distance o the obj ect can be calculated. When measuring the distance o a star rom the Earth ( in parsec) a similar technique is employed. Parallax is based on the act that nearby obj ects will appear to cross distant obj ects when viewed rom dierent positions. This can be seen rom inside a moving car when ence posts by the roadside appear to speed by but distant hills hardly move. As the Earth orbits the Sun, the stars that are quite close to us appear to move across the distant fxed stars as shown in fgure 8.
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fxed stars in background
July image
parallax angle p
star or which distance is being measured
position o Earth sun January image in July baseline = diameter o Earths orbit position o = 2AU Earth in January
Figure 8 Stellar parallax. When the image o a nearby star is recorded in both January and July the star appears to have moved across the fxed stars in the background. Using the equation 1 d= _ p gives the distance d in parsecs when p is the parallax angle in arcsecond. This simple relationship is used or defning the parsec: when a star is at a distance o 1 pc rom the Earth the parallax angle given by the equation will be one arcsecond. In a circle there are 3 60 degrees. In every degree there are 60 arcminutes and 60 arcseconds in every arcminute. Thus 1 arcsecond is very small 1 being j ust ____ o a degree. 3600 There is a limit to the distance that can be measured using stellar parallax parallax angles o less than 0.01 arcsecond are difcult to measure rom the surace o the Earth because o the absorption and scattering o light by the atmosphere. Turbulence in the atmosphere also limits the resolution because it causes stars to twinkle. Using the 1 parallax equation, gives a maximum range o d = ____ = 1 00 pc. 0.01
Figure 9 The Gaia mission.
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In 1 989, the satellite Hipparcos ( an acronym or High Precision Parallax C ollecting S atellite) was launched by the European Space Agency ( ES A) . B eing outside the atmosphere, Hipparcos was able to measure the parallaxes o 1 1 8 000 stars with an accuracy o 0.001 arcsecsond ( to distances 1 000 pc) ; its mission was completed in 1 993. Gaia ( fgure 9) , Hipparcoss successor, was launched in 2 01 3 and is charged with the task o producing an accurate three- dimensional map showing the positions o about a billion stars in the Milky Way. This is about one per cent o the total number o stars in the galaxy! Gaia is able to resolve a parallax angle o 1 0 microarcsecond measuring stars at a distance o 1 00 000 pc. Amateur astronomers taking images will be working at approximately 1 arcsecond per pixel.
D .1 S TE LL AR Q U AN TI TI E S
Luminosity and apparent brightness of stars In Sub-topic 4.3 we looked at the intensity o a wave at a point distance r rom the source. Intensity was defned as the power emitted by a source divided by the area o the sphere over which the energy is spread equally. P I = _2 4 r For stars P is called the luminosity ( L) o the star and represents the total energy emitted by the star per second in watt. I in this context is the apparent brightness ( b) and is measured in watts per square metre ( W m 2 ) ; the distance rom the star is usually denoted by d. This version o the equation is written as: L b = _2 4 d The luminosity o a star is a very important quantity in establishing the nature o a star. Hence, the equation is particularly useul because, although we are not in a position to measure the luminosity o the star, we can measure its apparent brightness (or example, by using a telescope and a charge-coupled device) . I we also calculate the distance o the star we can then work out its luminosity. Alternatively, when we know the luminosity o a star (because it is similar to other stars o known luminosity) we are able to use the measurement o its apparent brightness to estimate the stars distance.
Worked example a) For a star, state the meaning o the ollowing terms: ( i) Luminosity ( ii) Apparent brightness. b) The spectrum and temperature o a certain star are used to determine its luminosity to be approximately 6. 0 1 0 31 W. The apparent brightness o the star is 1 . 9 1 0 - 9 W m 2 . These data can be used to determine the distance o the star rom Earth. C alculate the distance o the star rom Earth in parsec.
Solution a) ( i) The luminosity is the total power emitted by the star. ( ii) The apparent brightness is the incident power per unit area received at the surace o the Earth. L b) Using b = _2 we rearrange to get 4 d ___
d=
L = ___ 4 b
__________
6.0 1 0 = 5 .0 ____________ 4 1 . 9 1 0 31
-9
1 0 19 m
As 1 pc = 3 .2 6 ly and 1 ly = 9. 46 1 0 1 5 m then 1 pc = 3 .2 6 9.46 1 0 1 5 m
c) D istances to some stars can be measured by using the method o stellar parallax.
= 3 .08 1 0 1 6 m. 5 .0 1 0 or 1 62 3 pc 5 .0 1 0 1 9 m is ________ 3 .08 1 0 1 600 pc 19
( i) O utline this method. (ii) Modern techniques enable the measurement rom Earths surace o stellar parallax angles as small as 5 .0 1 0 3 arcsecond. Calculate the maximum distance that can be measured using the method o stellar parallax.
16
c)
( i) The angular position o the star against the background o fxed stars is measured at six month intervals. The distance d is then 1 ound using the relationship d = __ p ( this is shown in fgure 6.) 1 ( ii) d = _______ = 2 00 pc. 5 1 0 -3
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Black-body radiation and stars
Worked example Some data or the variable star B etelgeuse are given below. Average apparent brightness = 1 .6 1 0 7 Wm 2 Radius = 790 solar radii EarthB etelgeuse separation = 1 3 8 pc The luminosity o the S un is 3 .8 1 0 26 W and it has a surace temperature o 5 800 K. a) C alculate the distance between the Earth and B etelgeuse in metres. b) D etermine, in terms o the luminosity o the Sun, the luminosity o B etelgeuse. c) C alculate the surace temperature o B etelgeuse.
In Sub-topic 8.2 we considered black bodies as theoretical objects that absorb all the radiation that is incident upon them. B ecause there is no reection or re-emission they appear completely black as their name suggests. Such bodies would also behave as perect emitters o radiation, emitting the maximum amount o radiation possible at their temperature. All objects at temperatures above absolute zero emit black-body radiation. This type o radiation consists o every wavelength possible but containing dierent amounts o energy at each wavelength or a particular temperature. Although stars are not perect black-bodies they are capable o emitting and absorbing all wavelengths o electromagnetic radiation. Figure 1 0 shows the black-body radiation curves or the Sun, the very hot star Spica, and the cold star Antares. B ecause each o the stars will produce dierent intensities, the curves have been normalized by dividing the intensity emitted at a given wavelength by the maximum intensity that the star yields this means the vertical scale has no unit and the maximum value it can take is 1 .00. The maximum intensity o radiation emitted by the Sun has a wavelength o just over 5 00 nm making it appear yellow; the peak intensity or Spica is in the ultraviolet region, but there is sufcient intensity in the blue region or it to appear blue; the peak or Antares is in the near inra-red but, with plenty o red light emitted, it appears to be red to the naked eye.
a) As 1 pc = 3 .1 1 0 1 6 m, 1 3 8 pc = 1 3 8 3 .1 1 0 1 6 = 4.3 1 0 1 8 m 2 L b) b = ____ L = 4 d b 2
4 d
= 4[4.3 1 0 ] 1 .6 1 0 18 2
= 3 .7 1 0
31
-7
3 .7 1 0 31 ________ = 9.7 1 0 4. 26
So Betelgeuse has a luminosity o 9.7 1 0 4 L Sun c) As L = 4 R T , by taking ratios we get 2
4
2
4
L Sun 4 R S u n T S u n _______ = _______________ 2 4 L 4 R B e te lge u s e T B e te lge u s e
_________
Sun
=
4
4
2
L B e te lge u se R S u n _________ L S u n R B e te lge u se 2
_______
9.8 1 04 _______ = 0. 63 7902
TBetelgeuse = 0.63 5 800 K = 3 700 K
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1.00 0.75
the Sun (5800 K)
0.50
Antares (3400 K)
0.25 0 0
500
1000 wavelength/nm
1500
2000
Figure 10 Black-body radiation curves for three stars.
3.8 1 0
T B e te lge u s e ______ = T
Spica (23,000 K)
W
D ividing by the luminosity o the S un gives
B e te lg u e u s e
normalized intensity
Solution
For a star the S teanB oltzmann law is written as L = AT4 where L is the luminosity in watt, A the surace area o the star in square metres, and T the temperature in kelvin. is the S teanB oltzmann constant = 5 .67 1 0 8 Wm 2 K 4. When we assume that a star is spherical we can use this equation in the orm: L = 4 R 2 T4 where R is the radius o the star. We can see that the luminosity o a star depends on its temperature and its size ( measured here by its surace area) . In the next sub- topic we will see how the balance between temperature and surace star size is used to categorize star type.
D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N
D.2 Stellar characteristics and stellar evolution Understandings Stellar spectra HertzsprungRussell (HR) diagram Massluminosity relation or main sequence stars
Applications and skills Explaining how surace temperature may be
Cepheid variables Stellar evolution on HR diagrams Red giants, white dwars, neutron stars, and
black holes Chandrasekhar and OppenheimerVolko limits
Nature of science In 1859, the physicist Gustav Kirchho and chemist Robert Bunsen worked at the University o Heidelberg in Germany. Having developed the frst spectroscope, they repeated Foucaults experiment o passing sunlight through a bright sodium ame to fnd the absorption lines seen in the solar spectrum. They then repeated their experiment with other alkali metals and were able to show that the solar absorption spectra were the reverse o emission spectra. Kirchho went on to provide strong evidence or the presence o iron, magnesium, sodium, nickel, and chromium in the atmosphere o the Sun. Their experiments paved the way or our understanding o many o the properties o stars.
obtained rom a stars spectrum Explaining how the chemical composition o a star may be determined rom the stars spectrum Sketching and interpreting HR diagrams Identiying the main regions o the HR diagram and describing the main properties o stars in these regions Applying the mass luminosity relation Describing the reason or the variation o Cepheid variables Determining distance using data on Cepheid variables Sketching and interpreting evolutionary paths o stars on an HR diagram Describing the evolution o stars o the main sequence Describing the role o mass in stellar evolution
Equations Wiens law: m a x T = 2.9 10
-3
mK luminosity-mass relationship: L M 3 .5
Introduction The magnitude o the distance between stars is unimaginable with the nearest star (ignoring the Sun) being so distant that, even with the astest rocket ever built, it would take almost a hundred thousand years or us to reach it. The curiosity o the human race knows no bounds and we have striven to fnd out all we can about stars their lie cycles and their distances rom the Earth. In this sub-topic we investigate how it is possible to estimate the distance o stars more than 1 00 pc rom the Earth, and explore the lives o stars o dierent mass, radius, temperature, and colour.
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Stellar spectra In Sub- topic 7.1 we considered emission and absorption spectra occurring as electrons make transitions between energy levels. S uch spectra provide important inormation about the chemical composition, density, surace temperature, rotational and translational velocities o stars. When we observe the spectrum o stars using a spectrometer we fnd that nearly all stars show a continuous spectrum which is crossed by dark absorption lines; some stars also show bright emission lines. S eeing absorption lines across a continuous stellar spectrum tells us that the stars have a hot dense region ( to produce the continuous spectrum) surrounded by cooler, low- density gas ( to produce the absorption lines) . In general, the density and temperature o a star decreases with distance rom its centre. B ecause its temperature is so high, a stars core has to be composed o high- pressure gases and not o molten rock, unlike the cores o some planets.
Composition of stars Absorption o certain wavelengths is apparent when we observe the intensitywavelength relationship or stars. The smooth theoretical black- body curve is modifed by absorption dips as can be seen in fgure 1 . The graph shows the variation with wavelength o the intensity or the Sun and Vega ( which is much hotter than the S un having a surace temperature o 9600 K) . In the case o Vega, the cooler hydrogen in the stars outer layers ( the photosphere) absorbs the photons emitted by hydrogen. The clear pattern between the wavelengths absorbed and those o the visible part o the hydrogen absorption spectrum shown in fgure 2 strongly suggests that Vegas photosphere is almost entirely made up o hydrogen.
intensity/arbitrary units
1.5
Vega
1
Sun
0.5
0
-0.5 400
450
500
550
600 650 wavelength/nm
Figure 1 Intensitywavelength relation for Vega and the Sun.
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700
750
800
D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N The absorption lines correspond to those o hydrogen shown in the spectrum o fgure 2 . hydrogen absorption spectrum
400 nm
700 nm
Figure 2 Atomic hydrogen absorption spectrum. It also implies that the transitions occur in agreement with the B almer series rom level n = 2 to higher levels. O n the other hand, the S un has some o the visible hydrogen lines. B ecause the Sun is cooler than Vega, many o its hydrogen atoms are in the ground state producing absorption lines that are in the ultraviolet region o the spectrum ( corresponding to transitions rom level n = 1 in the Lyman series) . S tars that are even hotter than Vega tend not to produce hydrogen absorption lines in the visible spectrum because their high temperatures mean that the hydrogen in the photosphere is ionized ( and thereore has no electron to become excited by the absorption o a photon) .
Nature of science Spectral classes S tars were originally categorized in terms o the strength o the hydrogen absorption lines with the stars producing the darkest absorption lines being called type A and those with successively weaker lines type B and C etc. Within the last century astronomers recognized that the line strength depended on temperature o the stars rather than their composition. Subsequently, the
whole system o classifcation was revised with the result that many categories were abandoned and many others reordered as shown in the table below. This orm o star classifcation will not be included in your IB D iploma Programme Physics examinations, but is included here or historical interest and to indicate how many scientists have a reluctance to abandon a well-loved system!
Main sequence star properties Colour
Class Solar masses Solar diameters
Nature Oof science 20100
bluest
Spectral classes bluish B
420
1225 412
S tars were originally categorized in terms o the blue-white A 24 1.54 strength o the hydrogen absorption lines with the stars producing the darkest absorption lines white F 1.052 1.11.5 being called type A and those with successively weaker lines type etc. Within the last yellow-white G B , C0.81.05 0.851.1 century astronomers recognized that the line orange depended K on temperature 0.50.8 0.60.85 strength o the stars rather than their composition. Subsequently, the
red
M
0.080.5
0.10.6
Temperature/K Prominent lines 40 000
ionized helium
18 000
neutral helium, neutral hydrogen
whole system o classifcation was revised with 10result 000 thatneutral the manyhydrogen categories were abandoned and many others reordered as shown in the 7000 neutral hydrogen, ionized calcium table below. This orm o star classifcation will not5500 be included in your IB Physics neutral hydrogen, strongestexaminations, ionized calcium but is included here or historical interest and to 4000 how neutral metals (calcium, , ionized to indicate many scientists have iron) a reluctance calcium abandon a well-loved system!
3000
molecules and neutral metals
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A S T R O P H YS I C S Using the absorption spectrum to determine the chemical elements in a particular star is not easy as the spectra o many elements ( dierently ionized) are superimposed on one another. Although the lines present in an absorption spectrum tell us quite a lot about the temperature o the photosphere ( as described above) , they are difcult to interpret in terms o the abundance o elements. The movement o the gas atoms in the star causes the photons o light emitted to undergo both red and blue D oppler shits. In hotter stars, the atoms move aster than in cooler stars and, thereore, the D oppler broadening is more pronounced. The rotation o the stars themselves means that the light reaching us will come rom dierent parts o the star one edge moving towards us, the other moving away rom us and the central region rotationally stationary. This, o course, adds to the thermal D oppler broadening.
Wiens displacement law and star temperature B y treating a star as a black body it is possible to estimate its surace temperature using Wiens law as discussed in S ub- topic 8.2 . S tars range in surace temperature rom approximately 2 000 K to 40 000 K. The temperature T o the star in kelvin is given by: 2 .9 1 0 - 3 mK T = __ max where max is the wavelength in metres at which the black- body radiation is o maximum intensity and T is in kelvin.
Worked example a) Explain the term black-body radiation. The diagram is a sketch graph o the blackbody radiation spectrum o a certain star.
Solution a) B lack-body radiation is that emitted by a theoretical perect emitter or a given temperature. It includes all wavelengths o electromagnetic waves rom zero to infnity.
intensity
intensity
b) and c)
b) C opy the graph and label its horizontal axis. c) O n your graph, sketch the black-body radiation spectrum or a star that has a lower surace temperature and lower apparent brightness than this star. d) The star B etelgeuse in the O rion constellation emits radiation approximating to that emitted by a black-body radiator with a maximum intensity at a wavelength o 0. 97 m. C alculate the surace temperature o B etelgeuse.
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wavelength
The red line intensity should be consistently lower and the maximum shown shited to a longer wavelength. -3
2 .9 1 0 mK 2 .9 1 0 - 3 d) T = __ = __ max 0. 97 1 0 - 6 = 2 .99 1 0 3 K 3 000K
D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N
Cepheid variables
luminosity/arbitary units
C epheid variables are extremely luminous stars that undergo regular and predictable changes in luminosity. B ecause they are so luminous it means that very distant C epheids can be observed rom the Earth. In 1 784, the periodic pulsation o the supergiant star, D elta C ephei, was discovered by the E nglish amateur astronomer, John Goodricke. This star has a period o about 5 .4 days as seen in fgure 3 .
period 5.4 days 4
6 8 10 12 time ( days) Figure 3 Lum inositytime relationship for Delta Cephei.
In 1 908, the American astronomer, Henrietta Leavitt, was working at the Harvard C ollege Observatory. She published an article describing the linear relationship between the luminosity and period o pulsation or 25 stars, all practically the same distance rom the Earth, in the Small Magellanic C loud a galaxy that orbits the Milky Way. This relationship allows us to estimate the luminosity o a given star by measuring its period o pulsation. It is now known that there are many more o these variable stars collectively known as Cepheids. The period o these stars varies between twelve hours and a hundred days. Although the period is regular it is not sinusoidal and it takes less time or the star to brighten than it does to ade.
C epheid variable stars are known as standard candles because they allow us to measure the distances to the galaxies containing C epheid variable stars. The distances, d, o the C epheids can be calculated rom the apparent brightnessluminosity equation: L b = _2 . 4 d The apparent brightness is measured using a telescope and C C D and the luminosity is calculated rom a measurement o the period o the C epheid.
10 3
type II (W vignis) Cepheids
10 2 RR Lyrae
Figure 4 shows the relative luminosityperiod relationship or C epheid stars. The relative luminosity is the ratio o the luminosity o the star to that o the S un. The S uns luminosity is conventionally written as L C epheid stars are stars that have completed the hydrogen burning phase and moved o the main sequence ( see later or an explanation o this) . The variation in luminosity occurs because the outer layers within the star expand and contract periodically. This is shown diagrammatically in fgure 5 . I a layer o an element loses hydrostatic equilibrium ( between the gas and radiation pressure and that due to gravity) and is pulled inwards by gravity ( 1 ) , the layer becomes compressed and less transparent to radiation ( 2 ) ; this means that the temperature inside the layer increases, building up the internal pressure ( 3 ) and causing the layer to be pushed outwards ( 4) . D uring expansion the layer cools, becoming less dense ( 5 ) and more transparent, allowing radiation to escape and letting the pressure inside all ( 6) . S ubsequently the layer alls inwards under gravity ( 1 ) and the cycle repeats causing the pulsation o the radiation emitted by the star.
type I (classical) Cepheids
10 4
L /L
2
1 0.5
1
3 5 10 period/days
30 50
100
Figure 4 Relative luminosity period relationships for Cepheid stars.
2
3
4
1
6
5
Figure 5 Cycle in a Cepheid star.
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TOK Women in science, technology, and engineering Over the ages there have been ew women scientists, and physics has been no exception. Astronomy, however, is one o the areas in which women have shone. Many women have made signifcant contributions. For example, in astrophysics, we have the German British astronomer Caroline Herschel (who discovered several comets), Agnes Clerk, Maria Mitchell, Annie Jump Cannon, and Henrietta Leavitt. Even today, there continues to be ewer women ollowing careers in science, and ewer opportunities or them to gain promotion to senior roles. Why should women be so under- represented in science, technology, and engineering?
Worked example a) D efne ( i) luminosity ( ii) apparent brightness.
Solution
b) State the mechanism or the variation in the luminosity o the C epheid variable.
a) ( i) Luminosity is the total power radiated by a star.
The variation with time t, o the apparent brightness b, o a C epheid variable is shown below. 1.3
A
b/10 -10 W m -2
1.2
( ii) Apparent brightness is the power rom a star received by an observer on the Earth per unit area o the observers instrument o observation. b) O uter layers o the star expand and contract periodically due to interactions o the elements in a layer with the radiation emitted.
1.1 1.0
c) ( i) The radius is larger ater two days (point A) because, at this time the luminosity is higher and so the stars surace area is larger.
0.9 B 0.8 0
1
2
3
4 5 6 time/da s
7
8
9
10
Two points in the cycle o the star have been marked A and B . c) ( i) Assuming that the surace temperature o the star stays constant, deduce whether the star has a larger radius ater two days or ater six days. ( ii) Explain the importance o C epheid variables or estimating distances to galaxies. d) ( i) The maximum luminosity o this C epheid variable is 7.2 1 0 29 W. Use data rom the graph to determine the distance o the C epheid variable. ( ii) C epheids are sometimes reerred to as standard candles. Explain what is meant by this.
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(ii) Cepheid variables show a regular relationship between period o variation o the luminosity and the luminosity. By measuring the period the luminosity can be calculated and, by L using the equation b = _ , the distances 4 d 2 to the galaxy can be measured. This assumes that the galaxy contains the Cepheid star. L 7.2 1 0 29 d) ( i) b = _2 thus 1 . 2 5 1 0 1 0 = _ 4 d 4 d 2 _______________
d=
7.2 1 0 __ 4 1 .2 5 1 0
d = 2 .1 4 1 0
29
-10
19
m
( ii) A standard candle is a light source o known luminosity. Measuring the period o a Cepheid allows its luminosity to be estimated. From this, other stars in the same galaxy can be compared to this known luminosity.
D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N
Hertzsprung-Russell (HR) diagram We saw in S ub- topic D .1 that the luminosity o a star is proportional both to its temperature to the ourth power and to its radius squared. C learly, large hot stars are the most inherently bright, but how do other combinations o temperature and radius compare to these? In the early 1 900s, two astronomers, Ej nar Hertzsprung in D enmark and Henry Norris Russell in America, independently devised a pictorial way o illustrating the dierent types o star. B y plotting a scattergram o the luminosities o stars against the stars temperatures, clear patterns emerged. These scattergrams are now known as Hertzsp rung-Russell ( HR) diagrams. In general, cooler red stars tend to be o relatively low luminosity, while hotter blue stars tend to be o high luminosity. With high temperature conventionally drawn to the let o the horizontal axis, the maj ority o stars create a diagonal stripe which goes rom top let to bottom right this is known as the main sequence. A small number o stars do not ollow the main sequence pattern but, instead, orm island groups above and below the main sequence. The vertical axis is commonly modifed to show the ratios o star luminosity to that o the S un ( denoted as L ) as shown in fgure 6 in this case the axis is logarithmic and has no unit. The temperature axis is also logarithmic and doubles with every division rom right to let. The HR diagram shows the position o many stars o dierent ages; during the lietime o a star its position will move on the diagram as its temperature and luminosity changes. We know rom black-body radiation that the luminosity depends on the size o a star and its temperature. Small, cool stars will be dim and be positioned to the bottom right o the diagram rom Wiens law we know they will be red. Large, hot stars will be o high luminosity and blue or blue-white in colour thus placing them at the top let o the diagram. A modifcation o the HR diagram to include the dierent star classes is shown in fgure 7. Main sequence stars are ordinary stars, like the S un, that produce energy rom the usion o hydrogen and other light nuclei such as helium and carbon. Nearly 90% o all stars ft into this category. 10 6 10 6 10 4
10 4
cool bright
hot bright
supergiants
10 2 L star L
10
2
L star L
10 -2 10 -4
giants 10 0
Sun
10 0
main sequence Sun
10 2 main sequence hot dim 40 000
20 000
cool dim 10 000 T/K
5000
Figure 6 HertzsprungRussell diagram.
2500
10 4 40 000
white dwars 20 000
10 000 5000 surace temperature/K
2500
Figure 7 HertzsprungRussell diagram showing the diferent classes o stars.
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A S T R O P H YS I C S Red giants are cooler than the Sun and so emit less energy per square metre o surace. However, they have a higher luminosity, emitting up to 1 00 times more energy per second than the S un. This means that they must have a much greater surace area to be able to emit such large energies. They, thereore, have a much larger diameter than the S un making them giant stars. S up ergiant stars are gigantic and very bright. A supergiant emitting 1 00 000 times the energy per second and at the same temperature o the Sun must have a surace area 1 00 000 times larger. This leads to a diameter that is over 3 00 times the diameter o the S un. O nly about 1 % o stars are giants and supergiants. White dwarfs are the remnants o old stars and constitute about 9% o all stars. Although they were very hot when they fnally stopped producing energy, they have a relatively low luminosity showing them to have a small surace area. These very small, hot stars are very dense and take billions o years to cool down.
Massluminosity relation for main sequence stars Not all main sequence stars are the same as the S un some are smaller and cooler while others are larger and hotter. High mass stars have shorter lietimes a star with a mass o 1 0 times the solar mass might only live 1 0 million years compared to the expected lietime o around 1 0 billion years or the Sun. O bservations o thousands o main sequence stars have shown there to be a relationship between the luminosity and the mass. For such stars this takes the orm L M 3.5 where L is the luminosity in W ( or multiples o the S uns luminosity, L ) and M is the mass in kg ( or multiples o the S uns mass, M ) . Because mass is raised to a positive power greater than one, this means that even a slight dierence in the masses o stars results in a large dierence in their luminosities. For example, a main sequence star o 1 0 times the mass o the Sun has a luminosity o (1 0) 3.5 3200 times that o the Sun. For a star to be stable it needs to be in hydrostatic equilibrium, where the pressure due to the gravitational attraction o inner shells is equalled by the thermal and radiation pressure acting outwards. For a stable star o higher mass there will be greater gravitational compression and so the core temperature will be higher. Higher temperatures make the usion between nuclei in the core more probable giving a greater rate o nuclear reaction and emission o more energy; thus increasing the luminosity. The mass o a star is undamental to the stars lietime those with greater mass have ar shorter lives.
Stellar evolution Formation of a star We saw in S ub- topic D .1 that the initial process in the ormation o a star is the gravitational attraction o hydrogen nuclei. The loss o potential energy leads to an increase in the gas temperature. The gas becomes
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D . 2 S TE LL AR CH ARACTE RI S TI CS AN D S TE LL AR E VO LU TI O N
denser and, when the protostar has sufcient mass, the temperature becomes high enough or nuclear usion to commence. The star moves onto the main sequence where it remains or as long as its hydrogen is being used into helium this time occupies most o a stars lie. E ventually when most the hydrogen in the core has used into helium the star moves o the main sequence.
The fate of stars All stars collapse when most o the hydrogen nuclei have used into helium. Gravity now outweighs the radiation pressure and the star shrinks in size and heats up. The hydrogen in the layer surrounding the shrunken core is now able to use, raising the temperature o the outer layers which makes them expand, orming a giant star. Fusion o the hydrogen adds more helium to the core which continues to shrink and heat up, orming heavier elements including carbon and oxygen. The very massive stars will continue to undergo usion until iron and nickel (the most stable elements) are ormed. What happens at this stage depends on the mass o the star.
A. Sun-like stars For stars like the S un o moderate mass ( up to about 4 solar masses) the core temperature will not be high enough to allow the usion o carbon. This means that, when the helium is used up, the core will continue to shrink while still emitting radiation. This blows away outer layers orming a planetary nebula around the star. When the remnant o the core has shrunk to about the size o the Earth it consists o carbon and oxygen ions surrounded by ree electrons. It is prevented rom urther shrinking by an electron degeneracy p ressure. Paulis exclusion principle prevents two electrons rom being in the same quantum state and this means that the electrons provide a repulsive orce that prevents gravity rom urther collapsing the star. The star is let to cool over billions o years as a white dwarf. Such stars are o very high density o about 1 0 9 kg m 3 . Figure 8 shows S irius B , the companion star o S irius A, and the frst white dwar to be identifed.
Figure 8 Sirius A and B.
The probable uture or the Sun is shown as the purple line on the HR diagram fgure 9. 10 6 core star shrinks. ejected gases thin and orm planetary nebula
10 4
yellow giant
10 2
burning helium in core
L star L
core star cools
10 0
10 2
no uel available or burning, star cools
10 4 40 000
star ejects outer layers red supergiant 12.2 billion years helium in core ignites: helium fash red giant
12 billion years burning hydrogen Sun now 4.5 billion years in shell around core burning hydrogen in core
main sequence white dwar
20 000
10 000 5000 surace temperature/K
2500
Figure 9 HertzsprungRussell diagram showing the Suns path.
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B. Larger stars
Figure 10 Supernova 1987A with the right hand image the region of the sky taken just before the event
The core o a star that is much bigger than the S un undergoes a dierent evolutionary path rom that o a S un-like star. When such stars are in the red giant phase, the core is so large that the resulting high temperature causes the usion o nuclei to create elements heavier than carbon. The giant phase ends with the star having layers o elements with proton numbers that decrease rom the core to the outside ( much like layers in an onion see S ub-topic D . 4, fgure 8) . The dense core causes gravitational contraction which, as or lighter stars, is opposed by electron degeneracy pressure. E ven with this pressure, massive stars cannot stabilize. The C handrasekhar limit stipulates that it is impossible or a white dwar to have a mass o more than 1 . 4 times the mass o the Sun. When the mass o the core reaches this value the electrons combine with protons to orm neutrons emitting neutrinos in the process. The star collapses with neutrons coming as close to each other as in a nucleus. The outer layers o the star rush in towards the core but bounce o it in a huge explosion a supernova. This blows o the outer layers and leaves the remnant core as a neutron star. In a quantum mechanical process similar to that o electron degeneracy, the neutrons provide a neutron degeneracy p ressure that resists urther gravitational collapse. The O p p enheimerVolkoff limit places an upper value on a neutron star or which neutron degeneracy is able to resist urther collapse into a black hole. This value is currently estimated at between 1 . 5 and 3 solar masses. life cycle of a low-mass star
protostar
main sequence star
red giant
helium burning star
double-shell burning red giant
planetary nebula
white dwarf
ssupernova pernova
neutron star or black hol hole
life cycle of a high-mass star
protostar
blue main sequence star
red supergiant
helium multiple-shell hell burning burning superupersupergiant gaint
Figure 11 The evolution of Sun-like and more massive stars.
Black holes B lack holes are discussed in more detail in Sub-topic A.5 but here we discuss their importance in astrophysics. It is not possible to orm a neutron star having a mass greater than the OppenheimerVolko limit instead the remnant o a supernova orms a black hole. Nothing can escape rom a black hole including the astest known particles, photons. For this reason it is impossible to see a black hole directly but their existence can be strongly inerred by the ollowing.
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The X- rays emitted by matter spiralling towards the edge o a black hole and heating up. X-ray space telescopes, such as NASAs C handra, have observed such characteristic radiation.
Giant j ets o matter have been observed to be emitted by the cores o some galaxies. It is suggested that only spinning black holes are sufciently powerul to produce such j ets.
The unimaginably strong gravitational felds have been seen to inuence stars in the vicinity, causing them to eectively spiral. A black hole has been detected in the centre o the Milky Way and it has been suggested that there is a black hole at the centre o every galaxy.
Worked example A partially completed HertsprungRussell ( HR) diagram is shown below.
b) State and explain the change in the luminosity o the Sun that occurs between positions S and I. c) E xplain, by reerence to the C handrasekhar limit, why the fnal stage o the evolutionary path o the Sun is at F.
I
d) O n the diagram, draw the evolutionary path o a main sequence star that has a mass o 3 0 solar masses.
luminosity
Solution a) Most o the S uns hydrogen has used into helium. S
b) B oth the luminosity and the surace area increase as the Sun moves rom S to I. c) White dwars are ound in region F o the HR diagram. Main sequence stars that end up with a mass under the C handrasekhar limit o 1 . 4 solar masses will become white dwars.
F
temperature
The line indicates the evolutionary path o the S un rom its present position, S , to its fnal position, F. An intermediate stage in the S uns evolution is labelled by I.
d) The path must start on the main sequence above the Sun. This should lead to the super red giant region above I and either stop there or curve downwards towards and below white dwar in the region between F and S.
a) S tate the condition or the S un to move rom position S .
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D.3 Cosmology Understandings
Applications and skills
The Big Bang model Cosmic microwave background (CMB) radiation Hubbles law
Describing both space and time as originating
The accelerating universe and redshit (Z) The cosmic scale actor (R)
Nature of science Cosmology and particle physics When we look at astronomical objects we see them as they were in the past. The Sun is always viewed as it was 8 minutes earlier and the most distant galaxies appear as they were more than 10 billion ( 10 1 0 ) years ago. The urther away the galaxy, the urther back in time we are looking. Cosmology is a way o studying the history o the universe. Particle physicists study the universe in a diferent way, but they have the same objectives. They use particle accelerators, such as the Large Hadron Collider (LHC) or the Relativistic Heavy Ion Collider (RHIC) , to attempt to recreate events that mimic conditions in the very early universe.
with the Big Bang Describing the characteristics o the CMB radiation Explaining how the CMB radiation is evidence or a hot big bang Solving problems involving z, R, and Hubbles law Estimating the age o the universe by assuming a constant expansion rate
Equations ___vc the redshit equation: z = _______ 0
relation between redshit and cosmic scale actor: R z = _____ -1 R0
Hubbles law: v = H 0 d 1 age o the universe estimate: T ______ H 0
Introduction S ir Isaac Newton believed that the universe was infnite and static. In his model o the universe he argued that the stars would exert equal gravitational attractions on each other in all directions, and this would provide a state o equilibrium. In 1 82 3 , the German astronomer, Heinrich O lbers, suggested that Newtons view o an infnite universe conicted with observation; when we look up into the night sky we see darkness but, in an infnite universe, we should be able to see a star in every direction and, thereore, the night sky should be uniormly bright on a cloudless night. In 1 848, the America author, E dgar Allan Poe, suggested that the universe, although infnite, was simply not old enough to have allowed the light to travel rom the most ar- ung regions to reach us. Although Poe was a naive scientist, he had suggested a theory that was a orerunner o the B ig B ang model. In this sub- topic we will consider how the ( inationary) B ig B ang model has developed into the most probable explanation o the beginning o the universe.
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The redshift and Hubbles law In the 1 92 0s, the American astronomer, Edwin Hubble, was working at the Mount Wilson O bservatory in C aliornia. He set out to fnd an additional way o gauging the distance o remote galaxies that would supplement using C epheid stars. Following on rom the work o others Hubble started to compare the spectra o distant galaxies with their E arth-bound equivalents. He ound that the spectra rom the galaxies invariably appeared to be redshited in line with the D oppler eect. Such consistent results could only mean that all the galaxies were moving away rom the Earth. Figure 1 shows the typical absorption spectra or a range o sources.
Note
In the case of infra-red or
microwaves the radiation already has wavelengths that are longer than red light. For these, the term redshift is ambiguous because redshift changes their wavelengths to even longer wavelengths and not towards the wavelength of red light.
very distant galaxy
distant galaxy
Stars within the Milky Way
and, therefore, relatively close to the Earth might actually be moving towards us and could show a blueshift this is simply a local phenomenon.
nearby galaxy nearby star labratory reference 500
600
700
wavelength/nm
Figure 1 Redshifted absorption spectra. For optical spectra the wavelengths are moved towards the red end o the spectrum. The shit applies to all waves in the spectrum, so the absorption lines in the spectrum can be seen to have shited. In addition to recognizing the consistent redshit, Hubble showed that the urther away the galaxy the greater the redshit. To do this he used the standard candles available to him, C epheid variables. Although his data had large uncertainties ( see fgure 2 ) he suggested that the recessional speed o a galaxy is proportional to its distance rom Earth. Hubbles law is written as
velocity of recession/km s -1
400
1000
500
0 0
1 distance/Mpc
2
Figure 2 Recreation of Hubbles original 1929 data.
where v is the velocity o recession and d is the distance o the galaxy ( both measured rom Earth) , H 0 is the Hubble constant. v is usually being measured in km s 1 and d in Mpc, H 0 is usually measured in km s 1 Mp c 1 . From the gradient o the graph o Hubbles data you will see that it gives a value or H 0 o about 5 00 km s 1 Mpc 1 . Now that we have more reliable data we can see that Hubbles intuition was valid but, by using 1 3 5 5 galaxies, data give a modern value or H 0 that is much closer to 70 km s 1 Mpc 1 . However, it can be seen in fgure 3 that there is still uncertainty in the value o the Hubble constant.
velocity of recession/km s -1
v = H0d 10000
5000
0 0
50 distance/Mpc
100
Figure 3 More recent velocity distance plot.
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Worked example a) The value o the Hubble constant H 0 is accepted by some astronomers to be in the range 60 km s 1 Mpc 1 to 90 km s 1 Mpc 1 . ( i) S tate and explain why it is difcult to determine a precise value o H 0 . ( ii) S tate one reason why it would be desirable to have a precise value o H 0 . b) The line spectrum o the light rom the quasar 3 C 2 73 contains a spectral line o wavelength 75 0 nm. The wavelength o the same line, measured in the laboratory, is 660 nm. Using a value o H 0 equal to 70 km s 1 Mpc 1 , estimate the distance o the quasar rom Earth.
determine how ar away they are. This is because o the difculty o both locating a standard candle, such as fnding a Cepheid variable within the galaxy, and the difculties o accurately measuring its luminosity. ( ii) Having a precise value o H 0 would allow us to gain an accurate value o the rate o expansion o the universe and to determine an accurate value to distant galaxies. It would also allow us to determine a more reliable value or the age o the universe. b) From the D oppler shit equation ( see Topic 9 and later in this topic) ___ _vc 0
Solution a) ( i) The Hubble constant is the constant o proportionality between the recessional velocity o galaxies and their distance rom Earth. The urther galaxies are away (rom Earth) the more difcult it is to accurately
0
TOK
The Big Bang model and the age of the universe
Values for H 0
Hubbles conclusion that the galaxies are moving urther apart provides compelling evidence that they were once much closer together. According to the S tandard Model, about 1 3 . 7 billion years ago the universe occupied a space smaller than the size o an atom. At that instant the entire universe exploded in a B ig B ang, undergoing an immense expansion in which both time and space came into being. 32 S tarting o at a temperature o 1 0 K, the universe rapidly cooled so that one second ater the B ig B ang it had allen to 1 0 1 0 K. In the time since the B ig B ang, the universe has continued to cool to 2 .7 K. In this time there has been an expansion o the abric o space the universe has not been expanding into a vacuum that was already there. As the galaxies move apart the space already between them becomes stretched. This is what we mean by expansion and why we believe the redshit occurs. The space through which the electromagnetic radiation travels is expanding and it stretches out the wavelength o the light. The urther away the source o the light, the greater space becomes stretched, resulting in a more stretched- out wavelength and increasing the cosmological redshift not to be conused with local redshit due to the D oppler eect.
There is still much debate about the value o the Hubble constant. Using better data has brought Hubbles original value down by a actor o almost 10 but the nature o measurement o intergalactic distances means that there are inevitable uncertainties. In December 2012, NASAs Wilkinson Microwave Anisotropy Probe (WMAP) gave a value or H 0 o (69.32 0.80) km s 1 Mpc 1 but, in March 2013, the Planck Mission provided a value o (67.80 0.77) km 1 s 1 Mpc . With two very credible scientifc projects providing values that barely overlap at their extreme values, should we believe scientists claims?
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= 90 1 0 9 m 90 1 0 - 9 v = 3 1 0 8 __ 660 1 0 - 9 7 = 4.1 1 0 m s 1 4 4. 1 1 0 v _ d = __ = = 5 90 Mpc H 70
C osmologists have spent a great deal o time considering what happened between the time that the B ig B ang occurred and the present day. One o the key questions is how did the universe appear in the distant past? Hubbles law certainly suggests that the galaxies were closer together than they are now and, logically, it ollows that there must have been a point in time when they were all in the same place that time being the B ig B ang. It is possible to estimate the age o the universe using Hubbles law.
D . 3 CO SM O LO G Y
Assuming that Hubbles law has held true for all galaxies at all times, the light from the most distant star ( at the edge of the observable universe) has taken the age of the universe to travel to us. If the light was emitted immediately after the time of the B ig B ang, the space between the galaxy and the Earth must have expanded at slightly less than the speed of light for the light to have j ust reached us. This makes the recessional speed of the galaxy ( almost) that of the speed of light, c. The distance that light has travelled from the galaxy = c T where T is the age of the universe v = H0 d c H 0 cT 1 T _ H0 Using a value for H 0 of 70 km s 1 Mpc 1 from the IB Physics data booklet 1 ly = 9.46 1 0 1 5 m and 1 pc = 3 .2 6 ly so 1 Mpc = 1 0 6 9.46 1 0 1 5 3 . 2 6 3 .1 1 0 22 m 3 .1 1 0 22 1 _ = 4.4 1 0 1 7 s = _ H0 70 1 0 3 10 1 . 4 1 0 yr ( or 1 4 billion years) The derivation of the age of the universe equation assumes that the galaxy and the Earth are moving at a relative constant speed of c and that there is nothing in their way to slow them down. In Sub-topic D.5 we consider a range of possibilities for the continued expansion of the universe.
The importance of the cosmic microwave background (CMB) Until the 1 960s there were two competing theories of the origin of the universe. One was the B ig B ang theory and the other was known as the steady state theory. One aspect of the B ig B ang theory is that it suggested a very high temperature early universe that cooled as the universe expanded. In 1 948, Gamow, Alpher, and Herman predicted that the universe should show the spectrum of a black-body emitter at a temperature of about 3 K. 5 In the Big B ang model, at approximately 4 1 0 years after the formation of the universe, the temperature had cooled to about 3000 K and the charged ion matter was able to attract electrons to form neutral atoms. This meant that space had become transparent to electromagnetic radiation, allowing radiation to escape in all directions (previously, when matter was ionic, it had been opaque to radiation) . The expansion of the universe has meant that each of the photons emitted at this time has been shifted to a longer wavelength that now peaks at around 7 cm in the microwave region of the spectrum. At earlier times the photons would have been much more energetic and of far shorter wavelengths, peaking in the visible or ultraviolet region of the electromagnetic spectrum. The C MB in the sky looks essentially the same in all directions ( it is isotropic) and does not vary with the time of day; this provides compelling support for the B ig B ang model. With the discovery of C MB , the advocates of the steady state theory were forced to concede to the strength of evidence. The Wilkinson Microwave Anisotropy
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A S T R O P H YS I C S Probe ( WMAP) was able to show that, on a fner scale, there are some uctuations in the isotropy o the C MB ; this indicates the seeds o galaxies in the early universe.
Figure 4 The CMB sky.
Figure 4 is an image produced by the European Space Agencys Planck mission team showing an all-sky image o the inant universe created rom 1 5 .5 months o data. The radiation is essentially uniorm but shows tiny variations in temperature ( 400 000 M ) . Given such mass, it is no surprise that stars orm in clusters in these regions individual stars have a mass between 0.1 M and 1 5 0 M . In even cooler regions o space, with higher gas densities o 1 0 1 1 particles per cubic metre and temperatures o 1 0 K, the Jeans mass alls to approximately 5 0 M . With these masses, short-lived giant stars are ormed. The atmospheric number density close to the surace o the Earth is o the order o 1 02 5 molecules per cubic metre. B ut, with the high temperature and low mass o gas, you need not worry that a star will orm in the atmosphere!
Nuclear fusion We have seen that the energy generated by a star is the result o thermonuclear usion reactions that take place in the core o the star. The process that occurs during the main sequence is known as hydrogen burning; here hydrogen uses into helium. For Sun-like stars the process advances through the p rotonproton chain but stars o greater than our solar masses undergo a series o reactions known as the C NO cycle.
stage 1
ra ma
p p
stage 3
stage 2 ga m ga m ma
p n p n
y
ra y
ga p p n
a mm
p np
p
total reaction
ra y
p np
p p n pn p
p p p p
gamma ray p n pn
gamma ray
Figure 2 The protonproton chain.
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A S T R O P H YS I C S The protonproton chain has three stages:
1 1
H + 11 H 21 H + 01 e + 00
[two protons ( hydrogen- 1 nuclei) fuse into a hydrogen- 2 nucleus ( plus a positron and a neutrino) ]
1 1
H + 21 H 32 He +
[a third proton fuses with the hydrogen-2 to form a helium-3 nucleus + a gamma-ray photon]
3 2
He + 32 He 42 He + 11 H + 11 H
[two helium- 3 nuclei fuse to produce helium- 4 and two hydrogen- 1 nuclei]
Thus, in order to produce a helium nucleus, four hydrogen nuclei are used in total ( six are used in the fusion reactions and two are generated) . p
12 6
The C NO process occurs in the larger stars with a minimum core temperature of 2 1 0 7 K and involves six stages:
C
1 1
13 7
1 1
H+
13 6
C
14 7
N+
[carbon- 1 3 fuses with proton to give nitrogen- 1 4 + a gamma- ray photon]
1 1
H+
14 7
N
15 8
O+
[nitrogen- 1 4 fuses with proton to give unstable oxygen- 1 5 + a gamma- ray photon]
15 8
1 1
gamma ray 13 7
N
neutrino e+
13 6
p
C
positron gamma ray
14 7
p
N gamma ray
15 8
O
15 7
N
12 6
13 6
N
15 7
O
H+
C
15 7
13 7
N+
C + 01 e + 00
N + 01 e + 00
N
12 6
C + 42 He
neutrino e+
p
H+
positron 4 2
He
[proton fuses with carbon-1 2 to give unstable nitrogen-1 3 + a gamma-ray photon] [nitrogen- 1 3 undergoes positron decay into carbon-1 3 ]
[oxygen- 1 5 undergoes positron decay into nitrogen- 1 5 ] [nitrogen- 1 5 fuses with proton to give carbon-1 2 ( again) and helium- 4]
Again, four protons are used to undergo the fusion process; carbon- 1 2 is both one of the fuels and one of the products. Two positrons, two neutrinos and three gamma- ray photons are also emitted in the overall process. The fusing of hydrogen into helium takes up the maj ority of a stars lifetime and is the reason why there are far more main sequence stars than those in other phases of their life- cycle.
Figure 3 CNO chain.
Fusion after the main sequence Note The CNO cycle is part of the main sequence and no heavy elements are synthesized in this process.
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We saw in Sub-topic D.2 that, once the hydrogen in a stars core is used up, the core converts into helium. The lack of radiation pressure causes the core to shrink and heat up. The hot core instigates the fusion and expansion of the hydrogen surrounding it, causing the red giant phase. Eventually, the cores rising temperature becomes high enough to make the star move off the main sequence. Helium now fuses into the unstable beryllium which then fuses with a further helium nucleus to produce carbon and then oxygen:
4 2
He + 42 He 84B e [two helium nuclei fuse to produce unstable beryllium- 8]
4 2
He + 84B e
4 2
He +
12 6
C
12 6
16 8
C
[a helium nucleus quickly fuses with beryllium- 8 to produce carbon-1 2 ]
O
[a further helium nucleus fuses with carbon- 1 2 to produce oxygen- 1 6]
D . 4 S TE LL AR PRO CE SS E S AH L
You can see rom these reactions that nucleosynthesis (the production o dierent nuclides by the usion o nuclei) is not a simple process. Eventually, both the carbon and oxygen produced will undergo usion and orm nuclei o silicon, magnesium, sodium, and so on until iron-56 is reached. This element represents one o the most stable o all the elements (nickel-62 is the most stable nuclide but is ar less abundant in stars than iron-56) . These nuclides have the highest binding energy per nucleon. Energy cannot be released by urther usions o these elements to produce even heavier nuclides, instead energy will need to be taken in to allow usion to occur. I nickel-62 is the most stable nuclide then how are heavier nuclides made? This is where neutron capture comes in. Neutrons, being uncharged, do not experience electrostatic repulsion and can approach so close to nuclei that the strong nuclear orce is able to capture them. The capture o a neutron increases the nucleon number by one and so does not produce a new element, just a heavier isotope o the original element (isotopes o X in fgure 4) . The newly created isotope will be excited and decay to a less energetic sibling by emitting a gamma-ray photon. The neutron in the newly ormed nucleus might be stable or it could decay into a proton, an electron and antineutrino (by negative beta decay) . This raises the proton number o the nucleus by one and produces the nucleus o a new element (Y in fgure 4) . The new nuclide will initially be excited and releases a gamma-ray photon in decaying into a less energetic nuclide. The hal-lie o beta decay depends solely on the nature o the particular parent nuclide. Whether or not there is sufcient time or a nucleus to capture a urther neutron depends on the density o neutrons bombarding the nuclei. -particle neutron antineutrino
target A ZX
neutron capture
excited nucleus A+1 Z X*
A+1 ZX
beta decay
gamma radiation A+1 Z+1 Y
A+1 Z+1 Y*
gamma radiation
Figure 4 Neutron capture. In a massive star, heavy nuclides ( up to bismuth- 2 09) can be produced by slow neutron capture or the s-p rocess. These stars provide a airly small neutron ux as a by- product o carbon, oxygen, and silicon burning. This means that there is time or nuclides to undergo beta decay beore urther neutron captures build up their nucleon number producing successively heavier isotopes o the original element.
ep Wn
In rapid neutron capture, or the r-process, there is insufcient time or beta decay to occur so successively heavier isotopes are built up very quickly, one neutron at a time. Type II supernovae produce a very high neutron ux and orm nuclides heavier than bismuth-209 in a matter o minutes (and well beore there is any likelihood o beta decay occurring) . This is something
neutrino interacts with neutron p + en+
Figure 5 Neutrino interacts with neutron.
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A S T R O P H YS I C S that does not appear to happen in the massive stars. There is also a high neutrino ux in a supernova and this has the eect o causing neutrons to convert into protons through the weak interaction, orming new elements. This weak interaction is shown in the Feynman diagram, fgure 5.
Worked example a) S uggest why nuclear usion processes inside stars can only synthesize elements with a nucleon number less than 63 . b) O utline how heavier elements could be produced by stars.
Solution a) Energy is released when the binding energy per nucleon increases. The binding energy per nucleon is a maximum at nucleon number 62
(nickel) and so urther usions would require energy to be supplied. This means that, in a star producing heavier elements, usion is no longer energetically avourable. b) With a strong neutron ux, nuclei can absorb neutrons. This can either be a slow process in massive stars, which are capable o producing nuclei no more massive than bismuth- 2 09, or it can be a rapid process in a supernova, that is able to produce still more massive nuclei.
Lifetimes of main sequence stars It might appear that the more matter a star contains the longer it can exist. However, this takes no account o stars having dierent luminosities and, in act, the more massive a star is, the shorter its lietime! Massive stars need higher core temperatures and pressures to prevent them rom collapsing under gravity. This means that usion proceeds at a aster rate than in stars with lower mass. Thereore, massive stars use up their core hydrogen more quickly and spend less time on the main sequence than stars o lower mass. In S ub- topic D .2 we saw that the luminosity o a main sequence star is related to its mass by the relationship: L M 3.5 Luminosity is the total energy E released by the star per unit time while hydrogen is being used or E L=_ t While usion occurs, the energy emitted is accompanied by a loss o mass. This will amount to a proportion o the total star mass during its lietime so a star o mass M loses a mass M. Using Einsteins mass 2 energy relationship E = mc , this makes the energy emitted during the hydrogen burning phase o the stars lie E = Mc 2 . Its average luminosity will be given by: Mc2 L=_ where is the lietime o the star. This can be rearranged to give: Mc 2 =_ L and because L M 3.5
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D . 4 S TE LL AR PRO CE SS E S AH L we can deduce that M _ M 3.5 or M - 2.5 It is useful to compare the mean lifetime of a star with that of the S un and so, using the normal notation for solar quantities, we have:
( )
M _ _ = M
- 2.5
The Sun is expected to have a lifetime of approximately ten billion years ( = 1 0 1 0 years) . A bigger star of, say, 1 0 solar masses would start with ten times the S uns hydrogen but it would have a much shorter lifetime given by
(
1 0M star _ = _ 10 M 10
)
- 2.5
= 3 . 2 1 0 -3 7
This gives star = 3 .2 1 0 years. We can see from this that although it has ten times the Suns mass it only lives for 0.3% of the lifetime of the Sun.
Note You will not be tested on this relationship in your IB Physics examinations, but it is included here to help explain the basic ideas, as these might be tested.
Figure 6 is a HR diagram showing the masses and lifetimes of a number of stars. This HR diagram is for interest only and there is no need to try to learn it; however, the various regions should make sense to you. 10 6 60 m Sun 30 m Sun 10 sol ar r adi i 10 4
lifetime 10 7 yrs 1s ola r ra diu s
10 2 0 .1
L star L 10
sol
ar r
adi
us
0
1 0 2
sol
ar r
adi
10 2 1 0 3
sol
ar r
us
adi
10 4
40 000
10 2
sol
10 3 ar r
adi
i
Deneb Rigal
sol
ar r
adi
i
Supergiants
Spica 10 m Sun
Betelgeuse Canopus Antares Bellatrix 6 m Sun Polaris ma Giants i n s Achernar lifetime e q u Arcturus Aldebaran e n c 3 m Sun 10 8 yrs e Vega Pollux Sirius 1.5 m Sun Procyon lifetime 1 m Sun 10 9 yrs Sun
Sirius b
us
Ceti lifetime 0.3 m Sun 10 10 yrs white lifetime 0.1 m Sun dwarfs 10 11 yrs Barnards star Ross 128 Wolf 359 Proxima centauri Procyon b
20 000
10 000 T/K
Dx cancri
5 000
2500
Figure 6 HR diagram.
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Worked example a) Explain why a star having a mass o 5 0 times the solar mass would be expected to have a lietime o many times less than that o the Sun. b) B y reerring to the massluminosity relationship, suggest why more massive stars will have shorter lietimes.
Solution a) The more massive stars will have much more nuclear material ( initially hydrogen) . Massive stars have greater gravity so equilibrium is reached at a higher temperature at which the outward pressure due to radiation and the hot gas will balance the inward gravitational
pressure. This means that usion proceeds at a aster rate than in stars with lower mass meaning that the nuclear uel becomes used up ar more rapidly. b) As the luminosity o the star is the energy used per second, stars with greater luminosity are at higher temperatures and will use up their uel in shorter periods o time. The luminosity o a star is related to its mass by the relationship L M 3.5 . Thereore, increasing the mass raises the luminosity by a much larger actor which in turn means the temperature is much higher. At the higher temperature the uel will be used in a much shorter time.
Supernovae We now delve into supernovae a little more because o their importance to astrophysics. Supernovae are very rare events in any given galaxy but, because there are billions o galaxies, they are detected quite regularly. They appear as very bright stars in positions that were previously unremarkable in brightness. The advent o high-powered automated space telescopes has meant that astrophysicists are no longer dependent on observing supernovae as random events within or close to the Milky Way. Until recently amateur astronomers had discovered more supernovae than the proessionals. With several automated surveys, such as the C atalina Real-Time Survey, more and more supernovae are being detected so that the number detected in the last ten years is greater than the total detected beore ten years ago. Supernovae are classifed as being Type I or Type II in terms o their absorption spectra (Type I have no hydrogen line but Type II do. This is because Type I are produced by old, low-mass stars and Type II by young, massive stars.) . Type I supernovae are subcategorized as Ia, Ib, Ic, etc. depending on other aspects o their spectra.
Figure 7 An artists impression of the accretion of stellar matter leading to a Type Ia supernova. The white dwarf is the star on the right, while the left-hand star is a red giant.
Type Ia supernovae result rom accretion o matter between two stars in a binary star. O ne o the stars is a white dwar and the other is either a giant star or a smaller white dwar. The ormation o these supernovae show up as a rapid increase in brightness ollowed by a gradual tapering o. Type II supernovae have been discussed briey in Sub- topic D .2 and consist o single massive stars in the fnal stages o their evolution. These classes o supernovae produce light curves with dierent characteristics as shown in fgure 9.
Type Ia supernovae These are very useul to astrophysicists as they always emit light in a predictable way and behave as a standard candle or measuring the distance o the galaxy in which the supernova occurs. Given the immense density o the material within a white dwar, the gravitational feld is unimaginably strong and attracts matter rom the companion star. When the mass o the
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growing white dwar exceeds the Chandrasekhar limit o 1 .4 solar masses, the star collapses under gravity. The usion o carbon and oxygen into nickel generates such radiation pressure that the star is blown apart, reaching a luminosity o 1 0 1 0 times that o the Sun. Because the chain reaction always occurs at this mass we know how bright the supernova actually is and, by comparing it with the apparent brightness observed on Earth, we can estimate the distance o the supernovas galaxy up to distances o 1 000 Mpc. Ater the explosion the ejected material continues to expand in a shell around the remnant or thousands o years until it mixes with the interstellar material giving the potential to orm a new generation o stars.
Type II supernovae Ater approximately 1 0 million years (or stars o 81 0 solar masses) all the hydrogen in the core has converted into helium and hydrogen usion can now only continue in a shell around the helium core. The core undergoes gravitational collapse until its temperature is high enough or usion o helium into carbon and oxygen. This phase lasts or about a million years until the cores helium is exhausted; it will then contract again under gravity, causing it to heat up and allowing the usion o carbon into heavier elements. It takes about 1 0 thousand years until the carbon is exhausted. This pattern continues, with each heavier element lasting or successively shorter lengths o time, until silicon is used into iron-56 taking a ew days (see fgure 8) . At this point the star is not in hydrostatic equilibrium because there is now little radiation pressure to oppose gravity. On reaching the Chandrasekhar limit o 1 .4 M , electron degeneracy pressure is insufcient to oppose the collapse and the star implodes producing neutrons and neutrinos. The implosion is opposed by a neutron degeneracy pressure that causes an outward shock wave. This passes through the outer layers o the star causing usion reactions to occur. Although this process lasts just a ew hours it results in the heavy elements being ormed. As the shock wave reaches the edge o the star, the temperature rises almost instantly to 20 000 K and the star explodes, blowing material o as a supernova. non-burning hydrogen
10 9
hydrogen fusion helium fusion carbon fusion
relative 10 8 luminosity (L/L ) 10 7
Type Ia Type II
oxygen fusion neon fusion magnesium fusion
10 6
0 100 200 300 days after maximum brightness
Figure 9 Light curves for class Ia and II supernovae.
silicon fusion iron ash
Figure 8 The onion model of a massive star before it goes supernova.
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A S T R O P H YS I C S Type Ia and Type II supernovae can be distinguished by observers on Earth rom the manner in which the stars emit light. The Type Ia emits light up to 1 0 1 0 times the luminosity o the Sun; this rapidly reaches a maximum and then gradually tails o over six months or so. The Type II emits light up to 1 0 9 times the luminosity o the Sun; however, the burst alls a little beore reaching a slight plateau where it stays or some days beore alling away more rapidly. The light curves or these supernovae are shown in fgure 9.
Worked example a) Outline the dierence between a Type Ia and a Type II supernova. b)
( i) What is meant by a standard candle? ( ii) Explain how a Type Ia supernova can be used as a standard candle.
Solution a) A Type Ia supernova results rom a white dwar in a binary star system accreting material rom its companion giant or smaller white dwar star. Eventually, when the mass o the white dwar passes the Chandrasekhar limit, the star collapses and is blown apart as a supernova by the huge temperature generated. Type II supernovae occur in stars having masses between 8 and 50 times the mass o the Sun. When the core has changed into inert iron and nickel, with no urther usion occurring, gravity collapses the core. Again, on reaching the Chandrasekhar limit o 1 .4 M , electron degeneracy pressure is insufcient to oppose the collapse and the star implodes producing neutrons and neutrinos. The implosion is opposed by a neutron degeneracy pressure that causes an outward shock wave blowing away the surrounding material as a supernova. b)
( i) A standard candle is a star o known luminosity that, when compared with its apparent brightness, can be used to calculate its distance. ( ii) As all Type Ia supernova occur when the mass reaches the C handrasekhar mass they are all ( essentially) o the same peak luminosity. B y measuring the apparent brightness, the distance o the supernova ( and the galaxy in which it is a L member) can be calculated (rom b = ____ ). 4d2
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D.5 Further cosmology AHL Understanding The cosmological principle Rotation curves and the mass of galaxies Dark matter
Applications and skills Describing the cosmological principle and its
Fluctuations in the CMB The cosmological origin of redshift
Critical density Dark energy
Nature of science How constant is the Hubble constant? ...or when is a constant not a constant? The Hubble constant H0 is a quantity that we have used in Subtopic D.3 to estimate the age of the universe. This is a very important cosmological quantity, indicating the rate of expansion of the universe. The current value is thought to be around 70 km s 1 Mpc 1 , but it has not always been this value and will not be in the future. So it is a constant in space but not in time and it would be more appropriately named the Hubble parameter. The zero subscript is used to indicate that we are talking about the present value of the constant in general use we should omit the subscript. In the IB Physics course, H0 is used to indicate all values of the Hubble constant.
role in models of the universe Describing rotation curves as evidence for dark matter Deriving rotational velocity from Newtonian gravitation Describing and interpreting the observed anisotropies in the CMB Deriving critical density from Newtonian gravitation Sketching and interpreting graphs showing the variation of the cosmic scale factor with time Describing qualitatively the cosmic scale factor in models with and without dark energy
Equations velocity of rotating galaxies: v =
____
4G r _________ 3
3H2 critical density of universe: c = _______ 8G
Introduction In Sub-topic D .3 we saw that the B ig B ang should correspond to the simultaneous appearance o space and time (spacetime) . Hubbles law and the expansion o the universe tell us that the observable universe is certainly larger than it was in the past and that it can be traced back to something smaller than an atom, containing all the matter and energy currently in the universe. There is no special place in the universe that would be considered to be the source o the B ig B ang, it expanded everywhere in an identical manner. It is not possible to use the B ig B ang model to speculate about what is beyond the observable universe it should not be thought o as expanding into some sort o vacuous void. In this sub-topic we will consider ways in which the universe might continue to expand and look at possible models or fat, open, and closed universes.
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Nature of science Philosophy or cosmology? Although there is substantial evidence leading us to believe in the B ig B ang model nobody actually knows what instigated the B ig B ang. We have explained that this was the beginning o space and time and so we cannot ask what happened beore the B ig B ang? There are a
number o theories that reect on why the B ig B ang occurred such as uctuations in gravity or quantum uctuations but these theories stimulate other questions such as what caused this? As o yet these theories cannot be put to the test.
The Cosmological Principle Buoyed up by the success o his general theory o relativity in 1 91 5, Einstein sought to extend this theory to explain the dynamics o the universe or cosmos. In order to make headway, because he recognized that this would be a complex matter, he made two simpliying assumptions these have subsequently been shown to be essentially true on a large scale:
the universe is homogenous
the universe is isotropic
The frst o these requirements simply says that the universe is the same everywhere which, when we ignore the lumpiness o galaxies, it is. The second requirement is that the universe looks the same in all directions. Although this may not seem too dierent rom the homogeneity idea it actually is! It really says that we are not in any special place in the universe and ties in with the theory o relativity that says there is no special universal reerence rame. Imagine that we are positioned towards the edge o a closed universe and we look outward there would be a limited number o galaxies to send photons to us. I we look inward there would be an immense number o galaxies to send us photons. The two situations would appear very dierent. The isotropicity says that this isnt the case. Jointly, these two prerequisites o Einsteins theory are known as the Cosmological Principle. These assumptions underpin the Big Bang cosmology and lead to specifc predictions or observable properties o the universe. Figure 1 shows an image produced by the Automated Plate Measurement ( APM) Galaxy survey o around 3 million galaxies in the S outhern Hemisphere sky. The image shows short- range patterns but, in line with the cosmological principal, on a large scale the image shows no special region or place that is dierent rom any other. Using the cosmological principle and the general theory o relativity it can be shown that matter can only distort spacetime in one o three ways. This is conventionally shown diagrammatically by visualizing the impact o the third dimension on a at surace; however, the ourth dimension o time is also involved and this makes visualization even more complex.
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Figure 1 APM Galaxy survey image.
The at surace can be positively curved into a spherical shape o a fnite size. This means that, by travelling around the surace o the sphere, you could return to your original position or, by travelling through the universe, you could return to your original position in spacetime.
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The at surace can be negatively curved like the shape o a saddle and have an infnite size. In this universe you would never return to the same point in spacetime.
The surace could also remain at and infnite as given in our everyday experiences. Again, you would never return to the same position in spacetime.
Knowledge o the amount o matter within the universe is essential when determining which model is applicable. There is a critical density ( c) o matter that would keep the universe at and infnite this density would provide a gravitational orce large enough to prevent the universe running away but j ust too little to pull it back to its initial state. With less than the critical density the universe would be open and infnite. With greater density than the critical value the universe would be closed and fnite with gravity pulling all matter back to the initial state o spacetime. The critical density o matter appears to be no greater than ten particles per cubic metre and current research suggests that the average density is very close to this critical value.
Figure 2 The visualizations o closed, open, and fat universes.
The implications of the density of intergalactic matter As can be seen in fgure 5 in Sub-topic D .3, theory suggests that, ater the initial inationary period ollowing the B ig B ang, the rate o expansion o the universe has been slowing down. Instrumental to the ate o the universe is the uncertainty about how much matter is available to provide a strong enough gravitational orce to reverse the expansion and cause a gravitational collapse. As discussed in Sub-topic D .3 , data rom Type Ia supernovae has suggested that the universe may actually be undergoing an accelerated expansion caused by mysterious dark energy. We can derive a relationship or the critical density using Newtonian mechanics: Imagine a homogenous sphere o gas o radius r and density . A galaxy o mass m at the surace o the sphere will be moving with a recessional speed v away rom the centre o the sphere along a radius as shown in fgure 3. B y Hubbles law the velocity o the galaxy is given by: v = H0 r The total energy o the galaxy is the sum o its kinetic energy and its gravitational potential energy ( relative to the centre o mass o the sphere o gas) . ET = E K + E P Mm 1 ET = __ mv2 - G _ 2 r
r
v
m
remembering that potential energy is always negative or obj ects separated by less than infnity. The mass M is that o the sphere o gas is given by 4 M = _ r3 3 4 __ r3 m 3 _ ET = 1 m( H0 r) 2 - G _ r 2
Figure 3 Critical density or the universe.
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A S T R O P H YS I C S The galaxy will continue to move providing that it has sufcient kinetic energy, thus making ET positive. In the limit ET = 0 this gives
Note
This equation contains
4 __ r3 cm 3 1 _ m( H0 r) 2 = G _ r 2
nothing but constants. Any value for the critical density of the material of the universe is dependent on how precisely the Hubble constant can be determined.
where c is the critical density o matter. S impliying this equation gives 3 H0 2 c = _ 8G
A more rigorous derivation
of this equation requires the use of general relativity.
The cosmic scale factor and time The ratio o the actual density o matter in the universe ( ) to the critical density is called the density parameter ( indicated in fgure 2 ) and is given the symbol 0 . 0 = _ c There are three possibilities ( shown in fgure 4) or the ate o the universe, depending on the density parameter o the universe: 1
I 0 = 1 ( or = c) the density must equal the critical density and must be the value or a at universe in which there is j ust enough matter or the universe to continue to expand to a maximum limit. However, the rate o expansion would decrease with time. This is thought to be the least likely option.
2
I 0 < 1 ( or < c) the universe would be open and would continue to expand orever.
3
I 0 > 1 ( or > c) then the universe would be closed. It would eventually stop expanding and would then collapse and end with a B ig C runch.
An accelerated expansion o the universe ( shown by the red line on fgure 4) might be explained by the presence o dark energy. This oers an interesting and, increasingly likely, prospect.
cosmic scale actor (R)
4
In S ub- topic D .3 we considered the cosmic scale actor ( R) . This is essentially the relative size, or radius, o the universe. Figure 4 shows how R varies with time or the dierent density parameters. Each o the models gives an 0 value that is based on the total matter in the universe. An explanation or the accelerated universe depends on the concept o the ( currently) hypothetical dark energy outweighing the gravitational eects o baryonic and dark matter.
accelerated universe
3
0< 1 0 =1
2 1
0> 1 0
-10
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now 10 time/10 9 years
20
30
Figure 4 Variation o R or diferent density parameters.
The cosmic scale factor and temperature The wavelength o the radiation emitted by a galaxy will always be in line with the cosmic scale actor (R) . So, as space expands, the wavelength will expand with it. We know rom Wiens law that the product o the maximum intensity wavelength and the temperature is a constant. Assuming that the spectrum o a black body retains its shape during the expansion this means that Wiens law has been valid rom the earliest
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times ollowing the B ig B ang. Thus, the wavelength and the cosmic scale actor are both inversely proportional to the absolute temperature. 1 1) T _ ( and T _ R
Worked example The diagram below shows the variation o the cosmic scale actor R o the universe with time t. The diagram is based on a closed model o the universe. The point t = T is the present time.
b)
R
R
T
T
t
a) Explain what is meant by a closed universe. b) O n a copy o the diagram, draw the variation based on an open model o the universe. c) Explain, by reerence to your answer to b) , why the predicted age o the universe depends upon the model o the universe chosen. d) ( i) What evidence suggests that the expansion o the universe is accelerating? ( ii) What is believed to be the cause o the acceleration?
Solution a) A closed universe is one that will stop expanding at some uture time. It will then start to contract due to gravity.
t
The graph should start at an early time ( indicating an older universe) and touch the closed universe line at T. It should show curvature but not fatten out as a fat universe would do. c) We only know the data or the present time so all curves will cross at T. B y tracing the curve back to the time axis, we obtain the time or the B ig B ang. This extrapolation will give a dierent time or the dierent models. d) ( i) The redshit rom distant type Ia supernovae has suggested that the expansion o the universe is now accelerating. ( ii) The cause o this is thought to be dark energy something o unknown mechanism but opposing the gravitational attraction o matter ( both dark and baryonic) .
Evidence for dark matter Let us imagine a star o mass m near the centre o a spiral galaxy o total mass M. In this region the average density o matter is . The star moves in a circular orbit with an orbital velocity v and radius r. B y equating Newtons law o gravitation to the centripetal orce we obtain Mm mv2 G_ = _ 2 r r C ancelling m and r M 2 G_ r = v
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A S T R O P H YS I C S In terms o the density and taking the central hub to be spherical, this gives 4 __ r3 3 _ G r = v2
_____
This means that v =
G4 r _ 3
or
v = constant r
From this we can see that the velocity is directly proportional to the radius.
What i the star is in one o the less densely populated arms o the galaxy? In this case we would expect the star to behave in a similar manner to the way in which planets rotate about the S un. The galaxy would behave as i its total mass was concentrated at its centre; the stars would be ree to move with nothing to impede their orbits. This gives
Figure 5 The spiral galaxy M81.
M 2 G_ r = v and so 1_ v _ r When the rotational velocity is plotted against the distance rom the centre o the galaxy, we would expect to see a rapidly increasing linear section that changes to a decaying line at the edge o the hub. This is shown by the broken line in fgure 6. What is actually measured ( by measuring the speed rom the redshits o the rotating stars) is the upper observed line. This is surprising because this at rotation curve shows that the speed o stars, ar out into the region beyond the arms o the galaxy, are moving with essentially the same speed as those well inside the galaxy. O ne explanation or this eect is the presence o dark matter orming a halo around the outer rim o the galaxy ( as shown in fgure 6) . This matter is not normal luminous or baryonic matter and emits no radiation and, thereore, its presence cannot be detected. v/km s -1 observed 100
expected from luminous disc
50
dark matter halo
5 luminous matter
Figure 6 Dark matter halo surrounding a galaxy.
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Figure 7 The rotation curve for the spiral galaxy M33.
10
r/kpc
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In fgure 8, the experimental curve has been modelled by assuming that the halo adds sufcient mass to that o the galactic disc. This maintains the high rotational speeds well away rom the galactic centre. 200
Vcir /km s -1
150 halo 100
50 disc
0
0
10
20
30
40
60
radius/kpc
Figure 8 distribution of dark matter in NGC 3198.
O ther evidence or the presence o dark matter comes rom:
the velocities o galaxies orbiting each other in clusters these galaxies emit ar less light than they ought to in relation to the amount o mass suggested by their velocities
the gravitational lensing eect o radiation rom distant obj ects ( such as quasars) because the radiation passes through a cluster o galaxies it becomes much more distorted than would be expected by the luminous mass o the cluster
the X-ray images o elliptical galaxies show the presence o haloes o hot gas extending well outside the galaxy. For this gas to be bound to the galaxy, the galaxy must have a mass ar greater than that observed up to 90% o the total mass o these galaxies is likely to be dark matter.
At the moment no one knows the nature o dark matter but there are some candidates:
MAC HO s are MAssive C ompact Halo O bj ects that include black holes, neutron stars, and small stars such as brown dwars. These are all high density ( compact) stars at the end o their lives and might be hidden by being a long way rom any luminous obj ects. They are detected by gravitational lensing, but it is questionable whether or not there are sufcient numbers o MAC HO s to be able to provide the amount o dark matter thought to be in the universe.
WIMPs are Weakly Interacting Massive Particles subatomic particles that are not made up o ordinary matter (they are non-baryonic) . They are weakly interacting because they pass through ordinary, baryonic, matter with very little eect. Massive does not mean big, it means that these particle have mass (albeit very small mass) . To produce the amount o mass needed to make up the dark matter there would need to be
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A S T R O P H YS I C S unimaginably large quantities o WIMPs. In 1 998, neutrinos with very small mass were discovered and these are possible candidates or dark matter; other than this the theory depends on hypothetical particles called axions and neutralinos that are yet to be discovered experimentally.
Dark energy In 1 998, observations by the Hubble S pace Telescope ( HS T) o a very distant supernova showed that the universe was expanding more slowly than it is today. Although nobody has a defnitive explanation o this phenomenon its explanation is called dark energy. ES As Planck mission has provided data that suggest around 68% o the universe consists o dark energy ( while 2 7% is dark matter leaving only 5 % as normal baryonic matter see fgure 9) . dark matter 22.7%
dark matter 26.8%
ordinary matter 4.5%
ordinary matter 4.9%
dark energy 72.8%
dark energy 68.3%
before Planck
after Planck
Figure 9 The mass/energy recipe for the universe.
It has been suggested that dark energy is a property o space and so, with the expansion o the universe as space expands, so too does the amount o dark energy, i.e. more dark energy coming into existence along with more space. This orm o energy would subsequently cause the expansion o the universe to accelerate. Nobody knows i this model is viable. It is possible that an explanation or the accelerated expansion o the universe requires a new theory o gravity or a modifcation o Einsteins theory such a theory would still need to be able to account or all the phenomena that are, at the moment, correctly predicted by the current theory.
Nature of science The Dark Energy Survey (DES) With dark energy being one o the most upto- the- minute research topics in the whole o science, it is unsurprising that international collaboration is prolifc. The D E S is a survey with the aim o gaining the best possible data or the rate o expansion o the universe. This is being carried out by observing around 3 0 0 0 distant supernovae, the most distant o which exploded when the universe was about hal
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its current size. Using the Victor M. B lanco Telescope at C erro Tololo Inter- American O bservatory ( C TIO ) in C hile, 1 2 0 collaborators rom 2 3 institutions in 5 countries are using a specially developed camera to obtain images in the near inra- red part o the visible spectrum. The survey will take fve years to complete and will add to the sky- based research o missions such as WMAP and Planck.
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Anisotropies in the CMB In S ub- topic D . 3 we discussed the importance o the cosmic microwave background with regard to the B ig B ang. In this model, the universe came into being almost 1 3 . 8 billion years ago when its density and temperature were both very high ( this is oten reerred to as the hot Big Bang) . S ince then the universe has both expanded and cooled. The Planck satellite image in S ub- topic D . 3 , fgure 4 showed that, although the C MB is essentially isotropic, there are minute temperature uctuations; these variations are called anisotrop ies. In the early 1 9 9 0s, the C osmic B ackground E xplorer ( C O B E ) satellite provided the frst evidence o these anisotropies but, with the launch o NAS As WMAP in 2 001 and the E S As Planck satellite in 2 009 , the resolution has been improved dramatically. B oth these missions have shown signifcant low- level temperature uctuations. It is thought that these uctuations appear as the result o tiny, random variations in density, implanted during cosmic ination the period o accelerated expansion that occurred immediately ater the B ig B ang. When the universe was 3 8 0 000 years old and became transparent, the radiation emitted rom the B ig B ang was released and it has travelled outwards through space and time, including towards the E arth. This radiation was in the red part o the electromagnetic spectrum when it was released, but its wavelength has now been stretched with the expansion o the universe so that it corresponds to microwave radiation. The pattern shown in the variation demonstrates the dierences present on the release o the radiation: uctuations that would later grow into galaxies and galaxy clusters under the inuence o gravity. The inormation extracted rom the Planck sky map shows isotropy on a large scale but with a lack o symmetry in the average temperatures in opposite hemispheres o the sky. The S tandard Model suggests that the universe should be isotropic but, given these dierences, this appears not to be the case. There is also a cold spot ( circled in S ub- topic D . 3 , fgure 4) extending over a patch o sky and this is larger than WMAP had previously shown. The concepts o dark matter and dark energy have already been added to the S tandard Model as additional parameters. The evidence rom the C MB anisotropies may require urther tweaks to the theory or even a maj or re- think because o this. The Planck data also identiy the Hubble constant to be 67. 1 5 km s 1 Mpc 1 ( signifcantly less than the current standard value in astronomy o around 1 00 km s 1 Mpc 1 ) . The data imply that the age o the universe is 1 3 . 82 billion years. O ver the next ew years this value may well be modifed because o additional data being collected by this mission and rom orbiting telescopes including the James Webb space telescope ( fgure 1 0) and the j oint NAS A/E S A E uclid mission. There has, arguably, never been a more productive time in the history o cosmology.
Figure 10 The James Webb space telescope artists impression.
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A S T R O P H YS I C S
Questions 1
b) O utline why:
( IB)
( i) white dwar stars cannot have a greater mass than 1 .4 M
a) The star Wol 3 5 9 has a parallax angle o 0.41 9 arcsecond. ( i)
( ii) it is possible or a main sequence star with a mass equal to 8 M to evolve into a white dwar. ( 6 marks)
D escribe how this parallax angle is measured.
( ii) C alculate the distance in light- year rom Earth to Wol 3 5 9. ( iii) S tate why the method o parallax can only be used or stars at a distance less than a ew hundred parsecond rom E arth. b) The ratio ap p are nt b righ tne ss o Wo l 3 5 9 ______________________ is 3 .7 1 0 1 5 . ap p are nt b rightn e ss o the S u n
4
(IB) a)
D efne luminosity.
b) The sketch- graph below shows the intensity spectrum or a black body at a temperature o 6000 K.
( 1 1 marks)
2
The average intensity o the Suns radiation at the surace o the E arth is 1 .3 7 1 0 3 W m - 2 . C alculate ( a) the luminosity and ( b) the surace temperature o the S un. The mean separation o the Earth and the S un = 1 .5 0 1 0 1 1 m, radius o the S un = 6.96 1 0 8 m, S teanB oltzmann constant = 5 .67 1 0 - 8 W m 2 K 4. ( 4 marks)
3
(IB) The diagram below is a ow chart that shows the stages o evolution o a main sequence star such as the Sun. ( Mass o the S un, the solar mass = M )
main sequence star mass M
red giant nebula
planetary nebula
white dwar
a) C opy nad complete the boxes below to show the stages o evolution o a main sequence star that has a mass greater than 8 M . main sequence star mass > 8M
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0 0
wavelength
O n a copy o the axes, draw a sketch-graph showing the intensity spectrum or a black body at 8000 K. c) A sketch o a HertzsprungRussell diagram is shown below.
luminosity
lu m ino sity o Wo l 3 5 9 ________________ is 8.9 1 0 4. lu m ino sity o the S u n
intensity
Show that the ratio
temperature
C opy the diagram above and identiy the: ( i)
main sequence ( label this M)
( ii) red giant region ( label this R) ( iii) white dwar region ( label this W) . d) In a HertzsprungRussell diagram, luminosity is plotted against temperature. Explain why the diagram alone does not enable the luminosity o a particular star to be determined rom its temperature. ( 8 marks)
QUESTION S 5
7
(IB)
a) In an observation of a distant galaxy, spectral lines are recorded. Spectral lines at these wavelengths cannot be produced in the laboratory. Explain this phenomenon.
The diagram below shows the grid of a HertzsprungRussell ( HR) diagram on which the positions of the S un and four other stars A, B , C and D are shown. 10 6
A
b) Describe how Hubbles law is used to determine the distance from the Earth to distant galaxies.
B
10 4
c) Explain why Hubbles law is not used to measure distances to nearby stars or nearby galaxies (such as Andromeda) . (6 marks)
luminosity (L) (Sun L = 1)
10 2 Sun
1
8
10 -2
C D
00 40 0 30 0 00
00
a) Describe any differences between this galaxy and the Milky Way.
50
60
0
00
80
00 10
00
0
10 -6
surface temperature (T/K)
Hubbles law predicts that NGC 5 1 2 8 is moving away from Earth.
a) Name the type of stars shown by A, B, C, and D.
b) (i) State Hubbles law.
b) Explain, using information from the HR diagram and without making any calculations, how astronomers can deduce that star B is larger than star A.
(ii) State and explain what experimental measurements need to be taken in order to determine the Hubble constant. c) A possible value for the Hubble constant is 68 km s 1 Mpc 1 . Use this value to estimate:
c) Using the following data and information from the HR diagram, show that star B is at a distance of about 700 pc from Earth.
(i) the recession speed of NGC51 28 (ii) the age of the universe.
Apparent brightness of the S un = 1 .4 1 0 3 W m 2 9
Apparent brightness of star B = 7.0 1 0 8 W m 2
(IB) O ne of the most intense radio sources is the Galaxy NGC 5 1 2 8. Long exposure photographs show it to be a giant elliptical galaxy crossed by a band of dark dust. It lies about 1 .5 1 0 7 light years away from Earth.
10 -4
25
(IB)
(1 0 marks)
a) D escribe what is meant by a nebula. b) E xplain how the Jeans criterion applies to star formation. (3 marks)
Mean distance of the S un from Earth = 1 .0 AU 1 parsec = 2 .1 1 0 5 AU
6
1 0 O utline how hydrogen is fused into helium in: ( 1 1 marks)
(IB)
a) stars of mass similar to that of the S un b) stars of mass greater than ten solar masses. (6 marks)
a) State what is meant by cosmic microwave background radiation. b) D escribe how the cosmic microwave background radiation provides evidence for the expanding universe. ( 5 marks)
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A S T R O P H YS I C S 1 1 a) ( i)
Explain what is meant by neutron capture.
( ii) Write a nuclear equation to show nuclide A capturing a neutron to become nuclide B . b) O utline the dierence between s and r processes in nucleosynthesis. ( 8 marks)
1 2 Explain why the lietimes o more massive main sequence stars are shorter than those o less massive ones. ( 4 marks)
a) D escribe the observational evidence in support o an expanding universe. b) Explain what is meant by the term critical density o the universe. c) D iscuss the signifcance o comparing the density o the universe to the critical density when determining the uture o the universe. ( 6 marks)
1 6 (IB)
1 3 B riey explain the roles o electron degeneracy, neutron degeneracy and the C handrasekhar limit in the evolution o a star that goes supernova. ( 6 marks)
a) Recent measurements suggest that the mass density o the universe is likely to be less than the critical density. S tate what this observation implies or the evolution o the universe in the context o the B ig B ang model.
1 4 (IB)
b) ( i)
a) E xplain the signifcance o the critical density o matter in the universe with respect to the possible ate o the universe. The critical density c o matter in the universe is given by the expression: 3 H0 2 c = _ 8G where H0 is the Hubble constant and G is the gravitational constant. An estimate o H0 is 2 .7 1 0 1 8 s 1 . b) ( i) C alculate a value or c. ( ii) Using your value or H0 , determine the equivalent number o nucleons per unit volume at this critical density. ( 5 marks)
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1 5 (IB)
O utline what is meant by dark matter.
( ii) Give two possible examples o dark matter. ( 5 marks)
I N TERN AL ASSESSM EN T Introduction In this chapter you will discover the important role of experimental work in physics. It guides you through the expectations and requirements of an independent investigation called the
internal assessment ( IA) . The investigation itself is likely to occur late in your second year, so you do not need to read this chapter until your teacher advises you to.
Advice on the internal assessment Understanding theory and experiment
Applications and skills to appreciate the interrelationship o theory
internal assessment requirements internal assessment guidance
internal assessment criteria
Nature of science Empirical evidence is a key to objectivity in science. Evidence is obtained by observation, and the details o observation are embedded in experimental work. Theory and experiment are two sides o the same coin o scientifc knowledge.
and experiment ability to plan your internal assessment understand teacher guidance appreciate the ormal requirements o an internal assessment to be critically aware o academic honesty
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I N T E R N AL A S S E S S M E N T
Theory and experiment The sciences use a wide variety o methodologies and there is no single agreed scientifc method. However, all sciences are based on evidence obtained by experiment. E vidence is used to develop theories, which then orm laws. Theories and laws are used to make predictions that can be tested in experiments. Science moves in a cycle that moves between theory and experiment. O bservations inorm theory. However, the refnement o a theory and improvements in instrumentation re-ocus on more observation. E xperimentation allows us to have confdence that a theory is not merely pure speculation. C onsider a amous analogy used by Albert Einstein o a man trying to understand the mechanism o a pocket watch. In the ollowing quote, Einstein illustrates that our scientifc knowledge can be tested against reality. He shows that we can confrm or deny a theory by experiment, but we can never know reality itsel. There is a continual dance between theory and experiment. Physical concepts are ree creations o the human mind, and are not, however it may seem, uniquely determined by the external world. In our endeavor to understand reality we are somewhat like a man trying to understand the mechanism o a closed watch. He sees the ace and the moving hands, even hears it ticking, but he has no way o opening the case i he is ingenious he may orm some picture o the mechanism which could be responsible or all the things he observes, but he may never be quite sure his picture is the only one which could explain his observations. He will never be able to compare his picture with the real mechanism and he cannot even imagine that possibility o the meaning o such a comparison. B ut he certainly believes that, as his knowledge increases, his picture o reality will become simpler and simpler and will explain a wider and wider range o his sensuous impressions. He may also believe in the existence o the ideal limit o knowledge and that it is approached by the human mind. He may call this ideal limit the objective truth. Albert Einstein and Leopold Infeld, The Evolution of Physics.
The internal assessment requirements E xperimental work is not only an essential part o the dynamic o scientifc knowledge it also plays a key role in the teaching and learning o physics. Experimental work should be an integral and regular part o your physics lessons consisting o demonstrations, hands- on group work, and individual investigations. It may include computer simulations, mathematical models, and online databases resources. It is only natural then that time should be allocated to you in order to ormulate, design, and implement your own physics experiment. You will produce a single investigation that is called an internal assessment. This means that your teacher will assess your report using IB criteria, and the IB will externally moderate your teachers assessment. Your investigation will consist o:
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selecting an appropriate topic
researching the scientifc content o your topic
defning a workable research question
adapting or designing a methodology
obtaining, processing, and analysing data
appreciating errors, uncertainties, and limits o data
writing a scientifc report 61 2 A4 pages long
receiving continued guidance rom your teacher.
AD VI CE O N TH E I N TE R N AL ASS E SS M E N T
Planning and guidance Ater your teacher introduces the idea o an internal assessment investigation, you will have an opportunity to discuss your investigation topic with your teacher. Through dialogue with your teacher you can select an appropriate topic, defne a workable research question, and begin by doing research into what is already known about your topic. You will not be penalized or seeking your teachers advice. It is your teachers responsibility to provide you with a clear understanding o the IA expectations, rules, and requirements. Your teacher will:
provide you with continued guidance at all stages o your work
help you ocus on a topic, then a research question, and then an appropriate methodology
provide quidance as you work and read a drat o your report, making general suggestions or improvements or completeness.
Your teacher will not, however, edit your report nor give you a tentative grade or achieving level or your report until it is fnally completed. Once your report is completed and ormally submitted you are not allowed to make any changes. Your teacher is responsible or your guidance, making sure you understand the IA expectations, and that your work is your own. As the student it is your resp onsibility to appreciate the meaning o academic honesty, especially authenticity and the respect o intellectual property. You are also responsible or initiating your research question with the teacher, seeking help when in doubt, and demonstrating independence o thought and initiative in the design and implementation o your investigation. You are also responsible or meeting the deadlines set by your teacher.
The internal assessment report There is no prescribed ormat or your investigation report. However, the IA criteria encourage a logical and j ustifed approach, one that demonstrates personal involvement and exhibits sound scientifc work. The style and orm o your report or the IA investigation should model a scientifc journal article. You should be amiliar with a number o high school level physics journal articles. For example, journals like the B ritish Physics Education publication ( http://iopscience.iop.org) or the American The Physics Teacher publication (http:// tpt.aapt.org) oten have articles that are appropriate or high school work. Moreover, many o these articles can provide good ideas or an investigation. There is no prescribed narrative mode, and your teacher will direct you to the style that they wish you to use. However, because a report describes what you have done, it is reasonable to write in the past tense. D escriptions are always clearer to understand i you avoid the use o pronouns (usually ' it' ) and reer specifcally to the relevant noun ( ' the wire' , ' the ammeter' , ' a digital caliper' , etc.) .
Academic honesty The IB learner profle ( page iii) describes IB students as aspiring to develop many qualities, including that o being principled. This means that you act with integrity and honesty, with a strong sense o airness and j ustice, and that you take responsibility or your actions and their consequences. The IA is your responsibility, and it is your work. Plagiarism and copying others work is not permissible. You must
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I N T E R N AL A S S E S S M E N T clearly distinguish between your own words and thoughts and those o others by the use o quotation marks ( or other methods like indentation) ollowed by an appropriate citation that denotes an entry in the bibliography. Although the IB does not prescribe reerencing style or in- text citation, certain styles may prove most commonly used; you are ree to choose a style that is appropriate. It is expected that the minimum inormation included is: name o author, date o publication, title o source, and page numbers as applicable.
Types of investigations Ater you have covered a number o physics syllabus topics and perormed a number o hands-on experiments in class, you will be required to research, design, perorm, and write up your own investigation. This project, known as an internal assessment, will count or 2 0% o your grade, so it is important that you make the most o your opportunity to do well. You will have 1 0 hours o class time, be able to consult with your teacher at all stages o your work, and research and write your report out o class. Your IA investigation may not be used as part o a physics extended essay. The variety and range o possible investigations is large, you could choose rom:
Traditional hands-on exp erimental work. You may want to measure the acceleration o gravity using Atwood' s approach, or determine the gas constant using standard B oyles law equipment.
D atabase investigations. You may obtain data rom scientifc websites and process and analyse the inormation or your investigation. Perhaps you fnd a pattern in the ebb and ow o the ocean tides, or process inormation on global warming, or use an astronomical database to confrm Keplers law.
S p readsheet. You can make use o a spreadsheet with data rom any type o investigation. You can process the data, graph the results, even design a simple model to compare textbook theory with your experimental values.
Simulations. It may not be easible to perorm some investigations in the classroom, but you may be able to fnd a computer simulation. The data rom a simulation could then be processed and presented in such a way that something new is revealed. Perhaps you might determine the universal gravitation constant through a simulation (an experiment too sensitive to perorm in most school laboratories) . Or you might investigate the eect o air resistance on projectile motion.
C ombinations o the above are also possible. The subj ect matter o your investigation is up to you. It may be something within the syllabus, something you have already studied or are about to study, or it can be within or outside the syllabus. The depth o understanding should be, however, commensurate with the course you are taking. This means that your knowledge o IB Physics ( either S L or HL) will be sufcient to achieve maximum marks when assessed.
The assessment criteria Your IA consists o a single investigation with a report 61 2 pages long. The report should have an academic and scholarly presentation, and demonstrate scientifc rigor commensurate with the course. There is the expectation o personal involvement, an understanding o physics, and that the study is set within a known academic context. This means you need to research your topic and fnd out what is already known about it. There are six assessment criteria, ranging in weight rom 82 5 % o the total possible marks. Each criterion reects a dierent aspect o your overall investigation.
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AD VI CE O N TH E I N TE R N AL ASS E SS M E N T
Criterion
Points
Weight
Personal engagement
02
8%
Exploration
06
25%
Analysis
06
25%
Evaluation
06
25%
Communication
04
17%
024
100%
Total
The IA grade will count or 2 0% o your total physics grade. The criteria are the same or standard and higher level students. We will now consider each criterion in detail. PE RS O NAL E NGAGE ME NT. This criterion assesses the extent to which you engage with the investigation and make it your own. Personal engagement may be recognized in dierent attributes and skills. These include thinking independently and/or creatively, addressing personal interests, and presenting scientifc ideas in your own way. For maximum marks under the personal engagement criterion, you must provide clear evidence that you have contributed signifcant thinking, initiative, or insight to your investigation: that you take the responsibility or ownership o your investigation. Your research question could be based upon something covered in class or an extension o your own interest. For example, you may be a keen music student and your teacher may have demonstrated resonance with a wine glass whilst studying sound. You could have been ascinated with this phenomenon and decide to design and perorm an investigation on the resonance o a wine glass. Personal signifcance, interest, and curiosity are expressed here. You may also demonstrate personal engagement by showing personal input and initiative in the design, implementation, or presentation o the investigation. Perhaps you designed an improved method or measuring the timing o a bouncing ball or devised an interesting method or the analysis o data. You are not to simply perorm a cookbook-like experiment. The key here is to be involved in your investigation, to contribute something that makes it your own. E XPLO RATIO N. This criterion assesses the extent to which you establish the scientifc context or your work, state a clear and ocused research question, and use concepts and techniques appropriate to the course you are studying. Where appropriate, this criterion also assesses awareness o saety, environmental, and ethical considerations. For maximum marks under the exploration criterion, your topic must be appropriately identifed and you must describe a relevant and ully ocused research question. B ackground inormation about your investigation must be appropriate and relevant, and the methodology must be suitable to address your research question. Moreover, or maximum marks, your research must identiy signifcant actors that may inuence the relevance, reliability, and sufciency o your data. Finally, your work must be sae and it must demonstrate a ull awareness o relevant environmental and ethical issues. The key here is your ability to select, develop, and apply appropriate methodology and produce a scientifc work.
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I N T E R N AL A S S E S S M E N T ANALYS IS . This criterion assesses the extent to which your report provides evidence that you have selected, processed, analysed, and interpreted the data in ways that are relevant to the research question and can support a conclusion. For maximum marks under the analysis criterion, your investigation must include sufcient raw data to support a detailed and valid conclusion to your research question. Your processing o the data must be carried out with sufcient accuracy. Moreover, your analysis must take a ull and appropriate account o experimental uncertainty. Finally, or maximum marks, you must correctly interpret your data, so that completely valid and detailed conclusions to the research questions can be deduced. The key here is to make an appropriate and j ustifed analysis o your data that is ocused on your research question. EVALUATIO N. This criterion assesses the extent to which your report provides evidence of evaluation of the investigation and results with regard to the research question and the wider world. For maximum marks under the evaluation criterion, you must describe a detailed and j ustifed conclusion that is entirely relevant to the research question, and ully supported by your analysis o the data presented. You should make a comparison to the accepted scientifc context i relevant. The strengths and weakness o your investigation, such as the limitations o data and sources o uncertainty, must be discussed and you must provide evidence o a clear understanding o the methodological issues involved in establishing your conclusion. Finally, to earn maximum marks or evaluation, you must discuss realistic and relevant improvements and possible extensions to your investigation. The key here is dierent rom the analysis criterion. The ocus o evaluation is to incorporate the methodology and to set the results within a a wider scientifc context while making reerence to your research topic. CO MMUNIC ATIO N. This criterion assesses whether the investigation is presented and reported in a way that supports effective communication of the investigations focus, process, and outcomes. For maximum marks under the communication criterion, your report must be clear and easy to ollow. Although your writing does not have to be perect, any mistakes or errors should not hamper the understanding, ocus, process, and outcomes o your investigation. Your report must be well structured and ocused on the necessary inormation, the process and outcomes must be presented in a logical and coherent way. Your text must be relevant and avoid wandering o onto tangential issues. Your use o specifc physics terminology and conventions must be appropriate and correct. Graphs, tables, and images must all be well presented. Your lab report should be 61 2 pages long. E xcessive length ( beyond 1 2 pages) will be penalized under the communication criterion. The key here is to demonstrate a concise, logical, and articulate report, one that is easy to ollow and is written in a scientifc context. This is not an assessed criterion but nevertheless is likely to be key to the ulflment o a successul IA. In conclusion, the IA represents a unique opportunity or you to take ownership o your physics learning by investigating something that matters to you. It is a chance or you to work independently and to ollow your own scientifc instincts. True, you should heed the advice and experience o your teacher and be guided so that you don' t go o down a blind alley; however you should be prepared to research your topic independently and approach your teacher ull o ideas and suggestions. Experience suggests that those students brim ull o proposals are likely to be successul, providing they stick to the physics skills and principles that have been encouraged to develop throughout the course.
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INDEX Page numbers in italics refer to questions. absorption 623 linear absorption coefficient 628 mass absorption coefficient 629 absorption spectra 271 3 Fraunhofer lines 272 accelerating universe 660, 665 acceleration 27, 334, 46 acceleration and displacement 1 1 5, 1 1 81 9 force, mass and acceleration 47 suvat/kinematic equations of motion 27, 368 accuracy 9 acoustic impedance Z 633 addition 8, 1 2 adiabatic changes 559, 563 air resistance 59 albedo 329, 3423 alpha particles 267, 273, 274, 2745 scattering of alpha particles 291 2, 4936 alternating current (ac) 1 94 ammeters 1 95, 21 4, 21 6 ampere (A) 3, 1 72, 1 84 definition of the ampere 238 amplitude 1 1 7, 1 25 interference by division of amplitude 373 analogue meters 1 94 angle of incidence 1 46 angular displacement 245, 246 angular magnification 593, 597 angular momentum 475, 483, 555 angular motion 554 angular speed 245, 2478, 354 angular velocity 245, 2478 anistrophies 683 approximations 1 Archimedes principle 570, 5723 assessment criteria 6901 analysis 692 communication 692 evaluation 692 exploration 691 2 personal engagement 691 asteroids 643 astronomical distances 641 , 6456 astronomical reflecting telescopes 61 41 5 cassegrain mounting 61 6 Newtonian mounting 61 5 astronomical refracting telescopes 61 21 4 astronomical unit (AU) 645 astrophysics 641 , 6846 astronomical distances 641 , 6456 luminosity and apparent brightness of stars 6478 objects that make up the universe 6423 stars 6435 atomic physics 267, 268, 3046 absorption of radiation 267, 2801 absorption spectra 271 3 background radiation 267, 27980 Bohr model 4834 emission spectra 2701 energy levels 267, 2689 half-life 267, 2767 nuclear radiation and safety 274 patterns in physics 270 probability of decay 267, 276 radioactive decay 267, 2736 transitions between energy levels 267, 26970 atoms 1 05 attenuation 6246 Avrogado constant 1 00, 1 023 Bartons pendulums 586 baryons 290, 295, 297 Becquerel, Henri 267 Bel scale 6223 Bernoulli equation 570, 575
applications 576 derivation 5756 flow out of a container 5789 Pitot static tubes 5778 Venturi tubes 5767 beta particles 267, 2734, 274, 275 Big Bang 660, 675 cosmic microwave background [CMB] 6634 redshift equation and the cosmic scale factor 6645 Type Ia supernovae 665 binary stars 643 black holes 5402, 5589 black-body radiation 329, 3367 black-body radiation and stars 648 building a theory 3401 emission spectrum from a black body 3379 grey bodies and emissivity 329, 33940 StefanBoltzmann law 329, 339 Wiens displacement law 329, 339 Bohr model 483 energies in the Bohr orbits 4834 Bohr, Niels 270, 483 Boyle, Robert 1 01 , 1 31 Brahe, Tycho 258 bridge circuits 4534 brightness, apparent 641 , 647-8 calculations 8, 1 21 4 adding and subtracting vectors 203 candela (cd) 3 capacitance 455, 457 capacitance of a parallel-plate capacitor 459 unit of capacitance 456 capacitors 455, 4567 capacitance of a parallel-plate capacitor 459 capacitors in practice 45961 charging a capacitor 46870 charging a capacitor with a constant current 4578 combining capacitors in parallel and series 4623 energy stored in a capacitor 458 Carnot cycle 559, 564-5 CAT scans (computed axial tomography) 631 CCDs (charge-coupled devices) 380 centripetal acceleration 245, 24950 centripetal force 245, 250 amusement park rides 2523 banking a bike 2534 centripetal or centrifugal? 251 2 how does speed change when motion is in a vertical circle? 2567 investigating how F varies with m, v and r 2501 moving in a vertical circle 2546 satellites in orbit 252 turning a car 253 Cepheid variables 649, 653-4 CERN 303, 492 Chandrasekhar limit 649, 658 charge 1 69, 405 charge and field 1 701 charge carrier drift speed 1 69, 1 846, 238 charge carriers 1 82 charge density 1 85 electric current 1 84 electronic or elementary charge 1 72 measuring and defining charge 1 725 moving charge 1 803 positive test charge 1 77 radial fields 1 77 charged parallel plates 3945, 407, 455 chromatic aberrations 606, 61 4 circle of least confusion 606 circuit diagrams 1 92, 1 93 circuit conventions 1 94
circuit symbols 1 934 constructing practical circuits from a diagram 1 956 practical measurements of current and potential difference 1 945 circular motion 245, 246, 2656 angular displacement 245, 246 angular speed 245, 2478 angular velocity 245, 2478 centripetal acceleration 245, 24950 centripetal force 245, 2507 circular motion and SHM 3545 linking angular and linear speeds 245, 2489 period and frequency 245, 248 radians or degrees 2467 climate change 329, 3489 international perspective 349 clock synchronization 521 collaboration 1 00, 303 collisions 734, 7980 elastic collisions 73, 767 inelastic collisions 73, 78 comets 643 communications and the advantages of optic fibres 620 coaxial cable 621 optic fibres 621 2 twisted pair 6201 compressions 1 26 conducting spheres 1 7980 conduction 91 , 1 81 conduction in gases and liquids 1 82 conduction in metals 1 81 2 models of conduction 1 83 conductors 1 789, 2302, 2368 conservation laws 532 conservation of charge 21 3 conservation of energy 61 , 63, 1 47, 21 3 conservation of linear momentum 73, 745 conservation of momentum 73, 745 constellations 644 convection 91 , 329, 3323 convection currents 332 convection in the Earth 333 modelling convection 3345 sea breezes 333 why the winds blow 3334 Copenhagen interpretation 486 Copernicus 258 Cosmic Background Radiation 1 33 cosmic microwave background [CMB] 660, 6634 anistrophies in the CMB 683 discovery of CMB 664 redshift equation and the cosmic scale factor 6645 cosmic scale factor [R] 660 cosmology 660, 6756 anistrophies in the CMB 683 Big Bang and the age of the universe 6625 cosmic scale factor and temperature 6789 cosmic scale factor and time 678 cosmological principle 6769 dark energy 6823 evidence for dark matter 67982 implications of the density of intergalactic matter 6778 redshift and Hubbles law 661 2 Coulombs law 1 69, 1 73, 1 77 coulombs 1 69, 1 72, 456 critical angle 1 45, 1 4850 CT scans (computer tomography) 631 current balances 238 damping 5834 dark energy 682 Dark Energy Survey {DES} 682
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INDEX dark matter 67982 MACHOs 681 WIMPs 681 2 database investigations 690 de Brglie wavelength 4801 decibels 623 deflection effect 540 degraded energy 31 4 density 570, 571 , 572 diatomic gases 1 03 dielectric materials 455, 4601 diffraction 1 45, 1 51 2 diffraction and resolution 377 diffraction and the satellite dish 3789 electron diffraction 481 2 single-slit diffraction 3647 diffraction grating 367, 371 2 appropriate wavelengths for effective dispersion 372 grating spacing and number of lines per mm 372 using a spectrometer 373 digital meters 1 95, 21 4, 21 6 diodes 2001 diode bridges 439, 4534 dioptres 603 direct current [dc] 1 69, 1 94 discharging a capacitor 4638 discrete energy 267 dispersion 6224 material dispersion 624 waveguide (or modal) dispersion 6245 displacement 27, 2830 acceleration and displacement 1 1 5, 1 1 81 9 angular displacement 245, 246 displacementdistance graphs 1 267 displacementtime graphs 1 27 relationship between displacement, velocity and acceleration 355 Wiens displacement law 329, 339, 654 distance 27, 2830 forcedistance graphs 61 , 65 division 8, 1 3 Doppler effect 381 astronomy 385 light 3845 measuring rate of blood flow 386 moving observer and stationary source 3823 radar 3856 sound waves 381 3 double-slit interference 1 524, 367, 368 intensity variation with the double-slit 3689 drag force 59 drift speed 1 69, 1 846, 238 Earth 262, 329, 333 energy balance in surfaceatmosphere system 329, 3423, 3479 gases in the atmosphere 41 9 greenhouse effect 329, 3437 orbit 341 2 Edison, Thomas 1 98, 453 efficiency 61 , 72 Einstein, Albert 48, 61 , 63, 79, 1 34, 284, 285, 340, 475, 507, 508, 51 2, 51 7, 688 Einsteins photoelectric equation 478 equivalence principle 5345 general theory of relativity 534, 538, 539, 540, 5423 elastic potential energy 61 , 702 electric cells 21 71 8, 2404 anodes and cathodes 21 8 capacity 2201 discharge 21 7, 221 2 electromotive force (emf) 21 7, 2236 internal resistance 21 7, 2236 lead-acid cells 21 920 Leclanch cells 21 81 9 power supplied by a cell 226 primary and secondary cells 21 7, 21 8221 , 223 recharging secondary cells 223 terminal potential difference 21 7, 221 , 222,
694
224, 225 electric current 1 69, 1 836, 2404 charge 1 84 chemical effect 1 92, 21 7 circuit diagrams 1 92, 1 936 conventional and electron current 1 88 conventional current 1 87, 2301 direction rules 2301 , 238, 431 eddy currents 446 Edison vs Westinghouse 453 effects 1 923 full-wave rectification 439, 451 2 half-wave rectification 439, 4501 heating effect 1 92, 21 1 1 2 high-voltage direct-current transmission (HVDC) 449 Kirchhoffs laws 1 92, 21 21 5 magnetic effect 1 92, 227 magnets 229 mechanism for electric current 1 81 2 rectifying ac 450 resistance 1 92, 1 96202 resistivity 1 92, 2021 2 Wheatstone bridge circuits 4534 Wien bridge circuits 454 electric fields 1 69, 1 75, 2404, 392, 406, 426 charge moving in magnetic and electric fields 4223 charged parallel plates 407 charges moving in electric fields 421 2 close to a conductor 1 789 comparison of gravitational and electric fields 4245 conducting sphere 1 7980 electric field strength 1 778, 392 electric field strength and potential gradient 4089 electric field strength and surface charge density 409 graphical interpretations of electric field strength and potential 41 01 1 plotting electric fields 1 76 potential difference 391 , 393 potential inside a hollow conducting charged sphere 41 21 3 electric lighting 1 98 electrical breakdown 1 82 electromagnetic force 290, 2989, 300 electromagnetic induction 427, 428, 471 4 applications 438 electromagnetic force 430 Lenzs law 427, 4302 magnetic flux 427, 4325 making a current 42830 electromagnetism 227, 2404 catapult field 235 Flemings left-hand rule 235, 4301 force between bar magnet field and currentcarrying wire 2346 force on a current-carrying conductor 2369 forces between two current-carrying wires 2334 magnetic field due to current in conductors 2302 magnetic field patterns 2279 motor effect 235 electromotive force [emf] 1 89, 21 7, 2236, 427 changing fields and moving coils 436 coil moving 437 inducing an emf 42830 magnetic field changes 4378 straight wire moving in uniform field 437 electron microscopes 61 0 electrons 2923 charge carriers 1 82 conduction electrons 1 82 electron degeneracy pressure 657 electron diffraction 481 2 free electrons 331 electronvolts 1 69, 1 901 electrostatics 1 701 , 391 , 392, 393 field between parallel plates 3945, 407
linking electrostatics and gravity 41 2 emissivity 329, 33940 empirical versus theoretical models 1 1 2 energy density 308, 31 1 1 2 energy sources 307, 3502 fossil fuels 308, 31 7 nuclear fuel 31 722 primary sources 308, 3091 0, 31 722 pumped storage 308, 3246 renewable and non-renewable energy sources 307, 3089 Sankey diagrams 308, 31 41 6 secondary sources 308 solar energy 308, 3268 specific energy and energy density 307, 31 1 13 types of energy sources 3091 1 wind generators 3224 energy transfers 61 doing work 61 , 635 efficiency 72 elastic potential energy 702 energy forms and transfers 623 energy moving between GPE and KE 6870 KE and GPE 667, 678 power 61 , 656 energy-momentum relation 531 engineering physics 549, 58992 engines 5647 entropy 559, 5678 entropy as a measure of disorder 5689 equilibrium position 1 1 6 equipotentials 391 , 396, 3989 field and equipotential in gravitation 399401 equivalence principle 5345 errors 8 error bars 8, 1 4 propagation of errors 1 7 random errors 8, 1 01 1 systematic errors 8, 91 0, 1 1 zero error 9 escape speed 405, 41 71 9 estimation 1 , 78 event horizon 541 exchange particles 290, 299300 experiment 688, 690 explosions 73, 80 exponential decay 466, 470 falsifiability 77 Faraday, Michael 427, 4356, 439 Faradays law of induction 427, 435 farads 456 Feynman diagrams 290, 3001 fibre optics 620 attenuation and dispersion 6224 structure and use of optic fibres 6202 field lines 391 , 394, 395 field and equipotential in gravitation 399401 field between a point charge and a charged plate 399 field due to a charged sphere 399 field due to a co-axial conductor 399 field due to a single point charge 398 field due to two point charges 399 measuring potential in two dimensions 3957 fields 391 , 392, 393, 405 comparison of gravitational and electric fields 4245 concept map for field theory 424 edge effects 395 field strength 392 radial fields 398 fluid dynamics 570, 573 continuity equation 570, 574 streamline or laminar flow 570, 5734 fluid resistance and terminal speed 27, 59 maximum speed of a car 601 skydiving 5960 fluids 570, 571 Archimedes principle 570, 5723 Bernoulli equation 570, 5759
measuring density of immiscible liquids using a U-tube 572 measuring the coefficient of viscosity of oil 580 Pascals principle 570, 572 static fluids density and pressure 571 Stokes law 57980 turbulent flow 5801 viscosity of fluids 579 flux 427, 4324, 4345 focal length 594, 599 focal point 594, 599 forces 44, 391 Aristotle and the concept of force 445 conservative forces 68 forcedistance graphs 61 , 65 forcetime graphs 73, 834 force, mass and acceleration 47 forces and inverse-square law behaviour 405, 4067 forces between charged objects 1 723 free-body force diagrams 44, 4951 identifying force pairs 44, 48 Newtons first law of motion 44, 456 Newtons second law of motion 44, 467, 1 24 Newtons third law of motion 44, 489, 1 24 representing forces as vectors 44, 53 resultant force 44, 52 translational equilibrium 44, 524 triangle of forces 53 fossil fuels 308, 309, 31 1 Fourier synthesis 1 58, 353 Franklin, Benjamin 1 70, 1 79, 1 88, 238 free-body force diagrams 44, 4951 frequency 1 1 5, 1 1 7, 1 23, 1 25, 354 angular speed or frequency [w] 354 Larmor frequency 635 natural frequency 1 47 period-frequency relationship 1 1 5, 245, 248 threshold frequency 477 friction 58 coefficients of friction 44, 55, 56, 57 dynamic friction 44, 56, 57 solid friction 44, 545 static friction 44, 556, 57 values for pairs of surfaces 56 full scale deflections (fsd) 1 94 full-wave rectification 439, 451 2 fusion (melting) 97 GM tubes 279, 502 galaxies 641 , 6445 Galileo Galilei 44, 45, 258 Galilean relativity and Newtons postulates 507, 5081 0 Galilean transformations 5091 0 gamma rays 1 33, 267, 2745 gases 91 , 1 00, 1 1 31 4, 1 69 Boltzmann constant 1 00, 1 09, 1 1 0 Boyles law 1 01 , 1 034 Brownian motion 1 06 Charless law 1 02, 1 045 differences between real and ideal gases 1 00, 1 1 1 1 2 diffusion 1 06 first law of thermodynamics 563 gas laws 1 01 5 ideal gas equation 1 00, 1 02, 1 078, 1 09, 1 1 0 kinetic model of an ideal gas 73, 1 00, 1 07, 1 1 2, 329 MaxwellBoltzmann distribution 1 1 01 1 molar mass 1 00, 1 03 mole and Avrogado constant 1 00, 1 023 pressure 1 00, 1 02 third gas law 1 02, 1 05 total internal energy of an ideal gas 1 09 gauge bosons 299 Geiger, Johannes 290, 291 2, 492 generators 439 alternating current [ac] generators 43941 measuring alternating currents and voltages 4434
modelling an ac generator 441 3 transformers 439, 4449 global warming 329, 3489 GPS (global positioning system) 521 , 537 gradient fields 6356 graphs 1 4 adding and subtracting vectors 203 calculating gradient of a graph 32 describing motion with graphs 27, 31 , 346 displacementdistance graphs 1 267 displacementtime graphs 1 27 drawing graphs manually 1 6 false origin 1 6 forcedistance graphs 61 , 65 forcetime graphs 73, 834 graphical interpretations of electric field strength and potential 41 01 1 linearizing graphs 1 7 SHM (Simple Harmonic Motion) 1 1 921 gravitation 245, 2656 field and equipotential in gravitation 399401 Newtons law of gravitation 257, 2601 , 263, 264, 402 universal gravitational constant 261 gravitational field strength 257, 258, 392 defining gravitational field strength 25960 field strength at a distance r from a point mass, M 261 2 field strength at a distance r from the centre outside a sphere of mass M 262 g and the acceleration due to gravity 260 inside the Earth 262 linking orbits and gravity 2623 gravitational fields 391 , 392, 405, 406, 426 comparison of gravitational and electric fields 4245 escaping the Earth 41 71 9 orbiting Earth 41 51 7 potential 41 31 5 potential energy 391 , 393 potential inside a planet 41 5 gravitational force 2989, 300 gravitational potential energy (GPE) 61 , 6870 gravitational time dilation 539 greenhouse effect 329, 3434 modelling climate balance 3457 why greenhouse gases absorb energy 3445 hadrons 290, 296 Large Hadron Collider (LHC) 492, 660 half-wave rectification 439, 4501 hard radiation 629 harmonics 1 58, 1 60 harmonics in pipes 1 623 harmonics on strings 1 61 Hawking, Stephen 542 heat pumps 5667 Heisenberg uncertainty principle 475, 4878, 489 HertzsprungRussell [HR] diagram 649, 6556 Higgs boson 290, 303 Hooke, Robert 70, 71 , 1 30 Hookes law 702 Hopper, Grace 623 Hubbles law 660, 661 2, 6623 Huygens, Christiaan 1 30, 1 34, 1 45 hydrostatic equilibrium 570 imaging 593, 63740 converging and diverging mirrors 593, 5948 converging and diverging thin lenses 593, 599606 imaging instrumentation 608 real images 593, 596 spherical and chromatic aberrations 593, 6067, 61 4 virtual images 593, 597 impulse 73, 823 inertia 45 inertial frame of reference 508, 51 0 infra-red radiation 1 32 interference 1 45, 367 constructive and destructive interference 1 52
double-slit interference 1 524, 3689 interference by division of amplitude 373 measuring wavelength of laser light using double slit 1 56 measuring wavelength of microwaves using double slit 1 57 multiple-slit interference 367, 36971 path difference and the double-slit equation 1 546 interferometer telescopes 61 71 8 internal assessment requirements 6889 academic honesty 68990 assessment criteria 6902 internal assessment report 689 planning and guidance 689 types of investigations 690 internal energy 91 , 95 invariant quantities 51 721 inverse-square relationships 1 38, 1 71 , 2601 forces and inverse-square law behaviour 405, 4067 isobaric change 559, 561 2 work done for non-isobaric changes 562 isothermal changes 559, 5623 isotopes 268 isovolumetric changes 559, 5634 Jeans criterion 6667 joule (J) 63, 1 69 kelvin (K) 3, 339 Kepler, Johannes 258, 260, 264 kilogram (kg) 3 kinetic energy (KE) 61 , 667, 6870 KE and momentum 80 Kirchhoffs laws 21 21 4 ideal and non-ideal meters 21 41 6 Large Hadron Collider (LHC) 492, 660 Larmor frequency 635 lasers 1 56 length contraction 5267 lenses 3745, 593, 599 aspherical shape 599 converging and diverging thin lenses 599 doublets lens 606 eyepiece lenses 6089, 61 3 finding approximate value for focal length of a convex lens 600 lens equation 6024 magnifying glasses 601 , 6046 objective lenses 608, 61 3 relationship between object and image for a thin convex lens 6002 same travel times 607 thin-lens theory 599 Lenzs law 427, 4302 leptons 290, 294, 297 light 1 45 interface between two media 1 47 polarization of light 1 42 reversibility of light 1 48 linear magnification 593, 597 light year (ly) 645 liquids 91 Lorentz transformations 51 3, 51 41 6 inverse Lorentz transformation 51 5 Lorentz factor 51 4 luminosity 641 , 6478 MACHOs (Massive Compact Halo Objects) 681 magic numbers 283 magnetic field lines 228 magnetic field patterns 2279 magnetic fields 227 charge moving in magnetic and electric fields 4223 charges moving in magnetic fields 41 921 magnetic field due to current in conductors 2302 magnetic field strength 2378 magnetic flux 427
695
INDEX flux density 4324 flux linkage 427, 4345 magnetic forces 227 force between bar magnet field and currentcarrying wire 2346 force on a current-carrying conductor 2369 forces between two current-carrying wires 2334 magnetic poles 2289 magnets 229, 232, 2346, 236 hard and soft magnetic materials 447 magnifying glasses 601 , 6046 magnitude, orders of 1 , 7 main sequence stars 649, 655, 656 lifetimes of main sequence stars 6702 Maluss law 1 33, 1 423 Marsden, Ernest 290, 291 2, 492 mass 46 force, mass and acceleration 47 inertial and gravitational mass 46 massspring system 361 , 362 matter 290 states of matter 91 matter waves 475, 4801 electron diffraction 481 2 Maxwell, James 1 77, 507, 51 2 measurements 1 4, 246 uncertainties in measurement 91 1 mechanics 27, 8790 medical imaging 626 computed tomography (CT) 631 nuclear magnetic resonance (NMR) in medicine 626, 6346 ultrasound in medicine 6324 X-ray images in medicine 62631 Meldes string 1 5960 melting (fusion) 97 Mendeleev, Dmitri 282 mesons 290, 295, 297 metabolism 561 metallic bond 1 81 2 meters 21 41 6 metre (m) 3, 63 metric multipliers 1 , 7 microscopes electron microscopes 61 0 make a microscope 61 0 normal adjustment 609 optical compound microscopes 6089 resolution of a microscope 61 01 2 microwaves 1 323, 1 57 Minkowski diagrams see spacetime diagrams mirrors 594 aberrations in a mirror 597598 caustic curves 597 images from a mirror 595 linear and angular magnification 593, 597 pole P 594 modelling 1 00, 224 mole (mol) 3 mole and Avrogado constant 1 00, 1 023 moment of inertia 549, 552, 55758 momentum 734 collisions and changing momentum 745 impulse 73, 823 KE and momentum 80 momentum and safety 73, 856 momentum and sport 73, 86 Newtons second law of motion 73, 845 relativistic momentum 529, 5301 two objects colliding 78 two objects when energy is gained 79 two objects with different masses 778 two objects with same mass 767 momentum conservation 745 helicopters 82, 85 recoil of gun 80 rocketry 82, 85 water hoses 81 2 Moon 343, 346, 41 9 motion 27 describing motion with graphs 27, 31 , 346
696
Newtons laws of motion 44, 456, 467, 489 objects and motion 445, 45 suvat/kinematic equations of motion 27, 368 MRI scans 6356 multiple-slit interference 367, 36971 multiplication 8, 1 3 muons 293 nanoseconds 623 near point 604 nebulae 644 Neumanns equation 427, 435 neutrinos 293, 499 neutron capture 669 r-process 66970 s-process 669 neutron stars 494 neutrons 293 neutron degeneracy pressure 658 thermal neutrons 31 8 newton (N) 4, 63 Newton, Isaac 44, 1 34, 1 45, 258, 61 4, 660 Newtons law of gravitation 257, 2601 , 263, 264, 402 Newtons laws of motion 44, 456, 467, 489, 73, 845, 1 24, 51 0 Principa Mathemetica 509 night vision 1 33 nuclear fusion 282, 2867 CNO cycle 667, 668 protonproton chain 6678 star formation 6678, 66870 nuclear interactions 73 nuclear magnetic resonance (NMR) in medicine 626, 634 basic NMR effect 6345 MRI modifications 6356 nuclear physics 267, 3046, 476, 493, 5046 alternative nuclear binding energy plot 289 energy levels in the nucleus 4689 fission chain reaction 2889 law of radioactive decay 5005 mass and energy units for nuclear changes 285 mass defect and nuclear binding energy 282, 2845 nuclear fission 2878 nuclear fusion 282, 2867 nuclides 274, 2823 Rutherford scattering and the nuclear radius 492, 4936 unified atomic mass unit 282, 2834 using electrons of higher energies 4968 variation of nuclear binding energy per nucleon 2856 nuclear power 31 722 fast breeder reactors 320 safety issues 321 society and nuclear power 321 2 nucleosynthesis 668, 671 nuclides 274, 282 patterns for stability 2823 Ohms law 1 92, 1 98200 OppenheimerVolkoff limit 649, 658 optic fibres 621 2 graded-index fibre 621 single-mode fibre 621 step-index fibre 621 optical centre 599 optical compound microscopes 6089 orbital energy 405 orbital motion 405 orbital speed 405, 41 51 6, 41 8 orbits 2634 satellites of Jupiter 264 oscillations 1 1 5, 1 648, 353 describing periodic motion 1 1 61 8 isochronous oscillations 1 1 61 8 oscillations and damping 5834 pair annihilation 293, 475, 489
pair production 293, 475, 4889 pair production and the Heisenberg uncertainty principle 489 X-rays 627 paradigm shifts 507, 51 2 parsec (pc) 645 particle acceleration 532 particle accelerators 531 particle physics 267, 290, 291 4, 3046, 660 annihilation and pair production 293 classification of particles the Standard Model 2948 conservation rules 2978 exchange particles 290, 299300 Feynman diagrams 290, 3001 fundamental forces 290, 2989, 300 international collaboration 303 particle properties 290 strangeness 290, 301 2 Pascals principle 570, 572 path difference 1 546 pendulums 35860, 586 period 1 1 5, 1 1 6, 1 23 period-frequency relationship 1 1 5 periodic motion 1 1 5, 1 1 61 8 Periodic Table 282 phase and phase difference 1 21 2 in phase 1 21 phase change 91 molecular exchange of phase change 99 photoelectric effect 475 demonstration of the photoelectric effect 4767 Einsteins photoelectric equation 478 explanation of the photoelectric effect 477 gold leaf experiment 4778 Millikans photoelectric experiment 479 wave theory and the photoelectric effect 480 X-rays 629 photons 269, 475, 5345 photovoltaic cells 308, 327 pitch 1 63 Planck, Max 340, 476 polarization 1 34, 1 41 2, 1 44 Maluss law 1 33, 1 423 partial polarization 1 41 polarimeters 1 43 polarization of light 1 42 Polaroids 1 42, 1 43 positrons 292 potential 405 graphical interpretations of electric field strength and potential 41 01 1 potential and potential energy 405, 41 3 potential inside a hollow conducting charged sphere 41 21 3 potential difference [pd] 1 69, 1 868, 1 90, 393, 405 electrical potential 391 , 404 electromotive force [emf] 1 89 gravitational potential 391 , 401 4 measuring pd 1 945 power, current and pd 1 8990 using a potential divider to give a variable pd 2091 1 potential gradient 405, 4089 electric field strength and potential gradient 4089 potentiometers 1 96 potters kiln 338 power 61 , 656, 439 power dissipation 1 92 power stations 308 baseload stations 325 control rods 31 9 enriched fuel 31 8 fossil fuels 31 7 heat exchangers 31 9 nuclear fuel 31 722 thermal power stations 31 21 3 precision 1 0, 1 1 pressure 570, 571
electron degeneracy pressure 657 gases 1 00, 1 02 neutron degeneracy pressure 658 principal axis 594, 599 principal focus 594 principle quantum number 484 projectile motion 27 falling (moving vertically) 3941 , 42 moving horizontally 423 proper length 5201 proper time 51 820 pumped storage hydroelectric systems 308, 3246 Q factor 5846 quantities 1 3 fundamental and derived quantities 2 invariant quantities 51 721 scalar quantities 1 81 9 vector quantities 1 81 9, 1 920, 21 , 223 quantum 269 quantum mechanics 484 quantum physics 476, 5046 Heisenberg uncertainty principle 4878 matter waves 4807 pair production and annihilation 48890 photoelectric effect 47680 quantum tunnelling 475, 4901 quarks 290, 2945, 500 quark confinement 290, 295 radar 1 33 radio telescopes 61 61 7 discoveries in radio astronomy 61 8 radio waves 1 32 radioactive decay 267, 2734, 4991 alpha decay 2745 decay constant and half-life 501 decay in height of water column 5034 gamma ray emission 2756 measuring long half-lives 5023 measuring radioactive decay 279 measuring short half-lives 501 modelling radioactive decay 2789 negative beta decay 275, 498 nuclide nomenclature 274 positron decay 275 raising to a power 8, 1 1 4 rarefactions 1 26 ratios 1 ray diagrams 593, 5957, 6002 predictable rays 595 Rayleigh criterion 340, 376, 377, 61 0, 61 6 Rayleigh scatter 6234 RayleighJeans law 476 rays 1 345 real-is-positive 603 red giants 649, 656 redshift 660, 661 2 cosmological redshift 662 gravitational redshift 5379 redshift equation and the cosmic scale factor 6645 reflection 1 45, 1 46 total internal reflection 1 45, 1 4850 refraction 1 45, 1 46 reflection and refraction of waves 1 46 refractive index 1 46, 1 47, 1 50 refrigerators 5667 Relativistic Heavy Ion Collider [RHIC] 660 relativistic mechanics 529 particle acceleration 532 photons 5323 relativistic momentum 529, 5301 total energy and rest energy 52930 relativity 507, 54448 applications of general relativity to the universe as a whole 5423 bending of light 53940 charges and currents a puzzle 51 01 2 clock synchronization 521 equivalence principle 5345
Galilean relativity and Newtons postulates 507, 5081 0 GPS (global positioning system) 521 , 537 gravitational redshift 5369 gravitational time dilation 537 invariant quantities 51 721 Lorentz transformations 51 3, 51 41 6 Maxwell and electromagnetism 507, 51 2 reference frames 507, 58 Schwarzchild black holes 5402 two postulates of special relativity 51 31 4 velocity addition 51 61 7 renewable energy 3089 repeaters 624 resistance 1 92, 1 967 diodes and thermistors 2001 how resistance depends on size and shape 202 measuring internal resistance of a fruit cell 225 Ohms law 1 92, 1 98200 resistance of metal wire 1 97 variation of resistance of a lamp filament 1 99 resistivity 1 92, 2023 combining resistors 2045 complicated networks 207 heating effect equations 21 1 1 2 potential dividers 208 practical resistors 204 resistivity of pencil lead 203 resistors in parallel 205, 2067 resistors in series 205, 2056 thermal and electric resistivity 331 2 using a potential divider to give a variable pd 2091 1 using a potential divider with sensors 2089 variable resistor or potential divider? 21 1 resistorcapacitor [RC] series circuits 455 resolution 376, 377, 38790 diffraction and resolution 377 resolution equation 378 resolution in a CCD (charge-coupled device) 380 resolvance of diffraction gratings 379 resonance 584 examples of resonance 587 resonance of a hacksaw blade 58788 rest mass 528 rheostats 21 0 root mean square (rms) 444 rotational dynamics 549, 550 angular acceleration 5501 equations of motion 551 uniform motion in a circle 550 Rutherford, Ernest 290, 291 2, 493 Sankey diagrams 308, 31 41 6 satellite dishes 3789 satellites 41 51 6, 41 71 9 geostationary orbit 41 7 geosynchronous satellites 41 61 7 GPS (global positioning system) 521 , 537 orbit shapes 41 81 9 polar orbit 41 6 scalars 1 8, 239 scalar quantities 1 81 9 scattering 291 2, 492, 493, 6234 deviations from Rutherford scattering 495 electron diffraction 4956 method of closest approach 4934 nuclear density 494 why is the sky blue? 624 Schrdingers equation 4856 Schwarzchild black holes 5402 Schwarzchild radius 541 scientific notation 1 , 57 second (s) 3 serendipity 44, 664 SHM (Simple Harmonic Motion) 1 1 5, 1 1 81 9, 353, 354 angular speed or frequency [] 354 circular motion and SHM 3545 energy changes in SHM 1 223, 353, 3623
graphing SHM 1 1 922 iteration with damped SHM 583 massspring system 361 , 362 modelling SHM with a spreadsheet 3557 relationship between displacement, velocity and acceleration 355 SHM equation and 2 357 simple pendulums 35860 velocity equation 353, 358 SI units 1 3 capital or lower case? 3 fundamental and derived SI units 1 , 34, 1 84 sign conventions 607 significant figures 1 , 45 simulations 690 single-slit diffraction 364 graph of intensity against angle 364 single slit with monochromatic and white light 3667 single-slit equation 3656 small angle approximation 365 Snells law 1 45, 1 46, 1 47 soap films 3756 solar constant 329, 341 2 solar energy 308 solar heating panels 3267 solar photovoltaic panels 3278 solar system 6423 solids 91 sound waves 1 23 bell jar experiment 1 301 measuring speed of sound 1 2930 spacecraft 41 9 Earths rotation 41 9 slingshots 41 9 spacetime diagrams 5223 simultaneity 5245 time dilation and length contraction 5257 twin paradox 52728 spacetime interval 51 71 8 time travel 51 8 specific energy 308, 31 1 1 2 specific heat capacity 91 , 957 specific latent heat 91 , 979 spectrometers 373 Bainbridge mass spectrometer 423 speed 27, 30 instantaneous and average values 323 terminal speed 27, 60 spherical aberrations 6067 spreadsheet models 34, 3557 spreadsheets 692 springs 70, 1 1 71 8, 1 245 extension 71 2 massspring system 361 , 362 spring constant 71 stars 641 , 643 binary stars 643 black-body radiation and stars 648 Cepheid variables 649, 6534 composition of stars 6502 fate of stars 657 formation of a star 6567 groups of stars 6434 HertzsprungRussell [HR] diagram 649, 6556 Jeans criterion for star formation 6667 larger stars 658 luminosity and apparent brightness of stars 641 , 64749 main sequence stars 649, 655, 656, 6702 massluminosity relation for main sequence stars 649, 656 spectral classes 651 stellar evolution 649, 65659 stellar parallax 641 , 6456 stellar spectra 649, 6502 Sun-like stars 657 Wiens displacement law and star temperature 652 stellar clusters 643 globular clusters 644 open clusters 6434
697
INDEX stellar evolution 649, 65659 stellar parallax 641 , 6456 stellar processes 666 fusion after the main sequence 66872 Jeans criterion for star formation 6667 nuclear fusion 666, 66769 supernovae 6724 stellar spectra 649, 6501 strangeness 290 conservation of strangeness 301 2 strong nuclear force (strong interaction) 290, 2989, 300 subtraction 8, 1 2, 1 3 Sun 329, 341 2, 660 supergiant stars 656 supernovae 6724, 673 Type Ia supernovae 665, 666, 6723 Type II supernovae 666, 67374 superposition 1 58 symbols 2 symmetry 290, 296 telescopes astronomical reflecting telescopes 61 41 6 astronomical refracting telescopes 61 21 4 Earth and satellite-borne telescopes 61 81 9 interferometer telescopes 61 71 8 making a telescope 61 3 radio telescopes 61 61 8 temperature 91 , 92 absolute temperature 91 , 945 conversion from Celsius to kelvin 91 temperature and energy transfer 92 thermometers 934 tesla (T) 237 theory 688 thermal energy transfer 91 , 32930, 3502 atomic vibration 331 convection 91 , 329, 3325 thermal and electric resistivity 331 2 thermal conduction 329, 3301 thermal physics 91 , 1 1 31 4 thermal radiation 91 , 329, 335 black and white surfaces 3356 radiators 336 saucepans 336 thermistors 93, 94, 2001 thermodynamics 55960 cycles and engines 559, 5647 first law of thermodynamics 559, 5604 second law of thermodynamics 559, 56769 thermal efficiency 559 thermometers 934 thin film interference 367, 3734 coating of lenses 3745 vertical soap films 3756 waves reflected by the film 374 waves transmitted through the film 374 thin lenses see lenses tidal effects 538 time constant 455, 467 time dilation 5256, 537 torque 549, 553 angular momentum 555 conservation of angular momentum 555 couples 5534 Newtons first law for angular motion 554 Newtons second law for angular motion 554 Newtons third law for rotational motion 555 rolling and sliding 5567 rotational kinetic energy 556 total energy and rest energy 52930 trajectories 40, 523 transformers 439, 4446 choosing core material 447 choosing wire in the coils 447 core design 447 efficiency of transformer 446 laminating the core 4467 step-up and step-down transformers 446 transformer action 4478 transformer rule 446
698
transformers in action 4489 twin paradox 52728 ultrasound 628, 6346 ultraviolet catastrophe 476 ultraviolet radiation 1 32 uncertainties 8, 246 absolute, fractional and percentage uncertainty 8, 1 1 1 2 gradient and intercepts 8, 1 41 5 propagation of uncertainties 8, 1 2 zone of uncertainty 1 4 unified atomic mass unit 282, 2834 binding energy 284 values 1 vaporization 989 vectors 1 8, 239 adding vector quantities at right angles 21 adding vector quantities not at right angles 223 combination and resolution 1 8 horizontal and vertical components of vector A 18 representing vector quantities 1 920 resolving vectors 21 2 resultant vectors 20 scale drawing [graphical] approach 20 subtraction of vectors 23 vector quantities 1 81 9 velocity 27, 30 angular velocity 245, 2478 relationship between displacement, velocity and acceleration 355 velocity addition 51 61 7 velocity equation 353, 358 vibrations 582 forced vibrations 584 free vibrations 582 voltmeters 1 95, 21 4, 21 6 watt (W) 1 90 Watt, James 66 wave function 475 wave phenomena 353, 38790 waves 1 23, 1 24, 1 456, 1 648 boundary conditions 1 58, 1 63 comparison of travelling waves and stationary waves 1 58, 1 63 critical angle and total internal reflection 1 48 50 describing waves 1 256 displacementdistance graphs 1 267 displacementtime graphs 1 27 electromagnetic waves 1 23, 1 24, 1 31 3 HuygensFresnel principle 1 37 intensity 1 378 longitudinal waves 1 23, 1 25 measuring the refractive index 1 50 mechanical waves 1 24 motion of wavefronts 1 357 nodes and antinodes 1 589 plane polarized waves 1 41 polarization 1 34, 1 41 2 reflection and refraction of waves 1 46 refractive index and Snells law 1 46, 1 47 reversibility of light 1 48 standing waves 1 58 standing waves in pipes 1 61 3 standing waves on strings 1 5861 superposition 1 34, 3940 transverse waves 1 23, 1 245 travelling waves 1 23, 1 245 wave equation 1 23, 1 289 wave theory and the photoelectric effect 480 waveparticle duality 480, 482 wavefronts and rays 1 345 wavelength 1 23, 1 25 weak nuclear force (weak interaction) 290, 2989, 300, 301 white dwarfs 649, 656, 657 Wiens displacement law 329, 339, 652
WIMPs (Weakly Interacting Massive Particles) 681 2 wind generators 308, 3224 women scientists 654 work 61 , 634 rate of doing work 66 work done against resistive force 645 work function 477 worldlines 5227 X-rays 1 33 coherent scattering 627 collimation 630 Compton scattering 627 computed tomography (CT) 631 contrast 631 pair production 627 photoelectric effect 627 soft X-rays 629 X-ray images in medicine 62630 zone of stability 282
PH YSI CS
2 0 1 4 ED I TI O N
Supporting the latest syllabus at SL and HL, this 2014 edition was developed with the IB to most closely embody the IB way of teaching. The holistic approach to all aspects of the syllabus, including Nature of Science, encourages an inquiring, active approach to learning whilst integrated worked examples support exceptional achievement. Oxford course books are the only DP resources developed with the IB. This means that they are: The most accurate match to IB specifications Written by expert and experienced IB examiners and teachers Packed with accurate assessment support, directly from the IB Truly aligned with the IB philosophy, challenging learners with fresh and topical TOK Experiments strengthen practical understanding, cementing comprehension
Supporting resources: Online Course Book 978 0 19 830773 0 Print and Online Course Book pack 978 0 19 830776 1
Unrivalled support for the new concept-based approach to learning
Kerboodle Online Resources 978 0 19 839074 9
Study Guide 978 0 19 839355 9
Online
How to get in contact: web www.oxfordsecondary.co.uk/ib email [email protected] tel +44 (0)1536 452620 fax +44 (0)1865 313472
Authors David Homer Michael Bowen-Jones
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