Beginning physics II ware electromanetism OPtics and modern physics

SCHAUM'S OUTLINE OF THEORY AND PROBLEMS of B E G I N " G PHYSICS 11 Waves, Electromagnetism, Optics, and Modern Physics ALVIN HALPERN, Ph.D Professor of Physics Brook Zy n CoZleg e City University of New York ERICH ERLBACH Professor Emeritus of Physics City College City University of New York SCHAUM'S OUTLINE SERIES McGRAW-HILL New York San Francisco Washington, D.C Auckland Bogotu Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto ALVIN HALPERN, Ph.D., Professor of Physics at Brooklyn College of the City University of New York Dr Halpern has had extensive teaching experience in physics at all college levels elementary through doctoral He was chairman of the physics department at Brooklyn College for ten years, and Vice President for Research Development at the Research Foundation of CUNY for four years He presently is Acting President of the Research Foundation and University Dean for Research ERICH ERLBACH, Ph.D., is Professor Emeritus of Physics at The City College of the City University of New York He has had over 35 years of experience in teaching physics courses at all levels Dr Erlbach served as chairman of the physics department at City College for six years and served as Head of the Honors and Scholars Program at the College for over ten years Schaum’s Outline of Theory and Problems of BEGINNING PHYSICS I1 : Waves, Electromagnetism, Optics, and Modern Physics Copyright 1998 by The McGraw-Hill Companies, Inc All rights reserved Printed in the United States of America Except as permitted under the Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher 10 11 12 13 14 15 16 17 18 19 20 PRS PRS 9 ISBN 0-07-025707-8 Sponsoring Editor: Barbara Gilson Production Supervisor: Sherri Souffrance Editing Supervisor: Maureen B Walker Library of Congress Catalogiag-in-PublicationData Halpern, Alvin M Schaum’s outline of theory and problems of beginning physics 11: waves, electromagnetism, optics, and modern physics/Alvin Halpern, Erich Erlbach p cm.-(Schaum’s outline series) Includes index ISBN 0-07-025707-8 Physics I Erlbach, Erich 11 Title QC23.H213 1998 5304~21 McGraw -Hi22 98-24936 CIP E A Division of TheMcGmwHill Companies This book is dedicated to Edith Erlbach, beloved wife of Erich Erlbach and to the memory of Gilda and Bernard Halpern, beloved parents of Alvin Halpern This page intentionally left blank Preface Beginning Physics I I : Waves, Electromagnetism, Optics and Modern Physics is intended to help students who are taking, or are preparing to take, the second half of a first year College Physics course that is quantitative in nature and focuses on problem solving From a topical point of view the book picks up where the first volume, Beginning Physics I : Mechanics and Heat leaves off Combined with volume I it covers all the usual topics in a full year course sequence Nonetheless, Beginning Physics I I stands alone as a second semester follow on textbook to any first semester text, or as a descriptive and problem solving supplement to any second semester text As with Beginning Physics I , this book is specifically designed to allow students with relatively weak training in mathematics and science problem solving to quickly gain quantitative reasoning skills as well as confidence in addressing the subject of physics A background in High School algebra and the rudiments of trigonometry is assumed, as well as completion of a first course covering the standard topics in mechanics and heat The second chapter of the book contains a mathematical review of powers and logarithms for those not familiar or comfortable with those mathematical topics The book is written in a “user friendly” style so that those who were initially terrified of physics and struggled to succeed in a first semester course can gain mastery of the second semester subject matter as well While the book created a “coaxing” ambiance all the way through, the material is not “ watered down ” Instead, the text and problems seek to raise the level of students’ abilities to the point where they can handle sophisticated concepts and sophisticated problems, in the framework of a rigorous noncalculus-based course In particular, Beginning Physics I I is structured to be useful to pre-professional (premedical, predental, etc.) students, engineering students and science majors taking a second semester physics course It also is suitable for liberal arts majors who are required to satisfy a rigorous science requirement, and choose a year of physics The book covers the material in a typical second semester of a two semester physics course sequence Beginning Physics I I is also an excellent support book for engineering and science students taking a calculus-based physics course The major stumbling block for students in such a course is not calculus but rather the same weak background in problem solving skills that faces many students taking non-calculus based courses Indeed, most of the physics problems found in the calculus based course are of the same type, and not much more sophisticated than those in a rigorous noncalculus course This book will thus help engineering and science students to raise their quantitative reasoning skill levels, and apply them to physics, so that they can more easily handle a calculus-based course ALVINHALPERN ERICHERLBACH This page intentionally left blank To the Student The Preface gives a brief description of the subject matter level, the philosophy and approach, and the intended audience for this book Here we wish to give the student brief advice on how to use the book Beginning Physics I consists of an interweaving of text with solved problems that is intended to give you the opportunity to learn through exploration and example The most effective way to gain mastery of the subject is to go through each problem as if it were an integral part of the text (which it is) The last section in each chapter, called Problemfor Review and Mind Searching, gives additional worked out problems that both review and extend the material in the book It would be a good idea to try to solve these problems on your own before looking at the solutions, just to get a sense of where you are in mastery of the material Finally, there are supplementary problems at the end of the chapter which given only numerical answers You should try to as many of these as possible, since problem solving is the ultimate test of your knowledge in physics If you follow this regime faithfully you will not only master the subject but you will sense the stretching of your intellectual capacity and the development of a new dimension in your ability Good luck This page intentionally left blank Contents Chapter I WAVE MOTION Chapter SOUND Chapter Propagation of a Disturbance in a Medium Continuous Traveling Waves Reflection and Transmission at a Boundary Superposition and Interference Problems of Review and Mind Stretching 1.1 1.2 1.3 1.4 tion and Interference 2.3 Human Perception of Sound 2.4 Other Sound Wave Phenomena Problems for Review and Mind Stretching 2.1 Mathematical Addendum-Exponential and Logarithmic Functions 2.2 Propagation of Sound-Velocity Wave.Fronts Reflection Refraction Diffrac- COULOMB’S LAW AND ELECTRIC FIELDS 1 13 18 30 37 37 42 50 53 58 64 Chapter ELECTRIC POTENTIAL AND CAPACITANCE 101 Chapter SIMPLE ELECTRIC CIRCUITS 5.1 Current Resistance Ohm’sLaw 138 Introduction Electric Charges Coulomb’s Law The Electric Field-Effect The Electric Field-Source The Electric Field-Gauss’ Law Problems for Review and Mind Stretching 3.1 3.2 3.3 3.4 3.5 3.6 Potential Energy and Potential Potential of Charge Distributions The Electric Field-Potential Relationship Equipotentials Energy Conservation Capacitance Combination of Capacitors Energy of Capacitors Dielectrics Problems for Review and Mind Stretching 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.2 5.3 Resistors in Combination EMF and Electrochemical Systems 64 64 68 70 72 80 90 101 103 105 110 114 117 120 123 125 128 138 143 146 518 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS [CHAP 18 interaction is called “cbromodynamics” because the label “color” has been commonly used for this index Neither quarks nor gluons have ever been seen alone It is now believed that quarks can only exist in bound states so that one will never see a free quark This is a consequence of the nature of the strong force provided by the gluons to the quarks In contrast, the electromagnetic force allows charged particles to exist in both bound states (electrons in an atom) or free states (electrons or protons moving through space) Much of our present understanding of elementary particles and the world built out of them is explained by a comprehensive theory called the “Standard Model” Even so, much more needs to be understood, such as the reason the fundamental particles have the rest masses that they do, and whether all the forces of nature can be understood to be different aspects of the same interaction 18.4 SOLID-STATE PHYSICS Another area in which the concepts of quantum mechanics have made a major impact is in the area of solid-state physics This refers to the study of materials in which the atoms are firmly fixed in place and not free to move as they are in gases and liquids The simplest case is the situation when the atoms are arranged in a fixed symmetric pattern which is repeated throughout the solid Such a system is called a crystal The results of analyzing this special case can be generalized to more complicated cases of solids with varying degrees of disorder (amorphous solids) and even to liquids Together all the classes of material are called “condensed matter” We have seen that atoms are described by quantum mechanics in terms of energy levels Each energy level can hold two electrons, one for each spin direction If one has two, or more, atoms that are far apart, then each has its own identical energy level structure, and electrons can be specified as being in a particular level of a particular atom As one brings two atoms closer together, the wave functions begin to overlap, and the electron can no longer be considered as confined to only one atom The result is that the individual levels of each of the two overlapping atoms become two closely spaced shared levels in each of which the electron is shared by the two atoms As one adds more and more atoms, the energy levels split into more and more of these closely spaced levels (3 atoms, levels; atoms, levels; etc.), and are best described as an energy “band” with allowed energies extending from the bottom to the top of the band Again, each level in a band allows for two electrons, one for each spin direction The number of possible electrons in each band is therefore equal to twice the number of atoms that created this band Each energy level of the individual atoms become converted to a band able to accommodate two electrons per atom This status is depicted in Fig 18-18, where the original energy levels are shown as they convert into bands Electrons are contributed to these bands from all the atoms in the solid The first electron from each atom is contributed to the lowest band 1, and will use up half of the states in that band The second electron (with opposite spin) on each atom also goes into band 1, filling that band The third electrons fill half of band 2, and the fourth electrons fill that band It is therefore clear that if there are an odd number of electrons per atom, then the last band to be filled will not be full If there are an even number of electrons per atom, then one expects the highest band to be filled However, if the bands overlap, as bands and in Fig 18-18, then the highest band containing electrons may not be filled even for an even number of electrons per atom There is an “energy gap” between some of the bands which, in Fig 18-18 is large between bands and 2, and is small between bands and At the energies within these gaps, there cannot be any electrons in the solid This basic idea permits us to understand why certain solids act as insulators, while others act as conductors, and still others act as “semiconductors”.When one imposes a voltage between the ends of a wire, an electric field tries to accelerate the electrons in the material This acceleration adds small amounts of energy to the electrons If there are energy states available to the electrons at this small added energy, then the electrons can be accelerated by the electric field, and the material will be a conductor This is the case if the highest band containing electrons is not filled The electrons that can move as a result of the imposed voltage are called “free” electrons, and the band in which they move is called the conduction band The lower bands are called valence bands If the band is filled, then there are no energy levels available for the highest energy electrons, and they will be unable to accelerate We will then have an CHAP 181 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS 519 Atom Band 5 Band Fig 18-18 insulator, which has a filled upper valence band If one can excite electrons from this filled valence band to the empty next higher band, that band would constitute a conduction band, and these excited electrons can move in an electric field Additionally, there is now room in the valence band for electrons to accelerate into the vacated level left by the excited electron A vacated level leaves a net positive charge in the region where the electron was excited and conducted away Under the external electric field another electron in that same level can hop to the new location, neutralizing that location, but leaving a net positive charge where it came from The net effect is as if a positive charge moved in the direction opposite to that of the electrons We describe the apparent motion of a positive charge as follows: when an electron excites into the conduction band we say that there is a positive “hole” created in the valence band and that the positive holes act just like positively charged particles and provide conduction in this band, just as the negative electrons provide conduction in the conduction band The material will be conducting as long as these excited electrons and the holes they leave behind persist This type of excitation can occur if the temperature is sufficiently high that the thermal energy of the electrons allows them to jump to the conduction band For a large energy gap this can occur only at very high temperatures, but if the gap is small it occurs at normal temperatures Materials such as these, with small energy gaps, are called semiconductors They can change from insulators to conductors as the temperature increases In practice, we can get electrons in the conduction band of these semiconductors, or holes in their valence band in a different manner If one inserts an “impurity” into the materisrl, that is, one inserts atoms which have one extra electron than the atoms of the base material, then there will be one more electron per impurity atom than is needed to fill the valence band These added electrons can partially fill the conduction band, and provides conduction appropriate to negative electrons We call this material an n-type semiconductor On the other hand, if one inserts an impurity with one less electron than the host material, then there will be one state in the valence band that may be unfilled This “hole” acts to provide conduction appropriate to a positive charge, and this semiconductor is called ptype We see that we have great flexibility in producing the type of material we want by “doping” the base semiconductor with the appropriate type and amount of impurity The most common semiconductor in current use is silicon 520 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS [CHAP 18 It is also possible to excite electrons in an insulator or semiconductor to the conduction band by absorbing a photon The incident photon that is absorbed will make it possible for the material to conduct electricity This process is the basis for constructing photodetectors, since the absorption of a photon is signaled by a change in the conductivity of the material It is also the basis for converting solar energy into electrical energy Of course, only photons with sufficient energy (large enough frequency) can excite these electrons, so different materials will be sensitive to different wavelengths Problem 18.40 Silicon has a band gap of 1.1 eV (a) At what temperature is the average thermal energy of the electrons equal to the band gap energy? (b) At what maximum wavelength can photons excite electrons from the valence band into the conduction band? (c) What is the maximum band gap that one can have in a material if one wants to build a detector for infrared light at 9.5 p? Solution (a) The thermal energy of a free electron equals ($)kT The band gap is 1.1 eV = l.l(l.6 x 10 ’’)) J = (3M1.38 x 10-23 J/K)T, giving T = 8.5 x 103 K (b) The energy of the photon is hc/A This will equal or exceed the band gap energy if li is less than or equal to: A = hc/E,,, = (6.625 x 10-34J s)(3.0 x 108m/s)/(l.l eV)(1.6 x 10- l J,/eV)= 1.13 x 10-’m = 1.13 p * (c) Again we use that A = hc/E,,, and x 10-6 m) = 2.09 x O P O J = 0.13 eV solve for EBaP= h(./R = 6.625 x 10-”(3.0 x 108),’(9.5 Semiconductor devices are presently the components of all electronics systems, including radios, TV, radar and computers There are many different ways in which semiconductors can be used to provide the properties needed to be useful in electronic applications All these devices make use of the fact that one can make materials in which one or more parts are n-type and other parts are p-type By applying voltages to different parts one can vary the current flow through the material Some simple junctions function as diodes which conduct current primarily in only one direction Some function as transistors that have the same properties as the old vacuum tubes and can amplify a signal greatly Some act as data storage elements that can take on one of two different states, that can be viewed as “zero” and “one”, and are used in computers extensively Some act as lasers, others as light emitting diodes Devices have been built as photovoltaic cells, and as photodetectors With modern technology it is possible to build millions of these elements in a small volume This is known as building integrated circuits, and for even more densely packed devices, as VLSI (very large scale integration) There is no aspect of modern day technology that does not make use of some of these devices Problems for Review and Mind Stretching Problem 18.41 Consider the case of a Bohr atom of hydrogen Assume that the radiation that is emitted is the result of transitions between the bound levels or from E = (the minimum energy of an unbound, or ionized, electron) to one of the levels (a) What is the minimum and the maximum wavelength of the light emitted in the first series ending on the n = level? CHAP 181 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS 52 (6) What is the minimum and the maximum wavelength of the ight emitted in the second series ending on the n = level? (c) What is the minimum and the maximum wavelength of the light emitted in the third series ending on the n = level? (d) What is the minimum and the maximum wavelength of the light emitted in the fourth series ending on then = 4level? (e) Between which series is there overlap in wavelength? Solution We use Eq (18.9), l/A, = (E,,, - E,)/hc = 1.097 x 107Z2(l/n2- l/m2) m-', with n = and Z = The maximum wavelength occurs for m = (the nearest level corresponding to the least energy photon), and the minimum wavelength is for m = CO (the E = state) Thus l/A,, = 1.097 x 107(1- i2) = 8.23 x 106 m-', or A,, = 1.22 x 10-7 m Furthermore, 1/Amin = 1.097 x 107(1- 0) = 1.097 x 107 m-', or Amin = 9.11 x 10-* m Thus, the wavelength ranges from 91.1 to 122 nm We use Eq (28.9), l/d, = (E,,, - EJhc = 1.097 x 107Z2(l/n2- l/rn2) m-', with n = and Z = The maximum wavelength occurs for m = (the nearest level corresponding to the least energy photon), and the minimum wavelength is for m = CQ (the furthest away in energy) Thus l/A,,,,, = 1.097 x 107(i2- 4') = 1.523 x 106 m-', or A,, = 6.56 x 10-7 m Furthermore, 1/Amin = 1.097 x 107(4) = 2.74 x 106 m-', or Amin = 3.64 x lO-' m Thus, the wavelength ranges from 364 to 656 nm We again use Eq (28.9) with n = and = The maximum wavelength occurs for m = and the minimum wavelength is for m = CO Thus 1/dmax= 1.097 x 107(i2= 5.33 x 105 m-', or A,, = 1.875 x 10-6 m Furthermore, 1/Amin = 1.097 x lO'(4) = 1.22 x 106 m-', or Amin = 8.20 x 10-7 m Thus, the wavelength ranges from 820 to 1875 x nm a2) We again use Eq (28.9) with n = and = The maximum wavelength occurs for m = and the minimum wavelength is for m = CQ Thus l/A,,, = 1.097 x lO7(a2- j )= 2.47 x 10' m-', or A,, = 4.05 x O w m Furthermore, l/A,,n = 1.097 x 107(&)= 6.86 x 105 m-', or Amin = 1.46 x 10-6 m Thus, the wavelength ranges from 1460 to 4050 nm We tabulate the minimum and maximum wavelengths for each series Series Amin (nm) 91.1 364 820 1460 A,, (nm) 122 656 1875 4050 The minimum of each higher series is above the maximum of the previous series except for the last series Thus series and overlap Problem 18.42 Certain atoms in the ground state can absorb radiation at wavelengths of 150 nm, 450 nm, 600 nm and 700 nm (a) From this data deduce the energies of some of the levels of this atom above the ground state (b) After absorbing radiation at the given wavelength the atom can radiate at these same wavelengths as well as at additional wavelengths What are the additional wavelengths? Solution (a) The atom will have energy levels above the ground state with energies equal to the energy of the photon that is absorbed The longest wavelength corresponds to the nearest energy level, which we call E, The energies that can be absorbed are E = hc/A Using the wavelengths given, these energies are: 522 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS E4 = 1.325 x 10-l8 J [CHAP 18 = 8.28 eV E, = 4.42 x 10-l9 J = 2.76 eV E, = 3.31 x 10-l9 J = 2.07 eV E, = 2.84 x 10-l9 J = 1.77 eV These are the energies of the levels above the ground state (b) If the atom is in one of these states and makes a transition to the ground state it will radiate at the wavelengths given However, the atom in level can also make a transition to level 3,2, or 1, the one in level can make a transition to levels or 1, and the one in level can make a transition to level The wavelengths emitted in these transitions correspond to the energy differences between the levels These are: -+ -+ 1, = hc/(E4 - E , ) = hc/(13.25 - 2.84) x 10-19 = 1.91 x loh7 m 2, R = hc/(E4 - E,) = hc/(13.25 - 3.31) x 10-19 = 2.00 x 10-' m + 3, = hc/(E4 - E , ) = hcM13.25 - 4.42) x 10-l9 = 2.25 x 10-7 m -+ 1, = hc/(E, - E,) = hc/(4.42 - 2.84) x 10-l9 = 1.26 x 10-6 m -+ 2, = hc/(E, - E,) = hc/(4.42 - 3.31) x 10-l9 = 1.79 x 10-6 m -+ 1, = hc/(E, - E , ) = hc/(3.31 - 2.84) x 10-19 = 4.23 x 10-6 m Problem 18.43 The characteristic K, X-rays from W (tungsten, = 74) have a wavelength of 2.18 x 10-" m What is the characteristic wavelength of the same line from the next element Re (Rhenium, = 75)? solution The characteristic X-rays for high atoms have energies proportional to (2 - l),, or wavelengths proportional to 1/(Z - l), Thus L,J1, = (73/74), = 0.973, and A,, = 0.973(2.18 x 10-" m) = 2.12 x lO-" m Problem 18.44 The binding energy of an a-particle is 28.24 MeV The mass of a neutron is 1.008,665 U and of ,H'is 1,007,825 U (a) What is the mass of ,He4 in U? (b) What minimum frequency photon is needed to completely dissociate an a-particle into its nucleons? (c) If the a-particle is broken up into two deuterons (1H2), 23.86 MeV must be supplied What is the binding energy of a deuteron? (6) If one wanted to get fusion power by creating a particles from deuterons, how many would one have to form per second to get W of power? a particles Solution (a) We know that the binding energy is the difference between the mass of the constituents and the mass of the nucleus, converted to energy Thus, converting to atomic mass units, B.E = (28.24 MeV/ (932 MeV/u) = (2m, + 2mH - mHc) Thus 28.24/932 = 2(1.008,625) + 2(1.007,825)- mHc This results in mHs = 4.002,60 U (b) The energy we need is 28.24 MeV This requires a photon of minimum frequency hf= 28.24 MeV = 28.24 x 106(1.6 x 10-l9) J, or!= 6.82 x 10,' Hz (c) The reaction is EK + ,He4 2,H2 Since there are two protons and two neutrons on each side, their masses will cancel, leaving only the binding energies Then EK - B.E.H, = -2 B.E., = 23.86 MeV - 28.24 MeV = -4.38 MeV Then B.E., = 4.38/2 = 2.19 MeV -+ CHAP 181 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS 523 (d) If one fuses two deuterons into He, one releases 23.9 MeV of energy For a power of W one must generate J/s = (2/1.6 x 10-l9) eV/s = 1.25 x 1013 MeV/s Since each fusion releases 23.9 MeV, we need 1.25 x 1013/23.9= 5.23 x 10" fusions/s Problem 18.45 A certain sample has 5.1 x 10" particles which decay with a half life of 1.5 x 10-2 s (a) What is the initial activity of the sample? (b) How many particles have decayed after a time of 2.0 x 10-6s? (c) How many particles have decayed after a time of 5.2 x 10A2s? Solution (a) The activity, I A N / A t 1, is AN, where A = (In 2)/2 Thus, initially, the activity is A N / A t I = [0.693/( 1.5 x 10-2 s)][5.1 x 10"] = 2.36 x 1013decay+ = 6.38 x 102 Ci (b) As long as the time during which the decays occur is small compared to the half life, the number of particles in the sample remains essentially constant Then we can use AN = ANAt = (2.36 x 1013s-')(2 x 10-6 s) = 4.7 x 10' particles (c) In this case the time during which decays occur is not much less than the half life, and we cannot use A N = ANAt since N changes during the time we are considering We therefore calculate the number of particles remaining after the time t, using the equation N = N o 2(-''') = Noe - = 5.1 x 10'' exp [-(46.21)(-5.2 x 10-2 s)] = 4.61 x 10" particles The number that have decayed is therefore 5.1 x 10" - 4.6 x 10" = 4.64 x 10" particles Problem 18.46 The /I-decay of 15P32has a half life of 14.3 days and 1.72 MeV is released in the decay 15P32 has a mass of 31.973,907 U (a) Write down the reaction equation for the process (b) What is the mass of the daughter atom? (c) If one has 2.3 x 10- l kg of phosphorus, what is the activity of the sample? Solution (a) 15P32 -+ 16s32 + fl- + V (b) Using conservation of energy, mp = m, - 0.001,83 = 31.97,206 U + E, Therefore, m, = 31.973,907 - 1.72/932 = 31.973,907 (c) The activity equals A N / A t = AN = 0.693 N / z We are given the mass, m, of our sample and we need the number, N , of atoms this corresponds to If one has N atoms, then there are n = N / N A moles, where NA is Avogadro's number Each mole has a mass of M = 31.97 g = 0.031,97 kg Thus, n = m/M = (2.3 x 10-" kg)/(0.031,97 kg) = N/(6.02 x 1023 atoms/mol) -+ N = 4.33 x 1014 atoms The activity is 0.693(4.33 x 1014)/[(14.3 d)(24 h/d)(60 min/h)(60 s/min)] = 2.43 x 108 Bq = (2.43 x 108/3.7 x 10") Ci = 6.6 x 10-3 Ci Problem 18.47 The R particle can decay into a A' and a K - : f2- + A' serves (a) charge, (b)energy, (c) baryons, and (d) leptons + K- Show that this con- (e) Show that the strangeness changes by one What does this imply about the decay? Solution (a) There is one negative charge before the decay and one after the decay, conserving charge (b) The mass of the R - is 3272m, while the masses of the Ao and K- after the decay are(2184 + 967)m, = 3151m, Thus the initial mass is greater than the final mass and the extra mass is converted into kinetic energy This will satisfy conservation of energy 524 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS [CHAP 18 (c) There was one baryon before the decay (the Q-) and there is one baryon after the decay (the AO) The kaon is not a baryon Therefore, we have conservation of baryons (d) There are no leptons before or after the decay, so leptons are conserved IR- has a strangeness of -3, while the A' and the K- each have a strangeness of - Thus the total strangeness after the decay is -2, which is one less than the strangeness before the decay The decay cannot occur via the strong interaction since strangeness is not conserved, and we therefore expect the half life to be relatively long (e) The Supplementary Problems Problem 18.48 A hydrogen atom is in the level n = (a) What wavelength photon is needed to cause a transition to n = 3? (b) What maximum wavelength is needed to ionize the atom from n = 2? Am (a) 6.56 x i O - m; (b) 3.65 x 10-' m Problem 18.49 A hydrogen atom is in a state with n = 3, = 2, rn, = - 1, m, hydrogen atom state is given by E = - 13.6 eV/n2 = Assume that the energy of a (a) What is the orbital angular momentum of the atom in this state? (b) If the atom changes to the state n = 1,l = 1, mI= 0, m, = 4, what is the frequency of the photon emitted? (c) List all the possible states of the atom with n = (d) If light of wavelength 200 nm is absorbed by the atom (with n = 3), what is the kinetic energy of the emitted electron? Ans (a) 1.49 x 10-34 kg - m2/s; (b) 2.93 x lOI5 Hz; (c) I = 0; rn, = 0; m, = I = 1;m,=O, + ; m , = I = 2; m,= 0, + I , k ; m, = + j (a total of 18 states); (d) 4.70 eV ++ ++ Problem 18.50 Write down all the possible states of an f electron (for a given n) Ans = 3; m,= 0, + 1, _+2,_+3;rn, = _+ (a total of 14 states) Problem 18.51 An atom has a ground state at an energy of 25 eV below ionization Other levels are at - 1.5, -6, and - 10 eV The atom is in the state at -6 eV (a) What wavelengths can be emitted by the atom? (b) What wavelengths can be absorbed by the atom? Ans (a) 3.11 x 10-7 m, 6.53 x 10-7'8 m; (b) 2.76 x 10-7 m, 2.07 x 10-7 m and below Problem 18.52 A certain atom is in the state with n = and ml = - (a) What are the possible values of for this state? (b) What are the possible values of angular momentum for this state? (c) What are the possible angles that the angular momentum can make with the z axis for his state? Ans (a) I = 1, or 3; (b) L = 1.49 x 10-34, 2.58 x 10-34, or 3.65 x 10-34 kg m2/s; (c) 135", 114" or 106" CHAP 18) MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS 525 Problem 18.53 List the electrons that fill the ground state of the following elements: (a) silicon with 14 electrons; (b) titanium with 22 electrons; (c) copper with 29 electrons; and (d)selenium with 34 electrons Ans (a) ( ~ ) ~ ( ~ ) ~ ( p ) ~ ( ~(b) ) ’ ((1~)~(2~)~(2p)~(3~)’(3p)~(4~)~(3d)~; 3p)~; (c) (1~)~(2~)~(2p)~(3~)~(3p)~(4~)~(3d) (4(1s)2(2s)2(2P)6(3s)2(3P)6(4S)2(3d)10(4P)4 Problem 18.54 What must be the difference in energy of the levels responsible for producing the laser light in the case of: (a) an ultraviolet laser at 100 nm; and (b) an X-ray laser at 1.0nm? Ans (a) 12.4eV; (b) 1.24keV Problem 18.55 An X-ray laser operates at a wavelength of 1.0 nm How many photons are emitted per second if the power of the laser is: (a) 2.0W; (b) 10 W; and (c) 10 kW? Ans (a) 1.01 x 1OI6/s; (b) 5.03 x 1OI6/s;(c) 5.03 x 1019/s Problem 18.56 Lead, with atomic number 82, has isotopes with mass number (a) 204, (b) 206, (c) 207, (d)208, (e) 210 and cf)214.Determine the number of neutrons, of protons and the symbol for each of these isotopes Ans (a) N = 122, Z = 82, 82Pb204;(b) N = 124, Z = 82, 82Pb206;(c) N = 125, Z = 82, 82Pb207; (d)N = 126,Z = 82, 82Pb208;(e) N = 128,Z = 82,,,Pb2’O; cf)N = 132,Z = 82,82Pb214 Problem 18.57 What is the binding energy per nucleon for each of the isotopes in the previous problem, if the atomic masses of the isotopes are: m204 = 203.973,037U; m206= 205974,455 U; ~ = 206.975,885U; m208= 207.976,641 U; mz10 = 209.984,178U; m214 = 213.999,764U? (The mass of a neutron is 1.008,665U and of ,HI is 1.007,825u.) Ans (a) 7.88 MeV/A; (b) 7.88 MeV/A; (c) 7.87 MeV/A; (d) 7.87MeV/A; (e) 7.84MeV/A; (f) 7.78 MeV/A Problem 18.58 The binding energy for an electron in a hydrogen atom is 13.6 eV The binding energy for a nucleon in a nucleus of heavy water is 2.25 MeV What fraction of the original mass of the constituents is lost in forming the atom and in forming the nucleus? Ans (a) 1.4 x 10-6%; (b)0.12% Problem 18.59 At a certain time, a sample is decaying at the rate of 9.4 x 104 Ci The decay constant is measured to be 1.3 x 104 s-’ (a) How many particles are there in the sample at this time? (b) At what time will the activity be reduced to of its initial value? Ans (a) 2.7 x 10” particles; (b) 5.3 x 10-5s Problem 18.60 Particles decay with a half life of 155 years How long does it take till only 0.10% remain? ~ n s 1.54 x 103 y Problem 18.61 The half life for the a-decay of Bi2I2 is 60.6 s How much mass of bismuth is needed for an activity of 3.3 x lO-’ Ci? Ans 3.76 x lO-”kg 526 MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS [CHAP 18 Problem 18.62 Element X with 180 nucleons, decays via a-emission, and releases 4.1 MeV in the process The half life for the decay is 3.0 x 104 s (a) If one started with 10,' nuclei of X, how many will remain after 1.2 x 105 s? (b) The binding energy of X is 5.9 MeV/A, and of the a-particle is 2.5 MeV/A What is the binding energy per nucleon of the daughter nucleus? Ans (a) 6.25 x lOZ3;(b) 5.95 MeV/A Problem 18.63 The energy released in the a-decay of P ~ is 05.3 MeV The mass of this isotope of polonium is 209.982,876 U and the mass of a helium atom (,He4) is 4.002,603 U (a) Identify the daughter nucleus (b) What is the mass of the daughter isotope? Ans (a) 82Pb206;(b) 205.9746 U Problem 18.64 The nucleus of 3Li7 has a mass defect of 0.0422 ,H' is 1.007,825 u.) U (The mass of a neutron is 1.008,665 U and of (a) What is the binding energy per nucleon for this nucleus? (b) What is the mass of 3Li7? (c) A photon of large enough frequency can cause the emission of a neutron from ,Li7 What is the resulting nucleus? If the isotope of this nucleus has an atomic mass of 6.015 U, what is the minimum frequency of the photon needed to cause the neutron emission? Ans (a) 5.62 MeV/A; (b) 7.0159 U; (c) 3Li6, 1.75 x 1OZ1Hz Problem 18.65 83Bi214can decay to 82Pb2loeither by a-emission followed by /I-emission or in the reverse order Write the reaction equation for both paths Ans 83B12'4+ 81T1210+ a, E1T12'o EzPb2'o+ /I + f ; (b) 8JBi214* P O l + /I + f, 84PO2l4 82pb2l0 (U) + -+ Problem 18.66 A slow neutron causes a nucleus of 94Pu239to fission, producing two neutrons and two identical fragments The plutonium has a binding energy per nucleon of 7.62 MeV/A, and 207 MeV of kinetic energy is given to the fission fragments and outgoing neutrons + (a) Fill in the missing information for X in the reaction equation, 94Pu239 on' + + 2zXA 2,n' (b) What is the B.E./A for the isotope X? A m (a) 4,Ag119; (b) 8.52 MeV/A Problem 18.67 A fission reactor, using U235,produces energy at the rate of 160 MW Each fission release energy of 208 MeV How much uranium is used during a period of three years? Ans 177 kg Problem 18.68 A nuclear reaction converts nuclide X into nuclide Y in the reaction X(n, a)Y If the incident neutron has an energy of 1.2 MeV, the products have a kinetic energy of 5.3 MeV The binding energy of the a-particle is 28.3 MeV What is the difference in binding energy of the two nuclides, B.E., - B.E.,? Ans 24.2 MeV + + Problem 18.69 In the fusion reaction, ,He3 ,He3 + ,He4 2,H', calculate how much energy is released and given to the products as kinetic energy (Atomic masses are: ,He3 = 3.016,030 U, ,He4 = 4.002,603 U, ,H' = 1.007,825 u.) CHAP 18) MODERN PHYSICS: ATOMIC, NUCLEAR AND SOLID-STATE PHYSICS 527 Ans 12.86 MeV Problem 18.70 Which conservation laws are violated in each of the following hypothetical reactions? (a) p + n + e + + v (b) n- +2y n+Ko +no (6) A o + n + z o (e) n++c(++ e + + e (c) Ans (a) energy; (b)charge; (c) baryons; (d) strangeness; (e) leptons Problem 18.71 In the strong interaction: R(a) + p + -+ A’ + KO What are the quark combinations before the reaction? (b) What are the quark ,combinations after the reaction? (c) Noting that the quantum numbers of a quark-antiquark pair cancel each other out, what are the “net” quarks before and after the reaction? Ans (a) (ad), (uud); (b) (uds), (dS); (c) u,d,d before and after Problem 18.72 A crystalline material turns from insulating to conducting when bathed in light of wavelength less than 260 nm What is the band gap between the valence band and the conduction band? Ans 4.78 eV This page intentionally left blank Index Aberrations, 354 Absorption (of light), 410 Accommodation, 374 Airy disk, 407 Ammeters, 149 Ampere, 139 Ampere’s law, 201, 202, 12, 16 Angle of: deviation, 335 diffraction, 408 incidence, 33 reflection, 33 refraction, 33 Angular magnification, 372 Anode, 147 Anti-nodes, 19 Apparent depth, 342 Atomic number, 493 Compass, 180 Composites, 516 Compound microscope, 376 Compton scattering, 460 Compton wavelength, 461 Concave, 353 Conductivity, 142 Conductors, 65 Control rods, 510 Converging, 361 Convex, 353 Coulomb, 65 Coulomb’s Law, 68 Cross-section (likelihood of fusion), 509 Curie, 500 Current : direct, 139, 243 alternating, 243 Back EMF, 259,266 Balmer series, 478 Baryons, 501 Battery, 138 Beat, 54 Becquerel, 500 Binary, 38 Binding energy, 438,495 Birefringence (sec speed of propagation), 41 Black body, 453 Bohr atom, 475,476 Breeder reactors, 10 Bremsstrahlung radiation, 457 Brewster angle, 41 Bulk modulus, 42 DeBroglie wavelength, 462 Decay : alpha, 502 beta, 504 gamma, 506 Decibel scale, 50 Decimal, 38 Diamagnetic, 217 Diatonic scale, 53 Dichroism (see absorption), 410 Dielectric, 126 Dielectric constant (K), 126, 317 Diffraction, 390 Diffraction grating, 408 Diffraction limit of resolution, 407 Dipoles, 126 Dirac’s equation, 465 Direction of propagation, 323 Discharging, 148 Dispersion, 337 Displacement current, 13 Displacement current density, 313 Diverging, 361 Doppler shift, 54 Camera, 373 Capacitance, 117 Capacitive reactance, 292 Capacitor, 118 Cathode, 147 Chain reaction, 509 Chromodynamics, 18 Circular motion, 168 Coaxial cable, 203 Coherent light, 409 Einstein’s special theory of relativity, 422 529 530 Electric charge, 64 Electric displacement, 316 Electric field, 70,71 Electric field lines, 80 Electric force, 64 Electric generator, 138 Electromagnetic motor, 175 Electromagnetic waves, 317 Electromotive force (EMF), 138,227 Electron probability cloud, 487 Electron-volt, 114 Emissivity (E), 453 Energy density, 124 Equally tempered scale, 53 Equipotential surface, 111 Excited state, 492 Exponential decays, 279 Exponential function, 37 Eye, 374 Eyepiece, 376 f-number, 374 Faraday’s law, 235,316 Farad (F), 118 Farsighted (hyperopia), 374 Ferromagnetism, 18 Field lines, 200 Fission, 508 Flux, 82 Flux density, 82 Focal length, 361 Focal point, 361 Free electrons, 518 Fringes (interference),392 Fundamental, 23 Fundamental particles, 516 Fuse, 144 Galvanometer, 149 Gamma (y)-rays, 438 Gauss, 164 Gauss’ law, 84,109,315 Gaussian surface, 85 Generator, 230 Geometrical optics, 329 Gluons, 517 Gradient, 113 Greenhouse effect, 510 Ground state, 477,492 INDEX [adron, 514 [alf-life,281, 499 [alf-silveredmirror, 393 [all effect, 171 Iarmonics, 23 :eavy water, 493,510 Ienry, 255 Iuygen’s wavelets, 391 :yperopic (farsighted),374 Image, 348 Impedance, 295 In phase, 387 Index of refraction, 330 Induced EMF, 227,232 Inductive reactance, 298 Insulating materials, 125 Interference, 49,390 constructive, 388, 390 destructive, 388 fringes, 392 patterns, 18 Internal resistance, 147 Inverted, 350 Ionized atom, 477 Isotopes, 493 Kaon, 515 Kilowatt-hour (kW-h), 157 Lags, 291 Law of Biot and Savart, 188 Law of conservation of change, 64 Leads, 291 Left-right handedness, 350 Lensmaker’s equation, 367 Lenz’s Law, 236 Leptons, 513 Light year, 325 Logarithmic function, 39 Longitudinal wave, Lyman series, 478 Magnetic field, 164,200,217 Magnetic field lines, 200 Magnetic flux, 232 Magnetic force, 164 Magnetic intensity, 316 INDEX Magnetic permeability, 317 Magnetic susceptibility, 22 Magnetization, 217,220 Magnetized, 220 Magnifying glass, 372 Mass defect, 438 Mass number, 493 Matter waves, 462 Maxwell equations, 315,465 Meson, 514 Michelson interferometer, 393 Moderator (of fission), 510 Motational EMF, 227 Muons, 426 Mutual inductance, 260,261 Myopic (near sighted), 375 Naperian base (e), 40 Natural logarithm (loge,ln),40 Nearsighted (myopia),374 Neutrino, 505 Newton’s laws, 435 Newton’s rings, 401 Newton’s second law, 435 Newton’s third law, 325 Nodes, 19 Non-reflecting coating, 398 North pole, 219 Nuclear magnetic resonance, 486 Null measurement, 154 Object, 348 Objective, 376 Ohm (Q), 140 Ohm’s law, 140 Open circuit EMF, 147 Orbital motion, 217 Order (of the spectrum),408 Out of phase, 388 Overtones, 23 Parallel connection, 144 Paramagnetic, 17 Paschen-Bach series, 478 Pauli’s exclusion principle, 489,495 Permanent magnet, 218 Permeability, 221 Permettivity, 126 Permettivity of free space ( ), 126 531 Phase angle, 295 Photo-electric effect, 454 Photons : electromagnetic wave, 513 virtual, 513 Physical optics, 329 Pion, 514 Pitch, 52 Planck’s constant, 453 Plane wave, 46 Polarization, 409 Polarized, 125 Polarized light, 409 Polaroid, 10 Positive, 291 Positron, 438 Potential difference, 102, 138 Potential energy, 102 Power, 155 Poynting vector, 323,460, (See also direction of propagation) Primary coil, 267 Principle of superposition, 18 Probability amplitude, 465 Projector, 375 Propagation constant, Quantum, 451 Quantum numbers, 482 Quarks, 516 Radiation pressure, 325 Rainbow, 339 Real, 348 Recharging, 148 Reflected, 13 Reflection (of light), 410 Refraction, 49 Refractive power of the lens, 368 Relativistic mass, 437 Relativity, 425 Remanent magnetization, 218 Resistance, 140 Resistivity, 140 Resistor, 140 Resonance frequency, 302 Resonant frequencies, 22 Rest mass, 437 Rest mass energy, 437 Reverberation time, 51 532 Right-hand rule, 167 Roentgen, 457 Root-mean-square (RMS), 290 Rydberg constant, 477 Scattered light, 41 Schroedinger’s equation, 465 Secondary coil, 267 Self inductance, 257 Semiconductors, 17 , s18 n-type, 519 p-type, 519 Series connection, 144 SHM (simple harmonic motion), Snell’s law, 331 Solenoid, 199,204 toroidal, 206 South pole, 219 Spectrum, 337 order of, 408 Speed of light, 17 Speed of propagation (birefringence),41 Spin, 217 Spontaneous emission, 492 Standing wave, 19 Stationary solutions, 48 Stefan-Boltzman law, 452 Stimulated emission, 492 Superconductor, 223 Supersonic, 57 Telescope, 376 reflecting, 377 refracting, 377 Terminals, 147 INDEX Tesla, 164 Thick lens, 368 Thin lens, 366 Total energy, 435,437 Total reflection, 334 Transformation equations : Galilean, 43 Lorentz, 43 Transformer, 267 Transmitted, 16 Transverse wave, Twin paradox, 427 Uncertainty principle, 466 Uniform field, 199 Unpolarized light, 409 Valence bands, 18 Velocity selector, 171 Virtual, 14, 348 Volt, 102 Voltmeters, 149 Wave function, 468 Wave-front, 45 Wavelength, Wheatstone bridge, 154 Work function, 455 X-rays, 457 Zero point energy, 466 ... theory and problems of beginning physics 11: waves, electromagnetism, optics, and modern physics/ Alvin Halpern, Erich Erlbach p cm.-(Schaum’s outline series) Includes index ISBN 0-07-025707-8 Physics. .. College for over ten years Schaum’s Outline of Theory and Problems of BEGINNING PHYSICS I1 : Waves, Electromagnetism, Optics, and Modern Physics Copyright 1998 by The McGraw-Hill Companies, Inc...SCHAUM'S OUTLINE OF THEORY AND PROBLEMS of B E G I N " G PHYSICS 11 Waves, Electromagnetism, Optics, and Modern Physics ALVIN HALPERN, Ph.D Professor of Physics Brook Zy n CoZleg e City