INSTRUCTOR SOLUTIONS MANUAL THIRD EDITION FOR SCIENTISTS AND ENGINEERS a strategic approach randall d knight Larry Smith Brett Kraabel Snow College PhD-Physics, University of Santa Barbara Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Publisher: James Smith Senior Development Editor: Alice Houston, Ph.D Senior Project Editor: Martha Steele Assistant Editor: Peter Alston Media Producer: Kelly Reed Senior Administrative Assistant: Cathy Glenn Director of Marketing: Christy Lesko Executive Marketing Manager: Kerry McGinnis Managing Editor: Corinne Benson Production Project Manager: Beth Collins Production Management, Illustration, and Composition: PreMediaGlobal, Inc Copyright ©2013, 2008, 2004 Pearson Education, Inc All rights reserved Manufactured in the United States of America This publication is protected by Copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1900 E Lake Ave., Glenview, IL 60025 For information regarding permissions, call (847) 486-2635 Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps MasteringPhysics is a trademark, in the U.S and/or other countries, of Pearson Education, Inc or its affiliates ISBN 13: 978-0-321-76940-4 ISBN 10: 0-321-76940-6 www.pearsonhighered.com Contents Preface v PART I Newton’s Laws Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Concepts of Motion 1-1 Kinematics in One Dimension 2-1 Vectors and Coordinate Systems 3-1 Kinematics in Two Dimensions 4-1 Force and Motion 5-1 Dynamics I: Motion Along a Line 6-1 Newton’s Third Law 7-1 Dynamics II: Motion in a Plane 8-1 PART II Conservation Laws Chapter Chapter 10 Chapter 11 Impulse and Momentum 9-1 Energy 10-1 Work 11-1 PART III Applications of Newtonian Mechanics Chapter 12 Chapter 13 Chapter 14 Chapter 15 Rotation of a Rigid Body 12-1 Newton’s Theory of Gravity 13-1 Oscillations 14-1 Fluids and Elasticity 15-1 PART IV Thermodynamics Chapter 16 Chapter 17 Chapter 18 Chapter 19 A Macroscopic Description of Matter 16-1 Work, Heat, and the First Law of Thermodynamics 17-1 The Micro/Macro Connection 18-1 Heat Engines and Refrigerators 19-1 © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher iii iv CONTENTS PART V Waves and Optics Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Traveling Waves 20-1 Superposition 21-1 Wave Optics 22-1 Ray Optics 23-1 Optical Instruments 24-1 PART VI Electricity and Magnetism Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 Chapter 31 Chapter 32 Chapter 33 Chapter 34 Chapter 35 Electric Charges and Forces 25-1 The Electric Field 26-1 Gauss’s Law 27-1 The Electric Potential 28-1 Potential and Field 29-1 Current and Resistance 30-1 Fundamentals of Circuits 31-1 The Magnetic Field 32-1 Electromagnetic Induction 33-1 Electromagnetic Fields and Waves 34-1 AC Circuits 35-1 PART VII Relativity and Quantum Physics Chapter 36 Chapter 37 Chapter 38 Chapter 39 Chapter 40 Chapter 41 Chapter 42 Relativity 36-1 The Foundations of Modern Physics 37-1 Quantization 38-1 Wave Functions and Uncertainty 39-1 One-Dimensional Quantum Mechanics 40-1 Atomic Physics 41-1 Nuclear Physics 42-1 © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Preface This Instructor Solutions Manual has a twofold purpose First, and most obvious, is to provide worked solutions for the use of instructors Second, but equally important, is to provide examples of good problem-solving techniques and strategies that will benefit your students if you post these solutions Far too many solutions manuals simply plug numbers into equations, thereby reinforcing one of the worst student habits The solutions provided here, by contrast, attempt to: • • • • Follow, in detail, the problem-solving strategies presented in the text Articulate the reasoning that must be done before computation Illustrate how to use drawings effectively Demonstrate how to utilize graphs, ratios, units, and the many other “tactics” that must be successfully mastered and marshaled if a problem-solving strategy is to be effective • Show examples of assessing the reasonableness of a solution • Comment on the significance of a solution or on its relationship to other problems Most education researchers believe that it is more beneficial for students to study a smaller number of carefully chosen problems in detail, including variations, than to race through a larger number of poorly understood calculations The solutions presented here are intended to provide a basis for this practice So that you may readily edit and/or post these solutions, they are available for download as editable Word documents and as pdf files via the “Resources” tab in the textbook’s Instructor Resource Center (www.pearsonhighered.com/educator/catalog/index.page) or from the textbook’s Instructor Resource Area in MasteringPhysics® (www.masteringphysics.com) We have made every effort to be accurate and correct in these solutions However, if you find errors or ambiguities, we would be very grateful to hear from you Please contact your Pearson Education sales representative © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher v vi PREFACE Acknowledgments for the First Edition We are grateful for many helpful comments from Susan Cable, Randall Knight, and Steve Stonebraker We express appreciation to Susan Emerson, who typed the word-processing manuscript, for her diligence in interpreting our handwritten copy Finally, we would like to acknowledge the support from the Addison Wesley staff in getting the work into a publishable state Our special thanks to Liana Allday, Alice Houston, and Sue Kimber for their willingness and preparedness in providing needed help at all times Pawan Kahol Missouri State University Donald Foster Wichita State University Acknowledgments for the Second Edition I would like to acknowledge the patient support of my wife, Holly, who knows what is important Larry Smith Snow College I would like to acknowledge the assistance and support of my wife, Alice Nutter, who helped type many problems and was patient while I worked weekends Scott Nutter Northern Kentucky University Acknowledgments for the Third Edition To Holly, Ryan, Timothy, Nathan, Tessa, and Tyler, who make it all worthwhile Larry Smith Snow College I gratefully acknowledge the assistance of the staff at Physical Sciences Communication Brett Kraabel PhD-University of Santa Barbara © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher CONCEPTS OF MOTION Conceptual Questions 1.1 (a) significant figures (b) significant figures This is more clearly revealed by using scientific notation: sig figs P 0.53 = 5.3 × 10−1 (c) significant figures The trailing zero is significant because it indicates increased precision (d) significant figures The leading zeros are not significant but just locate the decimal point 1.2 (a) significant figures Trailing zeros in front of the decimal point merely locate the decimal point and are not significant (b) significant figures Trailing zeros after the decimal point are significant because they indicate increased precision (c) significant figures (d) significant figures Trailing zeros after the decimal point are significant because they indicate increased precision 1.3 Without numbers on the dots we cannot tell if the particle in the figure is moving left or right, so we can’t tell if it is speeding up or slowing down If the particle is moving to the right it is slowing down If it is moving to the left it is speeding up 1.4 Because the velocity vectors get longer for each time step, the object must be speeding up as it travels to the left The acceleration vector must therefore point in the same direction as the velocity, so the acceleration vector also points to the left Thus, a x is negative as per our convention (see Tactics Box 1.4) 1.5 Because the velocity vectors get shorter for each time step, the object must be slowing down as it travels in the Ϫ y direction (down) The acceleration vector must therefore point in the direction opposite to the velocity; namely, in the +y direction (up) Thus, a y is positive as per our convention (see Tactics Box 1.4) 1.6 The particle position is to the left of zero on the x-axis, so its position is negative The particle is moving to the right, so its velocity is positive The particle’s speed is increasing as it moves to the right, so its acceleration vector points in the same direction as its velocity vector (i.e., to the right) Thus, the acceleration is also positive 1.7 The particle position is below zero on the y-axis, so its position is negative The particle is moving down, so its velocity is negative The particle’s speed is increasing as it moves in the negative direction, so its acceleration vector points in the same direction as its velocity vector (i.e., down) Thus, the acceleration is also negative © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-1 1-2 Chapter 1.8 The particle position is above zero on the y-axis, so its position is positive The particle is moving down, so its velocity is negative The particle’s speed is increasing as it moves in the negative direction, so its acceleration vector points in the same direction as its velocity vector (i.e., down) Thus, the acceleration is also negative Exercises and Problems Section 1.1 Motion Diagrams 1.1 Model: Imagine a car moving in the positive direction (i.e., to the right) As it skids, it covers less distance between each movie frame (or between each snapshot) Solve: Assess: As we go from left to right, the distance between successive images of the car decreases Because the time interval between each successive image is the same, the car must be slowing down 1.2 Model: We have no information about the acceleration of the rocket, so we will assume that it accelerates upward with a constant acceleration Solve: Assess: Notice that the length of the velocity vectors increases each step by approximately the length of the acceleration vector © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Concepts of Motion 1-3 1.3 Model: We will assume that the term “quickly” used in the problem statement means a time that is short compared to 30 s Solve: Assess: Notice that the acceleration vector points in the direction opposite to the velocity vector because the car is decelerating Section 1.2 The Particle Model 1.4 Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point The size and shape of the object will not be considered This is a reasonable approximation of reality if (i) the distance traveled by the object is large in comparison to the size of the object and (ii) rotations and internal motions are not significant features of the object’s motion The particle model is important in that it allows us to simplify a problem Complete reality—which would have to include the motion of every single atom in the object—is too complicated to analyze By treating an object as a particle, we can focus on the most important aspects of its motion while neglecting minor and unobservable details (b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance (c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies, or how water flows through a pipe Section 1.3 Position and Time Section 1.4 Velocity 1.5 Model: We model the ball’s motion from the instant after it is released, when it has zero velocity, to the instant before it hits the ground, when it will have its maximum velocity Solve: © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-4 Chapter Assess: Notice that the “particle” we have drawn has a finite dimensions, so it appears as if the bottom half of this “particle” has penetrated into the ground in the bottom frame This is not really the case; our mental particle has no size and is located at the tip of the velocity vector arrow 1.6 Solve: The player starts from rest and moves faster and faster 1.7 Solve: The player starts with an initial velocity but as he slides he moves slower and slower until coming to rest Section 1.5 Linear Acceleration 1.8 G G Solve: (a) Let v0 be the velocity vector between points and and v1 be the velocity vector between points and Speed v1 is greater than speed v0 because more distance is covered in the same interval of time (b) To find the acceleration, use the method of Tactics Box 1.3: Assess: The acceleration vector points in the same direction as the velocity vectors, which makes sense because the speed is increasing 1.9 G G Solve: (a) Let v0 be the velocity vector between points and and v1 be the velocity vector between points and Speed v1 is greater than speed v0 because more distance is covered in the same interval of time (b) Acceleration is found by the method of Tactics Box 1.3 Assess: The acceleration vector points in the same direction as the velocity vectors, which makes sense because the speed is increasing © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-5 42.15 Solve: From Figure 42.8, the nuclear potential energy at r = 1.0 fm is U nuclear = −50 MeV = −(50 MeV)(1.6 × 10−19 J/eV) = −8.0 × 10−12 J The gravitational potential energy is (6.67 × 10−11 Nm /kg )(1.67 × 10−27 kg) Gm =− = −1.86 × 10−49 J U grav = − −15 r 1.0 × 10 m ⇒ U grav U nuclear = 1.86 × 10−49 J 8.0 × 10 −12 J = 2.3 × 10−38 Section 42.4 The Shell Model 42.16 Solve: (a) The A = 10 nuclei listed in Appendix C are 10Be, 10B, and 10C Their nuclear energy level diagrams are shown in the figure below 10 Be has Z = 4, so N = 10 B has Z = 5, N = 5, and 10C has Z = 6, N = These three nuclei are called the A = 10 isobars (b) 10 B is a stable nucleus, but 10 Be and 10C are radioactive 10 Be undergoes beta-minus decay and 10C undergoes beta-plus decay In beta-minus decay, a neutron within the nucleus changes into a proton and an electron In beta-plus decay, a proton changes into a neutron and a positron Assess: The above decays for 10 Be and 10C are consistent with the fact that the line of stability follows the N = Z line for Z < 16 © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 42-6 Chapter 42 42.17 Solve: (a) The A = 14 nuclei listed in Appendix C are 14C, 14 N, and 14O 14C has Z = 6, so N = 14 N has Z = 7, N = 7, and 14O has Z = 8, N = These nuclei are the A = 14 isobars (b) 14 N is a stable nucleus, but 14C and 14O are radioactive 14C undergoes beta-minus decay and 14O undergoes beta-plus decay In beta-minus decay, a neutron within the nucleus changes into a proton and an electron In beta-plus decay, a proton changes into a neutron and a positron Assess: The above decays for 14C and 14O are consistent with the fact that the line of stability follows the N = Z line for Z < 16 Section 42.5 Radiation and Radioactivity 42.18 Model: The number of radioactive atoms decreases exponentially with time ⎛1⎞ Solve: The number of remaining 226 Ra nuclei at time t is N = N ⎜ ⎟ ⎝2⎠ t/t1/2 (a) After 200 years, N = (1.0 × 1010 )(1/2)200 years/1600 years = 9.17 × 109 (b) After 2000 years, N = (1.0 × 1010 )(1/2)2000 years/1600 years = 4.20 × 109 (c) After 20,000 years, N = (1.0 × 1010 )(1/2) 20,000 years/1600 years = 1.73 × 106 Assess: Do not think that if half the nuclei decay during one half-life, then all will decay in two half-lives 42.19 Model: The number of radioactive atoms decreases exponentially with time Solve: The mass of remaining 131Ba nuclei at time t is ⎛1⎞ N = N0 ⎜ ⎟ ⎝2⎠ t/t1/2 (a) After day, N = (250 μ g)(1/2)1 day/12 days = 236 μ g (b) After 10 days, N = (250 μ g)(1/2)10 days/12 days = 140 μ g (c) After 100 days, N = (250 μ g)(1/2)100 days/12 days = 0.775 μ g Assess: Do not think that if half the nuclei decay during one half-life, then all will decay in two half-lives 42.20 Model: The number of radioactive atoms decreases exponentially with time Solve: (a) 3H is called tritium and has Z = and N = 3H undergoes beta-minus decay and gives rise to the daughter nucleus 3He, that is, H →3He + β − (b) From Equation 42.18, the decay rate is 0.693 0.693 year = × = 1.83 × 10−9 s −1 r= = τ t1/2 12 years 3.15 × 107 s The time constant is τ = 5.46 × 108 s © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-7 42.21 Model: The number of atoms decays exponentially with time Solve: The number of remaining radioactive atoms at t = 50 and t = 200 is ⎛1⎞ N50 = N ⎜ ⎟ ⎝ 2⎠ t/t1/2 ⎛1⎞ = (1.0 × 1010 ) ⎜ ⎟ ⎝ 2⎠ (50 min)/(100 min) t/t = 7.07 × 109 atoms (200 min)/(100 min) ⎛ ⎞ 1/2 ⎛1⎞ N 200 = N ⎜ ⎟ = (1.0 × 1010 ) ⎜ ⎟ = 2.50 × 109 atoms 2 ⎝ ⎠ ⎝ ⎠ Thus the number that decay between 50 and 200 is N decay = N50 − N 200 = 4.6 × 109 atoms Each decay emits an alpha particle, so there are 4.6 × 109 alphas emitted 42.22 Solve: The activity of a radioactive sample is R = rN From Appendix C, the half-life of 60Co is t1/2 = 5.27 yr = 1.663 × 108 s Thus the decay rate is r= τ = ln 0.693 = = 4.17 × 10−9 s −1 t1/2 1.663 × 108 s We know that the sample’s activity is R = 3.50 × 109 Bq = 3.50 × 109 decays/s, so the number of atoms in the sample is N= R 3.50 × 109 decays/s = = 8.40 × 1017 atoms r 4.17 × 109 s −1 The mass of the sample is M = mN = (60 × 1.661 × 10−27 kg)(8.40 × 1017 ) = 8.36 × 10−8 kg = 83.6 μ g 42.23 Model: The activity R of a radioactive sample is the number of decays per second Solve: The rate of decay is R= ⇒ t1/2 = dN N (0.693)N = rN = = dt t1/2 τ (0.693)N (0.693)(5.0 × 1015 atoms) day = = 6.93 × 106 s × = 80.2 days ≈ 80 days R 86,400 s 5.0 × 108 Bq Section 42.6 Nuclear Decay Mechanisms 42.24 Solve: (a) The decay is 230Th → X + 4He, so X = 226Ra (b) The decay is 35S → X + e − + v , so X = 35Cl (c) The decay is X → (d) The decay is 24 40 K + e+ + v, so X = Na → 24 40 Ca − Mg + e + v → X + γ , so X = 24 Mg 42.25 Solve: (a) The decay is X → 224Ra + 4He, so X = 228Th (b) The decay is X → 207 Pb + e − + v , so X = − 207 Tl (c) The decay is Be + e → X + v, so X = Li (d) The decay is X → 60 Ni + γ , so X = 60 Ni 42.26 Solve: (a) From Appendix C, the A = 17 isobars are 17 N, 17O, and 17 F (b) The isotope 17O is stable but rare (c) The daughter nucleus for both of the unstable isotopes is 17O 17 N decays by beta-minus decay 17 F decays by electron capture © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 42-8 Chapter 42 42.27 Solve: (a) From Appendix C, the A = 19 isobars are 19O, 19F, and 19 Ne (b) The 19F isotope is stable (c) The daughter nucleus for both of the unstable isotopes is 19F 19O decays by beta-minus 19 Ne decays by beta-plus 42.28 Model: The decay is 239Pu → 235 U + 4He Solve: Using Appendix C to find the masses, the energy released is E = Δmc = [m(239Pu) − m( 235U) − m( 4He)]c = [239.052157 u − 235.043924 u − 4.002602 u]c = (0.00563 u)c × 931.49 MeV/c u = 5.24 MeV Assess: Essentially all of this energy goes into the alpha particle’s kinetic energy 42.29 Model: Assume the energy released goes into the alpha particle’s kinetic energy so that Kα = 5.52 MeV Visualize: Use Equation 42.15: Kα = (mX − mY − mHe )c = 5.52 MeV We are also given mX + mY = 452 u Solve: (mX − mY − mHe )c = 5.52 MeV ⎛ ⎞ 1u (mX − mY − mHe ) = 5.52 MeV/c ⎜⎜ 2⎟ ⎟ = 0.00593 u ⎝ 931.49 MeV/c ⎠ mX − mY = 0.00593 u + mHe = 0.00593 u + 4.00260 u = 4.00853 u We now have a system of two equations in two unknowns which we will add together to solve for mX ⎧ mX − mY = 4.00853 u ⎨ ⎩ mX + mY = 452 u 2mX = 456 u mX = 228 u Looking in Appendix C for a nucleus with a mass of 228 u which decays by α decay, we find 228Th Assess: 228Th is a step in the 232Th decay series 238Th has a half-life of 1.9 yr and decays by α radiation releasing 5.52 MeV of energy 42.30 Model: The decay is 3H → 3He + e− + v Solve: Beta-minus decay leaves the daughter atom as a positive ion However, the mass of the ion plus the mass of the escaping electron is the mass of a neutral atom, which is what is tabulated in Appendix C Thus the mass loss is the mass difference between the two neutral atoms In Appendix C we find that m(3H) = 3.016049 u, and m(3He) = 3.016029 u The energy released in the beta-minus decay corresponds to a mass change of 0.000020 u The energy released is E = Δmc = (0.000020 u)(931.49 MeV/u) = 0.0186 MeV Assess: The energy E is shared between the electron and the antineutrino 42.31 Model: The decay is 24 Na → 24Mg + e− + v Solve: Beta-minus decay leaves the daughter atom as a positive ion However, the mass of the ion plus the mass of the escaping electron are the mass of a neutral atom, which is what is tabulated in Appendix C Thus the mass loss is the mass difference between the two neutral atoms In Appendix C we find that m( 24 Na) = 23.990961 u and m( 24Mg) = 23.985042 u The energy released in the beta-minus decay corresponds to a mass change of 0.005919 u The energy released is 931.49 MeV/c = 5.51 MeV 1u Assess: This energy is shared between the electron and the antineutrino E = Δmc = (0.005919 u)c ì â Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-9 42.32 Model: The decay is n → p + + e− + v Solve: Use Table 42.2 The mass lost in this decay is Δm = m(n) − m(p + ) − m(e− ) = 939.57 MeV/c2 − 938.28 MeV/c2 − 0.51 MeV/c2 = 0.78 MeV/c2 The energy released is the energy equivalent of 0.78 MeV/c , or E = (0.78 MeV/c )(c ) = 0.78 MeV Assess: This energy is shared between the electron and the neutrino Section 42.7 Biological Applications of Nuclear Physics 42.33 Model: The rad (radiation absorbed dose) measures the energy deposited in an irradiated material Solve: Because Gy = 1.00 J/kg of absorbed energy, 1.5 Gy corresponds to a dose of 1.5 J/kg This means a 150 g tumor absorbs an amount of energy equal to (1.50 J/kg)(0.150 kg) = 0.225 J 42.34 Model: The gray measures the energy deposited in an irradiated material Solve: The energy to be absorbed by the 100 g tumor is 0.20 J Because Gy is defined as 1.00 J/kg of absorbed energy, the dose given is 0.20 J Gy × = 2.0 Gy 100 g 1.00 J/kg 42.35 Model: The radiation dose in units of rems is a combination of deposited energy and biological effectiveness The RBE for beta radiation is 1.5 Solve: Gy is defined as 1.00 J/kg of absorbed energy In the case of the 50 kg worker, the energy absorbed per kg is 20 × 10−3 J = 4.0 × 10−4 J/kg 50 kg This corresponds to a dose of 4.0 × 10−4 J/kg × Gy = 4.0 × 10−4 Gy 1.00 J/kg The dose equivalent in rems is (4.0 × 10−4 Gy)(1.5) = 0.060 Sv = 60 mrem 42.36 Solve: For α -particles, the RBE = 20 The dose in Sv of the alpha radiation is 30 Gy × 20 = 600 Sv For γ -rays, the RBE = The dose in Gy of 600 Sv of γ -rays is 600 Gy 42.37 Model: Assume the 197 Au nucleus remains at rest Visualize: The radii of the alpha particle and the gold nucleus are rα = (1.2 fm)(4)1/3 = 1.90 fm rAu = (1.2 fm)(197)1/3 = 6.98 fm If the alpha just touches the surface of the gold nucleus, the distance between their centers is rf = 1.90 fm + 6.98 fm = 8.88 fm © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 42-10 Chapter 42 Solve: (a) The energy conservation equation K f + U f = Ki + U i is 0J+ (2e)(79e) = mvi2 + J 4πε 8.88 × 10−15 fm 2(9.0 × 109 N m /C2 )(158)(1.60 × 10−19 C) ⇒ vi = (4 × 1.661 × 10−27 kg)(8.88 × 10−15 fm) = 3.5 × 107 m/s (b) The energy of the alpha particle is K = 12 mvi2 = 12 (4 × 1.661 × 10−27 kg)(3.5 × 107 m/s) = 4.09 × 10−12 J × (1 MeV/1.60 × 10−13 J) = 26 MeV 42.38 Model: Assume the 207 Pb nucleus remains at rest Visualize: The radii of the proton and the lead nucleus are rp = (1.2 fm)(1)1/3 = 1.20 fm rPb = (1.2 fm)(207)1/3 = 7.10 fm At the momentum of impact on the lead nucleus, the distance between their centers is rf = 1.20 fm + 7.10 fm = 8.30 fm Solve: The energy conservation equation K f + U f = Ki + U i is 20 MeV + ⇒ Ki = 20 MeV + (e)(82e) = Ki + J 4πε 8.30 × 10−15 fm (9.0 × 109 N m /C2 )(82)(1.60 × 10−19 C)2 8.30 × 10−15 fm × MeV 1.60 × 10−13 J = 34.2 MeV 42.39 Solve: (a) The sun’s mass of 1.99 × 1030 kg is unchanged, but it assumes the density of nuclear matter, which we found to be ρ nuc = 2.3 × 1017 kg/m3 The volume of the collapsed sun is M 1.99 × 1030 kg = 8.65 × 1012 m3 V = π r3 = S = ρnuc 2.3 × 1017 kg/m3 Thus its radius is 1/3 ⎛ 3(8.65 × 1012 m3 ) ⎞ r =⎜ ⎟⎟ ⎜ 4π ⎝ ⎠ = 12,700 m = 12.7 km (b) We can use the conservation of angular momentum to find the new rotational period: (I ω )after = (I ω )before ⇒ ωafter = M R2 I before S S ωbefore = ω M r before I after S 2 ⎛ 1.27 × 104 m ⎞ 2π ⎛ R ⎞ 2π −4 ⇒ =⎜ S⎟ ⇒ Tafter = ⎜ ⎟⎟ (27 days) = 7.8 × 10 s = 780 μs ⎜ Tafter ⎝ r ⎠ Tbefore ⎝ 6.96 × 10 m ⎠ © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-11 42.40 Visualize: Let x1 be the fraction of 69Ga with atomic mass 68.92 u and x2 be the fraction of 71Ga with atomic mass 70.92 u We know that x1 + x2 = or x2 = − x1 We seek x1 Solve: x1 (68.92 u) + x2 (70.92 u) = 69.72 x1 (68.92 u) + (1 − x1)(70.92 u) = 69.72 ⇒ x1 = 69.72 u − 70.92 u = 0.60 68.92 u − 70.92 u The percent abundance of 69Ga is 60% Assess: While Ga does not appear in Appendix C, this result can be verified in other tables of isotopes of the elements 42.41 Solve: (a) The binding energy of the electron in a hydrogen atom is B = 13.6 eV That is, the mass decreases by the equivalent of 13.6 eV when an electron and proton form a hydrogen atom Since B = Δmc , Δm = 13.6 eV c = 13.6 eV c × 1u 931.49 MeV/c = 1.46 × 10−8 u As a percentage of the hydrogen mass, the mass decrease is Δm 1.46 × 10−8 u = = 1.45 × 10−6 % 1.007825 u 1.007825 u (b) The mass decrease is Δm = 2mp + 2mn − mHe nuc These are nuclear masses, but Appendix C tabulates atomic masses Add and subtract the mass of two electrons: Δm = 2(mp + me ) + 2mn − (mHe nuc − 2me ) = 2mH + 2mn − mHe where mH is the mass of a hydrogen atom and mHe is the mass of a helium atom Using Appendix C, Δm = 2(1.007825 u) + 2(1.008665) − 4.002602 = 0.0304 u As a percentage of the helium mass, the mass decrease is 0.0304 u = 0.0076 = 0.76% 4.0026 u (c) Although mass does change in chemical reactions, the change is an incredibly small fraction of the mass of the atoms No experiment will be sensitive to changes of ≈ × 10−6 %, so this small change in mass is easily neglected Not so in nuclear reactions, where the mass change can be ≈1% of the particle masses Not only is this mass change easily detectable, it is essential for understanding nuclear reactions 42.42 Visualize: Please refer to Figure 42.6 for the graph of the binding energy versus mass number Solve: For a nucleus of mass number of 240, the binding energy per nucleon is ≈7.8 MeV Hence, the total binding is (7.8 MeV)(240) = 1870 MeV For a nucleus of mass number of 120, the binding energy per nucleon is ≈8.5 MeV and the total binding energy is (8.5 MeV)(120) = 1020 MeV When a nucleus with mass number 240 fissions into two nuclei with mass number 120, the total energy released is 2(1020) MeV − 1870 MeV = 170 MeV 42.43 Visualize: Please refer to Figure 42.6 for the graph of the binding energy versus mass number Solve: For a He nucleus, the binding energy per nucleon is 7.2 MeV Because A = 4, the binding energy of the He nuclei is 7.2 MeV × = 28.8 MeV Three He nuclei thus have a total energy of 28.8 MeV × = 86.4 MeV For the 12 C nucleus, the binding energy per nucleon is 7.7 MeV Because A = 12, the binding energy of 12C is 7.7 MeV × 12 = 92.4 MeV When three He nuclei fuse together to form a 12C nucleus, the total energy released is 92.4 MeV − 86.4 MeV = 6.0 MeV © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 42-12 Chapter 42 42.44 Visualize: Please refer to Figure 42.6 for the graph of the binding energy versus mass number Solve: For a 56 Fe nucleus, the binding energy per nucleon is 8.8 MeV Because A = 56, the total binding energy is 8.8 MeV × 56 = 493 MeV For a 28Al nucleus, the binding energy per nucleon is 8.4 MeV and the total binding energy is 28 × 8.4 MeV = 235 MeV The total binding energy for two 28Al nuclei is × 235 MeV = 470 MeV Because the 56 Fe is more tightly bound than two 28Al nuclei, 56 Fe cannot fission spontaneously It could perhaps be forced to fission with the input of 493 MeV − 470 MeV = 23 MeV 42.45 Solve: The radius of a 238U nucleus is r = r0 A1/3 = (1.2 fm)(238)1/3 = 7.436 fm For the de Broglie wavelength to be equal to the diameter of the 238 U nucleus, p = h/λ = h/2r Hence, K= p ⎛ h ⎞ ⎛ 6.63 × 10−34 J s ⎞ 1 eV =⎜ ⎟ =⎜ × = 0.93 MeV ⎟ 2m ⎝ 2r ⎠ 2m ⎝⎜ × 7.436 × 10−15 m ⎠⎟ 2(4 × 1.67 × 10−27 kg) 1.6 × 10−19 J 42.46 Model: The 14C/12C ratio in living matter is 1.3 × 10−12 After death, this ratio decays exponentially with time Solve: The decay equation in terms of N, the number of radioactive atoms, can equally well be written in terms of the ratio R = N/N ref where N ref is a stable reference In dating experiments, the number of 12C atoms is a reference to which the number of 14C atoms can be compared Thus ⎛1⎞ R = R0 ⎜ ⎟ ⎝2⎠ t/t1/2 ⎛1⎞ = (1.3 × 10−12 ) ⎜ ⎟ ⎝ 2⎠ t/5730 yr where R0 = 1.3 × 10−12 is the ratio at the time of death and the half-life of 14C is known to be 5730 yr We can find the time at which the ratio has decreased to R by taking the logarithm of both sides: ⎛1⎞ ⎜ ⎟ ⎝2⎠ t/5730 yr = R 1.3 × 10−12 ⇒ t R ln (1.3 × 10−12 /R ) ⎛1⎞ ⎛ ⎞ ln ⎜ ⎟ = ln ⎜ t ⇒ = ⋅ 5730 yr ⎟ 5730 yr ⎝ ⎠ ln (2) ⎝ 1.3 × 10−12 ⎠ This sample has R = 1.65 × 10−13 , so its age is t= ln (1.3 × 10−12 /1.65 × 10−13 ) ⋅ 5730 yr = 17,100 yr ln (2) 42.47 Solve: From Appendix C, the half-life of a 133Cs sample is t1/2 = 30 years × 3.15 × 107 s = 9.45 × 108 s year The activity of a radioactive sample is R = rN The decay rate is 0.693 0.693 r= = = = 7.33 × 10−10 s −1 t1/2 τ 9.45 × 108 s ⇒N= That is, at t = s, we have 2.7 × 1017 137 R0 2.0 × 108 Bq = = 2.7 × 1017 atoms −10 −1 r 7.33 × 10 s Cs atoms and after a very long time all will have decayed Thus 2.7 × 1017 beta particles will have been emitted 42.48 Model: The number of radioactive tracers decays exponentially Solve: (a) The activity of a radioactive sample decays as ⎛1⎞ R = R0 ⎜ ⎟ ⎝2⎠ t /t1/2 16 hours/t1/2 ⎛1⎞ ⇒ 0.95 mCi = 1.15 mCi ⎜ ⎟ ⎝ 2⎠ ⎛ 95 ⎞ ⎛ 16 hours ⎞ ⎛ ⎞ ⇒ ln ⎜ ⎟ ln ⎜ ⎟ ⇒ t1/2 = 58.0 hours ⎟=⎜ ⎝ 115 ⎠ ⎝ t1/2 ⎠ ⎝ ⎠ © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-13 (b) Let t be the time it takes the activity to drop from 0.95 mCi to 0.10 mCi Using the results of part (a), ⎛1⎞ 0.10 mCi = 0.95 mCi ⎜ ⎟ ⎝ 2⎠ t /58.0 hours ⎞ t ⎛ 10 ⎞ ⎛ ⇒ ln ⎜ ⎟ = ⎜ ⎟ ln 95 58.0 hours ⎝ ⎠ ⎝ ⎠ ( 12 ) ⇒ t = 188 hours 42.49 Model: The decay is 223Ra → 219 Rn + 4He Solve: (a) The energy released in the above decay is E = [m( 223Ra) − m( 219Rn) − m( 4He)] × 931.49 MeV/u = [223.018499 u − 219.009477 u − 4.002602 u] × 931.49 MeV/u = 5.98 MeV That is, each α -particle is released with an energy of 5.98 MeV = 5.98 × 106 eV × 1.6 × 10−19 J/eV = 9.57 × 10−13 J The amount of energy needed to raise the temperature of 100 mL of water at 18°C to 100°C is ⎛ kg ⎞ Q = mcΔt = (0.10 L) ⎜ ⎟ (4190 J/kg K)(100 K − 18 K) = 34,400 J ⎝ 1L ⎠ The number of decays we need to generate this amount of energy is 34,400 J 9.57 × 10−13 J = 3.59 × 1016 The total number of radium atoms in the cube at t = s is N0 = 1g × 6.02 × 1023 atoms/mol = 2.70 × 1021 atoms 223 g/mol The number of needed decays is very small compared to N , so we can write dN ΔN ≈ = − rN dt Δt ⇒ Δt = − t ΔN ΔN ΔN = −τ = − 1/2 rN N ln N where we used r = 1/τ ΔN = −3.59 × 1016 , with the negative sign due to the fact that the number of radium atoms decreases Since t1/2 = 11.43 days = 9.876 × 105 s, the time for 3.59 × 1016 decays is Δt = − 9.876 × 105 s (−3.59 × 1016 ) = 18.9 s ≈ 19 s ln 2.70 × 1021 (b) It’s possible that an alpha-particle collision will break the molecular bond in a very small number of H 2O molecules, causing H gas and O gas to bubble out of the water The water that remains in the container has not changed or been altered 42.50 Model: The number of radioactive atoms decreases exponentially with time Solve: (a) If 90% of the sample has decayed, 10% is still present This happens at time t such that ⎛1⎞ N = 0.10 N = N ⎜ ⎟ ⎝2⎠ t/t1/2 t/t t ⎛ ⎞ 1/2 ⎛1⎞ ⇒⎜ ⎟ = 0.10 ⇒ ln ⎜ ⎟ = ln (0.10) t1/2 ⎝ ⎠ ⎝ 2⎠ ln (0.10) t= t1/2 = 3.32t1/2 ln (0.50) That is, 90% decay in 3.32 half-lives (b) When 99% of the sample has decayed, 1% is left The analysis is the same as in part (a), giving t= ln (0.010) t1/2 = 6.64t1/2 ln (0.50) That is, 99% decay in 6.64 half-lives © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 42-14 Chapter 42 42.51 Model: The number of radioactive atoms decays exponentially with time Solve: At t = hours, the number of A and B atoms is equal Thus ⎛1⎞ N A = (N A )0 ⎜ ⎟ ⎝2⎠ (2.0 h)/(0.5 h) = (N A )0 ⎛1⎞ = N B = (N B )0 ⎜ ⎟ 16 ⎝2⎠ (2.0 h)/t1/2 Initially, (N A )0 = 5(N B )0 With this information, ⎛1⎞ =⎜ ⎟ 16 ⎝ ⎠ (2.0 h)/t1/2 42.52 Model: The number of ⇒ ln (5/16) = 2.0 h ln (1/2) ln (1/2) ⇒ t1/2 = ⋅ h = 1.19 h ≈ 1.2 h t1/2 ln (5/16) 131 Ba and 47Ca atoms decays exponentially Solve: From Equation 42.8, the number of radioactive atoms is N = N 0e −t/τ At time t = s, (N )Ca = 2(N ) Ba The ratio of atoms at time t is N Ca (N )Ca e −t/τ Ca = N Ba (N ) Ba e −t/τ Ba Relating the half-life to the time constant, τ Ca = (t1/2 )Ca 4.5 days = = 6.492 days ln ln ⇒ τ Ba = (t1/2 ) Ba 12 days = = 17.31 days ln ln N Ca e −(2.5)7 days/6.492 days 2(0.0675) = (2) −(2.5)7 days/17.31 days = = 0.371 N Ba 0.3639 e 42.53 Model: The number of 40K atoms decays exponentially with time Solve: Suppose the number of 40 K atoms is N at the time the lava solidifies There are no 40 Ar atoms at that time As time passes, 11% of the 40 K that decays goes into 40 Ar, but the total number of atoms locked inside the lava is unchanged That is, N Ar = (0.11)(N , − N K ) where N K is the remaining number of 40 K atoms Dividing by N K , we have ⎛N ⎞ 0.013 N Ar N N = (0.11) ⎜ − 1⎟ ⇒ +1 = ⇒ K = = 0.894 NK 0.11 NK N 1.118 ⎝ NK ⎠ That is, 89.4% of the 40 K remains at a time when the Ar/K ratio is 0.013 The 40 K decays as t/ t t ⎛ ⎞ 1/2 N K = N0 ⎜ ⎟ ln (1/2) = ln (N K /N ) ⇒ t ⎝ ⎠ 1/2 ln(0.894) ⇒ t = (1.28 billion years) = 0.21 billion years = 210 million years ln(0.50) 42.54 Model: The number of Solve: The 235 235 U atoms decays exponentially with time U decays as ⎛1⎞ N U = (N U )0 ⎜ ⎟ ⎝2⎠ t/t1/2 The ratio of 235U atoms when the earth formed to the number now present is (N U )0 = = (2)t/t1/2 = (2)(4500 my)/(700 my) = 86.1 ≈ 86 NU (1/2)t/t1/2 That is, the abundance of 235U was 86 times larger then than it is now © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-15 42.55 Solve: Since the dose equivalent in Sv is the dose in Gy times the RBE, the dose is 0.30 × 10−3 Sv 1.00 J/kg = 3.53 × 10−4 Gy × = 3.53 × 10−4 J/kg 0.85 Gy Since only one-fourth of the body received x-ray exposure, the total amount of energy received is 3.53 × 10−4 J/kg × × 60 kg = 5.29 × 10−3 J The energy of each photon is 10 keV × 1.6 × 10−19 J = 1.6 × 10−15 J eV Thus, the number of x-ray photons absorbed by the body is 5.29 × 10−3 J 1.6 × 10 −15 J = 3.3 × 1012 42.56 Model: The rate of removal of a tracer is the rate of decay plus the rate of excretion Solve: The rate at which the number of tracer atoms in the body decays is dN ⎛ dN ⎞ ⎛ dN ⎞ =⎜ +⎜ = − rd N − re N = −(rd + re )N = − reff N ⎟ ⎟ dt ⎝ dt ⎠decay ⎝ dt ⎠excrete where rd and re are the decay rate and the excretion rate The effective loss rate is the sum of these two rates The effective half-life is t1/2 = ⎡ ln ln ln ⎤ = = =⎢ + ⎥ reff rd + re ln 2/(t1/2 )d + ln 2/(t1/2 )e ⎣ (t1/2 )d (t1/2 )e ⎦ −1 The effective half-life is the inverse of the sum of the inverses of the half-lives of the individual processes Note that this is exactly like finding the equivalent resistance of two parallel resistors Here the two removal processes run “in parallel.” Using the known values, ⎡ 1 ⎤ t1/2 = ⎢ + ⎥ 9.0 days 6.0 days ⎦ ⎣ −1 = 3.6 days 42.57 Model: The number of 239Pu atoms decays exponentially Solve: (a) The number of 239Pu atoms in a 1.0-μm-diameter particle is ρ ⎛ m ⎞ (ρV ) N =⎜ NA = ⎟ NA = MA ⎝ MA ⎠ = (19.8 kg/m ) ( ) (0.5 ×10 4π ( 43π ) r 3N A MA −6 m)3 (6.02 × 1023 mol−1 ) 239 × 10−3 kg/mol = 2.61 × 107 ≈ 2.6 × 107 (b) The activity of the particle is R= dN ln 0.6931 N= = rN = × 2.61 × 107 = 2.39 × 10−5 Bq ≈ 2.4 × 10−5 dt t1/2 24,000 years × 3.15 × 107 s/year (c) The volume of the 50-μm diameter sphere of tissue around the particle is 4π (25 × 10−6 m)3 = 6.545 × 10−14 m3 This volume of the tissue has a mass of m = ρV = (1000 kg/m3 )(6.545 × 10−14 m3 ) = 6.545 ì 1011 kg â Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 42-16 Chapter 42 In one year, the activity changes insignificantly Thus the number of decays per year is dN Δt = (2.39 × 10−5 Bq)(3.15 × 107 s) = 7.54 dt Since each decay creates an α -particle with energy 5.2 MeV, the total energy received per year by the tissue is (7.529 × 105 × 5.2 MeV) × (1.6 × 10−19 J/eV) = 6.264 × 10−10 J Dividing this energy by the tissue’s mass, the dose received by the tissue is 6.264 × 10−10 J/6.545 × 10−11 kg = 9.586 J/kg = 9.586 Gy The dose per year in mSv is (9.586 Gy)(RBE) = (9.586 Gy)(20) = 191.7 Sv ≈ 1.9 × 105 mSv (d) This is a very high dose to a very small volume of body mass Table 42.5 gives a typical exposure (in units of mrem/year) from various radiation sources The background radiation from various natural-occurring sources is about mSv/year The exposure to this tissue is much higher than the background level 42.58 Model: Although individual radon atoms decay, the radon is always being replaced such that the radon concentration remains constant with a constant activity of pCi/L Solve: (a) Activity is R = rN, so the number of atoms giving rise to a known activity is N = R/r The half-life of 222 Rn is 3.82 days = 330,000 s, so its decay rate is r= ln ln = = 2.1 × 10−6 s −1 t1/2 330,000 One liter is L = 1000 cm3 = 0.0010 m3 , so m3 of air is 1000 L At pCi/L, the activity in m3 is R = 4000 pCi = × 10−9 Ci × 3.7 × 1010 Bq = 148 Bq Ci Thus the number of 222Rn atoms in m3 of air is R 148 Bq = = 7.05 × 107 atoms ≈ × 107 atoms r 2.10 × 10−6 s −1 (b) We first need to find how many 222Rn atoms decay within cm of the person in year The person is a 25-cm-diameter cylinder 180 cm tall The decays must be inside a 31-cm-diameter cylinder (i.e., 25 cm + cm + cm) 183 cm tall The volume of air in this shell is N= V = π (0.155 m) (1.83 m) − π (0.125 m) (1.80 m) = 0.0498 m3 The concentration of 222 Rn is 7.05 × 107 m −3 , so the number of 222 Rn atoms in this shell is N = (0.0498 m3 ) × (7.05 × 107 m −3 ) = 3.51 × 106 The total number of decays per second is R = rN = (2.10 × 10−6 s −1 )(3.51 × 106 ) = 7.4 decays per second Only half of these decays direct the alpha particle toward the person, so the number of alphas striking the person in year is Nα = × (7.4 decays/s) × (3.15 × 107 s/yr) = 1.2 × 108 alphas/yr Each alpha deposits 5.50 MeV = 8.8 × 10−13 J of energy, so the energy absorbed in year is E = (1.2 × 108 alphas) × (8.8 × 10−13 J/alpha) = 1.06 × 10−4 J The dose of alpha radiation (with RBE = 20 ) received by a 65 kg person in a year is dose = 1.06 × 10−4 J Gy × × 20 = 3.3 × 10−2 mSv 65 kg 1.00 J/kg (c) This dose is roughly 1% of the natural background of radioactivity Although an exposure increase of 1% may seem too small to worry about, we assumed in using a 65 kg mass that the radiation is uniformly received But alpha radiation is not very penetrating, so the skin and outer extremities would receive a dose considerably higher than 2.4 mrem Since some people are housebound and really would receive this yearly dose, an upper limit of pCi/L seems prudent © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-17 42.59 Model: Assume the alpha particle is a point particle Solve: From Equation 18.3, the mean free path of a molecule moving through a gas is λ= 2π (N/V )r Treating the alpha particle as a point particle, the “collision cylinder” of radius 2r in the derivation of Equation 18.3 becomes a cylinder of radius r Since the radius is squared, the in the denominator of the expression vanishes In Example 18.1, the mean free path of nitrogen molecules in nitrogen gas is calculated to be 225 nm A point particle will have a mean free path times greater, that is, 900 nm The stopping distance of the alpha particle is the number of collisions times the mean free path The number of collisions is the energy of the alpha particle divided by the energy lost per collision The stopping distance is estimated to be 5.0 MeV × 900 nm = 1.667 × 105 × 900 × 10−9 m = 15 cm 30 eV 42.60 Model: The decay is AX Z → AYZ −1 + e+ + v Solve: (a) This decay is energetically possible only if the mass of the X nucleus exceeds the mass of the Y nucleus plus the mass of the positron The X nucleus contains Z protons and N neutrons while the Y nucleus has (Z − 1) protons and (N + 1) neutrons Thus the mass requirement is Zmp + Nmn > (Z − 1)mp + (N + 1)mn + me Tabulated masses are atomic masses, which include the electron masses Atom X has Z electrons and atom Y has (Z − 1) electrons Add the term Zme to both sides, obtaining Zmp + Zme + Nmn > (Z − 1)mp + Zme + (N + 1)mn + me ⇒ Z (mp + me ) + Nmn > [(Z − 1)(mp + me ) + (N + 1)mn ] + 2me The term on the left is the mass of an AX Z atom and the term in square brackets on the right is the mass of an A Yz −1 atom Thus the threshold condition for beta-plus decay is m( AX z ) > m( AYz −1) + 2me (b) From Appendix C, m(13 N) = 13.005738 u and m(13C) = 13.003355 u Thus m(13C) + 2me = 13.004455 u This is less than m(13 N), so beta-plus decay is allowed The energy released is E = Δmc = (0.001283 u)(931.49 MeV/u) = 1.20 MeV 42.61 Model: The number of 235U and 238U atoms decays exponentially with time Solve: From Appendix C, the half-life of 235 U is (t1/2 ) 235 = 7.04 × 108 yr and that of 238 U is (t1/2 ) 238 = 4.47 × 109 yr Today, N 238 /N 235 = 0.9928/0.0072 = 137.9, but it is thought that (N 238 )0 /(N 235 )0 ≈ immediately after the supernova that created these elements The decays are such that N 238 (N 238 )0 (1/2)t/(t1/2 )238 (1/2)t/(t1/2 )238 ⎛ ⎞ = ⇒ =⎜ ⎟ t t /( ) N 235 (N 235 )0 (1/2) 1/2 235 (1/2)t/(t1/2 )235 ⎝ ⎠ t/(t1/2 ) 238 −t/(t1/2 ) 235 = 137.9 To solve for t, take the logarithm of both sides: ⎡⎛ t ⎞ ⎡⎛ ⎞ ⎛ t ⎞ ⎤ ⎛ ⎞ ⎤ ⎢⎜ ⎟ −⎜ ⎟ ⎥ ln (1/2) = t × ⎢⎜ ⎟ −⎜ ⎟ ⎥ ln (1/2) = ln137.9 ⎢⎣⎝ t1/2 ⎠238 ⎝ t1/2 ⎠235 ⎥⎦ ⎢⎣⎝ t1/2 ⎠ 238 ⎝ t1/2 ⎠ 235 ⎥⎦ ln137.9 ⇒t = = 5.9 × 109 years ≈ billion years [(1/t1/2 )238 − (1/t1/2 )235 ] ln (1/2) The supernova occurred approximately billion years ago Assess: Our sun and solar system are ≈4.5 billion years old, so the supernova occurred ≈1.5 billion years before the debris coalesced to form the solar system © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 42-18 Chapter 42 42.62 Model: If we ignore the antineutrino, we can model beta decay as n → p + + e− Solve: (a) The mass difference is Δm = mn − mp − me = 1.00866 u − 1.00728 u − 0.00055 u = 0.00083 u This “mass loss” is transformed into the kinetic energy of the proton and electron, so K = (0.00083 u) × (931.40 MeV/u) = 0.773 MeV (b) The neutron is initially at rest, with γ n = 1, so the energy conservation equation is mn c = γ p mpc + γ e mec The c cancels, so we can also write the equation as Δm = mn − γ p mp − γ e me (c) The neutron momentum is zero, so the sum of the proton and electron momenta must be zero: = γ p mpvp − γ e meve or γ p mpvp = γ e meve (d) Using the definition of γ , γ= ⇒ v2 =1− = γ −1 c γ −1 ⇒v= ⇒ γ v = (γ − 1)1/2c γ γ γ c − v /c (e) Each side of the momentum equation has a γ v term Using the result of part (d), the momentum equation becomes (γ p2 − 1)1/2mpc = (γ e2 − 1)1/2mec ⇒ mp2 (γ p2 − 1) = me2 (γ e2 − 1) ⇒ mp2γ p2 − mp2 = me2γ e2 − me2 where, in the second step, we squared both sides of the equation We can use the energy equation to find meγ e in terms of γ p : meγ e = mn − mpγ p ⇒ me2γ e2 = (mn − mpγ p ) = mn2 − 2mn mpγ p + mp2γ p2 ⇒ me2γ e2 − me2 = mn2 − me2 − 2mn mpγ p + mp2γ p2 Substitute this expression for the right side of the momentum equation, giving mp2γ p2 − mp2 = mn2 − me2 − 2mn mpγ p + mp2γ p2 The term involving γ p2 cancels Solving for γ p then gives γp = mn2 + mp2 − me2 2mn mp = (1.00866 u)2 + (1.00728 u) − (0.00055 u) = 1.000000788 2(1.00866 u)(1.00728 u) With γ p known, γ p is found from the energy equation: γe = mn − γ p mp me = (1.00866 u) − (1.000000788)(1.00728 u) = 2.508 (0.00055 u) Finally, using the next-to-last step of part (d) to find v from γ , we obtain vp = γ p2 − c γp = 0.00126c and ve = γ e2 − c = 0.917c γe The electron is ejected at nearly the speed of light, but the proton’s speed is sufficiently small that we can treat the proton non-relativistically (f) The kinetic energies are K p = (γ p − 1)mp = (0.000000788)(1.00728 u) = 7.94 × 10−7 u × 931.59 MeV/u = 0.0007 MeV K e = (γ e − 1)me = (1.508)(0.00055 u) = 8.29 × 10−4 u × 931.59 MeV/u = 0.7722 MeV The total kinetic energy is K p + K e = 0.773 MeV, the same as we found in part (a) © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Nuclear Physics 42-19 (g) The proton has almost no recoil energy Since nucleons are bound inside the nucleus by many MeV, a recoil energy of 0.0007 MeV is far too small to eject the proton from the nucleus To say that the electron is ejected because it is moving so fast—0.917c—is close to being correct, but not quite Although the electron is ejected with ≈0.8 MeV of kinetic energy, the electric potential energy of an electron at the surface of a nucleus is typically −5 MeV A classical electron, even one moving at 0.917c, does not have enough kinetic energy to escape the attractive electric force of the nucleus However, the electron is not a classical particle A simple calculation finds that the de Broglie wavelength of an electron with v = 0.917c is ≈1000 fm This is ≈100 times the diameter of a nucleus The uncertainty principle prevents a particle from being confined in a region much smaller than its de Broglie wavelength Consequently, the electron cannot be confined within the nucleus 42.63 Model: A quantum particle can tunnel through a classically forbidden region Solve: (a) Kinetic energy is K = E − U The alpha particle has E = MeV whether it is inside or outside the nucleus Inside, where U = −60 MeV, the kinetic energy is Kin = 65.0 MeV Outside, where U = MeV, the kinetic energy is K out = 5.0 MeV (b) The particle’s speed is found from mv 2 = Kin ⇒ v = 2(65 × 106 eV)(1.60 × 10−19 J/eV) × (1.661 × 10 −27 = 5.60 × 107 m/s kg) (Although this is > 0.1c, the spirit of this simple model of alpha decay is best served with a classical calculation rather than a more complex relativistic calculation of the speed.) The time needed to move from one side of the potential well to the other, a distance of 15 fm, is 15 × 10−15 m Δt = = 2.7 × 10−22 s 5.60 × 107 m/s The particle collides with one wall or the other of the potential-energy barrier each time it moves across the potential well, so the rate of collisions is 1 = = 3.7 × 1021 collisions/s R= Δt 2.7 × 10−22 s per collision (c) From Chapter 41, the tunneling probability is Ptunnel = e −2 w/η , where w is the width of the barrier and η is the penetration distance into the classically forbidden region: η= = 1.05 × 10−34 J s = 2m(U − E ) 2(4 × 1.661 × 10−27 kg)(25 × 106 eV × 1.60 × 10−19 J/eV) = 4.55 × 10−16 m = 0.455 fm The probability of tunneling through the 20-fm-wide barrier is Ptunnel = e −2(20 fm)/(0.455 fm) = 6.6 × 10−39 (d) Although the probability of tunneling is extremely small, the alpha particle collides with the barrier a very large number of times per second The probability that the particle is still inside the nucleus after N collisions is Pin ≈ − NPtunnel The number of collisions needed to reduce Pin to 0.50, at which time half of a sample of nuclei would have decayed, is N= − Pin 0.50 = = 7.6 × 1037 collisions Ptunnel 6.6 × 10−39 At a collision rate of 3.7 × 1021 collisions/s, we can estimate the half-live to be t1/2 = 7.6 × 1037 collisions 21 3.7 × 10 collisions/s = 2.05 × 1016 s × year 3.15 × 107 s = 6.5 × 108 years = 650 million years Assess: The tunneling probability is very sensitive to the value of η If you round your results slightly differently and get a slightly different value of η, your answer for t1/2 could differ by more than 100 million years Even so, heavy nuclei that decay by the emission of alpha particles with K ≈ MeV usually have t1/2 of several hundred million years © Copyright 2013 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... understood calculations The solutions presented here are intended to provide a basis for this practice So that you may readily edit and/or post these solutions, they are available for download... Senior Development Editor: Alice Houston, Ph .D Senior Project Editor: Martha Steele Assistant Editor: Peter Alston Media Producer: Kelly Reed Senior Administrative Assistant: Cathy Glenn Director... Saturday After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday Absent-mindedly he missed his exit and stopped after one hour of driving