Modern Physics Kenneth S. Krane 3rd Edition Instructor Solution Manual, Exercises of Advanced Physics

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Instructor’s Manual

to accompany

Modern Physics, 3

rd

Edition

Kenneth S. Krane

Department of Physics

Oregon State University

ii

• Chapter 1…………………………………………………….. Table of Contents
• Chapter 2……………………………………………………
• Chapter 3…………………………………………………....
• Chapter 4…………………………………………………....
• Chapter 5…...……………………………………………….
• Chapter 6……………………………………………….….
• Chapter 7…………………………………………………..
• Chapter 8………………………………………………..…
• Chapter 9…………………………………………………..
• Chapter 10…………………………………………………
• Chapter 11…………………………………………..……..
• Chapter 12…………………………………………………
• Chapter 13………………………………………………....
• Chapter 14……………………………………………..…..
• Chapter 15…………………………………………..……..

Suggestions for Additional Reading

Some descriptive, historical, philosophical, and nonmathematical texts which give good background material and are great fun to read: A. Baker, Modern Physics and Anti-Physics (Addison-Wesley, 1970). F. Capra, The Tao of Physics (Shambhala Publications, 1975). K. Ford, Quantum Physics for Everyone (Harvard University Press, 2005). G. Gamow, Thirty Years that Shook Physics (Doubleday, 1966). R. March, Physics for Poets (McGraw-Hill, 1978). E. Segre, From X-Rays to Quarks: Modern Physicists and their Discoveries (Freeman, 1980). G. L. Trigg, Landmark Experiments in Twentieth Century Physics (Crane, Russak, 1975). F. A. Wolf, Taking the Quantum Leap (Harper & Row, 1989). G. Zukav, The Dancing Wu Li Masters, An Overview of the New Physics (Morrow, 1979).

Gamow, Segre, and Trigg contributed directly to the development of modern physics and their books are written from a perspective that only those who were part of that development can offer. The books by Capra, Wolf, and Zukav offer controversial interpretations of quantum mechanics as connected to eastern mysticism, spiritualism, or consciousness.

Materials for Active Engagement in the Classroom

A. Reading Quizzes

1. In an ideal gas at temperature T , the average speed of the molecules: (1) increases as the square of the temperature. (2) increases linearly with the temperature. (3) increases as the square root of the temperature. (4) is independent of the temperature.
2. The heat capacity of molecular hydrogen gas can take values of 3 R /2, 5 R /2, and 7 R / at different temperatures. Which value is correct at low temperatures? (1) 3 R /2 (2) 5 R /2 (3) 7 R /

Answers 1. 3 2. 1

B. Conceptual and Discussion Questions

1. Equal numbers of molecules of hydrogen gas (molecular mass = 2 u) and helium gas (molecular mass = 4 u) are in equilibrium in a container. (a) What is the ratio of the average kinetic energy of a hydrogen molecule to the average kinetic energy of a helium molecule? K (^) H / K (^) He = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/

(b) What is the ratio of the average speed of a hydrogen molecule to the average speed of a helium molecule? v H (^) / v (^) He = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4 (C) (c) What is the ratio of the pressure exerted on the walls of the container by the hydrogen gas to the pressure exerted on the walls by the helium gas? P H (^) / P He (^) = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/

2. Containers 1 and 3 have volumes of 1 m^3 and container 2 has a volume of 2 m^3. Containers 1 and 2 contain helium gas, and container 3 contains neon gas. All three containers have a temperature of 300 K and a pressure of 1 atm.

(a) Rank the average speeds of the molecules in the containers in order from largest to smallest. (1) 1 > 2 > 3 (2) 1 = 2 > 3 (3) 1 = 2 = 3 (4) 3 > 1 > 2 (5) 3 > 1 = 2 (6) 2 > 1 > 3 (b) In which container is the average kinetic energy per molecule the largest? (1) 1 (2) 2 (3) 3 (4) 1 and 2 (5) 1 and 3 (6) All the same

3. (a) Consider diatomic nitrogen gas at room temperature, in which only the translational and rotational motions are possible. Suppose that 100 J of energy is transferred to the gas at constant volume. How much of this energy goes into the translational kinetic energy of the molecules? (1) 20 J (2) 40 J (3) 50 J (4) 60 J (5) 80 J (6) 100 J (b) Now suppose that the gas is at a higher temperature, so that vibrational motion is also possible. Compared with the situation at room temperature, is the fraction of the added energy that goes into translational kinetic energy: (1) smaller? (2) the same? (3) greater?

Answers 1. (a) 4 (b) 3 (c) 4 2. (a) 2 (b) 6 3. (a) 4 (b) 1

He He^ Ne

3. One mole of N 2 gas (molecular mass = 28 u) is confined to a container at a temperature of 387 K. At this temperature, you may assume that the molecules are free to both rotate and vibrate. (a) What fraction of the molecules has translational kinetic energies within ±1% of the average translational kinetic energy? (b) Find the total internal energy of the gas.

Answers: 1. 11.25 2. 0.0066 3. (a) 0.0093 (b) 11.2 kJ

Problem Solutions

1. (a) Conservation of momentum gives px (^) ,initial = px ,final, or

m v H H,initial (^) + m He (^) v He,initial (^) = m v H H,final (^) + m He (^) v He,final

Solving for v He,final with v He,initial (^) = 0 , we obtain

H H,initial H,final He,final He 27 7 6 6 27

(1.674 10 kg)[1.1250 10 m/s ( 6.724 10 m/s)] 4.527 10 m/s 6.646 10 kg

m v v v m − −

× × − − ×

= = ×

×

(b) Kinetic energy is the only form of energy we need to consider in this elastic collision. Conservation of energy then gives K initial (^) = K final, or

1 2 1 2 1 2 1 2 2 m v H^ H,initial^^ +^2 m He^^ v He,initial^^ =^2 m v H^ H,final^^ + 2 m He^^ v He,final

Solving for v He,final with v He,initial (^) = 0 , we obtain

2 2 H H,initial H,final He,final He 27 7 2 6 2 6 27

(1.674 10 kg)[(1.1250 10 m/s) ( 6.724 10 m/s) ] 4.527 10 m/s 6.646 10 kg

m v v v m − −

× × − − ×

= = ×

×

2. (a) Let the helium initially move in the x direction. Then conservation of momentum gives: ,initial ,final He He,initial He He,final He O O,final O ,initial ,final He He,final He O O,final O

: cos cos : 0 sin sin

x x y y

p p m v m v m v p p m v m v

From the second equation,

27 6 He He,final He (^6) O,final (^26) O O

sin (^) (6.6465 10 kg)(6.636 10 m/s)(sin 84.7 ) 2.551 10 m/s sin (2.6560 10 kg)[sin( 40.4 )]

m v v m

− −

× × °

= − = − = ×

× − °

(b) From the first momentum equation,

5. (a) The kinetic energy of the electrons is

1 2 1 31 6 19 K i (^) 2 mv i (^) 2 (9.11 10 kg)(1.76 10 m/s) 14.11 10 J = = × −^ × = × −

In passing through a potential difference of Δ V = V f (^) − V i = +4.15 volts , the potential energy of the electrons changes by

Δ U = q Δ V = −( 1.602 × 10 −^19 C)( + 4.15 V) = −6.65 × 10 −^19 J

Conservation of energy gives K (^) i + U (^) i = K f (^) + U f, so

19 19 19 f i i f i

19 f^6 f (^31)

( ) 14.11 10 J 6.65 10 J 20.76 10 J

2 2(20.76 10 J)

2.13 10 m/s 9.11 10 kg

K K U U K U

K

v m

− − −

− −

= + − = − Δ = × + × = ×

×

= = = ×

×

(b) In this case Δ V = −4.15 volts, so Δ U = +6.65 × 10 −^19 J and thus

19 19 19 f i

19 f 6 f (^31)

14.11 10 J 6.65 10 J 7.46 10 J

2 2(7.46 10 J)

1.28 10 m/s 9.11 10 kg

K K U

K

v m

− − −

− −

= − Δ = × − × = ×

×

= = = ×

×

6. (a) Δ x (^) A = v Δ t (^) A = (0.624)(2.997 × 10 8 m/s)(124 × 10 −^9 s) =23.2 m

(b) (^) Δ x (^) B = v Δ t (^) B = (0.624)(2.997 × 108 m/s)(159 × 10 −^9 s) =29.7 m

7. With T = 35 C° = 308 Kand P = 1.22 atm = 1.23 × 105 Pa,

5 25 3

1.23 10 Pa 2.89 10 atoms/m (1.38 10 J/K)(308 K)

N P

V kT

×

= = = ×

×

so the volume available to each atom is (2.89 × 1025 /m^3 )−^1 = 3.46 × 10 −^26 m^3. For a spherical atom, the volume would be

4 3 4 10 3 30 3

3 π^ R 3 π^ (0.710^10 m)^ 1.50^10 m

= × −^ = × −

The fraction is then 30 5 26

− − −

×

= ×

×

8. Differentiating N ( E ) from Equation 1.22, we obtain

1 1/ 2^ /^ 1/ 2^ / 3/ 2 2

dN N (^) E e E kT (^) E e E kT

dE π kT kT

= ⎡^ −^ −^ + ⎛^ − ⎞ − ⎤

⎣ ⎝^ ⎠ ⎦

To find the maximum, we set this function equal to zero:

1/ 2 / 3/ 2

N (^) E e E kT E

π kT^ kT

− − ⎛^ − ⎞=

Solving, we find the maximum occurs at E = 12 kT. Note that E = 0 and E = ∞ also satisfy the equation, but these solutions give minima rather than maxima.

9. For this case kT = (280 K)(8.617 × 10 −^5 eV/K) = 0.0241 eV. We take dE as the width of the interval (0.012 eV) and E as its midpoint (0.306 eV). Then

1/ 2 (0.306 eV)/(0.0241 eV) 6 3/ 2

( ) (0.306 eV) (0.012 eV) 6.1 10 (0.0241 eV)

N

dN N E dE e N

= = −^ = × −

10. (a) From Eq. 1.31,

5 5 3 Δ E (^) int (^) = 2 nR Δ T = 2 (2.37 moles)(8.315 J/mol K)(65.2 K)⋅ = 3.21 10× J

(b) From Eq. 1.32,

5 7 3 Δ E (^) int (^) = 2 nR Δ T = 2 (2.37 moles)(8.315 J/mol K)(65.2 K)⋅ = 4.50 × 10 J

(c) For both cases, the change in the translational part of the kinetic energy is given by Eq. 1.29:

3 5 3 Δ E (^) int (^) = 2 nR Δ T = 2 (2.37 moles)(8.315 J/mol K)(65.2 K)⋅ = 1.93 × 10 J

11. After the collision, m 1 moves with speed v 1 ′ (in the y direction) and m 2 with speed v 2 ′

(at an angle θ with the x axis). Conservation of energy then gives E initial = E final:

1 2 1 2 1 2 2 2 2 2 m v 1 1^ =^2 m v 1 1^ ′ +^2 m v 2 2^ ′^ or^ v^ =^ v 1^ ′ +^3 v 2 ′

Conservation of momentum gives:

where v 2 may be positive or negative. The initial velocity v is

3 - 5

2 2(40.0 10 eV)(1.602 10 J/eV) 9.822 10 m/s (8.00 u)(1.6605 10 kg/u)

K

v m

× ×

= = = ×

×

The energy available to the two helium atoms after the decay is the initial kinetic energy of the beryllium atom plus the energy released in its decay:

1 2 1 2 1 2 1 2 K + 92.2 keV = 2 m v 1 1 (^) + 2 m v 2 2 (^) = 2 m v 1 1 (^) + 2 m 2 (^) (2 v − v 1 )

where the last substitution is made from the momentum equation. Solving this quadratic equation for v 1 , we obtain v 1 (^) = 2.47 × 106 m/s or − 0.508 × 106 m/s. Because we identified m 1 as the helium moving in the positive x direction, it is identified with the positive root and thus (because the two heliums are interchangeable in the equation) the second value represents the velocity of m 2 :

6 6 v 1 (^) = 2.47 × 10 m/s, v 2 = −0.508 × 10 m/s

(b) Suppose we were to travel in the positive x direction at a speed of v = 9.822× 105 m/s, which is the original speed of the beryllium from part (a). If we travel at the same speed as the beryllium, it appears to be at rest, so its initial momentum is zero in this frame of reference. The two heliums then travel with equal speeds in opposite directions along the x axis. Because they share the available energy equally, each helium has a kinetic energy of 46.1 keV and a speed of 2 K / m = 1.49 × 106 m/s, as we found in Problem 4. Let’s represent these velocities in this frame of reference as v 1 ′ = +1.49 × 10 6 m/s and v 2 ′= −1.49 × 106 m/s. Transforming back to the original frame, we find

6 5 6 1 1 6 5 6 2 2

1.49 10 m/s 9.822 10 m/s 2.47 10 m/s

1.49 10 m/s 9.822 10 m/s 0.508 10 m/s

v v v

v v v

= ′+ = × + × = ×

= ′+ = − × + × = − ×

14. (a) Let the second helium move in a direction at an angle θ with the x axis. (We’ll

assume that the 30° angle for m 1 is measured above the x axis, while the angle θ for

m 2 is measured below the x axis. Then conservation of momentum gives:

,initial ,final 1 1 2 2 1 2

,initial ,final 1 1 2 2 1 2

: cos 30 cos or 2 cos 2 1 : 0 sin 30 sin or sin 2

x x

y y

p p mv m v m v v v v

p p m v m v v v

We can eliminate the angle θ by squaring and adding the two momentum equations:

2 2 2 4 v + v 1 (^) − 2 3 vv 1 (^) = v 2

The kinetic energy given to the two heliums is equal to the original kinetic energy 1 2 2 mv^ of the beryllium plus the energy released in the decay:

1 2 1 2 1 2 1 2 1 2 2 2 2 1 1 2 2 2 2 1 1 2 2 1 1 1 2 2 1 2 2 1 2 1 2 1 2 2

92.2 keV (4 2 3 )

( ) 3 (2 92.2 keV) 0

mv m v m v m v m v v vv

m m v m vv m v mv

Solving this quadratic equation gives

6 6 v 1 (^) = 2.405 × 10 m/s, − 0.321 10× m/s

Based on the directions assumed in writing the momentum equations, only the positive root is meaningful. We can substitute this value for v 1 into either the

momentum or the energy equations to find v 2 and so our solution is:

6 6 v 1 (^) = 2.41 10× m/s, v 2 = 1.25 × 10 m/s

The angle θ can be found by substituting these values into either of the momentum

equations, for example 6 1 1 1 6 2

2.41 10 m/s sin sin 74. 2 2(1.25 10 m/s)

v v

θ −^ −^

×

×

(b) The original speed of the beryllium atom is v = 2 K / m = 1.203 × 106 m/s. If we

were to view the experiment from a frame of reference moving at this velocity, the original beryllium atom would appear to be at rest. In this frame of reference, in which the initial momentum is zero, the two helium atoms are emitted in opposite directions with equal speeds. Each helium has a kinetic energy of 46.1 keV and a

speed of v 1 ′ = v 2 ′ = 1.49 × 10 6 m/s. Let φ represent the angle that each of the helium

atoms makes with the x axis in this frame of reference. Then the relationship between the x components of the velocity of m 1 in this frame of reference and the original frame of reference is

v 1 cos 30 ° = v 1 ′cos φ+ v

and similarly for the y components

v 1 sin 30 ° = v 1 ′sin φ

We can divide these two equations to get

Chapter 2

This chapter presents an introduction to the special theory of relativity. It is written assuming that students have not yet seen a full presentation of then topic, even though they might have seen selected bits in their introductory courses. I have chosen not to introduce the speed parameter β = v / c and the Lorentz

factor γ = (1 − β 2 )−1/ 2, as I have found from practical experience in teaching the subject

that while they may render equations more compact they also can make an intimidating subject more obscure. Students seem more comfortable with equations in which the velocities appear explicitly. For similar reasons I have also chosen not to rely on spacetime (Minkowski) diagrams for the presentation of the space and time aspects of relativity, although in this edition I give a short introduction to their use in analyzing the twin paradox, where they do serve to enhance the presentation. Any presentation of special relativity offers the instructor an opportunity to dwell on elucidating the new ways of thinking about space and time engendered by the theory. Some references to the resulting logical paradoxes are listed below. In terms of applicability, however, the remainder of the textbook relies more heavily on the more straightforward applications of relativistic dynamics. The Lorentz transformation, for example, does not reappear beyond this chapter, nor does reference to clock synchronization. Relativistic time dilation and Doppler shift do appear occasionally. Approximately 2/3 of this chapter deals with aspects of space and time, while only 1/ deals with the more applicable issues of relativistic mass, momentum, and energy. In terms of division of class time, I try to divide the two topics more like 50/50, being especially careful to make sure that students understand how to apply momentum and energy conservation to situations involving high-speed motion.

Supplemental Materials

Time dilation: http://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/Flash/TimeDilation.html http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/lightclock.swf Physlet Quantum Physics , Section 2.

Length contraction: http://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/Flash/LengthContract.html http://science.sbcc.edu/physics/flash/relativity/LengthContraction.html

Simultaneity: http://faraday.physics.utoronto.ca/GeneralInterest/Harrison/SpecRel/Flash/Simultaneity.html http://science.sbcc.edu/physics/flash/relativity/Simultaniety.html

Twin paradox: http://webphysics.davidson.edu/physletprob/ch10_modern/default.html Physlet Quantum Physics, Sections 2.8 and 3.

Other relativity paradoxes: The best collection I know is that of Taylor and Wheeler’s Spacetime Physics (2nd edition, 1992). See especially space war (pp. 79-80), the rising manhole (p. 116), the pole and barn paradox (p. 166), and the detonator paradox (pp. 185-186).

Suggestions for Additional Reading

Special relativity has perhaps been the subject of more books for the nontechnical reader than any other area of science: L. Barnett, The Universe and Dr. Einstein (Time Inc., 1962). G. Gamow, Mr. Tompkins in Paperback (Cambridge University Press, 1967). L. Marder, Time and the Space Traveler (University of Pennsylvania Press, 1971). N. D. Mermin, It’s About Time: Understanding Einstein’s Relativity (Princeton University Press, 2009). B. Russell, The ABC of Relativity (New American Library, 1958). L. Sartori, Understanding Relativity: A Simplified Approach to Einstein’s Theories (University of California Press, 1996). J. T. Schwartz, Relativity in Illustrations (New York University Press, 1962). R. Wolfson, Simply Einstein: Relativity Demystified (Norton, 2003).

Gamow’s book takes us on a fanciful journey to a world where c is so small that effects of special relativity are commonplace. Other introductions to relativity, more complete mathematically but not particularly more difficult than the present level, are the following: P. French, Special Relativity (Norton, 1968). H. C. Ohanian, Special Relativity: A Modern Introduction (Physics Curriculum and Instruction, 2001) R. Resnick, Introduction to Special Relativity (Wiley, 1968). R. Resnick and D. Halliday, Basic Concepts in Relativity (Macmillan, 1992).

For discussions of the appearance of objects traveling near the speed of light, see: V. T. Weisskopf, “The Visual Appearance of Rapidly Moving Objects,” Physics Today, September 1960. I. Peterson, “Space-Time Odyssey,” Science News 137 , 222 (April 14, 1990).

Some other useful works are: L. B. Okun, “The Concept of Mass,” Physics Today, June 1989, p. 31. C. Swartz, “Reference Frames and Relativity,” The Physics Teacher, September 1989, p. 437. R. Baierlein, “Teaching E = mc^2 ,” The Physics Teacher, March 1991, p. 170.

Okun’s article explores the history of the “relativistic mass” concept and the connection between mass and rest energy. The article by Swartz gives some mostly classical descriptions of inertial and noninertial reference frames. Bayerlein’s article discusses some of the common misconceptions about mass and energy in special relativity.

6. In Tom's frame of reference, two events A and B take place at different locations along the x axis but are observed by Tom to be simultaneous. Which of the following statements is true? (Consider observer motion along the x axis only.) (1) No observers moving relative to Tom will find A and B to be simultaneous, but some may see A before B and others B before A. (2) No observers moving relative to Tom will find A and B to be simultaneous, but they all will observe events A and B in the same order. (3) All observers moving relative to Tom will also perceive A and B to be simultaneous. (4) Some observers moving relative to Tom will find A and B to be simultaneous, while others will not.
7. The quantity mc^2 represents: (1) the kinetic energy of a particle moving at speed c. (2) the energy of a particle of mass m at rest. (3) the total relativistic energy of a particle of mass m moving at speed c. (4) the maximum possible energy of a moving particle of mass m.
8. Which one of the following statements is true? (1) The laws of conservation of energy and momentum are not valid in special relativity. (2) The laws of conservation of energy and momentum are valid in special relativity only if we use definitions of energy and momentum that differ from those of classical physics. (3) According to special relativity, particles have energy only if they are in motion. (4) mc^2 represents the energy of a particle moving at speed c.

Answers 1. 3 2. 2 3. 2 4. 3 5. 3 6. 1 7. 2 8. 2

B. Conceptual and Discussion Questions

1. Rockets A and C move with identical speeds in opposite directions relative to B, who is at rest in this frame of reference. A, B, and C all carry identical clocks.

According to A: (1) B's clock and C's clock run at identical slow rates. (2) B's clock runs fast and C's clock runs slow. (3) B's clock runs slow and C's clock runs even slower. (4) B's clock runs fast and C's clock runs even faster. (5) B's clock runs slow and C's clock runs fast.

v B v

A C

2. Rockets A and C move with identical speeds v = 0.8 c in opposite directions relative to B, who is at rest in this frame of reference. A stick of length L 0 carried by A has length 0.6 L 0 according to B. What is the length of the stick according to C?

(1) L 0 (2) 0.6 L 0 (3) 0.36 L 0 (4) 0.22 L 0

3. Three identical triplets Larry, Moe, and Curly are testing the predictions of special relativity. Larry and Moe set out on round-trip journeys from Earth to distant stars. Larry's star is 12 light-years from Earth (as measured in the Earth reference frame), and he travels the round trip at a speed of 0.6 c. Moe's star is 16 light-years from Earth (also measured in the Earth frame), and he travels the round trip at a speed of 0.8 c. Both journeys thus take a total of 40 years, as measured by Curly who stays home on Earth. When Larry and Moe return, how do the ages of the triplets compare? (1) Larry = Moe > Curly (2) Moe < Larry < Curly (3) Larry < Moe < Curly (4) Larry = Moe = Curly (5) Moe > Larry > Curly (6) Larry = Moe < Curly
4. A star (assumed to be at rest relative to the Earth) is 100 light-years from Earth. (A light-year is the distance light travels in one year.) An astronaut sets out from Earth on a journey to the star at a constant speed of 0.98 c. (Note: At v = 0.98 c , 1 − v^2 / c^2 = 0.20) (a) How long does it take for a light signal from Earth to reach the star, according to an observer on Earth? (1) 100 y (2) 98 y (3) 102 y (4) 20 y (b) How long does it take for the astronaut to travel from Earth to the star, according to an observer on Earth? (1) 100 y (2) 98 y (3) 102 y (4) 20 y (c) According to the astronaut, what is the distance from Earth to the star? (1) 100 l.y. (2) 102 l.y. (3) 20 l.y. (4) 98 l.y. (d) According to the astronaut, how long does it take for the astronaut to travel from Earth to the star? (1) 100 y (2) 102 y (3) 20 y (4) 20.4 y (e) Light takes 100 years to travel from Earth to the star, but the astronaut makes the trip in 20.4 y. Does that mean that the astronaut travels faster than light? (1) Yes (2) No (3) Maybe
5. Two clocks, equidistant from O and at rest in the reference frame of O , start running when they receive a flash of light from the light source midway between them. According to O , the two clocks are synchronized (they start at the same time). According to O ′, who is moving with velocity u relative to O , clock 2 starts ahead of clock 1 by an amount Δ t ′.

v B^ v

A C