Tipler & Llewellyn Modern Physics 6th Solutions ISM

INSTRUCTOR SOLUTIONS MANUAL

Instructor Solutions Manual for

Modern Physics Sixth Edition Paul A. Tipler Ralph A. Llewellyn

Prepared by

Mark J. Llewellyn Department of Electrical Engineering and Computer Science Computer Science Division University of Central Florida

W. H. Freeman and Company New York

Instructor Solutions Manual to Accompany Tipler & Llewellyn Modern Physics, Sixth Edition © 2012, 2008, 2003 by W.H. Freeman and Company All rights reserved. Published under license, in the United States by W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com

Preface This book is an Instructor Solutions Manual for the problems which appear in Modern Physics, Sixth Edition by Paul A. Tipler and Ralph A. Llewellyn. This book contains solutions to every problem in the text and is not intended for class distribution to students. A separate Student Solutions Manual for Modern Physics, Sixth Edition is available from W. H. Freeman and Company. The Student Solutions Manual contains solutions to selected problems from each chapter, approximately one-fourth of the problems in the book. Figure numbers, equations, and table numbers refer to those in the text. Figures in this solutions manual are not numbered and correspond only to the problem in which they appear. Notation and units parallel those in the text. Please visit W. H. Freeman and Company’s website for Modern Physics, Sixth Edition at www.whfreeman.com/tiplermodernphysics6e. There you will find 30 More sections that expand on high interest topics covered in the textbook, the Classical Concept Reviews that provide refreshers for many classical physics topics that are background for modern physics topics in the text, and an image gallery for Chapter 13. Some problems in the text are drawn from the More sections. Every effort has been made to ensure that the solutions in this manual are accurate and free from errors. If you have found an error or a better solution to any of these problems, please feel free to contact me at the address below with a specific citation. I appreciate any correspondence from users of this manual who have ideas and suggestions for improving it.

Sincerely, Mark J. Llewellyn Department of Electrical Engineering and Computer Science Computer Science Division University of Central Florida Orlando, Florida 32816-2362 Email: [email protected]

Table of Contents

Chapter 1 – Relativity I

1

Chapter 2 – Relativity II

31

Chapter 3 – Quantization of Charge, Light, and Energy

53

Chapter 4 – The Nuclear Atom

79

Chapter 5 – The Wavelike Properties of Particles

109

Chapter 6 – The Schrödinger Equation

127

Chapter 7 – Atomic Physics

157

Chapter 8 – Statistical Physics

187

Chapter 9 – Molecular Structure and Spectra

209

Chapter 10 – Solid State Physics

235

Chapter 11 – Nuclear Physics

259

Chapter 12 – Particle Physics

309

Chapter 13 – Astrophysics and Cosmology

331

Chapter 1 – Relativity I 1-1.

(a) Speed of the droid relative to Hoth, according to Galilean relativity, uHoth , is uHoth  uspaceship  udroid  2.3 108 m / s  2.1108 m / s  4.4 108 m / s (b) No, since the droid is moving faster than light speed relative to Hoth.

1-2.

4 2 L 2  2.74 10 m  (a) t    1.83 104 s 8 c 3.00 10 m / s

(b) From Equation 1-6 the correction  t 

2L v 2  c c2

 t  1.83 104 s 104   1.83 1012 s 2

c

4 km / s  1.3 105 c 299, 796 km / s No, the relativistic correction of order 10-8 is three orders of magnitude smaller than

(c) From experimental measurements



the experimental uncertainty.

0.4 fringe



1.0 fringe

 v2 

1.0 2  29.9 km / s   2.22 103  v  47.1 km / s 0.4

1-3.

 29.8km / s 

1-4.

(a) This is an exact analog of Example 1-1 with L = 12.5 m, c = 130 mph, and v = 20

2

 v km / s 

2

mph. Calling the plane flying perpendicular to the wind plane #1 and the one flying parallel to the wind plane #2, plane #1 win will by Δt where

Lv 2 12.5mi  20mi / h  t  3   0.0023 h  8.2s 3 c 130mi / h  2

(b) Pilot #1 must use a heading   sin 1  20 /130   8.8 relative to his course on both legs. Pilot #2 must use a heading of 0 relative to the course on both legs.

Chapter 1 – Relativity I 1-5.

(a) In this case, the situation is analogous to Example 1-1 with L  3 108 m,

v  3 104 m / s, and c  3 108 m / s If the flash occurs at t = 0, the interior is dark until t =2s at which time a bright circle of light reflected from the circumference of the great circle plane perpendicular to the direction of motion reaches the center, the circle splits in two, one moving toward the front and the other moving toward the rear, their radii decreasing to just a point when they reach the axis 10-8 s after arrival of the first reflected light ring. Then the interior is dark again. (b) In the frame of the seated observer, the spherical wave expands outward at c in all directions. The interior is dark until t = 2s at which time the spherical wave (that reflected from the inner surface at t = 1s) returns to the center showing the entire inner surface of the sphere in reflected light, following which the interior is dark again.

1-6.

Yes, you will see your image and it will look as it does now. The reason is the second postulate: All observers have the same light speed. In particular, you and the mirror are in the same frame. Light reflects from you to the mirror at speed c relative to you and the mirror and reflects from the mirror back to you also at speed c, independent of your motion.

1-7.

N 

2 Lv 2 (Equation 1-10) Where λ = 590 nm, L = 11 m, and ΔN = 0.01 fringe c2

v2 

2 N  c 2   0.01 fringe   590 109 m  3.00 108 m / s  2 11m  2L

v  4.91103 m / s  5 km / s

1-8.

(a) No. Results depends on the relative motion of the frames. (b) No. Results will depend on the speed of the proton relative to the frames. (This answer anticipates a discussion in Chapter 2. If by “mass”, the “rest mass” is implied, then the answer is “yes”, because that is a fundamental property of protons.)

2

Chapter 1 – Relativity I (Problem 1-8 continued) (c) Yes. This is guaranteed by the 2nd postulate. (d) No. The result depends on the relative motion of the frames. (e) No. The result depends on the speeds involved. (f) Yes. Result is independent of motion. (g) Yes. The charge is an intrinsic property of the electron, a fundamental constant.

1-9.

The wave from the front travels 500 m at speed c + (150/3.6) m/s and the wave from the rear travels at c – (150/3.6) m/s. As seen in Figure 1-14, the travel time is longer for the wave from the rear. t  tr  t f 

500m 500m  8 3.00 10 m / s  150 / 3.6  m / s 3.00  10 m / s  150 / 3.6  m / s 8

 3 108  150 / 3.6  3 108  150 / 3.6         500   3 108   2 150 / 3.6   3 108   150 / 3.6 2     500

2 150 / 3.6 

 3 10 

8 2

 4.63  1013 s

*

1-10.

A

*

B

*

v

C

While the wavefront is expanding to the position shown, the original positions of A, B, and C have moved to the * marks, according to the observer in S.

(a) According to an S  observer, the wavefronts arrive simultaneously at A and B . (b) According to an S observer, the wavefronts do not arrive at A and C simultaneously. (c) The wavefront arrives at A first, according to the S observer, an amount Δt before arrival at C  , where

3

Chapter 1 – Relativity I (Problem 1-10 continued)

t 

BC  BA  cv cv

since BC   BA  L, Thus

c  v  c  v   2v  t  L   L  2 2  2 2   c v  c  v  1-11. β

  1/ 1   2 

1/ 2

0 0.2 0.4 0.6 0.8 0.85 0.90 0.925 0.950 0.975 0.985 0.990 0.995

1-12.

vx   t1    t1  20  c  

1 1.0206 1.0911 1.2500 1.6667 1.8983 2.2942 2.6318 3.2026 4.5004 5.7953 7.0888 10.0125 vx   t2    t2  20  c  

( from Equation 1-19)

vx vx   (a) t2  t1    t2  20  t1  20     t2  t1  c c  

(b) The quantities x1 and x2 in Equation 1-19 are each equal to x0 , but x1 and x2 in Equation 1-18 are different and unknown.

1-13. (a)

  1/ 1  v 2 / c 2 

1/ 2

 1/ 1   0.85c  / c 2    2

1/ 2

 1.898

x    x  vt   1.898 75m   0.85c   2.0  105 s    9.537  103 m y  y  18m z  z  4.0m t     t  vx / c 2   1.898  2.0 105 s   0.85c  75m  / c 2   3.756  105 s

4

Chapter 1 – Relativity I (Problem 1-13 continued) (b)

x    x  vt    1.898  9.537 103 m   0.85c   3.756 10 5 s    75.8m difference is due to rounding of  , x, and t . y  y  18m z  z  4.0m t    t   vx / c 2   1.898 3.756 105 s   0.85c   9.537 103 m  / c 2   2.0 10 5 s

1-14. To show that Δt = 0 (refer to Figure 1-8 and Example 1-1). t1 

L c v 2

2



L c v 2

2



2L c v 2

2



2L 1 c 1  v2 / c2

t2 , because length parallel to motion is shortened, is given by: t2 

L 1  v 2 / c 2 L 1  v 2 / c 2 2 Lc 1  v 2 / c 2   2 cv cv c 1  v 2 / c 2 

t2 

2L c



1  v2 / c2 1  v2 / c2



2



2L 1  t1 c 1  v2 / c2

Therefore, t2 – t1 = 0 and no fringe shift is expected. 1-15. (a) Let frame S be the rest frame of Earth and frame S  be the spaceship moving at speed v to the right relative to Earth. The other spaceship moving to the left relative to Earth at speed u is the “particle”. Then v = 0.9c and ux = −0.9c. u x  

ux  v 1  ux v / c 2

(Equation 1-22)

0.9c  0.9c 1.8c   0.9945c 2 1.81 1   0.9c  0.9c  / c

(b) Calculating as above with v  3.0 104 m / s  ux ux  1

3.0  104 m / s  3.0  104 m / s 6.0  104 m / s   6.0  104 m / s 8 4 4 1  10  3.0  10 m / s  3.0  10 m / s 

 3.0 10 m / s  8

2

5

Chapter 1 – Relativity I

1-16.

ax 

dux dt 

where u x 

And t     t  vx / c 2 

ux  v 1  ux v / c 2

(Equation 1-22)

(Equation 1-18)

dux   ux  v   vdu x / c 2 1  u x v / c 2   1  u x v / c 2  du x 2

1

 v  2  2   ux  v  du x  1  u x v / c  du x c   2 1  ux v / c2 

dt     dt  vdx / c 2   v  u  v  du x / dt   1  u x v / c 2   du x / dt   2  x  du c ax  x    3 dt   1  u v / c 2  x

 dux / dt  1  v 2 / c 2 



 1  u x v / c

ay 

du y dt 





2 3

where u y 

ax

 3 1  u x v / c 2  uy

3

(Equation 1-22)

 1  ux v / c 2 

du y   du y /   1  u x v / c 2    u y /   1  u x v 2 / c 2  du x 1

2

 du  1  u v / c    u v / c  du   1  u v / c  2

y

2

x

y

x

2 2

x

2 2 duy  du y / dt  1  u x v / c    u y v / c   du x / dt  ay   2 dt   1  u v / c 2   1  u v / c 2  x



x

a y 1  u x v / c 2   ax  u y v / c 2 

 2 1  u x v / c 2 

3

az is found in the same manner and is given by: az 

6

az 1  ux v / c 2   ax  u z v / c 2 

 2 1  ux v / c 2 

3

Chapter 1 – Relativity I 1-17. (a) As seen from the diagram, when the observer in the rocket ( S  ) system sees 1 c∙s tick by on the rocket’s clock, only 0.6 c∙s have ticked by on the laboratory clock. ct

ct  4

x 3

2 1 1

0 0

1

2

3

4

x

(b) When 10 seconds have passed on the rocket’s clock, only 6 seconds have passed on the laboratory clock.

1-18. (a)

y

u x  0 u y  0 c

y

x

v x

ux 

uy 

(b)

u x  v 0v  v 2 1 0 1  vu x / c u y

 1  vu x / c

2





(Equation 1-23)

c c   1  0  

u  ux2  u y2  v 2  c 2 /  2  v 2  c 2 1  v 2 / c 2   c 7

Chapter 1 – Relativity I 1-19. By analogy with Equation 1-23,

1-20

(a) ux 

ux  v 0.9c  0.9c 1.8c    0.9945c 2 2 1  vux / c 1   0.9c  0.9c  / c 1.81

(b) ux 

1.8c /1.81  0.9c  1.8   0.9 1.81 c  3.429 c  0.9997c ux  v  2 1  vux / c 1  1.8c /1.81 0.9c  / c 2 1.81  1.8  0.9  3.430 1



(a)

(Equation 1-16)

1  v2 / c2

 1  v / c 2

2  1  v  1     2  2  c



2 1/ 2

1 v2 3 v4  1   2 c2 8 c2

1

(b)



2

  1  3  1  v 2           2     2  2  2!  c 

1 v2  1 2 c2

 1  v 2 / c 2  1  v 2 / c 2 

1/ 2

2

2 2  1   v   1  1  1  v   1      2         2    2   c   2  2  2!  c 

 1

(c)

1 v2 1 v4   2 c2 8 c2

 1 

t  t 

1 v2 2 c2

1 v2 3 v4   2 c2 8 c4

  1  1

1-21.

 1

1





1

1





1 v2 1 v4  2 c2 8 c4

1 v2 2 c2

(Equation 1-26)

t  t  t   t  1 v2 1 v2    1     1  t  t  2 c2 2 c2 t  t  v  2c t  2

2

t  t    v  c 2  t   

1/ 2

8

 c  2  0.01

1/ 2

 0.14c

Chapter 1 – Relativity I S

1-22.

Orion (#2)

v = 1000 km/h

Lyra (#1)

S Earth

2.5 103 c y

2.5 103 c y

(a) Note that   1/ 1  v 2 / c 2  1 and 1 c y  c  3.15 107 s  From Equation 1-27: t  t2  t1  0 , since the novas are simultaneous in system S (Earth). Therefore, in S  (the aircraft) v t   t2  t1   2  x2  x1  c 6 10 m / h  2.5 103  2.5 103   c   3.15 107 s  2   3600s / h  c

 1.46 105 s  40.5h (b) Since t  is positive, t2  t1 ; therefore, the nova in Lyra is detected on the aircraft before the nova in Orion.

1-23. (a)

L  Lp / 

(Equation 1-28)

=Lp 1  v 2 / c 2  1.0m 1   0.6c  / c 2    2

1/ 2

 0.80m

(b) t  L / v  0.80m / 0.6c  4.4 109 s (c)

The projection OA on the x axis is L. The length OB on the ct axis yields t.

ct  ct back of meterstick passes x = 0

x B

meterstick

AA 0

x

9

Chapter 1 – Relativity I t 

t  t  

1-24. (a)



(Equation 1-26)

1  v2 / c2 2.6  108 s

1   0.9c  / c 2    2

1/ 2



2.6 108 s 0.19

 5.96  108 s

(b) s  vt   0.9   3.0 108 m / s  6.0 108 s   16.1m (c) s  vt   0.9   3.0 108 m / s  2.6 108 s   7.0m (d)

 s 

2

  ct    x  2

2

(Equation 1-31)

2  c  6.0  108    16.1m   324  259  65  s  7.8m 2

1-25. From Equation 1-28, L  Lp /   Lp 1  v 2 / c 2

where L  85m and Lp  100m

1  v 2 / c 2  L / Lp  85 /100 Squaring 1  v 2 / c 2  85 /100 

2

2  v 2  1   85 /100   c 2  0.2775c 2 and v  0.527c  1.58  108 m / s  

1-26. (a) In the spaceship the length L = the proper length Lp; therefore, ts 

Lp c



Lp c



2 Lp c

(b) In the laboratory frame the length is contracted to L  Lp /  and the round trip time is

tL  

L L L   c  v c  v c 1  v 2 / c 2  2 Lp / 

c 1  v 2 / c 2 



2 Lp 1  v 2 / c 2 c



1  v2 / c2



2



2 Lp c 1  v2 / c2

(c) Yes. The time ts measured in the spaceship is the proper time interval τ. From time dilation (Equation 1-26) the time interval in the laboratory tL   ; therefore, tL 

1

2 Lp

1  v2 / c2

c

which agrees with (b).

10

Chapter 1 – Relativity I

1-27. Using Equation 1-28, with LAp and LBp equal to the proper lengths of A and B and LA = length of A measured by B and LB = length of B measured by A.

LA  LAp /   100m 1   0.92c   39.2 m 2

LBp   LB  36 / 1   0.92c  / c 2  91.9m 2

1-28.

In S  : x  1.0m cos30  0.866m y   1.0m sin 30  0.500m where    30

In S : x  x 1   2  0.866m 1   0.8   0.520m 2

y  y   0.500m 0.500 where   tan 1  43.9 0.520 L

1-29. (a)

 x 

2

  y   2

 0.520m 

2

  0.500   0.721m 2

In S  : V   a  b  c   2m  2m  4m   16m3 In S: Both a and c have components in the x direction. ax  a sin 25   2m  sin 25  0.84m and cx  c cos 25   4m  cos 25  3.63m ax  ax 1   2  0.84 1   0.65   0.64m 2

cx  cx 1   2  3.634 1   0.65   2.76m 2

a y  ay  a cos 25  2 cos 25  1.81m and c y  cy  c sin 25  4sin 25  1.69m a  ax2  a y2 

 0.64   1.81

c  cx2  c y2 

 2.76   1.69 

2

2

2

 1.92m

2

 3.24m

b (in z direction) is unchanged, so b  b  2m

 (between c and xy-plane)  tan 1 1.69 / 2.76   31.5  (between a and yz-plane)  tan 1  0.64 /1.81  19.5 V  (area of ay face) b (see part[b]) V   c  a sin 78   b   3.24m 1.92m sin 78  2m   12.2m3

11

Chapter 1 – Relativity I (Problem 1-29 continued) (b) y

a

b 19.5◦

c

x

31.5◦

z

1-30.

 

c  f

c 1  1 

fo



2

   1 v / c     o  1  v / c

1 v / c o (Equation 1-36) 1 v / c 2

      1  v / c   1  v / c  o 

2 2 2     v     v 1     / o     1  1      Solving for v / c, c  o  c 1     / o 2   o   2 v 1   590nm / 650nm  o  650nm. For yellow    590nm.   0.097 c 1   590nm / 650nm 2

Similarly, for green    525nm  and for blue    460nm 

v  0.210 c

v  0.333 c

12

Chapter 1 – Relativity I 1-31.

 

c c 1 v / c   o f 1 v / c 1  fo 1 

(Equation 1-37)

1  1.85  10 m / s  /  3.00 10 m / s      o   1 v / c   1  1   8 o o 1 v / c  1  1.85  10 m / s  3.00 10 m / s   7



8

1/ 2

 1  0.064



1-32. Because the shift is a blue shift, the star is moving toward Earth. f 

1  f o where f  1.02 f o 1 

1.02 

2

1.02   1  0.0198  2 1.02   1 2

1   1 



v  0.0198c  5.94 106 m / s

1-33.

f 

1  fo 1 

 

1  1  o   656.3nm  1  1 

For   103 :    656.3nm 

1  103  657.0nm 1  103

For   10 :    656.3nm 

1  102  662.9nm 1  102

2

  101 :    656.3nm 

1  101  725.6nm 1  101

1-34. Let S be the rest frame of Earth, S  be Heidi’s rest frame, and S  be Hans’ rest frame. 1  S   1.1198 2 1   0.45c / c 

 S  

1 1   0.95c / c 

2

 3.206

When they meet, each will have traveled a distance d from Earth. Heidi: d  0.45c tHeidi Hans: d  0.95c tHans and tHans  tHeidi  1 in years.

Therefore,  0.95c  0.45c  tHeidi  0.95c and tHeidi  1.90 y; tHans  0.90 y (a) In her reference frame S  , Heidi has aged t  when she and Hans meet.

13

Chapter 1 – Relativity I (Problem1-34 continued)

v 0.45c     t    S   t  2 x   1.1198 1.90  2  0.45c 1.90   c c     2  1.198 1.90 y  1   0.45   1.697 y  

In his reference frame S  , Hans has aged t  when he and Heidi meet.

v 2   t    S   t  2 x   3.3026 1.90 y  1   0.95   0.290 y   c   The difference in their ages will be 1.697  0.290  1.407 y  1.4 y (b) Heidi will be the older. 1-35. Distance to moon = 3.85 108 m  R Angular velocity ω needed for v = c:

  v / R  C / R   3.00 108 m / s  /  3.85 108 m   0.78rad / s Information could only be transmitted by modulating the beam’s frequency or intensity, but the modulation could move along the beam only at speed c, thus arriving at the moon only at that rate.

1-36. (a) Using Equation 1-28 and Problem 1-20(b).

t   t /   t 1  v 2 / 2c 2   t  tv 2 / 2c 2 where t  3.15  107 s / y

v  2 RE / T   2   6.37 106 m  / 108min  60 s / min  v  6.177  103 m / s  2.06  105 c Time lost by satellite clock = tv 2 / c 2   3.15 10 7 s  2.06  105  / 2  0.00668s  6.68ms 2

(b)

1s  t  v 2 / 2c 2  t  2 /  v 2 / c 2   2 /  2.06 105   4.71109 s  150 y 2

14

Chapter 1 – Relativity I cos     (Equation 1-41) 1   cos   where    half-angle of the beam in S   30 cos 30  0.65 For   0.65, cos    0.97 or   14.1 1   0.65  cos 30

1-37.

cos  

A

0.75m 1.5m lamp

14.1 beam

The train is A from you when the headlight disappears, where

A

0.75m  3.0m tan14.1

1-38. (a) t  t0 For the time difference to be 1s, t  t0  1s

 1 t  t /   1  t 1    1   1 1 v2 Substituting  1  (From Problem 1-20)  2 c2 3.0 108    1 v2  2 2 t 1  1    1  t  2c / v  2 2 2 c2   1.5 106 / 3.6 103  2

 1.04 1012 s  32, 000 y (b)

t  t0  273  109 s. Using the same substitution as in (a). t 1  1/    273 109 and the circumference of Earth C  40, 000km, so 4.0 107 m  vt or t  4.0 107 / v, and 4.0 10 / v   2c / v 7

2

2

 273 10  , or v  9

2c 2  273 109  4.0 107

 1230 m / s

Where v is the relative speed of the planes flying opposite directions. The speed of each plane was 1230m / s  / 2  615 m / s  2210 km / h  1380 mph.

15

Chapter 1 – Relativity I

1-39. y

y

S (other)

c

S  (Earth)

v

θ

x

x (a)

cx  c cos   v c y  c sin  tan   

c y cx



c sin  sin   c cos   v cos   v / c

(b) If   90 , tan   

1 1  1 v/c 

 90     45 e.g., if v  0.5c,    63

1-40. (a) Time t for information to reach front of rod is given by:

ct 

Lp



 vt  t 

Lp

 (c  v )

Distance information travels in time t:

c (c  v)(c  v) / c 2 c 1  v2 / c2 cv ct   Lp  Lp  Lp 2  (c  v ) (c  v ) cv (c  v ) cLp

Since

(c  v) (c  v)  1 for v > 1, the distance the information must travel to reach

the front of the rod is  Lp ; therefore, the rod has extended beyond its proper length. (b)



1 1 v / c (ct  Lp )  1 Lp 1 v / c

16

Chapter 1 – Relativity I

Δ

0

0.11

0.29

0.57

0.73

1.17

2.00

2.50

3.36

5.25

8.95

v/c

0

0.10

0.25

0.40

0.50

0.65

0.80

0.85

0.90

0.95

0.98

Coefficient of extension vs. v/c 10

9

8

7

Delta

6

5

4

3

2

1

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

v/c

(c) As v  c,    , the maximum length of the rod   also.

1-41. (a) Alpha Centauri is 4 c∙y away, so the traveler went L  1   2 8c  y  in 6 y, or

8c  y 1  v 2 / c 2  v  6 y  1  v 2 / c 2  v  6 / 8c    3 / 4  v / c   1   2  3 / 4  2 2

2 3 / 4  2   2  1

 2  1/ 1  0.5625  v  0.8c (b)

t  t0    6 y  and   1/ 1   2  1.667 t  1.667  6 y   10 y or 4y older than the other traveler.

17

Chapter 1 – Relativity I (Problem 1-41 continued)

(c)

ct 

10

ct (c∙y)

8

x 6

4

2

Alpha Centauri

0

Earth

0

2

4

6

8

10

x

1-42. Orbit circumference  4.0 107 m . Satellite speed v  4.0 107 m /  90 min  60s / min   7.41103 m / s t  t0  tdiff 1  t  t /   tdiff  t 1  1 /    t   2  (Problem 1-20) 2  tdiff   3.16  107 s  1 / 2   7.41  103 / 3.0  108   0.0096s  9.6ms

1-43. (a) t   t 

v c2

x (Equation 1-20)

For events to be simultaneous in S , t   0.

v

v 1.5 103 m  2  c c 6 8 v   2 10 s  3 10 m / s  s /1.5 103 m

t 

2

x  2 106 s 

 1.2 108 m / s  0.4c

(b) Yes.

18

2

Chapter 1 – Relativity I (Problem 1-43 continued) ct 

ct

(c)

1000

x

500

B

0 500

1000

1500

x

0

A

(d)

 s 

2

  x    ct  2

2

(Equation 1-33)

 1.5  103    3  108 m / s  2  106 s    2.25  106  3.6  105  1.89  106 m2 s  1370m L  s  1370m 2

2

1-44. (a)   1/ 1   2  1/ 1   0.92   2.55 2

(b)   2.6 108 s

t  lab     2.55  2.6 108 s   6.63 108 s

(c) N  t   N0 et /  (Equation 1-29) L  1   2 L0  1   0.92   50m   19.6m 2

Where L is the distance in the pion system. At 0.92c, the time to cover 19.6m is: t  19.6m / 0.92c  7.0 108 s. So, for N0 = 50,000 pions initially, at the end of 50m

in the lab, N   5.0 104  e7.0/ 2.6  3,390 (d) 47

19

Chapter 1 – Relativity I

1-45.

 1 v2  L  Lp  L  Lp  Lp 1/    Lp 1  1/    Lp  2  (See Problem 1-20) 2c  For Lp  11m and v  3 104 m / s L  11 0.5 108   5.5 108 m "Shrinkage" 

5.5 108 m  550 atomic diameter 1010 m / atomic diameter

1-46. (a)

ct 

ct 1500

B

1000

500

A 0 0

500

1000

1500

2000

x

(b) Slope of ct  axis  2.08  1/  , so   0.48 and v  1.44 108 m / s (c)

ct    ct and   1/ 1   2 so ct  1   2  ct ct  877m t   1.5  877  / c  4.39 s

For ct   1000m and   0.48

(d) t  t   1.14t   t   5 s /1.14  4.39 s

1-47. (a) L  Lp /   Lp 1  u 2 / c 2  100m 1   0.85  52.7m 2

(b) u 

u u 0.85c  0.85c 1.70c    0.987c 2 2 1  uu / c 1.72 1   0.85

(c) L  Lp /    Lp 1  u2 / c2  100m 1  1.70 /1.72   16.1m 2

(d) As viewed from Earth, the ships pass in the time required for one ship to move its L 52.7m  2.1107 s own contracted length. t   8 u 0.85  3.00 10 m / s

20

Chapter 1 – Relativity I (Problem 1-47 continued) 0.987c

(e)

100 m

16 m

1-48. In Doppler radar, the frequency received at the (approaching) aircraft is shifted by approximately f / f0  v / c. Another frequency shift in the same direction occurs at the receiver, so the total shift f / f0  2v / c.

v   c / 2  8 107   120m / s .

1-49. #1

#2 For star #1: v  32km / s  3.2 104 m / s

cm

period  115d

L

c  c  3.2 104 c+

c−

Earth

c  c  3.2  104 Simultaneous images of star#1 in opposition will appear at Earth when L is at least as large as:

21

Chapter 1 – Relativity I (Problem 1-49 continued)

L L  57.4d  4 c  3.2 10 c  3.2  104 c  3.2  104 L L c  3.2  104  75.5 d  24 h / d  3600 s / h   c  3.2  104 c  3.2  104 c  3.2  104 c  3.2  104

 c  3.2 10  L   57.5d  24  3600   c  3.2 10  L c   3.2 10  c   3.2  10  4

4

4 2

2

4 2

2

2  c  3.2 104    c  3.2  104   L  c 2   3.2  104    57.5  24  3600     

c 2   3.2  104 2   57.5  24  3600    L  6.4 104 L  6.99  1018 m  739c y

1-50.

t2  t1    t2  t1  

v

 xb  xa  (Equation 1-20) c2 (a) t2  t1  0   t2  t1    v / c 2   xb  xa  

 0.5 1.0 y   v / c 2   2.0 1.0  c

Thus, 0.5   v / c   v  0.5c in the  x direction. (b) t     t  vx / c 2  Using the first event to calculate t  (because t  is the same for both events), 2 t   1 / 1   0.5  1 y   0.5c 1c y  / c 2   1.155 1.5  1.7 y  

(c)

 s 

2

  x    ct   1c y    0.5c y   0.75  c y  2

2

2

2

(d) The interval is spacelike. (e) L  s  0.866c y

22

2

 s  0.866c y

y

Chapter 1 – Relativity I 1-51. (a) ct (c∙y)

1.7y

1 2 1.0

x

x (c∙y)

-2.0

-1.0

1.0

2.0

Because events are simultaneous in S  , line between 1 and 2 is parallel to x axis. Its slope is 0.5   .

v  0.5c.

(b) From diagram t   1.7 y.

1-52.

xb  xr    xb  xr   v  tb  tr  (1) tb  tr    tb  tr   v  xb  xr  / c 2  (2)

Where xb  xr  2400m tb  tr  5 s xb  xr  2400m tb  tr  5 s Dividing (1) by (2) and inserting the values, 2400  v  5 106  400 2   2400  v  2400  / 5 106 c 2  2400  5 10 6 v 6 6 2 5 10 5 10  v  2400  / c   2400 2  6 v  5  10   4800  v  2.69 108 m / s in  x direction. 6 2  5 10 c 

1-53.

ux  0.85c cos50

u y  0.85c sin 50

  0.72   1/ 1   2  1.441

v  0.72c

23

Chapter 1 – Relativity I (Problem 1-53 continued)

ux 

ux  v 1  vux / c 2

ux 

0.85c cos 50  0.72c 0.1736c   0.286c 2 1   0.72c   0.85c cos 50 / c  1  0.3934

uy 

uy 

uy

 1  vux / c 2 

0.85c sin 50

1.441 1   0.72c   0.85c cos50 / c 2  

 0.745c

u  ux2  uy2  0.798c

tan    uy / ux  0.745 /  0.286    111 with respect to the  x axis.

1-54. This is easier to do in the xy and xy planes. Let the center of the meterstick, which is parallel

to

the

x-axis

and

x  y  x  y  0 at t  t   0.

moves

upward

with

speed

vy

in

S,

at

The right hand end of the stick, e.g., will not be at

t   0 in S  because the clocks in S  are not synchronized with those in S. In S  the

components of the sticks velocity are: uy 

ux 

uy

 1  vu x / c 2 



vy



because u y  v y and u x  0

ux  v  v because u x  0 1  vu x / c 2

When the center of the stick is located as noted above, the right end in S  will be at: x    x  vt   0.5 because t  0. The S  clock there will read: t     t  vx / c 2   0.5 v / c 2 Because t = 0. Therefore, when t   0 at the center, the right end is at xy given by: x  0.5 and    tan 1 For   0.65

v y  0.5 v    c 2  v  0.5 v  y  tan 1 y  2  / 0.5  tan 1  v y v / c 2  1   2 x   c  y   uy t  

    0.494v y / c 

24

Chapter 1 – Relativity I 1-55.

0  x 2  y 2  z 2   ct 

2

   x  vt     y 2  z 2  c  t   vx   / c 2  2

2

 x2   2  c 2 2 v 2 / c 4   y 2  z 2  t 2   2 v 2  c 2 2   x t   2 2 v  2vc 2 2 / c 2   x2  y 2  z 2   ct 

2

1-56. The solution to this problem is essentially the same as Problem 1-53, with the manhole taking the place of the meterstick and with the addition of the meterstick moving to the right along the x-axis. Following from Problem 1-53, the manhole is titled up on the right and so the meterstick passes through it; there is no collision. y

manhole x meterstick

1-57. (a)

t2    t2  vx2 / c 2  and t1    t1  vx1 / c 2  t2  t1   t2  t1  v  x2  x1  / c 2    T  vD / c 2 

(b) For simultaneity in S , t2  t1, or T  vD / c2

 v / c  cT / D.

Because v / c  1, cT / D is also 0 this is always positive because v/c < 1. Thus, t2  t1   T  vD / c 2  is always positive.

(d) Assume T  D / c with c  c. Then  c v T  vD / c 2   D / c    vD / c 2    D / c      c c 

This changes sign at v / c  c / c which is still smaller than 1. For any larger v still smaller than c)

t2  t1   T  vD / c 2   0 or t1  t2

25

Chapter 1 – Relativity I

1-58.



v  0.6c

1 1   0.6 

2

 1.25

(a) The clock in S reads   60min  75min when the S  clock reads 60 min and the first signal

from

S  is

sent.

At

that

time,

the

S  observer

is

at

v  75min  0.6c  75min  45c min. The signal travels for 45 min to reach the S

observer and arrives at 75 min + 45 min = 120 min on the S clock. (b) The observer in S sends his first signal at 60 min and its subsequent wavefront is found at x  c  t  60 min  .

The S  observer is at x  vt  0.6ct and receives the

wavefront when these x positions coincide, i.e., when

c  t  60 min   0.6ct 0.4ct  60c min t   60c min  / 0.4c  150 min x  0.6c  0 min   90c min The confirmation signal sent by the S  observer is sent at that time and place, taking 90 min to reach the observer in S. It arrives at 150 min + 90 min = 240 min. (c) Observer in S: Sends first signal Receives first signal Receives confirmation

60 min 120 min 240 min

The S  observer makes identical observations. 1-59. Clock at r moves with speed u  r , so time dilation at that clock’s location is: t0  t  tr  t0 1  u 2 / c 2  t0 1  r 2 2 / c 2  1  c, tr  t0 1  r 2 2 / c 2   2   1  t0 1  r 2 2 / c 2   t0 t  t0 r 2 2  2  And, r   t0 t0 2C 2

Or, for r

26

Chapter 1 – Relativity I

1-60.

vB

ux 

ux  v 1  ux v / c 2

uy 

(a) For vBA : v  vB , vAx  0, vAy  vA . So, vAx 

vAy

 B 1  vAx v / c

2





v A

where  B 

B

vBA  vAx2  vAy2  vB2   vA /  B    tan  BA

1  vB2 / c 2

2

vAy vA /  B v   A vAx  vB  B vB

vBy

 A 1  vBx v / c

vAB 

vAx  v  vB 1  vAx v / c 2

1

  (b) For vAB : v  vA , vBy  vB , vBx  0. So, vBx

  vBy

 1  ux v / c 2 

VA

Space station

vAy 

uy

2





vB

A

 v A    vB /  A 

  tan  AB

2

where  A 

vBx  v  vA 1  vBx v / c 2

1 1  vA2 / c 2

2

vB v  B  A  v A   A vA

(c) The situations are not symmetric. B viewed from A moves in the +y direction, and A viewed from B moves in the –y direction, so tan  A   tan  B  45 only if vA  vB and  A   B  1, which requires vA  vB  0.

27

Chapter 1 – Relativity I 1-61.

1-62. (a) Apparent time A  B  T / 2  t A  tB and apparent time B  A  T / 2  t A  tB where tA = light travel time from point A to Earth and tB = light travel time from point B to Earth. A B 

T L L T 2vL     2 2 2 cv cv 2 c v

B A

T L L T 2vL     2 2 2 cv cv 2 c v

(b) Star will appear at A and B simultaneously when tB  T / 2  t A or when the period is: L  4vL  L T  2 t B  t A   2    2 2  c  v c  v  c  v

1-63. The angle of u  with the x axis is:

uy uy 1  vux / c tan     ux ux  v  vu   1  2x  c   tan   

uy

  ux  v 



2



u sin  sin     u cos   v    cos   v / u 

28

Chapter 1 – Relativity I

#1

1-64.

d  vt cos 

θ d

xEarth  vt sin 

#2

xEarth L

L to Earth

(a) From position #2 light reaches Earth at time t2  L / c. From position #1 light reaches L d Earth at time t1    t. c c

L L d      t  c c c  vt cos    t c  t 1   cos  

t Earth  t2  t1  t Earth t Earth

(b)

Bapp 

vapp c



xEarth vt sin   sin    ctEarth ct 1   cos   1   cos 

(c) For   0.75 and vapp  c  app  1, the result in part (b) becomes

1

0.75sin   sin   cos   1/ 0.75 1  0.75cos 

Using the trigonometric identity

p sin   q cos   r sin   A where r  sin   cos   2 sin   45   1/ 0.75 sin   45   1/ 0.75 2

  45  70.6   25.6 29

p2  q2

sin A  p / r

cos A  q / r

Chapter 2 – Relativity II 2-1.

2 u yB  uo2 1  v 2 / c 2 

2 uxB  v2

2 1   u xB  uB2  / c 2  1  v 2 / c 2   uo2 / c 2 1  v 2 / c 2 

1  v



2

/ c 2 1  uo2 / c 2 

 1  v 2 / c 2 

1/ 2

p yB 

mu yB

2 2 1   u xB  u yB  / c2

1  u

2 o

/ c2 

1/ 2

muo 1  v 2 / c 2



1  v 2 / c 2 1  uo2 / c 2

 muo / 1  uo2 / c 2   p yA

2-2.

d  mu   m  ud    du 

  1  2u   u 2 3/ 2  u 2  1/ 2   m u    2  1  2   1  2   du   2  c   c   c    u 2  u 2  3/ 2  u 2  u 2  3/ 2   m  2 1  2   1  2 1  2   du  c  c   c  c    u2   m 1  2   c 

2-3.

(a)  

1 1 u / c 2

2

3/ 2

du

1



1   0.6c  / c 2 2



1  1.25 0.8

(b) p   mu    mc   u / c  / c  1.25  0.511 MeV  0.6  / c  0.383 MeV / c 2

(c) E   mc 2  1.25  0.511 MeV   0.639 MeV (d) Ek    1 mc2  0.25  0.511 MeV   0.128 MeV

31

Chapter 2 – Relativity II

2-4.

The quantity required is the kinetic energy. Ek    1 mc 2  1  u 2 / c 2  









2 (a) Ek   1   0.5 

2 (b) Ek   1   0.9  



1/ 2

2 (c) Ek   1   0.99  

2-5.

1/ 2



1/ 2

 1 mc 2 

 1 mc 2  0.155mc 2 

  1 mc 2  1.29mc 2 

1/ 2

  1 mc 2  6.09mc 2 

E  mc 2  m  E / c 2 

10 J

3.08 10 m / s  8

2

 1.11016 kg

Because work is done on the system, the mass increases by this amount.

2-6.

m(u)  m / 1  u 2 / c 2

(a)

(Equation 2-5)

m(u )  m  0.10  m 1 1 u / c 2

2

m / 1  u 2 / c2  m  0.10 m 1

 1  0.10 

1 u / c 2

2

 1.10  1  u 2 / c 2  1/ 1.10 

Thus, u 2 / c2  1  1/ 1.10   0.1736  u / c  0.416 2

(b) m(u)  5m m 1 u / c 2

2

 5m  1  u 2 / c 2  1/ 25

Thus, u 2 / c2  1 1/ 25  0.960  u / c  0.980 (c) m(u)  20m m 1 u / c 2

2

 20m  1  u 2 / c 2  1/ 400

Thus, u 2 / c2  1  1/ 400  0.9975  u / c  0.99870

32

2

Chapter 2 – Relativity II 2-7.

(a) v   Earth  moon distance  / time  3.8 108 m /1.5s  0.84c (b)

Ek   mc 2  mc 2  mc 2   1

(Equation 2-9)

  1 / 1   0.84   1.84 2

mc 2 (proton)  938.3MeV

Ek  938.3MeV 1.87  1  813MeV

(c) m(u ) 

m 1 u / c 2

2



938.3MeV / c 2 1   0.84 

2

 1.730 103 MeV / c 2  1.730GeV / c 2

1 2 1 2 mv   938.3MeV / c 2   0.84c   331MeV 2 2 813Mev  331MeV % error  100  59% 813MeV

(d) Classically, Ek 

2-8.

Ek  mc 2   1

(Equation 2-9)

Ek  u2   Ek  u1   W21  mc 2   u2   1  mc 2   u1   1

Or, W21  mc 2   u2     u1  1/ 2 1/ 2 (a) W21  938.3MeV 1  0.162   1  0.152    1.51MeV   1/ 2 1/ 2  1  0.852    57.6MeV (b) W21  938.3MeV 1  0.862   

(c) W21  938.3MeV 1  0.962  

2-9.

1/ 2

 1  0.952 

1/ 2

  3.35 103 MeV  3.35GeV 

E   mc2 (Equation 2-10) (a) 200GeV    0.938GeV  where mc2  proton   0.938GeV



1 1  v2 / c2

v 1  1 2 c 2



200GeV  213 0.938GeV

(Equation 2-40)

v 1  1  1  0.00001102 Thus, v  0.99998898c 2 c 2  213

33

Chapter 2 – Relativity II (Problem 2-9 continued) (b) E  pc for E

mc2 where E  200GeV 197  3.94  104 GeV (Equation 2-36)

p  E / c  3.94 104 GeV / c (c) Assuming one Au nucleus (system S  ) to be moving in the +x direction of the lab (system S), then u for the second Au nucleus is in the −x direction. The second Au’s energy measured in the S  system is:

E     E  vp x    213  3.94  104 GeV  v 3.94  104 GeV / c    213  3.94  104 GeV  1  v / c    213  3.94  104 GeV   2   1.68  107 GeV px    p x  vE / c 2    213  3.94  104 GeV  v 3.94  104 GeV / c 2     213  3.94  104 GeV   2   1.68  107 GeV / c 2-10. (a) E  mc 2  103 kg  c 2  9.0 1013 J (b) 1kWh  1103 J  h / s  3600s / h   3.6 106 J So, 9.0 1013 J / 3.6 106 J / kWh  2.5 107 kWh @$0.10 / kWh would sell for $2.5 106 or $2.5 million. (c) 100W  100 J / s, so 1g of dirt will light the bulb for:

9.0 1013 J 9.0 1011 s  9.0 1011 s   2.82 104 y 7 100 J / s 3.16 10 s / y

2-11.

E   mc2

(Equation 2-10)

where   1/ 1  u 2 / c 2  1/ 1   0.2   1.0206 2

E  1.0206 0.511MeV   0.5215 MeV Ek   mc 2  mc 2  mc 2    1

(Equation 2-9)

=  0.511MeV 1.0206  1  0.01054MeV

34

Chapter 2 – Relativity II (Problem 2-11 continued) E 2   pc    mc 2  2



2

(Equation 2-32)



2 1 2 1 2 2 E   mc 2   2  0.5215MeV    0.511MeV   2  c c  2 2  0.01087 MeV / c  p  0.104 MeV / c

p2 

2-12.

E   mc 2

(Equation 2-10)

  E / mc  1400MeV / 938MeV  1.4925 2

(a)

  1/ 1  u 2 / c 2  1.4925  1  u 2 / c 2  1/ 1.4925

2

u 2 / c 2  1  1/ 1.4925  0.551  u  0.74c 2

(b)

E 2   pc    mc 2  2

2

(Equation 2-32)

2 1 1 2 2 p   E 2   mc 2    1400MeV    938MeV    1040MeV / c     c c

2-13. (a)

pR   mS v

pN  mS v

  1/ 1  v 2 / c 2  1/ 1   2.5 105 / 3.0 108 

2

  1.00000035 3 3 pR  pN  mS  2.5 10   mS  2.5 10    1 3.5 107     3.5  107 3 pR  1.00000035  mS  2.5 10 

(b)

ER   mS c 2  mS c 2     1 mS c 2 1 mS v 2 2 2 2 ER  EN    1 c  0.5v  ER   1 c 2

EN 

3.5  107 c 2  0.5v 2  3.5  107 c 2  1

0.5  2.5 105  3.5 107 c 2

2

 0.0079

35

Chapter 2 – Relativity II

2-14.

u  2.2 106 m / s and   1/ 1  u 2 / c 2 (a) Ek  0.511MeV    1  0.511MeV 1/ 1  u 2 / c 2  1  0.5110  2.689  105 





 1.3741105 MeV 2 1 1 mu 2  mc 2  u 2 / c 2    0.5110MeV / 2   2.2 106 / c 2  2 2 5  1.374 10 MeV

Ek (classical) 

% difference =

1109 100  0.0073% 1.3741105

(b)

p

2 1 1 E 2   mc 2   c c

mc 2  mc   1  c 2

 mc    mc  2 2

2 2

2   6 8 2  1/ 1   2.2 10 / 3.0 10    1   

1/ 2

 0.5110MeV / c  7.33 103   3.74745  103 MeV / c p (classical)  mu 

mc 2  u  6 8    0.5110MeV / c   2.2  10 / 3.0  10  2  c c

 3.74733 10 3 MeV / c % difference =

1.2 107  100  0.0030% 2.74745 103

2-15. (a) 60W  60 J / s  3.16 107 s / y   1.896 109 J m  E / c 2  1.896 109 J /  3.0 108 m / s   2.1108 kg  2.1105 g  21 g 2

(b) It would make no difference if the inner surface were a perfect reflector. The light energy would remain in the enclosure, but light has no rest mass, so the balance reading would still go down by 21 μg.

2-16.

4

He  3H  p  e

Q  m  3H   mp  me  m  4 He  c 2

 2809.450MeV  938.280MeV  0.511MeV  3728.424MeV  19.827MeV

36

Chapter 2 – Relativity II 2-17.

H  2H  n Energy to remove the n = 22.014102u  2 H   1.008665u  n   3.016049u  3H  3

 0.006718u  931.5MeV / u  6.26MeV

2-18. (a) m 

E Eu 4.2eV  2  u  4.5 109 u 2 6 c uc 931.5 10 eV

(b) error =

2-19. (a)

m 4.5 109 u   7.7 1011  7.7 109 % m  Na   m  Cl  23u  35.5u

m  m  He   2m  H   4

2

m  4 He  c 2  2m  2 H  c 2

u uc 2  3727.409MeV  2 1875.628MeV  u / 931.5MeV  0.0256u

(b) E  m c 2   0.0256uc 2  931.5MeV / uc2   23.8MeV (c)

dN P 1W 1eV     2.62 1011 / s dt E 23.847MeV 1.602 1019 J

2-20.

v hf

M

(a) before photon absorbed:

1.01M

after photon absorbed:

E  hf  Mc2

E f  Ekinetic  1.01Mc 2

Conservation of energy requires: hf  Mc 2  Ekinetic  1.01Mc 2 Rearranging, the photon energy is:

hf  1.01Mc 2  Mc 2  Ekinetic hf  0.01Mc 2  Ekinetic  0.01Mc 2

(b) The photon’s energy is hf  0.01Mc2 because the particle recoils from the absorption of the photon due to conservation of momentum. The recoil kinetic energy (which is greater than 0) must be supplied by the photon.

37

Chapter 2 – Relativity II 2-21. Conservation of energy requires that Ei2  E 2f , or

 pi c 

2

  2m p c 2    p f c    2m p c 2  m c 2  and conservation of momentum requires that 2

2

2

pi  p f , so 4  m p c 2   4  m p c 2   2m p c 2  2m c 2   m c 2  2

2

0  2m p c 2  2m c 2   m c 2 

2

2

  m c2  m 0  m c 2  2   2   m c 2  2     2m p c  2m p  

  

Thus, m c 2  2  m / 2mp  is the minimum or threshold energy Ei that a beam proton must have to produce a π0.   m c 2  135  2 E  m c  2   135 MeV 2      280MeV   2m p c 2  2  938   

2-22.

E  mc 2  m  E / c 2 

2-23. (a)

200 106 eV  3.57 1025 g 32  5.6110 eV / g 

3 kT per atom 2 3 E  1.38 1023 J / K 1.50 107 K  2 E  3.105 1016 J / atom E

H atoms/kg  1kg /1.67 1027 kg / atom  6.0 1026 atoms Thermal energy/kg   3.1 1016 J / atom  6.0  1026 atoms   1.86  1011 J (b)

E  mc 2

 m  E / c2

m  1.86 1011 J / c 2  2.07 106 kg

38

Chapter 2 – Relativity II

p  E / c  1.0 J / s  / c

2-24. 1.0W  1.0 J / s 

(a) On being absorbed by your hand the momentum change is p  1.0 J / s  / c and, from the impulse-momentum theorem, F t  p where t  1s  F  1.0 J / s  / ct  1.0 / c  N  3.3 109 N This magnitude force would be exerted by gravity on mass m given by: m  F / g  3.3 109 N /  9.8m / s 2   3.4 1010 kg  0.34 g (b) On being reflected from your hand the momentum change is twice the amount in part (a) by conservation of momentum. Therefore, F  6.6 109 and m  0.68 g.

2-25. Positronium at rest:  2mc 2   Ei2   pi c  2

2

Because pi  0, Ei  2mc2  2  0.511MeV   1.022MeV After photon creation;  2mc 2   E 2f   p f c  2

2

Because pf  0 and energy is conserved,  2mc 2   E 2f  1.022MeV  or 2

2

2mc2  1.022MeV for the photons.

2-26.

E 2   pc 2    mc 2  2

E   pc    mc  2

2

(Equation 2-31)

 

1/ 2 2 2

 mc 1   p 2 / m2 c 2  

1/ 2

  pc 2   mc 1   2     mc   2

1/ 2

2

p2   2    mc 1  2m 2 c 2   

2  1  mc 1   p / mc    2 2

 mc 2  p 2 / 2m

2-27.

E 2   pc 2    mc 2  2

2

(Equation 2-31)

(a)  pc   E 2   mc 2    5MeV    0.511MeV   24.74 MeV 2 2

2

2

2

or, p  24.74 MeV / c  4.97MeV / c (b) E   mc 2

 1  u 2 / c 2   mc 2 / E 

   E / mc 2  1/ 1  u 2 / c 2

2 u / c  1   mc 2 / E    

1/ 2

1/ 2

2  1   0.511/ 5.0    

39

 0.995

2

Chapter 2 – Relativity II 2-28. 4

3

E/mc2

2

1

0 0

2-29.

1

E 2   pc 2    mc 2  2

1746MeV 

2

2

2

3

(Equation 2-31)

  500MeV    mc 2  2

2

1/ 2

2 2 mc 2  1746MeV    500MeV    

E   mc 2

 1673MeV

 m  1673MeV / c 2

   1/ 1  u 2 / c 2  E / mc 2

2 u / c  1   mc 2 / E    

1/ 2

2-30. (a) BqR  m u  p

B

p/mc

4

1/ 2

2  1  1673MeV /1746MeV    

 0.286  u  0.286c

(Equation 2-37)

m u and E   mc 2 which we have written as (see Problem 2-29) qR

2 u / c  1   mc 2 / E    

1/ 2

1/ 2

2  1   0.511MeV / 4.0MeV    

And   1/ 1  u 2 / c 2  1/ 1   0.992   7.83 2

9.1110 kg   7.83 0.992c   0.316T Then, B  1.60 10 C  4.2 10 m  31

19

2

(b)  m exceeds m by a factor of γ = 7.83.

40

 0.992

Chapter 2 – Relativity II 2-31. (a) p  qBR  e  0.5T  2.0  

3.0 108 m / s  300MeV / c c

2 2 Ek  E  mc 2   pc    mc 2    

1/ 2

(b)

 mc 2 1/ 2

2 2   300MeV    938.28MeV      46.8MeV

 938.28MeV

2-32. The axis of the spinning disk, system S  , is the z-axis in cylindrical coordinates. r  r, z  z,      t  dr  dr , dz  dz, d  d   dt Therefore, d  d    dt and d 2  d 2  2 d  dt   2 dt 2 . Substituting for

d 2 in Equation 2-43 yields ds 2  c2 dt 2  dr 2  r 2 (d 2  2 d  dt   2 dt 2 )  dz 2 

Simplifying, we obtain

ds 2  (c2  r 2 2 ) dt 2  (dr 2  r 2 d 2  2r 2 d  dt  dz 2 ) which is Equation 2-44.

2-33.   4GM / c2 R

(Equation 2-44)

Earth radius R  6.37 106 m and mass M  5.98 1024 kg



4  6.67 1011 N m2 / kg 2  5.98 1024 kg 

 3.00 10 m / s   6.37 10 m  8

2

6

 2.78 109 radians

  2.87 104 arc seconds

2-34. Because the clock furthest from Earth (where Earth’s gravity is less) runs the faster, answer (c) is correct.

41

Chapter 2 – Relativity II

2-35.

6  6.67 1011 N  m2 / kg 2 1.99 1030 kg  6 GM   2  2 c 1   2  R  3.00 108 m / s  1  0.0482  7.80 1011 m 

(Equation 2-51)

 3.64 108 radians/century  7.55 103 arc seconds/century

r 2 d  2-36. From Equation 2-45, dt  dt   2 2 2 c r 

2

2r 2 d dt   r 2 d   and dt  dt   2 2 2   2 2 2  . c r  c r   2

2

Substituting dt and dt 2 into Equation 2-44 yields 2  2r 2 d dt   r 2 d    2 ds  (c  r  )  dt   2 2 2   2 2 2   c r    c  r     2 2   r 2 d   2 2   dr  r d   2r  d   dt   2 2 2   dz 2  c r      2

2

2

2

 r  d    2r  d  dt   2

ds  (c  r  )dt  2

2

2

2

2

2

2

c 2  r 2 2

2 2    2 r  d    2 2 2 2 2   dr  r d   2r  d  dt   2 2 2  dz   c r   

Cancelling the 2r 2 d dt  terms, one of the 

 r  d    2

r d  2

2

2

c 2  r 2 2



2  r 2 d   c 2  r 2 2

2

terms, and noting that

c 2 r 2 d  2 c 2  r 2 2

we have that

ds 2  (c 2  r 2 2 )dt 2  (dr 2 

c 2 r 2 d  2  dz 2 ) which is Equation 2-46. 2 2 2 c r 

2-37. The transmission is redshifted on leaving Earth to frequency f , where f0  f  f0 gh / c 2 . Synchronous satellite orbits are at 6.623RE where

g

GM E

 6.623RE 

2



9.9m / s 2

 6.623

2

 0.223m / s 2

42

Chapter 2 – Relativity II (Problem 2-37 continued) h  6.623RE  6.623  6.37 106 m   4.22 107 m

f0  f   9.375 109 Hz  0.223m / s 2  4.22 107  /  3.00 108   0.980Hz 2

f  f0  0.980Hz  9.374999999 109 Hz

2-38. Earth

white dwarf

distant star

57 c∙y

35 c∙y

On passing “below” the white dwarf, light from the distant start is deflected through an angle:

  4GM / c R  2

4  6.67 1011 N  m2 / kg 2   3 1.99 1030 kg 

 3.00 10 m / s  10 m  2

8

7

(Equation 2-44)

 1.77 103 radians  0.051 or the angle between the arcs is 2  0.102

2-39. The speed v of the satellite is: v  2 R / T  2  6.37 106 m  /  90 min 60s / min   7.42 103 m / s Special relativistic effect: After one year the clock in orbit has recorded time t  t /  , and the clocks differ by: t  t   t  t /   t 1  1/    t  v 2 / 2c 2  , because v

c. Thus,

t  t    3.16 107 s  7.412 103  /  2  3.00 108 m   0.00965s  9.65ms 2

2

Due to special relativity time dilation the orbiting clock is behind the Earth clock by 9.65ms.

43

Chapter 2 – Relativity II (Problem 2-39 continued)

General relativistic effect: 2 5 f gh  9.8m / s  3.0 10 m     3.27 1011 s / s 2 8 f0 c2  3.0 10 m / s 

In one year the orbiting clock gains  3.27 1011 s / s  3.16 107 s / y   1.03ms . The net difference due to both effects is a slowing of the orbiting clock by 9.65−1.03 = 8.62 ms.

2-40. The rest energy of the mass m is invariant, so observers in S  will also measure m = 4.6kg, as in Example 2-9. The total energy E  is then given by:

 mc    E  pc  2 2

2

2

Because, p  0, E  mc 2  4.6kg   3.0 108 m / s   4.14 1014 J 2

2-41. (a)

E   me c 2

   E / mec 2  50 10 MeV / 0.511MeV  9.78 104

L  L0 /   102 m

L0  9.78 104 102 m   978m (length of one bundle) The width of one bundle is the same as in the lab. (b) An observer on the bundle “sees” the accelerator shortened to 978m from its proper length L0 , so L0    978  978 104  978  9.57 107 m.

(Note that this is about

2.5 times Earth’s 40,000km circumference at the equator.) (c) The e+ bundle is 10−2 m long in the lab frame, so in the e− frame its length would be measured to be: L  102 m  /   102 m / 9.78 104  1.02 107 m .

44

Chapter 2 – Relativity II 2-42.

Ek   mc2  mc2  mc2   1 If Ek  mc2  938MeV , then   2. (a)

 mc 

2 2

 pc 

2

 E 2   pc 

2

(Equation 2-32) Where E   mc 2  2  938MeV 

 E 2   mc 2    2  938    938   2.46 106 2

p   2.64 106 

1/ 2

2

2

/ c  1.62 103 MeV / c

(b) p   mu  u  p /  m  1.62 103 MeV / c  /  2   938MeV / c 2   0.866c

2-43. (a) The momentum of the ejected fuel is:

p   mu  mu / 1  u 2 / c 2  103 kg  c / 2  / 1   0.5  1.73 1011 kg  m / s 2

Conservation of momentum requires that this also be the momentum ps of the spaceship: ps  msus / 1  us2 / c 2  1.73 1011 kg  m / s

ms us / 1  us2 / c 2  1.73 1011 kg  m / s 

Or,

2

ms2 cs2  1  us2 / c 2 1.73 1011 kg  m / s   1.73 1011 kg  m / s 2    3.33 105 kg 2  us2 2

10 kg  6

Or,

2

us2   3.33 105 kg 2  us2  1.73 1011 kg  m / s 

2

us  1.73 1011 kg  m / s  /106 kg  1.73 105 m / s  5.77 10 4 c

(b) In classical mechanics, the momentum of the ejected fuel is: mu  mc / 2  103 c / 2, which must equal the magnitude of the spaceship’s momentum msus, so 103 kg  3.0 108 m / s  3 us  10  c / 2  / ms   5.0 104 c  1.5 105 m / s 6 2 10 kg 

(c) The initial energy Ei before the fuel was ejected is Ei  ms c 2 in the ship’s rest system. Following fuel ejection, the final energy Ef is:

E f  energy of fuel + energy of ship  mc2 / 1  u 2 / c 2   ms  m  c 2 / 1  us2 / c 2 Where u  0.5c and us c, so E f  1.155mc 2   ms  m  c 2  1.155  1 mc 2  ms c 2 The change in energy ΔE is: E  E f  Ei   0.155 103 kg  c 2  106 kg  c 2   106 kg  c 2  E  155kg  c 2 or 155 kg  E / c 2 of mass has been converted to energy.

45

Chapter 2 – Relativity II 2-44. The observer at the pole clock sees the light emission of the equatorial clock as transverse Doppler effect, measuring frequency f, where

f0 / f   1   cos  

(Equation 1-35a)

   / 2 for the equatorial clock, so f / f 0  1  v 2 / c 2  1 

1 v2 2 c2

f / f0  1  1.193 1012 (red shift)

The observer at the equatorial clock sees a gravitational blue shift for the pole clock and observes

f / f 0  1  gh / c 2

(Equation 2-45)

f / f 0  1  2.897 1012 (blue shift)

2-45. (a)

p  300 BR  q / e 

(Equation 2-38)

p  300 1.5T   6.37 106  1  2.87 109 MeV mc 2 , E  Ek and E  pc (Equation 2-32)  Ek  pc  2.87 109 MeV

For E (b)

For E  pc, u  c and T  2 R / c  2  6.37 106 m  / c  0.133s

2-46.

f  1  GM / c 2 R f0

(Equation 2-47)

The fractional shift is:

f0  f f  1   GM / c 2 R  7 104 f0 f0

The dwarf’s radius is:

R  GM / c  7  10 2

4



6.67  1011 N  m2 / kg 2  2  1030 kg 

 3.00 10 m / s   7 10  2

8

Assuming the dwarf to be spherical, the density is:



M 2 1030 kg   5.0 1010 kg / m3 V 4  2.12 106 m 3 / 3

46

4

 2.12  106 m

Chapter 2 – Relativity II 2-47. The minimum energy photon needed to create an e− − e+ pair is Ep = 1.022 MeV (see Example 2-13). At minimum energy, the pair is created at rest, i.e., with no momentum. However, the photon’s momentum must be p  E / c  1.022MeV / c at minimum. Thus, momentum conservation is violated unless there is an additional mass “nearby” to absorb recoil momentum.

2-48.

  1  vux / c 2     uy   m   py   muy   2  1  u 2 / c 2    1  vux / c  

Canceling γ and 1  vux / c 2  , gives: py  In an exactly equivalent way, pz  pz .

mu y 1  u 2 / c2

 py

2-49. (a) u   u  v  / 1  uv / c 2  where v  u, so u  2u / 1  u 2 / c 2  . Thus,

the speed of the particle that is moving in S  is: u  2u / 1  u 2 / c 2  from which we see that: 2 4u  u  1     1  2 / 1  u 2 / c 2  c c 2

2

 1  2u 2 / c 2  u 4 / c 4  4u 2 / c 2  / 1  u 2 / c 2   1  u 2 / c 2  / 1  u 2 / c 2  2

  u 2  And thus, 1       c  

1/ 2

2

1  u 2 / c2  1  u 2 / c2

(b) The initial momentum pi in S  is due to the moving particle, pi  mu / 1   u / c  where u and 1   u / c  were given in (a). 2

pi  m

2

2u 1  u 2 / c 2 

1  u

2

/c

2

1  u

2

/c

2



 2mu / 1  u 2 / c 2 

(c) After the collision, conservation of momentum requires that: p f  Mu / 1  u 2 / c 2 

1/ 2

 pi  2mu / 1  u 2 / c 2 

47

or M  2m / 1  u 2 / c 2 

1/ 2

Chapter 2 – Relativity II (Problem 2-49 continued) (d) In S: Ei  2mc 2 / 1  u 2 / c 2 and E f  Mc 2 (M is at rest.) Because we saw in (c) that M  2m / 1  u 2 / c 2 

1/ 2

, then Ei  E f in S .

In S : Ei  mc 2  mc 2 / 1   u / c  and substituting for the square root from (a), 2

Ei  2mc2 / 1  u 2 / c2  and E f  Mc2 / 1  u 2 / c 2 . Again substituting for M from (c), we have: Ei  E f .

2-50. (a) Each proton has Ek  mp c 2   1 , and because we want Ek  m p c 2 , then γ = 2 and u = 0.866c. (See Problem 2-40.) ux  v where u = v and ux =−u yields: 1  ux v / c 2 2  0.866c  2u ux    0.990c 2 2 1  u / c 1   0.866c 

(b) In the lab frame S  : ux 

(c) For u  0.990c,   1/ 1   0.99   7.0 and the necessary kinetic energy in the 2

lab frame S is: Ek  mp c 2   1  mp c 2  7  1  6mp c 2

2-51. (a) pi  0  E / c  Mv or v  E / Mc (b) The box moves a distance x  vt , where t  L / c,

so x   E / Mc  L / c   EL / Mc 2 (c) Let the center of the box be at x = 0. Radiation of mass m is emitted from the left end of the box (e.g.) and the center of mass is at: xCM 

M  0  m  L / 2 M m



mL 2  M  m

When the radiation is absorbed at the other end the center of mass is at: xCM 

M  EL / Mc 2   m  L / 2  EL / Mc 2  M m

48

Chapter 2 – Relativity II (Problem 2-51 continued) Equating the two values of xCM (if CM is not to move) yields: m   E / c 2  / 1  E / Mc 2 

Mc2 , then m  E / c 2 and the radiation has this mass.

Because E

2-52. (a) If v mass is 0:

E2   p c    m c  and Ev2   pv c   0 2

2

2

Ek   Ev  139.56755MeV  105.65839MeV m c 2    1  Ev  33.90916MeV



Ev  pv c  E2   m c 2 



2 1/ 2

 33.90916  m c 2    1

Squaring, we have

 m c    1   33.90916 

2 2

2

2

 2  33.90916   m c 2     1   m c 2     1 2

2

Collecting terms, then solving for    1 ,

 33.90916   1  2 2  m c 2   2  33.90916  m c 2 2

Substituting m c 2  105.65839MeV

  1  0.0390    1.0390 so, 1 2 2 1/ 2 Ek   4.12MeV and pu  109.78   105.66    29.8MeV / c  c Ev  29.8MeV and pv  29.8MeV / c (b)

If v mass = 190 keV , then Ev2   pv c  +  m v c 2  and 2

2

Ek   Ekv  139.56755MeV  105.65839MeV   0.190MeV  33.71916MeV Solving as in (a) yields E  109.78MeV ,

2-53.

f  1  Gm / c 2 R f0

p  29.8MeV / c, Ev  29.8MeV , and pv  29.8MeV / c

(Equation 2-47)

Since c  f  and c  f 0 0 ,

6.67 10 N  m / kg 1.99 10 kg     0  1  GM / c 2 R  1  2  c   3.00 108 m / s   6.96 106 m  c

11

0

 1  0.000212  0.999788

49

2

2

30

Chapter 2 – Relativity II (Problem 2-53 continued)

  0 / 0.999788  720.00nm / 0.999788  720.15nm     0  0.15nm u y   u y /   1  vu x / c 2 

2-54.

du y ay 

du y dt





1  vu

1

u y  vdx  2 2  2  1  vu x / c    c    dt  vdx / c 2 

/ c2   1

x

2 1 2 2 2   1   du y / dt  1  vu x / c    u y v / c   du x / dt  1  vu x / c   ay  2    1  v  dx / dt  / c2    2 ay ax u y v / c ay   2 3  2 1  vu x / c 2   2 1  vu x / c 2 

2-55. (a)

Fx 

dpx d  mv   dt dt

Fx  ma x because u x  0

Fx   m  dv / dt   mv d 1  v 2 / c 2   Fx  Fx 

ma x

1  v 2 / c2  ma x



m  v2 / c2  ax

1  v / c  1  v / c   m  v / c  a 1  v / c  1/2

2

 / dt 

2 3/2

2

2

2

1/2

2

2

x

2 3/2

Fx   3ma x Because u x  0, note from Equation 2-1 (inverse form) that a x  a x /  3 . Therefore, Fx   3ma x /  3  ma x  Fx

(b)

Fy 

dp y



d  mv y 

Fy  ma y because u y  u x  0 dt dt Fy   ma y because S  moves in +x direction and the instantaneous transverse impluse (small) changes only the direction of v. From the result of Problem 2-52 (inverse form) with u y  ux  0, a y  a y /  2 Therefore, Fy   ma y   ma y /  2  ma y /   Fy / 

50

Chapter 2 – Relativity II 2-56. (a) Energy and momentum are conserved. Initial system: E  Mc2 ,

p0

Invariant mass:  Mc 2   E 2   pc    Mc 2   0 2

2

2

Final system:

 2mc    Mc  2 2

2 2

0

For 1 particle (from symmetry)

 mc    Mc / 2 2 2

2

2

 p 2c 2   Mc 2 / 2    muc  2

2

 Mc 2  2   u / c  Rearranging, 1   2   2mc 

  2mc 2 2   Solving for u, u  1   2    Mc  

2

2

 Mc 2  1     2  2 2  2mc  1  u / c 2

1/ 2

c

(b) Energy and momentum are conserverd. Initial system: E  4mc 2 Invariant mass:

 Mc    4mc    pc  2 2

2 2

2

Final system: Invariant mass:  2mc 2    4mc 2    pc  2

2

 4mc 2 2   Mc 2 2  u pc     2 c E 4mc

1/ 2

2 2 2  Mc 2   u   4mc    Mc    1     2 2  c  4mc   4mc2  2

  Mc 2 2   u  1   2    4mc  

2

2

1/ 2

c

51

2

where  pc    4mc 2    Mc 2  2

2

2

Chapter 3 – Quantization of Charge, Light, and Energy 3-1.

The radius of curvature is given by Equation 3-2. R

  mu 2.5 106 m / s   m  3.911025 m / s  C  T   m 19 qB  1.60 10 C   0.40T  

Substituting particle masses from Appendices A and D: R (protron)  1.67 1027 kg  3.911025 m / s  C  T   6.5 102 m

R (electron)   9.111031 kg  3.911025 m / s  C  T   3.6 105 m R (deuteron)   3.34 1027 kg  3.911025 m / s  C  T   0.13m R (H 2 )   3.35 1027 kg  3.911025 m / s  C  T   0.13m R (helium)   6.64 1027 kg  3.911025 m / s  C  T   0.26m

3-2. u



s



r u

For small values of  , s  ; therefore,  =

mu 2 Recalling that euB  r

 r

mu eB

53

s  r r

 

mu / eB



eB mu

Chapter 3 – Quantization of Charge, Light, and Energy 3-3.

B

E , u

pc 

2 u pc  , and pc  E 2   mc 2  c E

 0.561MeV    0.511MeV  2

2

 0.2315MeV

u 0.2315MeV   0.41 c 0.561MeV 2.0 105V / m  B  1.63 103 T  16.3G 0.41c

3-4.

F  quB and FG  mp g





19 6 5 FB quB 1.6 10 C  3.0 10 m / s   3.5 10 T     1.03 109 27 2 FG m p g 1.67 10 kg 9.80m / s 

3-5.

(a)

mu  2 Ek / e  e / m   R  qB  e / m  B 

1/ 2

4 1 2 Ek / e 1   2   4.5 10 eV / e       B e/m 0.325T  1.76 1011 kg  

(b) frequency

f 

u 2 R



1/ 2

 2.2 103 m  2.2mm

 2 Ek / e  e / m  2 R

 2   4.5 104 eV / e 1.76 1011 C / kg     3 2  2.2 10 m 

1/ 2

 9.1109 Hz

period T  1/ f  1.11010 s

3-6.

(a) 1/ 2mu 2  Ek , so u 

 2Ek / e  e / m 

 u   2  2000eV / e  1.76 1011 C / kg 

(b)

t1 

1/ 2

x1 0.05m   1.89 109 s  1.89ns 7 u 2.65 10 m / s

54

 2.65 107 m / s

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-6 continued) (c) mu y  F t1  eEt1  u y   e / m  Et1  1.76 1011 C / kg  3.33 103V / m 1.89 109 s   1.11106 m / s

3-7.

3-8.

NEk  W  CV T 

3  4.186 J / cal   1.311014 CV T  5 10 cal / C   2C  N   Ek 2000eV 1.60 1019 J / eV 

Q1  Q2   25.41  20.64 1019 C  4.47 1019 C   n1  n2  e

Q2  Q3   20.64  17.47  1019 C  3.17 1019 C   n2  n3  e Q4  Q3  19.06  17.47  1019 C  1.59 1019 C   n4  n3  e Q4  Q5  19.06  12.70 1019 C  6.36 1019 C   n4  n5  e Q6  Q5  14.29  12.70 1019 C  1.59 1019 C   n6  n5  e where the ni are integers. Assuming the smallest Δn = 1, then Δn12 = 3.0, Δn23 = 2.0, Δn43 = 1.0, Δn45 = 4.0, and Δn65 = 1.0. The assumption is valid and the fundamental charge implied is 1.59 1019 C.

3-9.

For the rise time to equal the field-free fall time, the net upward force must equal the weight. qE  mg  mg  E  2mg / q.

3-10. (See Millikan’s Oil Drop Experiment on the home page at www.whfreeman.com/tiplermodernphysics6e.) The net force in the y-direction is mg  bv y  ma y . The net force in the x-direction is qE  bvx  max . At terminal speed ax  ay  0 and vx / vt  sin  .

sin  

vx  qE / b  qE   vt vt bvt

55

Chapter 3 – Quantization of Charge, Light, and Energy 3-11. (See Millikan’s Oil Drop Experiment on the home page at www.whfreeman.com/tiplermodernphysics6e.) (a) At terminal speed mg  bvt where m  4 / 3 a3 oil and b  6 a. Substituting gives  18  1.80 105 N  s / m 2  5.0 103 m / 20s   18   vt  2  a     a 4  oil g   4  0.75 1000kg / m3  9.8m / s 2     1.66 106 m  1.66  103 mm

1/ 2

m  4 1.66 106 m   750kg / m3  / 3  1.44 1014 kg 3

(b)

3-12.

19 5 FE  2  1.60 10 C  2.5 10 V / m  F  qE and FG  mg    0.57 FG 1.44 1014 kg  9.8m / s2 

mT  2.898 103 m  K (a) m 

2.898 103 m  K  9.66 104 m  0.966mm 3K

(b) m 

2.898 103 m  K  9.66 106 m  9.66 m 300 K

(c) m 

2.898 103 m  K  9.66 107 m  966nm 3000 K

3-13. Equation 3-4: R   T 4 . Equation 3-6: R 

1 cU . 4

From Example 3-4: U  8 5 k 4T 4  / 15h3c 2 

 

R 1/ 4  cU 1   c  8 5 k 4T 4  / 15h3c 2T 4  4 4 T T 4 2 5 1.38 1023 J / K 

15  6.63 10

34

4

J  s   3.00 10 m / s  3

8

2

56

 5.67 108W / m2 K 4

Chapter 3 – Quantization of Charge, Light, and Energy 3-14. Equation 3-18: u    

8 hc 5 ehc /  kT  1

u  f  df  u    d   u  f   u  f 

d Because c  f  , df

d c/ f 2 df

8 hc  f / c   c  8 f 2 hf u f    2  3 hf / kT hf / kT e 1  f  c e 1 5

3-15.

2.898 103 m  K (a) mT  2.898 10 m  K  m   1.07 103 m  1.07mm 2.7 K 3

(b) c  f   f 

c

m



3.00 108 m / s  2.80 1011 Hz 3 1.07 10 m

(c) Equation 3-6: R

1 c cU   8 5 k 4T 4 /15h3c3  4 4

 3.00 10 m / s 8 1.38 10 J / K   2.7    4 15  6.63 10 J  s   3.00 10 m / s  8

4

23

5

3

34

Area of Earth: A  4 rE2  4  6.38 106 m 

8

3

4

 3.01106 W / m2

2

Total power = RA   3.01106W / m2   4   6.38 106 m   1.54 109W 2

3-16.

mT  2.898 103 m  K 2.898 103 m  K  4140 K (a) T  700 109 m (b) T 

2.898 103 m  K  9.66 102 K 3 102 m

(c) T 

2.898 103 m  K  9.66 104 K 3m

57

Chapter 3 – Quantization of Charge, Light, and Energy 3-17. Equation 3-4: R1   T14

3-18. (a) Equation 3-17: E 

(b) E 

R2   T24    2T1   16 T14  16R1 4

hc / 10hc / kT  hc /  0.1kT   0.1  0.951kT hc /  kT hc / kT  / 10 hc / kT   e 1 e 1 e 1

hc /  0.1hc / kT  10kT hc /    10  4.59 104 kT hc /  kT hc / kT  /  0.1hc / kT   e 1 e 1 e 1

Equipartition theorem predicts E  kT . The long wavelength value is very close to kT, but the short wavelength value is much smaller than the classical prediction.

3-19. (a) mT  2.898 103 m  K  T1  R1   T14

2.898 103 m  K  107 K 27.0 106 m

and R2   T24  2R1  2 T14

 T24  2T14 or T2  21/ 4 T1   21/ 4  107 K   128K

2.898 103 m  K (b) m   23 106 m 128K

3-20. (a) mT  2.898 103 m  K

(Equation 3-5)

2.898 103 m  K m   1.45 107 m  145nm 4 2 10 K (b) m is in the ultraviolet region of the electromagnetic spectrum.

58

Chapter 3 – Quantization of Charge, Light, and Energy 3-21. Equation 3-4: R   T 4 Pabs  1.36 103W / m2  RE2 m2  where RE = radius of Earth Pemit   RW / m2  4 RE2   1.36 103W / m2  RE2 m2 

  RE2 R  1.36 103W / m2   2  4 RE T4 

 1.36 103 W  T 4  2 4 m 

1.36 103W / m2  T  278.3K  5.3C 4  5.67 108W / m2  K 4 

3-22. (a) mT  2.898 103 m  K  m 

f m  c / m 

2.898 103 m  K  8.78 107 m  878nm 3300K

3.00 108 m / s  3.42 1014 Hz 7 8.78 10 m

(b) Each photon has average energy E  hf and NE  40J / s. N

40 J / s 40 J / s   1.77 1020 photons / s 34 14 hf m  6.63 10 J  s 3.42 10 Hz 

(c) At 5m from the lamp N photons are distributed uniformly over an area A  4 r 2  100 m2 . The density of photons on that sphere is  N / A / s  m2 .

The area of the pupil of the eye is   2.5 103 m  , so the number of photons 2

entering the eye per second is:

n   N / A   2.5 10 m  3

2

1.77 10 

20

/ s     2.5 103 m  100 m2

 1.77 1020 / s     2.5 103 m   1.10 1013 photons / s 2

59

2

Chapter 3 – Quantization of Charge, Light, and Energy 3-23. Equation 3-18: u    

8 hc 5 A 5 Letting A   hc , B  hc / kT , and U     ehc /  kT  1 eB /  1

  5  1 e B /    B 2   du d  A 5  5 6     B/    A 2 B/ d  d   eB /   1 e  1 e  1     6 6 B /  A  B B/  A e B   e  5  e B /   1    5 1  e  B /     0 2  2  B/ B /    e  1     e  1  

The maximum corresponds to the vanishing of the quantity in brackets.

Thus,

5 1  e B /    B . This equation is most efficiently solved by iteration; i.e., guess at a

value for B/λ in the expression 5 1  e B /   , solve for a better value of B/λ; substitute the new value to get an even better value, and so on. Repeat the process until the calculated value no longer changes. One succession of values is: 5, 4.966310, 4.965156, 4.965116, 4.965114, 4.965114. Further iterations repeat the same value (to seven digits), so we have:

6.63 1034 J  s  3.00 108 m / s   hc hc  4.965114   mT   m  m kT  4.965114  k  4.965114  1.38 1023 J / K  B

mT  2.898 103 m  K

(Equation 3-5)

3-24. Photon energy E  hf  hc /  (a) For λ = 380nm: E  1240eV  nm / 380nm  3.26eV For λ = 750nm: E  1240eV  nm / 750nm  1.65eV (b) E  hf   4.14 1015 eV  s 100 106 s 1   4.14 107 eV

3-25. (a) hf  hc /   0.47eV .

max

 4.14 10 hc   4.87eV

15

eV  s  3.00 108 m / s  4.87eV

60

 2.55 107 m  255nm

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-25 continued) (b) It is the fraction of the total solar power with wavelengths less than 255nm, i.e., the area under the Planck curve (Figure 3-6) up to 255nm divided by the total area. The latter is: R   T 4   5.67 108W / m2  K 4   5800K   6.42 107W / m2 . 4

Approximating the former with u     with   127nm and   255nm :  8 hc 127  109 m 5    255  109 m   1.23  104 J / m3 u 127nm   255nm    9  ehc / kT 12710   1   

R  0  255nm  c 1.23  104 J / m 3    4 R 8 4 3  3.00  10 m / s 1.23  10 J / m  fraction = 1.4  104   4   6.42  107W / m2 

R  0  255nm  

3-26. (a) t 

hc





1240eV  nm  653nm, 1.9ev

ft 

 h



1.9eV  4.59 104 Hz 15 4.136 10 eV  s

1  hc  1  1240eV  nm  (b) V0         1.9eV   2.23V e   e  300nm  1  hc  1  1240eV  nm  (c) V0         1.9eV   1.20V e   e  400nm 

3-27. (a) Choose λ =550nm for visible light. nhf  E 

dn dE hf  P dt dt

 0.05  100W  550  109 m  dn P P     1.38  1019 / s 34 8 dt hf hc  6.63  10 J  s  3.00  10 m / s  (b) flux 

number radiated / unit time 1.38 1019 / s   2.75 1017 / m2  s 2 area of the sphere 4  2m 

61

Chapter 3 – Quantization of Charge, Light, and Energy 3-28. (a) hf    ft  (b) f  c /  

 h



4.22eV  1.02 1015 Hz 15 4.14 10 eV  s

3.00 108 m / s  5.36 1014 Hz 9 560 10 m

No.

Available energy/photon hf   4.14 1015 eV  s  5.36 1014 Hz   2.22eV . This is less than  .

3-29. (a) E  hf  hc /     hc / E For E  4.26eV :   1240eV  nm /  4.26eV   291nm and since f  c /  ,

f   3.00 108 m / s  /  291nm   1.03 1015 s 1

(b) This photon is in the ultraviolet region of the electromagnetic spectrum.

3-30. (a) First, add a row f 1014 Hz to the table in the problem, then plot a graph Ek ,max versus f . The slope of the graph is h/e; the intercept on the  Ek ,max axis is

work function. The graph below is a least squares fit to the data.

λ nm

544

Ek ,max eV

0.360 0.199 0.156 0.117 0.062

f 1014 Hz

5.51

594

5.05

62

604

4.97

612

4.90

633

4.74

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-30 continued)

Slope  h / e  3.90 1015 eV  s (b)

(3.90 1015  4.14 1015 ) eV  s  5.6 percent 4.14 1015 eV  s

(c) The work function is the magnitude of the intercept on the Ek ,max axis, 1.78 eV. (d) cesium

En

hc

3-32. (a)  

hc

3-31.







(b) Ek 



hc



 60   6.63 1034 J

s  3.00 108 m / s  9

550 10 m

1240eV nm  1.90eV 653nm

 

1240eV nm  1.90eV  2.23eV 300nm

63

 2.17 1017 J

Chapter 3 – Quantization of Charge, Light, and Energy 3-33. Equation 3-25: 2  1   

 6.63 10   9.1110

34

31



1

100 

J s  1  cos135 

kg  3.00 10 m / s 

3-36.

p

h



8

 4.14 1012 m  4.14 103 nm

4.14 103 nm 100  5.8% 0.0711nm

3-34. Equation 3-24: m 

3-35.

h 1  cos  mc

1.24 103 1.24 103 nm   0.016nm V 80 103V

hc c



(a) p 

1240eV nm 6.63 1034 J s  3.10eV / c   1.66 1027 kg m / s c  400nm  400 109 m

(b) p 

1240eV nm 6.63 1034 J s  1.24 104 eV / c   6.63 1024 kg m / s c  0.1nm  0.1109 m

(c) p 

1240eV nm 6.63 1034 J s 5  4.14  10 eV / c   2.211032 kg m / s 3 102 m c  3 107 nm 

(d) p 

1240eV nm 6.63  1034 J s  620eV / c   3.32  1025 kg m / s 9 c  2nm  2  10 m

2  1 

6.63 1034 J s  1  cos110   h 1  cos    3.26 1012 m   31 8 mc  9.1110 kg  3.00 10 m / s 

6.63 1034 J s  3 108 m / s   hc 1    2.43 1012 m E1  0.511106 eV 1.60 1019 J / eV 

2  1  3.26 1012 m   2.43  3.26 1012 m  5.69 1012 m

64

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-36 continued)

E2 

hc

2



1240eV nm  2.18 105 eV  0.218MeV 3 5.69 10 nm

Electron recoil energy Ee  E1  E2 (Conservation of energy)

Ee  0.511MeV  0.218MeV  0.293MeV . The recoil electron momentum makes an angle θ with the direction of the initial photon. PE θ

h/λ1

110° h/λ2 20° h

2

cos 20  pe sin   1/ c  E 2   mc 2  sin  2

 3.00 10 m / s  6.63 10 8

sin  

 5.69 10

12

(Conservation of momentum) 34

J s  cos 20

2 2 m   0.804MeV    0.511MeV    

1/ 2

1.60 10

13

J / MeV 

 0.330 or   19.3

3-37. (a) First, add a row ( 1  cos  ) to the table in the problem, then plot a graph of  versus ( 1  cos  ). The slope of the graph is the Compton wavelength of the electron.  pm  degrees 1  cos 

0.647 1.67 2.45 3.98 4.80 45 75 90 135 170 0.293 0.741 1.000 1.707 1.985

65

Chapter 3 – Quantization of Charge, Light, and Energy

h (4.5  1.0) nm  slope   2.43nm mc (1.88  0.44)

(b) The graph above is a least squares fit to the data. The percent difference is (2.43  2.426) nm 0.004 nm 100  100  0.15percent 2.426 nm 2.426 nm

3-38.

  2  1   

1  100 

3-39. (a) E1 

h 1  cos   100 0.00243nm 1  cos90  0.243nm mc

hc

1



(b) 2  1  (c) E2 

h 1  cos   0.011 Equation 3-25 mc

hc

2

1240eV nm  1.747 104 eV 0.0711nm

h 1  cos   0.0711nm   0.00243nm 1  cos180   0.0760nm mc



1240eV nm  1.634 104 eV 0.0760nm

(d) Ee  E1  E2  1.128 103 eV

66

Chapter 3 – Quantization of Charge, Light, and Energy 3-40. (a)    h / mc 1  cos   From protons:    6.63 1034 J s  / 1.67 1027 kg  3.00 108 m / s  1  cos120    1.99 1015 m  1.99 106 nm

(b) Similarly, for electrons ( m  9.111031 kg )   2.43 1012 m  2.43 103 nm

(c) Similarly, for N2 molecules ( m  4.68 1026 kg )   4.72 1017 m  4.72 108 nm

3-41.

2  1 

h 1  cos   0.0711   0.00243nm 1  cos  mc

λ2 0.0750

0.0730

0.0710

0 Slope = 

1

1  cos 

θ

1  cos 

2 (nm)

0°

0

0.0711

45°

0.293

0.0718

90°

1

0.0735

135°

1.707

0.0752

2

 0.0745  0.0720  nm  2.381103 1.50  0.45 h  h   2.381103 nm  9.111031 kg  3.00 108 m / s   6.5110 34 J s mc

67

Chapter 3 – Quantization of Charge, Light, and Energy

3-42. (a) Compton wavelength =

electron:

h mc

h 6.63 1034 J s   2.43 1012 m  0.00243nm 31 8 mc  9.1110 kg  3.00 10 m / s 

h 6.63 1034 J s proton:   1.32 1015 m  1.32 fm 27 8 mc 1.67 10 kg  3.00 10 m / s  (b) E 

hc



(i) electron: E  (ii) proton: E 

1240eV nm  5.10 105 eV  0.510MeV 0.00243nm

1240eV nm  9.39 108 eV  939MeV 6 1.32 10 nm

3-43. Photon energy E  hf  hc /  allows us to rewrite Equation 3-25 as hc hc h   (1  cos  ) E2 E1 mc

Rearranging the above, hc h hc  (1  cos  )  E2 mc E1

Dividing both sides of the equation by hc yields 1 1 1 ( E1 / mc 2 )(1  cos  )  1  (1  cos  )   E2 mc 2 E1 E1

Or E2 

E1 ( E1 / mc )(1  cos  )  1 2

3-44. (a) eV0  hf    hc /    e  0.52V    hc / 450nm   

(i)

e 1.90V    hc / 300nm   

(ii)

68

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-44 continued) Multiplying (i) by 450nm / e and (ii) by 300nm / e , then subtracting (ii) from (i) and rearranging gives:

 300nm 1.90V    450nm  0.52V    2.24eV e 150nm



e  300 109 m   4.14V  hc (b)  1.90  2.24  h   6.63 1034 J s 8 e  300nm   3.00 10 m / s 

3-45. Including Earth’s magnetic field in computing y2, first show that y2 is given by

y2 

e  B 2 x1 x2 1 BE x22     m E 2 E 

where the second term in the brackets comes from Fy  euBE  ma y and y 

1 2 a yt . 2

e  B 2 x1 x2 1 BE x22   Thus, 1    The first term inside the brackets is the reciprocal of m  Ey2 2 Ey2 

0.7  1011 C, Thomson’s value for e/m. Using Thomson’s data (B = 5.5 104 T , E  1.5  104V / m,

x1  5cm,

y2 / x2  8 / 110 ) and the modern value for e/m =

1.76  1011 C / kg and solving for BE. 1 BE Bx22  8.20  1012. The minus sign means that B and BE are in opposite directions, 2 Ey2

which is why Thomson’s value underestimated the actual value.

BE 



 

 m

 8.20  1012  2  1.5  104V / m 8 / 110 

5.5 10 T 8 10 4

2

69

2

 3.1  105 T  31T

Chapter 3 – Quantization of Charge, Light, and Energy

3-46. (a) Q  Ne and cM T  N

mu 2 where N = number of electrons, c = specific heat of 2

the cup, M = mass of the cup, and u = electron’s speed.

Q 2cM T  e mu 2

N (b)  

eB mu



 u

e Qu 2  m 2cM T

eB m

Substituting u into the results of (a), e Q  eB / m   m 2cM T

2

Solving for e/m,

e 2cM  2 T  m QB 2 2

3-47. Calculate 1/λ to be used in the graph. 1/λ (106/m)

5.0

3.3

2.5

2.0

1.7

V0 (V)

4.20

2.06

1.05

0.41

0.03

5 4 3

V0 (V)

2 1 0 -1 -2

1

2

3

4

5

1/λ (106/m)

(a) The intercept on the vertical axis is the work function  .   2.08eV .

70

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-47 continued) (b) The intercept on the horizontal axis corresponds to the threshold frequency. 1

t

 1.65 106 / m

ft 

c

t

  3.00 108 m / s 1.65 106 / m   4.95 1014 Hz

(c) The slope of the graph is h/e. Using the vertical intercept and the largest experimental point.

4.20V   2.08V  h 1 V0    4.19  1015 eV / Hz 8 6 e c  1 /   3.00  10 m / s 5.0  10 / m  0







3-48. In the center of momentum reference frame, the photon and the electron have equal and opposite momenta. p  E / c   pe .



The total energy is: E  Ee  E  pe2c 2  m2c 4



1/ 2



 E  E2  m2c 4



1/ 2

By conservation of momentum, the final state is an electron at rest, pe  0 . Conservation of energy requires that the final state energy E  is

E   E  Ee





 mc  E  p c  mc 2





2 2

1/ 2

2  mc 2  E   p 2 c 2  mc 2   



Squaring yields, mc 2



2

2



2 1/ 2





2   E2  mc 2   



 2mc 2 E  E2  E2  mc 2

1/ 2



2

 mc 2 E  0. This can be true

only if E vanishes identically, i.e., if there is no photon at all.

3-49. Bragg condition:

m  2d sin .   (2)(0.28nm)(sin 20)  1.92  1010 m  0.192nm.

This is the minimum wavelength m that must be produced by the X ray tube.

m 

1.24  103 nm V

or

V

1.24  103  6.47  103V  6.47kV 0.192

71

Chapter 3 – Quantization of Charge, Light, and Energy









3-50. (a) E  100W  104 s  100 J / s  104 s  106 J The momentum p absorbed is p  (b)

 



E 106 J   3.33  103 J s / m 8 c 3.00  10 m / s









p  m v f  vi  2  103 kg v f  0  3.3  103 J s / m v 

3

3.33  10 J s / m  1.67m / s 2  103 kg





2  103 kg 1.67m / s  1 2 (c) E  mv f   2.78  103 J 2 2 2

The difference in energy has been (i) used to increase the object’s temperature and (ii) radiated into space by the blackbody. 3-51. Conservation of energy: E1  mc2  E2  Ek  mc 2  Ek  E1  E2  hf1  hf 2 From Compton’s equation, we have: 2  1  Thus,

h 1  cos  , mc

1 1 h   1  cos  f 2 f1 mc 2

1 1 h   1  cos   f2 f1 mc 2

f2 

f1mc 2 mc 2  hf1 1  cos 

Substituting this expression for f 2 into the expression for Ek (and dropping the subscript on f1 ): hfmc 2   hf  1  cos   hfmc 2 hfmc 2 Ek  hf  2   mc  hf 1  cos  mc 2  hf 1  cos  2

hf mc 2 1  hf 1  cos  

Ek has its maximum value when the photon energy change is maximum, i.e., when    so cos  1. Then Ek 

hf mc 2 1 2hf

72

Chapter 3 – Quantization of Charge, Light, and Energy

3-52. (a) mT  2.898  103 m K

(b) Equation 3-18:

 T

2.898  103 m K  3.50  104 K 9 82.8  10 m

 70nm  /  ehc /70nmkT  1  u 82.8nm   82.8nm 5 /  ehc / 82.8nmkT  1 5

u  70nm 







6.63  1034 J s 3.00  108 m / s hc   5.88 and where  70nm  kT 70  109 m 1.38  1023 J / K 3.5  104 K







 70nm  /  e5.88  1   0.929 u 82.8nm  82.8nm 5 /  e4.97  1 5

u  70nm 

hc  4.97 82.8nm  kT

Similarly,



100nm  /  e4.12  1   0.924 u 82.8nm  82.8nm 5 /  e4.97  1 5

u 100nm 

3-53. Fraction of radiated solar energy in the visible region of the spectrum is the area under the Planck curve (Figure 3-6) between 350nm and 700nm divided by the total area. The latter is 6.42  107W / m2 (see solution to Problem 3-25). Evaluating u     with

  525nm (midpoint of visible) and   700nm  350nm  350nm. u     







8 6.63  1034 J s 3.00  108 m / s  525nm 



 

5



350nm 

 6.63  10 J s 3.00  10 m / s   1 exp  23  1.38  10 J / k  5800 K  525nm  



R   350  700  

34



8



 0.389 J / m3



c u  3.00  108 m / s 0.389 J / m3 / 4  2.92  107W / m2 4







Fraction in visible = R  350  700 / R  2.92  107 W / m2 / 6.42  107 W / m2  0.455

3-54. (a) Make a table of f  c /  vs. V0 .



f 1014 Hz

V0 (V)



11.83

9.6

8.22

7.41

6.91

2.57

1.67

1.09

0.73

0.55

73

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-54 continued) 3

Li

2

Pb

1 0 2

−1

4

6

8

10

12

−2 −3 −4

The work function for Li (intercept on the vertical axis) is   2.40eV . (b) The slope of the graph is h/e. Using the largest V0 and the intercept on the vertical





 4.97V  1.60  1019 C h 2.57V   2.40V  axis,  or, h   6.89  1034 J s 14 14 11.53  10 Hz e 11.53  10 Hz  0 (c) The slope is the same for all metals. Draw a line parallel to the Li graph with the work function (vertical intercept) of Pb,   4.14eV . Reading from the graph, the threshold frequency for Pb is 9.8  1014 Hz; therefore, no photon wavelengths larger







than   c / ft  3.00  108 m / s 9.8  1014 Hz  306nm will cause emission of photoelectrons from Pb.

3-55. (a) Equation 3-18: u     gives u     (b)

8 hc 5 Letting C  8 hc and a  hc / kT ehc /  kT  1

C 5 ea /   1





  5  1 ea /  a 2  du d  C  5  5 6     a/   C 2 a/ d  d   ea /   1 e  1 e  1   6 6 a /  C  a a/  C e a   e  5 ea /   1    5 1  ea /    0 2  2    ea /   1   ea /   1  











74











Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-55 continued) The maximum corresponds to the vanishing of the quantity in brackets. Thus, 5 1  e a /   a





(c) This equation is most efficiently solved by trial and error; i.e., guess at a value for





a /  in the expression 5 1  e a /   a , solve for a better value of a /  ; substitute

the new value to get an even better value, and so on. Repeat the process until the calculated value no longer changes. One succession of values is 5, 4.966310, 4.965156, 4.965116, 4.965114, 4.965114. Further iterations repeat the same value (to seven digits), so we have

a

m

 4.965114 



hc m kT



6.63  1034 J s 3.00  108 m / s hc  (d) mT   4.965114 k  4.965114 1.38  1023 J / K



Therefore, mT  2.898  103 m K

3-56. (a) I 





Equation 3-5

P 1W  1  2  2 19 4 R 4 1m   1.602  10 J / eV

 17 2   4.97  10 eV / m s 

(b) Let the atom occupy an area of  0.1nm  . 2







dW 2  IA  4.97  1017 eV / m2 s  0.1nm  109 m / nm dt

(c) t 

 dW / dt





2

 4.97  103 eV / s

2eV  403s  6.71 min 4.97  103 eV / s

3-57. (a) The nonrelativistic expression for the kinetic energy pf the recoiling nucleus is

p 2 15MeV / c   1u  Ek    1.10  104 eV  2  2m 2  12u  931.5MeV / c  2

Internal energy U  15MeV  0.0101MeV  14.9899MeV

75

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-57 continued) (b) The nucleus must recoil with momentum equal to that of the emitted photon, about 14.98 MeV/c.

p 2 14.98MeV / c   1u  Ek    1.00  102 eV  2  2m 2  12u  931.5MeV / c  2

E  U  Ek  14.9899MeV  0.0100MeV  14.9799MeV

3-58. Derived in Problem 3-47, the electron’s kinetic energy at the Compton edge is

Ek 

hf 1  mc 2 / 2hf

hf E  520keV  1   511keV  / 2hf

 520keV 

2  hf 

2

2hf  511keV

Thus,  hf   520  hf    520  511 / 2  0 2

2

2 520   520    2  520  511    708keV Solving with the quadratic formula: hf  2

(only the + sign is physically meaningful). Energy of the incident gamma ray hf  708keV .

hc



 708keV



6.63  10 

34



J s 3.00  108 m / s

 708keV  1.60  10

16

J / keV



  1.76  10

3-59. (a) Ek  50keV and 2  1  0.095nm hc

1



hc

2

 5.0  104 eV

1



1

  

12   0.095nm 

2hc 5  104 eV



1 5.0  104 eV  1  0.095 hc

21  0.095 5.0  104 eV  12  0.0951 hc

 0.095nm  hc   1  5  104 eV  0 

 12  0.045411  2.36  103  0

76

12

m  1.76 pm

Chapter 3 – Quantization of Charge, Light, and Energy (Problem 3-59 continued) Applying the quadratic formula,





2 0.04541   0.04541  4 2.36  103    1  2

1/ 2

1  0.03092nm and 2  0.1259nm hc

(b) E1 

1



3-60. Let x 

kT







1240eV nm  40.1 keV 0.03092nm



Eelectron  9.90keV

hf in Equation 3-15: kT



   e 

f n  A e nx  A e0  e x  e x   n 0 n 0

2

3

x





  A 1  y  y 2  y3  

 1

Where y  e x . This sum is the series expansion of

1  y 

1

, i.e., 1  y   1  y  y 2  y3  1

. Then

f

 A 1  y   1 gives A  1  y. 1

n

Writing Equation 3-16 in terms of x and y. 





n 0

n 0

n 0

E   En Ae En / kT  A nhfe nhf / kT  Ahf  ne nx

Note that

 ne

 nx

 ne



 nx

   d / dx   e nx . But  e nx  1  y  , so we have 1

d d dy  1 2  2 e nx   1  y   1  y      y 1  y   dx dx  dx 

 

x dy d e Since   e x   y. dx dx

Multiplying this sum by hf and by A  1  y  , the average energy is 

E  hfA ne nx  hf 1  y  y 1  y   2

n 0

hfy hfe x  1  y 1  e x

Multiplying the numerator and the denominator by e x and substituting for x, we obtain

E

hf e

hf / kT

1

, which is Equation 3-17.

77

Chapter 4 – The Nuclear Atom

4-1.

1

mn

1   1  R  2  2  where R  1.097  107 m1 (Equation 4-2) n  m

The Lyman series ends on m =1, the Balmer series on m =2, and the Paschen series on 1 m =3. The series limits all have n  , so  0 n 1 1  R  2   1.097  107 m1 L 1  L  limit   1.097 107 m1  91.16 109 m  91.16nm

 1  R  2   1.097  107 m1 / 4 B 2  B  limit   4 / 1.097 107 m1  3.646 107 m  364.6nm 1

1  R  2   1.097  107 m1 / 9 P 3  1

P  limit   9 / 1.097  107 m1  8.204  107 m  820.4nm

4-2.

1

mn

1   1  R  2  2  where m  2 for Balmer series (Equation 4-2) n  m

1 1.097  107 m1  1 1    2  9 2 379.1nm 10 nm / m  2 n  1 1 109 nm / m  2   0.2405 4 n 379.1nm 1.097  107 m1





1  0.2500  0.2405  0.0095 n2 n2 

1 0.0095



n  1 / 0.0095

1/ 2

 10.3 

n  10  n  2

79

n  10

Chapter 4 – The Nuclear Atom

4-3.

1

mn

1   1  R  2  2  where m  1 for Lyman series (Equation 4-2) n  m 1 1.097  107 m1  1   1 2   9 164.1nm 10 nm / m  n  1 109 nm / m  1   1  0.5555  0.4445 n2 164.1nm 1.097  107 m1



n  1 / 0.4445

1/ 2



 1.5

No, this is not a hydrogen Lyman series transition because n is not an integer.

4-4.

1

mn

1   1  R 2  2  n  m

(Equation 4-2)

For the Brackett series m = 4 and the first four (i.e., longest wavelength lines have n = 5, 6, 7, and 8.

 1 1  1.097  107 m1  2  2   2.468  105 m1 45 4 5  1

45 

1  4.052  106 m  4052nm. Similarly, 5 1 2.68  10 m

46 

1  2.625  106 m  2625nm 3.809  105 m1

47 

1  2.166  106 m  2166nm 5 1 4.617  10 m

48 

1  1.945  106 m  1945nm 5 1 5.142  10 m

These lines are all in the infrared.

4-5.

None of these lines are in the Paschen series, whose limit is 820.4 nm (see Problem 4-1) and whose first line is given byL

1 1   R  2  2   34  1875nm. Also, none are in 34 3 4  1

the Brackett series, whose longest wavelength line is 4052 nm (see Problem 4-4). The Pfund series has m = 5. Its first three (i.e., longest wavelength) lines have n = 6, 7, and 8.

80

Chapter 4 – The Nuclear Atom (Problem 4-5 continued)

1 1  1.097  107 m1  2  2   1.341  105 m1 56 5 5  1

56 

1  7.458  106 m  7458nm. Similarly, 5 1 1.341  10 m

57 

1  4.653  106 m  4653nm 5 1 2.155  10 m

58 

1  3.740  106 m  3740nm 2.674  105 m1

Thus, the line at 4103 nm is not a hydrogen spectral line.

4-6.

(a) f   b2 nt (Equation 4-5)

For Au, n  5.90  1028 atoms / m3 (see Example 4-2) and for this foil t  2.0 m  2.0  106 m. 2 kq Q   2  79  ke 90  2  79 1.44eV nm  b cot  cot  m v 2 2 2 K 2 2 7.0  106 eV



5

 1.63 10 nm  1.63 10



f   1.63  1014 m

(b)

For   45,

14

m

 5.90 10 2



28





/ m3 2.0  106 m  9.8  105

b  45   b  90  cot 45 / 2  /  cot 90 / 2   b 90  tan 90 / 2  /  tan 45 / 2   3.92 105 nm  3.92  1014 m

and f  45   5.7  10 4 For   75,

b  75   b  90  tan 90 / 2  /  tan 75 / 2   2.12 105 nm  2.12  1014 m

and f  75   1.66  104

Therefore, f  45  75  5.7  104  1.66  104  4.05  104

81

Chapter 4 – The Nuclear Atom (Problem 4-6 continued) (c) Assuming the Au atom to be a sphere of radius r,

4 3 M 197 g / mole r   23 3 NA 6.02  10 atoms / mole 19.3g / cm3





 3 197 g / mole r 23  4 6.02  10 atoms / mole 19.3g / cm3







  



1/ 3

r  1.62 103 cm  1.62 1010 m  16.2nm

4-7.

N 

1

sin  / 2  4



A

(From Equation 4-6), where A is the product of the two

sin  / 2  4

quantities in parentheses in Equation 4-6.

N 10 

(a)

N 1 

N  30 

(b)

4-8.

N 1 

sin4  0.5  sin4 15 



sin4  0.5  sin4  5 

 1.01  104

 1.29  106

(Equation 4-3)

k 2e Ze  1.44MeV fm  Z  cot  cot 2 m v 2 Ek 2



4-9.



A / sin4 1 / 2 

kq Q  cot 2 m v 2

b

=



A / sin4 10 / 2 

rd 

1.44MeV

fm  79 

7.7 MeV

cot

kq Q ke2 2 79  Ek 1 / 2 m v 2

2  8.5  1013 m 2

(Equation 4-11)

1.44MeV

For Ek  5.0MeV :

rd 

For Ek  7.7MeV :

rd  29.5 fm

For Ek  12MeV :

fm  2  79 

5.0MeV

rd  19.0 fm

82

 45.5 fm

Chapter 4 – The Nuclear Atom

4-10.

rd 

kq Q ke2 2 79  Ek 1 / 2 m v 2

Ek 

4-11.

1.44MeV

fm  2 13

4 fm

xrms  N  

n

(Equation 4-11)

 9.4MeV

10  N  0.01   N  10 / 0.01   106 collisions 2

t 106 m  10  104 layers t 10 m

104 atomic layers is not enough to produce a deflection of 10 , assuming 1 collision/layer.

4-12. (a) f   b2 nt

(Equation 4-5)

For   25 (refer to Problem 4-6).

2  79  ke2  25  2  79 1.44eV b cot  2 K



2 7.0  10 eV

2

6

nm 



 25  cot    2 

 7.33 105 nm  7.33 1014 m



f   7.33  1014 m

 5.90 10 2

Because N  f  N  1000 For   45, b 



28



N  1000 / 1.992  103  5.02  105



 2  79 1.44eV

f   3.92  1014 m



nm 

2 7.0  10 eV 6

 5.90 10 2

28



/ m3 2.0  106 m  1.992  103



 45  cot   3.92  1014 m   2 





/ m3 2.0  106 m  5.70  104





Because N   45  f  N  5.70  104 5.02  105  286 (b)

N  25  45  1000  286  714

(c)

For   75 , b  b   25 tan 25 / 2 /  tan 75 / 2  2.12  1014 m



f  1.992  103 2.12  1014 m

 / 7.33  10 2

 1.992  103  2.12 / 7.33  1.67  104 2

83

14

m



2

Chapter 4 – The Nuclear Atom (Problem 4-12 continued) For   90, b  b   25 tan 25 / 2 /  tan 90 / 2  1.63  1014 m



f  1.992  103 1.63  1014 m

 / 7.33  10 2

14

m



2

 1.992  103 1.63 / 7.33  9.85 105 2





N  f  N  9.85  105 5.02  105  49 N   75  90   84  49  35 n2 a0 4-13. (a) rn  Z r6 

(Equation 4-18)

62  0.053nm  1





(b) r6 He 

4-14.

a0 

mke2

E1 

2

(Equation 4-19)

mk 2 e4 2 2

(from Equation 4-20)

 

mc 2 ke2 2  c

2

2

2

mc 2  ke2  1 2 2     mc  2 2  c

c 0.00243nm   0.053nm 2 2 1 / 137 

 1 1   Z 2R  2  2   n f ni     1

 0.95nm

 c c 1 1 h 1   2    2  c 2 2 mcke mc ke / c 2 mc ke / c 2

a0 

4-15.

62  0.053nm 

2





 1.91nm

E1 

(Equation 4-22)

84

1 2 2 5.11  105 eV mc    13.6eV 2 2 2 137 

Chapter 4 – The Nuclear Atom (Problem 4-15 continued)

1 1   n2  1   R  2  2  =R  i 2  ni  1 ni   ni  1

ni 



ni2

 n i2   91 . 17 nm    2  1.0968  107 m n i2  1  ni  1 

=

n i2

 

R ni2  1





4 9  91.17nm   121.57nm 3   91.17nm   102.57nm 3 8 16 4   91.17nm   97.25nm   91.17 nm 15 None of these are in the visible; all are in the ultraviolet.

2 

∞

4

3

2

λ, nm 80

4-16.

90

L  mvr  n

100

110

120

(Equation 4-17) vE  2 r / 1y  2 r / 3.16 107 s

mE  5.98  1024 kg







n  m 2 r / 3.16  107 s r /  2 mr 2 / 3.16  107 s 





2 5.98  1024 kg 1.50  1011 m



3.16  10 s 1.055  10 7

mv  n / r

130



34

J s





2

 2.54  1074

E   mv  / 2m   n / r  / 2m 2





2

(from Equation 4-17)

 



2 1.055  1034 J s 2.54  1074 1   1 E     0.210  1040 J  2nn   2 11 24 r 2 m   1.50  10 m 5.98  10 kg 2





This would not be detectable.

n  E 



  2.54  10   r   5.34  10  r   5.98  10 kg 

34  2r  1.055  10 J s   2m  r 3  1.50  1011 m 2

40

or  r  0.210  10



2

74

2

33

2

24

J / 5.34  1033 J / m  3.93  1075 m

The orbit radius r would still be 1.50  1011 m. 85

Chapter 4 – The Nuclear Atom

4-17.

f rev  

mk 2 Z 2e4 2 3n3

(Equation 4-29)

 

mc 2 Z 2 ke2

2 n3  c 

2

2

2

 ke2  cZ 2 cZ 2 2    h / mc  n3  c  c n3

3.00 10 m / s  1  1   8.22  10 Hz   0.00243  10 m   2  137  N  f t  8.22  10 Hz 10 s   8.22  10 revolutions 2

8

2

14

9

3

8

14

6

rev

4-18. The number of revolutions N in 10-8 s is: N  108 s /  time/revolution   108 s /  circumference of orbit/speed 

N  108 s /  C / v   108 s /  2 r / v  The radius of the orbit is given by: r

2 n2 a0 4  0.0529nm   Z 3

so the circumference of the orbit C  2 r is C  2 42  0.0529nm  / 3  1.77nm  1.77  109 m

The electron’s speed in the orbit is given by



8.99  10 N m / mr   9.11 10 9

v  kZe 2

2

2

   kg 1.77  10 m 

/ C 2  3 1.60  1019 C

31

2

9

v  6.54  105 m / s

Therefore, N  108 s /  C / v   3.70  106 revolutions In the planetary analogy of Earth moving around the sun, this corresponds to 3.7 million “years”.

4-19. (a) au  (b) E 

2

 ke2



 k 2e4 2

2

2 e e 9.11  1031 kg  a   0.0529nm   2.56  104 nm  e ke2  0 1.69  1028 kg



  e k 2 e 4   1.69  1028 kg  E  13.6eV   2520eV e 2 2 e 0 9.11  1031 kg

86

Chapter 4 – The Nuclear Atom (Problem 4-19 continued) (c) The shortest wavelength in the Lyman series is the series limit ( ni  , n f  1). The photon energy is equal in magnitude to the ground state energy  E .

 

hc 1240eV nm   0.492nm E 2520eV

(The reduced masses have been used in this solution.)

1/ 2

4-20.

E   Z  E0 / n 2

 n 2 E  Z=    E0 

2

 22  5.39eV     13.6eV  

1/ 2

 1.26

4-21. Energy (eV) n=∞ n=4 n=3

0 -2

(d) n=2

-4

(b)

(c)

-6 -8 -10 -12 n=1 (a) -14

(a) Lyman limit, (b) H  line, (c) H line, (d) longest wavelength line of Paschen series

4-22. (a)

 1 1   R 2  2   n f ni     1

For Lyman α:

EL 

hc

L



1 1  1.097373  107 m1  2  2   1 2  1



L  121.5023nm

E 1240eV nm  10.2056eV and pL  L  10.2056eV / c 121.5023nm c

87

Chapter 4 – The Nuclear Atom (Problem 4-22 continued) Conservation of momentum requires that the recoil momentum of the H atom

pH  pL and the recoil energy EH is: EH   pH  / 2mH   pH c  / 2mH c  2

2

2



10.2056eV / c 



2 1.007825uc 2 931.50  106 eV / uc 2

 5.55  108 eV

EH 5.5  108 eV (b)   5  109  EL  EH  10.21eV

4-23. (a) For C5+ (Z = 6) En  13.6 0

Z2 489.6  2 2 n n

n=∞ n=5 n=4 n=3

E∞ = 0 eV E5 = -19.6 eV E4 = -30.6 eV E3 = -54.4 eV E2 = -122.4 eV E1 = -489.6 eV

-100 n=2

En (eV) -200

-300

-400

n=1 -500

(b)  

hc hc 1240eV nm    18.2nm E E3  E2  54.4   122.4  eV

(c) 18.2nm lies in the UV (ultraviolet) part of the EM spectrum.

88

2



Chapter 4 – The Nuclear Atom 4-24. (a) The reduced mass correction to the Rydberg constant is important in this case. 1  R  R  1 m / M En  hcR / n2

 1 6 1   R  2   5.4869  10 m   

(from Equation 4-26)

(from Equations 4-23 and 4-24)







E1   1240eV nm  5.4869  106 m1 109 m / nm / 1  6.804eV 2

Similarly, E2  1.701eV and E3  0.756eV (b) Lyman α is the n  2  n  1 transition.

hc



 E2  E1



 

hc 1240eV nm   243nm E2  E1 1.701eV   6.804eV 

Lyman β is the n  3  n  1 transition.

 

hc 1240eV nm   205nm E3  E1 0.756eV   6.804eV 

4-25. (a) The radii of the Bohr orbits are given by (see Equation 4-18) r  n2 a0 / Z where a0  0.0529nm and Z  1 for hydrogen.

For n  600, r   600   0.0529nm   1.90  104 nm  19.0 m 2

This is about the size of a tiny grain of sand. (b) The electron‟s speed in a Bohr orbit is given by v2  ke2 / mr with Z  1

Substituting r for the n = 600 orbit from (a), then taking the square root,





  2



v 2  8.99  109 N m2 1.609  1019 C / 9.11 1031 kg 19.0  106 m

v2  1.33  107 m2 / s 2

 v  3.65  103 m / s

For comparison, in the n = 1 orbit, v is about 2  106 m / s

89



Chapter 4 – The Nuclear Atom

4-26. (a)

1  2 1  R  Z  1  2  2    n1 n2  1

1

 1  2 1 3   1.097  107 m1  42  1  2  2    6.10  1011 m  0.0610nm  1 3  





1

 1  2 1 4   1.097  107 m1  42  1  2  2    5.78  1011 m  0.0578nm  1 4  



(b) lim it

4-27.



1

 2 1    1.097  107 m1  42  1  2  0    5.42  1011 m  0.0542nm 1  





1  1  2 1 2 1  R  Z  1  2  2   R  Z  1  2  2  for K  1 2   n1 n2  1

1/ 2

  1/ 2     1 1    Z 1     R 1  1     0.0794nm  1.097  102 / nm  3 / 4     4     Z  1  39.1  40 Zirconium





4-28. (a) Z  43; f 1 / 2  21 108 Hz1 / 2



f  4.4  1018 Hz

Z  61; f 1 / 2  30  108 Hz1 / 2



f  9.0  1018 Hz

Z  75; f 1 / 2  37 108 Hz1 / 2



f  1.4 1019 Hz

Note: f 1 / 2 for Z  61 and 75 are off the graph 4-19; however, the graph is linear and extrapolation is easy. (b) For Z = 43  

1

(Equation 4-37)

1   R  Z  1 1  2   n  2

where R  1.097  107 m1 and n  2



1

1.097  10 m  7

1

1  43  1 1    4 2

90

 6.89  1011 m  0.0689nm

Chapter 4 – The Nuclear Atom (Problem 4-28 continued) Similarly, For Z = 61, λ = 0.0327nm For Z = 75, λ = 0.0216nm

4-29.

n2 a0 (Equation 4-18) Z The n =1 electrons “see” a nuclear charge of approximately Z  1 , or 78 for Au. rn 







r1  0.0529nm / 78  6.8  104 nm 109 m / nm 1015 fm / m  680 fm , or about 100 times

the radius of the Au nucleus.

4-30.

En  13.6

Z2 eV n2

(Equation 4-20)

For Fe (Z = 26) E1

 26   13.6

2

12

 9.194keV

The fact that E1 computed this way (i.e., by Bohr theory) is approximate, is not a serious problem, since the Kα x-ray energy computed from Figure 4-19 provides the correct spacing between the levels. The energy of the Fe Kα x-ray is:

E  Fe K   hf where f 1 / 2  12.2  108 Hz1 / 2





E  Fe K   6.626  1034 J s 12.2  108 Hz1 / 2



2

 9.862  1016 J  6.156keV

Therefore, E2  E1  E  K    9.194  6.156   3.038keV The Auger electron energy E  K   E2  6.156  3.038  3.118keV

4-31.

E   me c 2 

511keV



1  2.25  10 / 3.00  10 m / s 8

8



2

 772.6keV

After emitting a 32.5 keV photon, the total energy is:

91

Chapter 4 – The Nuclear Atom (Problem 4-31 continued) E  740.1keV 

511keV 1 

2 v  1   511 / 740    

1/ 2

4-32. (a)  E1  E0 Z 2 / n2

 2  v 2 / c 2  1   511 / 740 



2

2

c  2.17  108 m / s

(Equation 4-20)

 13.6eV  74  1 / 1  7.25  104 eV  72.5keV 2

(b)

2

 E1  E0  Z    / n2  69.5  103 eV  13.6eV  74    / 1 2

 74   

2

2

2

 69.5  103 eV / 13.6eV

  74   69.5  103 eV / 13.6eV 

1/ 2

 2.5

4-33. Element

Al

Ar

Sc

Fe

Ge

Kr

Zr

Ba

Z

13

18

21

26

32

36

40

56

E (keV)

1.56

3.19

4.46

7.06

10.98

14.10

17.66

36.35

6.14

8.77

10.37

13.05

16.28

18.45

20.64

29.62



f 1 / 2 108 Hz1 / 2



60 Z 40

20

0

0

5

10

15

20

25



30

f 1 / 2 108 Hz1 / 2

92



Chapter 4 – The Nuclear Atom (Problem 4-33 continued)

slope =

58  10  1.90  108 Hz 1 / 2 8  30  4.8  10

slope (Figure 4-19) =

30  13  2.13  108 Hz 1 / 2 8 5  7  10  

The two values are in good agreement.

4-34. (a) The available energy is not sufficient to raise ground state electrons to the n =5 level which requires 13.6 − 0.54 = 13.1eV. The shortest wavelength (i.e., highest energy) spectral line that will be emitted is the 3rd line of the Lyman series, the n = 4 → n = 1 transition. (See Figure 4-16.) (b) The emitted lines will be for those transitions that begin on the n = 4, n = 3, or n = 2 levels. These are the first three lines of the Lyman series, the first two lines of the Balmer series, and the first line of the Paschen series.

60

4-35.

60 15.7eV

50 E (eV)

44.3

40 14.7eV

Average transition energy = 15.7 eV 30

29.6 16.6eV

20 13.0

10

93

Chapter 4 – The Nuclear Atom

4-36.

E 

hc





1240eV nm  1.610eV . The first decrease in current will occur when the 790nm

voltage reaches 1.61V. 4-37. Using the results from Problem 4-24, the energy of the positronium Lyman α line is

E  E2  E1  1.701eV   6.804eV   5.10eV . The first Franck-Hertz current decrease would occur at 5.10V, the second at 10.2V.

4-38. In an elastic collision, both momentum and kinetic energy are conserved. Introductory physics texts derive the following expression when the second object (the Hg atom here)

 m  m2  is initially at rest: v1 f   1  v1i . The fraction of the initial kinetic energy lost by  m1  m2  the incident electron in a head-on collision is: 2

f 

KEei  KEef KEei



v12i  v12f v12i

 m  m2  2 v  1  v m1  m2  1i   v12i 2 1i

 

2  0.511MeV  200uc 2 931.5MeV / uc 2  m1  m2  = 1    1   0.511MeV  200uc 2 931.5MeV / uc 2  m1  m2  

   

2

= 1.10  105 If the collision is not head-on, the fractional loss will be less. 4-39. (a) Equation 4-24: En   E0 / n2  13.6 / n2 eV  1 1 1   1 En1  En  13.6   2  eV  13.6  2  2  eV=2.89 104 eV 2  46 45   (n  1) n 

(b) Ionization energy  En  13.6 / n2  13.6 / 452  6.72 103 eV (c) E  hf  hc /   f  E / h  (2.89 104 eV)(1.60 1019 J/eV)/6.63 1034 J  s

f  6.97 1010 Hz   c / f  (3.00 108 m/s)/(6.96 1010 Hz)  4.30 103 m=4.30mm 94

Chapter 4 – The Nuclear Atom (Problem 4-39 continued) (d) Equation 4-18: rn  n2 a0 / Z r45  452 a0 /1  107 nm  1.07 104 mm , or 2025× the radius of the

For hydrogen:

hydrogen atom ground state.

1   7 1 4-40. (a) Equation 4-26: R  R   where R  1.0973732 10 m  1 m / M    1 Rd  R   31 27  1  9.1094 10 kg / 3.3436 10 kg  Rd  1.0970743 107 m 1   1 Rt  R   31 27  1  9.1094 10 kg / 5.0074 10 kg  Rt  1.0971736 107 m 1

(b) Equation 4-22:

 1 1   R  2  2  with Z  1 . n    f ni  1

 1 1   1 1 5 The Balmer α transition is n  3  n  2 .  2  2    2  2   n   f ni   2 3  36

d  t 

36  1 1    5.3978 1011 m  5.3978 102 nm  5  Rd Rt 

(c) Computed as in (b) above with RH  1.096762 107 m1 ,

H  t 

36  1 1    2.4627 1010 m  2.4627 101 nm  5  RH Rt 

95

Chapter 4 – The Nuclear Atom

4-41.

N  I 0  2 b  db where b  and db 

kq Q  cot 2 m v 2

(Equation 4-3)

kq Q    csc  d 2  2m v  2 2

 kq Q   1    N  I 0 2   2   cot  csc 2  d 2  2  m v   2 Using the trigonometric identities:

csc 2 

1 sin  / 2 2

and cot

 2



sin sin sin   2 2 1  cos 1  cos  / 2   sin  / 2  2 sin2  / 2 

   kq Q   1   sin 1 N  I 0 2   2     d  2 2  2 sin  / 2   sin  / 2   m v 2         2

and inserting 2e  q and Ze  Q, 2

 kZe2  sin d N  I 0 2  2  4  m v  sin  / 2  4-42. Those scattered at   180 obeyed the Rutherford formula. This is a head-on collision where the α comes instantaneously to rest before reversing direction. At that point its kinetic energy has been converted entirely to electrostatic potential energy, so k  2e  79e  1 where r = upper limit of the nuclear radius. m v 2  7.7 MeV  2 r r

4-43. (a)

k  2  79  e2 7.7 MeV

i  qf rev =e



2  79 1.440MeV fm  7.7 MeV

Z 2 mk 2 e4 e 2 3n3

  1

mc 2 ke 2 2

2

 c  1 2

1.602  10 =

19

2

3

 29.5 fm

(from Equation 4-28) 2

ec  ke 2  ec 2   c  h / mc   c 





C 3.00  1017 nm / s  1 2  1.054  103 A   0.00243nm  137 

96

Chapter 4 – The Nuclear Atom (Problem 4-43 continued)

(b)

 emk 2 e4   2  e  = 3   2   2   mke  2m

  iA  i a02  

1.602  10 C 1.055  10 2  9.11  10 kg  19

=

34

J s

31

  9.28  10

24

A m2

or



 

= 1.054  103 A  0.529  1010 m



2

 9.27  1024 A m2

4-44. Using the Rydberg-Ritz equation (Equation 4-2), set-up the columns of the spreadsheet to carry out the computation of λ as in this example (not all lines are included here).

4-45.

C=m2 D=n2

1/λ

λ (nm)

0.96

10534572

94.92

16

0.9375

10287844

97.20

1

9

0.888889

9754400

102.52

2

1

4

0.75

8230275

121.50

2

6

4

36

0.222222

2438600

410.07

2

5

4

25

0.21

2304477

433.94

2

4

4

16

0.1875

2057569

486.01

2

3

4

9

0.138889

1524125

656.11

3

7

9

49

0.090703

995346.9

1004.67

3

6

9

36

0.083333

914475

1093.52

3

5

9

25

0.071111

780352

1281.47

3

4

9

16

0.048611

533443.8

1874.61

m

n

1

5

1

25

1

4

1

1

3

1

  1 1     R  2  2    n f ni  

Because R  ,

1

1/C−1/D

d      R 2 d





dR / d   R / .    R

97

2



1

 1 1  dR   2  2   n f ni  d 



 1 1   2  2   n f ni 

1

 R /        /  

Chapter 4 – The Nuclear Atom (Problem 4-45 continued)

H  





me m p me  m p

D 

me md me  md

me  md  m p  m m /  me  md  m /  me  md  D  H D  1  e d 1  d 1  H H m p  me  md  me m p /  me  m p  m p /  me  m p 

If we approximate md  2mp and me

     /      656.3nm 

md , then







me and 2m p

0.511MeV  0.179nm 2  938.28MeV 

4-46. For maximum recoil energy for the Hg atoms, the collision is „head-on‟. (a) Before collision

Ek 

kinetic energy

1 2 mvei 2

pei  mvei

momentum

After collision Ek 

1 2 mvef 2

EHg 

1 2 MvHg 2

pef  mvef pHg  M vHg

Conservation of momentum requires: mvei  mvef  MvHg  vHg 



m vei  vef M



Therefore, the maximum Hg recoil kinetic energy is given by: 2

2 1 1 m m2 2 2 MvHg  M    vei  vef   vei  2vei vef  vef2 2 2 M  2M 2 m  4vei2 since m M , vei  vef 2M 4m  1 2  4m  mvei   Ek M  2  M



 

98



Chapter 4 – The Nuclear Atom (Problem 4-46 continued)

(b) Since the collision is elastic, kinetic energy is conserved, so the maximum kinetic energy gained by the Hg atom equals the maximum kinetic energy lost by the electron. If Ek = 2.5eV, then the maximum lost is equal to:

9.11  1031 kg  2.5eV  m 4  2.5eV   4  2.7  105 eV 27 M 201 u 1 . 66  10 kg / u  



4-47. (a) En   E0 Z 2 / n2



(Equation 4-20)

For Li++, Z = 3 and En  13.6eV  9  / n2  122.4 / n2eV The first three Li++ levels that have the same (nearly) energy as H are:

n  3, E3  13.6eV

n  6, E6  3.4eV

n  9, E9  1.51eV

Lyman α corresponds to the n = 6 → n = 3 Li To the n = 9 → n = 3 Li

++

++

transitions. Lyman β corresponds

transition.

(b) R  H   R 1 / 1  0.511MeV / 938.8MeV    1.096776  107 m1 R  Li   R 1 / 1  0.511MeV / 6535MeV    1.097287  107 m1

For Lyman α: 1   R  H  1  2   1.096776  107 m1 109 m / nm  3 / 4   121.568nm   2 



1



For Li++ equivalent: 1 1 1 1  2  R  Li   2  2  Z 2  1.097287  107 m1 109 m / nm     3  3 6   9 36 



1

  121.512nm

  0.056nm

2

1  I A nt   kZe2  (Equation 4-6) 4-48. N   0 SC    2  r   2 EK  sin4  2 3 where ASC  0.50cm r  10cm t  106 m

99



Chapter 4 – The Nuclear Atom (Problem 4-48 continued)

10.5g / cm  6.02 10 n  Ag   3

23

atoms / mol



107.5 g / mol  5.88  10 atoms / cm3  5.88 1028 atoms / m3 22

EK  6.0MeV

 1 I 0  1.0nA  109 C / s   2 1.60  1019 C 









   3.18  109 alphas / s  

(a) At θ = 60°









 3.13  109  / s 0.50cm 2 5.88  1028 / m3 106 N    102 cm 2 

   

  9  109 N m 2 / C 2 1.60  1019 C 2  47     1  =  468 / s  13  2  6.0 MeV  1.60  10 J / MeV   4 60     sin 2 











60   4 120   (b) At θ = 120°: N  N  60   sin4 / sin  52 / s 2   2  

4-49.

En   E0 Z 2 / n2

(Equation 4-20)

For Ca, Z = 20 and E1  13.6eV  20  / 1  5.440keV 2

2

The fact that E1 computed this way is only approximate is not a serious problem because the measured x-ray energies provide us the correct spacings between the levels.

E2  E1  3.69keV  5.440  3.69  1.750keV E3  E2  0.341keV  1.750  0.341  1.409keV E4  E3  0.024keV  1.409  0.024  1.385keV These are the ionization energies for the levels. Auger electron energies E   En where E  3.69keV . Auger L electron: 3.69keV  1.750keV  1.94keV Auger M electron: 3.69keV  1.409keV  2.28keV Auger N electron: 3.69keV  1.385keV  2.31keV

100

Chapter 4 – The Nuclear Atom 4-50. (a) E  hc /   1240eV nm / 0.071nm  17.465keV E  hc /   1240eV nm / 0.063nm  19.683keV

(b) Select Nb (Z = 41) The Kβ Mo x-rays have enough energy to eject photoelectrons, producing 0.693 keV electrons. The Kα Mo x-rays could not produce photoelectrons in Nb.

 180   4-51. (a) b  R sin   R sin  2 

    R cos 2 

(b) Scattering through an angle larger than θ corresponds to an impact parameter smaller than b. Thus, the shot must hit within a circle of radius b and area πb2. The rate at which this occurs is I 0 b 2  I 0 R 2 cos2

 2

  (c)    b    R cos    R 2 2  2

2 0

(d) An α particle with an arbitrarily large impact parameter still feels a force and is scattered. 4-52. For He: En  13.6eV Z 2 / n2  54.4eV / n2

(Equation 4-20)

(a) ∞

0 5

4 3 -10 2

Energy (eV) -20

-30

E1 = -54.4 eV E2 = -13.6 eV E3 = -6.04 eV E4 = -3.04 eV E5 = -2.18 eV E∞ = 0 eV

-40

-50 1

101

Chapter 4 – The Nuclear Atom (Problem 4-52 continued)

(b) Ionization energy is 54.5eV. (c) H Lyman α:   hc / E  1240eV nm / 13.6eV  3.4eV   121.6nm H Lyman β:   hc / E  1240eV nm / 13.6eV  1.41eV   102.6nm He+ Balmer α:   hc / E  1240eV nm / 13.6eV  6.04eV   164.0nm He+ Balmer β:   hc / E  1240eV nm / 13.6eV  3.40eV   121.6nm   42.4nm   19.0nm (The reduced mass correction factor does not change the energies calculated above

to three significant figures.) (d) En  13.6eV Z 2 / n2 because for He+, Z = 2, then Z 2 = 22. Every time n is an even number a 22 can be factored out of n 2 and cancelled with the Z 2 = 22 in the numerator; e.g., for He+, E2  13.6eV 22 / 22  13.6eV

(H ground state)

E4  13.6eV 2 / 4  13.6eV / 2

(H  1st excited state)

E6  13.6eV 22 / 62  13.6eV / 32

(H  2nd excited state)

2

2

2

etc. Thus, all of the H energy level values are to be found within the He+ energy levels, so He+ will have within its spectrum lines that match (nearly) a line in the H spectrum. 4-53. Element

P

Ca

Co

Kr

Mo

I

Z

15

20

27

36

42

53

Lα λ(nm)

10.41

4.05

1.79

0.73

0.51

0.33

1.70

2.72

4.09

6.41

7.67

9.53



f 1 / 2 108 Hz









where f 1 / 2   3.00  108 m / s 109 nm / m /  

Slope =

1/ 2

50  15  4.62  108 Hz 1 8 9 . 15  1 . 58  10 Hz  

Slope (Figure 4-19) =

74  46  4.67  108 Hz 1 8 14  8  10 Hz

102

Chapter 4 – The Nuclear Atom (Problem 4-53 continued)

The agreement is very good. 50

40 Z 30

20

10



f 1 / 2 108 Hz1 / 2 1

2

3

4

5

6

7

8

9



10

The f 1 / 2  0 intercept on the Z axis is the minimum Z for which an Lα X-ray could be emitted. It is about Z = 8.

4-54. (a) En  

ke2 ke2  2 2rn 2n ro

hf  En  En 1  

En 1  

ke2 2  n  1 ro 2

 ke2  ke2     2n2 ro  2  n  12 ro   





2 2 ke2  1 1  ke2 n  n  2n  1  f    2 2hro   n  12 n 2  2hro n 2  n  1  



(b) f rev 

ke2 2n  1 ke2  for n 2hro n2  n  12 ro hn3 v 2 r



2 f rev 

1

v2 1 mv 2 1 ke2 ke2    4 2 r 2 4 2 mr r 4 2 mr r 2 4 2 mro3n6

103

Chapter 4 – The Nuclear Atom (Problem 4-54 continued)

(c) The correspondence principle implies that the frequencies of radiation and revolution are equal. 2

2

 ke2  ke2 2 f    f rev 3  2 3 6 4 mro n  ro hn 

2 ke2  hn3  h2 ro      4 2 mn6  ke2  4 2 mke2 mke2

2

which is the same as ao in Equation 4-19.

4-55.

kZe2 mv 2  r r



kZe2  mv   r2 mr

1/ 2

 kZe2  v     mr 



2

(from Equation 4-12)

v 1  2

 2  kZe2    kZe2  c 2  2  kZe2  2  Therefore,  c         1   2  mr   mr    mr  

1  kZe2    2  c  mao  2



  0.0075Z 1 / 2



v  0.0075cZ 1 / 2  2.25  106 m / s  Z 1 / 2

 1  kZe2 kZe2 2 E  KE  kZe / r  mc    1   mc   1  2 r r   1   2

2

And substituting   0.0075 and r  ao   1  E  511  10 eV  1  28.8Z eV 2     1   0.0075 3

 14.4eV  28.8Z eV  14.4Z eV

104

Chapter 4 – The Nuclear Atom 4-56. (The solution to this problem depends on the kind of calculator or computer you use and the program you write.)

4-57.

Energy (eV)

23

22

Levels constructed from Figure 4-26.

21

20

0

4-58. Centripetal acceleration would be provided by the gravitational force: Mm mv 2 FG  G 2  r r L  mvr  n

rn 



n m  GM / rn 

1/ 2

1/ 2

 GM  M  proton mass and m  electron mass, so v     r  r  n / mv or



The total energy is: E 

rn2 

n2 2 rn n2 2 and, r  n m2GM GMm2

1 2  GMm  1  GM mv     m 2 r  2  r 



 GMm  GMm GMm En    2rn 2n 2 2

2

  G M 2

2

2n 2

11

N m2 / kg 2

GMm2

m3

G 2 M 2 m3  1 1   2 2  22 32  2

27

34

105

2

GMm     2r 

 1.67 10 kg  9.11 10 2 1.055  10  2

ao 

2

The gravitational Hα line is: E  E2  E3 

 6.67  10 E 



2

31

kg

  0.1389 3

Chapter 4 – The Nuclear Atom (Problem 4-58 continued)  5.85 1098 J  3.66 1079 eV

E 5.85  1098 J f    8.28  1065 Hz 34 h 6.63  10 J s For the Balmer limit in each case,

E  3.66  1079 eV  0.250 / 0.1389   6.58  1079 eV f  6.58  1079 eV / h  1.59 1064 Hz These values are immeasurably small. They do not compare with the actual H values.

4-59. Refer to Figure 4-16. All possible transitions starting at n = 5 occur. n = 5 to n = 4, 3, 2, 1 n = 4 to n = 3, 2, 1 n = 3 to n = 2, 1 n = 2 to n = 1 Thus, there are 10 different photon energies emitted. ni

nf

fraction

no. of photons

5

4

1/4

125

5

3

1/4

125

5

2

1/4

125

5

1

1/4

125

4

3

1/ 4 1 / 3

42

4

2

1/ 4 1 / 3

42

4

1

1/ 4 1 / 3

42

3

2

1 / 2 1 / 4  1 / 4 1 / 3

83

3

1

1 / 2 1 / 4  1 / 4 1 / 3

83

2

1

1 / 2 1 / 4  1 / 4 1 / 3   1 / 4 1 / 3  1 / 4  

250 Total = 1,042

Note that the number of electrons arriving at the n = 1 level (125+42+83+250) is 500, as it should be.

106

Chapter 4 – The Nuclear Atom

4-60.

m k 2e4 Z2 En   Eo 2 where Eo  n 2 2

m  1.88  1028 kg



Eo  4.50  1016 J  2.81 103 eV





Z2 Thus, for muonic hydrogen-like atom: En   2.81  10 eV 2 n 3

(a) muonic hydrogen 0

n=∞ n=5 n=4 n=3

-500 n=2

En (eV)

E∞ = 0 eV E5 = -112 eV E4 = -176 eV E3 = -312 eV E2 = -703 eV E1 = - 2.81 103 eV

-1000

-1500

-2000

-2500 n=1 -3000

2 2 n2 2 n2 4 n (b) rn    2.56  10 nm mkZe2 mke2 Z Z

For H, Z = 1: r1  2.56  104 nm For He1+ , Z = 2: r1  1.28  104 nm For Al12+ , Z = 13: r1  1.97  105 nm For Au 78+ , Z = 79: r1  3.2  106 nm

107

(Equation 4-18)

Chapter 4 – The Nuclear Atom (Problem 4-60 continued) (c) Nuclear radii are between about 1 and 8  105 m , or 1  8  106 nm . (See Chapter 11.) The muon n = 1 orbits in H, He1+, and Al12+ are about roughly 10nm outside the nucleus. That for Au78+ is very near the nucleus’ surface. (d) hf  hc /   E2  E1 For H:  

1  2



  hc /  E2  E1 

hc  5.89  1010 m  0.589nm 3 2.81  10 1  1 / 4 



Similarly, For He1+:   0.147nm For Al12+:   0.00349nm For Au78+:   9.44  105 nm

108



where En   2.81  103 eV

 Zn

2 2

Chapter 5 – The Wavelike Properties of Particles

5-1.



 

(b) v 



h 6.63  1034 J s   6.6  1029 m / s  2.1  1021 m / y 3 2 m 10 kg 10 m







h h hc 1240MeV fm     12.4 fm p E/c E 100MeV

5-2.



5-3.

 hc  p2 Ek  eVo   2m 2mc 2 2

5-4.



1240eV nm  1 Vo   940V e 2 5.11  105 eV  0.04nm 2

2

h  p

h 2mEk

2



hc



(b) For a proton:  



(from Equation 5-2)

2mc 2 Ek

(a) For an electron:  

1240eV nm



 2  0.511  106 eV 





4.5  103 eV 

1/ 2

1240eV nm



 2  983.3  106 eV 

(c) For an alpha particle:  

5-5.



6.63  1034 J s 3.16  107 s / y h h (a)      2.1  1023 m 3 p mv 10 kg 1m / y 

 4.5 10 eV  3

1/ 2

 0.0183nm

 4.27  104 nm

1240eV nm



 2  3.728  109 eV 

  h / p  h / 2mEk  hc / 2mc 2 1.5kT 

1/ 2

 4.5 10 eV  3

1/ 2

 2.14  104 nm

(from Equation 5-2)

Mass of N2 molecule = 2  14.0031u 931.5MeV / uc 2  2.609  104 MeV / c 2  2.609  1010 eV / c 2





109

Chapter 5 – The Wavelike Properties of Particles (Problem 5-5 continued)



5-6.

5-7.



1240eV nm









 2  2.609  1010 eV 1.5 8.617  105 eV / K  300 K    

h  p

h

hc



2mEk



2

2mc Ek

1/ 2

 0.0276nm

1240eV nm



 2 939.57  106 eV 



 0.02eV 

1/ 2

 0.202nm

(a) If there is a node at each wall, then n   / 2  L where n  1, 2, 3,... or   2 L / n . E  p 2 / 2m   hn / 2L  / 2m  h 2n 2 / 8mL2

(b) p  h /   hn / 2L En

 hc  

2

2

n2

8mc 2 L2

For n = 1: E1 



1240eV

nm  1

8 938  106 eV

2

2

  0.01nm 

 2.05eV

2

For n = 2: E2  2.05eV  2   8.20eV 2

5-8.

(a)  / c  102 is a nonrelativistic situation, so 

 / c   hc Ek 

2mc 2 Ek

mc 2 2   c 

2





 hc mc    mc 2

0.511  106 eV

 

2 102

2

2

2Ek



1/ 2

 25.6eV

(b)  / c  0.2 is a relativistic for an electron, so   h  mu   u  h  m .

uc 1  u / c 

u c  2 1  u / c  2

2



 h  c mc 

   c  

2



u c

c /  1/ 2

1    /  2  c  

110

Chapter 5 – The Wavelike Properties of Particles (Problem 5-8 continued)

u c

1 0.2   0.981 1  1 0.2 2   

   5.10

Ek  mc2   1  0.511MeV   1  2.10MeV (c)  c  103

1 10  u c 1  1 10     3

3

2

1/ 2

 0.9999



  1000

Ek  mc2   1  0.511MeV  999   510MeV

5-9.

Ek  mc 2   1

p   mu

(a) Ek  2GeV

mc 2  0.938GeV

  1  Ek mc2  2GeV 0.938GeV  2.132 Thus,   3.132 Because,   1

 

1  u / c 

2

where u / c  0.948

h h hc   2 p  mc  u / c   mc  u / c 

1240eV nm  4.45  107 nm  0.445 fm 6  3.132 938  10 eV  0.948





(b) Ek  200GeV

  1  Ek mc2  200GeV 0.938GeV  213 Thus,   214 and u / c  0.9999



5-10.

1240eV nm  6.18  103 fm  214 938MeV  0.9999 

n  D sin (Equation 5-5) n n hc sin   (see Problem 5-6) D D 2mc 2 Ek

111

Chapter 5 – The Wavelike Properties of Particles (Problem 5-10 continued)

 5.705eV  1 1240eV nm   1/ 2 0.215nm  2 5.11  105 eV  Ek Ek  

1/ 2





(a)

 5.705eV  sin 

(b)

 5.705eV  sin 

1/ 2

75eV

 0.659

  sin1  0.659   41.2

 0.570

  sin1  0.570   34.8

1/ 2

5-11.



100eV

h  p

h 2m p Ek

 0.25nm

Squaring and rearranging,

 hc   1240eV nm  h2 Ek    0.013eV 2 2 2m p  2 m p c 2  2 2 938  106 eV  0.25nm  2



2







n  D sin

 sin  n / D  1 0.25nm  /  0.304nm 

sin  0.822

   55

5-12. (a) n  D sin  D 



(b) sin 

n nhc  sin sin 2mc 2 Ek

11240eV



nm 



 sin 55.6 2 5.11 105 eV  50eV 

1/ 2

11240eV nm  n   0.584 D  0.210nm   2 5.11  105 eV 100eV 1 / 2  





  sin1  0.584   35.7

112

 0.210nm

Chapter 5 – The Wavelike Properties of Particles 5-13.

d  t cos 42

n  t  d  t 1  cos 42  0.30nm 1  cos 42

42°

For the first maximum n = 1, so   0.523nm t

h   p

h 2mEk

 hc  h2 Ek   2 2m 2mc 2 2 2



d

Ek 

5-14.



1240eV

2 939  106 eV

nm 



2

 0.523nm 

 3.0  103 eV

D sin (Equation 5-6) n For 54eV electrons λ =0.165nm and sin   0.165nm  n / 0.215nm  0.767n



For n = 2 and larger sin  1 , so no values of n larger than one are possible.

5-15.

sin  n / D

(Equation 5-6)

  h / p  h / 2mEk  hc / 2mc 2 Ek 

1240eV nm





 2 0.511  106 eV  350eV   

1/ 2

 0.0656nm

sin  n  0.0656nm  /  0.315nm   0.208n For n = 1,  = 12°. For n = 2,  = 24.6°. For n = 3,  = 38.6°. For n = 4,  = 56.4°. This is the largest possible  . All larger n values have sin  1.

1 1   105 s  10 s 1 f 100, 000s 1 1 1  f    1.59  104 Hz (b) f t  2 2t 2  105 s

5-16. (a) t 

113

Chapter 5 – The Wavelike Properties of Particles 5-17. (a)

y  y1  y2

 0.002m cos 8.0 x / m  400t / s   0.002m cos  7.6 x / m  380t / s 

1 1   2  0.002m  cos  8.0 x / m  7.6 x / m    400t / s  380t / s  2 2  1 1   cos  8.0 x / m  7.6 x / m    400t / s  380t / s  2 2   0.004m cos  0.2 x / m  10t / s   cos  7.8x / m  390t / s 

(b) v 



(c) vs 

k



390 / s  50m / s 7.8 / m

 20 / s   50m / s k 0.4 / m

(d) Successive zeros of the envelope requires that 0.2x / m   , thus  2 x   5 m with k  k1  k2  0.4m1 and x   5 m. 0.2 k

5-18. (a) v  f  Thus,



dv df dv df  2 d  f  , multiplying by ,    f  2 v d d d d 2 d 

 2 d dv  v Because k  2 / , dk    2 /  2  d  and 2 d  d

d dv  vs  v   dk d

(b) v decreases as λ decreases, dv/dλ is positive.

5-19. (a) c  f    / T

 T   / c  2  102 m / 3  108 m / s  6.7  1011 s / wave



 m  3.73  10   74.6m

The number of waves = 0.25 s / 6.7  1011 s / wave  3.73  103 Length of the packet =    # of waves   2  102 (b)



3



f  c /   3  108 m / s / 2  102 m  1.50  1010 Hz

(c)  t  1 

  1 / t  1 / 0.25  106 s  4.0  106 rad / s  637kHz

114

Chapter 5 – The Wavelike Properties of Particles 5-20.

 t  1 

  1 / t  1 / 0.25s  4.0rad / s or f  0.6Hz

5-21.

 t  1 

 2f  t  1

5-22. (a)  

h  p

h 2mEk

Thus, t  1 /  2  5000Hz   3.2  105 s

hc



2mc 2 Ek



1240eV nm





 2 0.511  106 eV  5eV   

1/ 2

 0.549nm

d sin   / 2 For first minimum (see Figure 5-17).

d

 2 sin



0.549nm  3.15nm slit separation 2 sin 5

(b) sin 5  0.5cm / L where L = distance to detector plane L 

0.5cm  5.74cm 2 sin 5

5-23. (a) The particle is found with equal probability in any interval in a force-free region. Therefore, the probability of finding the particle in any interval ∆x is proportional to ∆x. Thus, the probability of finding the sphere exactly in the middle, i.e., with ∆x = 0 is zero. (b) The probability of finding the sphere somewhere within 24.9cm to 25.1cm is proportional to ∆x =0.2cm. Because there is a force free length L = 48cm available to the sphere and the probability of finding it somewhere in L is unity, then the probability that it will be found in ∆x = 0.2cm between 24.9cm and 25.1cm (or any interval of equal size) is: Px  1 / 48 0.2cm   0.00417cm.

5-24. Because the particle must be in the box Let u   x / L;

 A  L /   sin 2

0

L

0

0

2 2  * dx  1   A sin  x / L  dx  1

x  L  u   and dx   L /   du , so we have

x  0  u  0;



L



2

udu  A  L /    sin2 udu  1 2

0





u sin 2u    L /   A2  / 2   LA2 / 2  1  L /   A2  sin2 udu   L /  A2    4 0 2 0  A2  2 / L  A   2 / L 

1/ 2

115





Chapter 5 – The Wavelike Properties of Particles 5-25. (a) At x  0 : (b) At x   : (c) At x  2 :

Pdx    0, 0  dx  Ae0 dx  A2 dx 2

2

Pdx  Ae

2

/ 4 2

Pdx  Ae4

2

2

/ 4 2

2

dx  Ae1 / 4 dx  0.61A2 dx 2

2

dx  Ae1 dx  0.14 A2 dx

(d) The electron will most likely be found at x = 0, where Pdx is largest.

5-26. (a) One does not know at which oscillation of small amplitude to start or stop counting. N t

f 

(b)  

N 1  t t

x 2 2 N 2 n 2 and k   , so k   N  x x x

5-27. E t 

5-28. xp 

f 

 E  / t 



1.055  1034 J s  6.6  109 eV 7 19 10 s 1.609  10 J / eV





p  mv 

2x 1.055  1034 J s  v    5.3  1027 m / s 3 2 3 2mx 2  10 kg 10  10 m 2



5-29.

E t 





E  / t 



6.58  1016 eV s  1.99  1021 eV 4 3.823d 8.64  10 s / d





The energy uncertainty of the excited state is ∆E, so the α energy can be no sharper than ∆E.

5-30.

xp 

 p  h

 p  h / . Because   h / p, p  h / ; thus, p  p.

5-31. For the cheetah p  mv  30kg  40m / s   1200kg m / s. Because p  p (see Problem 5-30), x  / p  50 J s / 1200kg m / s  4.2  102 m  4.2cm

116

Chapter 5 – The Wavelike Properties of Particles 5-32. Because c  f  for photon,   c / f  hc / hf  hc / E, so hc

E





and p 

1240eV nm  2.48  105 eV 5.0  103 nm

E 2.48  105 eV   8.3  107 eV s / m c 3  108 m / s

For electron:

p 

h 4.14  1015 eV s   8.3  104 eV s / m x 5.0  1012 m

Notice that ∆p for the electron is 1000 times larger than p for the photon.

5-33. (a) For

48

Ti:

E  upper state   E  lower state  

t

t



1.055  1034 J s  4.71  1010 MeV 14 13 1.4  10 s 1.60  10 J / MeV



1.055  1034 J s  2.20  1010 MeV 12 13 3.0  10 s 1.60  10 J / MeV









E  total   EU  EL  6.91 1010 MeV ET 6.91  1010 MeV   5.3  1010 E 1.312MeV (b) For Hα: EU 

1.055  1034 J s  6.59  108 eV 8 19 10 s 1.60  10 J / eV





and EL  6.59  108 eV also. ET  1.32  107 eV is the uncertainty in the Hα transition energy of 1.9eV.

5-34.

 t  1 

2f t  1

For the visible spectrum the range of frequencies is f   7.5  4.0   1014  3.5  1014 Hz The time duration of a pulse with a frequency uncertainty of f is then:

t 

1 1   4.5  1016 s  0.45 fs 14 2f 2  3.5  10 Hz

117

Chapter 5 – The Wavelike Properties of Particles 5-35. The size of the object needs to be of the order of the wavelength of the 10MeV neutron.

  h / p  h /  mu.  and u are found from: Ek  mnc2   1 or   1  10MeV / 939MeV

  1  10 / 939  1.0106  1 / 1  u 2 / c 2 

1/ 2

or u  0.14c

h hc 1240eV nm    9.33 fm 2  mu  mc  u / c  1.0106  939  106 eV  0.14    Nuclei are of this order of size and could be used to show the wave character of 10MeV neutrons.

Then,  





5-36. (a) E  135MeV , the rest energy of the pion. (b) E t 

2

6.58  1016 eV s t    2.44  1024 s 6 2E 2  135  10 eV

5-37.

ps L  rps

s

φ r



s r

s r L   rps  s / r   ps s 

L  r p

 

In the Bohr model, L  n and may be known to within L  0.1 . Then   /  0.1

  10rad.

This exceeds one revolution, so that  is completely

unknown.

5-38.

E  hf Et  h



E  hf E



f t  1 where t  0.85ns

118

Chapter 5 – The Wavelike Properties of Particles (Problem 5-38 continued)

f  1 / 0.85ns  1.18  109 Hz For   0.01nm

f  c/ 

3.00  108 m / s  109 nm / m  3.00  1019 Hz 0.01nm

f 1.18  109 Hz   3.9  1011 19 f 3.00  10 Hz

5-39.

Et 

 t  2 2E 16 6.58  10 eV s t   1.32  1024 s 6 2  250  10 eV

5-40. In order for diffraction to be observed, the aperture diameter must be of the same order of magnitude as the wavelength of the particle. In this case the latter is



h 6.63 1034 J  s   1.66 1033 m 3 p (4 10 kg)(100 m/s)

The diameter of the aperture would need to be of the order of 1033 m. This is many, many orders of magnitude smaller than even the diameter of a proton or neutron. No such apertures are available.

5-41. The kinetic energy of the electron needed must be no larger than 0.1 nm. The minimum kinetic energy of the electrons needed is then given by:

h2 2me  2



h h  p 2me E

E

(6.6334 J  s) 2  2.411017 J  151eV 31 9 2 2(9.1110 kg)(0.110 nm)

 E

119

Chapter 5 – The Wavelike Properties of Particles 5-42. (a) For a proton or neutron: xp  v 

and p  mv assuming the particle speed to be non-relativistic.

2



2mx

1.055  1034 J s  3.16  107 m / s  0.1c (non-relativistic) 27 15 2 1.67  10 kg 10 m











1.67  1027 kg 3.16  107 m / s 1 2 (b) Ek  mv  2 2



2

 8.34  1013 J  5.21MeV

(c) Given the proton or neutron velocity in (a), we expect the electron to be relativistic, in which case, Ek  mc 2   1 and p 

2x

  mv

 v 

2mx

For the relativistic electron we assume v  c

 

2mcx



1.055  1034 J s  193 2 9.11  1031 kg 3.00  108 m / s 1015 m











Ek  mc 2   1  9.11 1031 kg 3.00  108 m / s

E  hf  

5-43. (a) E 2  p2c2  m2c4

v

 k

(b) vs 



 k

2



d d  dk dk

 192  1.58 10 2

p  h/  /k

k 2 c 2  m2 c 4  c 1  m2c 2 / k

k 2c 2  m2c 4  k



2

k2  c

c2k k 2c 2  m2c 4 2



5-44.

c2k





c2 k





c2 p u E

2 y 1 2 y  (Equation 5-11) x 2 v 2 t 2

(by Equation 2-41)

y3  C1 y1  C2 y2

 1  2 y1   1  2 y2   2 y3  2 y1  2 y2  C  C  C  C   1 2 1 2 2 2 2 2  x 2 x 2 x 2  v t   v t  120

2 

2

11

J  98MeV

2

k 2c 2  m2c 4

Chapter 5 – The Wavelike Properties of Particles (Problem 5-44 continued)



1 2 1  2 y3 C y  C y    1 1 2 2 v 2 t 2 v 2 t 2

5-45. The classical uncertainty relations are  t  2f t  1 (Equation 5-18)

and x  (a) f 

2 2

(Equation 5-20)

1 1   0.0541Hz 2t 2  3.0s 

(b) Length of the wave traing L  vt , where v = speed of sound in air = 330m/s.

L   330m / s  3.0s   990m (c)  

2 where x = length of the wave train  990m from (b) 2x

 0.13m   2.72  106  2.72 m   2  990m  2

and   0.13m from (d) .

(d) v  f 



5-46. (a) n   / 2   L



v  330m / s    0.13m  13cm f  2500 Hz 

   2L / n. Because   h / p  h / 2mE , then

h2 h2 h2 n2 E   2m 2 2m  L / n 2 8mL2

h2 n2 If E1  h / 8mL , then En   n 2 E1 2 8mL 2

2

 hc   1240eV nm  h2 (b) For L  0.1nm, E1   2 2 2 8mL 8mc L 8 0.511  106 eV  0.1nm 2 2

2



E1  37.6eV and En  37.6n 2eV

121



Chapter 5 – The Wavelike Properties of Particles (Problem 5-46 continued) n E (eV)

1000

5

800

600

4

400

3

200

0

(c)

f  E / h

2 1 0



L

c /   E / h





x

hc E

For n  2  n  1 transition, E  112.8eV and  

1240eV nm  11.0nm 112.8eV

(d) For n  3  n  2 transition, E  188eV and  

1240eV nm  6.6nm 188eV

(e) For n  5  n  1 transition, E  903eV and  

1240eV nm  1.4nm 903eV

5-47. (a) For proton: E1 

 hc 

2

8m p c 2 L2

1240MeV fm  E1  2 8  938MeV 1 fm 

from Problem 5-46.

2

 205MeV and En  205n2 MeV

 E2  820MeV and E3  1840MeV (b) For n  2  n  1 transition,  =

hc 1240MeV fm   2.02 fm E 615MeV

(c) For n  3  n  2 transition,  =

hc 1240MeV fm   1.22 fm E 1020MeV

(d) For n  3  n  1 transition,  =

hc 1240MeV fm   0.76 fm E 1635MeV

122

Chapter 5 – The Wavelike Properties of Particles 5-48. (a) E 

2

/ 2mL2 (Equation 5-28) and E 

2

/ 2mA2

(b) For electron with A  1010 m :

E

 c

2

2mc 2 A2





197.3eV

nm 



2 1

2 0.511  10 eV 10 nm 6



2

 3.81eV

For electron with A  1cm or A  102 m :



 / 10 nm  3.8110 eV 1.055 10 J s   2 100  10 g  10 kg / g  2  10 

E  3.81eV 101

(c) E 

2

2

7

16

34

2 2

2mL

3

2

3

2

2

 1.39  1061 J  8.7  1043 eV

5-49. p  mv  m  0.0001 500m / s   0.05m For proton: xp 







x  / p  6.58  1016 eV s /  0.05m / s  938  106 eV



 1.40  1023 m  1.40 108 fm









For bullet: x  1.055  1034 J s /  0.05m / s  10  103 kg  2.1 1031 m

5-50.

2 y 1 2 y  (Equation 5-11) where y  f   and   x  vt. x 2 v t 2

y f  y 2 f  2   2 f    and      x  x x 2  x 2 x  2 x y f  y 2 f  2   2 f    and      t  t t 2  t 2 t  2 t  2 Noting that 2  0, x

  1, x

 2   0, and  v, we then have: 2 t t

 f 2 f 1  f 2 f  0  1  2  1  2   0   v   2    v     v    

2 f 2 f   2  2

123

Chapter 5 – The Wavelike Properties of Particles 5-51. (a)   h / p The electrons are not moving at relativistic speeds, so

  h / mv  6.63  1034 J s /  9.111031 kg 3 106 m / s   2.43 1019 m  0.243nm (b) The energy, momentum, and wavelength of the two photons are equal. 1 2 1 1  mv  mc 2  mc 2 v 2 / c 2  mc 2  mc 2  v 2 / c 2  1 2 2 2 



E



1  0.511  106 eV  3  106 / 3  108 2



(c)









  1  0.511MeV 

2

p  E / c  0.511MeV / c

(d)   hc / E  1240eV nm / 0.511106 eV  2.43 103 nm

5-52. (a) Q  mp c 2  mnc 2  m c 2  1.007825uc2  1.008665uc2  139.6MeV

 938.8MeV  939.6MeV  139.6MeV  140.4MeV E  140.4MeV

(b) Et 



t  / E  6.58  1016 eV s / 140.4  106 eV  4.7  1024 s





(c) d  ct  3  108 m / s 4.7  1024 s  1.4  1015 m  1.4 fm 5-53. hf   mc 2

  

hf 1  2 2 mc 1 v

c2 1/ 2

2 2   mc 2 2    mc 1 v 2    v  1     c  c  hf  hf      2 Expanding the right side, assuming mc hf , 2

2

4

v 1  mc 2  1  mc 2   1       c 2  hf  8  hf 

and neglecting all but the first two terms,

2

v 1  mc 2   1   Solving this for m and inserting deBroglie’s assumptions that c 2  hf 

v  0.99 and   30m, m is then: c





1  0.99  2 6.63  1034 J s m  1.04  1044 kg 8 3.00  10 m / s  30m  1/ 2





124

Chapter 5 – The Wavelike Properties of Particles

5-54. (a)

m

xp 



mxvx  1/ 2

 2y  1 y0  gt 2  t   0  2  g 



vx  / mx 1/ 2

 2y  1 X  vx t  vx  0  2  g  1/ 2

 2y  2  0 g    mx

y0 1/ 2

 2y  X  2vx  0   g 

∆x

(b) If also yp y 

 v y  / my and

t  v y / g  / mg y so, X 

5-55.

1 X  x  t  t  where vy  g t or 2

2  1/ 2 2 y0 / g   / mg y    mx 

1 3 m v 2  kT 2 2

vrms





23 3kT  3 1.381  10 J / K  300 K      m  56u 1.66  1027 kg / u 

f   fo 1  v / c 





1/ 2



hf   hf o 1  v / c 

E  hf   hf o  hf o v / c 

1eV  366m / s   1.2  106 eV 3.0  108 m / s

This is about 12 times the natural line width.

10 eV  366m / s   1.2eV E  hf v / c  6

o

 366m / s

3.0  108 m / s

This is over 107 times the natural line width.

125

Chapter 5 – The Wavelike Properties of Particles

5-56. recoil    E / c (a) Erecoil 

1eV 

Erecoil 2





2 56uc 2

   recoil

2

2m



E2 2mc 2

uc 2  9.6  1012 eV 6 931.5  10 eV

This is about 10-4 times the natural line width estimated at 10-7eV. (b) Erecoil

1MeV  



2

2 56uc 2



uc 2  9.6eV 931.5  106 eV

This is about 108 times the natural line width.

126

Chapter 6 – The Schrödinger Equation

6-1.

d d 2  kAekx t  k  and  k 2 2 dx dx Also,

d   . The Schrödinger equation is then, with these substitutions, dt

 2 k 2  / 2m  V   i  . Because the left side is real and the right side is a pure Imaginary number, the proposed  does not satisfy Schrödinger’s equation.

6-2.

For the Schrödinger equation:

 2    ik  and  k 2 . Also,  i . 2 x t x

Substituting these into the Schrödinger equation yields: 2

k 2  / 2m  V    , which is true, provided  

2

k 2 / 2m  V , i.e., if E  Ek  V .

For the classical wave equation: (from Equation 6-1)

2  2  2   k  and also   2 . Substituting into Equation 6-1 2 2 x t (with  replacing E and v replacing c) k 2   1/ v2  2  , which is true for From above:







v   / k.

6-3.

(a)

d d 2   x / L2  and dx dx 2





 x  x   2   2  L  L

x2 1  1      2  2  4 L L  L 

Substituting into the time-independent Schrödinger equation, 2 2 2 2   x    V  x   E    4 2  2mL2  2mL 2mL 

Solving for V(x), V  x  

2 2 2 2 2   x x 1      kx 2   2 4 2 4 2mL  2mL 2mL  2mL 2 2

127

Chapter 6 – The Schrödinger Equation (Problem 6-3 continued) where k 

2

/ mL4 . This is the equation of a parabola centered at x = 0. V(x)

0

x

(b) The classical system with this dependence is the harmonic oscillator.

6-4.

(a) Ek  x   E  V  x  

2

/ 2mL2 

x / 2mL4 

2 2



2



/ 2mL2 1  x 2 / L2



(b) The classical turning points are the points where E  V  x  or Ek  x   0. That occurs when x2 / L2  1, or when x   L. (c) For a harmonic oscillator V  x   m 2 x 2 / 2, so 2

x2   2 x2 / 2   2  2mL4 Thus, E 

6-5.

2

/ m2 L4    / mL2

1    2   2mL  mL  2 2 2

2

(a)   x, t   A sin  kx  t 

   A cos  kx  t  t i

  i  A cos  kx  t  t

2  k 2 A sin  kx  t  x 2  2 2  k 2 A   sin  kx  t   i 2 2m x 2m t

128

Chapter 6 – The Schrödinger Equation (Problem 6-5 continued) (b)   x, t   A cos  kx  t   iA sin  kx  t  i

  i  A sin  kx  t   i 2  A cos  kx  t  t

  A cos  kx  t   i  A sin  kx  t  

2 2 2 k A  cos  kx  t   2 2m x 2m 2

k2  A cos  kx  t   iA sin  kx  t   2m   t

i

(a)

ik 2 A sin  kx  t  2m

2



6-6.

2

2

if

k2   it does. (Equation 6-5 with V = 0) 2m

For a free electron V(x) = 0, so 

2 d 2 d 2  E     2.5  1010  2 2 2m dx dx



2



Substituting into the Schrödinger equation gives: 2.5  1010

 2



2



and, since E  Ek  p 2 / 2m for a free particle, p 2  2m 2.5  1010



p  2.5  1010

(b)





/ 2m   E

 2

2

 2.64  1024 kg m / s









E  p 2 / 2m  2.64  1024 kg m / s /  2  9.11 1031 kg  3.82  1018 J



2





 3.82  1018 J 1 / 1.60  1019 J / eV  23.9eV

(c)

6-7.

  h / p  6.63 1034 J s 2.64 1024 kg m / s  2.511010 m  0.251nm

  x   Ce x (a)

2

/ L2

and E  0

d 2  V  x   0 2m dx 2 d 2x d 2  4 x 2 2    2   x  and   2  dx L dx 2  L4 L  2 2  4x2 2   2 x2     V x   0 so V x   1 And       4 2  2  2 2m  L L  mL  L 



2

129



/ 2m and

Chapter 6 – The Schrödinger Equation (Problem 6-7 continued) V(x)

(b)

x

6-8.

a

a

a

a

 i  kx t  i kx t  2 e  dx  1    dx  A  e a

A

2

 dx  A x 2

a a

 A2  2a   1

a

 A

1

 2a 

1/ 2

Normalization between −∞ and +∞ is not possible because the value of the integral is infinite.

6-9.

(a) The ground state of an infinite well is E1  h2 / 8mL2   hc  / 8mc 2 L2 2

For m  m p , L  0.1nm : E1 

(b) For m  m p , L  1 fm : E1 





1240MeV

8 938.3  106 eV

1240MeV

2

  0.1nm

fm 



fm 

2

 0.021eV

2

8 938.3  106 eV 1 fm 

2

 205MeV

6-10. The ground state wave function is (n = 1)  1  x   2 / L sin  x / L  (Equation 6-32) The probability of finding the particle in ∆x is approximately: P  x  x 

2 2x 2   x  x  sin2  x  sin    L L  L   L 

130

Chapter 6 – The Schrödinger Equation (Problem 6-10 continued) 2  0.002 L  2   L  L  (a) For x  and x  0.002 L, P  x  x  sin   0.004 sin2  0.004  2 L 2  2L  2  0.002 L  2  2 L  2L 2 (b) For x  and P  x  x  sin   0.004 sin2  0.0030  3 L 3  3L  (c) For x  L and P  x  x  0.004 sin2   0

6-11. The second excited state wave function is (n = 3)  3  x   2 / L sin  3 x / L  (Equation 6-32). The probability of finding the particle in ∆x is approximately: P  x  x 

2  3 x  sin2   x L  L 

(a) For 2  0.002 L  2  3 L  L 3 x  and x  0.002 L, P  x  x  sin   0.004 sin2  0.004  2 L 2  2L  2L  6 L  2 (b) For x  and P  x  x  0.004 sin2    0.004 sin 2  0 3 3 L    3 L  (c) For x  L and P  x  x  0.004 sin2   0.004 sin2 3  0   L 

6-12.

2 1 n2 2 2  1 2   2mL   mvL  2 E  mv 2  (Equation 6-24) n  mv 2   2 2      2 2mL2     

n

mvL



10 

9





kg 103 m / s 102 m

 1.055  10

34

J s



  3  10

19

  p  0.0001 p   0.0001 10 kg 10

6-13. (a) x  0.0001L   0.0001 102 m  106 m 9

(b)

xp

10 m10  6

16

1.055  10

kg m / s

34

J s

3



m / s  1016 kg m / s

  9  10

11

131

2

Chapter 6 – The Schrödinger Equation 6-14

(a) This is an infinite square well with width L. V(x) = 0, and E  Ek  p 2 / 2m. From uncertainty principle: Ekmin  pmin  p  / x  / L and 2 Emin  pmin / 2m 

2

/ 2mL2  h2 / 8 2 mL2

(b) The solution to the Schrödinger equation for the ground state is: 1/ 2  1  x    2 / L  sin  x / L  d 2 1    2  and      2 dx  L L

1/ 2

2

  sin  x / L       1 L 2

h2     E  or E  1 1 1 2m  L  8mL2 The result in (a) is about 1/10 of the computed value, but has the correct 2

2

So,

dependencies on h, m, and L.

6-15. (a) For the ground state, L   / 2, so   2L. (b) Recall that state n has n half-wavelengths between x = 0 and x = L, so for n = 3, L  3 / 2, or   2L / 3.

6-16.

(c)

p  h /   h / 2L in the ground state.

(d)

p 2 / 2m  h2 / 4L2 / 2m  h2 / 8mL2 , which is the ground state energy.

En 

h2 n2 h2 and  E  E  E  n 2  2n  1 n n 1 n 2 2 8mL 8mL







or, En   2n  1  3 h  so, L     8mc 

1/ 2



h2 hc  2 8mL 

 3 hc   2   8mc 

1/ 2

1/ 2

 3  694.3nm 1240eV nm     6   8 0 . 511  10 eV  



6-17. The uncertainty principle requires that E 



 0.795nm

2

2mL2

for any particle in any one-dimensional

box of width L (Equation 5-28). For a particle in an infinite one-dimensional square well:

En 

n2 h2 8mL2 132

Chapter 6 – The Schrödinger Equation (Problem 6-17 continued)

h2  0. This violates Equation 5-28 and, hence, the 8mL2

For n = 0, then E0 must be 0 since exclusion principle.

6-18. (a) Using Equation 6-24 with L = 0.05 nm and n = 92, the energy of the 92nd electron in the model atom is E92 given by: En  n 2

2

2

 E92  (92)2

2mL2

(6.63 1034 J  s)2 8(9.111031 kg)(0.05 109 m)2

1eV  1.28 106 eV  1.28 MeV 19 1.60 10 J (b) The rest energy of the electron is 0.511 MeV. E92  2.5  the electron’s rest energy. E92  2.04 1013 J 

6-19. This is an infinite square well with L = 10cm.





2.0  103 kg  20nm / y  h2 n2 1 2 En   mv  2 8mL2 2 2 3.16  107 s / y n2 







  20 10 m   0.1m  2  3.16  10 s   6.63  10 J s  2

8 2.0  103 kg

7

n



9

2

2

34

  3.16 10 s  6.63 10 J s 

2 2.0  103 kg 20  109 m  0.1m  34

7

6-20. (a)  5  x    2 / L 

1/ 2

2

2

2

 3.8  1014

sin  5 x / L  dx

0.4 L

P

  2 / L  sin 5 x / L  dx 2

0.2 L

Letting 5 x / L  u, then 5 dx / L  du and x  0.2L  u   and x  0.4L  u  2 , so

133

Chapter 6 – The Schrödinger Equation (Problem 6-20 continued) 2

x   sin 2 x   2 L 1  2  L     2 P    sin2 udu          4 5  L  5    L  5      2

(b) P   2 / L  sin2

5  L / 2  L

 0.01L   0.02 where 0.01L  x

1240MeV fm  E1  2 8  0.511MeV 10 fm  2

6-21. (a) For an electron:

1240MeV fm  E1  2 8  938.3Mev 10 fm 

 3.76  103 MeV

2

(b) For a proton:

 2.05MeV

(c) E21  3E1 (See Problem 6-16) For the electron: E21  3E1  1.13  104 MeV For the proton: E21  3E1  6.15MeV

6-22.

F  dE / dL comes from the impulse-momentum theorem F t  2mv where t  L / v.

So, F

mv2 / L

E / L. Because E1  h2 / 8mL2 , dE / dL  h2 / 4mL3 where the minus

sign means “on the wall”. So F  h / 4mL  2

3

The weight of an electron is mg  9.11 1031

 6.63 10

34

J s



2

  m  kg  9.8m / s   8.9  10 N 31

4 9.11  10 kg 10 2

10

30

3

 1.21  107 N

which is

minuscule by comparison.

2 n x sin L L L  n x   m x  To show that  sin   sin  L  dx  0 L     0

6-23.  n  x  

Using the identity 2sin A sin B  cos  A  B   cos  A  B , the integrand becomes





1 cos  n  m   x / L   cos  n  m   x / L  2

134

Chapter 6 – The Schrödinger Equation (Problem 6-23 continued)

L sin  n  m   x / L and similarly for the second   n  m Term with (n + m) replacing (n – m). Since n and m are integers and n ≠ m, the sines both The integral part of the first term is

vanish at the limits x = 0 and x = L.

 n x   m x    sin   sin  L  dx  0 for n  m. L     0 L

6-24. (a)

4 x L

(b)

P

x L

6-25. Refer to MORE section “Graphical Solution of the Finite Square Well”. If there are only two allowed energies within the well, the highest energy E2  V0 , the depth of the well. From Figure 6-14, ka   / 2 , i.e., ka 

2mE2

a  2

where a  1 / 2 1.0 fm   0.5 fm and m  939.6MeV / c 2 for the neutron. Substituting above, squaring, and re-arranging, we have:   E2  V0    2  2  2 939.6MeV / c 2  0.5 fm  2

2





   c  V0  2 8  939.6MeV   0.5 fm  106 nm / fm  2

2

V0  2.04  108 eV  204MeV

135

  197.3eV 2





8 939.6  106 eV

nm 

2

 0.5 10

6

nm



2

Chapter 6 – The Schrödinger Equation 0.5eV and for a finite well also En

6-26. Because E1

n2 E1 , then n = 4 is at about 8eV, i.e.,

near the top of the well. Referring to Figure 6-14, ka 2mE

ka

L

L 2

2 / 7.24 109

6-27. For V2

E

7.24 109 m

1

L

2 .

2

8.7 10 10 m 0.87nm

V1 : x1 is where V

0

:

V1 and x2 and x2 is where V1

V2

From

to 0 and x2 to

is exponential

0 to x1:

is oscillatory; Ek is large so p is large and λ is small; amplitude is small because Ek is large, hence v is large.

x1 to x2:

is oscillatory; Ek is small so p is small and λ is large; amplitude is large

because Ek is small, hence v is small.

6-28. (a) 6 5 4 3 2 1

x -10

+10 0

136

Chapter 6 – The Schrödinger Equation (Problem 6-28 continued)

6-29.

* 3

px L

x

s dx

2 3 x sin L L i x

0

L

2 Li Let

3

i x

3 x L

sin 0

3 x L

y Then x

cos

0

(Equation 6-48)

2 3 x sin dx L L 3 x L y

0,

3 dx L x

L

y

3 , and

3 dx L

dy

dx

L dy 3

Substituting above gives:

2 L Li 3

px

2 Li

3

sin y cos ydy 0

3 L

3

sin y cos ydy 0

2 sin2 y Li 2

3

2 0 0 Li

0

0

Reconiliation: px is a vector pointing half the time in the +x direction, half in the –x direction. Ek is a scalar proportional to v 2, hence always positive.

6-30. For n = 3,

3

2/ L

1/ 2

sin 3 x / L

L

(a)

x 2 / L sin2 3 x / L dx

x 0

137

Chapter 6 – The Schrödinger Equation (Problem 6-30 continued) 3 x / L, then x

Substituting u

x

0

u

0 and x

L

u

Lu / 3 and dx

L/3

du. The limits become:

3

3

2/ L L/3

x

u sin2 udu

1/ 3 0

2

2/ L L/3 2

2 / L 1/ 3

u2 4

u sin 2u 4 2

3

/4

cos 2u 8

3

0

L/2

L

x2

(b)

x 2 2 / L sin2 3 x / L dx 0

Changing the variable exactly as in (a) and noting that: 3 2

u3 6

2

u sin udu 0

u2 4

1 1 3 18

We obtain x 2

L2

2

3

u cos 2u 4

1 sin 2u 8

0

0.328L2

6-31. (a) Classically, the particle is equally likely to be found anywhere in the box, so L

P(x) = constant. In addition,

P x dx 1 so P x

1/ L.

0 L

(b)

L

x / L dx

x

L / 2 and x

2

x 2 / L dx

0

0

2

6-32.

d2 2m dx 2

L2 / 3

1 d 2m i dx

V x d i dx

1 pop pop 2m Multiplying by

x

E x

x

(Equation 6-18)

E V x

x

E V x and integrating over the range of x,

138

Chapter 6 – The Schrödinger Equation (Problem 6-32 continued) 2 pop

dx

2m

E V x

p2 2m

E V x

dx

p2

or

2m E V x x does not vanish and vice versa.

For the infinite square well V(x) = 0 wherever

0 and p 2

Thus, V x

6-33.

x

L2 3

2

x

x

L2 2 2 2

2mE

2m

n2 2 2 2mL2

L2 3

x

L2 2 2

L2 4

1/ 2

for n 1

L2

L 2

(See Problem 6-30.) And x 2

2 2

1 L 12

1/ 2

1

0.181L

2

2

2 2

p

2 x

p

P

6-34.

0

x

x

2

p

A0e A02 xe

x

A02

x

0.)

2

(See Problem 6-32) 1/ 2

2

0

L2

m x2 /

L

. And

dx Letting u 2

ue

m x2 /

/m

/m

u2

du

x

0.181L

p

/L

m x2 / 1

udu

1/ 2

/m

and x

u

xdx; limits are unchanged.

0 (Note that the symmetry of V(x) would also tell us that

dx

3/ 2

3/ 2

u 2e

0.568

1/ 4

m /

where A0

2 xdx . And thus, m /

A02 x 2 e

2 A02

0

2

/m

A02

p

m x2 / 2

m /

2udu

x2

and

L2

u2

du

/4

2 A02

m /

/m 1/ 2

139

3/ 2

/m

u 2e 3/ 2

u2

du

/2

/ 2m

Chapter 6 – The Schrödinger Equation

6-35.

p2 2m

1 m 2

n 1/ 2

x

2 m 2

x2

m

2 2

p2 m

1

6-36. (a)

/2

p 2 / 2m

or

2m

x,t

0

. For the ground state (n = 0),

m /

1/ 4

x2

and

m x2 / 2

e

m

p2 m

1

e

p2 m2 2

1 2

/ 2m

1 m 2

p2

i t/2

2

(b) px op

p

i x

0

A0 m x /

x

2

x, t

0

1/ 4

e

m x2 / 2

e

0

i x

i t/2

2 0

x

2

m x/

A0

p2

2

A02 m /

2

A02 m /

m x/

Letting u p2

2

m x/

m /

m x2 /

1 e

m x2 / 1/ 2

e

m

2

1/ 2

2

m /

m /

m x2 /

m x2 /

1/ 2

u 2e

2

u 2e

u2

1/ 2

1/ 2

m /

m

u2

du

0

2

m x2 / 2

e

e

i t/2

dx

dx

e

du

e

m x2 / 2

x , then

A02 m /

A02 m /

x, t dx

e

u2

du

0

1/ 2

2

/2

140

4

2

u2

du

dx

(See Problem 6-34)

Chapter 6 – The Schrödinger Equation 6-37.

0

x

m x2 / 2

C0e

(a)

0

2

x

(Equation 6-58) 2

dx 1 2

C0

(b) x 2

0

m

(c)

6-38.

1

x

C1 xe

(a)

1

3

1

1 4

with

x

2

x2

dx

3

m /

with

1 m 2

1 2m

m x2 /

2

2

1 4

1 2 m3

2

3

3

dx

C1

with

2

2I 2

m /

3 3

1/ 4

3

dx

m /

1 2m

2

4m3

C1

x

m x2 /

2

C1 x 2e

C1

x

1 2

2

(Equation 6-58)

C1

(b)

e

1 m 2

x2

dx 1

2

=

3

m

m x2 / 2

2

x

m

3

2

m

x2

2I2

1 m 2

V x

2

1/ 4

dx

1 m 2

C0

m

m

2

x2

2I0

dx

2

C0

C0

m x2 /

C0 e

x

3

4m3

3 3

1/ 2

e

m x2 /

141

dx

0

1 4

Chapter 6 – The Schrödinger Equation (Problem 6-38 continued) x

(c)

2

x

2

2 1

4m3

dx

x

6-39. (a)

3

1 m 2

x p

m 2

1 m 2

(1) E

5

/ x

2 0

2

En

2m

n 1/ 2 1

En

limn

En En

2

2 2

2

1 m 2

x2

2

3 8

5

where

m /

2

3 2m

3 4

/ 8mA2

E02 2mA2 4 2 4

m 2

En

E02 4 Ek

4 Ek

1

/4

2 Ek

n 3/ 2

En n 3 / 2 n 1

n 1/ 2

En En

x 2 dx

/ x2 is computed in Problem 6-36(b). Using that quantity,

Ek

En

2

/ 2 also E0

2

6-40.

1/ 2

3 3

2

A2

m x2 /

/ 2A

Because E0 (2)

4m3

h / 2 A / 2m 2

e

3 2m

5

1 m 2

x2

p p 2 / 2m

(b) Ek

5

3

1/ 2

3 3

2I4

3

(d) V x

4m3

1/ 2

3

3 m3 2

2

limn

1 n 1/ 2 1 n 1/ 2

0. In agreement with the correspondence principle.

142

Chapter 6 – The Schrödinger Equation 6-41. (a) Harmonic oscillator ground state is n = 0. x

0

A0e

m x2 / 2

(Equation 6-58) A02

Therefore, for x: x Let m x 2 /

y2

m x2 /

xe

x

dx

m y

dx

m dy

A02

Substituting above yields: x

m

ye

y2

dy

0

by inspection of Figure 6-18, integral tables, or integration by parts. For x 2 :

x2

A02

m x2 /

x 2e

dx

Substituting as above yields: x 2

A02

The value of the integral from tables is

x2

Therefore,

A02

2

m

3/ 2

m

y 2e

y2

dy

2.

3/ 2

(b) For the 1st excited state, n = 1, A1 m / xe

x

1

A12

x

x3 m

e

m x2 / 2

m x2 /

(Equation 6-58) dx

Changing the variable as in (a), x

A12 m A12

3/ 2

m

m

y 3e

y2

dy

y 3e

y2

m

1/ 2

dy

0

by inspection of Figure 6-18, integral tables, or integration by parts. x2

A12

x4 m

e

m x2 /

dx

changing variables as above yields:

143

Chapter 6 – The Schrödinger Equation (Problem 6-41 continued) x2

A12 m A12

2

m 3/ 2

m

y 4e

y 4e

y2

y2

m

1/ 2

dy

The value of the integral from tables is 3

x2

Therefore,

6-42. (a)

A12 3

2 /T

2 f 1 E0 2

4

2 / 1.42s 34

1.055 10

dy

4.

3/ 2

m

4.42rad / s

J s 4.42rad / s / 2

2.33 10

34

J 499.9 mm

500.0 mm

(b) A

500.0

2

499.9

2

10mm 0.1mm

E

n 1/ 2

1 / 2m

2

A

2

A

n 1 / 2 1 / 2 0.010kg 4.42rad / s 10 2 m

2.1 1028 or n

(c)

6-43.

0

f

x

1.055 10

34

J s

2.1 1028

0.70 Hz

2

x2 / 2

A0e

2

1

x

A1

m

xe

m x2 / 2

From Equation 6-58. Note that

0

is an even function of x and

It follows that

0

1

dx

0

144

1

is an odd function of x.

Chapter 6 – The Schrödinger Equation 2

6-44. (a) For x > 0, So, k2 (b) R

k22 / 2m V0 1/ 2

2mV0 2

k1 k2

2

E

k12 / 2m

. Because k1

k1

k2

2

2V0 1/ 2

4mV0

, then k2

k1 / 2

(Equation 6-68)

2

2

1 1/ 2

1 1/ 2

0.0294, or 2.94% of the incident particles are

reflected. (c) T

1 R 1 0.0294 0.971

(d) 97.1% of the particles, or 0.971 106

9.71 105 , continue past the step in the +x

direction. Classically, 100% would continue on.

6-45

(a) Equation 6-76:

T

16

E E 1 e V0 V0

2 a

where

2 2m p (V0

E) /

and a barrier width . 2 a

2(938 MeV/c 2 )(50 44)MeV / 6.58 10

2

(b) decay rate N

44 MeV 44 MeV 1 e 50 MeV 50 MeV

T

16

T

0.577

22

MeV s

10

15

1.075

N T where 2 44 MeV 1.60 10 13 J/MeV 1.67 10 27 kg

v proton 2R

0.577 4.59 1022 s

decay rate

(c) In the expression for T, e

1.075

e

2.150

1/2

1

1 2 10 15 m

2.65 1022 s

, and so T

0.577

4.59 1022 s

So, k2

2

k22 / 2m V0

6mV0

1/ 2

E

2

k12 / 2m

. Because k1

T

0.197 . The decay

2V0

4mV0

145

1/ 2

, then k2

1

1

rate then becomes 9.05 10 21 s 1 , a factor of 0.34× the original value.

6-46. (a) For x > 0,

1.075

3 / 2k1

Chapter 6 – The Schrödinger Equation (Problem 6-46 continued) (b) R

k1 R

k2 k1

2

k2

k1 2

k2 k1

2

2

2

k2

2

3/ 2

1

1

3/ 2

0.0102

Or 1.02% are reflected at x = 0. (c) T

1 R 1 0.0102

0.99

(d) 99% of the particles, or 0.99 106

9.9 105, continue in the +x direction.

Classically, 100% would continue on.

6-47. (a)

E

4eV

2m V0

9eV =V0

E /

2 0.511 106 eV / c2 5eV / 0.6 nm = a

5.11 106 eV 2260eV 197.3eV nm

and Since

a

0.6nm 11.46nm 1 a is not

eV / c 11.46nm

1

6.87

1, use Equation 6-75:

The transmitted fraction

T

Recall that sinh x

T

1

sinh2 a 1 4 E V0 1 E V0 ex

e

x

1

81 sinh2 6.87 80

1

2,

81 e6.87 e 1 80 2

6.87

2

1

4.3 10

(b) Noting that the size of T is controlled by

6

is the transmitted fraction.

a through the sinh2 a and increasing T

implies increasing E. Trying a few values, selecting E = 4.5eV yields T or approximately twice the value in part (a).

146

8.7 10

6

Chapter 6 – The Schrödinger Equation

6-48.

B

E1 / 2 E

E V0

1/ 2

E V0

1/ 2 1/ 2

A

For E

V0 , E V0

denominator are complex conjugates. Thus, B hence T

6-49.

A B

1 R

1/ 2

2

is imaginary and the numerator and 2

A and therefore R

B

2

A

2

0.

C and k1 A k1B

k 2C

Substituting for C, k1 A k1 B B

k1 k1

A

k1 k2 A C k1 k2

(Equations 6-65a & b)

k2 A k2 B and solving for B,

k2 A B

k2 A, which is Equation 6-66. Substituting this value of B into Equation 6-65(a), k2

A

k1

k2 k1 k2 k1 k2

or C

2k1 , which is Equation 6-67. k1 k2

6-50. Using Equation 6-76,

6-51.

T

16

T

16

R

1,

E E 1 e V0 V0 2.0 6.5

k1

k2

k1

k2

1

2 a

2.0 e 6.5

where E 2 10.87 0.5

2.0eV , V0 6.5 10

5

6.5eV , and a

0.5nm.

(Equation 6-75 yields T

6.6 10 5 .)

2

and T

2

1 R

(Equations 6-68 and 6-70)

(a) For protons:

k1

2mc 2 E / c

k2

2mc 2 E V0 / c

R

1.388 0.694 1.388 0.694

40MeV / 197.3MeV fm 1.388

2 938MeV

2

2 938MeV 10MeV / 197.3MeV fm 0.694 2.082

2

0.111 And T

147

1 R

0.889

0.694

Chapter 6 – The Schrödinger Equation (Problem 6-51 continued) (b) For electrons: 1/ 2

0.511 1.388 938

k1

0.0324

0.0324 0.0162 0.0324 0.0162

R

k2

0.511 0.694 938

1/ 2

0.0162

2

0.111 And T

1 R

0.889

No, the mass of the particle is not a factor. (We might have noticed that

m could

be canceled from each term.

6-52. (a) En

E1

(b)

n2 h2 8mL2 hc

The ground state is n = 1, so 2

1240MeV fm

8 mc 2 L2

2000

2

8 938.3MeV 1 fm

2

204.8MeV

(c)

1844 MeV

E21

21

n =3 21

1500 En (MeV)

(d) 1000

hc /

32

1240MeV fm 819 205 MeV

1240MeV fm 1844 819 MeV

2.02 fm

1.21 fm

819 MeV

(e)

500 205 MeV

0

148

31

1240 MeV fm 1844 205 MeV

0.73 fm

Chapter 6 – The Schrödinger Equation 2 x 2 / L sin2 x / L. 6-53. (a) The probability density for the ground state is P x The probability of finding the particle in the range 0 < x < L/2 is:

L/2

P

2L L

P x dx 0 L/3

(b) P

2L L

P x dx 0

/2

sin2 udu

2

sin2 udu

2

4

0 /3

sin 2 / 3 4

6

0

1 where u 2

0

1 3

x/L 3 4

0.195

(Note: 1/3 is the classical result.) 3L / 4

(c) P

2L L

P x dx 0

3 /4

2 3 8

sin2 udu 0

sin 3 / 2 4

3 4

1 2

0.909

(Note: 3/4 is the classical result.)

6-54.

(a) En So,

2

n2 h2 8mL2 En

1

and En En

n2

En

2 n

En En

1

En

n 1 h2 1

8mL2

2n 1 n 2 n2

2n 1 n2

2 1/ n For large n,1 / n n

(b) For n = 1000 the fractional energy difference is

2 1000

0.002

2 and

0.2%

(c) It means that the energy difference between adjacent levels per unit energy for large n is getting smaller, as the correspondence principle requires.

6-55. The n = 2 wave function is 2

Ek

op

2

2m x 2

Therefore, 2

Ek

2

x

2 2 x sin and the kinetic energy operator L L

x

2

2

2 2 x sin L L

x dx

2

2m x 2 2

2

2m x

2

2 2 x sin dx L L

149

Chapter 6 – The Schrödinger Equation (Problem 6-55 continued) L

2

2 L

2m

0

2 2 L 2m

Let

2 x L

2 L

y, then x

2 L

2 x dx L

y

0 and x

0

0

sin ydy 0

6-56. (a) The requirement is that x

x

or

2 2

0

2 2 L 2m

Therefore, Ek

2 L

2

2 x dx L

L

2 2 L 2m

2

sin 2 y 4

y 2

2

sin

sin2

Substituting above gives: Ek 2

2

2 x 2 sin L L

2

2

2

2 L

0

dy

dx

L dy 2

2

L 2

sin2 ydy 0

0 0

x

h2 2mL2

x

x . This can only be true if:

x .

x

(b) Writing the Schrödinger equation in the form of this 2nd order differential equation are: where k

2 dx L

2 and

4 2 2 2mL2

L 2

x

y

d2 dx 2

2mE 2

, the general solutions

A sin kx and

x

x

2mE . Because the boundaries of the box are at x

A cos kx L / 2, both

solutions are allowed (unlike the treatment in the text where one boundary was at

L / 2 provided that an integral number

x = 0). Still, the solutions are all zero at x

L / 2 and x

of half wavelengths fit between x n

x

2/ L

n

x

2/ L

1/ 2

1/ 2

L / 2. This will occur for:

cos n x / L when n 1, 3, 5,

. And for

sin n x / L when n

.

2, 4, 6,

The solutions are alternately even and odd. (c) The allowed energies are: E

2

k 2 / 2m

150

2

n /L

2

/ 2m

n 2 h 2 / 8mL2 .

Chapter 6 – The Schrödinger Equation 6-57.

0

(a)

x2 / 2 L2

Ae d 0 dx

d 1 dx

So,

x 2 / 2 L2

x / L2 Ae

And

1/ L

d2 1 dx 2

0

1/ L d 2 x / L3

and

x/L d

0

L

1

/ dx

0

1/ L d

0

2

/ 2m 3m / L3

x3 / L5

0

2

will make

1

1

0

E

0

0

x 2 / 2mL4 , the Schrödinger equation 2 3

x / 2mL5

0

x/L

0

E

0

L

0

x

2

2 2 2n x x sin dx L L

2 L L n 2 L L n

2

3

x/L

0

or,

. Thus, choosing E appropriately

3

2

/ 2mL2 , or three times the ground state energy.

plotted looks as below. The single node indicates that

x2

0

/ dx 2

1

is the first excited state.

(The energy value in [b] would also tell us that.)

6-58.

x/L

a solution.

(b) We see from (a) that E (c)

x3 / L5

3 2 x / 2mL3

simplifying:

x/ L d2

/ dx

Recalling from Problem 6-3 that V x becomes

x 2 / 2 L2

x / L2 Ae

L

/ dx

x / L3

0

d 0 dx

L n u3 6

Letting u

n x / L, du

n

u 2 sin2 udu 0

u2 4

n

1 sin 2u 8

u cos 2u 4

151

0

n / L dx

Chapter 6 – The Schrödinger Equation (Problem 6-58 continued) 3

2 L L n

6-59.

T

16

2m V0

n 4

0

6

E E 1 e V0 V0

(a)

3

n

2 a

2 m0c 2 V0

E

T

16

10 25

16

10 25

x, t

6-60. (a) For

d2 dx 2

1

k2

1

1

29.68

10 e 25

2.968

A sin kx

and

k2 A sin kx 2m

4.95 10 1

1nm.

c

197.3eV nm 19.84nm

19.84nm

a

2

25eV , and

E

19.84;

1nm

10 e 25

0.1nm :

(b) For a T

19.84nm

a

L2 2n 2

where E 10eV , V0

2 0.511 106 eV 15eV And

L2 3

0

2 a

1

29.68

13

0.1nm

1.984

0.197

t

A cos kx

t

t so the Schrödinger equation becomes:

2

V x A sin kx

t

t

i

Because the sin and cos are not proportional, this for

x, t

A cos kx

(b) For

x, t

A cos kx

d2 dx 2 k2 2m

k2

and

t

cos kx

cannot be a solution. Similarly,

t , there are no solutions. i kx

t

t

i sin kx

i

. And the Schrödinger equation becomes:

t

Ae

2

V x

t

2

for

152

k 2 / 2m V .

, we have that

Chapter 6 – The Schrödinger Equation 6-61. V

mgz E

Classically allowed: 0 < z < z0 Ek = E - mgz

0

z0

z

The wave function will be concaved toward the z axis in the classically allowed region and away from the z axis elsewhere. Each wave function vanishes at z = 0 and as z → ∞. The smaller amplitudes are in the regions where the kinetic energy is larger.

6-62. Writing the Schrödinger equation as: Ek have: Ek

x

E V x

x

x 2

V x

x

E

x from which we

/ 2m d 2 / dx 2 . The expectation value of

153

Chapter 6 – The Schrödinger Equation (Problem 6-62 continued) Ek is Ek

Ek

x dx. Substituting Ek

x

x from above and reordering

2

multiplied quantities gives: Ek

6-63. (a)

p x

x dx .

m v x /m a

v

d2 2m dx 2

x

34

1.055 10

v 1.6 108 m / s

9.11 10

J s

31

10

12

m

0.39c

(b) The width of the well L is still an integer number of half wavelengths, L and deBroglie’s relation still gives: L

nh / 2 p . However, p is not given by: E2

2mEk , but by the relativistic expression: p

p

nhc

Substituting this yields: L 2 E2

En

nhc 2L

hc 4 L2

(c) E1

/2 ,

n

2

mc 2

E2

1/ 2

mc 2

mc 2

2

2

1/ 2

c.

nhc / 2 L

2

1/ 2

2

mc 2

2

1/ 2

2

mc 2

2

1/ 2

1240eV nm 3

4 10 nm

2

2

0.511 106 eV

2

8.03 105 eV

(d) Nonrelativistic: E1

h2 8mL2

hc 2

2

1240eV nm 2

8 mc L

6

2

3

8 0.511 10 eV 10 nm

2

3.76 105 eV

E1 computed in (c) is 2.14 times the nonrelativistic value.

6-64. (a) Applying the boundary conditions of continuity to

and d / dx at x = 0 and x = a,

where the various wave functions are given by Equation 6-74, results in the two pairs of equation below:

154

Chapter 6 – The Schrödinger Equation (Problem 6-64 continued) At x

0:

At x

a : Feika

A B C D and ikA ikB a

Ce

De

a

C

and ikFeika

D Ce

a

De

a

Eliminating the coefficients C and D from these four equations, a straightforward 2

but lengthy task, yields: 4ik A

2

a

ik e

ik e

a

Feika

The transmission coefficient T is then: 2

F

T

A

2

4ik

2

2

eika

1 e 2

Recalling that sinh

2

a

ik e

ik e

a

and noting that

e

complex conjugates, substituting k

2mE

ik and

and

2m V0

ik are E

, T then

1

can be written as T

sinh2 a

1 4

a

(b) If

E E 1 V0 V0

1, then the first term in the bracket on the right side of the * equation in part

(a) is much smaller than the second and we can write:

F A

ik

Or T

6-65.

2 II

ik a

4ik e

16

2

C e

E V0

2 a

F A

and T

2

1

E e V0

2

6

2

k 2e

2

2 a

k2

2

2 a

(Equation 6-72) 2

Where C

2 E1 / 2

2

2m V0

E1 / 2

E

E V0

1/ 2

A

2 0.5V0

2

0.5V0

2 mp c 2 20MeV

155

1/ 2

2

1/ 2

0.5V0

1/ 2

2.000

Chapter 6 – The Schrödinger Equation (Problem 6-65 continued) 2 938.3MeV

x ( fm)

20MeV

e

197.3MeV fm

0.982 fm

2

2 x II

2

C e

1

0.1403

0.5612

2

0.0197

0.0788

3

2.76×10-3

1.10×10-2

4

3.87×10-4

1.55×10-3

5

5.4×10-5

2.2×10-4

156

1

2 x

Chapter 7 – Atomic Physics

7-1.

2

n 2mL 2

En1 n2 n3  E311 

2 1

2

2

3 2mL

 n22  n32



(Equation 7-4)



2

 12  12  11E0 where E0 

2

2



2

2

2mL2







E222  E0 22  22  22  12E0 and E321  E0 32  22  12  14E0

The 1st, 2nd, 3rd, and 5th excited states are degenerate.

E (×E0)

14

321

12

222 311

10 221 8

6

211

4 111 2 0

 2  n12

2 2 n22 n32    2 n22 n32    =    n1    2m  L12 L22 L23  2mL12  4 9  2

7-2.

En1 n2 n3

n1  n2  n3  1 is the lowest energy level. E111  E0 1  1 / 4  1 / 9   1.361E0 where E0 

The next nine levels are, increasing order,

157

2

2

2mL12

(Equation 7-5)

Chapter 7 – Atomic Physics (Problem 7-2 continued)

7-3.

n1

n2

n3

E  E0 

1

1

2

1.694

1

2

1

2.111

1

1

3

2.250

1

2

2

2.444

1

2

3

3.000

1

1

4

3.028

1

3

1

3.360

1

3

2

3.472

1

2

4

3.778

(a)  n1 n2 n3  x, y, z   A cos

nz n1 x ny sin 2 sin 3 L L L

(b) They are identical. The location of the coordinate origin does not affect the energy level structure.

7-4.

7-5.

 111  x, y, z   A sin

x

 121  x, y, z   A sin

x

 113  x, y, z   A sin

x

En1 n2 n3 

L1 L1 L1

sin sin sin

y 2 L1

y L1

y 2 L1

z

sin sin

3L1

z 3L1

sin

 112  x, y, z   A sin

x

 122  x, y, z   A sin

x

L1 L1

sin sin

y 2 L1

y L1

sin

2 z 3L1

sin

2 z 3L1

z L1

 2  n12

2 2 n32  n22   2 n22 n32   2   =  n1    2m  L1  2 L1 2  4 L1 2  2mL12  4 16    2

2 2  n2 n2   En1 n2 n3   n12  2  3  where E0  4 16  2mL12 

158

(from Equation 7-5)

Chapter 7 – Atomic Physics (Problem 7-5 continued) (a)

n1

n2

n3

E  E0 

1

1

1

1.313

1

1

2

1.500

1

1

3

1.813

1

2

1

2.063

1

1

4

2.250

1

2

2

2.250

1

2

3

2.563

1

1

5

2.813

1

2

4

3.000

1

3

1

3.313

(b) 1,1,4 and 1,2,2

7-6.

7-7.

 111  x, y, z   A sin

x

 113  x, y, z   A sin

x

 114  x, y, z   A sin

x

 123  x, y, z   A sin

x

 124  x, y, z   A sin

x

E0 

2

2

2mL2

L1 L1 L1 L1

L1

sin sin sin sin

sin

y 2 L1

y 2 L1

y 2 L1

y L1

y L1

z

 112  x, y, z   A sin

x

3 z 4 L1

 121  x, y, z   A sin

x

z

 122  x, y, z   A sin

x

3 z 4 L1

 115  x, y, z   A sin

x

z

 116  x, y, z   A sin

x

sin

4 L1

sin sin sin

sin

L1

L1

1.055  10 J s   kg  0.10  10 m  1.609  10 2

34





2 9.11  1031

9

E311  E111  E  11E0  3E0  8E0  301eV

159

2

L1 L1 L1 L1

L1

sin sin sin sin

sin

y 2 L1

y L1

y L1

y 2 L1

y 2 L1

sin sin

J / eV



2 L1

z 4 L1

sin

z 2 L1

sin

5 z 4 L1

sin

3 z 2 L1

2

19

z

 37.68eV

Chapter 7 – Atomic Physics (Problem 7-7 continued)

E222  E111  E  12E0  3E0  9E0  339eV E321  E111  E  14E0  3E0  11E0  415eV

7-8.

(a) Adapting Equation 7-3 to two dimensions (i.e., setting k3 = 0), we have

 n n  A sin 1 2

n1 x ny sin 2 L L

(b) From Equation 7-5, En1 n2 

2

2

2

2mL

n

2 1

 n22



(c) The lowest energy degenerate states have quantum numbers n1 = 1, n2 = 2, and n1 = 2, n2 = 1.

7-9.

 0, 1, 2

(a) For n = 3, (b) For

 0, m  0 . For

 1, m  1, 0,  1 . For

 2, m  2,  1, 0,  1,  2 .

(c) There are nine different m-states, each with two spin states, for a total of 18 states for n = 3.

7-10. (a) For

4



L

 1  4  5  20

m 4

min  cos1 (b) For

4 20

 min  26.6

2

L 6

m 2

min  cos1



2 6

 min  35.3







7-11. (a) L  I  105 kg m2  2  735min1 1min / 60s   7.7  104 kg m2 / s (b) L 



 1  7.7  104 kg m2 / s

160

Chapter 7 – Atomic Physics (Problem 7-11 continued)

 7.7  10 kg m / s    1  1.055  10 J s  4

2

34

2

2

7.7  104 kg m2 / s   7.3  1030 34 1.055  10 J s

7-12. (a) +1

1 L  2 0

−1

(b) +2

+1

2 L  6

0

−1 −2

161

Chapter 7 – Atomic Physics (Problem 7-12 continued) (c) +4 +3 +2

4

+1

L  20

0 −1 −2 −3 −4

(d) L 

7-13.



 1

(See diagrams above.)

L2  L2x  L2y  L2z  L2x  L2y  L2  L2z  (a)

L

2 x

 L2y



(b)

L

 L2y



2 x

min

max









 6  22

 6  02

(c) L2x  L2y   6  1

2

5

2

2

2

2

6



 1

2

 m



L

(b) For  1,

m  1, 0,  1



 6  m2



2

2

Lx and Ly cannot be determined separately.

 1  2  1.49  1034 J s

(a) For  1,

2

2

(d) n = 3

7-14



162

Chapter 7 – Atomic Physics (Problem 7-14 continued) (c)

Z

+1ћ

L  2 0

−1ћ

(d) For  3,

L



 1  12  3.65  1034 J s and m  3,  2,  1, 0,1, 2, 3 Z 3ћ 2ћ 1ћ 0

−1ћ −2ћ −3ћ

7-15.

L= r× p

dL dr dp   p+ r dt dt dt

dr dp  p  v  mv = mv  v  0 and r   r  F . Since for V =V(r), i.e., central forces, dt dt

F is parallel to r, then r  F = 0 and

7-16. (a) For (b) For

dL 0 dt

 3, n = 4, 5, 6, … and m = −3, −2, −1, 0, 1, 2, 3

 4, n = 5, 6, 7, … and m = −4, −3, −2, −1, 0, 1, 2, ,3 ,4

163

Chapter 7 – Atomic Physics (Problem 7-16 continued) (c) For

 0, n = 1 and m = 0

(d) The energy depends only on n. The minimum in each case is: E4  13.6eV / n2  13.6eV / 42  0.85eV E5  13.6eV / 52  0.54eV

E1  13.6eV 7-17. (a) 6 f state: n  6,  3 (b) E6  13.6eV / n2  13.6eV / 62  0.38eV (c) L  (d) Lz  m



 1  3  3  1  12  3.65  1034 J s

Lz  3 ,  2 ,  1 , 0, 1 , 2 , 3

7-18. Referring to Table 7-2, R30 = 0 when

 2r 2r 2   0 1  2   3a0 27a0  Letting r a0  x, this condition becomes x2  9 x  13.5  0 Solving for x (quadratic formula or completing the square), x = 1.90, 7.10. Therefore, r a0  1.90, 7.10 . Compare with Figure 7-10(a). 2

 kZe2   7-19. Equation 7-25: En     2   2n

Using SI units and noting that both Z and n are unitless, we have: 2

 (N  m2 / C2 )  C2  J   kg J  s   2

 N  m2  Cancelling the C and substituting J  N  m on the right yields J     kg . N  m  s   2

164

Chapter 7 – Atomic Physics (Problem 7-19 continued) 2

m Cancelling the N and m gives J     kg  J , since kg  m2 / s2 are the units of kinetic s

energy.

7-20. (a) For the ground state n = 1,

 100  R10Y00  (b)  2 

2 a03

e r / a0

 0, and m = 0. 1 4



2 4 a02

e r / a0 

4 a03

at r  a0

1 2 r / a0 1 e  3 e2 at r  a0 3  a0  a0

(c) P  r    2  4 r 2 

4 2 e at r  a0 a0





7-21. (a) For the ground state, P  r  r   2 4 r 2 r 

4r 2 2 r / a0 e r a03

For r  0.03a0 , at r  a0 we have P  r  r  (b) For r  0.03a0 , at r  2a0 we have P  r  r 

7-22.

2e1

4a02 2 e  0.03a0   0.0162 a03

4  2a0  3 0

a

2

e 4  0.03a0   0.0088

P  r   Cr 2e2 Zr / a0 For P(r) to be a maximum,   2Z  2 Zr / a0  dP 2Zr  a0   C r 2    2re2 Zr / a0   0  C   r  e2 Zr / a0  0 e  dt a0  Z    a0  

This condition is satisfied with r = 0 or r = a0 /Z. For r = 0, P(r) = 0 so the maximum P(r) occurs for r = a0 /Z.

165

Chapter 7 – Atomic Physics   2

7-23.

2 2 2  d      r sin drd d  1 0 0 0 

 4  r dr  4 C 2 2

2 210

0

2



 Zr  2  Zr / a 0  a0  r e 0 dr  1

  Z 2 r 4   Zr / a0 2  4 C210 0  a02  e dr  1

Letting x  Zr / a0 , we have that r  a0 x / Z and dr  a0dx / Z and substituting these above, 2  4 a03C210 4 x  d  Z 3 0 x e dx 2

Integrating on the right side 

x e

4 x

dx  6

0 1/ 2

Solving for C

7-24.  200

Z    32  a0  1

P  r  r   200

2



3/ 2

2 210

yields:

2 210

C

 r   r / 2 a0 1   e  a0 

 Z3  Z3   C   210 3  24 a03  24 a0 

(Z = 1 for hyrdogen) 2

1 1 r  4 r r  1   e r / a0 4 r 2 r 3  32 a0  a0  2







(a) For r  0.02a0 , at r  a0 we have P  r  r 

4 1 1 2 1  1 e1a02  0.02a0    0  e1  0.02   0 3  32 a0 8

(b) For r  0.02a0 , at r  2a0 we have P  r  r 

4 1 1 2 1 e2 a02  0.02a0   1 e2  0.02   3.4  104 3  32 a0 8

166

Chapter 7 – Atomic Physics 7-25.  210  C210

Zr  Zr / 2 a0 e cos a0

P  r    210

2

 4 r   4 r

 4 C210

2

2

Z

2

(Equation 7-34)

2

C210

2

Z 2 r 2  r / a0 e cos2  a02



/ a02 r 4e r / a0 cos2 

 Ar 4e r / a0 cos2 

where A  4 C210

7-26.  200

1    32  a0  1

2

Z

3/ 2

(a) At r  a0 ,  200

2



/ a02 , a constant.

 r   r / 2 a0 2  e 2a0  

(Z = 1 for hyrdogen)

 1 0.606  1  1 / 2    3   2  1 e   32  a0  32  a0  1

 1  1 0.368 1  3 e  32 a03 32  a0  1

(b) At r  a0 ,

 200 

(c) At r  a0 ,

P  r    200

2

3/ 2

2

 4 r  2

4 0.368a02 0.368   32 a03 8a0

7-27. For the most likely value of r, P(r) is a maximum, which requires that (see Problem 7-25)   Z  dP  A cos2   r 4    e Zr / a0  4r 3e Zr / a0   0 dr   a0  





For hydrogen Z = 1 and A cos2  r 3 / a0  4a0  r  e r / a0  0 . This is satisfied for r = 0 and r = 4a0. For r = 0, P(r) = 0 so the maximum P(r) occurs for r = 4a0.

7-28. From Table 7-1, Y21  ,   From Table 7-2, R32  r  

15 sin cos ei 8

r 2  r 3a0 e 2 81 30a02 a0 4

167

Chapter 7 – Atomic Physics (Problem 7-28 continued)

 321  r, ,  

4

1 2 81 30a03 a0

15 2  r 3a0 r e sin cos ei 8

To be sure that  is normalized, we do the usual normalization as follows, where C is the normalization constant to be compared with the coefficient above.   2

  2

0 0 0

0 0 0

2  2 4 2 r 3a0 C32 sin2  cos2  r 2 sin dr d d  1 1       dr  C32 1    r e 2

 d  2 , we have

Noting that

0

2 C

2 32 1



r e

6 2 r 3 a0

0



dr  sin3  cos2  d  1 0

Evaluating the integrals (with the aid of a table of integrals) yields: 4 2 4  2 C32 a07     1 1  1.23  10 15 





C32 1  0.00697

a07

This value agrees with the coefficient of  32 1 above.

7-29.  100 

2

e r / a0

4 a

3 0

Because  100 is only a function of r, the angle derivatives in Equation 7-9 are all zero.

d  dr r2

d  dr

2

e r / a0

4 a

3 0

 1  2  r / a0  r e 4 a03  a0  2

d  2 d  r  dr  dr 

1 d  2 d r r 2 dr  dr

  1  2  1     r     2r  e r / a0   4 a03  a0    a0 

  

2

 1   2 1   r / a0 Substituting into Equation 7-9,    e 4 a03  a0   r a0  2

168

Chapter 7 – Atomic Physics (Problem 7-29 continued)



 1 2   2  100  V 100  E 100 2  a0 a0 r  2

For the 100 state r  a0 and 2 a0    2 / k or a0  1 / k, so

 1 2   1 2  1 2  2    2  2    2  k a0  a0 a0 r   a0 a0  Thus,  2

k2

2

2 2  1 2  k and we have that  2  2  a0 a0 r  2 2

 V  E, satisfying the Schrödinger equation.

7-30. (a) Every increment of charge follows a circular path of radius R and encloses an area

 R 2 , so the magnetic moment is the total current times this area. The entire charge Q rotates with frequency f   / 2 , so the current is i  Qf  q / 2

  iA   Q / 2   R2   Q R2 / 2 L  I 

g

1 MR 2 2

2M  2MQ R 2 / 2  2 QL QMR 2 / 2

(b) The entire charge is on the equatorial ring, which rotates with frequency f   / 2 . i  Qf  Q / 2

  iA   Q / 2   R2   Q R2 / 2 g

2M  2MQ R 2 / 2   5 / 2  2.5 QL QMR 2 / 5

169

Chapter 7 – Atomic Physics 7-31. Angular momentum S  I   2 / 5 mr 2  v / r  or v   5 / 2  S 1 / mr   5S / 2mr  5  3 / 4 

1/ 2

1.055  10 2  9.11  10 kg 10

5 3 / 4

1/ 2



34

31

J s

15

m



/ 2mr

  2.51  10

11

m / s  837c

 0 , so two lines due to spin of the single s electron would

7-32. (a) The K ground state is be seen.

 0 with two s electrons whose spins are opposite resulting

(b) The Ca ground state is

in S = 0, so there will be one line. (c) The electron spins in the O ground state are coupled to zero, the orbital angular momentum is 2, so five lines would be expected. (d) The total angular momentum of the Sn ground state is j = 0, so there will be one line.

7-33.

Fz  ms g L B  dB / dz   mAg az

(From Equation 7-51)

and az  mS g L B  dB / dz  / mAg Each atom passes through the magnet’s 1m length in t = (1/250)s and cover the additional 1m to the collector in the same time. Within the magnet they deflect in the z direction an amount z1 given by: z1  1 / 2  az t 2  1 / 2  ms g L B  dB / dz  / mAg  1 / 250  and leave 2

the magnet with a z-component of velocity given by vz  az t . The additional z deflection in the field-free region is z2  vz t  az t 2 . The total deflection is then z1  z2  0.5mm  5.0  104 m.

5.0  104 m  z1  z2   3 / 2  az t 2   3 / 2  ms g L B  dB / dz  / mAg  1 / 250 or, 2







5.0  104 m  250  mAg dB  dz ms g L  B 2

  2

5.0 10 m  250s  1.79 10 kg   2  0.805T / m  3 1 / 2 1  9.27  10 J / T  4

1

2

25

24

170

Chapter 7 – Atomic Physics 7-34. (a) There should be four lines corresponding to the four mJ values −3/2, −1/2, +1/2, +3/2. (b) There should be three lines corresponding to the three m values −1, 0, +1.

7-35. (a) For the hydrogen atom the n = 4 levels in order of increasing energy are: 4 2S1 / 2 , 4 2 P1 / 2 , 4 2 P3 / 2 , 4 2 D3 / 2 , 4 2 D5 / 2 , 4 2 F5 / 2 , 4 2 F7 / 2

(b) 2j + 1

7-36. For

 2,



L

 1  6  2.45 ,

j   1 / 2  3 / 2, 5 / 2 and J 

For j  3 / 2,

J

3 / 23 / 2  1

 15 / 4  1.94

For j  5 / 2,

J

5 / 25 / 2  1

 35 / 4  2.96

j  j  1

7-37. (a) j   1 / 2  2  1 / 2  5 / 2 or 3 / 2 (b) J 

5  5 / 2  1  2.96 2

j  j  1 

or

3 3 / 2  1  1.94 2

(c) J  L  S J Z  LZ  SZ  m  ms  m j where m j   j,  j  1,..., j  1, j. For j = 5/2 the z-components are 5 / 2,  3 / 2,  1/ 2,  1/ 2,  3 / 2,  5 / 2. For j = 3/2 the z-components are 3 / 2,  1 / 2,  1 / 2,  3 / 2.

7-38. (a) d5 / 2

L   cos1 s  3/ 4

ms =1/2

θ

(b) d3 / 2

L

θ   cos1

ms =−1/2

s  3/ 4 171

1/ 2 3/ 4

 54.7

 1 / 2   125.3 3/ 4

Chapter 7 – Atomic Physics 7-39.

m

n

−3, −2, −1, 0, 1, 2, 3 −2, −1, 0, 1, 2 −1, 0, 1 0 −2, −1, 0, 1, 2 −1, 0, 1 0 −1, 0, 1 0

3 2 1 0 2 1 0 1 0

4

3 2

ms 1 / 2 for each m state

1 / 2 for each m state 1 / 2 for each m state

7-40. (a) L  L1  L2 

1



2

, 

1



 1,...,

2

1



2

 1  1, 1  1  1, 1  1  2, 1, 0

(b) S  S1  S2

s   s1  s2  ,  s1  s2  1,..., s1  s2  1 / 2  1 / 2 , 1 / 2  1 / 2   1, 0 (c) J  L  S

j    s , For



 s  1 ,...,  s

 2 and s  1, j  3, 2, 1  2 and s  0, j  2

For

 1 and s  1, j  2, 1, 0  1 and s  0, j  1

For

 0 and s  1, j  1  0 and s  0, j  0

(d) J1  L1  S1

J 2  L2  S2 (e) J  J1  J 2

j1  j2 

1

 1 / 2  3 / 2, 1 / 2

2

 1 / 2  3 / 2, 1 / 2

j   j1  j2  ,

 j1  j2  1,...,

For j1  3 / 2 and j2  3 / 2, j  3, 2, 1, 0 j1  3 / 2 and j2  1 / 2, j  2, 1 For j1  1 / 2 and j2  3 / 2, j  2, 1 j1  1 / 2 and j2  1 / 2, j  1, 0

These are the same values as found in (c).

172

j1  j2

Chapter 7 – Atomic Physics 7-41. (a)

E  hf  f  E / h  (4.372 106 eV)(1.602 1019 J/eV) / 6.63 1034 J  s=1.056 109 Hz (b) c  f     c / f  (3.00 108 m/s) / (1.056 109 Hz)  0.284m  28.4cm (c) short wave radio region of the EM spectrum

7-42. (a) E3 / 2 

hc



E3 / 2 

Using values from Figure 7-22, 1239.852eV nm  2.10505eV 588.99nm

E1 / 2 

1239.852eV nm  2.10291eV 589.59nm

(b) E  E3 / 2  E1 / 2  2.10505eV  2.10291eV  2.14  103 eV (c) E  2 B B  B 

7-43.  12    x1, x2   C sin



E 2.14  103 eV   18.5T 2 B 2 5.79  104 eV / T



 x1 L

sin

2 x2 L

Substituting into Equation 7-57 with V = 0,

  2 12  2 12   2  2    1  4        2  12  E 12 2m  x12 x22   2m  L  2

Obviously,  12 is a solution if E 

7-44.



En 

n 2 2 2 2mL2

5 2 2 2mL2

Neutrons have antisymmetric wave functions, but if spin is ignored then

one is in the state n = 1 state, but the second is in the n = 2 state, so the minimum energy





is: E  E1  E2  12  22 E1  5E1 where

E1

 hc  

2

2

2mc 2 L2

197.3  2  2 2  939.6  2.0  2

 51.1MeV

173

E  5E1  255MeV

Chapter 7 – Atomic Physics 7-45. (a) For electrons: Including spin, two are in the n = 1 state, two are in the n = 2 state, and one is in the n = 3 state. The total energy is then:

E  2 E1  2 E2  E3 where E1

 hc  

2

2

2me c 2 L2

where En 





n2 2 2 2mL2

197.3

2



2



2 0.511  106 1.0 

 



E  2 E1  2 22 E1  32 E1  19 E1

2

 0.376eV

E  19 E1  7.14eV

(b) Pions are bosons and all five can be in the n = 1 state, so the total energy is: E  5E1 where E1 

0.376eV  0.00142eV 264

E  5E1  0.00712eV

7-46. (a) Carbon: Z  6; 1s 2 2s 2 2 p 2 (b) Oxygen: Z  8; 1s 2 2s 2 2 p 4 (c) Argon: Z  18; 1s 2 2s 2 2 p6 3s 2 3 p6

7-47. Using Figure 7-34: Sn (Z = 50)

1s 2 2s 2 2 p6 3s 2 3 p6 3d 10 4s 2 4 p6 4d 10 5s 2 5 p 2 Nd (Z = 60)

1s 2 2s 2 2 p6 3s 2 3 p6 3d 10 4s 2 4 p6 4d 10 5s 2 5 p6 4 f 4 6s 2 Yb (Z = 70)

1s 2 2s 2 2 p6 3s 2 3 p6 3d 10 4s 2 4 p6 4d 10 4 f 14 5s 2 5 p6 6s 2 Comparison with Appendix C. Sn: agrees Nd: 5 p6 and 4 f 4 are in reverse order Yb: agrees 7-48. Both Ga and In have electron configurations  ns   np  outside of closed shells 2

 n  1, s   n  1, p   n  1, d  2

6

10

. The last p electron is loosely bound and is more easily

removed than one of the s electrons of the immediately preceding elements Zn and Cd. 174

Chapter 7 – Atomic Physics 7-49. The outermost electron outside of the closed shell in Li, Na, K, Ag, and Cu has

 0. The

ground state of these atoms is therefore not split. In B, Al, and Ga the only electron not in a closed shell or subshell has

 1, so the ground state of these atoms will be split by the

spin-orbit interaction.

7-50.

En  

Z eff2 E1

Z eff  n

n2

(Equation 7-25)

 En 5.14eV 3  1.84 E1 13.6eV

7-51. (a) Fourteen electrons, so Z = 14. Element is silicon. (b) Twenty electrons, so Z = 20. Element is calcium.

7-52. (a) For a d electron, (b) For an f electron,

 2, so Lz  2 ,  1 , 0, 1 , 2  3, so Lz  3 ,  2 ,  1 , 0, 1 , 2 , 3

7-53. Like Na, the following atoms have a single s electron as the outermost shell and their energy level diagrams will be similar to sodium’s: Li, Rb, Ag, Cs, Fr. The following have two s electrons as the outermost shell and will have energy level diagrams similar to mercury: He, Ca, Ti, Cd, Mg, Ba, Ra.

7-54. Group with 2 outer shell electrons: beryllium, magnesium, calcium, nickel, and barium. Group with 1 outer shell electron: lithium, sodium, potassium, chromium, and cesium.

7-55. Similar to H: Li, Rb, Ag, and Fr. Similar to He: Ca, Ti, Cd, Ba, Hg, and Ra.

175

Chapter 7 – Atomic Physics 7-56. n

j

4

0

1/2

4

1

1/2

4

1

3/2

5

0

1/2

3

2

3/2

3

2

5/2

5

1

1/2

5

1

3/2

4

2

3/2

4

2

5/2

6

0

1/2

4

3

5/2

4

3

7/2

Energy is increasing downward in the table.

7-57. Selection rules:   1 j  1, 0 Transition

∆ℓ

∆j

Comment

4S1/2 → 3S1/2

0

0

ℓ - forbidden

4S1/2 → 3P3/2

+1

+1

allowed

4P3/2 → 3S1/2

−1

−1

allowed

4D5/2 → 3P1/2

−1

−2

j – forbidden

4D3/2 → 3P1/2

−1

−1

allowed

4D3/2 → 3S1/2

−2

−1

ℓ - forbidden

5D3/2 → 4S1/2

−2

−1

ℓ - forbidden

5P1/2 → 3S1/2

−1

0

allowed

7-58. (a) E1  13.6eV  Z  1  13.6eV  74  1  7.25  104 eV  72.5keV 2

2

(b) E1  exp   69.5keV  13.6eV  Z     13.6eV  74  1 2



74    69.5  103 eV / 13.6eV



1/ 2

 71.49

  74  71.49  2.51

176

2

Chapter 7 – Atomic Physics 7-59.

j  1, 0

 no j  0  j  0

(Equation 7-66)

The four states are 2 P3 / 2 , 2 P1 / 2 , 2 D5 / 2 , 2 D3 / 2 Transition

∆ℓ

∆j

Comment

D5/2 → P3/2

−1

−1

allowed

D5/2 → P1/2

−1

−2

j - forbidden

D3/2 → P3/2

−1

0

allowed

D3/2 → P1/2

−1

−1

allowed

7-60. (a) E  hc / 

E  3P1 / 2   E  3S1 / 2  

1240eV nm  2.10eV 589.59nm

E  3P1 / 2   E  3S1 / 2   2.10eV  5.14eV  2.10eV  3.04eV E  3D   E  3P1 / 2  

1240eV nm  1.52eV 818.33nm

E  3D   E  3P1 / 2   1.52eV  3.04eV  1.52eV  1.52eV (b) For 3P :

Z eff  3

3.04eV  1.42 13.6eV

For 3D :

Z eff  3

1.52eV  1.003 13.6eV

(c) The Bohr formula gives the energy of the 3D level quite well, but not the 3P level.

7-61. (a) E  gm j B B

(Equation 7-73) where s  1 / 2,  0 gives j  1 / 2 and

(from Equation 7-73) g  2.



m j  1 / 2.



E   2  1 / 2  5.79  105 eV / T  0.55T   3.18  105 eV

The total splitting between the m j  1 / 2 states is 6.37  105 eV . (b) The m j  1 / 2 (spin up) state has the higher energy. (c) E  hf  f  E / h  6.37  105 eV / 4.14  1015 eV s  1.54  1010 Hz This is in the microwave region of the spectrum. 177

Chapter 7 – Atomic Physics

7-62.

E

hc



 E 

7-63. (a) E 

dE hc 2    2     E d  hc





e B  5.79  105 eV / T  0.05T   2.90  106 eV 2m

(b)  

2 hc

 579.07nm   2.90  106 eV  2

E 

1240eV nm

 7.83  104 nm

(c) The smallest measurable wavelength change is larger than this by the ratio

0.01nm 7.83 104 nm  12.8. The magnetic field would need to be increased by this same factor because B  E  . The necessary field would be 0.638T.

7-64.

  13.6eV  Z

 2   5.39eV

En  13.6eV Zeff2 n2 E2

2 eff

2

Zeff  2  5.39 13.6 

1/ 2

7-65.  100

1 Z      a0 

 1.26

3/ 2

e Zr / a0

* P  r   4 r 2 100  100

(Equations 7-30 and 7-31)

(Equation 7-32)

Z 3  Zr / a0 4Z 3 2 2 Zr / a0  4 r e  3 r e  a03 a0 2





4Z 3 3 2 Zr / a0 r   rP  r  dr   3 r e dr a0 0 0 3

a0   2Zr  2 Zr / a0 a 3a  d  2Zr / a0   0  3!  0  e  4Z 0  a0  4Z 2Z

178

Chapter 7 – Atomic Physics

7-66. (a) E 



 1

2

: 0  1: 1 E : 0

2I

30E1

E5 = 30E1

20E1

E4 = 20E1

1 2 2 3 1E1 6E1

3 4 4 5 12E1 20E1

5 … 6 … 30E1 …

E3 = 12E1

10E1

E2 = 6E1 2

E1 = 0

/I

E0 = 0

(b) E 1  E 

2

  1  2   2I 





2

  1  2  2I 

The values of

 1

 

2

I



 1    1 E1

 0, 1, 2, … yield all the positive integer multiples of E1.

2 2 c 1 2 2 (c) I  m p r 2  E1    2 2 I mp r mpc2r 2 2



(d)  

2 197.3eV nm 

2

938.28  10 eV   0.074nm  6

2

 1.52  102 eV

hc 1.24  106 eV nm   8.18  105 m  81.8 m E1 1.52  102 eV

7-67. (a) Fz  ms g L B  dB / dz  (From Equation 7-51) From Newton’s 2nd law, Fz  mH az  ms g L B  dB / dz 







az  ms g L  dB / dz  / mH  1 / 2 1 9.27  1024 J / T  600T / m  / 1.67  1027 kg  1.67  106 m / s 2

179



Chapter 7 – Atomic Physics (Problem 7-67 continued)



(b) At 14.5km / s  v  1.45  104 m / s, the atom takes t1  0.75m / 1.45 104 m / s



 5.2 105 s to traverse the magnet. In that time, its z deflection will be:





z1  1 / 2  az  t12  1 / 2  1.67  106 m / s 2 5.2  105 s



2

 2.26  103 m  2.26mm

Its vz velocity component as it leaves the magnet is vz  az t1 and its additional z deflection before reaching the detector 1.25m away will be:



z2  vz t2   az t1  1.25m 1.45  104 m / s 











 1.67  106 m / s 2 5.2  105 s 1.25 1.45 104 m / s



 7.49  103 m  7.49mm

Each line will be deflected z1  z2  9.75mm from the central position and, thus, separated by a total of 19.5mm = 1.95cm.



7-68. min  cos1 m 

cosmin 



or, sin min  1  2

 1  with m  . 

 1 . Thus, cos2 min  2



 1





 1 



2

2



 1

 2



 1  1  sin2 min

  2   1

1/ 2

 1  And, sinmin     1

For large , min is small. 1/ 2

 1  Then sinmin  min     1



1

 

1/ 2

7-69. (a) E1  hf  hc / 1  1240eV nm / 766.41nm  1.6179eV

E2  hf  hc / 2  1240eV nm / 769.90nm  1.6106eV (b) E  E1  E2  1.6179eV  1.6106eV  0.0073eV (c) E / 2  gm j B B  B 

E 0.0073eV   63T 2 gm j B 2  2 1 / 2  5.79  105 eV / T



180



Chapter 7 – Atomic Physics

7-70.

P r 

4Z 3 2 2 Zr / a0 re a03

(See Problem 7-65)

For hydrogen, Z = 1 and at the edge of the proton r  R0  1015 m. At that point, the exponential factor in P(r) has decreased to: e2 R0 / a0  e



2 1015

 0.52910

10

m

  e3.7810   1  3.78  105  1 5

Thus, the probability of the electron in the hydrogen ground state being inside the nucleus, to better than four figures, is: r0

4r 2 P r   3 a0

P   P  r dr  0

R0

4r 2 4  3 3 a0 a0

 0

R0

4 r3 r dr  0 a03 3





j  j  1  s  s  1 

7-71. (a) g  1 

2 j  j  1

For 2 P1 / 2 :

g  1

j  1 / 2,



 1



g  1

 9.0  1015

 1, and s  1 / 2

2  1 / 2 1 / 2  1

j  1 / 2,

3

(Equation 7-73)

1 / 2 1 / 2  1  1 / 2 1 / 2  1  11  1

For 2S1 / 2 :

0

3

4 1015 m 4  R03   3  a0  3  3 0.529  1010 m



R0

2

 1

3/ 4  3/ 4  2  2/3 3/ 2

 0, and s  1 / 2

1 / 2 1 / 2  1  1 / 2 1 / 2  1  0 2  1 / 2 1 / 2  1

 1

3/ 4  3/ 4 2 3/ 2

The 2 P1/ 2 levels shift by: E  gm j B B 

2 1 1   B B   B B  3 2 3

The 2 S1/ 2 levels shift by:  1 E  gm j B B  2     B B   B B  2

181

(Equation 7-72)

Chapter 7 – Atomic Physics (Problem 7-71 continued) To find the transition energies, tabulate the several possible transitions and the corresponding energy values (let Ep and Es be the B = 0 unsplit energies of the two states.): Transition

Energy

P1 / 2,1 / 2  S1 / 2,1 / 2

1 2    E p  3 B B    Es  B B   E p  Es  3 B B  

P1 / 2, 1 / 2  S1 / 2,1 / 2

1 4    E p  3 B B    Es  B B   E p  Es  3 B B  

P1 / 2,1 / 2  S1 / 2, 1 / 2

1 4    E p  3 B B    Es  B B   E p  Es  3 B B  

P1 / 2, 1 / 2  S1 / 2, 1 / 2

1 2    E p  3 B B    Es  B B   E p  Es  3 B B  



 





 



Thus, there are four different photon energies emitted. The energy or frequency spectrum would appear as below (normal Zeeman spectrum shown for comparison).

anomolous

normal

(b) For 2 P3 / 2 :

g  1

j  3 / 2,

 1, and s  1 / 2

3 / 2  3 / 2  1  1 / 2 1 / 2  1  11  1 2  3 / 2  3 / 2  1

 1

15 / 4  3 / 4  2  4/3 30 / 4

These levels shift by: E  gm j B B 

4 1 2   B B   B B  3 2 3

182

E 

4 3  B B  2 B B 3  2 

Chapter 7 – Atomic Physics (Problem 7-71 continued) Tabulating the transitions as before: Transition

Energy

P3 / 2, 3 / 2  S1 / 2,1 / 2

E

P3 / 2, 3 / 2  S1 / 2, 1 / 2

forbidden, m j  2

P3 / 2,1 / 2  S1 / 2,1 / 2

2 1    E p  3 B B    Es  B B   E p  Es  3 B B  

P3 / 2,1 / 2  S1 / 2, 1 / 2

2 5    E p  3 B B    Es  B B   E p  Es  3 B B  

P3 / 2, 1 / 2  S1 / 2,1 / 2

2 5    E p  3 B B    Es  B B   E p  Es  3 B B  

P3 / 2, 1 / 2  S1 / 2, 1 / 2

2 1    E p  3 B B    Es  B B   E p  Es  3 B B  

P3 / 2, 3 / 2  S1 / 2,1 / 2

forbidden, m j  2

P3 / 2, 3 / 2  S1 / 2, 1 / 2

E

p







 2B B   Es  B B   E p  Es  B B















p









 2B B   Es  B B   E p  Es  B B

There are six different photon energies emitted (two transitions are forbidden); their spectrum looks as below:

anomolous

normal

7-72. (a) Substituting   r,  into Equation 7-9 and carrying out the indicated operations yields (eventually)



2

2

  r,   2 / r 2  1 / 4a02  

2

2

  r,   2 / r 2   V  r,   E  r, 

183

Chapter 7 – Atomic Physics (Problem 7-72 continued) Canceling   r,  and recalling that r 2  4a02 (because  given is for n = 2) we have 

 1 / 4a   v  E 2 2

2 0

The circumference of the n = 2 orbit is: C  2  4a0   2  a0   / 4  1 / 2k. Thus, 

(b) or

2 2 1  k    V  E  V  E  2  2   4 / 4k  2 2

p2  v  E and Equation 7-9 is satisfied. 2m 2



 r  r / a 2 2 0  dx   A  a0  e 0 cos  r sin drd d  1 2

2

2

 2  r  A    e r / a0 r 2 dr  cos2  sin d  d  1 a 0 0  0 0 

2

Integrating (see Problem 7-23),

 

A2 6a03  2 / 3 2   1

A2  1 / 8a03  A  1 / 8a03

7-73.

   g L B L

(Equation 7-43)

(a) The 1s state has The 2p state has

 0, so it is unaffected by the external B.  1, so it is split into three levels by the external B.

(b) The 2 p  1s spectral line will be split into three lines by the external B. (c) In Equation 7-43 we replace B with k  e / 2mk , so

kz   11 e / 2mk   B  me / mk 

(From Equation 7-45)

Then E  B  me / mk  B









 5.79  105 eV / T  0.511 106 MeV / c 2 / 497.7MeV / c2  1.0T   5.94  108 eV

184

Chapter 7 – Atomic Physics (Problem 7-73 continued) 







hc

E (From Problem 7-62) where λ for the (unsplit) 2p → 1s transition

is given by

  hc Ek and Ek  E2  E1  13.6eV  mk / me 1  1 / 4  9.93  103 eV and   1240eV nm 9.93 103 eV  0.125nm 

and

7-74.



E    B 





0.125nm 5.94  108 eV 1240eV nm

12

ke2 S L where, for n  3, r  a0 n 2  9a0 r 3m mc 2



For 3P states S L  E 

  5.98  10





2

  6.58 10 9  0.053nm   0.511  10 eV  2

1.440eV nm 3.00  108 m / s  109 nm / m 3

For 3D states S L 

2

6

16

eV s



2

 1.60  104 eV

2

/3

E  1.60 104 eV / 3  0.53 104 eV

 = B  L  2 S 

7-75. (a) J = L + S

J 

(Equation 7-71)

  B  L  2 S  /    L + S      B  L L  2 S S  3S L  J J J

J



B J

L

2

 2S 2  3S L



(b) J 2  J J   L + S   L + S   L L + S S + 2 S L  S L  (c)  J  

B 

(d) Z   J



1 2 J  L2  S 2 2

 3  L2  2S 2  J 2  L2  S 2    B 3J 2  S 2  L2  J 2 2 J 



JZ  J    B 3J 2  S 2  L2 Z   B 3J 2  S 2  L2 J 2 J J 2 J











185







Chapter 7 – Atomic Physics (Problem 7-75 continued)

 J 2  S 2  L2  J Z    B 1   2J 2   (e) E  Z B

(Equation 7-69)

 j  j  1  s  s  1     B B 1  2 j  j  1   gm j B B



 1   mj 

(Equation 7-72)

 j  j  1  s  s  1  where g  1  2 j  j  1 



 1   

(Equation 7-73)

7-76. The number of steps of size unity between two integers (or half-integers) a and b is b – a. Including both values of a and b, the number of distinct values in this sequence is b – a + 1. For F = I + J, the largest value of f is I + J = b. If I < J, the smallest value of f is J – I = a. The number of different values of f is therefore (I + J) – (J – I) + 1 = 2I + 1. For I > J, the smallest value of f is I – J = a. In that case, the number of different values of f is (I + J) – (I – J) + 1 = 2J + 1. The two expressions are equal if I = J.

7-77. (a)  N  B



e  5.05  1027 J / T 2m p

2km  2km  2.8 N  2km  2.8 N    r3 r3 a03







2 107 H / m  2.8  5.05  1027 J / T

 0.529  10



10

m



3

  0.0191T



(b) E  2B B  2 5.79  104 eV / T  0.0191T   2.21 106 eV

(c)  

hc 1.24  106 eV m   0.561m  56.1cm E 2.21  106 eV

186

Chapter 8 – Statistical Physics 1/ 2

8-1.

(a) vrms

3RT  3 8.31J / mole K  300 K      M  2 .0079  103 kg / mole 



 1930m / s







2 2 1.0079  103 kg / mole 11.2  103 m / s Mvrms (b) T   3R 3 8.31J / mole K 

8-2.

8-3.

2

 1.01  104 K

2 13.6eV  2 Ek   1.05  105 K 5 3k 3 8.617  10 eV / K

(a) Ek 

3 kT 2

(b) Ek 

3 3 kT  8.67  105 eV / K 107 K  1.29keV 2 2

T 













3RT M

vrms 

(a) For O2: vrms  (b) For H2: vrms 

1/ 2

3 8.31J / K mol  273K  32  103 kg / mol 3 8.31J / K mol  273K  2  103 kg / mol

  J / mole K  K     kg / mole  

1/ 2

 461m / s

 1840m / s

1/ 2

 kg m2 / s 2    kg  

8-4.

 3RT   M   

8-5.

3 3 (a) EK  n  RT  1 mole  8.31J / mole K  273  3400 J 2 2

 m/ s

(b) One mole of any gas has the same translational energy at the same temperature.

187

Chapter 8 – Statistical Physics

8-6.

v2

v

2

1 2 v n v dv N0 m 2 kT

4

I4

v

2

vrms

8-7.

3 8

1/ 2

m 2 kT

4

v2

m 2 kT

4

3/ 2 v2

v 4e

dv where

m / 2kT

0

3/ 2

I 4 where I 4 is given in Table B1-1.

3 8

5/ 2

3/ 2

3 8

1/ 2

1/ 2

m / 2kT 5/ 2

2kT m

5/ 2

3kT m

3RT mN A

3RT M

3RT M

v

8kT m

8 1.381 10

vm

2kT m

2 1.381 10

n v

4 N m / 2 kT

23

27

1.009u 1.66 10

At the maximum:

23

dn dv

v 2e

27

mv 2 / kT

1/ 2

2220m / s

kg / u

(Equation 8-28)

0

4 N m / 2 kT

0

ve

mv 2 / 2 kT

2510m / s

kg / u

J / K 300 K

1.009u 1.66 10 3/ 2

1/ 2

J / K 300 K

3/ 2

2v v 2

mv / kT e

mv 2 / 2 kT

2 mv 2 / kT

The maximum corresponds to the vanishing of the last factor. (The other two factors give minima at v = 0 and v = ∞.) So 2 mv 2 / kT

188

0 and vm

2kT / m

1/ 2

.

Chapter 8 – Statistical Physics 8-8.

8-9.

n v dv

4 N

dn dv

A v2

A

2mv3 2kT

m 2 kT

2vm 2kT

2v e

3/ 2

v 2e

2v e mv2 / 2 kT

mv 2 / 2 kT

mv2 / 2 kT

dv

(Equation 8-8)

The v for which dn / dv

0 is vm .

0

Because A = constant and the exponential term is only zero for v → ∞, only the quantity in [] can be zero, so or v 2

2kT m

vm

2mv3 2kT 2kT m

2v

0

(Equation 8-9)

8-10. The number of molecules N in 1 liter at 1 atm, 20°C is: N

1 1g mol / 22.4

N A molecules / g mol

Each molecule has, on the average, 3kT/2 kinetic energy, so the total translational kinetic

189

Chapter 8 – Statistical Physics (Problem 8-10 continued) 23 6.02 1023 3 1.381 10 J / K 293K 22.4 2

energy in one liter is: KE

8-11.

n2 n1

g2e g1e

e

T

E2 / kT

g2 e g1

E1 / kT

g2 g1

E2 E1 / kT

E2

163J

E2 E1 / kT

n1 n2

E1 / kT

E2

ln

E1

g2 g1

n1 n2

10.2eV 8.617 10 5 eV / K ln 4 106

k ln g 2 / g1 n1 / n2

7790 K

4 10 3 eV

8-12.

n2 n1

g2 e g1

E2 E1 / kT

3 e 1

8.617 10 5 eV / K 300 K

2.57

8-13. There are two degrees of freedom, therefore,

8-14.

Cv

2 R/2

cv

3R / M

(a) Al: cv (b) Cu: cv (c) Pb: cv

R, C p

R

R

3 1.99cal / mole K 27.0 g / mole 3 1.99cal / mole K 62.5 g / mole 3 1.99cal / mole K 207 g / mole

2 R, and

2R / R

0.221cal / g K

2.

0.215cal / g K

0.0955cal / g K

0.0920cal / g K

0.0288cal / g K

0.0305cal / g K

The values for each element shown in brackets are taken from the Handbook of Chemistry and Physics and apply at 25° C.

190

Chapter 8 – Statistical Physics

8-15.

2 N

n( E )

kT

3/ 2

E1 / 2e

E / kT

dn dE

0

At the maximum:

(Equation 8-13)

2 N kT 1/ 2

E

1 2

E / kT

e

1 E 2

3/ 2

1/ 2

E1 / 2

1 kT

e

E / kT

E / kT

The maximum corresponds to the vanishing of the last factor. (The vanishing of the other two factors corresponds to minima at E = 0 and E = ∞.)

1/ 2 E / kT

0

E 1/ 2kT.

3/ 2

m 4 N 2 kT

8-16. (a) n v

4 N 3/ 2

v 2e

v 2e

v 2 / vm2

n v

v

v 2 m / 2 kT

2kT / m

where vm 2

4N 1 v vm vm N

(Equation 8-8)

3/ 2

m 2kT

4 N v2 e vm3

mv 2 / 2 kT

v / vm

e

2

4N A 1 v vm vm

1.36 1022 v vm 2

(b)

N

1.36 1022 0 e

(c)

N

1.36 1022 1 e

(d)

N

1.36 1022 2 e

(e)

N

1.36 1022 8 e

2

0

2

e

2

v / vm

e

v / vm

2

0.01vm

2

0

1

5.00 1021

2

2

2

8

2

9.96 1020

2

1.369 10

191

4

(or no molecules most of the time)

Chapter 8 – Statistical Physics

mk 2e4 1 2 2 n2

8-17. For hydrogen: En

13.605687 eV using values of the constants accurate n2

to six decimal places. E1

13.605687eV

E2

3.401422eV

E2

E1

10.204265eV

E3

1.511743eV

E3

E1

12.093944eV

(a)

(b)

n2 n1

g2 e g1

n3 n1

g3 e g1

n2 n1

0.01 4e

E2 E1 / kT

E3 E1 / kT

(c)

n3 n1

10.20427 / 0.02586

18 e 2

4e

12.09394 / 0.02586

10.20427 / kT

10.20427 / kT

T

8 e 2

e

ln 0.0025

12.09394 / 8.61734 10

5

19,760

9e

4 10

468

10.20427 / kT

172

9 10

203

0

0

0.0025

5.99146

10.20427eV 5.99146 8.61734 10 5 eV K 9e

395

19, 760 K

0.00742

0.7%

B field

8-18.

m

1

no field

E

m

E

m

hf

0 1

E

ground state

Neglecting the spin, the 3p state is doubly degenerate: levels equally populated. E

hf

hc /

1.8509eV

670.79nm

192

0,1 hence, there are two m = 0

Chapter 8 – Statistical Physics (Problem 8-18 continued) e B 2me

E

2.315 10 4 eV

(a) The fraction of atoms in each m-state relative to the ground state is: (Example 8-2) n1 n

e

1.8511 / 0.02586

n0 n

2 e

n0 n1

e

71.58

e

1.8509 / 0.02586

1.8507 / 0.02586

10 71.57

2e 71.56

e

31.09

8.18 10

2 10

10

31.08

31.08

32

1.64 10

8.30 10

31

32

(b) The brightest line with the B-field “on” will be the transition from the m = 0 level, the center line of the Zeeman spectrum.

With that as the “standard”, the relative

intensities will be: 8.30 / 16.4 / 8.18

N h3 V 2 2 me kT

8-19. (a) e

N V

e

2 2 me kT

(Equation 8-44)

3/ 2

3/ 2

e

h3

2 2 me c 2 kT hc

5.11 105 eV

2

1240eV nm

8-20. (a) e

O2

3/ 2

3/ 2

3

2.585 10 2 eV

1 2

N h3 V 2 MkT

0.51/ 1.00 / 0.50

3

1/ 2

107 nm 1cm

3

2.51 1019 / cm3

(Equation 8-44)

3

hc NA VM 2 Mc 2 kT

3/ 2

6.022 1023 / mole 22.4 103 cm3 / mole

1.24 10 4 eV cm 2

3

32uc 2 931.5 106 eV / u 8.617 10 5 eV / K 273K

1.75 10 7

193

Chapter 8 – Statistical Physics (Problem 8-20 continued) (b) At temperature T, e 7

1.75 10

1

273K

1.75 10

T

7

1.75 10

O2

2/3

3/ 2

/ T 3/ 2

7

273

T 3/ 2

3/ 2

/ T 3/ 2

1.75 10

7

273K

3/ 2

8.5mK

273K

8-21. Assuming the gasses are ideal gases, the pressure is given by: P

2N E for classical, 3 V

FD, and BE particles. PFD will be highest due to the exclusion principle, which, in effect, limits the volume available to each particle so that each strikes the walls more frequently than the classical particles. On the other hand, PBE will be lowest, because the particles tend to be in the same state, which in effect, is like classical particles with a mutual attraction, so they strike the walls less frequently.

1

8-22. (a) f BE

e e

E / kT

1 e

E / 5800 k

1

1 1

For e E / 5800 k

0.35V ,

1 0.35 / kT

1

0.5

0.5

= 0 and f BE

e0.35 / kT

3

0.35eV 8.62 10 5 eV / K T

T

8-23.

h p

ln 2

0.347eV

(b) For E

e

5800 K

2

E 5800 K 8.62 10 5 eV / K

E

1, at T

= 0 and f BE

0.35eV ln 3 8.62 10 5 eV / K

h

h

2m E

2m 3kT / 2

ln 3

3700 K

h 3mkT

1/ 2

The distance between molecules in an ideal gas V / N

194

1/ 3

is found from

Chapter 8 – Statistical Physics (Problem 8-23 continued) PV

nRT

nRT N A N A

and equating this to

V /N

NkT

above, kT / P

1/ 3

kT / P

h

1/ 3

3mkT

h3

kT P

3mkT

3/ 2

2/5

Ph

T

3/ 2

k 3mk

8-24.

N0 N

1

(a) For T

(b) For T

(c) For T

(d) For T

T TC

N0 N

TC / 2

N0 N

TC / 4

N0 N

TC / 8

N0 N

which for small

TC

J s

kg 1.38 10

23

2/5

3

4.4 K

5/ 2

J/K

(Equation 8-52)

3TC / 4

h2 2mk 2

27

34

3/ 2

3/ 2

8-25. For small values of

8-26.

101kPa 6.63 10 3 2 1.67 10

1/ 2

P h3 k 3mk

and solving for T , yields: T 5 / 2

3

1/ 3

,e

1

3T 4TC

1

T 2TC

1

T 4TC

1

T 8TC

1

3/ 2

0.351 3/ 2

0.646 3/ 2

0.875 3/ 2

0.956

2

/ 2!

values becomes: N0 1

2/3

N 2.315 V

(Equation 8-48)

The density of liquid Ne is 1.207 g/cm3, so

195

and N0

1

N0

1 e

1

N0 e

1 or N0

1

1

1

Chapter 8 – Statistical Physics (Problem 8-26 continued)

1.207 g / cm3 6.022 1023 molecules / mol 106 cm2 / m3

N V T

3.601 1028 / m3

20.18 g / mol 34

6.626 10 27

2 20u 1.66 10 20

Thus, TC at which

2

J s

23

kg / u .381 10

2/3

3.601 1028 m3 2 2.315

J /K

0.895K

Ne would become a superfluid is much lower than its freezing

temperature of 24.5K.

8-27. Power per unit area R arriving at Earth is given by the Stefan-Boltzmann law: R is Stefan’s constant. For a 5% decrease in the Sun’s temperature,

where

T4

R 0.95T

R T

0.95T

R T

8-28.

E

T

hf e

hf / kT

(a) For T

10hf / k;

(b) For T

hf / k;

(c) For T

0.1hf / k;

hf hf

kT / 10 kT

hf

E 10kT

According to equipartition E

8-29.

CV e hf / kT

CV

0.95

1

4

4

0.186 , or a decrease of 18.6%.

(Equation 8-60)

1

hf 3N A k kT

4

2

1

1 / 10

e

hf 1

e

1

kT 1.718

1

hf

E

10

e

1

kT / 10 0.1051

0.951kT

0.582kT 10kT 2.20 104

As T

2

, hf / kT gets small and

1 hf / kT

hf 3N A k kT

2

1 hf / kT hf / kT

4.54 10 4 kT

kT in each case.

ehf / kT ehf / kT

hf

E

2

3N A k

The rule of Dulong and Petit.

196

3N A R / N A

3R

T4

Chapter 8 – Statistical Physics

8-30.

CV

3R

2

hf kT

ehf / kT ehf / kT

Writing hf / kT CV

3R Af

1

(Equation 8-62)

2

h / kT

Af where A

e Af

2 2

e Af

CV

3R / e Af

e Af

8-31.

CV

3R

3 8.31

ln

f

hf kT

ehf / kT e

hf / kT

1

3R 1

2

e1 e1 1

2

22.95K / K mol

, so

8.97 1011 Hz

13

2.46 1012 Hz

(Equation 8-62)

2

At the Einstein temperature TE CV

2

(From Figure 8-13)

1 / 2.40 10

13.8

2

13

13.8 J / K mol

For Si, CV 200 K

1 / e Af

(From Figure 8-13)

1 / 2.40 10

20.1

2

ln 3R / CV 1 / A

f

20.1J / K mol

3 8.31

ln

f

Af

0 and e Af dominates Af

3R / CV

For Al, CV 200 K

e

1

Because Af is “large”, 1 / e Af

200 K ,

when T

1

2

eR Af

2e Af

13

2.40 10

hf / k,

3R 0.9207

3 8.31J / K mol 0.9207

5.48cal / K mol

3/ 2

8-32. Rewriting Equation 8-69 as

V

E1 / 2

8mc 2

n E 2

hc

2

e

E EF / kT

1

Set up the equation on a spreadsheet whose initial and final columns are E(eV) and n(E)/V (eV•nm3)-1, respectively.

197

Chapter 8 – Statistical Physics (Problem 8-32 continued) E eV

n E /V

eV nm3

4.5

14.4

4.6

14.6

4.7

14.5

4.8 (=EF)

7.46

4.9

0.306

5

0.0065

5.1

0.00014

1

The graph of these values is below.

From the graph, about 0.37 electrons/nm3 or 3.7 1026 electrons / m3 within 0.1eV below EF have been excited to levels above EF.

198

Chapter 8 – Statistical Physics 8-33. The photon gas has the most states available, since any number of photons may be in the ground state. In contrast, at T = 1K the electron gas’s available states are limited to those within about 2kT

2 8.62 10 5 eV K 1K

1.72 10 4 eV of the Fermi level. All

other states are either filled, hence unavailable, or higher than kT above the Fermi level, hence not accessible.

8-34. From the graph. TE Au TE Be

136 K 575K

TE Al

243K

TE Diamond

off the graph (well over 1000K)

8-35. Approximating the nuclear potential with an infinite square well and ignoring the Coulomb repulsion of the protons, the energy levels for both protons and neutrons are given by En

n2h2

8mL2 and six levels will be occupied in

protons and six levels with 12 neutrons. EF protons

5

2

1240MeV fm

2

8 1.0078u 931.5MeV / u 3.15 fm

199

2

516MeV

22

Ne , five levels with 10

Chapter 8 – Statistical Physics (Problem 8-35 continued) 6

EF neutrons

2

1240 MeV fm

2

8 1.0087u 931.5MeV / u 3.15 fm 3 / 5 EF

E protons

742 MeV

310MeV

3 / 5 EF

E neutrons

2

445MeV

As we will discover in Chapter 11, these estimates are nearly an order of magnitude too large. The number of particles is not a large sample.

8-36.

E1

h 2 / 8mL2 . All 10 bosons can be in this level, so E1 total

8-37. (a) f FD E

1 e

E EF / kT

e

E EF / 0.1EF

e

E EF / 0.5 EF

(Equation 8-68)

1

1

(b) f FD E

1 1

10 E EF / EF

e

1

1

1 1

e

2 E EF / EF

200

1

10h 2 / 8mL2 .

Chapter 8 – Statistical Physics

8-38.

NO N

3/ 2

T Tc

1

N (a) O N

N (b) O N

(Equation 8-52) 3/ 2

1

Tc / 2 Tc

1

Tc / 4 Tc

3/ 2

1

1 2

1

1 4

3/ 2

0.646 3/ 2

0.875

n2 h2

8-39. For a one-dimensional well approximation, En

8mL2 . At the Fermi level EF,

n=N/2, where N = number of electrons. 2

N / 2 h2

EF

2

h2 N 32m L

8mL2

where N / L = number of electrons/unit length,

i.e., the density of electrons. Assuming 1 free electron/Au atom,

N L EF

6.02 1023 electrons / mol 19.32 g / cm3 102 cm / m

1/ 3

NA M

3

197 g / mol 2

6.63 10

34

J s

3.81 109 m

32 9.11 10

31

kg 1.602 10

19

1

1/ 3

3.81 109 m

1

2

1.37eV

J / eV

This is the energy of an electron in the Fermi level above the bottom of the well. Adding the work function to such an electron just removes it from the metal, so the well is

1.37eV

4.8eV

8-40. (a) At T

6.2eV deep.

850 K

v

vrms

vm

8kT m 3kT m

1/ 2

1/ 2

2kT m 4

3 2

/2

2 1.3807 10

23

J / K 850 K

6.94 10

1/ 2

vm

207.5m / s

1/ 2

vm

225.2m / s

201

25

kg

1/ 2

183.9m / s

Chapter 8 – Statistical Physics (Problem 8-40 continued) The times for molecules with each of these speeds to travel across the 10cm diameter of the rotating drum is:

0.10m 183.9m / s

t vm

5.44 10 4 s

t v

0.10m 207.5m / s

4.82 10 4 s

t vrms

0.10m 225.2m / s

4.44 10 4 s

The drum is rotating at 6250rev/min = 104.2rev/s or 9.600 10 3 s / rev. The fraction of a revolution made by the drum while molecules with each of these three speeds are crossing the diameter is:

5.44 10 4 s 9.600 10 3 s / rev

for vm :

0.05667rev

for v :

4.82 10 4 s 9.600 10 3 s / rev

0.05021rev

for vrms :

4.44 10 4 s 9.600 10 3 s / rev

0.04625rev

Assuming that point A is directly opposite the slit s2 when the first (and fastest) molecules enter the drum, molecules with each of the three speeds will strike the plate at the following distances from A: (The circumference of the drum C vm :

0.05667 rev 0.314159m / rev

0.01780m 1.780cm

v :

0.05021rev 0.314159m / rev

0.01577m 1.577cm

vrms :

0.04625rev 0.314159m / rev

0.01453m 1.453cm

0.10 m .)

(b) Correction is necessary because faster molecules in the oven will approach the oven’s exit slit more often than slower molecules, so the speed distribution in the exit beam is slightly skewed toward higher speeds. (c) No. The mean speed of N2 molecules at 850K is 710.5m/s, since they have a smaller mass than Bi2 molecules.

Repeating for them the calculations in part (a),

N2

molecules moving at vm would strike the plate only 0.4cm from A. Molecules moving at v and vrms would be even closer to A.

202

Chapter 8 – Statistical Physics

8-41.

1 m vescape 2

EK ( escape)

0

1 2 3 mv v e 2

0

1 2 mv F v dv 2 mv 2 / 2 kT

v 3e

mv 2 / 2 kT

3

m

dv

1 I5 m 2 I3

dv

0

1 m 2

8-42. (a) f u du 1

(b)

8-43.

Ce

2

/2

E / kT

du

f u du

Ce

Ce

2C

1/ 2

C

kT / A

C

Au 2

vx

2kT

du

du

2C e

/ 2 where

du

A / kT

Au 2 A / kT e

A A / kT 2

1 A A / kT kT / A 2

3/ 2

e

Au 2 / kT

A / kT

Au 2 f u du

1/ 2

m 2kT

where

(from Equation 8-5)

A A / kT 2I 2

m / 2 kT

f vx

2kT m

Au 2 / kT

Au 2 / kT

2CI 0

E

m

mvx2 / 2 kT

/4

Au 2 / kT

du

3/ 2

where

A / kT

1 kT 2

(from Equation 8-6)

vx f vx dvx 0

vx m / 2 kT

1/ 2

e

mvx2 / 2 kT

vx m / 2 kT

dvx 0

2 m / 2 kT

1/ 2

vx e

mvx2 / 2 kT

dvx

0

2 m / 2 kT

1/ 2

I1

with

m / 2kT

203

1/ 2

e

mvx2 / 2 kT

dvx

Chapter 8 – Statistical Physics (Problem 8-43 continued) 1/ 2

2 m / 2 kT

8-44.

1

f FD

e

E EF / kT

E EF / kT

1

f FD

N

1

EF , e

For E

8-45.

1

E / kT

e e

e

e

2kT m

1/ 2

1

1 and

4

EF / kT

e

3/ 2

2me h

V

e

=

E / kT

E1 / 2 e

3

EF kT

where

1

E EF / kT

2kT m

E / kT

1 e e E / kT

dE

fB

(Equation 8-43)

0

Considering the integral, we change the variable: E / kT E

kTu 2 , E1 / 2

E1 / 2 e

E / kT

kT

dE

1/ 2

2 kT

0

u, and dE

3/ 2

u 2e

u2

kT 2u du. So,

du

0

/ 4, so

The value of the integral (from tables) is

N

u 2 , then

e

4

3/ 2

2me

V 2 kT

h3

8-46. (a) N

ni

3/ 2

or e

4

f 0 E0

f1 E1

2 2me kT

3/ 2

Nh3

(with g0

g1

1)

i

Ce0

So, C (b)

E

Ce

N 1 e 0 n0

/ kT

/ kT

/ kT

n1 N

C 1 e

Ce N

/ kT

N e 1 e

/ kT / kT

204

N

e 1 e

/ kT / kT

V

, which is Equation 8-44.

Chapter 8 – Statistical Physics (Problem 8-46 continued) / kT

As T

0,

e

As T

,

e

(c) CV

1/ e

/ kT

1/ e

d N E

dE dT

1 e 2

Nk

kT

/ kT

2

/ kT

e

0, so E

e

/ kT

2

e

/ kT

1 e

/ kT

0

0, so E

/2

/ kT

d N e dT 1 e

dT

N 2 kT 2

/ kT

/ kT

/ kT

1 e

/ kT

2

(d) T

/k

CV

Nk

8-47. (a) p

k

L

2 x

n12

k

2 y

n22

1

k

2 z

0.1

0.25

0.5

1.0

2.0

3.0

0.005

0.28

0.42

0.20

0.06

0.03

1/ 2

n32

1/ 2

n1 L

2

n2 L

N L 205

2

n3 L

2

1/ 2

Chapter 8 – Statistical Physics (Problem 8-47 continued)

E

pc

c N L

(b) Considering the space whose axes are n1, n2 , and n3 . The points in space correspond to all possible integer values of n1, n2 , and n3 , all of which are located in the all positive octant. Each state has unit volume associated with it. Those states between N and N + dN lie in a spherical shell of the octant whose radius is N and whose thickness is dN. Its volume is 1 / 8 4 N 2 dN . Because photons can have two polarizations (spin directions), the number of possible state is 2

1 / 8 4 N 2 dN

N 2 dN .

(c) This number of photon states has energy between E and E+dE, where N

EL / hc.

The density of states g(E) is thus: g E dE

number of photon states at E

dE

number of photon states at N

dN

N 2dN EL

c

8 L3 2

c

2

L

c dE

8 L3

2

3

E dE

hc

3

E 2 dE

The probability that a photon exists in a state is given by:

f BE E

1 e e

E / kT

1 1

e

E / kT

1

(Equation 8-24)

The number of photons with energy between E and E+dE is then:

n E dE

f BE E g E dE

3

L / hc E 2 dE

8

e E / kT

1

(d) The number of photons per unit volume within this energy range is n E dE / L3. Because each photon has energy E, the energy density for photons is:

u E dE

8 E 3dE

3

E n E dE / L

hc

3

e E / kT

1

which is also the density of photons with wavelength between λ and λ+dλ, where 206

Chapter 8 – Statistical Physics (Problem 8-47 continued)

hc / E d

u

E

d dE dE

d

hc / . So, 2

hc dE E2

u E dE

hc

8

dE 3

hc / hc

dE

3

hc /

ehc /

kT

207

hc 2

2

1

d

d

8 hc 3d ehc / kT 1

Chapter 9 – Molecular Structure and Spectra

9-1.

(a) 1

19 23 eV eV    1.609  10 J  6.022  10 molecules   1    molecule  molecule   eV mole  

J  1cal    96472 mole    4.184 J

cal kcal    23057 mole  23.06 mole 

eV   23.06kcal / mole  (b) Ed   4.27   98.5kcal / mole molecule    1eV / molecule  eV   1eV / molecule  (c) Ed  106   1.08eV / molecule molecule    96.47kJ / mole 

9-2.

Dissociation energy of NaCl is 4.27eV, which is the energy released when the NaCl molecule is formed from neutral Na and Cl atoms. Because this is more than enough energy to dissociate a Cl2 molecule, the reaction is exothermic. The net energy release is 4.27eV – 2.48eV = 1.79eV.

9-3.

From Cs to F: 3.89eV – 3.40eV = 0.49eV From Li to I: 5.39eV – 3.06eV = 2.33eV From Rb to Br: 4.18eV – 3.36eV = 0.82eV

9-4.

Ed  U C   CsI :

NaF :

LiI :



ke2  Eion r0

ke2 1.440eV nm  Eion     3.89eV  3.06eV   Ed  3.44eV r0 0.337nm



ke2 1.440eV nm  Eion     5.14eV  3.40eV   Ed  5.72eV r0 0.193nm

ke2 1.440eV nm   Eion     5.39eV  3.06eV   Ed  3.72eV r0 0.238nm

While Ed for CsI is very close to the experimental value, the other two are both high. Exclusion principle repulsion was ignored.

209

Chapter 9 – Molecular Structure and Spectra

9-5.

(a) Total potential energy: U  r   

ke2  Eex  Eion r

(Equation 9-1)

ke2 1.44eV nm attractive part of U  r0      5.39eV r0 0.267nm

(b) The net ionization energy is:

Eion   ionization energy of Rb    electron affinity of Cl   4.18eV  3.62eV  0.56eV

Neglecting the exclusion principle repulsion energy Eex ,

dissociation energy  U  r0   5.39eV  0.56eV  4.83eV (c) Including exclusion principle repulsion,

dissociation energy  4.37eV  U  r0   5.39eV  0.56eV  Eex Eex  5.39eV  4.37eV  0.56eV  0.46eV

9-6.

Uc  

ke2 1.440eV nm  Eion     4.34eV  3.36eV   4.13eV r0 0.282nm

The dissociation energy is 3.94eV. Ed  U c  Eex  3.94eV  4.13eV  Eex

Eex  0.19eV at r0  0.282nm

9-7.

Eex 

A rn

(Equation 9-2)

0.19eV 

A

 0.282nm 

n

At r0 the net force on each ion is zero, so we have (from Example 9-2) U c  r0  r0 n



ke2 nA n A n  18.11eV / nm  n1   n   0.19eV  2 r0 r0 r0 r0 r0

18.11eV / nm  0.282nm   26.9  27 0.19eV

A  Eex r0n   0.19eV  0.282nm   2.73  1016 eV nm27 27

210

Chapter 9 – Molecular Structure and Spectra 9-8.

Ed  3.81eV per molecule of NaBr (from Table 9-2)





1eV / molecule  1eV / molecule  1.609  1019 J / eV 

 6.02 10

23



molecules / mol / 1cal / 4.186 J   23.0kcal / mol

Ed  NaBr    3.81eV / molecule  23.0kcal / mol  / 1eV / molecule   87.6kcal / mol

9-9.

For KBr : U C 

1.440eV nm   4.34eV  3.36eV   4.13eV 0.282nm

Ed  3.94eV  UC  Eex  4.13eV  Eex Eex  0.19eV For RbCl : U C 

1.440eV nm   4.18eV  3.62eV   4.60eV 0.279nm

Ed  4.37eV  UC  Eex  4.60eV  Eex

Eex  0.23eV

9-10.

H 2 S , H 2Te, H3 P, H3Sb

9-11. (a) KCl should exhibit ionic bonding. (b) O2 should exhibit covalent bonding. (c) CH4 should exhibit covalent bonding. 9-12. Dipole moment pionic  er0

(Equation 9-3)





 1.609  1019 C  0.0917nm   1.47 1020 C nm 109 m / nm  1.47  1029 C m

if the HF molecule were a pure ionic bond. The measured value is 6.64 1029 C m , so



the HF bond is 6.40  1030 C m

 1.47 10

211

29



C m  0.44 or 44% ionic.

Chapter 9 – Molecular Structure and Spectra 9-13.

pionic  er0  1.609 1019 C  0.2345 109 m 

(Equation 9-3)

 3.757 1029 C m, if purely ionic. The measured value should be:





pionic  measured   0.70 pionic  0.70 3.757  1029 C m  2.630  1029 C m

9-14.

pionic  er0  1.609 1019 C  0.193 109 m 

(Equation 9-3)

 3.09  1029 C m

The measured values is 2.67  1029 C m, so the BaO bond is

 2.67 10

29

Cm

 3.09 10

29



C m  0.86 or 86% ionic.

9-15. Silicon, germanium, tin, and lead have the same outer shell configuration as carbon. Silicon and germanium have the same hybrid bonding as carbon (their crystal structure is diamond, like carbon); however, tin and lead are metallic bonded. (See Chapter 10.)

9-16.

p = p1  p2 and p  6.46  1030 C m and p  p1 cos 52.25  p2 cos 52.25

If bonding were ionic, pionic  er0  1.609 1019 C  0.0956 109 m   1.532 1029 C m

p1  actual   p / 2  cos 52.25  6.46  1030 C m / 2 cos 52.25   5.276 1030 C m Ionic fraction = fraction of charge transferred =

5.276  1030 C m  0.34 or 34% 1.532  1029 C m

9-17. U   k 2 p12 / r 2 (Equation 9-10) (a) Kinetic energy of N2  0.026eV , so when U  0.026eV the bond will be broken.

1.110 0.026eV  r

6

1.110 

37

37



m C 2 / N 9  109 N m2 / C 2

  6.46 10 2

30

Cm



2

r6



  6.46  10 J / eV 

m C 2 / N 9  109 N m2 / C 2



0.026eV 1.60  10

r  6.7  1010 m  0.67nm

212

19

2

30

Cm



2

 8.94  1056 m6

Chapter 9 – Molecular Structure and Spectra (Problem 9-17 continued) (b) U 

ke2 1.440eV nm  U  0.026eV   r  55nm r r

(c) H2O-Ne bonds in the atmosphere would be very unlikely. The individual molecules will, on average, be about 4nm apart, but if a H2O-Ne molecule should form, its U  0.003eV at r  0.95nm , a typical (large) separation. Thus, a N2 molecule with

the average kinetic energy could easily dissociate the H2O-Ne bond. 9-18. (a) E  0.3eV  hc /   1240eV nm /     1240eV nm / 0.3eV  4.13  103 nm (b) Infrared (c) The infrared is absorbed causing increased molecular vibrations (heat) long before it gets to the DNA.

9-19. (a) NaCl is polar. The Na+ ion is the positive charge center, the Cl− ion is the negative charge center. (b) O2 is nonpolar. The covalent bond involves no separation of charges, hence no polarization of the molecule.

9-20. For N 2

E0 r  2.48  104 eV 

2.48  104 eV  2 I   r02 

2

/ 2I where I 

1 2 mr0 and m  14.0067u 2

2

2

 2.48 10

4



eV 14.0067u 





2  1.055  1034 J s r0    2.48  104 eV 1.60  1019 J / eV 14.0067u  1.66  1027 kg / u 







 1.611010 m  0.161nm

213





   

1/ 2

Chapter 9 – Molecular Structure and Spectra

9-21.

E0 r 

E0 r 

2

(Equation 9-14) where I 

2I 2 2 0

mr



 c

197.3eV

2

2 2 0

mc r



16uc   931.5  10 2

9-22. For Co: f  6.42 1013 Hz

EV   v  1 / 2  hf

1 2 mr0 for a symmetric molecule. 2

6

nm 

eV / uc

2

2

  0.121nm 

2

 1.78  104 eV

(See Example 9-6)

(Equation 9-20)

(a) E1  E0  3hf / 2  hf / 2  hf





 4.14  1015 eV s 6.42  1013 Hz



 0.27eV

(b)

n1  E  E / kT  e  1 0 n0 0.01  e

(from Equation 8-2)





 0.27  / 8.62105 T





ln  0.01    0.27eV  8.62  105 eV / K T

T

  0.27eV 



ln  0.01 8.62  105 eV / K



T  680K

9-23. For LiH: f  4.22  1013 Hz

(from Table 9-7)







(a) EV   v  1 / 2  hf  E0  hf / 2  4.14  1015 eV s 4.22  1013 Hz / 2

E0  0.087eV (b)  

 (c) f 

m1m2 m1  m2

(Equation 9-17)

 7.0160u 1.0078u   0.8812u  7.0160u   1.0078u  1 2

K



(Equation 9-21)

214

Chapter 9 – Molecular Structure and Spectra (Problem 9-23 continued)













K   2 f     2  4.22  1013 Hz  0.8812  1.66  1027 kg / u 2

2

K  117 N / m

(d) En  n2 h2 8mr02  r02  n2 h2 8mEn r0  h 8mE0 

1/ 2

r0 

6.63  1034 J s





8  0.8812u  1.66  1027 kg / u  0.087eV  1.60  1019 J / eV   

1/ 2

r0  5.19  1011 m  0.052nm

1.0078u   0.504u mm 9-24. (a) For H2:   1 2  m1  m2 2 1.0078u  2

14.0067u   7.0034u  2 14.0067u  2

(b) For N2:

(c) For CO:  

12.0111u 15.9994u   6.8607u

(d) For HCl:  

9-25. (a)  

12.0111u  15.9994u

1.0078u 35.453u   0.980u 1.0078u  35.453u

 39.1u  35.45u   18.6u m1m2  m1  m2 39.1u  35.45u

(b) E0 r 

2

2I

E0 r   r0 

(Equation 9-14) I   r02 2

2 r02



 c

2

2 c 2 r02

c

 2 c E 

1/ 2

2

V



r  2 0

 c

2

2 c 2 EV 197.3eV nm









1/ 2

 2 10.6uc 2 931.5  106 eV / uc 2 1.43  105 eV   

r0  0.280nm

215

Chapter 9 – Molecular Structure and Spectra

9-26.

1 2

f 

K

(Equation 9-21)



(a) For H35Cl: μ = 0.980u and f  8.97  1013 Hz .



K   2 f     2  8.97  1013 Hz 2

2

(b) For K79Br:   2

1 2

K



2

27



kg / u  517 N / m .

 39.102u  78.918u   26.147u and f  6.93 1012 Hz m1m2  m1  m2 39.102u  78.918u



K   2 f     2  6.93  1012 Hz

9-27. 1. f 

  0.980u  1.66 10

2

  26.147u  1.66 10 2

27



kg / u  82.3N / m

(Equation 9-21) Solving for the force constant, K  (2 f )2 

2. The reduced mass  of the NO molecule is



mN mO (14.01u)(16.00 u)   7.47 u mN  mO 14.01u  16.00 u

3. K  (2  5.63 1013 Hz)2  7.47 u 1.66 1027 kg/u  1.55 103 N/m ( Note: This is equivalent to about 8.8 lbs/ft, the force constant of a moderately strong spring.)

9-28.

E0r 

2

2I Treating the Br atom as fixed,





I  mH r02  1.0078u  1.66  1027 kg / u  0.141nm  E0 r 



1.055  10

34



J s



2

2



2 1.0078u  1.66  1027 kg / u  0.141nm  109 m / nm 2

 1.67  1022 J  1.04  103 eV

E 



 1 E0r for  0, 1, 2,

(Equation 9-13)

216



2

Chapter 9 – Molecular Structure and Spectra (Problem 9-28 continued) 14

The four lowest states have energies:

E0  0

E

3

12

10

3

eV



10

3

E1  2E0r  2.08  10 eV

8

3

E2  6E0r  6.27  10 eV

2

6

E3  12E0r  12.5  103 eV 4

9-29.

2

1

0

0

E  hf where f  1.05  1013 Hz for Li. Approximating the potential (near the bottom)

2  2 with a square well, E  2  1  22  1   2  hf  2  mr0



For Li2: r02 



3 2 1 3  2 2 f  4 f

 3  1.055  1034 J s r     4  1.05  1013 Hz  6.939u  1.66  1027 kg / u







 4.53  1011 m  0.045nm

9-30.

E0 r 

2

2I

where I   r02

(Equation 9-14)

For K35Cl:  

 39.102u  34.969u   18.46u

For K37Cl:  

 39.102u  34.966u   19.00u

39.102u  34.969u 39.102u  34.966u

r0  0.267nm for KCl.

217



  

1/ 2

Chapter 9 – Molecular Structure and Spectra (Problem 9-30 continued)





E0 r K Cl  35



1.055  10

34

J s





2

2 18.46u  1.66  1027 kg / u 0.267  109 m



2

 2.55 1024 J  1.59 105 eV





E0 r K 37Cl 



1.055  10

34

J s





2

2 19.00u  1.66  1027 kg / u 0.267  109 m



2

 2.48 1024 J  1.55 105 eV E0r  0.04  105 eV

9-31. (a) NaF – ionic

(b) KBr – ionic

(c) N2 – covalent

9-32.

E0,1

 

 1 I

(d) Ne – dipole-dipole

2



2

 r02

(a) For NaCl: r0  0.251nm (from Table 9-7)



m  Na  m





 22.9898 34.9689  13.8707u m  Na   m  35Cl   22.9898   34.9689 

E0,1 

35

Cl



1.055 10

13.8707u  1.66 10

27

 kg / u  0.251  10 m  34

J s

2

9

2

E0,1  7.67  1024 J  4.80  105 eV (b) E0,1  hf  f  E0,1 / h 

7.67  1024 J 6.63  1034 J s

f  1.16  1010 Hz

  c / f   3.00  108 m / s 1.16  1010 Hz   0.0259nm  2.59cm

218

Chapter 9 – Molecular Structure and Spectra 9-33. (a)   2400nm  E  hc / 2400nm 

E2  E1  3.80eV

1240eV nm  0.517eV 2400nm

E3  E2  0.500eV

E4  E3  2.9eV

E5  E4  0.30eV

The E3  E2 and E5  E4 transitions can occur. (b) None of these can occur, as a minimum of 3.80eV is needed to excite higher states. (c)   250nm  E  1240eV nm 250nm  4.96eV . All transitions noted in (a) can occur. If the temperature is low so only E1 is occupied, states up to E3 can be reached, so the E2  E1 and E3  E2 transitions will occur, as well as E3  E1 . (d) E4  E3  2.9eV  hc  or   1240eV nm 2.9eV  428nm

E4  E2  3.4eV  hc  or   1240eV nm 3.4eV  365nm E4  E1  7.2eV  hc  or   1240eV nm 7.2eV  172nm

9-34.

A21  ehf / kT  1 B21u  f 

(Equation 9-42)

For the Hα line λ=656.1nm At T = 300K,

hf hc 1240eV nm    73.1 kT  kT  656.1nm  8.62  105 eV / K  300 K 





ehf / kT  1  e73.1  1  5.5  1031

Spontaneous emission is more probable by a very large factor!

9-35.

n  E1 

n  E0 

e E1 / kT   E0 / kT i.e., the ratio of the Boltzmann factors. e

For O2: f  4.74  1013 Hz and







E0  hf 2  4.14  1015 eV s 4.74  1013 Hz 2  0.0981eV

E1  3hf 2  0.294eV





At 273K, kT  8.62  105 eV / K  273K   0.0235eV

219

Chapter 9 – Molecular Structure and Spectra (Problem 9-35 continued)

n  E1 

n  E0 



e0.0294 / 0.0235 e12.5  4.17  2.4  104 0.0981 / 0.0235 e e

Thus, about 2 of every 10,000 molecules are in the E1 state. Similarly, at 77K,

9-36.



E

n  E0 

 1.4  1013

 1 E0r for  0, 1, 2,

Where E0 r  E0 r 

n  E1 

2

2I

(Equation 9-13)

and I   r02 with   m / 2



1.055 10

2 18.99u  1.66  10

27

 kg / u  0.14  10 m  34

J s

2

9

2

 1.80  1023 J  1.12  104 eV

(a) E0  0 E1  2E0r  2.24  104 eV

E1  E0  2.24  104 eV

E2  6E0r  6.72  104 eV

E2  E1  4.48  104 eV

E3  12E0r  13.4  104 eV

E3  E2  6.72  104 eV

14

3

12

E

10

4

eV



10 8

2

6 4

1 0

2 0

220

Chapter 9 – Molecular Structure and Spectra (Problem 9-36 continued)

E  hc     hc E

(b)   1

For E1  E0 :  

1240eV nm  5.54  106 nm  5.54nm 4 2.24  10 eV

For E2  E1 :  

1240eV nm  2.77  106 nm  2.77nm 4.48  104 eV

For E3  E2 :  

1240eV nm  1.85  106 nm  1.85nm 4 6.72  10 eV







9-37. (a) 10MW  107 J / s  E  107 J / s 1.5  109 s  1.5  102 J (b) For ruby laser:   694.3nm, so the energy/photon is:

E  hc   1240eV nm 694.3nm  1.786eV

1.5  10 J  Number of photons = 1.786eV  1.60  10 2

19

9-38.

J / eV



 5.23  106

4mW  4  103 J / s

E  hc  

1240eV nm  1.960eV per photon 632.8nm

4  103 J / s  1.28  1016 / s Number of photons = 19 1.960eV  1.60  10 J / eV







9-39. (a) sin  1.22 / D  1.22 600  109 m

 10 10 m  7.32 10 2

6

   7.32 106 radians

  S / R where S  diameter of the beam on the moon and R  distance to moon.







S  R  3.84  108 m 7.32  106 radians  2.81 103 m  2.81km



(b) sin  1.22 600  109 m



 1m  7.32 10



7



radians

S  R  3.84  108 m 7.32  107 radians  281m

221

Chapter 9 – Molecular Structure and Spectra

9-40. (a)

n  E2  n  E1 



e E2 / kT E  E / kT  e 2 1   E1 / kT e

At T  297 K,

E2  E1  hc   1240eV nm 420nm  2.95eV





kT  8.61 105 eV / K  297 K   0.0256eV

n  E2   n  E1  e2.95 / 0.0256  2.5  1021 e115  2  1029  0





(b) Energy emitted = 1.8  1021  2.95eV / photon   5.31  1021 eV  850 J

9-41. (a) Total potential energy: U  r   

ke2  Eex  Eion r

the electrostatic part of U  r  at r0 is 

ke2 1.44eV nm   6.00eV r0 0.24nm

(b) The net ionization energy is:

Eion   ionization energy of Na    electron affinity of Cl   5.14eV  3.62eV  1.52eV

dissociation energy of NaCl =4.27eV (from Table 9-2)

4.27eV  U  r0   6.00eV  1.52eV  4.67eV  Eex Eex  6.00eV  4.27eV  1.52eV  0.21eV (c) Eex 

A rn

(Equation 9-2)

At r0  0.24nm, Eex  0.21eV At r0  0.14nm, U  r   0 and Eex  At r0 : Eex  0.21eV 

A

 0.24nm 

At r  0.14nm : Eex  8.77eV 

n

ke2  Eion  8.77eV r

 A   0.21eV  0.24nm  A

 0.14nm 

n

 A  8.77eV  0.14nm 

Setting the two equations for A equal to each other:

 0.24nm  n  0.14nm 

2

2

 0.24   8.77eV     0.14   0.21eV

n    1.71  41.76 

222

n

n

Chapter 9 – Molecular Structure and Spectra (Problem 9-41 continued) n log1.71  log 41.76

n  log 41.76  log1.71  6.96 A  0.21eV  0.24nm   0.21eV  0.24nm  n

9-42. (a)



sin  1.22 / D  1.22 694.3  109 m

6.96

 1.02  105 eV nm6.96

 0.01m  8.47 10

5

   8.47  105 radians (b) E photon  hc /   1240eV nm / 694.3nm  1.786eV / photon For 1018 photons / s :





Etotal  1.786eV / photon  1.602  1019 J / eV 1018 photons / s



 0.286 J / s  0.286W

Area of spot A is: A   d 2 / 4   8.47cm  / 4 2

2 and E  Etotal / A  0.286W /  8.47cm  / 4  5.08  103W / cm2  

9-43. (a) U att  

ke2 1.440eV nm   5.39eV r 0.267nm

(b) To form K  and Cl  requires Eion  4.34eV  3.61eV  0.73eV

 ke2  Ed  U C      Eion   5.39eV  0.73eV  4.66eV  r  (c) Eex  4.66eV  4.43eV  0.23eV at r0

9-44.

E0 r 



2

2I

where I   r02 with r0  0.267nm and

 39.102u  35.453u   18.594u m1m2  m1  m2 39.102u  35.453u

E0 r 



1.055 10

2 18.594u  1.66  10

27

 kg / u  0.267  10 m  34

J s

2

9

223

2

 2.53  1024 J  1.58  105 eV

Chapter 9 – Molecular Structure and Spectra

9-45. (a) Ed 

kp1 where p1  qa, being the separation of the charges  q and  q of the dipole x3

(b) U   p E and p  E  p =  E So the individual dipole moment of a nonpolar molecule in the field produced by p1 is p2   Ed   kp1 x3 and U   p2 Ed    kp1  x6 2

Fx  



dU d    k 2 p12 dx dx



x6   6 k 2 p12 x7

9-46. 1. The energies E of the vibrational levels are given by Equation 9-20: 1 E  (  )hf for   0, 1, 2, 3, 2 The frequencies are found from Equation 9-21 and requires first that the reduced mass

for each of the molecules be found using Equation 9-17. mH mH (1.01u)(1.01u)   0.51u 2 mH  mH 1.01u  1.01u HD  0.67 u

H  Similarly,

D  1.01u 2

2. The vibrational frequencies for the molecules are then: f H2 

Similarly,

1 2

K

H

 2

1 2

580 N/m  1.32 1014 Hz 27 (0.51u)(1.66 10 kg/u)

f HD  1.15 1014 Hz f D2  9.36 1013 Hz

3. The energies of the four lowest vibrational levels are then: For H 2 : 1 1 hf H2  (6.63 1034 J  s)(1.32 1014 Hz)  4.38 1020 J 2 2 4.38 1020 J E0   0.27 eV 1.60 1019 J/eV E0 

224

Chapter 9 – Molecular Structure and Spectra (Problem 9-46 continued) E1  0.82 eV

Similarly,

E2  1.37 eV E3  1.91eV

For HD: 1 1 hf HD  (6.63 1034 J  s)(1.15 1014 Hz)  3.811020 J 2 2 3.811020 J E0   0.24 eV 1.60 1019 J/eV And again similarly, E1  0.72 eV E0 

E2  1.19 eV E3  1.67 eV

For D 2 : 1 1 E0  hf D2  (6.63 1034 J  s)(9.36 1013 Hz)  3.10 1020 J 2 2 3.10 1020 J E0   0.19 eV 1.60 1019 J/eV And once again similarly, E1  0.58eV E2  0.97 eV E3  1.36 eV 4. There are three transitions for each molecule:

  3   2;   2   1;   1   0 For H 2 : E  hf  hc /     hc / E  1240eV  nm / E eV E  E 3  E 2  (1.91  1.37)eV  0.54eV 32  1240eV  nm / 0.54eV  2.30 103 nm Similarly,

21  2.25 103 nm 10  2.25 103 nm

For HD:

E  E 3  E 2  (1.67  1.19)eV  0.48eV 32  1240eV  nm / 0.48eV  2.58 103 nm

225

Chapter 9 – Molecular Structure and Spectra (Problem 9-46 continued)

21  2.64 103 nm

And again similarly,

10  2.58 103 nm

For D 2 :

E  E 3  E 2  (1.36  0.97)eV  0.39eV 32  1240eV  nm / 0.39eV  3.18 103 nm And once again similarly,

21  3.18 103 nm 10  3.18 103 nm





9-47. (a) E3  hc   1240eV nm  0.86mm  106 nm / mm  1.44  103 eV

   1240eV nm  2.59mm  10 nm / mm   4.79  10

E2  1240eV nm 1.29mm  106 nm / mm  9.61 104 eV E1

6

4

eV

These are vibrational states, because they are equally spaced. Note the v  0 state at the ½ level spacing.

18

v3

16 14

v2

12



Ev 104 eV



10

v 1

8 6 4

v0

2 0

226

Chapter 9 – Molecular Structure and Spectra (Problem 9-47 continued) (b) Approximating the potential with a square well (at the bottom), E1  4.79  104 eV  n 2

2

2

2 mr02



  

 

2  22  12  2 1.055  1034 J s r0    2  28.01u  1.66  1027 kg / u 4.79  104 eV 1.60  1019 J / eV 





   

1/ 2

 2.15  1010 m  0.215nm

9-48. Using the NaCl potential energy versus separation graph in Figure 9-23(b) as an example (or you can plot one using Equation 9-1): The vibrational frequency for NaCl is 1.14  1013 Hz (from Table 9-7) and two vibrational levels, for example v = 0 and v = 10 yield (from Equation 9-20)

E0  1 / 2hf  0.0236eV

E10  11 / 2hf  0.496eV

above the bottom of the well. Clearly, the average separation for v10 > v0.

2

9-49. (a) E0 r 

2 r02



E0 r 

E 

where r0  0.128nm for HCl and

1.0079u  35.453u   0.980u m1m2  m1  m2 1.0079u  35.453u



1.055 10

2  0.980u  1.66  10



E0  0

27

34

J s





2

9

kg / u 0.128  10 m



2

 2.089  1022 J  1.303  103 eV

 1 E0 r E1  2E0r  2.606  103 eV

E01  E1  E0  2.606  103 eV

f 01   E01 h 

E2  6E0r  7.82  103 eV E12  E2  E1  5.214  103 eV

2.606  103 eV  0.630  1012 Hz 4.136  1015 eV s

227

Chapter 9 – Molecular Structure and Spectra (Problem 9-49 continued)

f12   E12 h 

5.214  103 eV  1.26  1012 Hz 4.136  1015 eV s

f01  f  f01  6.884  1014 Hz  0.63  1012  6.890  1014 Hz; 6.878  1014 Hz

01  c f01  435.5nm; 436.2nm f02  f  f02  6.884  1014 Hz  1.26  1012  6.897  1014 Hz; 6.871 1014 Hz

02  c f02  435.0nm; 436.6nm (b) From Figure 9-29: f01  0.6  1012 Hz and f12  1.2  1012 Hz The agreement is very good!

Ev   v  1 / 2  hf

9-50. (a) Li2 :







E1   3 / 2  4.14  1015 eV s 1.05  1013 Hz  0.0652eV  6.52  102 eV

E 



 1 E0 r





E1  2 8.39  105 eV  1.68  104 eV

(b) K 79 Br :

Ev   v  1 / 2  hf







E1   3 / 2  4.14  1015 eV s 6.93  1012 Hz  4.30 102 eV

E 



 1 E0 r





E1  2 9.1 106 eV  1.8  105 eV

9-51.

  HCl   0.980u

(See solution to Problem 9-49)

From Figure 9-29, the center of the gap is the characteristic oscillation frequency f : f  8.65  1013 Hz  E  0.36eV



K   2  8.65  1013 Hz 2

Thus, f 

  0.980u  1.66 10 2

228

27

1 2



K



or K   2 f  

kg / u  480 N / m

2

Chapter 9 – Molecular Structure and Spectra 9-52.

n En

g En e

En / kT

694.3nm

21

E2

E2

E1

hc

1.7860eV

E1

1240eV nm 694.3nm

21

0.0036eV

1.7860eV

1.7896eV

Where E2 is the lower energy level of the doublet and E2 is the upper. Let T = 300K, so kT = 0.0259eV. (a)

n E2

g E2

n E1

g E1

n E2

1 e 2

n E1

1.7896 / 0.0259

(b) If only E2

2 e 4

E2 E1 / kT

e

1.7896 / 0.0259

5.64 10

1 e 2

69

4.91 10

31

31

E1 transitions produce lasing, but E2 and E2 are essentially equally

populated, in order for population inversion between levels E2 and E1 , at least 2/3 rather than 1/2) of the atoms must be pumped. The required power density (see Example 9-8) is:

p

9-53. (a) Ev

so I (c) I

h/ 4

r 2 , where

mH mCl mH mCl

(Equation 9-20)

E 2

J s 4.32 1014 Hz

3 3 10 s

hf / 2

1,

34 3

v 1 / 2 hf

For v = 0, E0 (b) For

2 2 1019 cm3 6.63 10

2 N hf 3 ts

f

6.63 10 2

/I

34

J s 8.66 1013 Hz / 2

h f

6.63 10 34 J s 4 2 6 1011 Hz

2.8 10

is given by Equation 9-17.

0.973u

r

0.132nm

229

47

kg m 2

0.179eV

1273W / cm3

Chapter 9 – Molecular Structure and Spectra

9-54. (a)

dU dr

U0

12a12 r

For Umin,

dU / dr

(b) For U

U min ,

r

13

6a 6 r

2

0, so

12a12 r

a then U min

(c) From Figure 9-8(b): r0

7

6

U0

12a6 a a

0

12

0.074nm ( a )

2 U0

r a a

6

a

6

1 2 U0 32.8eV

6

r / r0

r0 / r

0.85

7.03

−5.30

+56.7

0.90

3.5

−3.8

−9.8

0.95

1.85

−2.72

−28.5

1.00

1

−2.0

−32.8

1.05

0.56

−1.5

−30.8

1.10

0.32

−1.12

−26.2

1.15

0.19

−0.86

−22.0

1.20

0.11

−0.66

−18.0

2 r0 / r

230

a

6

(d) 12

r

U

U0

Chapter 9 – Molecular Structure and Spectra

ke2 r

9-55. (a) U r

Eex

Eion

(Equation 9-1)

4.27eV and r0

For NaCl, Ed Eion Na

Eaff Cl

Eex

4.27

ke2 1.52 0.236

Ar

n

0.31eV

0.31eV

9-56. For H

4.27eV

0.31eV

ke2 r02

25.85eV / nm

n A r0 r0n

25.85eV / nm 0.236nm / 0.31eV

Solving for n: n

Ed

(Equation 9-2)

Following Example 9-2,

A

(Table 9-1).

5.14 3.62 1.52eV and U r0

Eion

(b) Eex

0.236nm

0.236nm

H system, U r

20

8.9 10

ke2 r

14

n 0.31eV r0 19.7

20

eV nm20

Eion

There is no Eex term, the two electrons of H are in the n = 1 shell with opposite spins.

Eion

ionization energy for H

electron affinity for H = 13.6eV

1.440eV nm 12.85eV r

U r

dU r dr

For U r to have a minimum and the ionic H

0.75eV

12.85eV .

1.440 r2 H molecule to be bound, dU / dr

As we see from the derivative, there is no non-zero or finite value of r for which this occurs.

9-57. (a)

u 35

u 37

35

1.007825u 34.968851u

0.979593u

1.007825u 34.968851u 1.007825u 36.965898u

0.981077u

1.007825u 36.965898u 1.52 10

3

(b) The energy of a transition from one rotational state to another is:

231

0.

Chapter 9 – Molecular Structure and Spectra (Problem 9-57 continued)

E,

2

1

f

hf

(Equation 9-15)

1 h2

hI f

4

2

4

2

1 h

2 0

h r

2

2

r

1 h

r02

f f

4

1.53 10

r02

1

2 0

1 h 4

2

4

1 h

df d

f f

(c)

/I

1

1

3

2

1

r02

from part (b). In Figure 9-29 the

f between the 35Cl

lines (the taller ones) and the 37Cl lines is of the order of 0.01 1013 Hz, so f / f

9-58. (a) For CO : I

r02

m1m2 m1 m2

0.113

6.861u 1.66 10 2

E0 r

r0

0.0012, about 20% smaller than

2I

(b) E

1.055 10

34

2 1.454 10

46

27

J s

0

E1

2E0r

4.78 10 4 eV

E2

6E0r

1.43 10 3 eV

E3

12E0r

2.87 10 3 eV

E4

20E0r

4.78 10 3 eV

E5

30E0r

7.17 10 3 eV

12.0112u 15.9994u 12.0112u 15.9994u

kg / u 0.113 10

9

2

1.454 10

46

2

kg m

2

3.827 10

1 E0 r

E0

/ .

232

23

J

2.39 10 4 eV

6.861u kg m2

Chapter 9 – Molecular Structure and Spectra (Problem 9-58 continued) (c) (See diagram) E54

7.17 4.78

10 3 eV

2.39 10 3 eV

E43

4.78 2.87

10 3 eV

1.91 10 3 eV

E32

2.87 1.43

10 3 eV

1.44 10 3 eV

E21

1.43 0.48

10 3 eV

0.95 10 3 eV

E10

4.78 10 4 eV

8 7

5

6

Ev

10 3 eV

5

4

4 3

3

2

2 1

1 0

0

hc / E

(d) 54

43

32

21

10

1240eV nm 2.39 10 3 eV

5.19 105 nm

0.519mm

1240eV nm 1.91 10 3 eV

6.49 105 nm

0.649mm

1240eV nm 1.44 10 3 eV

8.61 105 nm

0.861mm

1240eV nm 0.95 10 3 eV

13.05 105 nm 1.31mm

1240eV nm 4.78 10 4 eV

25.9 105 nm

2.59mm

All of these are in the microwave region of the electromagnetic spectrum.

233

Chapter 9 – Molecular Structure and Spectra

n 1, v 1,

#2

9-59.

1

# 3 n 1, v #1 n 1, v

0,

0

(a) E 1

1 hf H 2 2

1 E0 r

E 2

3 hf H 2 2

2 E0 r since

1

E 3

1 hf H 2 2

6 E0 r since

2

1 hf H since 2 2

A

E 2

E 1

3 hf H 2 2

2 E0 r

B

E 2

E 3

3 hf H 2 2

2 E0 r

Re-writing [A] and [B] with E0r A1

hf H 2

B1

hf H 2

2

2

0

1 hf H 2 2

h 1.356 1014 Hz

1 hf H 2 2 2

6 E0 r

h 1.246 1014 Hz

/ 2I :

h 1.356 1014 Hz

I 2

0,

I

h 1.246 1014 Hz

Subtracting [B1] from [A1] and cancelling an h from each term gives: 3h 4

I

2

2

4

3h 0.110 1014 Hz

2 0

(b) I

r

r0

I/

r0

0.110 1014 Hz

I

4.58 10

4.58 10

7.40 10 11 m

kg m2

1.007825

For H 2 : 1/ 2

48

2

2 1.007825 48

kg m2

0.503912u

0.5039u 1.66 10

27

kg / u

1/ 2

0.0740nm in agreement with Table 9-7.

Canceling an h from [B1] and substituting the value of I from (a) gives: 2

I 1.246 1014 Hz

f H2

2h 4

f H2

1.32 1014 Hz also in agreement with Table 9-7.

234

2

Chapter 10 – Solid State Physics 10-1. U  r0   

ke2  1  1 r0  n 

(Equation 10-6)

ke2  1  E  U  r0    1 r0  n 

1

1 Ed r0  741kJ / mol  0.257nm  1eV / ion pair     0.7844 n  ke2 1.7476 1.44eV nm  96.47kJ / mol n

1  4.64 1  0.7844

10-2. The molar volume is

M



 2 N A r03

 M   74.55 g / mole r0    23 3  2 N A    2 6.022  10 / mole 1.984 g / cm





10-3. The molar volume is



M



1/ 3

 3.15  108 cm  0.315nm

 2 N A r03

M 42.4 g / mole  3 23 2 N A r0 2 6.022  10 / mole 0.257  107 cm

10-4. (a) U att  



  





ke2 r0

(Equation 10-1)

ke2  1  1 (b) Ed  U  r0    r0  n 



3

 2.07 g / cm3

(Equation 10-6)

 1  8.01eV  1    7.12eV / ion pair  9

 96.47kJ / mol   1cal    7.12eV / ion pair      164kcal / mole  1eV / ion pair   4 / 186 J 

235

Chapter 10 – Solid State Physics (Problem 10-4 continued) (c) 1 

1 Ed r0 165.5kcal / mol  0.314nm  4.186 J  1eV / ion pair      0.8960 n  ke2 1.7476 1.44eV nm  1cal  96.47kJ / mol 

Therefore, n 

1  9.62 1  0.8960

10-5. Cohesive energy (LiBr)   1mol 1eV  = 788  103 J / mol   8.182ev / ion pair  23 19   6.02  10 ion pairs   1.60  10   4.09eV / atom

This is about 32% larger than the value in Table 10-1.

10-6. Molecular weight Na = 22.990 Molecular weight Cl = 35.453

 the NaCl molecule is by weight 0.3934 Na and 0.6066 Cl. Since the density of NaCl = 2.16 g/cm3, then



mol of Na / cm3   0.3934 2.16 g / cm3

  2.990g / mol   0.03696 mol / cm

3

mol of Cl / cm3  0.03696 mol / cm3 , also since there is one ion of each element per molecule. Number of Na ions / cm3  0.03696 N A Number of Cl ions / cm3  0.03696 N A





Total number of ions / cm3   0.07392  6.022  1023  4.45  1022 Nearest neighbor distance = equilibrium separation r0 .

 1 r0    4.45  1022 ions / cm3 102 cm / m 





  3  

1/ 3



236

 2.88  1010 m  0.288nm

Chapter 10 – Solid State Physics 10-7.

r0  KCl   0.315nm  3.15  1010 m





N ions / m3  1 / r03  3.20  1028 / m3  3.20  1022 / cm3

Half of the ions in 1cm3 are K and half are Cl , so there are 1.60 1022 / cm3 of each element. This number of ions equals: 1.60  1022 ions 1.60  1022 ions   0.02657 mol NA 6.022  1023 / mol

This is the moles of each ion in 1 cm3. Molecular weight of K = 39.102 g/mol. Molecular weight of Cl = 35.453 g/mol.

  Weight of Cl/cm3 =  35.453g / mol   0.02657mol / cm   0.942 g / cm Weight of K/cm3 =  39.102 g / mol  0.02657mol / cm3  1.039 g / cm3 3

3

 density of KCl  1.039  0.942  g / cm3  1.98 g / cm3

10-8. (a)

8.0 7.0

Cohesive Energy (eV)

6.0 5.0 4.0 3.0 2.0 1.0 500

1000

1500

2000

2500

Melting point, K

(b) Noting that the melting points are in kelvins on the graph, Co melting point = 1768 K, cohesive energy = 5.15 eV Ag melting point = 1235 K, cohesive energy = 3.65 eV Na melting point = 371 K, cohesive energy = 1.25 eV

237

Chapter 10 – Solid State Physics 2 2 2 2 2 2 10-9. U att  ke2         a 2a 3a 4a 5a 6a

2 1 2 1  U att  ke2  2  1      3 3 5 3 

  

  

The quantity in parentheses is the Madelung constant α. The 35th term of the series (2/35) is approximately 1% of the total, where α = 4.18.

10-10. (a)   

(b)

me v ne2



(Equation 10-13)

9.1110

8.47  10

28

 electrons / m 1.60  10

1/ 2

 100 K     300 K 

100  300

19

2

9

7

m

(from Equation 10-9) 1/ 2

100

  1.23  10 C   0.4  10 m 

kg 1.17  105 m / s 3

v   kT / me  v

31



1 3

3  7.00  108  m





4 103 A I I 10-11. (a) j    A  d 2 / 4  1.63  103 m





2

 479 A / m3

(from Equation 10-10)

I d 479 A / m2    3.53  108 m / s (b) vd  28 3 19 Ane ne 8.47  10 / m 1.602  10 C







 3.53  106 cm / s

10-12. (a) There are na conduction electrons per unit volume, each occupying a sphere of





volume 4 rs3 3 : na  4 rs3 / 3  1 rs3 

3 4 na



rs   3 / 4 na 

1/ 3

238

Chapter 10 – Solid State Physics (Problem 10-12 continued)

 3 (b) rs    4 8.47  1028 / m3





  

1/ 3

 1.41  1010 m  0.141nm

10-13. (a) n   N A / M for 1 electron/atom

10.5g / cm  6.022 10 n 3

23

/ mole

107.9 g / mole

19.3g / cm  6.022 10 n 3

(b)

23

/ mole

196.97 g / mole

  5.86 10

  5.90 10

22

22

/ cm3

/ cm3

Both agree with the values given in Table 10-3.

10-14. (a) n  2 N A / M for two free electrons/atom





2 1.74 g / cm3 6.022  1023 / mole

n

24.31g / mole





2 7.1g / cm3 6.022  1023 / mole

(b) n 

65.37 g / mole

  8.62  10

  13.110

22

22

/ cm3  8.62  1028 / m3

/ cm3  13.1  1028 / m3

Both are in good agreement with the values in Table 10-3, 8.61 1028 / m3 for Mg and 13.2 1028 / m3 for Zn.

10-15. (a)  

me v ne2



(Equation 10-13)

1





ne2 me v

(Equation 10-13)

9.1110 kg 1.08 10 m / s    1.22  10 8 . 47  10 m 1 . 602  10 C 0 . 37  10 m     31

28



1





3

5

19

2

1 1  8.17  106   m  7 1.22  10  m

239

9

7

m

Chapter 10 – Solid State Physics (Problem 10-15 continued)

(b)

v 

8kT  me

(Equation 10-9)

  200K     300K  v  200K 

v  300K 

   300K  200K / 300K 

1/ 2





 1.22  107  m  200 K / 300K 

  200 K  

1/ 2

1 1 1   1.00  107   m  8   200 K  9.96  10  m

(c)  100K     300K  v 100K 



v  300K 



 1.22  107  m 100K / 300K 

 100 K  

10-16. E 

3 EF 5

 9.96  108  m

1/ 2

 7.04  108  m

1 1 1   1.42  107   m  8  100 K  7.04  10  m

(Equation 10-22)

(a) for Cu:

E 

3  7.06eV   4.24eV 5

(b) for Li:

E 

3  4.77eV   2.86eV 5

EF

 hc  

2

N  3   2  2mc  8 V V 

1240eV

2/3





3 2 28 3 nm   3 5.90  10 m  109 m        8 2 5.11  105 eV   1nm  





240

1/ 3

 5.53eV

Chapter 10 – Solid State Physics 10-17. A long, thin wire can be considered one-dimensional.

 hc   N  h2  N  EF   32m  L  32mc 2  L  2

2



2

For Mg: N / L  8.61 1028 / m2



1/ 3

1240eV nm  10 m / nm  8.61 10  32  0.511  10 eV  2

9

EF

(Equation 10-15)

28

/ m3



2/3

 1.87eV

6

10-18. (a) For Ag: h 2  3N  EF  2m  8 V 

2/3

1240eV nm 10 m / nm   2  0.511  10 eV  9 6

2

 3  5.86  1028 m3    8  

2/3

 5.50eV

For Fe: Similarly, EF  11.2eV (b) For Ag: EF  kTF TF 

(Equation 10-23)

EF 5.50eV   6.38  104 K k 8.617  105 eV / K

For Fe: Similarly, TF  13.0  104 K Both results are in close agreement with the values given in Table 10-3.

10-19. Note from Figure 10-11 that most of the excited electrons are within about 2kT above the Fermi energy EF , i.e., E  2kT . Note, too, that kT

EF , so the number N of excited

electrons is: N  N  EF  n  EF  E  N  EF 1 / 2  2kT   N  EF  kT and 8 V  2m  N 3  h2 

3/ 2

EF3 / 2

(from Equation 10-20)

Differentiating Equation 10-19 gives: N  EF  

 V  8m 

N 2   Then, N 8 V 3

3/ 2

1/ 2  EF kT 3 h   kT EF 3/ 2 2  2m  3/ 2  2  EF h  2

241

 V  8m  2  h2 

3/ 2

EF1/ 2

Chapter 10 – Solid State Physics (Problem 10-19 continued) EF for Ag  5.35eV , so

1/ 2

 2E  10-20. uF   F   me 

N 3  8.617  105 eV / K  300 K  5.53eV  0.0070  0.1% 2 N





1/ 2

 2E   c  F2   me c 

(Equation 10-24)

 2  3.26eV   (a) for Na: uF  3.00  10 m / s   5  5.11  10 eV 





8

1/ 2

 1.07  106 m / s

 2  5.55eV   (b) for Au: uF  3.00  10 m / s   5  5.11  10 eV 





8

 2 10.3eV   (c) for Sn: uF  3.00  10 m / s   5  5.11  10 eV 



10-21.  

meuF ne2



8



(Equation 10-25)

1/ 2

 1.40  106 m / s

1/ 2

 1.90  106 m / s

meuF ne2 

9.11 10 kg 1.07 10 m / s  (a) for Na:    2.65 10 m 1.609 10 C   4.2 10 31

28

3

6

19

2

8

m



 3.42 108 m  34.2nm

9.11 10 kg 1.40  10 m / s  (b) for Au:   5.90  10 m 1.609  10 C   2.04  10 31

28

3

6

19

2

8

m



 4.14  108 m  41.4nm

9.1110 kg 1.90 10 m / s   14.8 10 m 1.609 10 C  10.6 10 31

(c) for Sn:

28

3

6

19

 4.31 108 m  43.1nm

242

2

8

m



Chapter 10 – Solid State Physics 10-22. Cv  electrons  

2

Cv  electrons  

2

assuming T

T

2

2

R

T TF

R

kT because EF  kTF . EF

(Equation 10-30)

TE (rule of Dulong and Petit):

0.10  3 2  EF

 k 2



 0.60  7.06eV 

 8.617  10 eV / K  5

2

Cv due to the lattice vibrations is 3R,

2 2

R

kT  0.10  3R  EF

 4.98  103 K

This temperature is much higher than the Einstein temperature for a metal such as copper.

10-23. U 

 kT  3 NEF   N   kT 5  EF 

(Equation 10-29)

 kT 3 Average energy/electron = U N  EF    5  EF

 3  2  kT  kT  E   F 5 4 EF 

2

For copper: E  7.06eV , so At T = 0K: U N 

3  7.06eV   4.236eV 5



5 3  2 8.61  10 eV / K At T = 300K: U N   7.06eV   5 4 7.06

 300K  2

2

 4.236eV

The average energy/electron at 300K is only 0.0002eV larger than at 0K, a consequence of the fact that 300K is very small compared to the TF for Cu (81,600K). The classical value of U N   3 / 2 kT  0.039eV , is far too small.

10-24. Cv  electrons  

2 2

R

T TF

(Equation 10-30)

Melting temperature of Fe = 1811K (from Table 10-1) TF for Fe  13  104 K (from Table 10-3)

The maximum Cv for the Fe electrons, which is just before Fe melts, is:

243

Chapter 10 – Solid State Physics (Problem 10-24 continued)

Cv  electrons  

2

 1811K  R  0.0219 R 2  13  104 K 

The heat capacity of solids, including Fe, is 3R (rule of Dulong and Petit, see Section 8-1). Cv  electrons 



Cv

10-25. P 

   M  B     kT

9.285  10 P 1.38  10

24

23

10-26.  

0.0291R  0.0073 3R

0 M



B

J /T



(Equation 10-35)

  2.0T   6.7  10

J / K  200 K 

3

0  2 kT 2

 N  1  J   1 

 units   2  3      A  m  T   J   

NJ 2 NJ NJ NJA2 m2    A2 m3T 2 J A2 m3 Wb / m 2 A2 m3  N / Am 2 A2 m3 N 2

J Nm   1 dimensionless Nm Nm

10-27. E  hc  (a) For Si:   hc E  1240eV nm 1.14eV  1.088  103 nm  1.09  106 m  1.09  103 nm (b) For Ge: Similarly,   1.722  103 nm  1.72  106 m  1.72  103 nm (c) For diamond: Similarly,   1.77  102 nm  1.72 107 m

10-28. (a) For Ge: All visible light frequencies (wavelengths) can excite electrons across the 0.72eV band gap, being absorbed in the process. No visible light will be transmitted through the crystal.

244

Chapter 10 – Solid State Physics (Problem 10-28 continued) (b) For insulator: No visible light will be absorbed by the crystal since no visible frequencies can excite electrons across the 3.6eV band gap. The crystal will be transparent to visible light. (c) Visible light wavelengths are 380-720nm corresponding to photon energies E  hf  hc /   1240eV nm / 

  380nm  E  3.3eV   720nm  E  1.7eV Visible light photons will be absorbed, exciting electrons for band gaps  3.3eV .





10-29. (a) E  hc   1240eV nm 3.35 m  103 nm /  m  0.37eV (b) E  kT  0.37eV T  0.37eV / k  0.37eV / 8.617  105 eV / K  4300K





2.33g / cm3 100nm  107 cm / nm mN A VN A 10-30. (a) N    M M 28 g / mol

  6.02  10 3

23

/ mol

 5.01107 Si atoms





(b) E  13eV 4  5.01 107  6.5  108 eV

2

1  ke2  m 1 10-31. (a) E1     2 2 2   1



(Equation 10-43)





2

 9  109 N m2 / C 2 1.60  1019 C 2  31 1    0.2  9.11  10 kg E1    2 2 2 11.8 1.055  1034 J s





 3.12  1021 J  0.0195eV

Ionization energy = 0.0195eV (b)





r1  a0 1 me m  2

(Equation 10-44)

 0.0529nm 1 / 0.2 11.8  3.12nm

(c) Eg ( Si )  1.11eV at 293K

245







Chapter 10 – Solid State Physics (Problem 10-31 continued) E1 / Eg  0.0195 / 1.11  0.0176 or about 2%.

2

1  ke2  m 1 10-32. (a) E1     2 2 2   1

(Equation 10-43)







2

 9  109 N m2 / C 2 1.60  1019 C 2   0.34  9.11 1031 kg 1   E1    2 2 2 15.9  1.055  1034 J s









 2.92  1021 J  0.0182eV

(b)





r1  a0 1 me m  2

(Equation 10-44)

 0.0529nm 1 / 0.34 15.9   2.48nm

10-33. Electron configuration of Si: 1s 2 2s 2 2 p6 3s 2 3 p 2 (a) Al has a 3s 2 3 p configuration outside the closed n = 2 shell (3 electrons), so a p-type semiconductor will result. (b) P has a 3s 2 3 p3 configuration outside the closed n = 2 shell (5 electrons), so an n-type semiconductor results.

 T  0.01eV / 8.617  105 eV / K  116K

10-34. E  kT  0.01eV

10-35. (a) VH  vd Bw 

dBw iB  nq qnt

(Equation 10-45 and Example 10-9)

The density of charge carriers n is:

n

(b) N 

 20 A 0.25T  iB   7.10  1028 carriers/m3 19 3 6 qtVH 1.60  10 C 0.2  10 m 2.2  10 V



 NA M





5.75 10 kg / m  6.02 10  3

3

118.7kg / mol

26

/ mol



  2.92 10

28

Each Sn atom contributes n / N  7.10  1028 / 2.92  1028  2.4 charge carriers

246

Chapter 10 – Solid State Physics





10-36. I net  I 0 eeVb / kT  1

(Equation 10-49)

(a) eeVb / kT  10, so eVb / kT  ln 10 . Therefore,

1.60 10 C V 1.38110 J / K  300K   ln 10 V  1.38  10 J / K   300K  ln 10  1.60  10 C   0.0596V  59.6mV 19

23

b

23

19

b

(b) eeVb / kT  0.1 Vb  0.0596V ln 0.1 / ln10   0.0596  59.6mV





10-37. I net  I 0 eeVb / kT  1

(Equation 10-49)

(a) eeVb / kT  5, so eVb / kT  ln  5. Therefore,

Vb 

kT ln  5

1.38  10 J / K   200K  ln 5  0.0278V  27.8mV  1.60  10 C  23

19

e

(b) eeVb / kT  0.5

eVb / kT  ln  0.5 Vb 

kT ln  0.5



1.38 10 J / K   200K  ln  0.5  0.0120V  12.0mV  1.60  10 C  23

19

e



10-38. I net  I 0 eeVb / kT  1

(Equation 10-49)

Assuming T = 300K, I  0.2V   I  0.1V  I  0.1V 





I0 e 

e 0.2V  / kT



 

 1  I0 e 

I0 e 

e 0.1V  / kT

e 0.1V  / kT



1

e

1

e 0.2V  / kT



e

e 

e 0.1V  / kT

e 0.1V  / kT



1

 47.7

10-39. (a) From Equation 10-49, exp  eVb / kT   10 Taking ln of both sides and solving for Vb,

1.38 10 J / K   77 K  ln 10  0.0153 volts  15.3mV   kT / e  ln 10   1.60  10 C  23

Vb

19

247

Chapter 10 – Solid State Physics (Problem 10-39 continued) (b) Similarly, for exp  eVb / kT   1; Vb  0





 I net =1mA 10  1  9mA

(c) For (a): I net  I 0 eeVb / kT  1 For (b): I net  0

10-40. E  hc / 

For   484nm

E  Egap

Egap  hc / 484nm  1240eV nm / 484nm  2.56eV

10-41. M  Tc  constant

(Equation 10-55)

First, we find the constant for Pb using the mass of natural Pb from Appendix A, Tc for Pb from Table 10-6, and α for Pb from Table 10-7. constant   207.19u 

0.49

 7.196K   98.20

For

206

Pb : Tc  constant / M   98.20  205.974u 

For

207

Pb : Tc  constant / M   98.20  206.976u 

For

208

Pb : Tc  constant / M   98.20  207.977u 

10-42. (a) Eg  3.5kTc

0.49

 7.217 K

0.49

 7.200K

0.49

 7.183K

(Equation 10-56)





Tc for I is 3.408K, so, Eg  3.5 8.617  105 eV / K  3.408K   1.028  103 eV

(b) Eg  hc / 

  hc / Eg  1240eV nm / 1.028  103 eV  1.206  106 nm  1.206  103 m  1.206mm

10-43. (a) Eg  3.5kTc



For Sn : Tc  3.722K



Eg  3.5 8.617  105 eV / K  3.722 K   0.0011eV

248

Chapter 10 – Solid State Physics (Problem 10-43 continued) (b) Eg  hc / 

  hc / Eg  1240eV nm / 6  104 eV  2.07  106 nm  2.07  103 m Eg  T  / Eg  0   0.95 where Eg  0   3.5kTc (Equation 10-56)

10-44. At T / Tc  0.5

So, Eg T   0.95  3.5 kTc  3.325kTc (a) For Sn : (b) For Nb : (c) For Al : (d) For Zn :

  E T   3.325 8.617  10 eV / K   9.25K   2.65  10 E T   3.325 8.617  10 eV / K  1.175K   3.37  10 E T   3.325 8.617  10 eV / K   0.85K   2.44  10

Eg T   3.325 8.617  105 eV / K  3.722K   1.07  103 eV 5

3

g

5

4

g

5

g

10-45. BC T  BC  0   1  T / TC 

2

4

eV

eV

2

(a) BC T  BC  0   0.1  1  T / TC 

T / TC 

eV

2

 1  0.1  0.9

T / TC  0.95 (b) Similarly, for BC T  BC  0   0.5

T / Tc  0.71 (c) Similarly, for BC T  BC  0   0.9

T / Tc  0.32

10-46. 1. Referring to Figure 10-56, notice that the length of the diagonal of a face of the cube is r; therefore, a 2  a 2  (4r )2  a  8 r 2. Volume of the unit cube Vcube  a3  83/2 r 3

249

Chapter 10 – Solid State Physics (Problem 10-46 continued) 3. The cube has 8 corners, each occupied by ⅛ of an atom’s volume. The cube has 6 faces, each occupied by ½ of an atom’s volume. The total number of atoms in the unit cube is then 8×⅛ + 6×½ = 4. Each atom has volume 4 r 3 / 3 . The total volume occupied by atoms is then Vatoms  4  (4 r 3 / 3) . 4. The fraction of the cube’s volume occupied by atoms is then: Vatoms 4  (4 r 3 )   0.74 Vcube 83/2 r 3

10-47. TF for Cu is 81,700K, so only those electrons within EF  kT of the Fermi energy could be in states above the Fermi level. The fraction f excited above EF is approximately:

f  kT / EF  T / TF (a) f  300K / 81, 700K  3.7 103 (b) f  1000K / 81, 700K  12.2 103

10-48. −e ● −6

−e ● −4

+e ● −5

−e ● −2

+e ● −3

−e ● 0

+e ● −1

+e ● 1

−e ● 2

+e ● 3

−e ● 4

+e ● 5

−e ● 6

r0 (a) For the negative ion at the origin (position 0) the attractive potential energy is: 2ke2  1 1 1 1 1 V  1      r0  2 3 4 5 6

(b) V  

ke2 , so the Madelung constant is: r0

 

1

1

1

1

1

  2 1       2 3 4 5 6

  

250

  

Chapter 10 – Solid State Physics (Problem 10-48 continued) Noting that ln 1  x   x 

x 2 x3 x 4    2 3 4

ln  2   1 

,

1 1 1    2 3 4

and   2 ln  2   1.386

10-49. Cv  TF 

2

R

2

2

R

2

T TF



(Equation 10-30)

T 3.74  10 J / mol K T

and EF  kTF 





4

2 2

R



kT 3.74  104 J / mol K T



 2 8.314 J / mol K  1.38  1023 J / K 



2 3.74  104 J / mol K







 1.51 1018 J 1 / 1.60  1019 J / eV  9.45eV

10-50. (a) N 

EF

EF

 g  E  dE   AE 0

(b) N  

1/ 2

dE  A  2 / 3 E

EF

0 EF



EF  kT

  2 A / 3 EF3 / 2

3/ 2 0

AE1 / 2 dE   2 A / 3  EF3 / 2   EF  kT  

3/ 2

  2 A / 3  EF3 / 2  EF3 / 2 1  kT / EF  

3/ 2

Because kT

 

 

EF for most metals,

1  kT / EF 

3/ 2

 1   3 / 2  kTEF1 / 2

N    2 A / 3  EF3 / 2  EF3 / 2   3 / 2  kTEF1 / 2   AkTEF1 / 2 AkTEF1 / 2 N 3kT   The fraction within kT of EF is then f  3/ 2 N  2 A / 3  EF 2 EF

(c) For Cu EF  7.04eV ; at 300K ,

f 

3  0.02585eV  2  7.04eV 

251

 0.0055

Chapter 10 – Solid State Physics

   0.72eV / pair   1.63 10  1.33  10 eV / photon   0.72eV / pair   1.85  10

10-51. (a) N1  1.17  106 eV / photon N2

6

6

pairs / photon

6

pairs / photon

(b) N1  N1  1.23  103  N1 / N1  7.8  104 N2  N2  1.36  103  N2 / N2  7.4  104

(c) Energy resolution

E N  E N

 E1 / E1  0.078% E2 / E2  0.074%

10-52.  

meff v

(Equation 10-13)

n e2



v  3kT / meff



1/ 2





 3 1.38  1023 J / K  300 K     0.2 9.11  1031 kg  



1/ 2



 2.61  105 m / s

Substituting into λ:







0.2 9.11  1031 kg 2.61  105 m / s

10

22

m

3

5  10

3



 m 1.60  10

19



C



2

 3.7  108 m  37nm

 2  7.06eV     31  9.11  10 kg 

1/ 2

For Cu: uF   2 EF / me 

1/ 2

 1.57  106 m / s

n  8.47  1028 m3 and   1.7  108  m (Example 10-5) Substituting as above,   3.9 108 m  39nm The mean free paths are approximately equal.

252

Chapter 10 – Solid State Physics 10-53. Compute the density of Cu (1) as if it were an fcc crystal and (2) as if it were a bcc crystal. The result closest to the actual measured density is the likely crystalline form. 1. The unit cube of an fcc crystal of length a on each side composed of hard, spherical atoms each of radius r contains 4 atoms. The side and radius are related by a  8 r and the volume of the cube is Vcube  a3  83/2 r 3 . (See problem 10-46 and Appendix B3.) The density of fcc Cu would be: mCu 4( M / N A )  Vcube 83/2 r 3 where M = molar mass (atomic weight) and N A = Avagadro’s number. d

d

4(63.55g/mol)  8.90g/cm3 8 3 23 8 (1.28 10 cm) (6.022 10 / mol) 3/2

2. For a bcc crystal (refer to Appendix B3), let the length of a side = a, the diagonal of a face = f, and the diagonal through the body = b. Then from geometry we have

a 2  f 2  b2  (4r )2 and a 2  a 2  f 2 . Combing these yields:

4r 3 The unit cube has 8 corners with ⅛ of a Cu atom at each corner plus 1 Cu atom at the a 2  f 2  3a 2  (4r )2  a 

center, or 8×⅛ + 1 = 2 atoms. The density of bcc Cu is then: d

2( M / N A ) m 2(63.55g/mol)  33/2    8.17 g/cm3 3 8 3 23 3 Vcube (4r / 3) 4 (1.28 10 cm) (6.022 10 /mol)

Based on the above density calculations, metallic Cu is most likely a face-centered cubic crystal.

253

Chapter 10 – Solid State Physics 10-54. (a) For small Vb (from Equation 10-49) eeVb / kT  1  eVb / kT , so I  I 0eVb / kT  Vb / R

R  Vb

 I0eVb / kT   kT





eI 0  0.025eV e  109 A  25.0M 

(b) For Vb  0.5V ;

R  Vb I  0.5 109 A  500M 

(c) For Vb  0.5V ;

I  109 A e0.5 / 0.025  1  0.485 A





Thus, R  Vb I  0.5 / 0.485  1.03 (d)

10-55. a0 

eI 0 eVb / kT dI  e dVb kT 

dVb kT  eVb / kT  e  25M e20  0.0515 dI eI 0

Rac 

 0 h2  o h2  2    me e2   me eff e2  me eff ke2

silicon: a0 

 kg  9  10 N m

 / C 1.602  10

12 1.055  1034 J s



0.2 9.11  1031

9

2

2

2

19

C



2

 3.17 109 m  3.17nm

This is about 14 times the lattice spacing in silicon (0.235nm) germanium: : a0 

 kg  9  10 N m

16 1.055  1034 J s



0.10 9.11  1031

9

2



2



/ C 2 1.602  1019 C



 8.46  109 m  8.46nm

This is nearly 35 times the lattice spacing in germanium (0.243nm)

2

1  ke2  m 1 10-56. (a) En     2 2 2   n

(Equation 10-44)



6 2 2 1  1.440eV nm   0.015 0.511  10 eV / c    2  6.582  1016 eV s   182 

or

254

  1  n

2

2

Chapter 10 – Solid State Physics (Problem 10-56 continued)





9 2 2 19  1 2 9  10 N m / C 1.60  10 C   2 6.63  1034 J s 





2

   

2



 0.015 9.11  1031 kg  182 

  1  n

2

1.01  1022 J  6.28  104 eV / n2 n2

 donor ionization energy = 6.28  104 eV (b) r1  a0

me  m

(Equation 10-45)

 0.0529nm 1 / 0.01518

r1  63.5nm (c) Donor atom ground states will begin to overlap when atoms are 2r1  127nm apart. 3

 109 nm / m  20 3 donor atom concentration     4.88  10 m  127nm / atom 

10-57. (a)  

meuF ne2

(Equation 10-25)

So the equation in the problem can be written as   m  i . Because the impurity increases m by 1.1 108  m, i  1.1 108  m and

i 



meuF 8

ne 1.18  10  m 2

and uF   2EF / me 

1/ 2



where n  8.47  1028 electrons / m3 (from Table 10-3)

 1.57  106 m / s. Therefore,

9.11 10 kg 1.57  10 m / s   8.47  10 / m 1.60  10 C  1.1 10 31

i

28

6

19

3

(b)   1 / na r 2 and d  2r

2

8

m



 6.00  108 m  60.0nm

(Equation 10-12)

So we have d 2  4 / nii where ni  1% of n  8.47 1026 / m3



 



d 2  4 / 8.47  1028 / m3  6.60  108 m  2.28  1022 m2  d  0.0151nm

255

Chapter 10 – Solid State Physics 10-58. (a) The modified Schrödinger equation is: 

d  2 dR  r    ke2  r    2mr 2 dr  dr   r 2



 1   R  r   ER  r  2m r 

2

 2

The solution of this equation, as indicated following Equation 7-24, leads to solutions of the form: Rn  r   a0 e r / a0 n r 1Ln  r / a0  , where a0 



 2 / ke2m 

2

(b) By substitution into Equation 7-25, the allowed energies are: 2

E1 1  ke2  m 1  ke2   En      where E     m 1 2    n2 n2 2  

(c) For As electrons in Si: m  0.2me (see Problem 10-31) and   Si   11.8,







2

 9  109 N m2 / C 2 1.60  1019 C 2  31 1   0.2 9.11  10 kg E1    2 2 2 11.8 1.055  1034 J s





 3.12  1021 J  0.0195eV





Energy  102 eV n 0 5 4 1.0

3 2

2.0

3.0

4.0

10-59. U  

F 

ke2 r0

 r 1  r n   0  0   r n  r  

(Equation 10-5)

 n  1 ke dU   Kr yields K   dr r03

2

(a) For NaCl:   1.7476, n  9.35, and r0  0.282nm and

256

1

Chapter 10 – Solid State Physics (Problem 10-59 continued)



m  Na  m  Cl 

 22.99u  35.45u   13.95u m  Na   m  Cl   22.99u    35.45u  

2 K 1    n  1 ke     M 2  13.95u  r03 

1/ 2

1 f  2





9 2 2 19  1  1.7476  8.35 9  10 N m / C 1.60  10 C  3 27 9 2  13 . 95 u  1 . 66  10 kg / u 0 . 282  10 m 











2

1/ 2

   

 1.28  1013 Hz



(b)   c / f  .00  108 m / s / 1.28  1013 Hz  23.4 m This is of the same order of magnitude as the wavelength of the infrared absorption bands in NaCl.

10-60. (a) Electron drift speed is reached for: dv  0  vd  eE / m dt

(b) Writing Ohm’s law as j   E and j  vd ne (from Equation 10-11)

j  eE ne / m  E ne2 / m, which satisfies Ohm’s law because j  E. Thus,

   ne2 / m and   1 /   m /  ne2. 10-61. (a) For r, s, and t all even  1



r  s t

 1 and the ion’s charge at that location is:



1 1.60  1019 C  1.60  1019 C.

Similarly, for any permutation of r, s even; t odd:  1

r  s t

 1, ion charge = 1.60  1019 C.

r even; s, t odd:  1

r  s t

 1, ion charge = 1.60  1019 C.

r, s, and t all odd:  1

r  s t

 1, ion charge = 1.60  1019 C.

257

Chapter 10 – Solid State Physics (Problem 10-61 continued) (b) U  

ke2 r

If the interatomic distance r = a, then a cube 2a on each side 4 4 2 4 4 4 4  U  ke2         2a a 2a 2a 3a 3a  a

U 

ke2  2.1335 where   2.1335. a

Similarly, for larger cubes (using spreadsheet). The value of α is approaching 1.7476 slowly.

10-62. (a) M        

M





M





     

e B / kT  e  B / kT   tanh   B / kT  e B / kT  e  B / kT

and M   tanh   B / kT  (b) For  B



0 M B

kT , T 

0 and tanh   B / kT    B / kT

0  B BkT



0  2 kT

258

Chapter 11 – Nuclear Physics 11-1. Isotope

Protons

Neutrons

F

9

9

Na

11

14

V

23

28

Kr

36

48

120

Te

52

68

148

Dy

66

82

175

W

74

101

222

Rn

86

136

18 25

51

84

11-2. The momentum of an electron confined within the nucleus is:

p  / x  1.055  1034 J s / 1014 m



 1.055  1020 J s / m  1 / 1.602  1013 J / MeV



 6.59  108 MeV s / m

The momentum must be at least as large as p, so pmin  6.59  108 MeV s / m and the electron’s kinetic energy is:







Emin  pminc  6.59  108 MeV s / m 3.00  108 m / s  19.8MeV .

This is twenty times the observed maximum beta decay energy, precluding the existence of electrons in the nucleus.

11-3. A proton-electron model of 6 Li would consist of 6 protons and 3 electrons. Protons and electrons are spin −1/2 (Fermi-Dirac) particles. The minimum spin for these particles in the lowest available energy states is 1 / 2 , so 6 Li  S  0  cannot have such a structure.

259

Chapter 11 – Nuclear Physics 14

11-4. A proton-electron model of

N would have 14 protons and 7 electrons. All are Fermi-

Dirac spin-1/2 particles. In the ground state the proton magnetic moments would add to a small fraction of the proton magnetic moment of 2.8 N , but the unpaired electron would give the system a magnetic moment of the order of that of an electron, about 1  B . Because  B is approximately 2000 times large than  N , the

14

N magnetic moment would

be about 1000 times the observed value, arguing against the existence of electrons in the nucleus. 11-5. The two proton spins would be antiparallel in the ground state with S  1/ 2  1/ 2  0. So the deuteron spin would be due to the electron and equal to 1 / 2 . Similarly, the proton magnetic moments would add to zero and the deuteron’s magnetic moment would be 1  B . From Table 11-1, the observed spin is 1

(rather than 1 / 2 found above) and the

magnetic moment is 0.857  N , about 2000 times smaller than the value predicted by the proton-electron model.

11-6. Isotopes (a)

18

(b)

208

(c)

120

17

F

Pb

206

Sn

119

F

Isotones 19

Pb

210

Sn

118

F

16

Pb

207

Sn

121

N

17

O

Tl

209

Sb

122

Bi

Te

11-7. Nuclide (a)

14 8

O6

(b)

63 28

(c)

236 93

Isobars 14 6

C8

Ni35

63 29

Np143

236 92

Cu34 U144

260

Isotopes 14 7

63 30 236 94

N7

Zn33

Pu142

16 8

O8

60 28 235 93

Ni32

Np142

Chapter 11 – Nuclear Physics

11-8.



mass  Au  A 1.66  1027 kg / u





volume   4 / 3  R 3   4 / 3  R0 A1 / 3

where R0  1.2 fm  1.2  1015 m





A 1.66  1027 kg / u mass density =   2.29  1017 kg / m3 3  15 volume  4 / 3  1.2  10 m A

11-9.





B  ZM H c2  NmN c2  M Ac2 (a)

9 4

Be5

(Equation 11-11)



     0.062443uc  931.5MeV / uc 

B  4 1.007825uc 2  5 1.008665uc 2  9.012182uc 2  0.062443uc 2

2

2

 58.2MeV B / A  58.2MeV / 9 nucleons  6.46MeV / nucleon (b)

13 6

C7



     0.104250uc  931.5MeV / uc 

B  6 1.007825uc 2  7 1.008665uc 2  13.003355uc 2  0.104250uc 2

2

2

 91.1MeV B / A  91.1MeV / 13 nucleons  7.47 MeV / nucleon 57 26

(c)

Fe31









B  26 1.007825uc 2  31 1.008665uc 2  56.935396uc 2





 0.536669uc 2  0.536669uc 2 931.5MeV / uc 2  499.9MeV B / A  499.9MeV / 57 nucleons  8.77 MeV / nucleon 11-10. R  R0 A1 / 3 where R0  1.2 fm O  R  1.2 fm 16 

1/ 3

(a)

16

(b)

56

(c)

197

(d)

238

(Equation 11-3)

 3.02 fm

Fe  R  1.2 fm  56 

1/ 3

 4.58 fm

Au  R  1.2 fm 197 

 6.97 fm

U  R  1.2 fm  238

 7.42 fm

1/ 3

1/ 3

261





3

Chapter 11 – Nuclear Physics



11-11. (a) B  M

3



He c 2  mn c 2  M



4



He c 2

 3.016029uc2  1.008665uc2  4.002602uc2





 0.022092uc 2 931.5MeV / uc 2  20.6MeV

(b) B  M

 Li  c 6

2

 Li  c

 mn c 2  M

7

2

 6.015121uc2  1.008665uc2  7.016003uc2





 0.007783uc 2 931.5MeV / uc 2  7.25MeV

(c) B  M

 N c 13

2

 mn c 2  M

 N c 14

2

 13.005738uc2  1.008665uc2  14.003074uc2





 0.011329uc 2 931.5MeV / uc 2  10.6MeV

11-12. B  a1 A  a2 A2 / 3  a3 Z 2 A1 / 3  a4  A  2Z  A1  a5 A1 / 2  c 2   2

(This is Equation 11-13

on the Web page www.whfreeman.com/tiplermodernphysics6e.) The values of the ai in MeV/c2 are given in Table 11-3 (also on the Web page). For 23Na:

B  15.67  23  17.23  23   184.9MeV M



2/3

 0.75 11  23 2



Na c 2  11m p c 2  12mnc 2  B

23





1 / 3

 93.2  23  2  11  23  0  23 2

1

(Equation 11-14 on the Web page)





 11 1.007825uc 2  12 1.008665uc 2   184.9 MeV M







Na  23.190055u  184.9MeV / c 2 1 / 931.5MeV / c 2 u

23



 23.190055u  0.198499u  22.991156u This result differs from the measured value of 22.989767u by only 0.008%. 11-13. R  1.07  0.02  A1 / 3 fm (Equation 11-5) (a)

16

(b)

63

O: Cu :

R  1.4 A1 / 3 fm (Equation 11-7)

R  1.07 A1 / 3  2.70 fm and R  1.4 A1 / 3  3.53 fm R  1.07 A1 / 3  4.26 fm and R  1.4 A1 / 3  5.57 fm

262

1 / 2

 c2 

Chapter 11 – Nuclear Physics (Problem 11-13 continued) (c)

208

11-14. U 

R  1.07 A1 / 3  6.34 fm and R  1.4 A1 / 3  8.30 fm

Pb :



3 1 e2 2 2 Z   Z  1 5 4 0 R



(Equation 11-2)

where Z = 20 for Ca and U  5.49MeV from a table of isotopes (e.g., Table of Isotopes 8th ed., Firestone, et al, Wiley 1998).



3 1 e2 2 R Z 2   Z  1 5 4 0 RU











 0.6 8.99  109 N m2 / C 2 1.60  1019 C 2.02  192 / 5.49  106 eV  6.13  1015 m  6.13 fm

11-15. (a) R  R0et  R0e at t  0 :

ln 2t / t1 / 2

(Equation 11-19)

R  R0  4000 counts / s

at t  10s :

R  R0e

ln 210 s  / t1 / 2

1000  4000e 1/ 4  e

 ln 2 10 s  / t1 / 2

 ln 2 10 s  / t1 / 2

ln 1 / 4    ln 2 10s  / t1/ 2  t1/ 2  5.0s

(b) at t  20s :

11-16. R  R0e

ln 2t / 2 min

R   4000counts / s  e

at t  0 :

ln 2  20 s  / 5 s

 200counts / s

R  R0  2000counts / s

(a) at t  4 min :

R   2000counts / s  e

(b) at t  6 min :

R   2000counts / s  e

(c) at t  8 min :

R   2000counts / s  e

 ln 2  4 min  / 2 min

 500counts / s

ln 2  6 min  / 2 min

 250counts / s

ln 2  8 min  / 2 min

 125counts / s

263



Chapter 11 – Nuclear Physics 11-17. R  R0et  R0eln 2t / t1/ 2 (a) at t  0 :

(Equation 11-19)

R  R0  115.0 decays / s

at t  2.25h :

R  85.2 decays / s

85.2 decays / s  115.0 decays / s  e

   2.25 h 

85.2 / 115.0   e  2.25h ln  85.2 / 115.0     2.25h 

   ln 85.2 / 115.0  / 2.25h  0.133h1 t1 / 2  ln 2 /   ln 2 / 0.133h1  5.21h (b)

dN  N dt



dN0  R0   N0 dt

(from Equation 11-17)





N0  R0 /   15.0 atoms / s  / 0.133h 1 1h / 3600s   3.11  106 atoms

11-18. (a)

226

t1 / 2  1620 y

Ra

R

dN ln 2 ln 2 N A m  N  N dt t1 / 2 t1 / 2 M







ln 2 6.022  1023 / mole 1g 

1620 y   3.16  10

7



s / y  26.025 g / mole 

1Ci  3.7  1010 s 1, or nearly the same.

(b) Q  M



226



Ra c 2   M



222



Rn c 2  M



4



He c 2 

 226.025402uc 2   222.017571uc 2  4.002602uc 2 





 0.005229uc 2  0.005229uc 2 931.5MeV / uc 2

 4.87MeV

264



 3.61  1010 s 1

Chapter 11 – Nuclear Physics 11-19. (a) R  

dN  ln 2 t / t  R0e   1 / 2 dt

when t  0,

R  R0  8000 counts / s

when t  10 min, e

(from Equation 11-19)

R  1000 counts / s  8000 counts / s  e

10 ln 2  / t1 / 2

10 ln 2  / t1 / 2

 1000 / 8000  1 / 8

10 ln( 2) / t1 / 2  ln 1 / 8 

t1 / 2 

10 ln  2  ln 1 / 8

 3.33 min

Notice that this time interval equals three half-lives. (b)   ln  2  / t1 / 2  ln  2  / 3.33 min = 0.208 min-1 (c) R  R0e

 t  ln 2  / t1 / 2

 R0et Thus, R  8000 counts / s  e

0.2081

 6500 counts / s

11-20. (a) and (b)

(c) Estimating from the graph, the next count (at 8 min) will be approximately 220 counts.

265

Chapter 11 – Nuclear Physics 11-21.

62

Cu is produced at a constant rate R0 , so the number of 62Cu atoms present is:





N  R0 /  1  e t (from Equation 11-26). Assuming there were no 62Cu atoms initially

present. The maximum value N can have is R0 /   N 0 ,



N  N 0 1  e  t



0.90 N 0  N 0 1  e e

 t  ln 2  / t1 / 2



 t  ln 2  / t1 / 2



 1  0.90  0.10

t ln  2  / t1 / 2  ln  0.10  t  10 ln  0.10  / ln  2   33.2 min 11-22. (a) t1 / 2  ln  2  /   ln  2  / 9.8  1010 y 1  7.07  108 y (Equation 11-22) (b) Number of

235

U atoms present is:







106 g 6.02  1023 atoms / mol 1.0 gN A N   2.56  1015 atoms M 235 g / mol







dN   N  9.8  1010 y 1 1 / 3.16  107 s / y 2.56  1015 atoms dt

 0.079 decays / s (c) N  N0et

(Equation 11-18)





N  2.56  1015 e





 9.81010 y 1 106 y

  2.558 1015

11-23. (a) t1 / 2  ln  2  /   ln  2  / 0.266 y 1  2.61y (Equation 11-22) (b) Number of N atoms in 1 g is:





23 1.0 gN A 1g  6.02  10 atoms / mol N   2.74  1022 atoms M 22 g / mol









dN   N  0.266 y 1 1 / 3.16  107 s / y 2.74  1022 atoms dt  2.3  1014 decays / s  2.3  1014 Bq

266





(Equation 11-17)

Chapter 11 – Nuclear Physics (Problem 11-23 continued) (c) 

dN   N 0 e  t dt

(Equation 11-19)









 0.266 y 1 1 / 3.16  107 s / y 2.74  1022 e

 0.266 y  3.5 y 

 9.1  1013 decays / s  9.1  1013 Bq

(d) N  N0et

(Equation 11-18)





N  2.74  1022 e

11-24. (a)

22





 0.266 y 1  3.5 y 

 1.08  1022

Na has an excess of protons compared with 23 Na and would be expected to decay

by β+ emission and/or electron capture. (It does both.) (b)

24

Na has an excess of neutrons compared with 23 Na and would be expected to decay

by β− emission. (It does.)

11-25. logt1 / 2  AE1 / 2  B

for t1 / 2  1010 s, for t1 / 2  1s,





(Equation 11-30)

E  5.4MeV

E  7.0MeV

log 1010  A  5.4 

1 / 2

from Figure 11-16

B

(i) 10  0.4303A  B log 1  A  7.0 

1 / 2

B

(ii) 0  0.3780 A  B  B  0.3780 A Substituting (ii) into (i),

10  0.4303A  0.3780 A  0.0523 A,

267

A  191, B  0.3780 A  72.2

Chapter 11 – Nuclear Physics 11-26. 237

Np

144 233

P

142

a 229

140

A

c

229

T

h

138

225

R

a 225

136

Ac

221

F

134

r

N

217

A

132

c 213

217

i

n

B

130 209

Tl

R

213

P

128

o 209

P

126

b 80

82

84

86

88

92

Z 11-27.

232 90

Th 

QM



228 88

Ra  



Th c 2  M

232



228



Ra c 2  M



4



He c 2

 232.038051uc2  228.031064uc2  4.002602uc2





 0.004385uc2 931.50MeV / uc 2  4.085MeV

The decay is a 2-particle decay so the Ra nucleus and the α have equal and opposite momenta.

  2m E  Ra  2mRa ERa where E  ERa  4.085MeV 2m E  2M Ra ERa  2M Ra  4.085  E  E 

228.031064  4.085MeV  M Ra   0.983 4.085MeV   4.01MeV  4.085MeV   M Ra  m 228.031064  4.002602

268

Chapter 11 – Nuclear Physics 11-28.

7 4

Be3  73 Li4  ve

(a) Yes, the decay would be altered. Under very high pressure the electrons are “squeezed” closer to the nucleus. The probability density of the electrons, particularly the K electrons, is increased near the nucleus making electron capture more likely, thus decreasing the half-life. (b) Yes, the decay would be altered. Stripping all four electrons from the atom renders electron capture impossible, lengthening the half-life to infinity.

11-29.

67

E .C .

Ga 

QM



67

Zn  ve



Ga c 2  M

67



67



Zn c 2

 66.9282uc2  66.972129uc2





 0.001075uc2 931.50MeV / uc 2  1.00MeV

11-30.

72

Zn  72 Ga     ve

QM



72



Zn c 2  M



72



Ga c 2

 71.926858uc2  71.926367uc2





 0.000491uc2 931.50MeV / uc 2  0.457MeV  457keV

11-31.

233

Np 

232

Np  n and

For n emission: Q  M



233

233

Np 

232

Up



Np c 2  M



232



Np c 2  mnc 2

 233.040805uc2  232.040022uc2  1.008665uc2

 0.007882uc2

Q  0 means M  products   M  For p emission: Q  M



233



Np c 2  M



233





U c2  mnc 2

232

 233.040805uc2  232.037131uc2  1.008665uc2

 0.004991uc2

269



Np ; prohibited by conservation of energy.

Chapter 11 – Nuclear Physics (Problem 11-31 continued)

Q  0 means M  products   M 

233





Np ; prohibited by conservation of energy.

11-32. 286

247

280

235

174

124

80

61

30

286

−

280

6

−

247

39

33

−

235

50

45

12

−

174

112

106

73

61

−

124

162

156

123

111

50

−

80

206

200

167

155

94

44

−

61

225

219

186

174

113

63

19

−

30

256

250

217

205

144

91

50

31

−

0

286

280

247

235

174

124

80

61

30

0

−

Tabulated γ energies are in keV. Higher energy α levels in Figure 11-19 would add additional columns of γ rays.

11-33. 8 Be  2 QM

 Be c 8

2

M



4



He c 2

 8.005304uc2  2  4.002602  uc 2





 0.000100uc2 931.50MeV / uc2  0.093MeV  93keV

11-34.

80

Br  80 Kr     ve and

For   decay: Q  M



80

80

Br 



Br c 2  M

80



Se     ve and

80

80

E.C .

Br 



Kr c 2

 79.918528uc2  79.916377uc2





 0.002151uc2 931.50MeV / uc 2  2.00MeV

270

80

Se  ve

Chapter 11 – Nuclear Physics (Problem 11-34 continued) For   decay: Q  M



80



Br c 2  M



80



Se c2  2mec 2

 79.918528uc2  79.916519uc2  2  0.511MeV 





 0.002009uc2 931.50MeV / uc 2  1.022MeV  0.85MeV

For E.C.: Q  M



80



Br c 2  M



80



Se c 2

 79.918528uc2  79.916519uc2





 0.002009uc2 931.50MeV / uc 2  1.87MeV

11-35. R  R0 A1 / 3 where R0  1.2 fm For 12C : R  1.2 12 

1/ 3

(Equation 11-3)

 2.745 fm  2.745 1015 m and the diameter = 5.490 10 15 m





9.00  109 1.6  1019 C ke2 Coulomb force: FC  2  2 r 5.490  1015 m

Gravitational force: FG  G



m2p r

2







2



 7.65 N

6.67  1011 1.67  1027 kg

5.490 10

15

m





2

 6.18  1036 N

2



The corresponding Coulomb potential is: UC  FC  r  7.65N 5.490  1015 m



 4.20  1014 J 1.60 1013 J / MeV  0.26MeV

The corresponding gravitational potential is:





UG  FG  r  6.18  1036 N 5.490  1015 m







 3.39  1050 J 1 / 1.60 1013 J / MeV  2.12 1037 MeV

The nuclear attractive potential exceeds the Coulomb repulsive potential by a large margin (50MeV to 0.26MeV) at this separation. The gravitational potential is not a factor in nuclear structure.

271

Chapter 11 – Nuclear Physics

11-36. The range R of a force mediated by an exchange particle of mass m is: R  / mc

(Equation 11-50)

mc2  c / R  197.3MeV fm / 5 fm  39.5MeV m  39.5MeV / c2

11-37. The range R of a force mediated by an exchange particle of mass m is: R  / mc

(Equation 11-50)

mc2  c / R  197.3MeV fm / 0.25 fm  789MeV m  789MeV / c 2

11-38. Nuclide

Last proton(s)

Last neutron(s)

 2s1 / 2 

0

1/2

1d3 / 2 

2

3/2

1

3/2

3

7/2

Si15

1d5 / 2 

Cl20

1d3 / 2 

71 31

Ga40

 2 p3 / 2 

3

 2 p1 / 2 

59 27

Co32

1 f7 / 2 

7

 2 p3 / 2 

73 32

Ge41

 2 p3 / 2 

29 14 37 17

6

j

4

2

4

4

1g9 / 2 

4

9/2

2

3/2

4

9/2

33 16 17

S

 2s1 / 2 

2

1d3 / 2 

81 38

Sr49

 2 p3 / 2 

1g9 / 2 

6

9

The nucleon configurations are taken directly from Figure 11-35, and the are those of the unpaired nucleon.

272

and j values

Chapter 11 – Nuclear Physics

11-39. Isotope

Odd nucleon

Predicted μ μ N

29 14

Si

neutron

−1.91

27 17

Cl

proton

+2.29

71 31

Ga

proton

+2.29

59 27

Co

proton

+2.29

73 32

Ge

neutron

−1.91

S

neutron

−1.91

Sr

neutron

−1.91

33 16 87 38

11-40. Nuclear spin of 14 N must be 1= because there are 3 mI states, +1, 0, and 1.

11-41.

36

S,

53

Mn,

82

Ge,

88

Sr,

94

Ru,

131

In,

145

Eu

11-42.

3

H

3

14

He

273

N

14

C

Chapter 11 – Nuclear Physics (Problem 11-42 continued)

n

11-43.

3 2

He,

15

40 20

Ca,

p

n

15

N

60 28

Ni,

124 50

204 82

Sn,

p

n

O

16

p

O

Pb

11-44. (a)

1p1/2

1p3/2

1s1/2 n

p 13

N

(b) j = ½ due to the single unpaired proton. (c) The first excited state will likely be the jump of a neutron to the empty neutron level, because it is slightly lower than the corresponding proton level. The j = 1/2 or 3/2, depending on the relative orientations of the unpaired nucleon spins.

274

Chapter 11 – Nuclear Physics (Problem 11-44 continued) (d)

1p1/2

1p3/2

1s1/2 n

p

First excited state. There are several diagrams possible.

11-45.

30 14

Si

j0

37 17

Cl

j  3/ 2

55 27

Co

j  7/2

90 40

Zr

j0

107 49

In

j  9/ 2

 H c  M  H c  M  H c  2  2.014102uc   3.016049uc  1.007825uc  0.004330uc  931.5MeV / uc   4.03MeV Q  M  He  c  M  H  c  M  He  c  M  H  c

11-46 (a) Q  M

 H c 2

2

M

2

2

3

2

3

1

2

2

2

(b)

2

2

2

2

2

2

4

2

1

2

 3.016029uc2  2.014012uc2  4.002602uc2  1.007825uc2





 0.019704uc2 931.5MeV / uc 2  18.35MeV

275

Chapter 11 – Nuclear Physics (Problem 11-46 continued)

 Li  c

(c) Q  M

6

2

 mnc 2  M

 H c 3

2

M



4



He c 2

 6.01512uc2  1.008665uc2  3.016049uc2  4.002602uc2





 0.005135uc 2 931.5MeV / uc 2  4.78MeV

 H c

11-47. (a) Q  M

3

2

 

 M 1H c 2  M

 He c 3

 mN c 2

2

 3.016049uc2  1.007825uc2  3.016029uc2  1.0078665uc2





 0.000820uc2 931.5MeV / uc 2  0.764MeV

(b) The threshold for this endothermic reaction is: Eth 

 (c) Eth 

11-48.

14

mM Q M

(Equation 11-61)

3.016049uc 2  1.007825uc 2 0.764MeV  3.05MeV 1.007825uc 2

1.007825uc 2  3.016049uc 2 3.016049uc 2

0.764MeV  1.02MeV

N  2 H  16O

O  16O  

O  14 N  2 H

16

O  15 N  p

16

16

Possible products:

O  12C  

16

11-49. (a)

O  15O  n

16

C  , p  15 N

12

QM

 Cc 12

2

M



4



He c 2  M

 N c 15

2

 mpc2

 12.000000uc2  4.002602uc2  15.000108uc2  1.007825uc2





 0.005331uc2 931.5MeV / uc2  4.97 MeV

(b)

O  d, p  17O

16

QM

 O c 16

2

M

 H c 2

2

M

 O c 17

2

 mpc2

 15.994915uc2  2.014102uc2  16.999132uc2  1.007825uc2





 0.002060uc2 931.5MeV / uc 2  1.92MeV

276

Chapter 11 – Nuclear Physics 75

11-50. The number of

N

As atoms in sample N is:







1cm  2cm  30 m  104 cm /  m 5.73g / cm3 6.02  1023 atoms / mol V  NA  M 74.9216 g / mol

 2.76  1020

75

As atoms 75

The reaction rate R per second per R I

As atoms is:

(Equation 11-62



 4.5  1024 cm2

75



As 0.95  1013 neutrons / cm2 s



 4.28  1011 s 1

Reaction rate = NR







 .76  1020 atoms 4.28  1011 / s atom  1.18  1010 / s

11-51. (a)

23

Ne  p, n  23 Na

22

(b)

11

B  , p  14C

(c)

29

Si  , d  31P

11-52. (a)

14

(e)

14

11-53.

C

N

(b) n 160

(f)

Ne  d, n  23 Na

20

14

N  n, p  14C

13

32

P  p, d  31P

32

Er

C  d , p  14C

Ni

(d) α

(g) 3 H

(i) p

(c)

58

Q  m p  mn  md c2  1.007276u  1.008665u  2.013553u

 0.002388u

(See Table 11-1.)





Q   0.002388u  931.5MeV / uc 2 c 2  224MeV

277

F  , n  23 Na

Si  n, d  31P



Chapter 11 – Nuclear Physics 11-54. P 

dW dN E dt dt

dN P 500  106 J / s 1eV     6 dt E 200  10 eV / fission  1.60  1019 J

 19   1.56  10 fissions / s 

11-55. The fission reaction rate is:

R  N   R  0 k N

(see Example 11-22 in More section)

k N  R  N  / R  0 N log k  log  R  N  / R  0 

N

log  R  N  / R  0   log k

(a) For the reaction rate to double R  N   2 R  0  : (b) For R  N   10 R  0  : (c) For R  N   100 R  0  :

N

N

log10  24.2 log1.1

N

log100  48.3 log1.1

(d) Total time t  N 1ms   N ms : (a) 7.27ms

(b) 24.2ms

(e) Total time t  N 100ms   100 N ms : (a) 0.727ms

11-56.

235 92

U143

log 2  7.27 log1.1

134  101 40 Zr61  52 Te82  n    101 Nb  133 Sb  2n  41 60 51 82 n  101 132  43 Tc58  49 In83  3n   102 130  45 Rh57  47 Ag83  4n

278

(c) 48.3ms

(b) 2.42ms

(c) 4.83ms

Chapter 11 – Nuclear Physics J  1MeV   1 fusion  11-57. 500MW   500   1.78  1014 fusions / s   13 s  1.60  10 J  17.6MeV  

Each fusion requires one 2 H atom (and one 3 H atom; see Equation 11-67) so 2 H must be provided at the rate of 1.78  1014 atoms / s.

238

11-58. The reactions per R I

U atom is:

(Equation 11-62)

 1m2   0.02  1024 cm2 atom 5.0  1011 n / m2  4 2   1.00  1018 / atom  10 cm 





238

The number N of

N



U atoms is:

 5.0 g   6.02  1023 atoms / mol 

Total

 1.26  1022

238.051g / mol

238

U atoms

239

U atoms produced = RN







 1.00  1018 / atom 1.26  1022 atoms  1.26  104

 

 

11-59. Q1  M 1H c 2  M 1H c 2  M

 H c 2

239

U atoms

2

 1.007825uc2  1.007825uc2  2.014102uc2





 0.001548uc2 931.50MeV / uc 2  1.4420MeV Q2  M



2



 

He c 2  M 1H c 2  M

 He  c 3

2

 2.014102uc2  1.007825uc2  3.016029uc2





 0.005898uc2 931.50MeV / uc 2  5.4940MeV

    2  3.016029uc   4.002602uc  2 1.007825uc   0.013806uc  931.50MeV / uc   12.8603MeV

Q3  M

 He c 3

2

M 2

2

 He c 3

2

M



4

He c 2  2m 1H c 2

2

2

2

Q  Q1  Q2  Q3  1.4420MeV  5.4940MeV  12.8603MeV  19.80MeV

279

Chapter 11 – Nuclear Physics 11-60. Total power = 1000MWe / 0.30  3333MW





 3.33  109 J / s 1MeV / 1.60  1013 J  2.08  1022 MeV / s

(a) 1 day  8.64  104 s





Energy/day  2.08  1022 MeV / s 8.64  104 s / day  1.80  1027 MeV / day

The fission of 1kg of 235

1kg

(b) 1kg

235

U provides 4.95 1026 MeV (from Example 11-19)





U / day  1.80  1027 MeV 4.95  1026 MeV / kg  3.64kg / day

U / year  3.64 kg

235

U / day  365days / year   1.33  103 kg / year

235

(c) Burning coal produces





3.15  107 J / kg 1MeV / 1.60  1013 J  1.97  1020 MeV / kg coal

Ratio of kg coal needed per kg of For 1 day: 3.64kg

235



235

U is:

4.95  1026 MeV / kg 235U  2.51  106 20 1.97  10 MeV / kg coal



U 2.51  106  9.1  106 kg This is about 10,000 tons/day,

the approximate capacity of 100 railroad coal hopper cars. For 1 year: 9.12  106 kg / day  365days / year   3.33 109 kg / year

11-61.   H 2O   1000kg / m3, so (a) 1000kg :







106 g 6.02  1023 molecules H 2O / mol  2H / molecule  0.00015 2 H 18.02 g / mol

 1.00  1025 2 H atoms Each fusion releases 5.49MeV.

   5.49  10 MeV 1.60  10

Energy release = 1.00  1025  5.49MeV   5.49  1025 MeV 25

13



J / MeV  8.78  1012 J

(b) Energy used/person (in 1999) = 3.58  1020 J 5.9 109 people

 6.07  1010 J / person y

280



Chapter 11 – Nuclear Physics (Problem 11-61 continued) Energy used per person per hour = 6.07  1010 J / person y 

1y 8760h

 6.93  106 J / person h At that rate the deuterium fusion in 1m3 of water would last the “typical” person

8.80  1012 J  1.27  106 h  145 y 6 6.93  10 J / person h

11-62. (a) Q  M



235



U c 2  mnc 2  M





Cd c 2  M

120



110



Ru c 2  5mnc 2

 235.043924uc2  1.008665uc2  119.909851uc2  109.913859uc2



5 1.008665uc 2







 1.186uc2 931.5MeV / uc 2  1.10  103 MeV

(b) This reaction is not likely to occur. Both product nuclei are neutron-rich and highly unstable.

11-63. The original number N0 of 14 C nuclei in the sample is:





N0  15g  6.78  1010 nuclei / g  1.017  1012 where the number of

of C was computed in Example 11-27. The number N of N  N 0 e  t  N 0 e



ln 2 t / t1 / 2 



 1.017  1012 e



C nuclei per gram

C present after 10,000y is:

(Equation 11-18)

ln 210000 / 5730 

R   N  ln 2 t1 / 2  N

14

14

 3.034  1011

(Equation 11-19)



 ln 2 5730 y  1y 3.16  107 s 3.034  1011



 1.16 decays / s

11-64. If from a living organism, the decay rate would be:

15.6decays / g min 175g   20.230decays / min

(from Example 11-27)

The actual decay rate is: 8.1decays / s  60s / min   486decays / min

281

Chapter 11 – Nuclear Physics (Problem 11-64 continued) 2

486decays / min 1  2   20, 230decays / min (from Example 11-27)  

2n  20, 230 / 486 n ln  2   ln  20, 230 / 486  n  ln  20, 230 / 486  ln(2)  5.379 half lives Age of bone =  5.379 half lives  5730 y / half life   30,800 y

t

11-65. t1 / 2



87

t1 / 2 ln 1  N D / N P  ln  2 

(Equation 11-92)



Rb  4.88  1010 y and N P / N D  36.5

t

4.88  1010 y ln 1  1 / 36.5  1.90  109 y ln  2 

11-66. The number of X rays counted during the experiment equals the number of atoms of interest in the same, times the cross section for activation  x , times the particle beam intensity, where I = proton intensity = 250nA   eC / proton   1.56  1012 protons / s 1

 x  650b  650  1024 cm2 m  mass  0.35mg / cm2  0.00001  3.5  106 mg / cm2 n  number of atoms of interest  mN A / A t  exposure time detector efficiency  0.0035   overall efficiency  0.60  detector efficiency

N  I x

mN A t A

  250  109 C / s 24 2 N   650  10 cm 19  1.60  10 C / proton 



282



Chapter 11 – Nuclear Physics (Problem 11-66 continued)







 0.35  103 g / cm2  0.00001 6.02  1023 mol 1   80 g / mol 

   

 15 min  60s/min  0.60  0.0035  5.06 104 counts in 15 minutes

t

11-67. t1 / 2



232

NP



232

ND



208

t1 / 2 ln 1  N D / N P  ln  2 

(Equation 11-92)



Th  1.40  1010 y



  1.066 10



4.11g 6.02  1023 atoms / mol



0.88 g 6.02  1023 atoms / mol

Th 

Pb 

232.04 g / mol



208 g / mol

22

  2.547  10

atoms

21

atoms

N D N P  2.547  1021 1.066  1022  0.2389

t

11-68. f 

1.40  1010 y ln 1  0.2389  4.33  109 y ln  2 

E 2   z  p B  h h

(a) For Earth’s field: f 







2  2.79 N   3.15  108 eV / T 1 N  0.5  104 T



4.14  1015 eV s

 2.13  103 Hz  2.12 kHz





(b) For B = 0.25T: f  2.12  103 Hz 0.25T 0.5  104 T  1.06  107 Hz  10.6 MHz





(c) For B= 0.5T: f  2.12  103 Hz 0.5T 0.5  104 T  2.12  107 Hz  21.2 MHz

283

Chapter 11 – Nuclear Physics

11-69. (a) N



14

12

C

3

12

C



12 10 

6



3 1.60  10

(b) mass

19

C



 1.50  1016

C ratio  1500 1.50  1016  1013

1.50 10 C

15

12



C / s 10 min  60s / min 





atoms / min  75 min 12  1.66  1027 kg



0.015

 1.49  107 kg  1.49  104 g  0.149mg (c) The

14

C

12

sample living

C ratio in living C is 1.35  1012. 14

C 14 C

C 1013 0.10  1     12 12 C 1.35  10 1.35  2  12

n

where n = # of half-lives elapsed. Rewriting as (see Example 11-28)

2n 

1.35  13.5 0.10

n ln  2  ln 13.5

 n  ln 13.5 / ln  2   3.75

age of sample = 3.75t1 / 2  3.75  5730 y   2.15 104 y

11-70. If from live wood, the decay rate would be 15.6 disintegrations/g•min. The actual rate is 2.05 disintegrations/g•min. 2

2.05decays / g min 1  2   15.6decays / g min (from Example 12-13)  

2n  15.6 / 2.05

n ln  2   ln 15.6 / 2.05 n  ln 15.6 / 2.05 ln  2   2.928 half lives of 14 C Age of spear thrower = nt1 / 2   2.928 5730 y   16,800 y

284

Chapter 11 – Nuclear Physics 11-71. Writing Equation 11-14 as:

M Z, A c2

Zm2p

a1 A a2 A2 / 3 a3 A

A Z mnc 2

1/ 3 2

z

a4 A 2Z

2

A

1

a5 A

1/ 2

c2

and differentiating,

M Z

mp

mn

0

mp

mn

2a3 A

0

mp

mn

4a4

mp

Z

mn

2a3 A

1/ 3

2a3 A

4a4 8a4 A

1/ 3

1/ 3

Z

Z

2a4 A 2Z

1/ 3

where a3

1

1

8a4 A 1Z

4a4

2a3 A

2 A

8a4 A

1

Z

0.75 and a4

93.2

1.008665 1.007825 931.5MeV / uc 2

(a) For A = 27: Z

2 0.75 27

1/ 3

8 93.2 27

4 93.2 1

13.2

Minimum Z = 13 (b) For A = 65: Computing as in (a) with A = 65 yields Z = 31.5. Minimum Z = 29. (c) For A = 139: Computing as in (a) with A = 139 yields Z = 66. Minimum Z = 57.

11-72. (a) R

0.31 E 3 / 2

(b) R g / cm2 (c) R cm

0.31 5MeV

R cm

3/ 2

3.47cm

3.47cm 1.29 10 3 g / cm3

R g / cm2

4.47 10 3 g / cm2

4.47 10 3 g / cm2

2.70 g / cm3

1.66 10 3 cm

11-73. For one proton, consider the nucleus as a sphere of charge e and charge density c

3e 4 R3. The work done in assembling the sphere, i.e., bringing charged shell dq

up to r, is: dU c

Uc

k

2 c

16 2 R5 15

k

c

4 r3 3

c

4 r 2 dr

1 and integrating from 0 to R yields: r

3 ke2 5 R

For two protons, the coulomb repulsive energy is twice Uc, or 6ke 2 5 R.

285

Chapter 11 – Nuclear Physics

11-74. The number N of

dN dt

144

Nd atoms is: N

N

11-75. R

ln 2

R0e

144 g / mol

2.25 1023 atoms

dN dt N 2.36s

t1 / 2

53.94 g 6.02 1023 atoms / mol

1

ln 2 1.05 10

t ln 2 / t1 / 2

2.25 1023 23

s

1

1.05 10

6.61 1023 s

23

s

1

2.09 1015 y

(from Equation 11-19)

(a) At t = 0: R = R0 = 115.0 decays/min At t = 4d 5h = 4.21d: R = 73.5 decays/min ln 2 4.21d / t1 / 2

73.5decays / min = 115.0decays / min e 73.5 / 115.0

e

ln 73.5 / 115.0

ln 2 4.21d / t1 / 2

ln 2 4.21d / t1 / 2 ln 2 2.41d / ln 73.5 / 115.0

t1 / 2

(b) R 10decays / min = 115.0decays / min e t

ln 10 / 115.0 6.52d ln 2

t

ln 2 t / 6.52 d

23.0d

2.5decays / min = 115.0decays / min e

(c) R

6.52d

ln 2.5 / 115.0 6.52d ln 2

ln 2 t / 6.52 d

36.0d

(because t = 0)

This time is 13 days (= 2t1/2) after the time in (b).

11-76. For

227

For

223

Th : t1 / 2 Ra : t1 / 2

18.72d

(nucleus A)

11.43d

(nucleus B)

At t = 0 there are 106 Th atoms and 0 Ra atoms (a) N A NB

N 0 Ae N0 A B

At

A

(Equation 11-18) e

At

e

Bt

N0Be

Bt

A

286

(Equation 11-26 on the Web page)

Chapter 11 – Nuclear Physics (Problem 11-76 continued) ln 2 15 d / 18.72 d

106 e

NA

5.74 105

106 ln 2 18.72d

NB

ln 2 1 / 11.43d 1 / 18.72d

(b) N A

N0 A

At

N B means N 0 Ae

B

ln 2 15 / 18.72

e

A

e

At

e

e

ln 2 15 / 11.43

0

2.68 105

Bt

A

Cancelling N0A and rearranging, B

A

1 e

A

t

B

A

ln 2 A

0.0370d

18.82d B

ln

A

1

ln 2

1 B

B

A

11.43d

0.0606d

1

t

A

B

ln

t

A

1

A

B

0.0606 0.0370 1 0.0370

ln

A

0.0370 0.0606

43.0d

/

11-77. (a)

6.582 10

hf

(b) Er

2

16

eV s 0.13 10 9 s

0.12939MeV

2Mc 2

2M

4.71 10 8 MeV

191

5.06 10 6 eV

2

(Equation 11-47)

I c2

4.71 10 2 eV

(c) (See Section 1-5) The relativistic Doppler shift Δf for either receding or approaching is: f

v c

f0

E

v c

v

c

E

h f

hf 0

E

5.06 10 6 eV

3.00 108 m / s

0.12939MeV 106 eV / MeV

287

0.0117m / s

1.17cm / s

Chapter 11 – Nuclear Physics 11-78. B

ZM 1H c 2

For 3 He :

B

Nmnc 2

M Ac 2

2 1.007825uc 2

1.008665uc 2

3.016029uc 2

0.008826uc 2 931.5MeV / uc 2

For 3 H :

B 1.007825uc 2

7.72MeV

2 1.008665uc 2

3.016029uc 2

0.009106uc 2 931.5MeV / uc 2

R Uc

R0 A1 / 3

1.2 fm31 / 3

ke2 / R

8.48MeV 15

1.730 fm 1.730 10 8.32 105 eV

ke / R eV

m

0.832MeV

or about 1/10 of the binding

energy.

11-79. For 47Ca : B

M

46

Ca c 2

mn c 2

M

47

Ca c 2

45.953687uc2 1.008665uc2 46.954541uc2 0.007811uc 2 931.5MeV / uc 2

7.28MeV

For 48Ca : B

M

47

Ca c 2

mn c 2

M

48

Ca c 2

45.954541uc2 1.008665uc2

47.952534uc2

0.010672uc 2 931.5MeV / uc 2

9.94MeV

Assuming the even-odd

Ca to be the “no correction” nuclide, the average magnitude of

47

the correction needed to go to either of the even-even nuclides 46Ca or

48

Ca is

approximately B – average binding energy of the odd neutron, 9.94MeV

7.28MeV / 2

MeV and for

8.61MeV . So the correction for 46Ca is 8.16 – 7.28 = 0.88

Ca is 9.94 – 8.16 = 1.78 MeV, an “average” of about 1.33 MeV. The

48

estimate for a5 is then: a5 A 1 / 2

1.33MeV

a5

30% below the accepted empirical value of a5 = 12.

288

1.33 / 48 1 / 2

9.2 . This value is about

Chapter 11 – Nuclear Physics 11-80. For a nucleus with I > 0 the α feels a centripetal force

mv2 / r

Fc

dV / dr where r

distance of the

from the nuclear center. The

ln r and becomes larger (i.e., more negative) as r

corresponding potential energy V

increases. This lowers the total energy of the α near the nuclear boundary and results in a wider barrier, hence lower decay probability.

11-81. (a) R

R0 A1 / 3 where R0

R

141

R

92

(b) V

1.2 f

Ba

1.2 fm 10

Kr

1.2 fm 10

15

15

(Equation 11-3)

m / fm 141

m / fm 92

1/ 3

1/ 3

6.24 10 5.42 10

15

8.998 109 N m2 / C 2 56 1.60 10

kq1q2 / r

6.24 10

2.49 108 eV

15

m 5.42 10

15

15

m

m

19

C 36 1.60 10

m 1.60 10

19

19

C

J / eV

249MeV

This value is about 40% larger than the measured value.

11-82. (a) In the lab, the nucleus (at rest) is at x = 0 and the neutron moving at vL is at x. xCM V

M 0 M

mx m

mx M m

vL

x and V dt

dxm dt

m x / dt M

m

mvL M m

(b) The nucleus at rest in the lab frame moves at speed V in the CM frame before the collision.

In an elastic collision in the CM system, the particles reverse their

velocities, so the speed of the nucleus is still V, but in the opposite direction. (c) In the CM frame in the nucleus velocity changes by 2V. This is also the change in the lab system where the nucleus was initially at rest. It moves with speed 2V in the lab system after the collision. (d)

1 M 2V 2

1/ 2

2mvL 1 M 2 M m

1 2 mvL 2

289

4mM M

m

2

Chapter 11 – Nuclear Physics (Problem 11-82 continued) Before collision: Ei

1 2 mvL 2 1 2 mvL 2

After collision: E

1 2 mvL 2

4mM M

m

11-83. At the end of the two hour irradiation the number of R0

N

For

32

t

1 e

M

P and

56

m

2

Mn atoms are given by

I (Equation 11-62)

P: 24

0.180 10

N0

R0t1 / 2 1 e ln 2 1.29 10

56

32

from Equation 11-26 where R0

R0

For

4mM

Ei 1

2

cm2 1012 neutrons / cm2 s

1.80 10

ln 2 t / t1 / 2

13

1.80 10

13

32

P atoms / s per 31 P

/ s 3600s / h 342.2h ln 2

9 32

1 e

ln 2 2 h / 342.2 h

P atoms / 31 P atom

Mn :

R0

N0

13.3 10

1.33 10

24

cm2 1012 neutrons / cm2 s

11

/ s 3600s / h 2.58h

1 e

ln 2 7.42 10

8 56

1.33 10

11

56

Mn atoms / s per 55 Mn

ln 2 2 h / 2.58 h

Mn atoms / 55 Mn atom

(a) Two hours after the irradiation stops, the activities are: dN dt

For

N 0e 32

56

N 0 ln 2 t1 / 2

e

ln 2 t / t1 / 2

P:

dN dt For

t

1.29 10

9

ln 2 4

14.26d 8.64 10 s / d

e

ln 2 2 h / 342.2 h

Mn :

290

7.23 10

16

decays / 31 P atom

Chapter 11 – Nuclear Physics (Problem 11-83 continued) 8

7.42 10

dN dt

ln 2

2.58h 3600s / h

e

ln 2 48 h / 2.58 h

1.39 10

17

decays / 56 Mn atom

The total activity is the sum of these, each multiplied by the number of parent atoms initially present.

11-84. Q E N

200MeV / fission. 7.0 1019 J

NQ

N 200MeV / fission 1.60 10

7.0 1019 J 200MeV / fission 1.60 10 235

Number of moles of

13

2.19 1030 6.02 1023

Fissioned mass/y = 3.63 106 moles 235 g / mole That is 3% of the mass of the

235

J / MeV

2.19 1030 fissions

J / MeV

U needed = N / N A

13

8.54 108 g

3.63 106 moles 8.54 105 kg

U atoms needed to produce the energy consumed.

Mass needed to produce 7.0 1019 J

8.54 105 kg / 0.03

2.85 107 kg.

Since the energy conversion system is 25% efficient: Total mass of

11-85. The number of

87

235

U needed = 1.14 108 kg.

Sr atoms present at any time is equal to the number of

have decayed, because

87

Rb nuclei that

Sr is stable.

N Sr

N 0 Rb

N Rb

0.010

N 0 Rb N Rb

N Sr

N Sr

87

N Rb

N Rb

N Sr

N Rb

N 0 Rb N Rb

1N

1 1.010

and also N Rb N0 Rb ln 2 t t1 / 2

t

e

ln 2 t / t1 / 2

1 / 1.010

ln 1 / 1.010

t1 / 2 ln 1 / 1.010 / ln( 2)

4.9 1010 y ln 1 / 1.010 / ln( 2)

291

7.03 108 y

Chapter 11 – Nuclear Physics 11-86. (a) Average energy released/reaction is: 3.27 MeV

P

E t

4W

4J / s

N 3.65MeV 1.60 10

4J / s 3.65MeV / reaction 1.60 10

N

13

4.03MeV / 2 13

3.65MeV

J / MeV

J / MeV

6.85 1012 reactions / s

Half of the reactions produce neutrons, so 3.42 1012 neutrons / s will be released. 0.10 3.42 1012

(b) Neutron absorption rate

3.42 1011 neutrons / s

Energy absorption rate = 0.5MeV / neutron 3.42 1011 neutrons / s 1.60 10

13

J / MeV

2.74 10 2 J / s

Radiation dose rate =

2.74 10 2 J / s / 80kg 3.42 10 2 rad / s 4

3.42 10 3 rad / s

100rad / J / kg 0.137 rem / s

493 rem / h

(c) 500 rem, lethal to half of those exposed, would be received in: 500rem / 492rem / h

11-87. R t

N0 I 1 e

For Co: N 0

For Ti: N 0

t

1.02h

(Equation 11-86) 35Bq

19 10

24

cm

2

12

3.5 10 / s cm

2

1 e

1.319 10

115 Bq 0.15 10

11-88. The net reaction is: 5 2 H

24

cm

2

3

12

3.5 10 / s cm

He

4

He

1

H

2

1 e

0.120 2

2

2.00 1017 atoms

1.03 1015 atoms

n 25MeV

Energy release / 2 H

5MeV (assumes equal probabilities)

4 water

2 1.007825

4000 g

6

15.994915 g / mol

222.1 moles

4 of water thus contains 2(222.1) moles of hydrogen, of which 1.5 10 4 is 2 H , or

Number of 2 H atoms = 292

Chapter 11 – Nuclear Physics (Problem 11-88 continued) 2 222.1 moles 6.02 1023 atoms / mol 1.5 10

Total energy release = 4.01 1022 5MeV

4

4.01 1022

2.01 1023 MeV

3.22 1010 J

Because the U.S. consumes about 1.0 1020 J / y, the complete fusion of the 2 H in 4 of water would supply the nation for about 1.01 10 2 s 10.1ms

2hc / Mc 2

11-89. (a)

hc

hc

E

2

2

2

hc

E 2 2hc E hc Mc 2

Ep

2E 2 Mc 2

E 2 Mc 2 E p / 2

E

Ef E

Ei

M

4mM

Ei 1

m

Mc 2 E p / 2

E

M

m

2

1 m/M

2

5.7MeV 938.28MeV / 2

(b) E

1/ 2

Ei

2

4m / M

4mM

Ei

E2 hc

4mM

Ei

M

m

2

which is Equation 11-82 in More section. 1/ 2

51.7 MeV

(c) O 14N (M)

x vL CM neutron (m)

The neutron moves at vL in the lab, so the CM moves at v

vL mN / mN

M

toward

the right and the 14N velocity in the CM system is v to the left before collision and v to the right after collision for an elastic collision. Thus, the energy of the nitrogen nucleus in the lab after the collision is:

E

14

N

1 M 2v 2

2

2Mv

2

mvL 2M m M

293

2

Chapter 11 – Nuclear Physics (Problem 11-89 continued)

2Mm mvL2 m M

4Mm

2

1 2 mvL 2

2

m M

4 14.003074u 1.008665u 1.008665u 14.003074u

5.7 MeV

2

= 1.43 MeV 14.003074uc 2 931.5MeV / uc 2 1.43MeV / 2

(d) E

1/ 2

96.5MeV

11-90. In the lab frame: photon

E

hv

p

hv / c

deuteron, M at rest

pc E /c

In CM frame:

For E

photon

deuteron, M

E p

EK

1 / 2Mv 2

p

2MEK

pc E/c

p 2 / 2M

pc in CM system means that a negligible amount of photon energy goes to

recoil energy of the deuteron, i.e.,

p2 2M E

pc pc

E 2Mc 2

or

pc 2Mc

2

pc

2

2 1875.6MeV

pc

2Mc 2

3751.2 MeV

(see Table 11-1)

In the lab, that incident photon energy must supply the binding energy B = 2.22 MeV plus the recoil energy EK given by: EK

p 2 / 2M 2.22 MeV

2

pc / 2Mc 2

2

B / 2Mc 2

2

2 1875.6 MeV

0.0013MeV

294

Chapter 11 – Nuclear Physics (Problem 11-90 continued) So the photon energy must be E

2.22MeV

0.001MeV

2.221MeV , which is much

less than 3751 MeV.

ZM 1H c 2

11-91. (a) B

For 7 Li :

Nmn c 2

M Ac 2

3 1.007825uc 2

B

(Equation 11-11)

4 1.008665uc 2

7.016003uc 2

0.042132uc 2 931.5MeV / uc 2 For 7 Be :

4 1.007825uc 2

B

39.25MeV

3 1.008665uc 2

0.040367uc 2 931.5MeV / uc 2

7.016928uc 2

37.60MeV

B 1.65MeV

For 11B :

B

5 1.007825uc 2

6 1.008665uc 2

11.009305uc 2

0.0081810uc 2 931.5MeV / uc 2 For 11C :

B

6 1.007825uc 2

76.21MeV

5 1.008665uc 2

11.011433uc 2

0.078842uc 2 931.5MeV / uc 2

B

73.44MeV

2.77MeV

For 15 N :

7 1.007825uc 2

B

8 1.008665uc 2

0.123987uc 2 931.5MeV / uc 2 For 15O :

B

8 1.007825uc 2

115.5MeV

7 1.008665uc 2

0.120190uc 2 931.5MeV / uc 2

B (b)

B

a3

For A

15.000108uc 2

15.003065uc 2 112.0MeV

3.54MeV Z2 A 7; Z

For A 11; Z

1/ 3

4: 6:

a3 a3 a3

B

Z2 A

1.65MeV 7 7

1/ 3

2.77 MeV 11 11

295

1/ 3

0.45MeV 1/ 3

0.56MeV

Chapter 11 – Nuclear Physics (Problem 11-91 continued) For A 15; Z

8:

1/ 3

3.54 MeV 15 15

a3

0.58MeV

0.53MeV

a3

0.75MeV .

These values differ significantly from the empirical value of a3

11-92. (a) Using M Z Z

mn

mp

2a3 A

1 2A

0 from Problem 11-71. 4

4

1/ 3

8a4 A

mn

mp

1

(b) & (c) For A

a3 A 29 :

1

where a3

2a4

Z

93.2MeV / c 2

a4

A 1 mn m p 4a4 2 1 a3 A2 / 3 4a4

4a4

1/ 3

0.75MeV / c 2 ,

29 1 2

1.008665 1.007276 931.5 / 4 93.2 1 0.75 29

The only stable isotope with A = 29 is

29 14

2/3

14

/ 4 93.2

Si

For A =59: Computing as above with A = 59 yields Z = 29. The only stable isotope with A = 59 is

59 27

Co

For A = 78: Computing as above with A = 78 yields Z = 38. stable. Stable isotopes with A = 78 are

78 34

The only stable isotope with A = 119 is

119 50

Se and

78 38

78 38

Sr is not

Kr .

Sn .

For A = 140: Computing as above with A = 140 yields Z = 69. stable. The only stable isotope with A = 140 is

140 69

Tm is not

140 58

Ce .

The method of finding the minimum Z for each A works well for A ≤ 60, but deviates increasingly at higher A values.

296

Chapter 11 – Nuclear Physics 11-93. (a)

(b) j = 3/2

n

1d5/2

1d5/2

1p1/2

1p1/2

1p3/2

1p3/2

1s1/2

1s1/2 n

p

11

Ground state B

(c) j = 5/2

1d5/2

1p1/2

1p3/2

1s1/2 n

2nd excited state 11B

j = 1/2

p

297

1st excited state 11B

p

Chapter 11 – Nuclear Physics (Problem 11-93 continued) (d)

17

n

O Ground state (j = 5/2)

1d5/2

1d5/2

1p1/2

1p1/2

1p3/2

1p3/2

1s1/2

1s1/2

p

n

(e) 2s1/2 1d5/2

1p1/2

1p3/2

1s1/2 n

17

p

O nd 2 excited state (j = 1/2)

298

17

O 1st excited state (j = 1/2)

p

Chapter 11 – Nuclear Physics 11-94. (a) Data from Appendix A are plotted on the graph. For those isotopes not listed in Appendix A, data for ones that have been discovered can be found in the reference sources, e.g., Table of Isotopes, R.B. Firestone, Wiley – Interscience (1998). Masses for those not yet discovered or not in Appendix A are computed from Equation 11-14 (on the Web). Values of M(Z, 151) computed from Equation 11-14 are listed below. Because values found from Equation 11-14 tend to overestimate the mass in the higher A regions, the calculated value was adjusted to the measured value for Z = 56, the lowest Z known for A = 151 and the lower Z values were corrected by a corresponding amount. The error introduced by this correction is not serious because the side of the parabola is nearly a straight line in this region. On the high Z side of the A = 151 parabola, all isotopes through Z = 70 have been discovered and are in the reference cited.

Z

N

M Z , 151 [Equation 11-14]

50

101

152.352638

151.565515

51

100

152.234612

151.447490

52

99

152.122188

151.335066

53

98

152.015365

151.228243

54

97

151.914414

151.127292

55

96

151.818525

151.031403

56

95

151.728507

150.941385*

* This value has been measured.

299

M Z , 151

[adjusted]

Chapter 11 – Nuclear Physics (Problem 11-94 continued)

(b) The drip lines occur for: protons: M Z , 151

M Z 1, 150

neutrons: M Z , 151

M Z , 150

mp mn

0 0

Write a calculator or computer program for each using Equation 11-14 (on Web page) and solve for Z.

11-95. (a) M Z , A

Zmp

Nmn

a1 A a2 A2 / 3

a3 A

1/ 3

a4 A 2Z

2

A

1

a5 A

1/ 2

from Equation 11-14 on the Web page.

15.67 310 M 126, 310

126m p 184mn

17.23 310

0.75 126

2

310

313.969022u

(b) For β− decay: (126, 310) → (127, 310) + β− + ve Computing M(127, 310) as in (a) yields 314.011614u.

300

1/ 3

93.2 310 2 126 12 310

M 126, 310

2/3

1

2

310

12 310

1/ 2

Chapter 11 – Nuclear Physics (Problem 11-95 continued) Q

M 126, 310 c 2

M 127, 310 c 2

313.969022uc2 314.011614uc2 0.042592u 931.5MeV / uc 2

39.7 MeV

Q < 0, so β− decay is forbidden by energy conservation. For β+ decay: (126, 310) → (125, 310) + β+ + ve Computing M(125, 310) as in (a) yields 313.923610u. Q

M 126, 310 c 2

M 125, 310 c 2

2mec 2

313.969022uc2 313.923610uc2 1.022MeV 41.3MeV β+ decay and electron capture are possible decay modes. For α decay: (126, 310) → (124, 310) + α Computing M(124, 310) as in (a) yields 309.913540u. 313.969022uc 2

Q

309.913540uc 2

4.002602uc 2

49.3MeV α decay is also a possible decay mode. 11-96. (a) If the electron’s kinetic energy is 0.782MeV, then its total energy is:

E E2

p

mec2

0.782MeV pc

E2

2

2

me c 2

me c 2

0.511MeV

1.293MeV

(Equation 2-32)

1/ 2

2

c 2

1.293MeV

0.782MeV

0.511MeV

2

1/ 2

c

1.189MeV / c (b) For the proton p 1.189MeV / c also, so Ekin

p 2 / 2m

pc

1.189MeV

2

2

2mc 2 2 938.28MeV

301

7.53 10 4 MeV

0.753keV

Chapter 11 – Nuclear Physics (Problem 11-96 continued) (c)

11-97.

7.53 10 4 MeV 100 0.782MeV

dN dt

Rp

(a)

N

N Rp

(Equation 11-17) N 0e

Rp /

N

0.0963%

t

1 e

Rp t

Rp e

t

Rp 1 e

(from Equation 11-17) R p / , its maximum value

At t = 0, N(0) = 0. For large t, N t

(b) For dN / dt Rp

N

N

t

0 N

Rp /

Rp / ln 2 / t1 / 2

100s 1 / ln 2 / 10 min 8.66 104

11-98. (a) 4n + 3 decays chain

100s

1

60s / min / ln 2 / 10 min

62

Cu nuclei

235 92

U143

207 82

Pb125 There are 12 α decays in the chain. (See graph

below.)

302

Chapter 11 – Nuclear Physics (b) There are 9 β− decays in the chain. (See graph below.)

(Problem 11-98 continued) (c) Q

235

U c2

M

207

M

235.043924uc 2

Pb c 2

7M

4

206.975871uc 2

0.049839uc 2 931.50MeV / uc 2

He c 2 7 4.002602 uc 2

46.43MeV

(d) The number of decays in one year is:

dN dt

N0e

t

ln 2 / t1 / 2

where

ln 2 / 7.04 108 y

1kg 1000 g / kg 6.02 1023 atoms / mol

N0

235 g / mol

dN dt

9.85 10

10

y

1

2.56 1024 e

1y

9.85 10

10

y

1

2.56 1024 atoms

2.52 1015 decays / y

Each decay results in the eventual release of 46.43 MeV, so the energy release per year Q is: Q

2.52 1015 decays / y 46.43MeV / decay

1.17 1017 MeV / y 1.60 10

13

J / MeV

1.87 104 J / y 1cal / 4.186 J

4.48 103 cal / y

The temperature change ΔT is given by:

Q

cm T or T

Q / cm where m 1kg 1000g

and the specific heat of U is c T

4.48 103 cal / y

0.0276cal / g C .

0.0276cal / g C 1000 g

303

162 C

Chapter 11 – Nuclear Physics (Problem 11-98 continued) 235

U

α

144 231

Th

β−

142

231

Pa

α 237

140

Ac

β−

α 223

138

α

β−

136 219

α 215

N

β

α − 215

Po

β−

215

At

211

α

Pb

β

128

− 211

Bi

β−

α 126

Ra

Rn

α

130

223

219

Bi

132

211

Po

207

Tl

α

β

− 207

Pb

124

80

82

84

86

88

90

Z 11-99. The reactions are: (1) 1H

1

(2) 1H

2

H H

2

H

3

ve

He

followed by (3) 3 He

3

Th

α

At

134

227

α

Fr

He

4

He

1

H

1

H or

304

92

Chapter 11 – Nuclear Physics (Problem 11-99 continued) (4) 1H

3

4

He

He

ve

(a) From 2(1) + 2(2) + (3):

61 H

2 2H

2 3He

2 2H

2 3He

4

He 2 1H

2

2ve

2

Cancelling 2 1H , 2 2 H , and 2 3 He on both sides of the sum,

41 H

4

He 2

2ve

2

From (1) + (2) + (4):

41 H

2

3

H

2

He

H

3

4

He

He 2

2ve

Cancelling 2 H , and 3He on both sides of the sum,

41 H (b) Q

4

He 2

4m 1 H c 2

M

4 938.280MeV

2ve 4

He c 2

2mec 2

3727.409MeV

2 0.511MeV

24.7MeV (c) Total energy release is 24.7 MeV plus the annihilation energy of the two β+: energy release

24.7 MeV

2 me c 2

24.7 MeV

2 1.022MeV

26.7MeV

Each cycle uses 4 protons, thus produces

26.7MeV / 4 6.68MeV / proton.

Therefore, 1H (protons) are consumed at the rate: dN dt

4 1026 J / s 1eV 6 6.68 10 eV 1.60 10

P E

19

J

3.75 1038 protons / s

The number N of 1H nuclei in the Sun is: N

1 / 2 2 1030 kg 1.673 10 27 kg

M M

1

H

5.98 1056 protons

which will last at the present consumption rate for t

N dN / dt

5.98 1056 protons 3.75 1038 protons / s

1.60 1018 s

1y 3.16 107 s

1.60 1018 s

5.05 1010 y

305

Chapter 11 – Nuclear Physics 11-100. At this energy, neither particle is relativistic, so 2 pHe 2mHe

EHe

2mHe EHe mHe

EHe En

pn2 2mn

En

pHe

2mn 17.7 MeV

2mn En

17.7 MeV mn

mn EHe

1.008665u 17.7 MeV 2.002602u 1.008665u 17.7 MeV

pn

EHe

EHe

Therefore, EHe

R 0

kN

(b) Because k

1.005 62.5 1

100

R N

1.005 1 1.005

0.00498

0.05

235 f

U

0.05 584b a

R 0

100

neutron flux, the fractional change in flux necessary is equal to the

11-102. (a) For 5% enrichment: f

R 0

62.5 generations

100 137%

fractional change in k:

k 1 k

17.7 MeV

k N (from Example 11-22 in More section)

100 where R N / R 0

1

mn

14.1MeV

Percentage increase in energy production =

1

mHe

5s 0.08s / gen

11-101. (a) The number N of generations is: N

R N

mn

3.56MeV

17.7 3.56 MeV

EHe

17.7MeV

En

0.05

235 a

0.05 97b

0.95

238 f

0.95 0

U

0.95

U

29.2b 238 a

0.95 2.75b

306

U 7.46b

Chapter 11 – Nuclear Physics (Problem 11-102 continued)

2.4 29.2b

f

2.4

k

f

1.91 (Equation 11-68 in More section)

29.2b 2.46b

a

(b) For 95% enrichment: 0.95

f

235 f

0.95 584b a

0.95

235 a

238

0.05

U

f

0.05 0

0.95 97b

554.8b 238

0.05

U

U

a

0.05 2.75b

U 92.3b R 0 kN.

The reaction rate after N generations is R N For the rate to double R N N 5%

ln 2 / ln 1.91

N 95%

kN

2 R 0 and 2

N

ln 2 / ln k .

1.07 generations

ln 2 / ln 2.06

0.96 generations

Assuming an average time per generation of 0.01s t 5%

1.07 10 2 s

0.96 10 2 s

t 95%

Number of generations/second = 1/seconds/generation In 1s: N 5%

93.5 and N 95%

104

One second after the first fission: R 5%

R 0 kN

1 1.91

93.5

1.9 1026

Energy rate = 1.9 1026 fissions / s 200MeV / fission 3.8 1028 MeV / s 1.60 10

6.1 1015 J / s R 95%

R 0 kN

13

J / MeV

6.1 1015W

1 2.06

104

4.4 1032

Energy rate = 4.4 1032 fissions / s 200MeV / fission 8.8 1034 MeV / s 1.60 10

1.4 1022 J / s 1.4 1022W

307

13

J / MeV

Chapter 12 – Particle Physics 12-1. (a) Because the two pions are initially at rest, the net momentum of the system is zero, both before and after annihilation. For the momentum of the system to be zero after the interaction, the momentum of the two photons must be equal in magnitude and opposite in direction, i.e., their momentum vectors must add to zero. Because the photon energy is E  pc, their energies are also equal. (b) The energy of each photon equals the rest energy of a   or a   . E  m c 2  139.6MeV (from Table 12-3)

(c) E  hf  hc /  Thus,  

hc 1240MeV fm   8.88 fm E 139.6MeV

12-2. (a) E  m c2  m c 2  2285MeV  139.6MeV  2424.6MeV (b) E  2mp c2  2  938.28MeV   1876.56MeV (c) E  2m c 2  2 105.66MeV   211.32MeV

12-3. (a)

e

e

e

e

γ

Z

e

e

e

e

(b)

γ

e

γ

e e

γ

e

γ

309

e+ e

Chapter 12 – Particle Physics 12-4. (a)

e

e

γ e

e

(b) See solution to Problem 12-3(a). (c) e e e

12-5. (a)

32

P  32S  e assuming no neutrino

QM

 P c 32

2

M

 S c 32

(electron’s mass is included in that of

2

32

S)

 31.973908uc2  31.972071uc2







 0.001837uc2 931.5MeV / uc 2  1.711MeV

To a good approximation, the electron has all of the kinetic energy

Ek  Q  1.711MeV (b) In the absence of a neutrino, the

32

S and the electron have equal and opposite

momenta. The momentum of the electron is given by:

 pc 

2



 E 2  mec 2



2

(Equation 2-32)



  m c 



  m c 

 Ek  me c 2  Q  mec 2

2

2

2

e

2

2

e

2

 Q 2  2Qmec 2

310

Chapter 12 – Particle Physics (Problem 12-5 continued) The kinetic energy of the

32

S is then:

 pc   Q 2  2Qmec 2 p2 Ek   2M 2Mc 2 2M 32 S c 2 2

 

1.711MeV  



2

 2 1.711MeV  0.511MeV 



2 31.972071uc 2 931.5MeV / uc 2



 7.85  105 MeV  78.5eV

(c) As noted above, the momenta of the electron and

32

S are equal in magnitude and

opposite direction.

 pc 

2

 Q2  2Qmec 2  1.711MeV   2 1.711MeV  0.511MeV  2

2 p  1.711MeV   2 1.711MeV  0.511MeV   

1/ 2

c

 2.16MeV / c

12-6. (a) A single photon cannot conserver both energy and momentum. (b) To conserver momentum each photon must have equal and opposite momenta so that the total momentum is zero. Thus, they have equal energies, each equal to the rest energy of a proton: E  mp c 2  938.28MeV (c) E  hv  hc /    

hc 1240MeV fm   1.32 fm E 938.28MeV

3.00  108 m / s (d) v    2.27  1023 Hz 15  1.32  10 m c

12-7. (a) Conservation of charge: +1 +1 → +1 −1 +1 −1 = 0. Conservation of charge is violated, so the reaction is forbidden. (b) Conservation of charge: +1 +1 → +1 −1 = 0. Conservation of charge is violated, so the reaction is forbidden.

311

Chapter 12 – Particle Physics



12-8. (a)   2 / m 5    2 / m 5



a b

c

Since the units of  must be seconds s, we have 5

 N m2  2 s  kg     e 2 C   1

 

5

 1     c

5 a b

c

having substituted the internal units of   ke2 / c. Re-writing the units, noting that

J  N m, gives: s 

1 C10 1  5 10  10 kg N m C

 J 5 s5 m5    s5  

a b

c

Noting that J  kg m2 / s 2 and cancelling yields, s   kg 

a 1

 m 2 a  b  s  a b

Since a  1 must be 0, a  1, and since  a  b  1, b  2.

   2 / mc2 5 (b)  





2 1.055  1034 J s 137 

9.11 10

31



5

kg 3.00  108 m / s



2

 1.24  1010 s  0.124ns

12-9. (a) Weak interaction (b) Electromagnetic interaction (c) Strong interaction (d) Weak interaction 12-10.  0     is caused by the electromagnetic interaction;       v is caused by the weak interaction. The electromagnetic interaction is the faster and stronger, so the  0 will decay more quickly; the   will live longer.

12-11. (a) allowed; no conservation laws violated. (b) allowed; no conservation laws violated. (c) forbidden; doesn’t conserve baryon number. (d) forbidden; doesn’t conserve muon lepton number.

312

Chapter 12 – Particle Physics 12-12. (a) Electromagnetic interaction (b) Weak interaction (c) Electromagnetic interaction (d) Weak interaction (e) Strong interaction (f) Weak interaction

12-13. For neutrino mass m = 0, travel time to Earth is t = d/c, where d =170,000c•y. For neutrinos with mass m  0, t   d / v  d /  c, where   v / c. t  t   t 

 

2 

1 1  2

 d 1   d1   1    c  c   (Equation 1-19)

1 1  2 1  1   1   

1  

1 1  2 since   1  1    2 2

Substituting into t, t 

d 1    c  2 2 

d  mc 2  t    2c  E 



E   mc 2   2  E / mc 2

2

(Equation 2-10)

2

  t  2cE 2  mc     d   2



1/ 2

 2tE 2     d /c 



1/ 2



2  2 12.5s  10  106 eV   170, 000c y / c  3.16  107 s / y 



m  22eV / c2

313



   

1/ 2

 21.6eV

Chapter 12 – Particle Physics 12-14. The  and  are members of a isospin multiplet, two charge states of the  hadron. Their mass difference is due to electromagnetic effects. The   and   are a particleantiparticle pair.

12-15. (a)

νμ

μ−

νe W−

W+

μ+

π−

e+

μ−

ντ W−

νμ

τ−

νμ

12-16. (See Table 12-8 in More Section.) (a) 30 MeV (b) 175 MeV (c) 120 MeV 12-17. (a) mp c 2   mn  me  c 2 Conservation of energy and lepton number are violated.





(b) mn c 2  mp  m c 2 Conservation of energy is violated. (c) Total momentum in the center of mass system is zero, so two photons (minimum) must be emitted. Conservation of linear momentum is violated. (d) No conservation laws are violated. This reaction, p p annihilation, occurs. (e) Lepton number before interaction is +1; that after interaction is −1. Conservation of lepton number is violated. (f) Baryon number is +1 before the decay; after the decay the baryon number is zero. Conservation of baryon number is violated.

314

Chapter 12 – Particle Physics 12-18. (a) The u and u annihilate via the EM interaction, creating photons.



 u

u

u

(b) Two photons are necessary in order to conserve linear momentum. (c)



 W

d

u

12-19. (a) The strangeness of each of the particles is given in Table 12-6. S  1 The reaction can occur via the weak interaction.

(b) S  2 This reaction is not allowed. (c) S  1 The reaction can occur via the weak interaction.

12-20. (a) The strangeness of each of the particles is given in Table 12-6. S  2 The reaction is not allowed.

(b) S  1 This reaction can occur via the weak interaction. (c) S  0 The reaction can occur via either the strong, electromagnetic, or weak interaction.

12-21. (a) n  n (b) n  p

1 1 T3     1 2 2 1 1 T3     0 2 2

T 1 T  1 or 0

315

Chapter 12 – Particle Physics (Problem 12-21 continued) 1 3  2 2

(c)    p

T3  1 

(d)    n

T3  1 

(e)    n

T3  1 

T

1 3  2 2

1 1  2 2

3 2

T T

3 2

1 3 or 2 2

12-22. (a)    e   Electron lepton number changes from 0 to 1; violates conservation of electron lepton number. (b)  0  e  e  ve  ve Allowed by conservation laws, but decay into two photons via electromagnetic interaction is more likely. (c)    e  e     v Allowed by conservation laws but decay without the electrons is more likely. (d) 0       Baryon number changes from 1 to 0; violates conservation of baryon number. Also violates conservation of angular momentum, which changes from 1/2 to 0. (e) n  p  e  ve Allowed by conservation laws. This is the way the neutron decays. 12-23. (a)   0   

   0   

(b)   p   0

  n   

(c) 0  p   

0  n   0

(d)  0     (e)       v

 0  e  e  e  e       0

12-24.    p   0      Because s have B = 0 and p has B = 1, conservation of B requires the  to have B = 1.   0   

The   has B = 0, so conservation of B requires that the 0 have B = 1.

316

Chapter 12 – Particle Physics 12-25. (a) 0  0  0  0

S is conserved.

(b) 2  0  1

S is not conserved.

(c) 1  1  0

S is conserved.

(d) 0  0  0  1

S is not conserved.

(e) 3  2  0

S is not conserved.

12-26. Listed below are the baryon number, electric charge, strangeness, and hadron identity of the various quark combinations from Table 12-8 and Figure 12-21. Quark Structure

Baryon Number

Electric Charge (e)

Strangeness

Hadron

(a)

uud

+1

+1

0

p

(b)

udd

+1

0

0

n

(c)

uuu

+1

+2

0

Δ++

(d)

uss

+1

0

−2

0

(e)

dss

+1

−1

−2



(f)

suu

+1

+1

−1



(g)

sdd

+1

−1

−1



Note that 3-quark combinations are baryons.

12-27. Listed be below are the baryon number, electric charge, strangeness, and hadron identity of the various quark combinations from Table 12-9 and Figure 12-21. Quark Structure

Baryon Number

Electric Charge (e)

Strangeness

Hadron

(a)

ud

0

+1

0



(b)

ud

0

−1

0



(c)

us

0

+2

+1



(d)

ss

0

0

+1

0

(e)

ds

0

0

−1

 0

* forms  and   along with uu and dd

317

Chapter 12 – Particle Physics 12-28.

u

K0

d

d

c

g

u

b

g

K0

d

b



12-29. (a) T3 = 0 (from Figure 12-20a) (b) T = 1 or 0 just as for ordinary spin. (c) uds

B = 1/3 + 1/3 + 1/3 = 1

S = 0 + 0 + −1 = −1

C = 2/3 – 1/3 – 1/3 = 0

The T = 1 state is the Σ0.

The T = 0 state is the Λ0.

12-30. The +2 charge can result from either a uuu, ccc, or ttt quark configuration. Of these, only the uuu structure also has zero strangeness, charm, topness, and bottomness. (From Table 12-5.) 12-31. The range R is R  c / mc2 (Equation 11-50). Substituting the mass of W + (from Table 12-4),

1.055  10 J s 3.00  10 m / s   2.44  10 R 81GeV / c 1.60  10 J / GeV  34 2

8

18

10



12-32.

u

d

  0        v

W

 ds   du 

v

s

d

0 318

m  2.44  103 fm

Chapter 12 – Particle Physics 12-33.



p

u 0 

p

u

d

d





 uds   uud   ud  Z0

Weak decay

u

12-34. n  p   

d

s

0

Q  mn c2  mp c 2  m c 2   939.6  938.3  139.6 MeV  138.3MeV

This decay does not conserve energy.



p

12-35.

u 0 

p



d

u

u

d



 uds   uud   ud  Z0 Strong decay

u

d

s

12-36. (a) B = 1, S = −1, C = 0, B  0 (b) Quark content is: uds 12-37. (a) The   has charge +1, B = 0, and S = +1 from Table 12-6. It is a meson (quarkantiquark) structure. us produces the correct set of quantum numbers. (From Table 12-5.)

319

Chapter 12 – Particle Physics (Problem 12-37 continued) (b) The  0 has charge 0, B = 0, and S = +1 from Table 12-6. The quark-antiquark structure tp produce these quantum numbers is d s . (From Table 12-5.) 12-38. (a) Being a meson, the D+ is constructed of a quark-antiquark pair. The only combination with charge = +1, charm = +1 and strangeness = 0 is the cd . (See Table 12-5.) (b) The D−, antiparticle of the D+, has the quark structure c d .

12-39. The  0 decays via the electromagnetic interaction whose characteristic time is

1020 s.

The  and  both decay via the weak interaction. The difference between these two being due to their slightly different masses.

12-40. If the proton is unstable, it must decay to less massive particles, i.e., leptons. But leptons have B = 0, so p  e  ve would have 1 = 0 + 0 = 0 and B is not conserved. The lepton numbers would not be conserved either; a “leptoquark” number would be conserved.





12-41. V  H 2O   0.75V  0.75 4 R 2 R , where R(Earth) = 6.37  106 m and R  1km  103 m.



V  H 2O   0.75  4 6.37  106 m 



 10   3.82 10 2

3

17



m3



M  H 2O   V  H 2O    3.82  1017 m3 1000kg / m3  3.82  1020 kg

Number of moles (H2O) = 3.82  1023 g / 18.02 g / mole  2.12  1022 moles Number of H2O molecules = N A   # of moles 





 6.02  1023 molecules / mole 2.12  1022 moles



 1.28  1046 molecules H2O

Each molecule contains 10 protons (i.e., 2 in H atoms and 8 in the oxygen atom), so the number of protons in the world’s oceans is N  1.28  1047.

320

Chapter 12 – Particle Physics (Problem 12-41 continued) The decay rate is

dN   N where   1 /   1 / 1032 y  1032 y 1 dt





 1032 y 1 1.28  1047 protons



 1.28  1015 proton decays / y  4  107 decays / s 12-42. (a) p  e  0  ve





 

Q  mp c 2  M 0 c 2  mec 2 MeV

  938.3  1116  0.511 MeV  178MeV Energy is not conserved. (b) p      Spin (angular momentum)

1  0  1  1. Angular momentum is not conserved. 2

(c) p      0 Spin (angular momentum)

1  0  0  0. Angular momentum is not conserved. 2

12-43. n, B = 1, Q = 0, spin = 1/2, S = 0

Quark strucure

u

d

d

B

1/3

+1/3

+1/3

=1

Q

2/3

−1/3

−1/3

=0

spin

1/2↑

1/2↑

1/2↓

= 1/2

0

+0

+0

=0

S

n , B = −1, Q = 0, spin = 1/2, S = 0 u

d

d

B

−1/3

−1/3

−1/3

= −1

Q

−2/3

+1/3

+1/3

=0

spin

1/2↑

1/2↑

1/2↓

= 1/2

0

+0

+0

=0

Quark strucure

S

321

Chapter 12 – Particle Physics (Problem 12-43 continued) (b)  0 , B = 1, Q = 0, spin = 1/2, S = −2 Quark strucure

u

s

s

B

1/3

+1/3

+1/3

=1

Q

2/3

−1/3

−1/3

=0

spin

1/2↑

1/2↓

1/2↑

= 1/2

0

−1

−1

= −2

S

(c)   , B = 1, Q = 1, spin = 1/2, S = −1 Quark strucure

u

u

s

B

1/3

+1/3

+1/3

=1

Q

2/3

2/3

−1/3

=0

spin

1/2↑

1/2↓

1/2↑

= 1/2

0

+0

−1

= −1

S

(d)  , B = 1, Q = −1, spin = 3/2, S = −3 Quark strucure

s

s

s

B

1/3

+1/3

+1/3

=1

Q

−1/3

−1/3

−1/3

= −1

spin

1/2↑

1/2↑

1/2↑

= 3/2

−1

−1

−1

= −3

S

(e)   , B = 1, Q = −1, spin = 1/2, S = −2 Quark strucure

u

d

d

B

1/3

+1/3

+1/3

=1

Q

−1/3

−1/3

−1/3

= −1

spin

1/2↑

1/2↓

1/2↑

= 1/2

0

−1

−1

= −2

S

322

Chapter 12 – Particle Physics 12-44. (a) Quark strucure

d

d

d

B

1/3

+1/3

+1/3

=1

Q

−1/3

−1/3

−1/3

= −3/2

spin

1/2

1/2

1/2

= 3/2, 1/2

0

+0

+0

=0

u

c

B

1/3

−1/3

=0

Q

2/3

−2/3

=0

spin

1/2

1/2

= 1, 0

0

+0

=0

u

b

B

1/3

−1/3

=0

Q

2/3

+1/3

=1

spin

1/2

1/2

= 1, 0

0

+0

=0

s

s

s

B

−1/3

−1/3

−1/3

= −1

Q

1/3

+1/3

+1/3

=1

spin

1/2

1/2

1/2

= 3/2, 1/2

1

+1

+1

=3

S

(b) Quark strucure

S

(c) Quark strucure

S

(d) Quark strucure

S

12-45. The Z 0 has spin 1. Two identical spin 0 particles cannot have total spin 1. 12-46. (a) The final products (p, γ, e−, neutrinos) are all stable. (b) 0  p  e  ve  v  v (c) Conservation of charge: 0  1  1  0  0  0  0 Conservation of baryon number: 1  1  0  0  0  0  1

323

Chapter 12 – Particle Physics (Problem 12-46 continued) Conservation of lepton number: (i) for electrons: 0  0  1  1  0  0  0 (ii) for muons: 0  0  0  0  1  1  0 Conservation of strangeness: 2  0  0  0  0  0  0 Even though the chain has S  2, no individual reaction in the chain exceeds

S  1, so they can proceed via the weak interaction. (d) No, because energy is not conserved. 12-47.  2, 1, 0, 1, 0   c u u

 0, 1,  2, 1, 0  c s s  0, 0, 1, 0,  1  b s

 0,  1, 1, 0, 0  s d u  0, 1,  1, 1, 0  c s d  1, 1,  3, 0, 0  s s s 12-48. (a)

t1  x / u1 t2  x / u2  t  t2  t1 

x x  u2 u1

 u  u  x u t  x  1 2   2 c  u1u2 

(b)

E

mc 2 1  u 2 / c2

(Equation 2-10) 1/2

2 (mc 2 )2 (mc 2 )2 u   mc 2   2 2 E   1 u / c    1     1  u 2 / c2 E2 c   E     Expanding the right side of the equation in powers of (mc 2 / E )2 and keeping only 2

the first term yields

324

Chapter 12 – Particle Physics (Problem 12-48 continued)

u 1  mc 2   1   c 2 E 

(c)

2

2  1  mc 2 2 1  mc 2   u1  u2  c 1    1    2  E2    2  E1  

c(mc 2 )2  1 1  c(mc 2 ) 2  E12  E22    2  2  2 2  2 2  E2 E1   E1 E2 

u  u1  u2 

c(2.2 eV / c 2  c 2 ) 2  (20 MeV) 2  (5 MeV) 2   (20 MeV) 2 (5 MeV) 2  2  

 c(2.2 eV) 2  (106 MeV/eV) 2  375  2 2 2 2  (20) (5) (MeV)  2 12 (2.2) (375) 10 c  2.7 105 m/s 2 2 2(20) (5) 

And therefore, t 

1.7 105 c  y  9.46 1015 m / c  y  2.7 105 m/s  0.48s (3.0 108 m/s)2

12-49. (a)    e  ve  v Electron lepton number: 0  1  1  0  0 Muon lepton number: 1  0  0  1  1 Tau lepton number: 0  0  0  0  0 (b)       v  v Electron lepton number: 0  0  0  0  0 Muon lepton number: 0  1  1  0  0 Tau lepton number: 1  0  0  1  1 (c)       v Electron lepton number: 0  0  0  0 Muon lepton number: 0  1  1  0 Tau lepton number: 0  0  0  0

325

Chapter 12 – Particle Physics 12-50.  0    

hf



E 2   pc   m c 2 2



In lab:

2

 

0

hf

Conservation of momentum requires that each carry half of the initial momentum, hence



the total energy: 2  hf / c  cos  p cos 





2 2 hf  E / 2   pc   m c 2   

1/ 2

p p  1/ 2 2 2 2  hf / c  2  pc   m c 2  2c  



pc





 pc 2  m c 2     2

1/ 2





850MeV  850MeV 2  135MeV 2   

1/ 2

 0.9876

  cos1  0.9876  9.02 12-51. (a) 0  p    Energy: 1116MeV   938  140  MeV  38MeV conserved. Electric charge: 0  1  1  0 conserved. Baryon number: 1  1  0  1 conserved. Lepton number: 0  0  0  0 conserved. (b)   n  p  Energy: 1197MeV   940  938 MeV  681MeV not conserved. Electric charge: 1  0  1  1 conserved. Baryon number: 1  1  1  0 not conserved. Lepton number: 0  0  0  0 conserved. This reaction is not allowed (energy and baryon conservation violated).

326

2

Chapter 12 – Particle Physics (Problem 12-51 continued) (c)    e  ve  v Energy: 105.6MeV  0.511MeV  105.1MeV conserved. Electric charge: 1  1  0  0  1 conserved. Baryon number: 0  0  0  0  0 conserved. Lepton number: (i) electrons: 0  1  1  0  0 conserved. (ii) muons: 1  0  0  1  1 conserved.

12-52. (a) The decay products in the chain are not all stable. For example, the neutron decays via n  p  e  ve

Only the e+ and e− are stable.

(b) The net effect of the chain reaction is:   p  3e  e  3ve  2v  2v (c) Charge: 1  1  3  1  1 conserved Baryon number: 1  1  0  0  0  0  0  0  1 conserved. Lepton number: (i) electrons: 0  0  3  1  3  1  0  0  0 conserved (ii) muons: 0  0  0  0  0  0  2  2  0 conserved Strangeness: 3  0  0  0  0  0  0  0  0 not conserved Overall reaction has S  3; however, none of the individual reactions exceeds

S  1, so they can proceed via the weak interaction.

12-53. The proton and electron are free particles. The quarks are confined, however, and cannot be separated. The gluon clouds give the u and d effective masses of about 330 MeV/c2, about 1/3 of the proton’s mass. 12-54. (a) 0  p   

 

Ekin   M 0  mp  m  c 2

 1116MeV / c 2  938.3MeV / c 2  139.6MeV / c 2  c 2  38.1MeV

327

Chapter 12 – Particle Physics (Problem 12-54 continued) (b) Because the  0 decayed at rest, the p and   have momenta of equal magnitudes and opposite direction. mp v p  m v  mp / m  v / v p

1 m v2 m  m 2 m 938.3 p p 2       6.72  Ekin  p  1 m v 2 m p  m  m 139.6 2 p p Ekin  

(c) Ekin  Ekin  p   Ekin    Ekin  p   6.72Ekin  p   7.72Ekin  p   38.1MeV

Ekin  p   38.1MeV / 7.71MeV  4.94MeV Ekin    6.72Ekin  p   33.2MeV

12-55. 0  0   (a) ET for decay products is the rest energy of the  0 , 1193MeV. (b) The rest energy of 0  116MeV , so E  1193MeV  1116MeV  77MeV and p  E / c  77MeV / c . (c) The  0 decays at rest, so the momentum of the  0 equals in magnitude that of the photon.

 





2 Ekin 0  p2 / 2M      77MeV / c  / 2 1116MeV / c 2 

 2.66MeV

small compared to E

(d) A better estimate of E and p are then E  77MeV  2.66MeV  74.3MeV and p  74.3MeV / c.

12-56. (a) t  t2  t1  t 

x x x  u1  u2    Note that u1u2  c 2 u2 u1 u1u2

x  u1  u2  c

2



xu c2

328

Chapter 12 – Particle Physics (Problem 12-56 continued) (b) E 

mc

2

1  u 2 / c2



1  m c 2 u o (Equation 2-10). Thus,   c  E2 



2

   

1/ 2

1  m c2   1  o  2 E 

2

2  1  m c 2 2 1  mo c 2   o (c) u1  u2  c 1    1    2 E    2  E1   2

2



c mo c 2 c  mo c 2  c  mo c 2         2  E2  2  E1  2



  



2

 E12  E22   2 2   E1 E2 



2 2 2 6 6 c  20eV   20  10 eV  5  10 eV    20  106 eV 2 5  106 eV 2  2  



2 2 2 c  20eV   20    5     20 2  52 106 eV 2 









   7.5  1012 c 2  



12 xu 170, 000c y  7.5  10 C T  2  1.28  106 y  40.3s c c2

(d) If the neutrino rest energy is 40eV, then u  3.00 1011 c and t  161s. The difference in arrival times can thus be used to set an upper limit on the neutrino’s mass. 12-57.    e  ve  v

      v  v    d  u  v The last decay is the most probable (three times as likely compared to each of the others) due to the three possible quark colors.

329

Chapter 12 – Particle Physics

12-58. (i )

(ii )

dp  ev  B dt

dE  0 which follows from the fact that the Lorentz force is  v; therefore, v  v = dt

constant and thus   v   constant. Equation (i) then becomes: For circular orbits

dp dv  m  ev  B dt dt

m v 2  evB or, re-writing a bit, R

m v  p  eBR and pc  ceBR 

1GeV  0.35BR GeV 1.60  1019 J

and finally, p  0.35BR GeV / c

330

Chapter 13 – Astrophysics and Cosmology vW

13-1.

vW  vE  4km / s. Assuming Sun’s rotation to be uniform, so that vW  vE , then vW  vE  2km / s.

r

Because v  2 / T , vE  2 r / T or ω T

vE

13-2.







2 6.96  105 km 2 r   2.19  106 s  25.3 days vE 2km / s



11 1.99  1030 2GM 2 2 6.67  10 U   R 6.96  108



2

J  7.59  1041 J

The Sun’s luminosity L  3.85  1026W

 tL 

U L



7.59  1041 J  1.97  1015 s  6.26  107 years 26 3.85  10 J / s

13-3. The fusion of 1H to 4 He proceeds via the proton-proton cycle. The binding energy of 4

He is so high that the binding energy of two 4 He nuclei excees that of 8 Be produced in

the fusion reaction: 4 He  4 He  8 Be   and the 8 Be nucleus fissions quickly to two 4

He nuclei via an electromagnetic decay. However, at high pressures and temperatures a

very small amount is always present, enough for the fusion reaction: 8

Be  4 He  12C   to proceed. This 3- 4 He fusion to 12C produces no net 8 Be and

bypasses both Li and B, so their concentration in the cosmos is low.

13-4. The Sun is 28, 000c y from Galactic center = radius of orbit



15 2 r 2 28, 000c y  9.45  10 m / c y  time for 1 orbit   v 2.5  105 m / s

 6.65  1015 s  2.11  108 yr

331



Chapter 13 – Astrophysics and Cosmology 1 H atom / m3

13-5. Observed mass (average) missing mass 500 photons / cm 3

9 1.67 10

kg / m3

1.56 10

26

27

kg / m3

10% of total mass (a)

kg / m3

500 106 photons / m 3, so the mass of each photon would be

1.50 10 26 kg / m 3 500 106 photons / m 3

3.01 10

3.01 10 35 kg 1.60 10 19 J / eV

or mv

27

1.67 10

35

kg

m2 s2

c2

16.9eV / c 2

13-6. 5

4

T 3 4 10 K 2

1

0 0

1

2

3

4 M /M

13-7.

1c s

c 1s

1c min 1c h 1c day

3.00 108 m / s 1s

c 1min 60s / min

6

7

8

3/ 8

3.00 108 m 3.00 105 km

3.00 105 km 60s 1.80 107 km

c 1h 3600s / h 1.08 109 km c 24h 3600s / h

13-8. (a) See Figure 13-16. 1AU

2.59 1010 km

1.496 1011 m. R 1pc when

332

1", so R

1AU 1"

Chapter 13 – Astrophysics and Cosmology (Problem 13-8 continued) or R

3600 " 1

1AU 1"

180 rad

3.086 1016 m 9.45 1015 m / c y

1 pc

3.26c y

0.01", R 100 pc and the volume of a sphere with that radius is

(b) When

4 3 R 3

V

3.086 1016 m 1 pc

4.19 106 pc3. If the density of stars is 0.08 / pc3, then the number

of stars in the sphere is equal to 0.08 / pc3 4.19 106 pc3

13-9.

L

4 r2 f

m1

m2

4 rp2 f p and LB

Thus, Lp rp2 f p

rB2 f B

rB2

12 pc,

Because rp

2.5 log f1 / f 2 4 rB2 f B and Lp

0.30 rp f p / f B

rB

f p / fB 1/ 2

13-10. (a) M

0.3M

Te

3300K

(b) M

3.0M

Te

13, 500K

(c) R

M

R

R M

Similarly, R3.0

3.0 R

tL

M

3

(d) tL 0.3 or tL 0.3

tL 0.3M

2.00

12 2

17.0 pc

5 10 2 L

L

L 102 L

1.93 1025W 3.85 1028W

R /M

M

0.3M

R0.3

LB

rp2 f p / f B

1.16 0.41 2.5

log f p / f B

3.4 105 stars.

3

2.09 108 m

0.3R

2.09 109 m 3

M

0.3M

tL / M

3

tL M 3

tL M 3 0.3M

37tL . Similarly, tL 3.0

333

3

0.3

0.04tL

3

tL

Chapter 13 – Astrophysics and Cosmology

S R

13-11. Angular separation 100 106 km 100c y

distance between binaries distance Earth 1011 m 100c y 3.15 107 s / y

1.057 10 7 rad

6.06 10 6 degrees 1.68 10 9 arcseconds

13-12. Equation 13-18: 56 26

13 24 He 4n. m56 Fe

Fe

Energy required: 13 m4 He 1u

4mn

931.49432 MeV / c 2

Equation 13-19:

4 2

He

Energy required: 2m1H

13-13.

55.939395u, m4 He

2n

2mn

m4 He

0.020277u

(a) r

1.5 c y ; assuming constant expansion rate, 2

m1H

Age of Shell (b) Lstar

R

M

Te

M 1/ 2

R mstar , M

Te star

or Rstar

Lstar L

28.3MeV

1.5 c y / 2 2.4 104 m / s

Te

M 1/ 2

Te

L

star

2.95 1011 s

9400 y

1.4Te

M4

L

M4

L /M4

Te 1/ 2 M star , 1/ 2 M

Using either the Te or L relations, Rstar

1.007825

12L

Te / M 1 / 2 ,

R /M , Rstar

120 MeV

2 1H

1.5 c y

M

1.008665u.

0.129104u.

m56 Fe

0.129104u

24 km / s

R

4.002603u, mn

Lstar M star R M

1/ 2

1.86 R

334

L 4 M star 4 M Te star Te

2

R

2

1.4 R

1.96 R

Chapter 13 – Astrophysics and Cosmology

2GM / c2

13-14. RS

(Equation 13-24)

2 6.67 10

(a) Sun RS

(b) Jupiter mJ

318mE

1.99 1030 / c2

2.9 103 m 3km

2.8m

RS

8.86 10 3 m

(c) Earth RS

13-15. M

11

9mm !

2M

(a) (Equation 13-22) R 1.6 1014 M (b) 0.5rev / s

1 I 2

2

2 MR 2 5 I

d

2

1 2 2M 2 5

1.01 104

d

1 / day 108

where

2 8.0 1038 J

d 8

1.99 1030 1011

(b) Its radius would be RS 11

1.8 1042 kg

2GM / c 2 (Equation 13-24).

1.8 1042 / c2

2.6 1015 m 17, 000 AU

72, 000 km / s.

(a) v

Hr

r

v H

72, 000km / s 21.2 km / s 106 c y

3.40 109 c y

(b) From Equation 13-29 the maximum age of the galaxy is: 1/ H

4.41 1017 s

L 1.85 1025W

10% of total

(a) Mass of a central black hole = 9 1.99 1041

13-17. v

8.0 1038 J

1011 stars of average mass M , therefore the visible mass =

1.99 10 41 kg

2 6.67 10

2

1.85 1025 J / s

5

10 d 8.64 10 s / d

13-16. Milky Way contains

RS

1.01 10 4 m

where for a sphere

I d

2

1/ 3

1.6 1014 2 M

rad / s

I (c) d

1/ 3

1.4 1010 y

335

Chapter 13 – Astrophysics and Cosmology (Problem 13-17 continued)

1/ H

r/v

1/ H

1/ H

r/v

1/ H

r r

10%

so the maximum age will also be in error by 10%.

13-18. The process that generated the increase could propagate across the core at a maximum rate of c, thus the core can be at most 1.5 y 3.15 107 s / y 3.0 108 m / s 1.42 1016 m

9.45 104 AU in diameter. The Milky Way diameter is

60, 000c y

3.8 109 AU .

13-19. Combining Hubble’s law (Equation 13-28) and the definition of the redshift (Equation 1327) yields

z

Hor c

0

r 5 106 c y

(a)

Hor 1 c

0

21.7

km 5 106 c y 1 656.3nm s 106 c y 3 108 m/s

656.5nm

(b) r

50 10 6 c y

Similarly, (c) r

(d) r

658.7nm

500 106 c y

Similarly,

680.0nm

5 109 c y

Similarly,

0

893.7nm

336

Chapter 13 – Astrophysics and Cosmology

13-20. Equation 13-33:

3H 2 8 G

c

3 8

1/ H

2

G

3 c

8

1.5 1010 y 3.15 107 s / y

8.02 10

2

6.67 10

11

27

kg / m3

Nm 2 / kg 2

(This is about 5 hydrogen atoms/m3 !)

13-21. Present size

1010 c y

Sp

1 T

1 with T T

Sp

2.7 K

2.7 1010 c yK

(a) 2000 years ago, S = Sp (b) 106 years ago, S = Sp (c) 10 seconds after the Big Bang, S

2.7 1010 c yK / 109 K 2.7 10 9 S p 2.7 1010 c yK / 5 109 K 5.4 10

(d) 1 second after the Big Bang, S

10

Sp

25c y 5c y

(e) 10-6 seconds after the Big Bang, S

13-22.

2.7 1010 c yK / 5 1012 K 5.4 10

proton

3 pl

10

1.67 10 10

osmium

5.5 10 8 kg

m pl

Planck time

15

3

27

kg

m

3

35

m

13

Sp

0.005c y

6.4 10 4 AU

5.5 1097 kg / m3

3

1.67 1018 kg / m3

3

2.45 104 kg / m3

13-23. Wien’s law (Equation 3-11):

max

2.898mm K T

2.898mm K 2.728K

1.062mm

(this is in the microwave region of the EM spectrum)

13-24. Muon rest energy

208me

106MeV / c2 . The universe cooled to this energy (average)

at about 10-3s (see Figure 13-34). 2.728K corresponds to average energy = 10-3 eV. Therefore, m

10 3 eV 1.6 10 c2

19

J / eV

337

1.8 10

39

kg

Chapter 13 – Astrophysics and Cosmology 13-25.

0

M / 4/3

r 3 t0

r t

R t r t0

t

(Equation 13-37) \

M 0

0

4/3 R3 t

r3 t

M / 4/3

r t /R t

M

R3 t

3

4/3

r2 t

t

13-26. B

If Hubble’s law applies in A, then vBA

FBC

vCA

HrCA .

From mechanics,

FBA

vBC

C FCA

HrBA ,

vBA

vCA

H rBA

rCA

HrBC

and Hubble’s lab applies in C, as well, and

A (Milky Way)

by extension in all other galaxies.

13-27. At a distance r from the Sun the magnitude of the gravitational force acting on a dust

GM m where m (4 / 3) a 3 . The force acting on r2 the particle due to the Sun’s radiation pressure at r is given by: (See Equation RP-9.) article of radius a is: F grav

Frad

a 2 Prad

a2

U 3

where

a2 is the cross sectional area of the particle and U

is the energy density of solar radiation at r. U is given by: (See Equation 3-6.)

U

4 R c

4 L c 4 r2

Therefore, Frad

a2

1 4L 3 4 r 2c

The minimum value of a is obtained from the condition that Fgrav

338

Frad :

Chapter 13 – Astrophysics and Cosmology (Problem 13-27 continued) GM m 1 4L a2 2 r 3 4 r 2c GM (4 / 3) a 3 1 4L a2 2 r 3 4 r 2c

Simplifying this expression yields:

L 4 cGM

a

3.84 1026 W 4 (3.00 108 m/s)(6.67 10 11 N m2 / kg 2 )(1.99 1030 kg)(5500 kg/m3 ) a 1.40 10 7 m or 1.40 10 5 cm Note that (i) a is very small and (ii) the magnitude of a is independent of r. a

13-28. (Equation 13-31) Robs M / 4/3

Z

Remit 1 Z

r3

0

Substituting for r0 in the 0

0

M / 4/3 Z / 1 Z

0

or

13-29. (a) H available for fusion = M (b) Lifetime of H fuel =

r 1 Z

M / 4/3

r03

equation:

r 1 Z 3

r0

3

M / 4/3 Z

0

1 Z

r3 1 Z

3

3

0.75 0.13 2.0 1030 kg 0.75 0.13 2.0 1029 kg

2.0 1029 kg 6.00 101 kg / s

3.3 1017 s

3.3 1017 s / 3.15 107 s / y

1.03 1010 y

(c) Start being concerned in 1.03 1010 y 0.46 1010 y

5.7 109 y

13-30. SN1987A is the Large Magellanic cloud, which is 170,000c•y away; therefore (a) supernova occurred 170,000 years BP.

339

Chapter 13 – Astrophysics and Cosmology (Problem 13-30 continued) (b) E

K

mo c 2

mo c 2 1

109 eV ,

v2 c2

mo2

9.28 108 eV

109

9.38 108

9.38 108

v2 1 2 c

0.875c

or v

Therefore, the distance protons have traveled in 170,000y =v

170, 000 y 149, 000c y. No, they are not here yet.

1.99 1030 kg.

13-31. M

(a) When first formed, mass of H = 0.7 M , m 1H number of H atoms

0.7 M 1.007825u 1.66 10

He; 4 1H

(b) If all H produced =

4

27

1.007825u 1.66 10

kg / u

27

8.33 1056

He 26.72eV . The number of He atoms

8.33 106 . 4

Total energy produced =

8.33 106 4

5.56 1057 MeV

26.72MeV

8.89 1044 J

(c) 23% of max possible = 0.23 8.89 1044 J 0.23 8.89 1044 L

tL

13-32. (a) F

Gm1m2 / r 2

v2 / r

ac m2

5.53 1017 s 1.7 1010 y

3.85 1026W

L

v 2 / r m2

Gm1 / r 2 and orbital frequency f

v/2 r

Substituting for f and noting that the period T or, T 2

kg / u, thus

1/ f , 4

4 2 r 3 / Gm1, which is Kepler’s third law.

(b) Rearranging Kepler’s third law in part (a),

340

2

f2

Gm1 / r 3

Chapter 13 – Astrophysics and Cosmology (Problem 13-32 continued)

mE

2 3 moon

4

r

/ GT

4

2

6.67 10

11

2

3.84 108 m

3

Nm2 / kg 2 27.3d 8.64 104 s / d

2

6.02 1024 kg

(c) T

2

11

2 T

12d

2 2

2

1.97 107

r2

2

2 12 24 3600

r2

6.06 10 6 / s 2

2

v2 r

G

m1m2 and r

2 3

m2

r G

from the graph v1

200km / s and v2

3.3 1010 m and, similarly, r2

100km / s

1.6 1010 m

4.9 1010 m

Assuming circular orbits,

m1

1.48 1022 kg

2

m1m2 , then m1 m2

1

5.44 103 s 1.5h

24

3

6.46d 3.1 10 s / d

, and

1/ 2

3

6.02 10

4

200 103 m / s 6.06 10 6 / s r1

11

Gm1m2 or m1 r2

r r

r1 1, v2

r

6.67 10

m2 : reduced mass

m1m2 m1 m2

r1

6.67 106

2

6.67 10

(b) For m1

(c) v1

1/ 2

4

(d) mcomb

13-33. (a) T

rsh3 GmE

m1v12 r1

6.63 1030 kg and m2

m2v22 and m1 r2

1.37 1031 kg

341

r1v22 m2 Substituting yields, r2v12

Chapter 13 – Astrophysics and Cosmology

13-34. E

1 2 mv 2

or GM m / r

GmM / r

1 2 mv 2

mv 2

1 GM m 2 r

E

FG

GM m / r 2

mv 2 / r

1 GM m / r 2

GM m r

1 2

GM m r

13-35. dV

20km / s 106 c y

H

universe V

Current average density = 1H atom / m3

4 3 R 3

V

dV

4 R 2 dR

dR The current expansion rate at R is:

v

dV

HR

20km / s 1010 c y 6 10 c y

dR

20 107 m / s 3.16 107 s / y

4 R 2 dR

1010

4

7.07 1074 m3 106 c y

Current volume V

2

20 104 km / s

20 107 m / s

106 y 106 y

9.45 1015 m / c y

2

20 107 m / s 3.16 107 s / y

106 y 106 y

# of H atoms to be added 106 c y

4 3

"new" H atoms =

1010

3

8.4 1077 m3

7.07 1074 atoms / 106 c y 8.4 1077 m3

342

0.001 "new" H atoms / m3 106 c y ; no

Chapter 13 – Astrophysics and Cosmology 13-36. (a) Equation 8-12: vrms

3RT / M is used to compute vrms vs T for each gas ® = gas

constant. M Gas

3

( 10 kg )

vrms (m/s) at T = :

3R / M 50K

200K

500K

750K

1000K

H2O

18

37.2

263

526

832

1020

1180

CO2

44

23.8

168

337

532

652

753

O2

32

27.9

197

395

624

764

883

CH4

16

39.5

279

558

883

1080

1250

H2

2

111.6

789

1580

2500

3060

3530

He

4

78.9

558

1770

1770

2160

2500

The escape velocities vsc

2 gR

2GM / R , where the planet masses M and

radii R, are given in table below. Planet

Earth

Venus

Mercury

Jupiter

Neptune

Mars

vesc (km/s)

11.2

10.3

4.5

60.2

23.4

5.1

vesc/6 (m/s)

1870

1720

750

10,000

3900

850

On the graph of vrms vs T the vesc/6 points are shown for each planet.

343

Chapter 13 – Astrophysics and Cosmology

(Problem 13-36 continued)

2GM / R

(b) vesc

2GM Pl / RPl

vPl

M Pl / RPl M E / RE

vPl vE

2GM E / RE

vE

M E / RE M E / RE

vPl

vE

M Pl / M E RPl / RE

(c) All six gases will still be in Jupiter’s atmosphere and Netune’s atmosphere, because vesc for these is so high. H2 will be gone from Earth; H2 and probably He will be gone from Venus; H2 and He are gone from Mars. Only CO2 and probably O2 remain in Mercury’s atmosphere.

13-37. (a) α Centauri d in pc

d

1AU sin 0.742 "

(b) Procyon d

Earth's orbit radius (in AU) sin p

2.78 105 pc

1AU sin 0.0286 "

9.06 105 c y

7.21 105 pc

2.35 106 c y

13-38. Earth is currently in thermal equilibrium with surface temperature T 4 and I

Earth radiates as a blackbody I is f

102 L then f

0.338 1.36 103 102

away. vrms

4

459W / m2 . The solar constant

1.36 103W / m 2 currently, so Earth absorbs 459/1360 = 0.338 of incident solar

energy. When L I

300

300K. Assuming

However, 3RT / M

102 f . If the Earth remains in equilibrium.

T 4 or T

the

vrms

3 8.31 949 18 10 3

solution to problem 14-26).

994 K

for

676 C sufficient to boil the oceans

H2 O

molecules

1146m / s 1.15km / s.

at

994K

is

The vesc = 11.2km/s (see

Because vrms ≈ 0.1 vesc, the H2O will remain in the

atmosphere.

344

Chapter 13 – Astrophysics and Cosmology

13-39. (a) a

n = grains/cm3 a

total scattering area =

dust grain

R

which is

R dx

photons N/mrs

N

R 2 na 2 dx

R2 a 2 n dx a2

R2 n dx of the

total area = fraction scattered = dN/N

d

dN N N0

n R 2 dx or N

N 0e

n R2d

0

From those photons that scatter at x = 0 (N0), those that have not scattered again after traveling some distance x = L is N L

N0e

n R2 L

. The average value of L (= d0) is

given by:

L d0

(b) I

0

dN L dL dL

dN L dL dL 0

I 0e

d / d0

1 n R2

Note:

near the Sun d0

1 n

n R 2 N 0e

n R2 L

R 10 5 cm

3000c y

3000c y 9.45 1017 cm / c y

(c)

dN L dL

10

n 1.1 10

2

5

12

/ cm3

2 gm / cm3

grains

mgrains 3

cm of space

mass in 300c y

2

4 3

10

5

3

1.1 10

9.41 10 27 gm / cm3 M 0.0012

12

9.41 10

27

9.45 1017 cm / c y

0.1%M

345

/ cm3

gm / cm3

3

300

Chapter 13 – Astrophysics and Cosmology

13-40. 56 11H

14 142 He

56 26

Fe 2

2e

14 4.002603 m56Fe

14 m4He

2

2e

2.04MeV / c 2

55.939395u

56.036442u 14 26.72 MeV 2.04MeV Net energy difference (release) =

90.40MeV 466.5MeV

56 2 26 Fe

2m

56

Fe

112 48

Cd

4

4e

4

2 55.939395u

m

Net energy required = 2m56Fe

13-41. (a) dt

4.08MeV / c 2

4e

112

111.902762u

Cd

0.023972u 4.08MeV

m112Cd

18.25MeV

1.024 104 2G 2 M dM hc 4 hc 4 1.024 1024

dM dt

rearranging, the mass rate of change is

2

G2M

Clearly, the larger the mass M, the lower the rate at which the black hole loses mass. (b) t t

(c) t

1.024 104

2

6.62 10

2

2.0 1030

2

hc 4 3.35 1044 s

1.024 104

1.06 1037 y far larger than the present age of the universe. 4

6.67 10

6.63 10 t

11

34

11

2

2.0 1030 1012

3.00 108

4

3.35 1068 s 1.06 1061 y

346

2