Tipler - Modern Physics 6th ed 2013 solutions

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: markl@cs.ucf.edu

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, u

Hoth

, is

u u u

Hoth spaceship droid

8 8

2.3 10 m / s 2.1 10 /

m s

8

4.410 m/

s

(b) No, since the droid is moving faster than light speed relative to Hoth.

1-2. (a)

4

2L

2 2.7410

m

t 1.8310

8

c 3.0010 m/

s

4

s

2L v

(b) From Equation 1-6 the correction t c

c

2

4 4 12

t 1.8310 s 10 1.8310

s

c

4 km / s

5

(c) From experimental measurements 1.3

10

c 299,796 km / s

No, the relativistic correction of order 10 -8 is three orders of magnitude smaller than

the experimental uncertainty.

2

2

1-3.

0.4 fringe 2

2

3

1.0 fringe 1.0 29.9 / 2.22 10 47.1 /

2 2

29.8 km / s v km / s

v km s v km s

0.4

1-4. (a) This is an exact analog of Example 1-1 with L = 12.5 m, c = 130 mph, and v = 20

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

12.5mi

20 mi / h

3

130 mi / h

2

Lv

t 0.0023 h 8.2s

3

c

1

(b) Pilot #1 must use a heading

2

sin 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.


8

1-5. (a) In this case, the situation is analogous to Example 1-1 with L3

10 m,

v

4

3 10 m/ s,

8

and c 3 10 /

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.

2

N

Lv

2

c

2

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

2

2 Nc

9 8

v fringe m m s m

2L

2

0.01 590 10 3.00 10 / 211

3

v 4.9110 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


(Problem 1-8 continued)

(c) Yes. This is guaranteed by the 2 nd 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.

500m

500m

t tr

t

f

8 8

3.0010 m / s 150 / 3.6 m / s 3.0010 m / s 150 / 3.6 m / s

500

8 8

310 150 / 3.6 310 150 / 3.6

8 8

310 2150 / 3.6310 150 / 3.6

2 150 / 3.6

500 4.6310

8

2

310

13

s

2

1-10.

A

*

B

*

C

*

v

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 Aand

B.

(b) According to an S observer, the wavefronts do not arrive at Aand

Csimultaneously.

(c) The wavefront arrives at Afirst, according to the S observer, an amount Δt before

arrival at C , where

3



BC

BA

t since BC BA L, Thus

c v c v

c v c v 2v

t L

L

2 2 2 2

c v c v

1-11.

2

β 1/ 2

1/ 1

0 1

0.2 1.0206

0.4 1.0911

0.6 1.2500

0.8 1.6667

0.85 1.8983

0.90 2.2942

0.925 2.6318

0.950 3.2026

0.975 4.5004

0.985 5.7953

0.990 7.0888

0.995 10.0125

1-12.

vx

0 vx

0

t1 t

1

t

2 2

t

2

( from Equation 1-19)

2

c

c

vx

c

vx

c

0 0

(a) t t t t t t

2 1 2 2 1 2

2 1

(b) The quantities x 1

and x 2

in Equation 1-19 are each equal to x

0

, but x 1 and x 2 in

Equation 1-18 are different and unknown.

1-13. (a)

2 2

1/ 2 2 2

1/ 1 v / c 1/ 1 0.85 c / c 1.898

y y 18m

5 3

x x vt 1.898 75m 0.85c 2.0 10 s 9.537 10 m

z z 4.0m

2 5 2 5

t t vx / c 1.898 2.010 s 0.85c 75 m / c 3.75610

s

1/ 2

4



(b)

3 5

x x vt 1.898 9.537 10 m 0.85c 3.756 10 s

75.8m

difference is due to rounding of , x, and t.

y y

18m

z z

4.0m

2 5 3 2 5

t t vx / c 1.898 3.75610 s 0.85c 9.537 10 m / c 2.010

s

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

L L 2L 2L

1

t1

c v c v c v c 1 v / c

t

t

2

2 2 2 2 2 2 2 2

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

L 1 v / c L 1 v / c 2Lc 1 v / c

c v c v c 1 v / c

2 2 2 2 2 2

2 2 2 2

2 2

2L 1 v / c 2L

1

2

t

2 2 2

1

2 2

t

c

1 v / c

c 1 v / c

Therefore, t 2 – t 1 = 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 u x = −0.9c.

u

u

v

x

x

2

1 uxv / c

(Equation 1-22)

0.9c 0.9c 1.8c

0.9945c

2

1 0.9c 0.9 c / c 1.81

(b) Calculating as above with

v m s u

4

3.0 10 /

x

4 4 4

3.0 10 m / s 3.010 m / s 6.0 10 m / s

ux

8

110

1

4 4

3.0 10 m / s3.0 10 m / s

8

2

3.0 10

m/

s

4

6.0 10 /

m s

5


1-16.

du

x

ux

v

a

x

where u

x

(Equation 1-22)

2

dt

1 u v / c

2

And t t vx / c (Equation 1-18)

x

2 2

2

/ 1 / 1 /

2 1

du u v vdu c u v c u v c du

x x x x x x

v

2

2 u vdu 1 u v / c du

c

2

2

1 u v / c

dt

dt vdx / c

du

a x

dt

2

x c

x x x x

v

2

x

2

u vdu / dt 1 u v / c du / dt

x x x x

1 u v / c

x

2

3

a

y

2 2

dux

/ dt1 v / c ax

2 3 2

1 uxv / c 1 uxv / c

duy

dt

3 3

uy

where uy

(Equation 1-22)

2

1 u v / c

2

2 2

/ 1 / / 1 /

2 2

1 / /

2

2

1 uxv / c

x

1 2

du du u v c u u v c du

y y x y x x

du u v c u v c du

y x y x

a

y

2 2

duy / dt1 uxv / c uyv / c dux

/ dt

2

1 / 2 1 /

2

x

x

duy

dt

u v c u v c

2 2

1 / /

2 2

3

1 uxv / c

a u v c a u v c

y x x y

a

z

is found in the same manner and is given by:

z

2 2

1 / /

2 2

3

1 uxv / c

az uxv c ax uzv c

a

6


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

4

3

ct

x

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

y

c

u 0

x

u 0

y

x

v

x

u

x

u x

v 0 v

v

2

1 vu

/ c 1

0

x

(Equation 1-23)

u

y

uy

c c

vu c

2

1 / 1

0

x

2 2 2 2 2 2 2 2 2

(b)

x y

/

1 /

u u u v c v c v c c

7


1-19. By analogy with Equation 1-23,

(a)

(b)

u v 0.9c 0.9c 1.8c

u 0.9945c

x

x

2

1 vux

/ c

2

10.9c0.9 c/ c 1.81

c c

u v 1.8 /1.81 0.9 1.8 0.9 1.81 3.429

u c c 0.9997c

x

x

2 2

1 vux

/ c 1 1.8 c /1.81 0.9 c / c 1.81

1.8 0.9 3.430

1-20 (a)

1

1 v / c

2 2

(Equation 1-16)

2 4 2

2 2 2

2 2

2

2 2

1/ 2 1 v 1 3 1 v

1 v / c 1

2

2

2 c 2 2 2! c

1 v 3 v 1 v

1 1

2 c 8 c 2 c

(b)

1

1 v / c 1 v / c

1/ 2

2 2 2 2

2 2

2

v v

2 2

1 1 1 1

1

2 c 2 2 2! c

1 v 1 v 1 v

1 1

2 c 8 c 2 c

2 4 2

2 2 2

(c)

1 v 3 v 1 1 v 1 v

1 1

2 c 8 c 2 c 8 c

2 4 2 4

2 4 2 4

1 1 v

1 1

2 c

2

2

1-21.

t

t

(Equation 1-26)

t t t t

1 v 1 v

1 1

t

t

2c 2c

2 2

2 2

1/ 2

2 2 t t

t t

1/ 2

v 2c v c 2 c 20.01 0.14c

t

t

8


1-22.

S

v = 1000 km/h

Orion (#2) Lyra (#1)

Earth

S

3

2.5

10 cy

3

2.5

10 cy

(a) Note that 1/ 1 v 2 / c 2 1 and 1c y c 3.15

10

7 s

From Equation 1-27: t t2t1 0, since the novas are simultaneous in system S

(Earth). Therefore, in S (the aircraft)

v

t t 2

t 1

2

x 2

x

1

c

6

10 m/

h

3 3 7

2 2.510 2.510 c3.1510

s

3600 s / h c

5

1.46 10 s 40.5

h

(b) Since t

is positive, t 2

t 1

; therefore, the nova in Lyra is detected on the aircraft

before the nova in Orion.

1-23. (a)

(b)

L

L p

/ (Equation 1-28)

2 2 2 2

=Lp

1 v / c 1.0m 1 0.6 c / c 0.80m

9

t L/ v 0.80 m/ 0.6c 4.4

10 s

1/ 2

(c)

ct

ct

The projection OA on

the x axis is L. The

length OB on the ct axis

yields t.

back of meterstick

passes x = 0

x

B

meterstick

A A

0

x

9


1-24. (a) t

t

t

1 v / c

2 2

(b)

8

8

(c)

8

8

(Equation 1-26)

8 8

2.610 s 2.610

s

5.9610

2

1/ 2

2

1

0.9 c

/ c 0.19

s vt 0.9 3.010 m/ s 6.010 s 16.1m

s vt 0.9 3.010 m/ s 2.610 s 7.0m

8

s

(d)

s ct x

2 2 2

8

2 2

(Equation 1-31)

c 6.010

16.1m 324 259 65 s 7.8m

1-25. From Equation 1-28,

2 2

1 / / 85/100

v c L L p

2 2

Squaring 1 v

/ c 85/100

L L L v c L m L m

2 2

p

/

p

1 / where 85 and

p

100

2

2 2 2 2 8

v 1 85/100 c 0.2775 c and v 0.527c 1.5810 m / s

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

t

s

L L 2L

c c c

p p p

(b) In the laboratory frame the length is contracted to L

L / and the round trip time is

t

L

L L L

c v c v c 1 v / c

2 2

c 1 v / c

2 2

2 2

2 Lp / 2Lp 1 v / c 2Lp

2 2 2

2 2

c 1 v / c c 1 v / c

p

(c) Yes. The time t s measured in the spaceship is the proper time interval τ. From time

dilation (Equation 1-26) the time interval in the laboratory t L

; therefore,

t

L

1

1 v / c

2 2

2L

c

p

which agrees with (b).

10


1-27. Using Equation 1-28, with L and L equal to the proper lengths of A and B and L A =

Ap

length of A measured by B and L B = length of B measured by A.

2

2 2

Bp

L L / 100m 1 0.92c 39.2m

A

Ap

L L 36 / 1 0.92 c / c 91.9m

Bp

B

1-28.

In S: x 1.0mcos30 0.866m

y 1.0msin30 0.500 m where 30

2

2

In : 1 0.866 1 0.8 0.520

S x x m m

y y

0.500m

0.500

0.520

1

where tan 43.9

2 2 2 2

L x y 0.520m 0.500 0.721m

1-29. (a)

In S : V ab c

2m 2m 4m 16m

In S: Both a and c have components in the x

direction.

a asin 25 2m sin 25 0.84 m and c ccos 25 4m cos 25 3.63m

x

x

x

x

x

2

2

1 0.84 1 0.65 0.64

a a

m

c

2

2

1 3.634 1 0.65 2.7

c

y y y y

2 2

x y

2 2

x y

2 2

2 2

1

(between c and xy-plane) tan 1.69 /

x

6m

a a acos 25 2cos 25 1.81 m and c c csin 25 4sin 25 1.69m

a a a 0.64 1.81 1.92m

c c c 2.76 1.69 3.24m

b

(in z direction) is unchanged, so b b2m

1

3

2.76 31.5

(between a and yz-plane) tan 0.64 /1.81 19.5

V (area of ay face) b (see part[b])

V c asin 78 b 3.24m 1.92m sin 78 2m 12.2m

3

11



(b)

y

c

a

b

19.5 ◦

x

31.5 ◦

z

1-30.

c c 1 v / c

o

(Equation 1-36)

f 1

1 v / c

fo

1

2 2

1 v/

c

1 v / c

1 v / c

o

1 v/

c o

v

v 1 /

o

Solving for v/ c, 1 1

c

o o c

1 /

o

2 2 2

v 1

590 nm / 650nm

o

650 nm. For yellow

590 nm. 0.097

2

c 1

590 nm / 650nm

v

Similarly, for green 525nm

0.210

c

v

and for blue 460nm

0.333

c

2

2

12


1-31.

c c 1 v / c

o

(Equation 1-37)

f 1

1 v / c

fo

1

7 8

m s m s

8

1 / 1 1.85 10 / / 3.00 10 /

o

v c

1 1

1 0.064

o

o

1 v/ c

1 1.8510 m/ s 3.0010 m/

s

1/ 2

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

1

f fo

where f 1.02

f

1

2 1

1.02 1

1.02 0.0198

2

1

1.02 1

6

v 0.0198c 5.9410 m / s

o

2

1-33.

1 1 1

f fo

o

656.3nm

1 1 1

110

110

3

3

For 10 : 656.3nm

657.0

3

110

110

2

2

For 10 : 656.3nm

662.9

2

110

110

1

1

10 : 656.3nm

725.6

1

nm

nm

nm

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

1

0.45 c/

c

S

2

1

c c 2

1

0.95 /

3.206

When they meet, each will have traveled a distance d from Earth.

Heidi: d 0.45c t Heidi

Hans: d 0.95c t Hans

and tHans

tHeidi

1 in years.

Therefore, 0.95c 0.45c tHeidi 0.95 c 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


(Problem1-34 continued)

v 0.45c

t

S t x 1.1198 1.90 0.45c

1.90

c c

In his reference frame S , Hans has aged

t

when he and Heidi meet.

v

t

3.30261.90 1 0.95 2

S t x y

0.290 y

2

c

The difference in their ages will be 1.697 0.290 1.407 y

1.4y

(b) Heidi will be the older.

2 2

y

2

1.198 1.90 1 0.45 1.697

y

1-35. Distance to moon =

8

3.8510 m

R

Angular velocity ω needed for v = c:

8 8

v / R C / R 3.0010 m/ s / 3.8510 m 0.78 rad / 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).

7

where t

3.1510 /

t

t / t 1 v / 2 c t tv / 2c

2 2 2 2

6

v 2 R / T 2 6.37 10 m / 108min 60 s / min

E

s y

3 5

v 6.177 10 m / s 2.0610

c

2 2 7 5

2

Time lost by satellite clock = / 3.1510 2.0610 / 2 0.00668 6.68

tv c s s ms

(b)

1 s t v / 2c

2 2

2 2 5

2

9

t 2 / v / c 2 / 2.0610 4.7110 s 150y

14


1-37.

cos

cos

(Equation 1-41)

1

cos

where half-angle of the beam in S 30

cos300.65

For 0.65, cos 0.97 or 14.1

1

0.65 cos30

A

0.75m

1.5m

lamp

14.1

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

beam

0.75m

A

3.0m

tan14.1

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

1s

(b)

1

t t / 1 t

1 1

2

1 1 v

Substituting 1 (From Problem 1-20)

2

2 c

2

8

1 v

3.010

2 2

t 11 1 t 2 c / v 2

2

2 c 1.510 / 3.610

12

1.04 10 s 32,000

t t s

0

y

2

6 3

9

273 10 . Using the same substitution as in (a).

t C km

9

1 1/ 273 10 and the circumference of Earth 40,000 , so

7 7

4.010 m vt or t 4.010 / v, and

2 9

2c

27310

v c v v

1230

m/

s

4.010

7

7 2 2 9

4.0 10 / 2 / 273 10 , or

2

Where v is the relative speed of the planes flying opposite directions. The speed of

each plane was 1230 m/ s/ 2 615 m/ s 2210 km / h 1380 mph.

15


1-39.

y

S (other)

y

S (Earth)

c

θ

v

(a)

(b) If

c c cos

v

c

x

y

csin

cy

csin

sin

tan

c c cos

v cos v / c

x

1 1

90 , tan

1

v/

c

90

45

e.g., if v 0.5 c,

63

x

x

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

Lp

Lp

ct vt t

( c

v)

Distance information travels in time t:

cLp

c ( c v)( c v) / c

ct L L L

( c v) ( c v) ( c

v)

c v

2 2

2

c 1 v / c c v

p

2

p p

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

L

1 v / c

p

16

Delta


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

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.

2

1-41. (a) Alpha Centauri is 4 c∙y away, so the traveler went

(b)

2 2

8 1 / 6

c y v c v y

t t y

2 2

1 / 6 / 8 3/ 4 /

v c v c v c

2 2 2

3/ 4

1

2 2 2

1

3/ 4

2

1/ 10.5625

v

0.8c

2

0

6 and 1/ 1 1.667

t 1.667 6y 10 y or 4 y older than the other traveler.

L 1 8 c y in 6 y, or

17



(c)

ct

10

ct

(c∙y)

8

6

x

4

2

Alpha Centauri

Earth

0

0

2 4 6 8 10

x

1-42. Orbit circumference

7

4.0 10 m.

7 3

v 4.010 m/ 90min 60 s / min 7.41

10 m/

s

Satellite speed

t t t

0

diff

1

t t t t t

2

t

diff

2

/

diff

1 1/ (Problem 1-20)

7 3 8

3.16 10 s1/ 27.41 10 / 3.0 10

0.0096s

9.6ms

2

v

1-43. (a) t

t x (Equation 1-20)

2

c

For events to be simultaneous in S, t0.

(b) Yes.

v

6 v

3

t x 210 s

2 2 1.510

m

c

c

6 8 3

v 210 s 310 m / s s /1.510

m

8

1.2 10 m / s 0.4

c

18



(c)

ct

1000

ct

500

B

x

0

0

500

1000 1500 x

A

(d)

s x ct

2 2 2

(Equation 1-33)

3 2

8

1.5 10 3 10 m / s2 10

6 s

2.2510 3.6 10 1.89 10

m

s

1370m

L s 1370m

6 5 6 2

2

2

1-44. (a) 2

1/ 1 1/ 1 0.92 2.55

(b)

8 8 8

2.610 s t lab 2.55 2.610 s 6.63

10 s

t

/

2

2

(c)

(d) 47

N t N e (Equation 1-29) L 1 L 1 0.92 50m 19.6m

0 0

Where L is the distance in the pion system. At 0.92c, the time to cover 19.6m is:

t m c s

8

19.6 / 0.92 7.0 10 . So, for N 0 = 50,000 pions initially, at the end of 50m

in the lab, N

4 7.0/ 2.6

5.0 10 e 3,390

19


2

1

v

L Lp L Lp Lp 1/ Lp 11/ Lp

(See Problem 1-20)

2

2

c

1-45.

For 11 and 3 10 / 11 0.5 10 5.5 10

4 8 8

Lp

m v m s L m

8

5.510

m

"Shrinkage" 550 atomic diameter

10

10 m / atomic diameter

1-46. (a)

ct

1500

ct

B

1000

500

A

0

0

500

1000 1500 2000

x

(b) Slope of

8

ct axis 2.08 1/ , so 0.48 and v 1.44 10 m/

s

(c)

2 2

ct ct and 1/ 1 so ct 1 ct

(d) t t 1.14t t

5 s /1.14 4.39s

For ct 1000 m and 0.48 ct 877m t1.5 877 / c 4.39s

L L / L 1 u / c 100m 1 0.85 52.7m

2 2

1-47. (a) 2

p

p

(b)

u u 0.85c 0.85c 1.70c

u 0.987c

1 uu / c

2 1

0.85 1.72

2

2 2

(c)

2

L L / L 1 u / c 100m 1 1.70/1.72 16.1m

p

p

(d) As viewed from Earth, the ships pass in the time required for one ship to move its

L 52.7m

7

own contracted length. t

2.110

s

8

u 0.853.00 10 m / s

20



(e)

0.987c

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 7

0

f / f 2 v / c. v c / 2 810 120 m/

s .

1-49.

#1

#2

L

c +

cm

Earth

c −

For star #1:

4

v 32 km / s 3.210 m / s

period 115d

c

c 3.210

4

4

c

c 3.210

Simultaneous images of star#1 in opposition

will appear at Earth when L is at least as large

as:

21



L

L

57.4d

c 3.210 c 3.210

4 4

c 3.210 L L c 3.210

75.5d 24 h / d 3600 s / h

c 3.210 c 3.210 c 3.210 c 3.210

c

4 4

4 4 4 4

4 4

c 3.210 L c 3.210

L

57.5d

24 3600

c

3.210 3.210

2 4

2

2 4

2

4 4 2 4

c 3.210 c 3.210 L c

3.210

2

57.5243600

L

c

2

3.2 10

57.5 24 3600

4

6.410

2 4

18

L 6.9910 m 739c y

v

2 1 2 1

(Equation 1-20)

2 b a

c

t

t 0 t t v / c 2 x x 0.5 1.0 y v / c 2 2.0 1.0 c y

1-50. t

t t t x x

(a)

2 1 2 1

0.5 v / c v 0.5 c in the x direction.

Thus,

2

(b) t

t vx / c

b

Using the first event to calculate t (because t is the same for both events),

2 2

t 1/ 1 0.5

1y 0.5c 1 c y / c

1.1551.5

1.7 y

2 2 2 2 2 2

(c)

a

s x ct 1c y 0.5c y 0.75 c y s 0.866c y

(d) The interval is spacelike.

(e) L s 0.866c y

22


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 0.5 c.

(b) From diagram t 1.7 y.

b r

b r b r

2

t b

t r

tb tr vxb xr

/ c

(2)

1-52. x x x x vt t

(1)

Where xb xr 2400m tb tr 5s x b

x r

2400m t b

t

r

5s

Dividing (1) by (2) and inserting the values,

6

400 2400 v

510

2 6 2 6

2400 v

6 6 2

2400 / 510 c 2400 510

v

510 510 v 2400 / c

2

2400

v v m s x

6 2

510

c

6 8

5 10 4800 2.69 10 / in direction.

1-53. u 0.85c cos50 u 0.85c

sin50

x

2

0.72 1/ 1 1.441 v 0.72

y

c

23



ux

v

u

ux

u

1 vu / c 1 vu / c

x

y

2 y

2

x

0.85c cos50 0.72c 0.1736c

ux

0.286c

1

0.72 0.85 cos50 / 1

0.3934

2

c c c

0.85c

sin50

uy

1.441

1

0.72 0.85 cos50 /

2

c c c

0.745c

u u u

0.798c

2 2

x y

tan u / u 0.745/ 0.286 111 with respect to the x

axis.

y

x

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

parallel to the x-axis and moves upward with speed v y in S, at

x y x y 0 at t t

0. 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 v

u because u v and u 0

y

y

y 2

y y x

1 vux

/ c

ux

v

ux

v because u 0

2

x

1 vu / c

x

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 0.5 v / c

Because t = 0. Therefore, when t 0 at the center, the right end is at xy given by:

x 0.5

vy

0.5

v

y

uy

t

2

c

and tan

y v 0.5

v

tan / 0.5 tan

x

c

v v / c 1

1 1 y

1 2 2

2

y

For 0.65

0.494 v / c

y

24

2 2


1-55.

0

2 2 2

x y z ct

2

/ 2 2 /

2

2 2 2

x vt y z c t vx / c

x c v c y z t v c xt

v vc c

2 2 2 2 2 4 2 2 2 2 2 2 2 2 2 2 2

2 2 2

x y z ct

2

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)

2 2

/ and /

t

t vx c t t vx c

2 2 2 1 1 1

2 2

2

1

2

1

2

1

/ /

t t t t v x x c T vD c

(b) For simultaneity in

S t t

T vD c v c cT D

2

,

2

1, or / / / .

Because v / c 1, cT / D is also <1 or D cT.

(c) If D < cT, then T vD c 2 T vcT c 2

T v c

2

positive because v/c < 1. Thus, t

2

t

1

T vD / c

(d) Assume T D/ cwith c c.

Then

/ / 1 / . For T > 0 this is always

c v

c

c

2 2

T vD / c D / c vD / c D / c

is always positive.

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

t t T vD / c 0 or t t

2

still smaller than c)

2 1 1 2

25


1-58.

1

v 0.6c 1.25

1

0.6

2

(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

v75min 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 ct

60min . The S observer is at x vt 0.6ct

and receives the

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

c t 60min 0.6ct

0.4ct

60c

min

t 60c min / 0.4c

150min

x 0.6c 0min 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:

t t t t 1 u / c t 1 r / c

2 2 2 2 2

0 r 0 0

1 2 2 2

Or, for r

c, tr

t0

1 r / c

2

1

t 1 r / c t

And,

2 2 2

0 0 2 2

tr

t

0 2

r

2

t0 t0

2C

26


1-60.

v B

Space station

V A

u

x

ux

v

u

u

1 u v / c

1 /

x

y

2 y

u 2

xv c

vAx

v

(a) For v : v v , v 0, v v . So, v

v

2

1 v v / c

BA B Ax Ay A Ax B

Ax

v v

vAy

where

Ay

A

2 B

2 2

B 1 vAxv / c

B

1 vB

/ c

v v

v

v v

/ 2

2 2 2

BA Ax Ay B A B

vAy vA /

B

vA

tan

BA

v

v v

Ax B B B

vBx

v

(b) For v : v v , v v , v 0. So, v

v

2

1 v v / c

AB A By B Bx Bx A

Bx

v v

v By

where

By

B

2 A

2 2

A 1 vBxv / c

A

1 vA

/ c

/

v v v

2 2

AB A B A

vB

vB

tan

AB

v

v

A A A A

(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

tan

45 only if v v

and 1, which requires v v 0.

A B A B

1

1

A B A B

27


1-61.

1-62. (a) Apparent time A B T / 2 t t and apparent time B A T / 2t t where

A B A B

t A = light travel time from point A to Earth and t B = light travel time from point B to

Earth.

T L L T 2vL

A B

2 c v c v 2 c v

2 2

T L L T 2vL

B A

2 c v c v 2 c v

2 2

(b) Star will appear at A and B simultaneously when t T /2 t or when the period is:

L L 4vL

T 2 t t

2

c v c v

c v

B A 2 2

B

A

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

u

1 /

y

u vu

y

x

c

tan

u

vu

x

x ux

v

1

2

c

uy

u sin

sin

tan

u v u cos v cos v / u

x

2

28


1-64.

#1

d

θ

#2

d vt

cos

x

Earth

vt

sin

x Earth

L

to Earth

L

(a) From position #2 light reaches Earth at time t 2

L / c.

From position #1 light reaches

L d

Earth at time t 1

t.

c

c

L L d

tEarth

t2 t1

t

c

c c

v tcos

tEarth

t

c

t

t

1

cos

Earth

(b)

B

app

vapp

xEarth

vtsin sin

c ct ct

1 cos 1

cos

Earth

(c) For 0.75 and v c 1, the result in part (b) becomes

app

app

0.75sin

1 sin

cos

1/ 0.75

1

0.75cos

Using the trigonometric identity

2 2

psin qcos r sin A where r p q sin A p / r cos A q / r

sin cos 2 sin 45 1/ 0.75

sin 45 1/ 0.75 2

45 70.6

25.6

29

Chapter 2 – Relativity II

2-1. 1 /

u u v c u v

2 2 2 2 2 2

yB o xB

2 2 2 2 2 2 2 2 2

uxB uB c v c uo

c v c

2 2 2 2

1 v / c 1 uo

/ c

2 2

1/ 2

2 2

1/ 2

1 v / c 1 uo

/ c

1 / 1 / / 1 /

p

yB

2 2

muyB

muo

1 v / c

1 u u / c 1 v / c 1 u / c

2 2 2

xB yB

mu / 1 u / c p

2 2

o o yA

2 2 2 2

o

2-2.

d mu m ud du

m u

2 c c c

2

3/ 2

2

1/ 2

1 2u u u

1 1

2 2 2

u u u u

m 1 1 1

2

2

2

2

c c c c

u

m1

c

2 2

3/ 2

2 2

3/ 2

2

2

3/ 2

du

du

du

2-3. (a)

1 1 1

1.25

1 u 2 / c 2 1

0.6 c 2 / c

2 0.8

2

(b)

p mu mc u / c / c 1.25 0.511 MeV 0.6 / c 0.383 MeV / c

2

(c)

E mc 1.25 0.511 MeV 0.639 MeV

2

(d)

E 1 mc 0.25 0.511 MeV 0.128 MeV

k

31


Ek

1 mc 1 u / c 1mc

2-4. The quantity required is the kinetic energy. 1/ 2

1/ 2

Ek

1 0.5 1

mc 0.155mc

(a)

1/ 2

(b)

2 2 2

Ek

1 0.9 1

mc 1.29mc

1/ 2

(c)

2 2 2

Ek

1 0.99 1

mc 6.09mc

2 2 2

2 2 2 2

2-5.

2 2 10J

16

E mc m E / c 1.110

kg

8

2

3.0810 m/

s

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

2-6.

m u m u c

2 2

( ) / 1 / (Equation 2-5)

(a)

2 2

m( u) m m / 1 u / c m

m

0.10 0.10

m

u / c 11/ 1.10 0.1736 u / c 0.416

2 2

Thus, 2

(b) m( u) 5m

m

1 1

2 2

1 0.10 1.10 1 u / c 1/ 1.10

2 2 2 2

1 u / c 1 u / c

2 2

5m 1 u / c 1/ 25

2 2

u c

1 /

2

Thus,

2 2

u c u c

/ 11/ 25 0.960 / 0.980

(c) m( u) 20m

m

2 2

20m 1 u / c 1/ 400

2 2

1 u / c

Thus,

2 2

u c u c

/ 11/ 400 0.9975 / 0.99870

32


v Earth moon distance / time 3.810 m/1.5s 0.84c

2-7. (a)

8

(b)

(c)

E mc mc mc

2 2 2

k

2

m 938.3 MeV / c

m( u) 1.73010 MeV / c 1.730 GeV / c

2 2 2

1 u

/ c 1

0.84

1 (Equation 2-9)

2

mc (proton) 938.3MeV

1/ 1 0.84 1.84

E 938.3MeV 1.87 1 813MeV

k

2

3 2 2

1 1

Classically, Ek

mv 938.3 MeV / c 0.84c 331MeV

2 2

813Mev

331MeV

% error 100 59%

813MeV

2 2

(d) 2

2

2-8. Ek

mc

1 (Equation 2-9)

k k

2

Or, W mc

u

u

E u E u W mc u 1 mc u 1

2 2

2 1 21 2 1

21 2 1

(a)

2

2

1/ 2 1/ 2

W21 938.3MeV 1 0.16 1 0.15 1.51MeV

(b)

2

2

1/ 2 1/ 2

W21 938.3MeV 1 0.86 1 0.85 57.6MeV

(c)

2

1/ 2

2

1/ 2

3

W21 938.3MeV 1 0.96 1 0.95 3.3510 MeV 3.35GeV

2-9.

E mc

2

(Equation 2-10)

2

(a)

200GeV 0.938 GeV where mc proton 0.938GeV

1 200GeV

213

2 2

1 v / c 0.938GeV

v 1

1 (Equation 2-40)

2

c 2

v

1 1 0.00001102 Thus, v 0.99998898c

c 2 213

1

2

33



(b)

E pc E mc E GeV GeV

2 4

for where 200 197 3.94 10 (Equation 2-36)

4

p E / c 3.94

10 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:

x

4 4

x 2133.94 10 3.94 10 /

E E vp GeV v GeV c

4

2133.94 10 GeV 1 v / c

4

2133.94 10 GeV 2

7

1.6810

GeV

2 4 4 2

x

/ 213 3.94 10 3.94 10 /

p p vE c GeV v GeV c

4

2133.94 10 GeV 2

7

1.6810 GeV /

2-10. (a)

2 3 2 13

E mc 10 kg c 9.0

10 J

1 1 10 / 3600 / 3.6 10

3 6

kWh J h s s h J

(b)

c

13 6 7

So, 9.010 J /3.610 J / kWh 2.5

10 kWh

6

@$0.10/ kWh would sell for $2.5

10 or $2.5 million.

(c) 100W 100 J / s, so 1 g of dirt will light the bulb for:

9.010 J

9.010

s

100 J / s

3.1610 s / y

13 11

11 4

9.010 s 2.8210

7

y

2-11.

E mc

2

(Equation 2-10)

2 2

where u c 2

1/ 1 / 1/ 1 0.2 1.0206

E 1.0206 0.511MeV 0.5215 MeV

2 2 2

E mc mc mc 1 (Equation 2-9)

k

= 0.511MeV 1.0206 1 0.01054MeV

34



2

2

(Equation 2-32)

1 1

E 2 mc

2

MeV MeV

E pc mc

2 2

2 2 2

2 2 2

p 0.5215 0.511

c

c

2 2

0.01087 MeV / c p 0.104 MeV / c

2-12.

E mc

2

(Equation 2-10)

E mc MeV MeV

2

/ 1400 / 938 1.4925

(a)

(b)

2 2 2 2

1/ 1 u / c 1.4925 1 u / c 1/ 1.4925

2 2

2

u / c 11/ 1.4925 0.551 u 0.74c

2

E pc mc

2 2

2

(Equation 2-32)

1 2

2 2 1 2 2

1400 938 1040 /

p E mc MeV MeV MeV c

c

c

2

2-13. (a)

(b)

p

m v p m v

R S N S

1/ 1 v / c 1/ 1 2.510 / 3.010

1.00000035

R

2 2 5 8

mS

3

mS

2.510

3 3

2.5 10 2.5 10 7

pR

p m

N S

1 3.510

3.510

p

1.00000035

E m c m c 1

m c

E

E

2 2 2

R S S S

N

R

1

mSv

2

E

E

R

N

2

1 c 0.5v

2 2

1

c

3.510 c 0.5v

7 2

3.510

c

2

7 2 2

5

0.5 2.510

1

0.0079

7 2

3.510

c

2

2

7

35


2-14.

6 2 2

u 2.210 m/ s and 1/ 1 u / c

(a)

E 0.511MeV 1 0.511MeV 1/ 1 u / c 1 0.5110 2.68910

k

1.374110

2 2 5

5

MeV

1 1

Ek

(classical) mu mc

2 2

u / c 0.5110 MeV / 2 2.210 / c

5

1.37410

MeV

2 2 2 2 6 2

9

110

% difference = 100 0.0073%

5

1.374110

2

(b)

1 1

p E mc mc mc

c

c

2 2

2

2

2

2

2

2

2

2 mc

6 8

2

mc 1 1/ 1 2.210 / 3.010 1

c

3 3

0.5110 MeV / c 7.33 10 3.74745 10 /

2

mc u

p(classical) mu 0.5110 MeV / c 2.2 10 / 3.0 10

2

c c

3.7473310

3

MeV c

1/ 2

6 8

MeV / c

7

1.210

% difference = 100 0.0030%

3

2.7474510

2-15. (a)

7 9

60W 60 J / s 3.1610 s / y 1.89610

J

2 9 8

2

8 5

m E / c 1.89610 J / 3.010 m / s 2.110 kg 2.110 g 21

g

(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 3

He H p e

p e

3 4 2

Q m H m m m He

c

2809.450 MeV 938.280 MeV 0.511MeV 3728.424 MeV 19.827MeV

36


2-17.

3 2

H H n

Energy to remove the n = 22.014102 u 2 H 1.008665 un 3.016049 u 3 H

0.006718 u 931.5 MeV / u 6.26MeV

2-18. (a)

(b)

E Eu 4.2eV

c uc 931.510

eV

9

m u 4.5 10 u

2 2 6

m

4.510

u

m Na m Cl 23u 35.5u

9

11 9

error = 7.710 7.710 %

2-19. (a)

2

4 2 2 2

m He c m H c

4 2

m m He 2m H

u

2

uc

3727.409MeV 21875.628 MeV u / 931.5MeV

0.0256u

2 2 2

(b)

E m c 0.0256uc 931.5 MeV / uc 23.8MeV

(c)

dN P 1W 1eV

19

dt E 23.847MeV 1.602 10

J

11

2.62 10 /

s

2-20.

hf

M

1.01M

v

(a) before photon absorbed:

E hf Mc

2

after photon absorbed:

E E 1.01Mc

f

kinetic

2

Conservation of energy requires:

hf Mc E 1.01Mc

2 2

kinetic

Rearranging, the photon energy is:

2 2

hf 1.01Mc Mc Ekinetic

hf 0.01Mc E 0.01Mc

2 2

kinetic

2

(b) The photon’s energy is hf 0.01Mc

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


2-21. Conservation of energy requires that

Thus, m 2

c 2 m

/ 2m

p

E

E

2 2

i f

2 2 2 2

pic mpc p

f

c mpc m

c

p

i

p

f

2 2

2

, or

2 2 and conservation of momentum requires that

, so

2 2 2 2 2

mpc mpc mpc mc mc

2 2 2

2

0 2mpc 2mc mc

2 2 2

4 4 2 2

2

2 m c

2 m

0 mc

2 m c 2

2

2mpc

2m

p

is the minimum or threshold energy E i that a beam proton

must have to produce a π 0 .

2

2 mc 135

E m

c 2 135MeV 2 280MeV

2

2mc

p

2938

2-22.

6

2 2 20010

eV

25

E mc m E / c 3.5710

g

32

5.6110 eV / g

2-23. (a)

(b)

3

E kT per atom

2

3

E 1.38 10 J / K 1.50 10 K

2

16

E 3.10510 J / atom

23 7

H atoms/kg 1 /1.6710 / 6.010

27 26

kg kg atom atoms

16 26 11

Thermal energy/kg 3.110 J / atom 6.010 atoms 1.8610

J

E mc m E / c

2 2

11 2 6

m 1.8610 J / c 2.0710

kg

38


2-24.

1.0W 1.0 J / s p E / c 1.0 J / s / c

p 1.0 J / s / c and,

(a) On being absorbed by your hand the momentum change is

from the impulse-momentum theorem,

Ft p where t 1s F 1.0 J / s / ct 1.0/ c N 3.3

10 N

9

This magnitude force would be exerted by gravity on mass m given by:

/ 3.3 10 9 / 9.8 / 2

m F g N m s 3.410 10 kg 0.34

g

(b) On being reflected from your hand the momentum change is twice the amount in part

9

(a) by conservation of momentum. Therefore, F 6.610 and m 0.68 g.

2

2 2

2-25. Positronium at rest: mc E p c

2

i i

2

Because p

i

0, E 2mc 2 0.511MeV 1.022MeV

i

2 2

2 2

After photon creation; 2mc E f

p

fc

2 2

Because p mc E MeV

f

2

2 2

0 and energy is conserved, 2 1.022 or

2

2mc

1.022 MeV for the photons.

f

2 2

2 2 2

2-26. E pc mc

(Equation 2-31)

E pc mc mc

2

2

1/ 2

2 2 2

pc

1

2

mc

1/ 2

2

2 2 2 2 2 1

2 2

p

mc 1 p / m c

mc

1 p / mc

mc 1

2 2

2

2mc

2 2

mc p /2m

1/ 2

2 2

2 2 2

2-27. E pc mc

(Equation 2-31)

2

(a)

2 2 2 2 2

2

pc E mc 5MeV 0.511MeV 24.74 MeV

or, p 24.74 MeV / c 4.97 MeV / c

2 2 2 2 2 2 2

(b) 2

E mc E / mc 1/ 1 u / c 1 u / c mc / E

2

2

1/ 2

2

1/ 2

u / c 1 mc / E 1 0.511/ 5.0 0.995

39


2-28.

4

3

2

E/mc 2 1

0

0 1 2 3 4

p/mc

2 2

2 2 2

2-29. E pc mc

(Equation 2-31)

1746MeV 2 500MeV 2 mc

2

2

2 2

1/ 2

2 2

mc 1746MeV 500MeV 1673MeV m 1673 MeV / c

E mc 1/ 1 u / c E / mc

2 2 2 2

2

2

1/ 2

2

1/ 2

u / c 1 mc / E 1 1673 MeV /1746MeV 0.286 u 0.286c

2-30. (a) BqR m

u p (Equation 2-37)

mu

B E mc

qR

2

and which we have written as (see Problem 2-29)

2

2

1/ 2

2

1/ 2

u / c 1 mc / E 1 0.511 MeV / 4.0MeV

0.992

1/ 1 u

/ c 1/ 1 0.992 7.83

2 2

And 2

Then,

B

31

9.1110 kg 7.830.992c

19 2

1.60 10 C4.210

m

0.316T

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

40


8

3.010 m/

s

p qBR e 0.5T 2.0 300 MeV / c

c

2-31. (a)

(b)

2

1/ 2

2 2 2

Ek

E mc pc mc mc

2

2 2

300MeV 938.28MeV 938.28MeV

46.8MeV

1/ 2

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

2 2 2 2

Therefore, d d

dt and d d

2 ddt dt . Substituting for

2

d in Equation 2-43 yields

2 2 2 2 2 2 2 2 2

ds c dt

dr r ( d

2 ddt

dt ) dz

Simplifying, we obtain

2 2 2 2 2 2 2 2 2 2

ds ( c r ) dt ( dr r d

2 r ddt dz ) which is Equation 2-44.

2-33.

2

4 GM / c R (Equation 2-44)

Earth radius

6 24

R 6.37 10 m and mass M 5.98

10 kg

11 2 2 24

N m kg kg

2

3.0010 8 m / s 6.3710

6 m

4 6.67 10 / 5.98 10

4

2.87 10 arc seconds

9

2.78 10 radians

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

answer (c) is correct.

41


2-35.

6

GM

11 2 2 30

6

6.67 10 N m / kg 1.99 10

kg

2

2 2

c 1

R

8 2 11

3.0010 m / s 1 0.048 7.8010

m

8 3

3.64 10 radians/century 7.55 10 arc seconds/century

(Equation 2-51)

2-36. From Equation 2-45,

2

r d

dt dt

2 2 2

c r

and dt

2 2

2r ddt r d

dt

2 2 2 2 2 2 .

c r c r

2 2

2

Substituting dt and

2

dt into Equation 2-44 yields

2 2

2

2 2 2 2 2 2r ddt r d

ds ( c r ) dt

2 2 2 2 2 2

c r c r

2

2 2 2 2 r d

2

dr r d 2r ddt

dz

2 2 2

c r

2

r d

2 2 2

c r

c r

2 2 2 2 2 2

ds ( c r ) dt 2r d

dt

Cancelling the

2 2

rd

we have that

2

2

2 r d

2 2 2 2

2

dr r d 2r ddt

dz

2 2 2

2

2

2r d dt

r 2 d c 2 r 2 d

2

2 2 2 2 2 2

c r c r

terms, one of the

2

2

2

2 r d

2 2 2

c r

terms, and noting that

2 2 2

2 2 2 2 2 2 c r d

2

ds ( c r ) dt

( dr dz )

2 2 2

c r

which is Equation 2-46.

2-37. The transmission is redshifted on leaving Earth to frequency f , where

Synchronous satellite orbits are at 6.623R E where

f f f gh c

2

0

0

/ .

2

GM

E

9.9 m/

s

g 0.223 m / s

6.623R

6.623

E

2 2

2

42



6 7

h 6.623R E

6.623 6.37 10 m 4.22

10 m

9 2 7 8

2

f0 f 9.37510 Hz 0.223 m/ s 4.2210 / 3.0010 0.980Hz

9

f f Hz Hz

0

0.980 9.374999999 10

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:

11 2 2 30

N m kg kg

2

3.0010 8 m / s 10

7 m

4 6.67 10 / 3 1.99 10

2

4 GM / c R

(Equation 2-44)

3

1.77 10 radians 0.051 or the angle between the arcs is 2

0.102

2-39. The speed v of the satellite is:

6 3

v 2 R / T 2

6.37 10 m / 90min 60 s / min 7.42 10 m/

s

Special relativistic effect:

After one year the clock in orbit has recorded time t

t/

, and the clocks differ by:

2 2

t t

t t / t 11/ t v / 2 c , because v c. Thus,

7 3

2

8

2

t t

3.1610 s 7.412 10 / 2 3.0010 m 0.00965s 9.65ms

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

9.65ms.

43



General relativistic effect:

2 5

9.8 m / s 3.010

m

8

2

3.010 /

f

gh

2

f0

c m s

11

3.27 10 /

s s

11 7

In one year the orbiting clock gains

3.27 10 s / s 3.1610 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:

2

mc E pc

2 2 2

2 8

2

14

Because, p 0, E mc 4.6kg 3.010 m / s 4.1410

J

2-41. (a)

E m c E / m c 5010 MeV / 0.511MeV

9.7810

2

L L0

/ 10

m

0

2 2 4

e

e

L m m

4 2

9.78 10 10 978 (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

L L m (Note that this is about

4 7

0, so

0

978 978 10 978 9.57 10 .

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:

2 2 4 7

L 10 m / 10 m/9.78 10 1.02 10

m

.

44


E mc mc mc 1 If E mc 938 MeV , then 2.

2-42.

k

(a)

2 2 2 2

k

2 2

mc 2 E 2 pc (Equation 2-32) Where E mc 2 2938MeV

2

pc E mc

6

1/ 2

3

p 2.6410 / c 1.6210 MeV / c

2 2 2 2 2

6

2938 938 2.4610

3 2

(b)

p mu u p / m 1.6210 MeV / c / 2 938 MeV / c 0.866c

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

2

2 2 3 11

p mu mu / 1 u / c 10 kg c / 2 / 1 0.5 1.7310 kg m/

s

Conservation of momentum requires that this also be the momentum p s of the

spaceship:

p m u u c kg m s

2 2 11

s

s s

/ 1

s

/ 1.7310 /

2 2 11

2

Or, msus / 1 us

/ c 1.7310 kg m / s

2

1 / 1.7310 / 1.7310 / 3.3310

6

2

2 5 2 2 11

2

10 kg us

3.3310 kg us

1.7310 kg m / s

11 6 5 4

Or, us

1.7310 kg m / s

/10 kg 1.7310 m / s 5.77 10

c

m c u c kg m s kg m s kg u

2 2 2 2 11 11 2 5 2 2

s s s s

3

(b) In classical mechanics, the momentum of the ejected fuel is: mu mc / 2 10 c / 2,

which must equal the magnitude of the spaceship’s momentum m s u s , so

3 8

10 kg 3.010 m / s

3

4 5

us

10 c / 2 / ms

5.010 c 1.5

10 m / s

6

2 10 kg

(c) The initial energy E i before the fuel was ejected is

system. Following fuel ejection, the final energy E f is:

E

2

i

msc

in the ship’s rest

Where 0.5 and , so 1.155 1.155 1

E energy of fuel + energy of ship mc / 1 u / c m m c / 1 u / c

2 2 2 2 2 2

f s s

u c u c E mc m m c mc m c

The change in energy ΔE is:

2 2 2 2

s f s s

0.15510 10 10

3 2 6 2 6 2

E E

f

Ei

kg c kg c

kg c

E kg c kg E

c

2 2

155 or 155 / of mass has been converted to energy.

45


2-44. The observer at the pole clock sees the light emission of the equatorial clock as transverse

Doppler effect, measuring frequency f, where

f 0

/ f 1 cos (Equation 1-35a)

2

2 2 1 v

/ 2 for the equatorial clock, so f / f0 1 v / c 1

2

2

c

12

f / f 11.193

10 (red shift)

0

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

observes

f f gh c

f

2

/

0

1 / (Equation 2-45)

12

/ f0

1

2.897 10 (blue shift)

2-45. (a)

(b)

BR q e

p 300 / (Equation 2-38)

6 9

p 300 1.5T 6.37 10 1 2.87 10

MeV

For E mc , E E and E pc (Equation 2-32) E pc 2.87 10

MeV

For E pc, u c

and

2 9

k

k

6

T 2 R / c 2

6.3710 m / c 0.133s

2-46.

f

f

0

2

1 GM / c R (Equation 2-47)

The fractional shift is:

The dwarf’s radius is:

f0

f f

1 GM / c R 7

10

f f

0 0

2 4

11 2 2 30

2

3.00 10 8

m/ s 7 10

4

6.67 10 N m / kg 2 10 kg

2 4 6

R GM / c 7 10 2.12 10

m

Assuming the dwarf to be spherical, the density is:

30

M 210

kg

5.010 kg / m

V

6

3

4

2.1210 m / 3

10 3

46


2-47. The minimum energy photon needed to create an e − − e + pair is E p = 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 1.022 MeV / c at minimum. Thus,

momentum conservation is violated unless there is an additional mass “nearby” to absorb

recoil momentum.

2-48.

2

1 vux

/ c u

y

py

muy

m

2 2

2

1 u / c 1 vux

/ c

Canceling γ and 1 vux

/ c

2

, gives:

In an exactly equivalent way, pz pz.

p

y

mu

y

1 u / c

2 2

p

y

2-49. (a) u

u v/ 1 uv / c 2 where v u, so u

2 u / 1 u 2 / c

2

.

the speed of the particle that is moving in S is: u 2 u / 1 u 2 / c

2

Thus,

see that:

2 2

u

4u

1 1 /

2 1 u / c

c c

2 2

2 2 4 4 2 2 2 2

1 2 u / c u / c 4 u / c / 1 u / c

2 2

2

2 2

2

1 u / c / 1 u / c

2

1/ 2

2 2

u

1 u / c

And thus, 1

c

1 u / c

2

2 2

from which we

(b) The initial momentum

pi

in S is due to the moving particle,

(c) After the collision, conservation of momentum requires that:

2 2 2 2 2 2

i

1/ 2 1/ 2

p Mu / 1 u / c p 2 mu / 1 u / c or M 2 m/ 1 u / c

f

2 2

2u 1 u / c

2 2 2 2

1 u / c 1 u

/ c

2 2

p mu / 1 u/ c where u and 1 u/ c were given in (a).

i

p m 2 mu / 1 u / c

i

2 2

47



(d) In S:

E 2 mc / 1 u / c and E Mc (M is at rest.) Because we saw in (c) that

i

2 2 2 2

f

2 2

1/ 2

M 2 m/ 1 u / c , then E E in S.

2 2

In

2

i

f

S : E mc mc / 1 u / c and substituting for the square root from (a),

i

i

E mc u c E Mc u c

2 2 / 1 2 / 2 and 2 / 1 2 / 2

f

. Again substituting for

M from (c), we have: E E .

i

f

2

2-50. (a) Each proton has E m c

1

u = 0.866c. (See Problem 2-40.)

k

, and because we want

p

E

k

2

m c , then γ = 2 and

p

ux

v

(b) In the lab frame S:

ux

where u = v and u

2

x =−u yields:

1 u v / c

2u

2 0.866c

ux

0.990c

2 2

1 u / c 1

0.866c

x

(c) For 2

u 0.990 c, 1/ 1 0.99 7.0 and the necessary kinetic energy in the

lab frame S is:

E m c 1 m c 7 1 6m c

2 2 2

k p p p

2-51. (a) p 0 E / c Mv or v E / Mc

i

(b) The box moves a distance x vt, where t L/ c,

2

so x E / Mc L / c EL / Mc

(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:

x

CM

0 / 2

M m L mL

M m 2 M m

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

x

CM

2 2

/ / 2 /

M EL Mc m L EL Mc

M m

48



Equating the two values of x

CM

2 2

/ / 1 /

m E c E Mc

Because

(if CM is not to move) yields:

2 2

E Mc , then m E / c and the radiation has this mass.

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

v v

2

2 2

2

2

E p c m c and E p c 0

E E 139.56755MeV 105.65839MeV

k

v

2

mc 1 Ev

33.90916MeV

v

v

Squaring, we have

1/ 2

2 2

2

2

33.90916

1

E p c E m c m c

2

2

2 2

2

mc 1 33.90916 233.90916mc 1 mc

1

2 2 2

Collecting terms, then solving for 1 ,

2

mc

2

33.90916

2

1 Substituting mc

105.65839

2 2

2 2 33.90916 mc

1 0.0390 1.0390 so,

1

2 2

1/ 2

Ek

4.12 MeV and pu

109.78 105.66

29.8 MeV / c

c

E 29.8 MeV and p 29.8 MeV / c

v

v

MeV

(b)

2

v v

2 2

2

If v

mass = 190 keV , then Ev

p c + m c and

E E 139.56755MeV 105.65839MeV 0.190MeV 33.71916MeV

k

kv

Solving as in (a) yields

E 109.78 MeV , p 29.8 MeV / c, E 29.8 MeV , and p 29.8 MeV / c

v

v

2-53.

f

f

0

2

1 Gm / c R (Equation 2-47)

Since c f and c f ,

c

0 0

0 0

2

1 GM / c R 1

1

0.000212 0.999788

11 2 2 30

2

3.0010 8 / 6.9610

6

6.67 10 N m / kg 1.99 10 kg

c

m s m

49



/ 0.999788 720.00 nm / 0.999788 720.15nm

0

0.15nm

0

2-54.

2

/ 1 /

u u vu c

y y x

1

du u vdx

duy

ay

dt dt vdx c

1

a

y

ay

y

2

1 y

2

2

1 vux

/ c 2 1 vux

/ c

c

2

/

2

2 2

duy / dt1 vux / c uyv / c dux / dt1 vux

/ c

1 vdx / dt

/ c

2 2

a

a u v / c

2

y x y

2 2

2

2 2

3

1 vux

/ c 1 vux

/ c

1 2

2-55. (a)

(b)

dp

d mv

x

Fx Fx max because ux

0

dt dt

1/2

Fx

m dv / dt mv d 1 v / c /

dt

F

F

F

x

x

x

Because u

2 2

2 2

ma mv / c a

x

x

2 2

1/2

2 2

3/2

v c v c

2 2 2 2

x1 / / x

2 2

3/2

1 v / c

1 / 1 /

ma v c m v c a

3

ma

x

x

3

0, note from Equation 2-1 (inverse form) that a a / .

Therefore, F ma / ma F

3 3

x x x x

dp d mv

y

y

Fy Fy may because uy ux

0

dt dt

F ma because S

moves in + x direction and the instantaneous transverse

y

y

impluse (small) changes only the direction of v. From the result of Problem 2-52

2

(inverse form) with u u 0, a a /

y x y y

2

Therefore, Fy ma

y

ma

y

/ ma

y

/ Fy

/

x

x

50


2-56. (a) Energy and momentum are conserved.

Initial system:

E Mc 2 , p 0

2 2

2 2 2 2

Invariant mass: Mc E pc Mc

Final system:

2 2

mc Mc

2 2

2 0

For 1 particle (from symmetry)

0

2 2 2 2 2

mc Mc / 2 p c Mc / 2

muc

2 2 2 2

Rearranging, 1 u/

c

2

2

2

2

2 2 Mc 1

2 2 2 2

Mc

2mc 2mc 1 u / c

Solving for u,

2

2

2mc

u 1

2

Mc

1/ 2

c

(b) Energy and momentum are conserverd.

Initial system: E 4mc

2

2 2 2

2 2

Invariant mass: Mc 4mc pc

Final system:

2 2 2 2

2 2 2 2

2 2

Invariant mass: 2mc 4 mc pc where pc 4mc Mc

2 2

4mc

Mc

u pc

c E 4mc

2 2

2 2

4mc

Mc

2

4mc

2

1/ 2

2 2

2 2

2

u

Mc

1

2 2

c

4mc

2

2

Mc

u 1

2

4mc

1/ 2

c

51

Chapter 3 – Quantization of Charge, Light, and Energy

3-1. The radius of curvature is given by Equation 3-2.

6

mu 2.510 m / s

25

R m

m 3.9110 m / s C T

Substituting particle masses from Appendices A and D:

27 25 2

R(protron) 1.67 10 kg 3.9110 m/ sC T 6.5

10 m

31 25 5

R(electron) 9.1110 kg 3.9110 m/ sC T 3.6

10 m

27 25

R(deuteron) 3.3410 kg 3.9110 m/ sC T 0.13m

2

19

1.60 10 0.40

qB

C T

27 25

R( H ) 3.3510 kg 3.9110 m/ sC T 0.13m

27 25

R(helium) 6.6410 kg 3.9110 m/ sC T 0.26m

3-2.

u

s

r

u

s

For small values of , s ; therefore, = r r

2

Recalling that euB mu r mu

eB

r eB mu / eB mu

53


3-3.

E u pc

B , , and pc E mc

u c E

2 2

2 2

pc 0.561MeV 0.511MeV 0.2315MeV

u 0.2315MeV

0.41

c 0.561MeV

5

2.010 V / m

3

B 1.63 10 T 16.3G

0.41c

2

3-4. F quB and F G

m p

g

19 6 5

1.610 C3.010 m / s 3.510

T

27 2

1.67 10 9.80 /

FB

quB

1.0310

F m g kg m s

G

p

9

3-5. (a)

2 Ek

/ ee / m

/

mu

R

qB e m B

1/ 2

24.510 4 eV / e

1/ 2

1 2 Ek

/ e 1

11

B e / m 0.325T 1.7610

kg

3

2.2 10 m 2.2

mm

(b)

2 E / ee / m

u

k

frequency f

2R

2R

period 1/ 1.1 10

4 11

24.510 eV / e1.7610 C / kg

3

2

2.210

m

9

9.110

Hz

10

T f s

1/ 2

2

3-6. (a) 1/ 2 mu E , so u 2 E / ee/

m

k

k

1/ 2

11 7

u 2 2000 eV / e 1.76 10 C / kg 2.65 10 m/

s

(b)

x 0.05m

u 2.6510 m / s

1

9

t1 1.89 10 s 1.89ns

7

54



(c) mu

y

Ft1 eE

t1

E

1

11 3 9 6

uy

e/ m t 1.76 10 C / kg 3.33 10 V / m 1.89 10 s 1.11 10 m/

s

3-7. NEk W CVT

3

510 cal / C 2C 4.186 J / cal

C T

N 1.3110

V

19

Ek

2000eV 1.6010 J / eV

14

3-8. 25.41 20.64 10 19

4.47 10

19

Q Q C C n n e

1 2 1 2

20.64 17.47 10 19

3.17 10

19

Q Q C C n n e

2 3 2 3

19.06 17.47 10 19

1.59 10

19

Q Q C C n n e

4 3 4 3

19.06 12.70 10 19

6.36 10

19

Q Q C C n n e

4 5 4 5

14.29 12.70 10 19

1.59 10

19

Q Q C C n n e

6 5 6 5

where the n i are integers. Assuming the smallest Δn = 1, then Δn 12 = 3.0, Δn 23 = 2.0,

Δn 43 = 1.0, Δn 45 = 4.0, and Δn 65 = 1.0. The assumption is valid and the fundamental

19

charge implied is 1.59

10 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 2 mg / 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 ma . The net force in the x-direction is qE bv ma

. At terminal speed

y

a a 0 and v / v

sin .

x y x t

y

v qE

/ b

x

qE

sin

v v bv

t t t

x

x

55


3-11. (See Millikan’s Oil Drop Experiment on the home page at

www.whfreeman.com/tiplermodernphysics6e.)

(a) At terminal speed

mg bv m a b a Substituting gives

3

t

where 4/3 oil

and 6 .

a

2

6 3

3

5 2 3

181.8010 N s / m 5.010 m / 20s

3 2

18 v

t

a

4 oil

g 4 0.75 1000 kg / m 9.8 m / s

1.6610 m 1.6610

mm

6 3 14

m 4

1.6610 m 750 kg / m / 3 1.4410

kg

1/ 2

(b)

19 5

21.60 10 C2.510 V / m

14 2

1.44 10 9.8 /

FE

F qE

and FG

mg 0.57

F kg m s

G

3-12.

T

3

m

2.898 10

m K

(a)

3

2.89810

mK

m

3K

4

9.66 10 m 0.966

mm

3

2.89810

mK

6

(b) m

9.6610 m

9.66m

300K

(c)

3

2.89810

mK

m

3000K

7

9.66 10 m 966

nm

3-13. Equation 3-4:

R

1

R cU.

4

U 8

5 k 4 T 4 / 15h 3 c

2

4

T

. Equation 3-6:

From Example 3-4:

R 1/ 4 cU 1

c k T h c T

4 4

T T 4

5 4 4 3 2 4

8 / 15

5 23

4

2

1.3810 J / K

34 8

J s m s

5.6710 W / m K

3 2

15 6.63 10 3.00 10 /

8 2 4

56


3-14. Equation 3-18: u

5

8hc

hc / kT

e

1

d

d

u f df u d u f u f Because c f , c / f

df

df

u f

5

2

hc f c c f hf

8 / 8

hf / kT 2

3 hf / kT

e 1 f c e 1

2

3-15.

3

3 2.89810

mK

3

(a) mT 2.89810 m K m

1.0710 m 1.07mm

2.7K

(b)

8

c 3.0010 m / s

11

c f f 2.8010

Hz

3

m

1.07 10

m

(c) Equation 3-6:

1 c

R cU 8 k T /15h c

4 4

5 4 4 3 3

8 5 23

3.0010 m / s8

1.3810 J / K 2.7

34 3

8

3

4156.6310 J s 3.0010

m / s

2 6

Area of Earth: 2

A 4r 4

6.38

10 m

E

Total power = 2

4 4

3.0110 W / m

6 2

6 2 6 9

RA 3.0110 W / m 4

6.3810 m 1.54

10 W

3-16.

T

3

m

2.898 10

m K

(a)

T

3

2.89810

mK

9

70010

m

4140K

(b)

(c)

T

T

3

2.89810

mK

9.6610

2

310

m

3

2.89810

mK

9.66

10

3m

2

4

K

K

57


3-17. Equation 3-4: 4

R T R T 2T 16

T 16R

4 4 4

1 1 2 2 1 1 1

3-18. (a) Equation 3-17:

hc / hc / 10 hc / kT 0.1kT

E 0.951kT

hc / kT hc / kT / 10 hc / kT 0.1

e 1 e 1

e 1

(b)

hc / hc / 0.1 hc / kT 10kT

E 4.5910

hc / kT hc / kT / 0.1 hc / kT 10

e 1 e

1

e 1

4

kT

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)

3

3

2.89810

mK

mT 2.89810 m K T1 107K

6

27.010

m

R T and R T 2R 2

T

4 4 4

1 1 2 2 1 1

4 4 1/ 4 1/ 4

2 1 2 1

T 2 T or T 2 T 2 107K 128K

(b)

3

2.89810

mK

m

23

10

128K

6

m

3-20. (a)

m T

3

2.898 10 m K (Equation 3-5)

3

2.89810

mK

m

4

210

K

7

1.45 10 m 145

nm

(b)

m

is in the ultraviolet region of the electromagnetic spectrum.

58


3-21. Equation 3-4:

R

T

3 2 2 2

4

P 1.36

10 W / m R m where R = radius of Earth

abs E E

2 2 3 2 2 2

/ 4

1.36 10 /

P RW m R W m R m

emit E E

R 1.3610

W

R 1.36 10 W / m T

2 3

3 2 E

4

2

2

4

RE

4 m

1.3610 W / m

3 2

4

T T 278.3K 5.3C

8 2 4

4 5.6710 W / m

K

3

3 2.89810

mK

7

3-22. (a) mT 2.89810 m K m

8.7810 m 878nm

3300K

8

3.0010 m/

s

14

fm

c / m

3.4210

Hz

7

8.7810

m

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

40 J / s

40 J / s

N

hf J s Hz

m

34 14

6.6310 3.4210

20

1.77 10 /

photons s

(c) At 5m from the lamp N photons are distributed uniformly over an area

A r m

N / A / s m .

2 2

2

4

100

. The density of photons on that sphere is

3

The area of the pupil of the eye is 2

entering the eye per second is:

2.5

10 m

, so the number of photons

3

/

2.5 10

n N A m

2

20 3

1.77 10 / s

2.510

m

100

m

20 3

2

13

1.7710 / s 2.510 m 1.1010

photons / s

2

2

59


5 5

8 hc A

hc / kT B/

e 1 e 1

3-23. Equation 3-18: u Letting A hc, B hc / kT,and

U

5 5 / 2

1 6

du d A

5

A

B/

B/

d

d

e 1 e 1

B

e

B

B /

2

e

1

6 6 B /

A

B B / B / A

e B

B

/

e 5

2 e 1

5

2 1 e 0

B/

B/

e 1

e 1

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

5

1

B/

e

B. This equation is most efficiently solved by iteration; i.e., guess at a

B/

value for B/λ in the expression 5

1

e

, 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:

3

2.898 10 m K (Equation 3-5)

34 8

6.6310 J s3.0010 m / s

23

B hc hc

4.965114 mT

kT 4.965114 k 4.965114 1.38 10 J / K

m

T

m

m

3-24. Photon energy E hf hc /

E 1240 eV nm / 380nm 3.26eV

(a) For λ = 380nm:

E 1240 eV nm / 750nm 1.65eV

For λ = 750nm:

(b)

15 6 1 7

E hf 4.1410 eV s 10010 s 4.14

10

eV

3-25. (a) hf hc / 0.47 eV.

max

4.1410 15 eV s3.0010 8 m / s

hc

4.87eV

4.87eV

7

2.55 10 m 255

nm

60



(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: 4

R T W m K K W m

4 8 2 4 7 2

5.67 10 / 5800 6.42 10 / .

Approximating the former with u with 127 nm and

255 nm :

9

5

m

9

8

hc 127 10

u127nm 255nm

25510 m 1.2310 J / m

hc/ kT 127 10

e 1

c

R nm J m

4 3

0 255 1.2310 /

4

8 4 3

3.00 10 m / s1.2310 J / m

7 2

46.42 10

W / m

R 0 255nm

R

9 4 3

fraction = 1.4 10

4

3-26. (a)

hc 1240eV nm 1.9eV

t

653 , 4.5910

1.9ev h 4.13610

eV s

4

nm ft

Hz

15

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

dn dE

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

dt dt

9

dn P P

0.05100W 550 10

m

19

1.3810 / s

34 8

dt hf hc 6.6310 J s 3.00 10 m / s

(b)

19

number radiated / unit time 1.3810 / s

17 2

flux 2.7510 / m s

2

area of the sphere

4

2m

61


3-28. (a)

h

4.22eV

4.1410

eV s

15

hf ft

1.0210

Hz

15

(b)

8

3.0010 m/

s

14

f c / 5.3610

Hz

9

56010

m

No.

15 14

Available energy/photon

This is less than .

hf 4.1410 eV s 5.3610 Hz 2.22 eV.

3-29. (a) E hf hc /

hc / E

For E 4.26 eV : 1240 eV nm / 4.26eV 291nm

and since f c / , f 3.0010 m/ s / 291nm 1.03

10 s

8 15 1

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

3-30. (a) First, add a row

f

14

10 Hz

to the table in the problem, then plot a graph

E versus

k,max

f . The slope of the graph is h/e; the intercept on the E k ,max

axis is

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

λ nm 544 594 604 612 633

E

k,max

eV 0.360 0.199 0.156 0.117 0.062

14

f 10 Hz 5.51 5.05 4.97 4.90 4.74

62



15

Slope h/ e 3.90 10 eV s

(b)

15 15

(3.9010 4.1410 )eV s

15

4.1410 eV s

5.6 percent

(c) The work function is the magnitude of the intercept on the

(d) cesium

Ek

,max

axis, 1.78 eV.

3-31.

hc

606.6310 34 J s3.0010 8 m / s

17

E n 2.1710

J

9

55010

m

hc 1240eV nm

3-32. (a) 1.90eV

653nm

(b)

hc 1240eV nm

Ek

1.90eV 2.23eV

300nm

63


h

2 1

1 cos

mc

3-33. Equation 3-25:

34

6.6310 Js1 cos135

4.1410 m 4.1410

1

31 8

9.1110 kg 3.0010 m / s

3

4.1410

nm

100 100 5.8%

0.0711nm

12 3

nm

3 3

1.2410 1.24 10


nm

0.016

m 3

V 8010

V

nm

3-35.

h hc

p

c

(a)

(b)

(c)

(d)

34

1240eV nm

6.6310

J s

27

p 3.10 eV / c 1.6610 kg m / s

9

c 400nm 40010

m

34

1240eV nm

4 6.6310

J s

24

p 1.2410 eV / c 6.6310 kg m / s

9

c 0.1nm 0.110

m

34

1240eV nm

5 6.6310

J s

32

p 4.1410 eV / c 2.2110 kg m / s

7

2

c 310

nm

310

m

34

1240eV nm

6.6310

J s

25

p 620 eV / c 3.32 10 kg m / s

9

c 2nm 2 10

m

3-36.

34

6.6310 Js1 cos110

h

1 cos 3.2610

mc kg m s

9.1110 3.0010 /

2 1 31 8

34 8

6.6310 J s310 m / s

0.51110 1.60 10 /

hc

2.4310

E eV J eV

1 6 19

1

12

m

12

m

12 12

12

2

1 3.2610 m 2.43 3.26 10 m 5.69

10 m

64



hc

1240eV nm

5

E2 2.1810 eV 0.218MeV

3

2

5.6910

nm

Electron recoil energy Ee

E1 E2

(Conservation of energy)

E 0.511MeV 0.218MeV 0.293MeV

. The recoil electron momentum makes an

e

angle θ with the direction of the initial photon.

P E

h/λ 1

θ

110°

h/λ 2

20°

h

2 2

cos 20 p sin 1/ 2

e

c E mc sin

(Conservation of momentum)

2

sin

8 34

3.0010 m / s6.6310 J scos 20

2 2

1/ 2

12 13

5.6910 m 0.804MeV 0.511MeV 1.6010 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 0.647 1.67 2.45 3.98 4.80

degrees 45 75 90 135 170

1 cos 0.293 0.741 1.000 1.707 1.985

65


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 0.004nm

100 100 0.15percent

2.426nm 2.426nm

h

2 1 1 cos 0.01 1

Equation 3-25

mc

3-38.

1 100 h 1 cos 1000.00243 nm1 cos90

0.243nm

mc

3-39. (a)

hc 1240eV nm

4

E 1

1.747 10 eV

0.0711nm

1

h

2

1 1 cos 0.0711nm 0.00243nm 1 cos180 0.0760nm

mc

(b)

(c)

E

2

hc 1240eV nm

4

1.634 10 eV

0.0760nm

2

(d)

3

Ee

E1 E2 1.128

10 eV

66


3-40. (a)

h / mc1 cos

From protons:

34

6.63 10 J s/ 1.67 10 27 kg 3.00 10 8 m/ s1 cos120

1.99 10 m 1.99

10

15 6

nm

31

(b) Similarly, for electrons ( m 9.11 10 kg )

2.4310 m 2.43

10

12 3

nm

26

(c) Similarly, for N 2 molecules ( m 4.68 10 kg )

4.72 10 m 4.72 10

17 8

nm

h

2 1

1 cos 0.0711 0.00243 1 cos

mc

3-41. nm

λ 2

0.0750

0.0730

0.0710

θ 1 cos

2

(nm)

0° 0 0.0711

45° 0.293 0.0718

90° 1 0.0735

135° 1.707 0.0752

0 1

1

cos

2

0.0745 0.0720 nm

Slope = 2.38110

1.50 0.45

h

mc

3

3 31 8 34

h 2.381 10 nm 9.11 10 kg 3.00 10 m / s 6.51 10 J s

67


3-42. (a) Compton wavelength =

h

mc

h 6.6310

J s

mc kg m s

34

12

electron: 2.4310 m 0.00243

31 8

9.1110 3.0010 /

nm

(b)

h 6.6310

J s

mc kg m s

34

15

proton: 1.3210 m 1.32

hc

E

(i)

(ii)

27 8

1.67 10 3.0010 /

1240eV nm

E eV MeV

0.00243nm

5

electron: 5.1010 0.510

1240eV nm

E eV MeV

1.3210

nm

8

proton: 9.3910 939

6

fm

3-43. Photon energy E hf hc / allows us to rewrite Equation 3-25 as

Rearranging the above,

hc hc h

(1 cos )

E E mc

2 1

hc h (1 cos )

hc

E mc E

2 1

Dividing both sides of the equation by hc yields

Or

2

( E1

/ mc )(1 cos ) 1

(1 cos )

2

2 1 1

1 1 1

E mc E E

E

E1

( E / mc )(1 cos ) 1

2 2

1

3-44. (a) eV0 hf hc /

e 0.52 V hc / 450 nm

(i)

e 1.90 V hc / 300 nm

(ii)

68



Multiplying (i) by 450 nm / e and (ii) by 300 nm / e , then subtracting (ii) from (i) and

rearranging gives:

300nm1.90V 450nm0.52V

2.24eV

e

150nm

(b)

hc

9

30010 4.14

e m V

1.90 2.24 h 6.6310

8

e 300nm 3.0010 m/

s

34

J s

3-45. Including Earth’s magnetic field in computing y 2 , first show that y 2 is given by

y

2

e B x x

m E

1 B x

2 E

2 2

1 2 E 2

where the second term in the brackets comes from

1 2

Fy euBE may and y ayt

.

2

Thus, 1

e B x x 1 B x

m Ey2 2 Ey2

2 2

1 2 E 2

The first term inside the brackets is the reciprocal of

11

0. 7 10 C,

Thomson’s value for e/m. Using Thomson’s data (B =

4

5. 5 10 T ,

E ) and the modern value for e/m =

4

1. 5 10 V / m, x1 5cm, y2 / x2

8/

110

11

1. 76 10 C / kg and solving for B E .

1 BE

Bx

2 Ey

2

2

2

8 20 10

12

. .

The minus sign means that B and B E are in opposite directions,

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

12 4

8. 20 10 21. 5 10 V / m8/

110

4 2

5.

510 T810

m

5

BE

3.

110 T 31T

2

69


3-46. (a)

2

mu

Q Ne and cM T N 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 Qu

2

N e mu 2

m 2 T

eB eB

(b) u

mu m

Substituting u into the results of (a),

e

m

/ m

2

Q eB

2cM T

Solving for e/m,

2

e 2cM T

m

QB

2 2

3-47. Calculate 1/λ to be used in the graph.

1/λ (10 6 /m) 5.0 3.3 2.5 2.0 1.7

V 0 (V) 4.20 2.06 1.05 0.41 0.03

5

4

V 0 (V)

3

2

1

0

-1

-2

1

2 3 4 5

1/λ (10 6 /m)

(a) The intercept on the vertical axis is the work function .

2.08eV .

70



(b) The intercept on the horizontal axis corresponds to the threshold frequency.

1

t

6

1.65 10 /

c

m

8 6 14

ft

3.0010 m / s 1.6510 / m 4.95

10 Hz

t

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

point.

h 1 V

4. 20V

2.

08V

e c 1/ 3. 0010 m / s 5. 010 / m 0

0

15

4. 1910

/

8 6

eV Hz

3-48. In the center of momentum reference frame, the photon and the electron have equal and

opposite momenta. p

E

/ c pe

.

1/ 2 1/

2

2 2 2 4 2 2 4

The total energy is: E Ee

E pec m c E E

m c

By conservation of momentum, the final state is an electron at rest, p

e

0 . Conservation

of energy requires that the final state energy E is

2 2

Squaring yields,

2 2 2 2 2 2

mc mc E E E mc mc E

2 0. This can be true

only if E

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

2

1/

2

2 2 2 2

E E E mc E p c mc

e

1/ 2 1/

2

2 2

2 2 2 2 2 2

mc E p c mc E mc

3-49. Bragg condition:

10

m 2d 2 0 28nm 20 1 9210 m 0 192nm

sin . ( )( . )(sin ) . . .

This is the minimum wavelength

m

that must be produced by the X ray tube.

3 3

1. 2410 1.

2410

3

m nm or V 6. 47 10 V 6.

47kV

V

0.

192

71


3-50. (a) /

E 100W 10 4 s 100J s 10 4 s 10

6 J

(b)

6

E 10 J

3

The momentum p absorbed is p 3. 3310

J s / m

8

c 3. 0010

m/

s

f i f

3

p m v v 2 10 kg v 0 3. 3 10

3 J s / m

3

3. 3310

J s / m

v

1. 67m / s

3

210

kg

(c)

210 3 kg 1. 67m / s 2

1 2 3

E mv

f

2.

78

10 J

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:

E mc E E mc E E E hf hf

2 2

1 2 k

k 1 2 1 2

h

2

1

1 cos ,

mc

From Compton’s equation, we have:

1 1 h

f f mc

Thus, 1 cos

2 1

2

2

1 1 h

f mc

1

cos

f

f f mc mc hf 1cos

1

2 2 2

2 1

1

Substituting this expression for f2

into the expression for

on

1

E

k

(and dropping the subscript

E

k

k

f ): 2

hfmc

hf

cos

2

1

cos

cos

2 2

hfmc hf hfmc

2 2

mc hf 1 mc hf 1

1 hf

hf

2

mc

1 cos

E has its maximum value when the photon energy change is maximum, i.e., when

hf

so cos 1.

Then Ek

2

mc

1

2 hf

72


3

3 2.

89810

mK

4

3-52. (a) mT 2. 89810 m K T 3.

5010

K

9

82.

810

m

(b) Equation 3-18:

5 hc / 70nm kT

u 70nm

70nm

/ e 1

u nm

nm e

5 hc / 82.

8nm kT

82. 8 82. 8 / 1

34 8

hc 6. 6310 J s3. 0010

m / s

where

9 23 4

70nm kT 7010 m1. 3810 J / K 3.

510

K

5.

88

and

5 5.

88

70nm / e

1

5 4.

97

hc

u 70nm

4. 97 0.

929

82. 8nm

kT u 82. 8nm 82. 8nm / e 1

Similarly,

5 4.

12

100nm / e

1

5 4.

97

u 100nm

u 82. 8nm 82. 8nm / e 1

0.

924

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

7 2

W m (see solution to Problem 3-25). Evaluating u

6. 42 10 /

525nm (midpoint of visible) and

700nm 350nm 350 nm.

u

34 8

5

. J s . m / s nm nm

34 8

6. 6310 J s3. 00 10

m / s

23

1. 3810 J / k 5800K 525nm

with

8

6 63 10 3 00 10 525 350

0. 389J / m

exp

1

c

R 350 700 u 3. 0010 m / s0. 389J / m / 4 2. 92

10 W / m

4

8 3 7 2

7 2 7 2

Fraction in visible = R R W m W m

350 700 / 2. 9210 / / 6. 4210 / 0.

455

3

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

f

14

10 Hz 11.83 9.6 8.22 7.41 6.91

V 0 (V) 2.57 1.67 1.09 0.73 0.55

73



3

Li

2

1

0

−1

−2

−3

−4

2

4 6 8 10 12

Pb

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 V 0 and the intercept on the vertical

axis,

h

e

2. 57V

2.

40V

14

11.

5310 Hz 0

or,

19

4. 97V1.

6010

C

h 6.

8910

14

11.

5310

Hz

(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

than / 8 14

t

. / .

34

J s

14

9. 8 10 Hz;

therefore, no photon wavelengths larger

c f 3 0010 m s 9 810 Hz 306nm

will cause emission of

photoelectrons from Pb.

3-55. (a) Equation 3-18:

gives

5

8hc

u Letting C 8 hc and a hc / kT

hc / kT

e 1

5

C

u

a /

e 1

(b)

5 5 2

1 6

du d C

5

C

a/

a/

d

d

e 1 e 1

a /

e

a

2

a /

e

1

6 6

a /

C

a a / a / C

e a

a /

e 5

2 e 1

5

2 1 e 0

a/

a/

e 1

e 1

74



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

a /

5

1e

a

Thus,

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

a /

a / in the expression 5

1e

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

a

hc

seven digits), so we have 4.

965114

kT

(d) T

m

m

34 8

6. 6310 J s3. 00 10

m / s

23

J K

hc

4. 965114 k 4. 965114 1. 38 10 /

m

Therefore,

m T

3

2.

898 10 m K Equation 3-5

3-56. (a)

I

P 1W

1

17 2

4. 97 10 eV / m s

2 2

19

4

R

4

1m

1. 602 10

J / eV

(b) Let the atom occupy an area of 01nm

2

. .

dW IA

17 2 9 3

4. 97 10 eV / m s 0. 1 nm 10 m / nm

4. 97 10 eV / s

dt

2

2

2eV

(c) t 403s

6.

71 min

3

dW / dt 4. 97 10

eV / s

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

E

k

MeV / c 2

2

p 15 1u

4

1.

10 10 eV

2m 2 12u

931 5MeV c

2

. /

Internal energy U 15MeV 0. 0101MeV 14.

9899MeV

75



(b) The nucleus must recoil with momentum equal to that of the emitted photon, about

14.98 MeV/c.

E

k

. MeV / c 2

2

p 14 98 1u

1.

00 10

2m 2 12u

931 5MeV c

2

. /

2

eV

E U E k

14. 9899MeV

0. 0100MeV 14.

9799MeV

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

E

k

hf

2

1 mc / 2

hf

hf

2hf

2

E 520keV 520keV

1 511keV / 2hf 2hf 511keV

2

Thus, hf

hf

520 520 511 / 2 0

Solving with the quadratic formula: hf

2

2

520

520 2 520 511

708keV

2

(only the + sign is physically meaningful). Energy of the incident gamma ray

hf 708 keV.

34 8

6. 6310 J s3. 00 10

m / s

16

708keV 1. 60 10

J / keV

hc

12

708keV 1. 76 10 m 1.

76 pm

3-59. (a) Ek

50 keV and 2 1

0.

095nm

hc hc 4

4 1 1 5.

010

eV

5.

010

eV

0.

095

1 2 1 1

1

2

1

0.

095

1

hc

4

2

0. 095 5.

010

eV

hc

2

2hc

0.

095nm hc

1 0.

095nm

4 1

0

4

5 10 eV

510

eV

0. 04541

2.

3610 0

2 3

1 1

76



Applying the quadratic formula,

1 /

2 3

2

0. 04541 0. 04541 4 2.

36 10

1

2

0. 03092 nm and 0.

1259nm

1 2

hc 1240eV nm

(b) E1

40. 1keV Eelectron

9.

90keV

0.

03092nm

1

hf

3-60. Let x in Equation 3-15:

kT kT

2 3

nx 0 x x x

2 3

fn

A e A

e e e e A y y y

n0 n0

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

1 1

1 1 2 3

1 y

, i.e., 1 y

1 y y y . Then 1

n

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

E n / kT nhf / kT nx

n

n0 n0 n0

E E Ae A nhfe Ahf ne

nx nx nx

Note that ne d / dxe . e y 1

f A 1 y 1 gives A 1 y.

But 1 , so we have

d d dy

dx dx

dx

Since

1 1 1

nx

nx

1 2 2

ne e y y y y

x

dy d e

x

e

y.

dx dx

Multiplying this sum by hf and by A 1

y

n0

, the average energy is

1 1

2

nx

hfy hfe

E hfA

ne hf y y y

1

y 1e

Multiplying the numerator and the denominator by

E e

hf

, which is Equation 3-17.

1

hf / kT

x

x

x

e and substituting for x, we obtain

77

Chapter 4 – The Nuclear Atom

4-1.

1 1 1

R R m

2 2

m n

mn

7 1

where 1.

097 10 (Equation 4-2)

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

7 1

R 1.

097 10 m

2

L

1

L

limit 1. 097 10 m

91. 1610 m 91.

16nm

7 1 9

1 1

7 1

R 1. 097 10 m / 4

2

B

2

B

limit 4/ 1. 097 10 m

3. 646 10 m 364.

6nm

7 1 7

1 1

R

2

3

P

7 1

1. 097 10 m / 9

7 1 7

P

limit 9/ 1. 097 10 m 8. 204 10 m 820.

4nm

4-2. 1 1 1

R where m 2 for Balmer series (Equation 4-2)

2 2

m n

mn

7 1

1 1.

097 10 m 1 1

9 2 2

379. 1nm 10 nm / m

2 n

9

1 1 10 nm / m

0.

2405

2 7 1

4 n 379. 1nm 1.

097 10

m

1

2

n

2 1

0. 2500 0. 2405 0.

0095

1 / 2

n n 1/ 0. 0095 10.

3 n 10

0.

0095

n10 n

2

79


4-3. 1 1 1

R where m 1 for Lyman series (Equation 4-2)

2 2

m n

mn

7 1

1 1.

097 10 m 1

1

9 2

164. 1nm 10 nm / m

n

9

1 10 nm / m

1 1 0. 5555 0.

4445

2 7 1

n 164. 1nm 1.

097 10

m

1 / 2

n 1/ 0. 4445 1.

5

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

1 1 1

4-4. R

m

2 n

2

mn

(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

1. 097 10 m

2.

468 10 m

4 5

45

7 1 5 1

2 2

1

45

5 1

2.

6810

m

1

46

5 1

3.

809 10

m

6

4.

052 10 m 4052 nm. Similarly,

6

2.

625 10 m 2625

1

47

5 1

4.

617 10

m

6

2.

166 10 m 2166

1

48

5 1

5.

142 10

m

These lines are all in the infrared.

6

1.

945 10 m 1945

nm

nm

nm

4-5. None of these lines are in the Paschen series, whose limit is 820.4 nm (see Problem 4-1)

1 1 1

and whose first line is given byL R

2 2 34

1875nm.

3 4

34

Also, none are in

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



1 1 1

1. 097 10 m

1.

341 10 m

5 5

56

7 1 5 1

2 2

1

56

5 1

1.

34110

m

1

57

5 1

2.

15510

m

6

7.

458 10 m 7458 nm. Similarly,

6

4.

653 10 m 4653

1

58

5 1

2.

674 10

m

6

3.

740 10 m 3740

nm

nm

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

4-6. (a)

f

2

b nt

(Equation 4-5)

28 3

For Au, n 5. 9010 atoms / m (see Example 4-2) and for this

foil t 2 0 m 2 010

m

6

. . .

2

279 ke 90 2791.

44eV nm

kq Q

b cot cot

2

mv 2 2K

2

6

2 7.

010

eV

1. 6310 nm 1.

63

10

5 14

2

m

f 1. 6310 m 5. 9010 / m 2. 010 m 9.

8

10

14 28 3 6 5

(b)

b cot / / cot /

b90tan 90 / 2 / tan 45

/ 2

For 45 , b 45 90 45 2 90

2

and f 45 5.

7 10

3. 92 10 nm 3.

9210

4

5 14

m

b tan / / tan /

For 75 , b 75 90 90 2 75

2

and f 75 1.

66 10

2. 12 10 nm 2.

1210

4

5 14

m

f

45 75 5. 710 1. 6610 4.

05

10

4 4 4

Therefore,

81



(c) Assuming the Au atom to be a sphere of radius r,

4 3 M

197g / mole

r

3 N 6 02 10 atoms mole 19 3g cm

A

23 3

. / . /

3 197g / mole

r

23 3

4

6. 02 10 atoms / mole19. 3g / cm

3 10

r 1. 62 10

cm 1. 6210 m 16.

2nm

1/

3

4-7.

1 A

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

2

2

sin 4 / sin 4 /

quantities in parentheses in Equation 4-6.

(a)

(b)

4 4

N

10 A/ sin 10 / 2 sin 0.

5

4 4

N

1 A/ sin 1 / 2 sin 5

1.

0110

4

N

30 sin 0.

5

1.

2910

4

N

1 sin 15

6

4

kq

Q

4-8. b cot (Equation 4-3)

2

mv 2

2

k 2e Ze 1.

44MeV fm Z

= cot

cot

m v 2 E

2

k

1.

44MeV fm79 2

13

cot 8.

5

10

7.

7MeV

2

m

4-9.

r

d

2

kq

Q ke 2 79

12 / m v E

2

k

(Equation 4-11)

1.

44MeV fm279

For E k

5. 0MeV : r d

45.

5 fm

50 . MeV

For E k

7. 7MeV : r

d

29.

5 fm

For E k

12MeV : r

d

19.

0 fm

82


4-10.

r

d

2

kq

Q ke 2 79

12 / m v E

2

k

1.

44MeV fm213

(Equation 4-11)

E k 94 . MeV

4 fm

4-11. 2 6

x N 10 N 0. 01 N 10 / 0.

01 10 collisions

rms

6

t 10 m

n

10

t

10 m

4

10 layers

10 4 atomic layers is not enough to produce a deflection of 10 , assuming 1

collision/layer.

4-12. (a)

f

2

b nt

(Equation 4-5)

For 25 (refer to Problem 4-6).

2

279 ke 2791.

44eV nm

25 25

b cot cot

6

2K

2 2 7.

0 10 eV 2

7. 3310 nm 7.

33

10

5 14

2

m

f 7. 3310 m 5. 9010 / m 2. 010 m 1.

992 10

Because

For 45

,

14 28 3 6 3

N f N 1000 N 1000 / 1. 992 10 5.

02

10

. eV nm

2 79 1 44 45

b cot 3.

9210

2 7.

0 10

2

6

eV

2

3 5

f 3. 9210 m 5. 9010 / m 2. 010 m 5.

70

10

14

14 28 3 6 4

Because N

f N . 4 .

5

(b) N

45 5 7010 5 0210 286

25 45 1000 286 714

(c) For 75

b b 25 tan 25 / 2 / tan 75 / 2 2.

12

10 m

14

,

2 2

3 14 14

f 1. 992 10 2. 12 10 m / 7.

3310

m

2

1. 992 10 2. 12 / 7. 33 1.

67 10

3 4

m

83



For 90 ,

b b 25 tan 25 / 2 / tan 90 / 2 1.

63

10 m

14

2 2

3 14 14

f 1. 992 10 1. 6310 m / 7.

3310

m

2

1. 992 10 1. 63/ 7. 33 9.

85

10

3 5

N f N

5 5

9. 85 10 5.

02 10 49

N

75 90 84 49 35

4-13. (a)

r

n

2

na

0

(Equation 4-18)

Z

2

6 0.

053nm

r6

1.

91nm

1

(b)

6

2

6 0.

053nm

r He 0.

95nm

2

4-14.

a

2

mke

0 2

(Equation 4-19)

c c 1 1 h 1 c

2 2 2 2

mcke mc ke / c 2

mc ke / c 2

2 4

mk e

E1 (from Equation 4-20)

2

2

2

2 2 2

2 2

mc ke mc ke

1 2 2

mc

2

2 c 2 c 2

5

c 0. 00243nm 1 2 2 5.

1110

eV

a0 0. 053 nm E1 mc 13.

6eV

2

2

2

1 137 2 2 137

/

4-15.

1

2 1 1

ZR

n

2 2

f n

i

(Equation 4-22)

84



2

1 1 1 ni

1

R =R

2 2 2

ni 1

ni ni

ni

2 2 2

n

n

i

n

i

i

= 91.

17nm 2

R n ni

1

2 7 2

i

1 1.

0968 10 mni

1

4 9

2 91. 17nm 121. 57 nm 3

91. 17nm

102.

57nm

3 8

16 4

91 . 17 nm

97 . 25 nm 91 . 17

nm

15

None of these are in the visible; all are in the ultraviolet.

∞

4 3 2

80 90 100 110 120 130

λ, nm

4-16. L mvr n (Equation 4-17)

m 5. 9810 kg v 2r / 1y 2r / 3.

16 10 s

E

24 7

E

7 2 7

2

/ 3. 16 10 / 2

/ 3.

16 10

2

24 11

. kg . m

74

2.

54 10

7 34

3. 1610 s1.

05510

J s

n m r s r mr s

2

5 98 10 1 50 10

2 2

mv n / r E mv / 2m n / r / 2 m (from Equation 4-17)

2

1. 055 10 34 Js 2.

54 10 74

1

2

. 11 .

24

2

1 2 0 210 10

40

E n n . J

r

2m 1 50 10 m 5 98 10

kg

This would not be detectable.

2 2 2

n 2r

1. 055 10 34 J s 2.

54 10

74

r

3

2

2m

r

11 24

1. 50 10 m 5.

98 10

kg

33

E

5.

34 10

or 0 210 10 5 34 10 3 9310

40 33 75

r . J / . J / m . m

11

The orbit radius r would still be 1. 50

10 m.

r

85


4-17.

f

rev

2 2 4

mk Z e

(Equation 4-29)

3 3

2

n

2

2 2 2 2

2 2 2 2

mc Z ke cZ ke cZ

3 2 3

3

2 n h / mc n c cn

c

2

8 2

3. 0010 m/

s 1 1

8.

22 10

3

137

rev

9

0.

0024310 m 2

N f t Hz s

14

Hz

8. 22 10 14 10 8 8.

22 10 6 revolutions

4-18. The number of revolutions N in 10 -8 s is:

N s s

10 8

/ time/revolution 10 8 / circumference of orbit/speed

N 10 8 s / C / v 10 8 s / 2 r / v

The radius of the orbit is given by:

2 2

4 0 0529

0

na

r

Z

. nm

3

so the circumference of the orbit C 2

r is

2

C 2

4 0. 0529nm / 3 1. 77nm 1.

77 10

9 m

The electron’s speed in the orbit is given by

2 2

v kZe / mr

5

v 6. 54

10 m/

s

Therefore, N s C v

9 2 2 19

8. 9910 N m / C 31.

6010

C

31 9

9. 1110 kg 1.

77 10

m

10 8 / / 3.

70 10 6 revolutions

In the planetary analogy of Earth moving around the sun, this corresponds to 3.7 million

“years”.

2

9.

1110

2 2 31

4

au

a

2 2 0

0. 0529nm 2.

5610

nm

28

ke eke

1.

6910

kg

e

e

4-19. (a)

E

E 13 6eV 2520eV

2 2 9 11 10

e

(b) .

kg

2 4 2 4

ke ke 1.

6910

28 kg

2 2 0

31

e

e

. kg

86



(c) The shortest wavelength in the Lyman series is the series limit ( n ,

n 1). The

photon energy is equal in magnitude to the ground state energy

hc 1240eV nm

0.

492nm

E 2520eV

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

i

E

.

f

4-20.

4-21.

1 / 2 2

1 /

. eV

2

2

2 2

nE

2 5 39

E Z

E0

/ n Z =

1.

26

E 13.

6eV

Energy (eV)

0

-2

-4

(b)

0

(c)

(d)

n=∞

n=4

n=3

n=2

-6

-8

-10

-12

-14

(a)

n=1

(a) Lyman limit, (b) H

line, (c) H

line, (d) longest wavelength line of Paschen series

1 1 1

4-22. (a) R n

2 2

f n

i

For Lyman α:

1 1 1

1 2

7 1

1. 097373 10 m

121.

5023

2 2

L

hc 1240eV nm

EL

E L

10. 2056 eV and pL

10. 2056eV / c

121.

5023nm

c

L

nm

87



Conservation of momentum requires that the recoil momentum of the H atom

p

H

p and the recoil energy EH

is:

L

2 2 10. 2056eV / c

2

EH pH / 2mH pHc

/ 2mHc

2 1 007825 931 50 10

5 55

10

8

. eV

2 6 2

. uc . eV / uc

2

E 5.

510

eV

510

E E 10.

21eV

8

H

(b)

L

H

9

4-23. (a) For C 5+ (Z = 6)

E

n

2

Z 489.

6

13.

6

2 2

n n

E n (eV)

0

-100

-200

n=5 n=∞

n=4

n=3

n=2

E ∞ = 0 eV

E 5 = -19.6 eV

E 4 = -30.6 eV

E 3 = -54.4 eV

E 2 = -122.4 eV

E 1 = -489.6 eV

-300

-400

-500

n=1

hc hc 1240eV nm

(b) 18.

2nm

E E3 E2

54. 4 122.

4eV

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

88


4-24. (a) The reduced mass correction to the Rydberg constant is important in this case.

1 1

R R R

m

1

m/

M

2

E

2

n

hcR / n (from Equations 4-23 and 4-24)

6 1

5.

4869 10 (from Equation 4-26)

6

1 9

2

E1 1240eV nm 5. 4869 10 m 10 m/ nm / 1 6.

804eV

Similarly, E2 1. 701 eV and E3

0.

756eV

(b) Lyman α is the n 2 n 1 transition.

hc 1240

E E hc eV nm 243nm

2 1

E2 E 1

1. 701eV 6.

804eV

Lyman β is the n 3 n 1 transition.

hc 1240eV nm

205nm

E E 0. 756eV 6.

804eV

3 1

4-25. (a) The radii of the Bohr orbits are given by (see Equation 4-18)

2

r n a0/ Z a0

. nm Z

where 0 0529 and 1 for hydrogen.

2 4

n 600 , r 600 0. 0529nm 1. 9010 nm 19.

0m

For

This is about the size of a tiny grain of sand.

(b) The electron‟s speed in a Bohr orbit is given by

2 2

v ke / mr Z

with 1

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

2

8. 99 10 1. 609 10 / 9. 11 10 19.

0 10

2 9 2 19 31 6

v N m C kg m

2 7 2 2 3

v 1. 3310 m / s v 3. 65

10 m/

s

For comparison, in the n = 1 orbit, v is about

6

2

10 /

m s

89


1 1 1

R Z 1

n1 n2

4-26. (a) 2 2 2

1

7 1 2 1 1

11

3 2 2

1. 097 10 m 42 1 6. 1010 m 0.

0610nm

1 3

1

7 1 2 1 1

11

4 2 2

1. 097 10 m 42 1 5. 7810 m 0.

0578nm

1 4

(b)

1

7 1 2 1

11

2

lim it

1. 097 10 m 42 1 0 5. 4210 m 0.

0542nm

1

1 R Z 2 1 1 2 1 1

1 R Z

1 for

2 2

2 2

n1 n2

1 2

K

1/

2

1

1

Z 1

2

1

R 1

0. 0794nm1. 097 10 / nm3/

4

4

Z 1 39.

1 40 Zirconium

4-27.

1/

2

4-28. (a)

1 2 8 1 2 18

Z 43 ; f / 2110 Hz / f 4.

4

10 Hz

1 2 8 1 2 18

Z 61 ; f / 3010 Hz / f 9.

0

10 Hz

1 2 8 1 2 19

Z 75 ; f / 37 10 Hz / f 1.

4

10 Hz

Note:

1/

f 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

R

where

1

1

n

Z

1 2 1

2

R

m n

7 1

1.

097 10 and 2

7 1

. m

(Equation 4-37)

1

11

6. 89 10 m 0.

0689

2 1

1 097 10 43 1 1

4

nm

90



Similarly,

For Z = 61, λ = 0.0327nm

For Z = 75, λ = 0.0216nm

4-29.

2

na

0

rn

(Equation 4-18)

Z

The n =1 electrons “see” a nuclear charge of approximately Z 1, or 78 for Au.

r nm nm m nm fm m fm , or about 100 times

4 9 15

1

0. 0529 / 78 6. 8 10 10 / 10 / 680

the radius of the Au nucleus.

4-30.

E

n

2

Z

13.

6 eV (Equation 4-20)

2

n

For Fe (Z = 26)

26 2

E1 13. 6 9.

194keV

2

1

The fact that E 1 computed this way (i.e., by Bohr theory) is approximate, is not a serious

problem, since the K α x-ray energy computed from Figure 4-19 provides the correct

spacing between the levels.

The energy of the Fe K α x-ray is:

1 2 8 1 2

where / 12.

2 10

/

E Fe K hf f Hz

/

2

34 8 1 2 16

E Fe K

6. 626 10 J s 12. 210 Hz 9. 862 10 J 6.

156keV

E2 E1 E K

9. 194 6. 156 3.

038keV

Therefore,

The Auger electron energy 2

6. 156 3. 038 3.

118

E K

E keV

4-31.

2

E mec 772.

6keV

2

8 8

511keV

1 2. 2510 / 3. 0010

m/

s

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

91



511keV

E keV v c

1

2

2 2 2

740. 1 / 1 511/

740

2

12 /

2 8

v 1 511 / 740 c 2. 17 10

m / s

4-32. (a)

E E Z n

2 2

1 0

/ (Equation 4-20)

2 2 4

13. 6eV 74 1 / 1 7. 2510 eV 72.

5keV

2 2 2

2 3

(b) E E Z / n . eV . eV /

1 0

69 5 10 13 6 74 1

2 3

74 69. 5 10 eV / 13.

6eV

1 /

3

eV eV

2

74 69. 510 / 13. 6 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

1 / 2 10

8 1 /

f Hz 2

6.14 8.77 10.37 13.05 16.28 18.45 20.64 29.62

60

Z

40

20

0

0

5

10

15

20

25 30

f 10 Hz

1 / 2 8 1 / 2

92



58 10

slope = 1.

9010

8

30 4.

8 10

Hz

8 1/

2

30 13

slope (Figure 4-19) = 2. 1310

Hz

8

5 7 10

The two values are in good agreement.

8 1/

2

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 3 rd 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.

4-35. 60

60

50

15.7eV

E

(eV)

40

30

44.3

29.6

14.7eV

Average transition energy = 15.7 eV

20

16.6eV

10

13.0

93


4-36.

hc 1240eV nm

E 1. 610eV

.

790nm

The first decrease in current will occur when the

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)

m1

m

2

is initially at rest: v v

m1

m2

.

1f 1i the incident electron in a head-on collision is:

m

m

v

KE KE v v m m

f

2 1 2 2

2 2 1i

1i

ei

ef 1i

1f

1

2

2 2

KEei v1 i

v1

i

The fraction of the initial kinetic energy lost by

2

2 2 2

m1 m 0. 511MeV 200uc 931. 5MeV / uc

2

= 1 1

2 2

m1 m2

0. 511MeV 200uc 931. 5MeV / uc

= 1.

1010

5

2

v

If the collision is not head-on, the fractional loss will be less.

4-39. (a) Equation 4-24:

E E n n

n

2 2

0

/ 13.6 / eV

E

1 1 1 1

( n1) n

46 45

4

n1 En

13.6 eV 13.6 eV=2.89 10 eV

2 2 2 2

2 2 3

(b) Ionization energy E 13.6 / n 13.6 / 45 6.72

10 eV

n

4 19 34

(c) E hf hc / f E / h (2.89 10 eV)(1.60 10 J/eV)/6.63 10 J

s

10

f 6.97 10 Hz

c

8 10 3

/ f (3.00 10 m/s)/(6.96 10 Hz) 4.30 10 m=4.30mm

94



(d) Equation 4-18:

For hydrogen:

2

r /

n

n a0 Z

r

45 a /1107nm 1.07 10 mm , or 2025× the radius of the

2 4

45 0

hydrogen atom ground state.

4-40. (a) Equation 4-26:

1

R R

1 m/

M where 7 1

R 1.0973732

10 m

R

R

d

d

t

1

R 31 27

1 9.109410 kg / 3.343610 kg

1.097074310 m

7 1

1

Rt

R 31 27

1 9.109410 kg / 5.007410 kg

R 1.0971736 10 m

7 1

1 1 1

(b) Equation 4-22: R n

2 2

f n

i

with Z 1.

1 1 1 1 5

The Balmer α transition is n 3n 2.

2 2 2 2

nf

n

i 2 3 36

36 1 1

d

t

5 Rd

Rt

11 2

5.3978 10 m 5.3978 10 nm

(c) Computed as in (b) above with

R H

7 1

1.096762 10 m ,

H

36 1 1

t

5 RH

Rt

10 1

2.4627 10 m 2.4627 10 nm

95


kq Q

N I 2 b db where b cot (Equation 4-3)

2

4-41.

0 2

mv

kq

Q

and db csc d

2

2mv

2

2

kq Q 1

2

N I02 cot csc d

2

mv

2 2

2

Using the trigonometric identities:

csc

2

1

sin sin sin

and cot

2 2 2 2

sin / 2 2 1 cos 1 cos / 2 sin / 2 2sin / 2

2

kq Q

1 sin

1

N I02 2 d

2 2

mv

2 2sin

/ 2 sin

/ 2

and inserting 2 e q and Ze Q,

N

2

kZe

I 2

0 2 4

mv

2

sind

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

1

k 2e 79e

2

m v 77

. MeV where r = upper limit of the nuclear radius.

2

r

2

279 2791.

440

k e MeV fm

r 29.

5 fm

7. 7MeV

7.

7MeV

4-43. (a)

Z mk e

i qfrev

e

3 3

2

n

2 2 4

1

2 3

c h / mc

(from Equation 4-28)

2

2 2 2 2

2 2

mc ke ec ke ec

= e

2

1

c c

. C . nm / s

19 17 2

1 602 10 3 0010 1

= 1.

054 10

0.

00243nm

137

3

A

96



(b)

2 4 2

2

emk e

0 3 2

e

iA i a

=

2

mke 2m

or

19 34

1. 60210 C1.

05510

J s

31

29.

1110

kg

= 9.

2810

Am

24 2

2

3 10 24 2

= 1.054 10 A 0. 529 10 m 9.

27 10

A m

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).

m n C=m 2 D=n 2 1/C−1/D 1/λ λ (nm)

1 5 1 25 0.96 10534572 94.92

1 4 1 16 0.9375 10287844 97.20

1 3 1 9 0.888889 9754400 102.52

1 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

1 1

1 1

d

1 1 dR

2 2

2 2

n

f

n

i

d

n

f

n

i d

2

4-45. R

R

1 1

Because R , dR / d R / .

R R / /

nf

n

i

1

2

2 2

97



mm

mm

H

m m m m

e p e d

D

e

p e

d

/ /

D

H memd m m m m

D

e

md md me m

d

1 1 1

m m / m m m / m m

m m m

e d p

H H e p e p p e p

p e d

me

If we approximate md 2 mp and me md

, then and

2m

0.

511MeV

/ 656. 3nm

0.

179nm

2 938.

28MeV

p

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

(a)

kinetic energy

Before collision

E

k

1

mv

2

momentum pei mvei

2

ei

After collision

1

E k

mv

2

E

Hg

p

p

ef

Hg

1

2

Mv

mv

2

ef

ef

M v

2

Hg

Hg

Conservation of momentum requires:

mv mv Mv v m v v

M

ei ef Hg Hg ei ef

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

2 2

2 m

m 2 2

Hg

ei

ef

ei

ei ef

ef

1 1

Mv M

2 2

v v v v v v

M

2M

2

m

2M

4m1 2 4m

mvei

E

M

2

M

2

2

2

4 vei since m M,

vei vef

k

98



(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 E k = 2.5eV, then the maximum lost is equal to:

31

m

9. 1110 kg 2.

5eV

4 2. 5eV

4 2.

7 10

M u kg u

27

201 1. 6610

/

5

eV

4-47. (a)

E E Z n

n

2 2

0

/ (Equation 4-20)

E 13. 6eV 9 / n 122. 4/

n eV

For Li ++ , Z = 3 and

2 2

n

The first three Li ++ levels that have the same (nearly) energy as H are:

n 3 , E 13. 6 eV n 6 , E 3. 4 eV n 9 , E 1.

51eV

3 6 9

Lyman α corresponds to the n = 6 → n = 3 Li ++ transitions. Lyman β corresponds

To the n = 9 → n = 3 Li ++ transition.

1/ 1 0. 511 / 938. 8

7

1.

096776 10

1

1/ 1 0. 511 / 6535

7

1.

097287 10

1

(b)

R H R MeV MeV m

R Li R MeV MeV m

For Lyman α:

1 1

7 1 9

RH 1 1. 096776 10 m

2

10 m / nm3/ 4

121.

568nm

2

For Li ++ equivalent:

1 1 1 1 1

R Li Z m m nm

2 2

3 6

9 36

2 7 1 9

1. 097287 10 10 / 3 2

121. 512 nm

0.

056nm

4-48.

2

2

I0ASCnt

kZe

1

2

r

2EK

4

N

(Equation 4-6)

sin

2

3 6

where A 0.

50 cm r 10 cm t 10

m

SC

99



n Ag

10 . 5 g / cm

3 6 . 02 10 23 atoms / mol

107. 5g / mol

5. 8810 atoms / cm 5. 88

10 atoms / m

22 3 28 3

9 1

9

EK

6. 0 MeV I0 1. 0nA 10 C / s

3. 1810

alphas / s

19

21.

6010

C

(a) At θ = 60°

9 2 28 3 6

3. 1310 / s0. 50cm 5. 8810 / m 10

N

2 2

10

cm

2

9 2 2 19

910 N m / C 1.

6010 C 47

1

=

468

/ s

13

26. 0MeV

1. 60 10 J / MeV

4 60

sin

2

60 120

N N 60 sin / sin 52

/ s

2

2

(b) At θ = 120°:

4 4

4-49.

E E Z n

n

2 2

0

/ (Equation 4-20)

2 2

E1 13. 6eV 20 / 1 5.

440keV

For Ca, Z = 20 and

The fact that E 1 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

E

n

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


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 Rsin Rsin

Rcos

2

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 πb 2 . The rate at

2 2 2

which this occurs is I0b I0R

cos

2

(c)

2

2 2

0

cos

b R R

2

(d) An α particle with an arbitrarily large impact parameter still feels a force and is

scattered.

4-52. For He:

(a)

Energy (eV)

E eV Z n eV n

n

0

-10

-20

-30

2 2 2

13. 6 / 54. 4 / (Equation 4-20)

5

3

2

∞

4

E 1 = -54.4 eV

E 2 = -13.6 eV

E 3 = -6.04 eV

E 4 = -3.04 eV

E 5 = -2.18 eV

E ∞ = 0 eV

-40

-50

1

101



(b) Ionization energy is 54.5eV.

(c) H Lyman α:

H Lyman β:

He + Balmer α:

He + Balmer β:

(d)

hc / E 1240eV nm / 13. 6eV 3. 4eV 121.

6nm

hc / E 1240eV nm / 13. 6eV 1. 41eV 102.

6nm

hc / E 1240eV nm / 13. 6eV 6. 04eV 164.

0nm

hc / E 1240eV nm / 13. 6eV 3. 40eV 121.

6nm

42. 4 nm

19.

0nm

(The reduced mass correction factor does not change the energies calculated above

to three significant figures.)

E eV Z n

2 2

n

13. 6 / because for He + , Z = 2, then Z 2 = 2 2 . Every time n is an even

number a 2 2 can be factored out of n 2 and cancelled with the Z 2 = 2 2 in the

numerator; e.g., for He + ,

E eV eV

2 2

2

13. 6 2 / 2 13.

6 (H ground state)

2 2 2 st

E4 13. 6eV 2 / 4 13. 6eV

/ 2 (H 1 excited state)

2 2 2 nd

E6 13. 6eV 2 / 6 13. 6eV

/ 3 (H 2 excited state)

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/

2 8

f 10 Hz 1.70 2.72 4.09 6.41 7.67 9.53

1/

2 8 9

where 3 0010 10

f . m / s nm / m /

50 15

Slope = 4.

6210

8

9. 15 1.

58 10

Hz

74 46

14 8 10

Hz

1/

2

8 1

8 1

Slope (Figure 4-19) =

8

4. 67 10

Hz

Hz

102



The agreement is very good.

50

Z

40

30

20

10

1

2

3

4

5

6

7

8

9

10

f

10 Hz

1/ 2 8 1/

2

The

1/

2

f 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)

E

n

ke ke ke

E

2 2 2

2

n1 2

2rn 2n ro 2 n1

ro

ke

hf En

E

2 2

n1 2

2

2nro

2 n1

ro

ke

2 2 2 2

n n n

ke 1 1 ke 2 1

f

2 2 2 2

2hro

n 1

n 2hr

o n n

1

2 2

ke 2n 1

ke

r hn

2hr

2

o n n1

2 3

o

for n 1

(b)

f

rev

2 2 2 2

v 2 v 1 mv 1 ke ke

frev

2 r 4 r 4 mr r 4 mr r 4

mr n

2 2 2 2 2 2 3 6

o

103



(c) The correspondence principle implies that the frequencies of radiation and revolution

are equal.

2 2

2 2 2 3 2 2

2

ke ke 2

ke hn h

3

2 3 6 rev o

2 6 2

2 2 2

rohn 4 mron 4 mn ke 4

mke mke

f f r

which is the same as a o in Equation 4-19.

4-55.

mv 2

2 2 2

kZe mv kZe


2

r r r mr

kZe

v

mr

2

1/

2

v

2

1

2 2 2 2 2

c kZe

2 2

kZe

kZe

Therefore, c

2

1 mr mr

mr

1 kZe

2

2 1 2 1 2 6 1 2

0. 0075Z / v 0. 0075cZ / 2. 25 10 m / s Z

/

2

c mao

kZe 1 kZe

E KE kZe / r mc

1

mc 1

r

2

1

r

2 2

2 2 2

And substituting 0 0075 and r a

.

o

3 1

E 51110 eV

1

28.

8Z eV

2

1

0.

0075

14. 4eV 28. 8Z eV 14.

4Z eV

104


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:

2

Mm mv GM

FG

G M proton mass and m electron mass, so v

2

r r

r

L mvr n r n / mv

or

n

n r

n

r r r a

m GM / r

m GM GMm GMm

2 2 2 2 2

2

n

n

and,

1/

2 n

2 n

2 o

2

The total energy is:

E

n

n

1 GMm 1 GM GMm

2

r

2

r

2r

2

E mv m

GMmGMm

2 2 2 3

GMm

G M m

2 2 2 2

2r 2n 2n

n

The gravitational Hα line is:

E

2 2 3

G M m 1 1

E E2 E3

2 2 2

2

2 3

2 2 3

11 2 2

27

6. 67 10 N m / kg 1. 67 10 kg 9. 1110 31 kg 0.

1389

2

34

21.

05510

12 /

105



5. 8510 J 3.

66

10

98 79

eV

98

E

5.

8510

J

f 8.

2810

34

h 6.

6310

J s

For the Balmer limit in each case,

65

Hz

E 3. 66 10 79

eV 0. 250 / 0. 1389 6.

58 10

79 eV

79 64

f 6. 58 10

eV / h 1.

59

10 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.

n i n f 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/ 21/ 4 1/ 41/

3

3 1 1/ 21/ 4 1/ 41/

3

2 1

83

83

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


4-60.

m k e

2

2 4

Z

n

o

where

2 o

2

E E E

n

2

m 1. 8810 kg E 4. 5010 J 2.

81

10 eV

28 16 3

o

Z

E eV n

3

Thus, for muonic hydrogen-like atom:

n

2.

81

10

(a) muonic hydrogen

2

2

0

n=5 n=∞

n=4

n=3

-500

E n (eV)

-1000

-1500

n=2

E ∞ = 0 eV

E 5 = -112 eV

E 4 = -176 eV

E 3 = -312 eV

E 2 = -703 eV

3

E 1 = - 2. 81 10 eV

-2000

-2500

n=1

-3000

(b)

r

n

2 2 2 2 2

n n 4

n

2.

56

10 nm (Equation 4-18)

2 2

mkZe mke Z Z

For H, Z = 1: r 2.

56

10

1

4

nm

For He , Z = 2: r 1.

28

10

1+ 4

1

For Al , Z = 13: r 1.

97 10

nm

12+ 5

1

For Au , Z = 79: r 3.

2

10

78+ 6

1

nm

nm

107



5

(c) Nuclear radii are between about 1 and 8 10

6

m , or 18

10 nm . (See Chapter

11.) The muon n = 1 orbits in H, He 1+ , and Al 12+ are about roughly 10nm outside the

nucleus. That for Au 78+ is very near the nucleus’ surface.

(d) hf hc / E E hc / E E

Z

E eV n

3

2

1

2

1

where

n

2.

81

10

hc

10

For H: 5. 8910 m 0.

589nm

2 3

1 2. 8110 11/

4

Similarly,

For He 1+ : 0. 147nm

For Al 12+ : 0. 00349nm

For Au 78+ 5

: 9. 44

10 nm

2

2

108

Chapter 5 – The Wavelike Properties of Particles

5-1. (a)

(b)

34 7

6. 6310 J s3. 1610

s / y

h h

2.

1

10

p mv kg m y

h 6.

6310

J s

m

kg m

3 2

10 10

3

10 1

/

34

29 21

v 6. 610 m / s 2. 1

10 m / y

23

m

5-2.

h h hc 1240MeV fm

12.

4 fm

p E / c E 100MeV

5-3.

2 2

hc 1 1240eV nm

2 2

5

. .

2

p

Ek eVo Vo

940V

2

2m 2mc e 2 5 11 10 eV 0 04nm

5-4.

h h hc

p

2

2mEk

2mc Ek

(a) For an electron:

(from Equation 5-2)

1240eV nm

20. 51110 6 eV 4.

510

3 eV

1 2

0 0183nm

/

.

(b) For a proton:

1240eV nm

4.

27 10

1/ 2

6 3

2983. 310 eV 4.

510

eV

4

nm

(c) For an alpha particle:

1240eV nm

2.

1410

1/ 2

9 3

23. 72810 eV 4.

510

eV

4

nm

5-5. 1 /

2

2

h / p h / 2mEk

hc /

2mc 1.

5 kT

(from Equation 5-2)

Mass of N 2 molecule =

214. 0031u 931. 5MeV / uc 2. 609 10 MeV / c 2. 609 10 eV / c

2 4 2 10 2

109



1240eV nm

10

2 2. 609 10 eV 1. 58. 617 10 5 eV / K 300K

1 2

0 0276nm

/

.

5-6.

h h hc 1240eV nm

0.

202nm

2

2 2

6

2939 57 10 0 02 1 /

p mE . .

2

k mc Ek

eV eV

5-7. (a) If there is a node at each wall, then

(b)

2

n / 2 L where n 1, 2, 3,... or 2L / n .

p h / hn / 2 L E p / 2m hn / 2L / 2m h n / 8mL

2 2 2 2

E

n

2 2

hc n

2 2

8mc L

For n = 1:

2 2

1240eV nm 1

6

eV . nm

E

8 938 10 0 01

1 2

2.

05eV

For n = 2: 2

E2 2. 05eV 2 8.

20eV

5-8. (a)

2

/ c

10 is a nonrelativistic situation, so

2 2 2 2

2 1 /

hc mc E hc mc mc E

2

/

c

k k

E

k

2 6

mc 0.

51110

eV

25.

6eV

2 2

2 2 10

2

c

(b) / 02 . is a relativistic for an electron, so h mu u h m

.

c

u c

u c 2

1 /

h c

mc

2 2

uc

c

c

/

uc

u/ c

/

2 1/

2

2

1 1

c

110



1 0.

2

. 2

uc 0. 981 5.

10

1 1 0 2

2

Ek

mc 1 0. 511MeV 1 2.

10MeV

(c)

c

10

3

3

1 10

uc 0.

9999 1000

1/ 2

2

3

1

1 10

.

2

Ek

mc 1 0 511MeV 999 510MeV

2

5-9.

1

E mc p mu

k

(a)

2

Ek

2 GeV mc 0.

938GeV

2

1 Ek

mc 2GeV 0. 938GeV

2. 132 Thus, 3.

132

Because, 2

1 1 u / c where u / c 0.

948

h h hc

p mc u c mc u c

2

/ /

1240eV nm

6

3. 132938 10 eV 0.

948

(b) E 200GeV

k

7

4. 45 10 nm 0.

445

2

1 Ek

mc 200GeV 0. 938GeV 213 Thus, 214 and u / c 0.

9999

1240eV nm

6.

18

10

214938MeV

0.

9999

3

fm

fm

5-10. n

Dsin

(Equation 5-5)

n

n hc

sin

(see Problem 5-6)

D D

2

2mc E

k

111



1 1240eV nm

0 215

25 11 10

1 /

5.

705eV

2

1/

2

. nm

5

. eV E E

k

k

(a)

(b)

1/

2

5.

705eV

sin

. sin . .

75eV

1

0 659 0 659 41 2

1/

2

5.

705eV

sin

. sin . .

100eV

1

0 570 0 570 34 8

5-11. h h

0.

25nm

p

2mE

Squaring and rearranging,

E

k

p

k

1240eV nm

.

2 2

2

h hc

0.

013eV

2

m mc eV nm

2 2 2

2

6

p

2

p

2 938 10 0 25

1 0 25 0 304

n Dsin sin n

/ D . nm / . nm

sin 0.

822 55

5-12. (a)

(b)

n

n

Dsin

D

nhc

sin

2

sin

2mc Ek

11240

eV nm

sin 55. 6 25.

1110 5 eV 50eV

1 2

11240

eV nm

0 210 25 11 10 5

100

1 /

. .

2

n

sin 0.

584

D nm

eV eV

sin 1 0 . 584 35 . 7

0 210nm

/

.

112


5-13.

t

42°

d

d tcos42

1 cos 42 0. 30 1 cos 42

n t d t nm

For the first maximum n = 1, so 0. 523nm

hc 2

2

h h h

Ek

2 2 2

p 2mE

2m

2mc

k

E

k

2

1240eV nm

6

eV . nm

3.

010

2 939 10 0 523

3

eV

Dsin

5-14. (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)

2

1240eV nm

h / p h / 2mEk

hc / 2mc Ek

0.

0656nm

1/

2

6

20.

51110 eV 350eV

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.

5-16. (a)

(b)

t 1 1

5

10 10

1

f 100,

000s

s s

1 1 1

4

f t f 1.

5910

Hz

5

2 2t

2

10

s

113


5-17. (a) y y1

y2

0. 002m cos 8. 0x / m 400t / s 0. 002m cos 7. 6x / m 380t / s

2 0 002 1 8 0 7 6 1

400 380

2 2

1 1

cos 8. 0x / m 7. 6x / m 400t / s 380t / s

2 2

0. 004m cos 0. 2x / m 10t / s cos 7. 8x / m 390t / s

. mcos . x / m . x / m t / s t / s

(b)

390 / s

v 50m / s

k 78 . / m

(c)

v

s

20 / s

50m / s

k

04 . / m

(d) Successive zeros of the envelope requires that 0. 2 x/

m,

thus

1

2

x 5 m with k k1 k2

0. 4 m and x 5m.

02 .

k

2

2

5-18. (a) Thus,

dv df , multiplying by ,

dv df

v f f f v

d

d d d d 2 d

2

d

dv

v k dk d

2 d d

2

Because 2 / , 2 / and

d

dv

vs

v

dk

d

(b) v decreases as λ decreases, dv/dλ is positive.

5-19. (a)

2 8 11

c f / T T / c 210 m/ 310 m/ s 6. 7

10 s / wave

The number of waves = s s wave

0. 25 / 6. 710 / 3.

73

10

11 3

2 3

Length of the packet = m

(b)

# of waves 210 3. 7310 74.

6m

8

f c / 310 m/ s / 210 2 m 1.

50

10

10 Hz

(c)

6 6

t 1 1/ t 1/ 0. 25 10 s 4. 0 10 rad / s 637kHz

114


5-20. t 1

1/ t 1/ 0. 25s 4. 0 rad / s or f 0.

6Hz

t 1 2f t 1 Thus, t 1/ 2

5000Hz 3.

2 10

s

5

5-21.

5-22. (a)

(b)

h h hc 1240eV nm

0.

549nm

2

2 2

6

20 511 10 5

1 /

p mE .

2

k mc Ek

eV eV

d sin /

2 For first minimum (see Figure 5-17).

0.

549nm

d 3.

15 nm slit separation

2sin

2sin5

05 . cm

sin 5 0. 5 cm / L where L = distance to detector plane L 5.

74cm

2sin5

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.

2 2

5-24. Because the particle must be in the box * sin

/

L

dx 1 A x L dx 1

0 0

Let / ; 0 0 ; and /

u x L x u x L u dx L du , so we have

A 2 L / sin 2 udu A 2 L / sin

2 udu

0 0

1

u sin2u

/ sin / / / 2 / 2 1

2 4

2 2 2

2 2

L A udu L A L A

LA

0

1 / 2

2

A 2/ L A 2/

L

0

L

115


5-25. (a) At 0 : 0,

0

2

2

0 2

x Pdx dx Ae dx A dx

(b) At

(c) At

2 2 2 2

/ 4

1/

4 2

x : Pdx Ae dx Ae dx 0.

61A dx

2 2 2 2

4 / 4

1 2

x 2

: Pdx Ae dx Ae dx 0.

14A 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 N

1

f f

t t t

2 2 2 2

(b)

x and k

N , so k

n

N x x x

5-27.

1.

05510

Js

10 s 1. 609 10

J / eV

34

E t E / t

9

6.

6 10 eV

7 19

5-28.

xp p mv

2 2 x

34

1.

05510

Js

27

v

5. 310

m / s

3 2 3

2mx 210 kg 10 10

m

5-29.

6.

5810

eV s

3. 823d

8. 6410

s / d

16

21

E t E / t 1.

99 10 eV

4

The energy uncertainty of the excited state is ∆E, so the α energy can be no sharper than

∆E.

5-30. xp p h p h / . Because h/ p, p h/ ; thus, p p.

p mv 30kg 40m / s 1200 kg m/ s.

Because p p

5-31. For the cheetah

(see Problem 5-30),

2

x / p 50J s / 1200kg m/ s 4. 2 10 m 4.

2cm

116


5-32. Because c f for photon, c / f hc / hf hc / E,

so

hc 1240eV nm

5

E 2.

4810

eV

3

5.

010

nm

5

E 2.

4810

eV

7

and p 8. 310

eV s / m

8

c 310

m / s

For electron:

15

h 4.

1410

eV s

p

8 310

12

x

5.

010

m

4

. /

eV s m

Notice that ∆p for the electron is 1000 times larger than p for the photon.

5-33. (a) For 48 Ti:

34

1.

05510

Js

E

upper state

4.

7110

14 13

t 1. 410 s 1. 6010

J / MeV

34

1.

05510

Js

E

lower state

2.

2010

12 13

t 3. 010 s 1. 6010

J / MeV

10

total 6.

91 10

E E E MeV

U

L

10

10

MeV

MeV

E

E

T

(b) For Hα:

10

6.

9110

MeV

5.

3

10

1.

312MeV

10

34

1.

05510

Js

EU

6.

5910

8 19

10 s 1. 6010

J / eV

8

eV

and

E

L

8

6.

59 10 eV also.

7

ET

1.

32 10 eV is the uncertainty in the Hα transition energy of 1.9eV.

5-34. t 1 2f t

1

f

7. 5 4. 0 10 3.

5

10

For the visible spectrum the range of frequencies is

14 14

The time duration of a pulse with a frequency uncertainty of

1 1

2f

2

3.

510

Hz

t 16

4. 5 10 s 0.

45 fs

14

f

is then:

Hz

117


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:

2

Ek

mnc 1 or 1 10MeV / 939MeV

1 /

2 2

2

110 / 939 1. 0106 1/ 1 u / c or u 0.

14c

h hc 1240eV nm

Then, 9.

33 fm

mu

mc 2 u / c 1. 0106939 10 6 eV 0.

14

Nuclei are of this order of size and could be used to show the wave character of 10MeV

neutrons.

5-36. (a) E

135 MeV , the rest energy of the pion.

(b)

Et

2

16

6.

5810

eV s

t

2.

4410

6

2E

213510

eV

24

s

5-37.

p s

φ

r

s

s

L rps

r

s

L

rp

r

L

rp s / r p s

s

s

In the Bohr model, L n and may be known to within L

01 . .

Then

unknown.

/ 0. 1 10rad.

This exceeds one revolution, so that is completely

5-38. E hf E hf

E

Et h f t 1 where t 0.

85ns

118



9

f 1/ 0. 85ns 1.

18 10 Hz

8 9

3. 0010 m / s 10

nm / m

19

For 0. 01nm f c / 3.

00

10 Hz

0.

01nm

9

f

1.

1810

Hz

11

3.

910

19

f 3.

0010

Hz

5-39.

Et

t

2 2 E

16

6.

5810

eV s

t

1.

3210

6

2250 10

eV

24

s

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

34

h 6.6310 J s

3

p (410 kg)(100m/s)

33

1.66 10 m

The diameter of the aperture would need to be of the order of

33

10 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:

2

h h h

E

2

p 2mE

2m

e

34 2

(6.63 J s)

E

31 9 2

2(9.1110 kg)(0.110

nm)

e

17

2.41 10 J 151eV

119


5-42. (a) For a proton or neutron:

xp and p m v assuming the particle speed to be non-relativistic.

2

(b)

1.

05510

Js

2mx 2 1 67 10 kg 10 m

34

7

v 3. 16 10 m / s 0.

1c

27 15

.

1. 67 10 27 kg 3. 1610

7 m / s 2

1

2 2

(non-relativistic)

2 13

Ek

mv 8. 3410 J 5.

21MeV

(c) Given the proton or neutron velocity in (a), we expect the electron to be relativistic,

2

in which case, E mc

k

1 and

p mv v

2x

2mx

For the relativistic electron we assume v

c

34

1.

05510

Js

193

2mcx 2 9 1110 kg 3 0010 m s 10 m

31 8 15

. . /

2

2 31 8 11

Ek

mc 1 9. 1110 kg 3. 0010 m/ s 192 1.

5810 J 98MeV

5-43. (a)

E p c m c E hf p h / / k k c m c

2 2 2 2 4 2 2 2 2 2 2 4

k c m c

k k k

2 2 2 2 4

2 2 2 2

v c 1 m c / k c

(b)

v

s

2 2 2 4 2

d

d k c m c c k

dk dk k k c m c

2 2 2 4

2

2 2 2

c k c k c p

u

E

(by Equation 2-41)

5-44.

2 2

y 1 y

(Equation 5-11) y C y C y

2 2 2

x v t

3 1 1 2 2

2 2 2 2 2

y3 y1 y

2

1 y

1

1 y

2

C

2 1

C

2 2

C

2 1 C

2 2

2 2 2

x x x v t v t

120



2

2

1

1 y3

2 2

C1 y1 C2 y2

2 2

v t v t

5-45. The classical uncertainty relations are

(a)

t 2f t

1 (Equation 5-18)

and

2

x

2

(Equation 5-20)

1 1

f 0.

0541Hz

2t

2

3.

0s

(b) Length of the wave traing L

/ .

L 330m s 3 0s 990m

v t , where v = speed of sound in air = 330m/s.

(c)

(d)

2

where x

= length of the wave train 990 m from (b)

2x

and 0. 13 m from (d) .

2

0.

13m

2

990m

v 330m/

s

v f 0.

13m 13cm

f 2500Hz

6

2. 72 10 2.

72

m

5-46. (a)

n / 2 L 2 L / n. Because h / p h / 2 mE , then

2 2 2 2

h h h n

E E h mL

2 2

2

2m

2m L / n 8mL

2 2

If

1

/ 8 , then

2 2

2

h hc 1240eV nm

(b) For L 01 . nm,

E

2 2 2

8mL 8mc L

6

8 0. 51110 eV 0.

1nm

2

E1 37. 6 eV and En

37.

6n eV

1 2

E

n

2 2

hn

2

8mL

2

n E1

121



E

(eV)

1000

800

n

5

600

4

400

3

200

2

0

0

1

L

x

(c) f E / h c / E / h

hc

E

1240eV nm

For n 2 n 1 transition, E 112. 8 eV and 11.

0nm

112.

8eV

(d)

1240eV nm

For n 3 n 2 transition, E 188 eV and 6.

6nm

188eV

(e)

1240eV nm

For n 5 n 1 transition, E 903 eV and 1.

4nm

903eV

5-47. (a) For proton:

E

hc 2

1 2 2

8m p

c L

from Problem 5-46.

2

1240MeV fm

MeV fm

2

E1 205 MeV and E 205

2

n

n MeV

8 938 1

E 820 MeV and E 1840MeV

2 3

(b)

(c)

(d)

hc 1240MeV fm

For n 2 n 1 transition, = 2.

02 fm

E

615MeV

hc 1240MeV fm


22 fm

E

1020MeV

hc 1240MeV fm


76 fm

E

1635MeV

122


5-48. (a)

E / 2 mL (Equation 5-28) and E / 2mA

2 2 2 2

(b) For electron with

(c)

A

10

10 m:

c 197.

3eV nm

2 2

E 3.

81eV

2 2

2

2mc A 2 0 511 10 eV 10 nm

For electron with

6 1

.

A cm A m

2

1 or 10 :

2 2

1 7 16

2

34

1.

05510

Js

3 3 2

/

E 3. 81eV 10 / 10 nm 3.

81

10 eV

2

61 43

E 1. 3910 J 8.

7 10

eV

2

2

2 mL 2 100 10 g 10 kg g 2 10

p mv m 0. 0001 500m / s 0.

05m

5-49.

For proton:

xp

6 58 10 16 6

0 05 938 10

x / p . eV s / . m/

s eV

1. 4010 m 1.

40

10

23 8

For bullet:

fm

x 1. 055 10 34 J s / 0. 05m / s 10 10 3 kg 2.

1 10

31

m

2 2

y 1 y

(Equation 5-11) where y f and x vt.

2 2

x v t

5-50.

2 2 2

y f y f f

and

2 2 2

x x x x x

x

2 2 2

y f y f f

and

2 2 2

t t t t t

t

Noting that

2 2

0 1 0 and

2 2

x x t t

, , , v,

we then have:

2 2

f f 1 f f

0 1 1 0

2 2

2

v

f

f

2 2

2 2

v v

123


5-51. (a) h/

p The electrons are not moving at relativistic speeds, so

34 31 6 19

h / mv 6. 63 10 J s / 9. 11 10 kg 3 10 m/ s 2. 43 10 m 0.

243nm

(b) The energy, momentum, and wavelength of the two photons are equal.

1 1 1

E mv mc mc v c mc mc

2 2 v c

2

/ /

2 2 2 2 2 2 2 2 2

1

2

2

6 6 8

0. 511 10 eV 3 10 / 3 10 1 0.

511

(c) p E / c 0. 511MeV / c

MeV

1

(d)

6 3

hc / E 1240eV nm / 0. 511 10 eV 2.

43 10 nm

5-52. (a)

Q m c m c m c

2 2 2

p n

(b)

2 2

1. 007825 uc 1. 008665 uc 139.

6MeV

938. 8MeV 939. 6MeV 139. 6MeV 140.

4MeV

E

140.

4MeV

16 6 24

E t t / E 6. 58 10

eV s / 140. 4 10 eV 4.

7 10 s

(c)

8 24 15

d ct 310 m/ s 4. 710 s 1. 410 m 1.

4 fm

5-53.

hf

hf

mc

2

mc

2 2

1

1

v

c

2

12 /

2 2

2 2

2

1 v

mc

v

mc

2 1

c

hf c

hf

2

Expanding the right side, assuming mc hf ,

2 4

2 2

v 1 mc 1 mc

1

c 2 hf 8 hf

2

and neglecting all but the first two terms,

2

v 1 mc

1 Solving this for m and inserting deBroglie’s assumptions that

c 2 hf

v

0. 99 and 30 m,

m is then:

c

1/

2

34

1 0. 992

6.

6310

Js

m

1.

0410

8

3. 0010

m / s30m

44

kg

124


5-54. (a)

y 0

m

xp mxv v / m

x

x

x

1/ 2 1/

2

1 2 2y0 1

2y0

y0

gt t X vx

t vx

2 g 2

g

0

1/

2

2

g

y0

x

X

2v

g

2

2y

mx

1/

2

∆x

1

y

y

and where vy

g t or

2

2

1 /

t v / / so, 2

2

y

g mgy X y0

/ g / mgy

mx

(b) If also y p v / my X xt t

5-55.

1 2 3

mv kT

2 2

v

rms

23

. J / K K

1/

2

3kT

3 1 381 10 300

366m / s

27

m

56u 1. 6610

kg / u

o

1 / 1

/

f f v c hf hf v c

o

1eV

366m / s 6

E hf hfo

hfov / c 1.

210

eV

8

3. 010

m/

s

This is about 12 times the natural line width.

10 6 eV 366m / s

E hfov / c 12 . eV

8

3. 010

m/

s

This is over 10 7 times the natural line width.

125


5-56.

recoil

2 2

E

recoil

E

/ c Erecoil

2m

2mc

2

2

1eV

uc

12

(a) Erecoil

9.

610

eV

2

6

256uc

931.

510

eV

This is about 10 -4 times the natural line width estimated at 10 -7 eV.

2

2

1MeV

uc

(b) Erecoil

96

. eV

2

6

2 56uc

931.

510

eV

This is about 10 8 times the natural line width.

2

126

Chapter 6 – The Schrödinger Equation

6-1.

d

dx

d

dx

2

kxt

2

kAe k and k

2

d Also, . The Schrödinger equation is then, with these substitutions,

dt

2 2

k / 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

2

and

Also,

2

Substituting these into the Schrödinger equation yields:

k / 2m V , which is true, provided

2 2

For the classical wave equation: (from Equation 6-1)

From above:

ik k . i

.

x

x

t

2 2

2 , i.e., if

k

k / m V E E V.

2 2

2 2

k

and also

. Substituting into Equation 6-1

2 2

x

t

k 2 1/ v 2

2 , which is true for

(with replacing E and v replacing c)

v

/ k.

2 2

d

d x x 1 x 1

and

2 2 2 2 4 2

dx

dx L L L L L

2

6-3. (a) x/

L

Substituting into the time-independent Schrödinger equation,

2 2 2 2

x

4 2 V x

E

2

2mL 2mL

2mL

Solving for V(x), V x

2 2 2 2 2 2

x x 1

2 kx

4 2

4

2mL 2mL 2mL 2mL

2

2

127



where

k

2 4

/ mL . 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) E 2 / 2 2 2 2 / 2 4 2 / 2 2 1

2 /

2

k

x E V x mL x mL mL x L

(b) The classical turning points are the points where E V x E x

or 0.

That occurs

k

when

2 2

x L 1, or when x L

/ .

V x m

x

(c) For a harmonic oscillator

2 2

x

2mL

/ 2, so

2 2

2 2 2 2 2 4 2

x / 2 / m L / mL

4

Thus,

2

1

E

2 2

2mL

mL 2 2

6-5. (a) x, t Asinkx t

Acoskx t

t

i i Acoskx t

t

2

2

k A kx t

sin

2

x

2 2 2

k A

sin

2

kx t

i

2m x 2m t

128



(b) x, t Acoskx t iAsinkx t

i i Asin

kx t i A kx t

t

2

cos

sin

Acos

kx t i A kx t

2 2 2 2 2 2

k A

ik A

cos

2

2m x 2m 2m

kx t sinkx t

2 2

k

Acoskx t iAsinkx t

2m

i

t

if

k

2m

2 2

it does. (Equation 6-5 with V = 0)

d

2m dx

d

dx

2 2 2

10

6-6. (a) For a free electron V(x) = 0, so E

2 2 2.

5

10

2

Substituting into the Schrödinger equation gives: 2 5 10 10 2 2

and, since

2

E Ek

p / 2m

2

. / m E

2

for a free particle, p 2 2m2. 5 10 10 2 / 2m

p 2 510 2 64

10

10 24

. . /

kg m s

2

(b)

and

2 24 31 18

E p / 2m 2. 6410 kg m/ s / 2 9. 1110 kg 3.

82

10

J

18 19

3. 8210 J 1/ 1. 6010 J / eV 23.

9eV

(c)

34 24 10

h/ p 6. 63 10 J s 2. 64 10 kg m/ s 2. 51 10 m 0.

251nm

6-7.

2 2

x

/ L

x Ce and E 0

2 2

d

(a) V 2

x

0

2m dx

2 2

d

2x d 4x

2

2

x

and

2

4 2

dx L dx L L

2 2 2 2

4x

2 2x

And 4 2

V x

0 so V x

1

2

2

2m L L mL L

129



(b)

V(x)

x

6-8.

a

a

a

2 ikxt ikxt

dx A e e dx

1

a

A

a

2 2 a

2

a

a

A dx A x A 2a

1

1

1 /

2a

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 2

E1 h / 8mL hc / 8mc L

2 2 2 2

1240MeV fm

For m mp, L 0. 1nm : E1 0.

021eV

6

2

8938. 310 eV 0.

1nm

2

1240MeV fm

(b) For m mp, L 1fm : E1 205MeV

6

2

8 938.

310 eV 1fm

2

6-10. The ground state wave function is (n = 1) x L x L

The probability of finding the particle in ∆x is approximately:

2 2 x 2x 2 x

Pxx sin x sin

L

L

L

L

1

2/ sin / (Equation 6-32)

130



L

20.

002L 2L

2

(a) For x and x 0. 002L, Pxx

sin 0. 004sin 0.

004

2 L

2L

2

2L

20.

002L 22L

2 2

(b) For x and Pxx

sin 0. 004sin 0.

0030

3 L

3L

3

x L and P x x 0. 004 sin 0

(c) For

2

6-11. The second excited state wave function is (n = 3) x / L sin x / L

3

2 3

(Equation 6-32). The probability of finding the particle in ∆x is approximately:

2 2 3

x

Pxx sin x

L

L

(a) For

L

20.

002L 23L

2 3

x and x 0. 002L, Pxx

sin 0. 004sin 0.

004

2 L

2L

2

2L

26

L

2

(b) For x and Pxx

0. 004sin 0. 004sin

2

0

3

3L

23

L

2

(c) For x L and Pxx

0. 004sin 0. 004sin

3

0

L

6-12.

2 2 2 2

1 n 2

2 1 2 2

(Equation 6-24)

mL

E mv n mv

mvL

2 2 2

2 2mL

2

9 3 2

10 kg 10 m / s10

m

34

1.

05510

Js

mvL

n 310

19

2

6-13. (a) . .

(b)

2 6

x 0 0001L 0 0001 10 m 10 m

9 3 16

p 0. 0001p 0. 0001 10 kg 10 m/ s 10

kg m/

s

6

10 m10

16 kg m / s

x p 9 10

34

1.

05510

Js

11

131


6-14 (a) This is an infinite square well with width L. V(x) = 0, and

uncertainty principle:

Ek min

pmin p / x / L and

E min

p / min

2m / 2mL h / 8 mL

2 2 2 2 2 2

(b) The solution to the Schrödinger equation for the ground state is:

1/

2

x / L sin x / L

1

2

2 1 2 2

2

d 2 /

sin /

2

1

dx

L

L

L

1

and

x L

2 2

2

2

E Ek

p 2m

/ . From

h

So,

1

E

1

or E1 2

2m

L

8mL

The result in (a) is about 1/10 of the computed value, but has the correct

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,

L3/ 2 , or 2L/ 3.

(c) p h/ h/

2 L in the ground state.

(d)

p 2 / 2m h 2 / 4L 2 / 2m h 2 / 8mL

2 , which is the ground state energy.

h n

h

8mL

8mL

2 2 2

2

6-16. En and E 2 n

En

1

En

2 n 2n

1

2

h

n

2 1 8

2

mL

or, E n

hc

so,

12

1 2 1 2

h hc

3694 3nm1240eV nm

/

/ /

.

2 6

mc mc . eV

3 3

L 0 795

8

. nm

8

80 511 10

6-17. The uncertainty principle requires that E 2

2mL

for any particle in any one-dimensional

box of width L (Equation 5-28). For a particle in an infinite one-dimensional square well:

2 2

nh

En

2

8mL

2

132



For n = 0, then E 0 must be 0 since

exclusion principle.

2

h

8mL . This violates Equation 5-28 and, hence, the

2

0

6-18. (a) Using Equation 6-24 with L = 0.05 nm and n = 92, the energy of the 92 nd electron in

the model atom is E

92

given by:

(6.6310 J s)

En

n E

2mL

8(9.1110 kg)(0.0510 m)

E

2 2 34 2

2 2

2 92

(92)

31 9 2

1eV

2.0410 J 1.2810 eV 1.28MeV

1.6010 J

13 6

92 19

(b) The rest energy of the electron is 0.511 MeV. E92 2.5 the electron’s rest energy.

6-19. This is an infinite square well with L = 10cm.

E

n

2 2

2

n

mv

2

2

3

2. 010 kg 20nm / y

hn 1

8mL

2 7

2 3. 16 10

s/

y

3 9

. kg m . m

2 2

7 34

23. 1610 s 6.

6310

J s

2 2 2

8 2 0 10 20 10 0 1

3 9

. kg m . m

7 34

3. 1610 s6.

6310

J s

2 2 0 10 20 10 0 1

n 3.

810

2

2

14

12 /

6-20. (a) / sin /

5

x 2 L 5 x L dx

04 . L

02 . L

2

2/ sin 5

/

P L x L dx

Letting 5 x / L u, then 5 dx / L du and x 0.

2 L u

and x 0. 4L u 2 ,

so

133



x

2

sin2x

2 L 2 2 L

2

1

P sin udu

L

5

L

5

4 5

(b)

L

/

5 2

2

P 2/ L sin 0. 01L 0. 02 where 0.

01L x

L

2

6-21. (a) For an electron:

(b) For a proton:

(c) E21 E1

2

1240MeV fm

. MeV fm

3

E1 3.

76

10 MeV

2

8 0 511 10

2

1240MeV fm

. Mev fm

E1 2.

05MeV

2

8 938 3 10

3 (See Problem 6-16)

For the electron:

4

E21 3E1

1.

13 10 MeV

For the proton: E21 3E1

6.

15MeV

6-22. F dE / dL comes from the impulse-momentum theorem Ft 2 mv where t L/ v.

So,

F mv 2 / L E / L . Because

sign means “on the wall”. So

E1 h 8mL dE dL h 4mL

The weight of an electron is

minuscule by comparison.

2 2 2 3

/ , / / where the minus

2

34

6.

6310

Js

31 10

. kg m

2 3 7

F h / 4mL 1.

2110

N

3

4 9 11 10 10

31 2

mg 9. 11 10 kg 9. 8m / s 8.

9 10

30 N

which is

6-23. x

n

To show that

2 n

x sin

L L

L

0

nx

mx

sin sin dx 0

L

L

Using the identity A B A B A B

1

2

2sin sin cos cos , the integrand becomes

cos n m x / L cos n m x

/ L

134



The integral part of the first term is

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

vanish at the limits x = 0 and x = L.

L

nx mx

sin sin dx 0 for n m .

L

L

0

6-24. (a)

4

L

x

(b)

P

L

x

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.,

2mE

2

ka a

a 1/ 2 1. 0 fm 0. 5 fm and m 939. 6MeV / c for the neutron.

where

2

Substituting above, squaring, and re-arranging, we have:

E

2 2

V

2

2 939 6 0 5

2

. MeV / c . fm

2 0 2

V

c

2

197.

3eV nm

2 2 2 2

8 939. 6MeV 0. 5 fm10 nm / fm 8 939. 610 eV 0.

510

nm

0 2 2

6 6 6

V eV MeV

8

0

2.

0410 204

135


6-26. Because E1 05 . eV and for a finite well also

near the top of the well. Referring to Figure 6-14, ka 2 .

2mE

L

ka m L

2

9 1

7.

24 10 2

9 10

L 2 / 7. 24 10 8. 7 10 m 0.

87nm

En

2

n E

1, then n = 4 is at about 8eV, i.e.,

6-27. For V2 E V1 : x1 is where V 0 V1 and x2 and x2 is where V1 V

2

From to 0 and x

2

to : is exponential

0 to x 1 : is oscillatory; E k is large so p is large and λ is small; amplitude is small because

E k is large, hence v is large.

x 1 to x 2 : is oscillatory; E k is small so p is small and λ is large; amplitude is large

because E k is small, hence v is small.

6-28. (a)

6

5

4

3

2

1

-10 +10

0

x

136



*

6-29. px

3

x

3

s dx (Equation 6-48)

i x

L

0

2 3 x 2 3 x

sin

sin dx

L L i x L L

L

2 3 x 3 x 3

sin cos

L i L L L

0

dx

Let 3 x

Then 0 0, 3 , and

3

L

y x y x L y dx dy dx dy

L

L

3

Substituting above gives:

3

2 L

3

px

sin ycos

ydy

L i 3

L

0

2

Li

3

0

sin ycos

ydy

3

2

2 sin y 2

L i 2

0

L i

0 0 0

Reconiliation: p x is a vector pointing half the time in the +x direction, half in the –x

direction. E k is a scalar proportional to v 2 , hence always positive.

6-30. For n = 3,

1/

2

3

2/ L sin 3 x / L

(a)

L

2

x x 2/ L sin 3 x / L dx

0

137



Substituting u 3 x / L, then x Lu / 3 and dx L / 3 du . The limits become:

x 0 u 0 and x L u 3

3

2

x 2/ L L / 3 1/ 3 u sin udu

0

2/ L L / 3

2

2

u u sin2u cos 2u

4 4 8

3

0

2 2

2/ L 1/ 3 3 / 4 L / 2

(b)

L

2 2 2

x x 2/ L sin 3 x / L dx

0

Changing the variable exactly as in (a) and noting that:

3 3 2

2 2

u u 1 u cos2u

u sin udu sin2u

6 4 8 4

0 0

3

We obtain

1 1

x L 0.

328L

3 18

2 2 2

2

6-31. (a) Classically, the particle is equally likely to be found anywhere in the box, so

P(x) = constant. In addition,

L

0

P x dx 1 so P x 1/ L .

(b)

L

0 0

L

2 2 2

2 and 3

x x / L dx L / x x / L dx L /

6-32.

2 2

d

V x x E x (Equation 6-18)

2

2m dx

1 d d

x E V x x

2m i dx i dx

1

pop

pop

E V x

2m

Multiplying by and integrating over the range of x,

138



2

p op

2m

dx E V x dx

2

p

2m

2

E V x or p 2m E V x

For the infinite square well V(x) = 0 wherever

x does not vanish and vice versa.

Thus,

n

V x p mE m n

2mL

L

2 2 2 2 2

2

0 and 2 2 for 1

2 2

6-33.

2 2

2

x L L (See Problem 6-30.) And x

L

2

3 2 2

x

p

2

x

1/ 2

2 2 2

1/

2

2 2 L L L 1 1

x x L 0.

181L

2 2

2 2

3 2 4 12 2

and p 0 (See Problem 6-32)

2

L

1/

2

2 2

2 2

P x p

p p 0 . And 0. 181L / L 0.

568

2

L

L

6-34.

2

m x / 2

1/

4

0

x A0e where A0

m /

2

2 m x /

2 2

1/

2

x A0 xe dx Letting u m x / and x / m u

2udu m / 2 xdx . And thus, m / udu xdx ; limits are unchanged.

1

2

2

u

0

/

0

x A m ue du (Note that the symmetry of V(x) would also tell us that

x

0.)

2 2 2

0

2

m x /

x A x e dx

3 2 2 3 2

2

2 2 2 2

0

/ / u 2

/

u

0

/

A m u e du A m u e du

2

0

3/ 2 1/ 2 3/

2

2A / m / 4 m / / m / 2 / 2m

139


6-35.

2

p 1

2m

2

2 2

m x n

12 / .

For the ground state (n = 0),

2

2 2

2 2

p

x / 2 p / 2 m and x / 2m

2 2 2

m m m

(See Problem 6-34)

2 2

p

p 1 2 1

1 or 1

p m

m m 2m m 2 2

6-36. (a)

0

1/ 4 2

m x / 2 i t / 2

x, t m / e e

(b)

2

2

px op

p

0

x, t

0

x,

t dx

i x i x

2

x

x

0

0

2

0

1/ 4 2

m x / 2 i t / 2

A m x / e e

0

2

m x / 2 i t / 2

A m x / m x / m / e e

2 2 2 2

0

/ / 1

2

m x /

p A m m x e dx

2 2

2 2 2 m x / m x / 2

0

/ /

A m m x e dx e dx

Letting

1/

2

u m x / x, then

2 2 2 1/

2 2

0

/

2 2

u

u

p A m m u e du e du

1 2

2 2

2 2 / 2 u

u

0

/ 2

0 0

A m u e du e du

2

1/ 2 1/

2

m / m /

2

4 2

2

m / 1/ 2 m / 2

140


6-37.

2

m x / 2

0

x C0 e (Equation 6-58)

(a)

2 2

0 0

2

m x /

x dx 1 C e dx

2 2 1

C0 2I0 C0

2 with m

2 x

/

C

0

C 0

m

2

m

1/

4

(b)

m

2

m x /

x x dx x e dx

2 2 2

2

0

m

m 1

2I2 2 with m

3

4

/

(c)

3

1 m 1

=

3 3

2 m 2 m

1 1 1 1 1

2 2 2 2 m 4

2 2 2 2

V x m x m x m

6-38.

2

m x / 2

1

x C1 xe (Equation 6-58)

(a)

2 2 2

2 m x /

2

1

x dx 1 C1 x e dx C1 2I

2

2 1

C1 2 with m

3

4

/

C

2

1

2

3

1 3 3

m

C

4m

3 3

1 3

1/

4

(b)

3 3

2 3

1 3

1/

2

4m

2

m x /

x x dx x e dx

0

141



(c)

1/

2

3 3

2

2

2 2 2 4m

m x / 2

1 3

x x dx x e x dx

1/ 2 1/

2

3 3 3 3

4m

4m

3

2I 3 4

2 where m

3 5

8

/

3 3 5

3 m

3

3 5 5

2 m 2

m

(d)

1 2 2 1 2 2 1 2 3 3

V x m x m x m

2 2 2 2 m 4

6-39. (a) x p p / x / 2A

(b)

2 2

2 2

E p / 2m h / 2A / 2m / 8mA

k

(1)

2

2 2 2

1 2 2 2 E02mA 4 E0

E m A

2 2

2 2 4 4E

k

Because E0 2 also E0

4E

/

k

(2)

2 2

0

/ x is computed in Problem 6-36(b). Using that quantity,

E

k

2

m

2m

2

/ 4 2E

k

6-40. En

n 1/ 2 En

1

n 3/

2

En 1

En En

n 3/

2 n 1

E E n 1/ 2 1 n 1/

2

n

n

En

1

limn

lim

n

0.

E

n 1/

2

n

In agreement with the correspondence principle.

142


6-41. (a) Harmonic oscillator ground state is n = 0.

2

m x / 2

0

x A0 e (Equation 6-58)

Therefore, for x:

2

0

2

m x /

x A xe dx

Let

2 2

m x / y x m y

dx m dy

Substituting above yields:

2

2

y

0

0

x A m ye dy

by inspection of Figure 6-18, integral tables, or integration by parts.

For

2 2 2 2

0

2

m x /

x : x A x e dx

Substituting as above yields:

2 2 3/ 2 2

2

y

0

x A m y e dy

The value of the integral from tables is 2.

Therefore,

x A m

2 2

0

2

3/

2

(b) For the 1 st excited state, n = 1,

2

m x / 2

1

x A1 m / xe (Equation 6-58)

2 3

1

2

m x /

x A x m e dx

Changing the variable as in (a),

2 3 / 2 3

2

y

1/

2

1

x A m m y e m dy

2

2 3 y

1

0

A m y e dy

by inspection of Figure 6-18, integral tables, or integration by parts.

2 2 4

1

2

m x /

x A x m e dx

changing variables as above yields:

143



2 2 2 4

2

y

1/

2

1

x A m m y e m dy

2 3/ 2 4

2

y

1

A m y e dy

The value of the integral from tables is 3 4.

Therefore,

2 2

x A m

1

3 4

3/

2

6-42. (a) 2 f 2 / T 2 / 1. 42s 4. 42rad / s

1

34 34

E0

1. 055 10 J s 4. 42rad / s / 2 2.

33 10 J

2

499.9 mm

500.0 mm

(b)

A

2 2

500. 0 499.

9 10mm

E n 1/ 2 1/

2m A

2 2

A

0.1mm

2

2 34

n 1/ 2 1/ 2 0. 010kg 4. 42rad / s 10 m 1.

055 10 J s

2. 1 10 or n 2.

1 10

28 28

(c) f 2 0.

70Hz

6-43.

x A e

0 0

2

x / 2

m

x A xe

1 1

2

m x / 2

From Equation 6-58.

Note that

0

is an even function of x and

1

is an odd function of x .

It follows that

0 1dx

0

144


6-44. (a) For x > 0,

k / 2m V E k / 2m 2V

2 2 2 2

2 0 1 0

So,

1/ 2 1/

2

k 2 mV . Because k 4 mV , then k k / 2

2 0 1 0 2 1

(b)

2 2

R k1 k2 k1 k

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

direction. Classically, 100% would continue on.

6 5

0. 971 10 9. 71 10 , continue past the step in the +x

6-45 (a) Equation 6-76:

T

E

16 1

V

E

V

0 0

e

2

a

where 2 2 mp( V0

E ) /

and a barrier width .

2 22 15

2 a 2 2(938MeV/c )(50 44)MeV / 6.58 10 MeV s 10 1.075

T

44 MeV 44 MeV

16 1

50 MeV 50 MeV

e

1.075

T

0.577

(b) decay rate N T where

N

13

1/2

v proton 2 44MeV 1.60 10 J/MeV 1

27 15

2R

1.67 10 kg 2 10 m

4.59 10 s

22 1

(c) In the expression for T,

rate then becomes

22 1 22 1

decay rate 0.577 4.59 10 s 2.65 10 s

1.075 2.150

e e , and so T 0.577 T 0.197 . The decay

21 1

9.05 10 s , a factor of 0.34× the original value.

6-46. (a) For x > 0,

k / 2m V E k / 2m 2V

2 2 2 2

2 0 1 0

1/ 2 1/

2

So, k 6 mV . Because k 4 mV , then k 3/

2k

2 0 1 0 2 1

145



(b)

2 2

R k k k k

1 2 1 2

2 2

2 2

R k1 k2 k1 k

2

1 3/ 2 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

Classically, 100% would continue on.

6 5

0. 99 10 9. 9 10 , continue in the +x direction.

6-47. (a)

E

4eV

9eV =V 0

2m V0

E /

0.6 nm = a

6 2

2 0 511 10 5

. eV / c eV /

6 eV

5. 11 10 eV / c

1

and

a nm nm

1

0. 6 11. 46 6.

87

Since a is not 1, use Equation 6-75:

The transmitted fraction

2260eV

197.

3eV nm

11.

46nm

T

1

2

1

sinh a

81 2

1 1 sinh 6.

87

4 E V 1 E V

80

0 0

x x

Recall that sinh x e e 2 ,

1

2

6. 87 6.

87

81 e e

6

T 1 4.

3 10 is the transmitted fraction.

80 2

(b) Noting that the size of T is controlled by

a through the

2

sinh

implies increasing E. Trying a few values, selecting E = 4.5eV yields T

or approximately twice the value in part (a).

a and increasing T

8.

7 10

6

146


6-48.

B

1/

2

E E V0

1/

2

E E V0

1/

2

1/

2

A

For

0 0

1/

2

E V , E V is imaginary and the numerator and

denominator are complex conjugates. Thus,

2 2 2 2

B A and therefore R B A 1,

hence T 1 R 0.

6-49. A B C and k1A k1B k2C

(Equations 6-65a & b)

Substituting for C, k1A k1B k2 A B k2A k2 B and solving for B ,

B

k

k

k

k

1 2

1 2

A,

which is Equation 6-66. Substituting this value of B into Equation 6-65(a),

k1 k2 k1 k2 k1 k2 2k1

A A C A or C ,

k k k k k k

1 2 1 2 1 2

which is Equation 6-67.

6-50. Using Equation 6-76,

E E

T e E . eV, V . eV, a . nm.

2 a

16 1 where 2 0

0

6 5 and 0 5

V0 V0

2. 0 2.

0 2 10. 87 0.

5 5

T 16 1 e 6.

5 10 (Equation 6-75 yields

6. 5 6.

5

5

T 6.

6 10 .)

6-51.

k

k

2

1 2

R T R

2

k

k

1 2

and 1 (Equations 6-68 and 6-70)

(a) For protons:

k mc E c MeV MeV MeV fm

2

1

2 / 2 938 40 / 197. 3 1.

388

k 2mc E V / c 2 938MeV 10MeV / 197. 3MeV fm 0.

694

2

2 0

2 2

1. 388 0. 694 0.

694

R 0. 111 And T 1 R 0.

889

1. 388 0. 694 2.

082

147



(b) For electrons:

1/ 2 1/

2

0. 511 0.

511

k1 1. 388 0. 0324 k

2

0. 694 0.

0162

938 938

2

0. 0324 0.

0162

R 0. 111 And T 1 R 0.

889

0. 0324 0.

0162

No, the mass of the particle is not a factor. (We might have noticed that

be canceled from each term.

m could

6-52. (a)

E

n

2 2

nh

2

8mL

The ground state is n = 1, so

E

hc

2 2

1240MeV fm

1 2 2

2

8 mc L 8 938.

3MeV 1fm

204.

8MeV

(b) (c) E21 hc /

21

2000

1500

E n

(MeV)

1000

1844 MeV

819 MeV

n = 3

(d)

32

21

1240MeV fm

819 205 MeV

1240MeV fm

1844 819 MeV

2.

02 fm

1.

21fm

500

205 MeV

(e)

31

1240MeV fm

1844 205 MeV

0.

73 fm

0

148


6-53. (a) The probability density for the ground state is

2 2

P x x 2 / L sin x / L .

The probability of finding the particle in the range 0 < x < L/2 is:

L / 2 / 2

2 L 2 2 1

sin 0 where /

P P x dx udu u x L

L

4 2

0 0

(b)

(c)

L / 3 / 3

2 L 2 2 sin 2 / 3 1 3

sin .

P P x dx udu

L

0 0

(Note: 1/3 is the classical result.)

6 4 3 4

0 195

3L

/ 4 3 / 4

2 L

2 2 3 sin 3 / 2 3 1

sin .

P P x dx udu

L

0 0

(Note: 3/4 is the classical result.)

8 4 4 2

0 909

6-54. (a)

E

n

2 2

2 2

nh

n 1 h

and E

2 n 1

2

8mL

8mL

So,

2 2

En

1

En

n 2n 1 n 2n 1 2 1/

n

2 2

En

n n n

For large n, 1/

n

2 and

E

n

1 n

2

E

n

E

n

2

(b) For n = 1000 the fractional energy difference is 0. 002 0. 2%

1000

(c) It means that the energy difference between adjacent levels per unit energy for large

n is getting smaller, as the correspondence principle requires.

2 2 x

6-55. The n = 2 wave function is

2

x sin

L L

E k op

2 2

2m x

2

Therefore,

2 2

Ek

2

x

2 2

x dx

2m x

and the kinetic energy operator

2 2

2 2 x

2 2

sin

sin

2

L L 2m x L L

x dx

149



2 L

2

2 2 x 2 2

x

sin

sin dx

L 2m L L L

0

2

2 L

2 2 2 2

x

sin dx

L 2m L L

0

Let 2 x y, then x 0 y 0 and x L y 2 and

2 dx dy dx L dy

L

L

2

Substituting above gives:

E

k

2 2

2

2

2 L

L 2m L 2

0

sin

2

ydy

2

sin

2

ydy

y sin2y

2

2 4 2

0 0

2

0 0 0

Therefore,

E

k

2 2

2 2 2

2 2 L 4 h

L 2m L 2 2mL 2mL

2 2

6-56. (a) The requirement is that

2 2

x x x x . This can only be true if:

x x or x x .

(b) Writing the Schrödinger equation in the form

2

d

dx

2mE

2 2

,

the general solutions

of this 2 nd order differential equation are: x Asinkx and x Acoskx

where k 2 mE . Because the boundaries of the box are at x L / 2,

both

solutions are allowed (unlike the treatment in the text where one boundary was at

x = 0). Still, the solutions are all zero at x L/

2 provided that an integral number

of half wavelengths fit between x L/ 2 and x L / 2.

This will occur for:

1/

2

n

x 2 / L cos n x / L when n 1, 3, 5,

. And for

1/

2

n

x 2 / L sin n x / L when n 2, 4, 6,

.

The solutions are alternately even and odd.

(c) The allowed energies are:

2 2 2 2

2 2 2

E k / 2m n / L / 2m n h / 8mL

.

150


6-57.

0

Ae

2 2

x / 2L

(a)

d

dx

d

dx

2 2 2 2

0 2 x / 2L 0

2 x / 2L

x / L Ae and

1

L L x / L Ae x / L

0

d

1

So,

dx

1/ L x / L d / dx

0 0

And

d

dx

2

1

2 2

1/ L d

2 0

/ dx 1/ L d

0

/ dx x / L d

0

/ dx

2 x / L x / L x / L

3 3 3 5

0 0 0

2 2 4

Recalling from Problem 6-3 that V x x / 2mL , the Schrödinger equation

becomes

simplifying:

will make

/ 2m 3m / L x / L x / 2mL E x / L or,

2 3 3 5 2 3 5

0 0 0

2 3

3 x / 2mL E x / L . Thus, choosing E appropriately

1 a solution.

0 0

(b) We see from (a) that

E

2 2

3 / 2mL , or three times the ground state energy.

(c) plotted looks as below. The single node indicates that 1 1

is the first excited state.

(The energy value in [b] would also tell us that.)

6-58.

L

2 2 2 2

n x

x x sin dx Letting u n x / L, du n / L dx

L L

0

2 n

2 2 L L 2 2

x u sin udu

L n n

0

3 3 2

2 L u u 1 u cos2u

sin2u

L n 6 4 8 4

n

0

151



3 3

2 2

2 L n n L L

0 0

L n 6 4 3 2n

2 2

6-59.

E E

T e E eV, V eV, nm.

(a)

2 a

16 1 where 10

0

25 and 1

V0 V0

2m V E 2 m c V E c

2

0 0 0

2 0. 511 10 eV 15eV 197. 3eV nm 19.

84nm

6 1

And

a nm nm a

1

19. 84 1 19. 84; 2 29.

68

(b) For

10 10

T 16 1 e 4.

95 10

25 25

29.

68 13

a nm a nm nm

1

0. 1 : 19. 84 0. 1 1.

984

10 10

T e

25 25

2.

968

16 1 0.

197

6-60. (a) For x, t Asin

kx t

d

dx

2

2

and cos

2

k A kx t

t

so the Schrödinger equation becomes:

2 2

k A sin kx t V x A sin kx t i cos kx t

2m

Because the sin and cos are not proportional, this

for x, t Acos

kx t , there are no solutions.

cannot be a solution. Similarly,

(b) For x, t A cos kx t i sin kx t Ae , we have that

d

dx

2

2

k and i .

2

2 2

k

2m

t

2 2

V x for k 2m V

i kx

And the Schrödinger equation becomes:

/ .

t

152


6-61.

V

mgz

E k = E - mgz

E

Classically allowed:

0 < z < z 0

0

z 0

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

x V x x E x from which we

have:

2 2 2

Ek

x E V x x / 2m d / dx . The expectation value of

153



E is E E x x dx.

Substituting E x from above and reordering

k k k k

multiplied quantities gives:

2 2

d

Ek

x x dx .

2

2m dx

6-63. (a) p x m v x

34 31 12

v / m a 1. 055 10 J s 9.

11 10 10 m

8

v 1. 6 10 m/ s 0.

39c

(b) The width of the well L is still an integer number of half wavelengths, L n / 2 ,

and deBroglie’s relation still gives: L nh / 2p . However, p is not given by:

p

2

k

mE , but by the relativistic expression:

1/

2

2

2 2

p E mc c .

Substituting this yields:

nhc

L E mc nhc L

2 E mc

2

2 2

2

/ 2

1/ 2

2

2 2

E

n

nhc

2L

2

mc

2

2

1/

2

(c)

1/

2

1/

2

2

2

hc

2 1240eV nm

2

2 6 5

1

0. 511 10 8.

03 10

2

2

3

E mc eV eV

4L

4 10 nm

(d) Nonrelativistic:

E

2 2

2

h hc

1240eV nm

8mL 8 mc L 8 0.

511 10 eV 10 nm

1 2 2 2 2

6 3

5

3.

76 10 eV

E 1 computed in (c) is 2.14 times the nonrelativistic value.

6-64. (a) Applying the boundary conditions of continuity to and d / dxat 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



At x 0 : A B C D and ikA ikB C D

At

x a : Fe Ce De and ikFe Ce De

ika a a ika a a

Eliminating the coefficients C and D from these four equations, a straightforward

but lengthy task, yields:

2 a 2 a ika

4ik A ik e ik e Fe

The transmission coefficient T is then:

T

F

2

4ik

2

ika 2 a 2 a

A e ik e ik e

2

Recalling that

1

sinh e e and noting that ik and

2

ik are

complex conjugates, substituting k 2 mE and 2 m V0

E , T then

1

can be written as

T

1

sinh

2

E

4 1

V

a

E

V

0 0

(b) If a 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:

ik a

2 2 2 2

F 4ik e F 6 k e

and T

A ik

A k

2 2

2 2

a

Or

T

E

16 1

V

E

V

0 0

e

2

a

6-65.

2 2 2

II

C e

a

(Equation 6-72)

Where

C

2 2

1/

2

1/

2

2 2E

2 2 0.

5V

0

A

1/

2

1/ 2 1/ 2 1/

2

E E V0 0. 5V 0

0.

5V

0

2.

000

2

2

0

2

p

20

m V E m c MeV

155



2 938. 3MeV 20MeV 197. 3MeV fm 0.

982 fm

1

x ( fm )

e

2 x

2 2 2 x

II

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

Chapter 7 – Atomic Physics

2 2

2 2 2

7-1. En n n

n1 n2 n3

1 2 3

(Equation 7-4)

2

2mL

E E E

2mL

2mL

2 2 2 2

2 2 2

311

3 1 1 11

2 0

where

0

2

2 2 2 2 2 2

E E 2 2 2 12 E and E E 3 2 1 14E

222 0 0 321 0 0

The 1 st , 2 nd , 3 rd , and 5 th excited states are degenerate.

E

(×E 0 )

14

12

10

321

222

311

221

8

6

211

4

2

111

0

7-2.

E

n1 n2 n3

2 2 2 2 2 2 2

2 2

n1 n2 n

3

2 n2

n

3

=

2 2 2 n

2 1

(Equation 7-5)

2m L1 L2 L3 2mL1

4 9

n1 n2 n3 1 is the lowest energy level.

E E 11/ 4 1/ 9 1.

361 E where E

2 2

111 0 0 0 2

2mL1

The next nine levels are, increasing order,

157



n

1

n

2

3

n

E E 0

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

7-3. (a) x y z

n1 n2 n3

n x n y

n3

z

L L L

1 2

, , Acos sin sin

(b) They are identical. The location of the coordinate origin does not affect the energy

level structure.

7-4. x, y, z

Asin sin sin

x y z

111

x y z

L 2L 3L

1 1 1

x, y, z

Asin sin sin

x y z

121

x y z

L L 3L

1 1 1

x y z

113

x, y, z

Asin sin sin

L 2L L

1 1 1

x y 2

z

112

, , Asin sin sin

L 2L 3L

1 1 1

x y 2

z

122

, , Asin sin sin

L L 3L

1 1 1

7-5.

E

n1 n2 n3

2 2 2 2 2 2 2

2 2

n1 n2 n

3

2 n2

n

3

= n

2 2 2

2 1

(from Equation 7-5)

2m L1 2L1 4L1

2mL1

4 16

n n

En 1 n2 n

n E

3

2 2

2 2

2 2 3

1 where

0 2

4 16 2mL1

158



(a)

n

1

n

2

3

n

E E 0

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. x, y, z

Asin sin sin

x y z

111

x y z

L 2L 4L

1 1 1

x, y, z

Asin sin sin

x y z

113

x y 3

z

L 2L 4L

1 1 1

x, y, z

Asin sin sin

x y z

114

x y z

L 2L L

1 1 1

x, y, z

Asin sin sin

x y z

123

x y 3

z

L L 4L

1 1 1

x, y, z

Asin sin sin

x y z

124

x y z

L L L

1 1 1

x y z

112

, , Asin sin sin

L 2L 2L

1 1 1

x y z

121

, , Asin sin sin

L L 4L

1 1 1

x y z

122

, , Asin sin sin

L L 2L

1 1 1

x y 5

z

115

, , Asin sin sin

L 2L 4L

1 1 1

x y 3

z

116

, , Asin sin sin

L 2L 2L

1 1 1

7-7.

E

2

34 2

1.

05510

Js

31 9 19

. . . /

2 2

37.

68eV

2mL 2 9 11 10 kg 0 10 10 m 1 609 10 J eV

0 2

2

E E E 11E 3E 8E 301eV

311 111 0 0 0

159



E E E 12E 3E 9E 339eV

222 111 0 0 0

E E E 14E 3E 11E 415eV

321 111 0 0 0

7-8. (a) Adapting Equation 7-3 to two dimensions (i.e., setting k 3 = 0), we have

n1x n2y

nn

Asin

sin

1 2

L L

2 2

2 2

(b) From Equation 7-5, Enn

2 n1 n2

1 2

2mL

(c) The lowest energy degenerate states have quantum numbers n 1 = 1, n 2 = 2, and n 1 = 2,

n 2 = 1.

7-9. (a) For n = 3, 0,,

1 2

(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 45

20

m 4

min

(b) For 2

cos min

26.

6

20

1 4

L 6 m 2

min

cos min

35.

3

6

1 2

7-11. (a)

5 2 1 4 2

L I 10 kg m 2 735min 1min / 60s 7. 7

10

kg m / s

L 1 7. 7

10 kg m / s

4 2

(b)

160



1

7 7 10

4 2

. kg m /

1.

05510

34

Js

s

2

4 2

7. 710

kg m / s

7.

310

34

1.

05510

Js

2

30

7-12. (a)

+1

1

L

2

0

−1

(b)

+2

2

L

6

+1

0

−1

−2

161



(c)

+4

+3

+2

4

L

20

+1

0

−1

−2

−3

−4

(d) L 1 (See diagrams above.)

7-13. L L 1 2

x

Ly Lz Lx Ly L Lz

m 6

m

2 2 2 2 2 2 2 2 2 2 2

(a) Lx

Ly

6 2 2

2 2 2 2 2

min

(b) Lx

Ly

6 0 6

2 2 2 2 2

max

L L 6 1 5 L and L cannot be determined separately.

(c)

(d) n = 3

2 2 2 2

x y x y

For 1, L 1 2 1.

49

10

34

7-14 (a)

(b) For 1, m 1, 0,

1

J s

162



(c)

Z

+1ћ

0

L

2

−1ћ

34

(d)

For 3, L 1 12 3. 6510 J s and m 3, 2, 1, 0, 1, 2,

3

Z

3ћ

2ћ

1ћ

0

−1ћ

−2ћ

−3ћ

7-15.

dL dr dp

L = r × p 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

dL

F is parallel to r, then r F = 0 and 0

dt

7-16. (a) For 3, n = 4, 5, 6, … and m = −3, −2, −1, 0, 1, 2, 3

(b) For 4, n = 5, 6, 7, … and m = −4, −3, −2, −1, 0, 1, 2, ,3 ,4

163



(c) For 0, n = 1 and m = 0

(d) The energy depends only on n. The minimum in each case is:

E eV n eV eV

2 2

4

13. 6 / 13. 6 / 4 0.

85

E eV eV

2

5

13. 6 / 5 0.

54

E1 13.

6eV

7-17. (a) 6 f state: n6,

3

(b)

E eV n eV eV

2 2

6

13. 6 / 13. 6 / 6 0.

38

34

(c)

L 1 3 3 1 12 3.

65

10

(d) L m L 3 , 2 , 1 , 0, 1 , 2 , 3

z

z

J s

7-18. Referring to Table 7-2, R 30 = 0 when

2

2r

2r

1 0

2

3a0 27a0

Letting r a0 x,

this condition becomes

x

2

9x13.

5 0

Solving for x (quadratic formula or completing the square), x = 1.90, 7.10.

Therefore, ra0 1. 90, 7.

10 . Compare with Figure 7-10(a).


E

n

kZe

2

2

2

2n

Using SI units and noting that both Z and n are unitless, we have:

Cancelling the

(N m / C ) C

J

Js

2 2 2

2

kg

2

C and substituting J Nm

on the right yields

2

Nm

J

N ms

2

kg

.

164



Cancelling the N and m gives

2

m

J kg J , since

s

2 2

kg m / s are the units of kinetic

energy.

7-20. (a) For the ground state n = 1, 0, and m = 0.

1

2 r / a 1 2

0 r / a 2e

0

100

R10Y 00

e e at r a

3 2 3

0

a 4 0

4a0 4a0

2 1 2

1

0

2

(b) r / a

e e at r a

3 3

a

a

0 0

4

P r

4 r e at r a

a

2 2 2

(c)

0

0

0

2

4r

P r r r r e r

a

2 2

2

/ 0

7-21. (a) For the ground state, 4

r a

(b) For

2

4a

0. 03 , at we have 0. 03 0.

0162

0 2

For r a r a Pr r e a

0 0 3

0

a0

42a

r 0. 03 a , at r 2 a we have P r r e 0. 03a

0.

0088

0 4

0 0 3

0

a0

3

0

2

P r

2 0

7-22. 2 Zr / a

Cr e

For P(r) to be a maximum,

dP

2

2Z

2Zr / a0 2Zr / a 2Zr

a

0 0 2Zr / a0

C r e 2re 0 C r e 0

dt a0 a

0

Z

This condition is satisfied with r = 0 or r = a 0 /Z. For r = 0, P(r) = 0 so the maximum

P(r) occurs for r = a 0 /Z.

165


7-23.

2

2 2 2

d r sindrd d

1

0 0 0

2

2 2 2

Zr

2 / 0

4

r dr 4

C Zr a

210

r e dr 1

a

0 0 0

2 4

2 Zr / 0

4

C Zr a

210

e dr 1

2

a

0 0

Letting x Zr / a0,

we have that r a0x / Z and dr a0dx / Z and substituting

these above,

3 2

2 4

aC

0 210 4 x

d

3

Z

0

x e dx

Integrating on the right side

4 x

x e dx

0

6

Solving for

2

C

210

yields:

C

Z

Z

3 3

2

210

C

3 210

3

24a0 24a0

1/

2

3/

2

1 Z r

2 0

7-24.

200

1 r / a

e

(Z = 1 for hyrdogen)

32

a0 a0

2 2 1 1 r

/ 0 2

Prr

200 4 r r 1 e r a

3 4

r r

32

a0 a0

(a) For r 0. 02 a0,

at r a0

we have

4

1 2 1 2 1 1

Prr 11 e a0 0. 02a0 0e

0.

02

0

32

a

8

(b) For r 0. 02 a0,

at r 2 a0

we have

3

0

4

1 1

Prr 1 e a 0 02a 1 e 0 02

3 4

10

32

a

8

3

0

2

2 2 2 2 4

0

.

0

. .

166


Zr Zr

/ 2a0

7-25. 210 C210

e cos

a

0

(Equation 7-34)

Zr

P r r r C e

2 2

2 2 2

210 210

cos

2

a0

2 2 r/

a0

4 4

r/

a0

2 2 2 4 2

210

/

0

cos

4

C Z a r e

Ar e

cos

4 r/

a0

2

where A 4 C 2 2 2

210

Z / a0

, a constant.

3/

2

1 1 r

2 0

7-26.

200

2 r / a

e

(Z = 1 for hyrdogen)

32

a0 2a0

1 1

1/ 2 0.

606 1

r a0 200

e

3

32

a0 32

a0

(a) At ,

2 1

(b) At

1 1 0.

368 1

r a e

2 1

0,

200 3

3

32

a0 32

a0

4 0. 368a

0.

368

2

2 2 0

0 200 3

32

a0 8a0

(c) At r a , Pr 4r

3/

2

7-27. For the most likely value of r, P(r) is a maximum, which requires that (see Problem 7-25)

dP

dr

Z

a0

4 0

2 4 Zr / a0 3 Zr / a0

Acos

r e r e

2 3

r/

a0

For hydrogen Z = 1 and

Acos r / a 4a r e 0. This is satisfied for r = 0

0 0

and r = 4a 0 . For r = 0, P(r) = 0 so the maximum P(r) occurs for r = 4a 0 .

7-28. From Table 7-1, Y

15

, sin cos

8

21

2

4 r

R r

e

r

3a0

From Table 7-2,

32

2 2

81 30a

a

0 0

i

e

167



, , 4 1 15 2 3 0

r r e

8

sin cos e

321

3 2

81 30a

a

0 0

r a i

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

2

2 4 2r

3a0

2 2 2

321 321

0 0 0 0 0 0

C dr C r e sin cos r sin dr d d

1

Noting that

2

0

d

2

, we have

2 6 2r

3a0

3 2

321 sin cos

0 0

2 C r e dr d

1

Evaluating the integrals (with the aid of a table of integrals) yields:

4

2

C

2

4

7

32 1

1.

23 10 a

0

1

15

C

7

321 0.

00697 a0

This value agrees with the coefficient of

32 1

above.

2 / 0

7-29.

r a

100

e

3

4

a

0

Because

100

is only a function of r, the angle derivatives in Equation 7-9 are all zero.

d

dr

2

4

a

3

0

e

r/

a0

r

d

dr

2 1

r e

4

a a0

2 2 r/

a0

3

0

d d

2 1 1

dr

dr a a

2 2

r/

a0

r r 2re

3

4

a0

0 0

1 d d

2 1 2 1

r e

r dr dr a r a

2

2

3

4

a0

0 0

r/

a0

Substituting into Equation 7-9,

168



2

1 2

V E

2

2

a0 a0r

100 100 100

For the 100 state r a0 and 2 a0 2 / k or a0

1 / k,

so

1 2 1 2 1

2 k

2 2

2

a0 a0r a0 a0 a0

2

Thus,

2 2 2

1 2 k

2

2a0 a0r

2

and we have that

2 2

k

V

E,

satisfying the Schrödinger equation.

2 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 / R Q

R /

1 2

L I

MR

2

2

2M

2MQR

/ 2

2 2

2 2

g 2

2

QL QMR / 2

(b) The entire charge is on the equatorial ring, which rotates with frequency f /

2

.

i Qf Q/

2

iA Q / R Q

R /

2

2M

2MQR

/ 2

2 2

2 2

g 5/ 2 2.

5

2

QL QMR / 5

169


7-31. Angular momentum S I

2/ 5mr 2

v / r

or

1 / 2

v 5/ 2 S 1/ mr 5S / 2mr 5 3/ 4 / 2mr

12 /

34

/ . Js

31 15

. kg m

5 3 4 1 055 10

2 9 11 10 10

11

2. 51 10 m / s 837

c

7-32. (a) The K ground state is 0 , so two lines due to spin of the single s electron would

be seen.

(b) The Ca ground state is 0 with two s electrons whose spins are opposite resulting

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. F m g dB / dz m a (From Equation 7-51)

z s L B Ag z

and

a m g dB / dz / m

z S L B Ag

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

2

amount z 1 given by:

2

z1 1/ 2 azt 1/ 2 ms gLB dB / dz / mAg

1/

250

and leave

the magnet with a z-component of velocity given by v z

a t . The additional z deflection

z

in the field-free region is

z v t a t

2

2

z

z

.

The total deflection is then

z1 z2 0 5mm 5 0

10 m

4

. . .

2

5. 0 10 3/ 2 3/ 2 / / 1/

250

dB

dz

4 2

m z1 z2

azt ms gL B

dB dz mAg

4

5. 010 m250 2

mAg

2

m g

s L B

2

4

1

5. 010 m 250s 1.

7910 25 kg 2

24

31/ 219. 27 10

J / T

0. 805T / m

or,

170


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 S , 4 P , 4 P , 4 D , 4 D , 4 F , 4 F

(b) 2j + 1

2 2 2 2 2 2 2

1/ 2 1/ 2 3/ 2 3/ 2 5/ 2 5 / 2 7 / 2

7-36. For 2, L 1 6 2. 45 , j 1/ 2 3/ 2, 5/

2 and J j j 1

For j J

3/ 2, 3/ 2 3/ 2 1 15/ 4 1.

94

For j J

5/ 2, 5/ 2 5/ 2 1 35 / 4 2.

96

7-37. (a) j 1/ 2 2 1/ 2 5/ 2 or 3/

2

5 3

1 5/ 2 1 2. 96 or 3/ 2 1 1.

94

2 2

(b) J j j

(c) J L S J L S m m m where m j, j 1,..., j 1, j.

For

Z Z Z s j j

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) d

5/

2

m s =1/2

L

θ

s

3/

4

1 12 /

cos 54.

7

3/

4

(b) d

3/

2

m s

=−1/2

L

θ

s

3/

4

171

1 12

/

cos 125.

3

3/

4


7-39.

n m m

s

4

3

2

3

2

1

0

2

1

0

1

0

−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

1/ 2 for each m state



7-40. (a) L L1L

2

, 1 ,..., 11 , 111 , 11 2, 1,

0

(b) S S1S

2

1 2 1 2 1 2

s s s , s s 1 ,..., s s 1/ 2 1/ 2 , 1/ 2 1/ 2 1,

0

(c) J L S

1 2 1 2 1 2

1

j s , s ,..., s

For 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

(d)

1

1

1

0 and s 0 , j 0

J L S j1

11/ 2 3/ 2 , 1/

2

J2 L2 S 2

2

(e)

1

2

j2 1/ 2 3/ 2 , 1/

2

J J J 1

1 2

1 2

j j j , j j ,..., j j

1 2 1 2 1 2

For j 3/ 2 and j 3/ 2 , j 3 , 2 , 1 , 0

j 3/ 2 and j 1/ 2 , j 2 , 1

For j 1/ 2 and j 3/ 2 , j 2 , 1

1 2

j 1/ 2 and j 1/ 2 , j 1 , 0

1 2

These are the same values as found in (c).

172


7-41. (a)

E hf f E h

6 19 34 9

/ (4.372 10 eV)(1.602 10 J/eV) / 6.63 10 J s=1.056 10 Hz

8 9

(b) c f c / f (3.00 10 m/s) / (1.056 10 Hz) 0.284m 28.4cm

(c) short wave radio region of the EM spectrum

hc

7-42. (a) E3/

2

Using values from Figure 7-22,

1239. 852eV nm

1239.

852eV nm

E3 / 2

2. 10505 eV E1 / 2

2.

10291eV

588. 99nm

589.

59nm

(b)

3/ 2

1/ 2

2. 10505 2. 10291 2.

14

10

3

E E E eV eV eV

(c)

3

E

2.

1410

eV

E 2BB B 18.

5T

4

2

2 5. 7910

eV / T

B

x

2x

x , x C sin sin Substituting into Equation 7-57 with V = 0,

L L

7-43.

1 2

12 1 2

2 2 2

2 2

12

12

2 2

1 4

E

2

2m x1 x2

2m L

2 2

5

Obviously, 12 is a solution if E

2

2mL

12 12

7-44.

E

n

2 2 2

n

Neutrons have antisymmetric wave functions, but if spin is ignored then

2

2mL

one is in the state n = 1 state, but the second is in the n = 2 state, so the minimum energy

2 2

is:

E E E 1 2 E 5 E where

1 2 1 1

2 2 2 2

hc 197.

3

2 2

2mc L 2939. 62.

0

E 51.

1 MeV E 5E 255MeV

1 2

1

173


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:

2 2 2

n

2 2

E 2E 2 E E where E E 2E 22 E 3 E 19E

2mL

where

1 2 3 n

2

1 1 1 1

2 2 2 2

hc 197.

3

2 2

2m 6

ec L 20. 51110 1.

0

E 0. 376 eV E 19E 7.

14eV

1 2

1

(b) Pions are bosons and all five can be in the n = 1 state, so the total energy is:

0.

376eV

E 5 E1 where E1 0. 00142 eV E 5E1

0.

00712eV

264

7-46. (a) Carbon:

(b) Oxygen:

Z 6;

1s 2s 2p

2 2 2

Z 8;

1s 2s 2p

2 2 4

(c) Argon:

Z 18;

1s 2s 2p 3s 3p

2 2 6 2 6

7-47. Using Figure 7-34:

Sn (Z = 50)

2 2 6 2 6 10 2 6 10 2 2

1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p

Nd (Z = 60)

2 2 6 2 6 10 2 6 10 2 6 4 2

1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p 4 f 6s

Yb (Z = 70)

2 2 6 2 6 10 2 6 10 14 2 6 2

1s 2s 2p 3s 3p 3d 4s 4p 4d 4 f 5s 5p 6s

Comparison with Appendix C.

Sn: agrees

6 4

Nd: 5 p and 4 f are in reverse order

Yb: agrees

2

7-48. Both Ga and In have electron configurations ns

n 1 s n 1 p n 1 d

2 6 10

np outside of closed shells

, , , . 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


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.

E

Z

n

eff

Z E

n

2

eff 1

(Equation 7-25)

2

En

5.

14eV

n 3 1.

84

E 13.

6eV

1

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, 2 , so 2 , 1 , 0 , 1 , 2

L z

(b) For an f electron, 3 , so 3 , 2 , 1 , 0 , 1 , 2 , 3

L z

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


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

4S 1/2 → 3S 1/2 0 0 l - forbidden

4S 1/2 → 3P 3/2 +1 +1 allowed

4P 3/2 → 3S 1/2 −1 −1 allowed

4D 5/2 → 3P 1/2 −1 −2 j – forbidden

4D 3/2 → 3P 1/2 −1 −1 allowed

4D 3/2 → 3S 1/2 −2 −1 l - forbidden

5D 3/2 → 4S 1/2 −2 −1 l - forbidden

5P 1/2 → 3S 1/2 −1 0 allowed

2 2 4

E1 13. 6eV Z 1 13. 6eV 74 1 7. 2510 eV 72.

5keV

7-58. (a)

2 2

(b) E exp . keV . eV Z

. eV

1

69 5 13 6 13 6 74 1

1 /

3

eV eV

2

74

69. 510 / 13. 6 71.

49

74 71. 49 2.

51

176


7-59. j , j j

1 0 no 0 0 (Equation 7-66)

The four states are 2 P , 2 P , 2 D ,

2 D

3/ 2 1/ 2 5 / 2 3/

2

Transition ∆l ∆j Comment

D 5/2 → P 3/2 −1 −1 allowed

D 5/2 → P 1/2 −1 −2 j - forbidden

D 3/2 → P 3/2 −1 0 allowed

D 3/2 → P 1/2 −1 −1 allowed

7-60. (a) E

hc /

1240eV nm

E 3P1 / 2 E 3S1 / 2 2.

10eV

589.

59nm

E 3P1 / 2

E 3S1 / 2

2. 10eV 5. 14eV 2. 10eV 3.

04eV

1240eV nm

E 3D E 3P1/ 2

1.

52eV

818.

33nm

E 3D E 3P1/ 2

1. 52eV 3. 04eV 1. 52eV 1.

52eV

(b)

3.

04eV

For 3P: Zeff

3 1.

42

13.

6eV

1.

52eV

For 3D: Zeff

3 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 B (Equation 7-73) where s 1/ 2, 0 gives j 1/

2 and

j

B

(from Equation 7-73) g 2. m j

1/ 2.

5

E 2 1/ 2 5. 79 10 eV / T 0. 55T 3.

18 10

5 eV

The total splitting between the

mj

1 2 states is 6 37 10

5

/ . eV.

(b) The m 12 / (spin up) state has the higher energy.

j

(c)

5 15 10

E hf f E / h 6. 37 10

eV / 4. 14 10 eV s 1.

54 10 Hz

This is in the microwave region of the spectrum.

177


7-62.

2

hc dE hc

E E

E

2

d hc

e

2m

7-63. (a)

(b)

5

E B 5. 79 10 eV / T 0. 05T 2.

90 10

6 eV

2

579. 07nm

2.

9010

6 eV

2

E

7.

83

10

hc

1240eV nm

4

nm

(c) The smallest measurable wavelength change is larger than this by the ratio

4

0 01nm

7 83 10 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.

2 2

7-64. En

13.

6eV Zeff

n

E eV Z eV

2 2

2

13. 6

eff

2 5.

39

1 / 2

Z 2 5. 39 13. 6 1.

26

eff

3/

2

1 Z

/ 0

7-65.

Zr a

100

e

a0

(Equations 7-30 and 7-31)

2

r

P r

4 (Equation 7-32)

*

100 100

Z 4Z

4

r e r e

3 3

2 Zr / a0 2 2Zr / a0

3 3

a0 a0

3

4 Z

/

3 2Zr a0

3

r rP r dr r e dr

a

0 0

0

3

0

2Zr

2Zr / a0

0

3

0

e d 2Zr / a0

3!

Z

a

0 0

Z Z

a a a

4

4 2

178


7-66. (a)

E

1 2

2I

: 0 1 2 3 4 5 …

1: 1 2 3 4 5 6 …

E : 0 1E 1 6E 1 12E 1 20E 1 30E 1 …

E 5 = 30E 1

20E 1

E 4 = 20E 1

10E 1

E 3 = 12E 1

E 2 = 6E 1

E 1 =

2 / I

30E 1

0

E 0 = 0

2

(b) E E

1

1 2 1

2I

2 2

1 2

1 1

2I

I

The values of 0, 1, 2, … yield all the positive integer multiples of E 1 .

E

1

(c)

1 2

c 2

2 2

2

I m r E1

2 2 2

2 p I mpr mpc r

2

2

2197.

3eV nm

6

938. 2810 eV 0.

074nm

1.

5210

6

hc 1.

2410

eV nm

5

(d) 8. 1810 m

81.

8m

2

E 1.

5210

eV

1

2

2

eV

7-67. (a) F m g dB / dz (From Equation 7-51)

z s L B

From Newton’s 2 nd law, F m a m g dB / dz

z H z s L B

/ / 1/ 219. 27 10 24

/ 600 / / 1.

67 10

27

a m g dB dz m J T T m kg

z s L H

1. 67 10 m/

s

6 2

179



(b) At 14. 5km / s v 1. 4510 4 m/ s, the atom takes t 4

1

0. 75m / 1. 45

10 m/

s

5

5. 2 10 s to traverse the magnet. In that time, its z deflection will be:

2

z 1/ 2 a t 1/ 2 1. 67 10 m/ s 5. 210 s 2. 2610 m 2.

26mm

2 6 2 5 3

1 z 1

Its v z velocity component as it leaves the magnet is vz

azt1

and its additional z

deflection before reaching the detector 1.25m away will be:

. .

4

z

/

z2 vzt2 a t1 1 25m

1 4510

m s

1 67 10 6 2 5 2 10

. m/ s . 5 s 1 . 25 1 . 45 10 4 m/

s

3

7. 49 10 m 7.

49

mm

Each line will be deflected z1z2 9.

75mm

from the central position and, thus,

separated by a total of 19.5mm = 1.95cm.

1

7-68. m

min

cos 1 with m .

cos 1 . Thus, cos 1

1

sin

or,

2 2 2

min min min

2

2 2 2

1

2

sin min

1

1 1 1

And,

1/

2

1

sin min

For large ,

1

min

is small.

Then sin

1/

2

1 1

1

min min 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

0.

0073eV

E / 2 gm

jBB B 63T

2gm

2 2 1 2 5 79 10

j

B

5

/ . eV / T

180


4Z

a

2 2Zr/

a0

7-70.

P r

3

0

3

r e

(See Problem 7-65)

15

For hydrogen, Z = 1 and at the edge of the proton r R0 10 m.

At that point, the

exponential factor in P(r) has decreased to:

15 10 5

2R

210 0 529 10 3 78 10

0 / a . m .

0

5

e e e 1 3.

7810 1

Thus, the probability of the electron in the hydrogen ground state being inside the nucleus,

to better than four figures, is:

2 r 0

0 R0 2 R

R

0

3

2

3 3 3

3

0 0 0 0 0 0 0

3

0

4r 4r 4 4 r

Pr

P P r dr r dr

a a a a

3

15

4 R

4 10 m

0

9 0 10

3 .

3

a

10

0

3

3 0.

529 10

m

3

15

7-71. (a)

1 ss

1 1

2j

j1

j j

g 1

(Equation 7-73)

For 2 P1/ 2: j 1/ 2, 1 , and s 1/

2

1/ 2 1/ 2 1 1/ 2 1/ 2 1 1 11 3/ 4 3/

4 2

g 1 1 2 / 3

2 1/ 2 1/ 2 1 3/

2

For 2 S1/ 2: j 1/ 2, 0 , and s 1/

2

1/ 2 1/ 2 1 1/ 2 1/ 2 1 0 3/ 4 3/

4

g 1 1 2

2 1/ 2 1/ 2 1 3/

2

The 2 P

12 /

levels shift by:

2 1 1

E gm

jBB BB BB

3

2

3

(Equation 7-72)

The 2 S

12 /

levels shift by:

1

E gm

jBB 2 BB BB

2

181



To find the transition energies, tabulate the several possible transitions and the

corresponding energy values (let E p and E s be the B = 0 unsplit energies of the two

states.):

Transition

Energy

P

1

2

S

Ep BB Es BB Ep Es BB

3

3

1/ 2, 1/ 2 1/ 2, 1/

2

P

1

4

S


3

3

1/ 2, 1/ 2 1/ 2, 1/

2

P S

1/ 2, 1/ 2 1/ 2, 1/

2

P

1/ 2, 1 / 2 1/ 2, 1 / 2

1 4


3

3

1

2

S


3

3

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: j 3/ 2, 1 , and s 1/

2

3/ 2 3/ 2 1 1/ 2 1/ 2 1 1 11 15/ 4 3/

4 2

g 1 1 4 / 3

2 3/ 2 3/ 2 1 30 / 4

These levels shift by:

4 1 2

4

3

E gm

jBB BB BB

3

2

E BB 2 BB

3

3

2

182



Tabulating the transitions as before:

Transition

Energy

P

S

Ep 2BB Es BB Ep Es BB

3/ 2, 3/ 2 1/ 2, 1/

2

P3 / 2, 3/ 2

S1 / 2, 1 / 2

forbidden, m j

2

P

2

1

S


3

3

3/ 2, 1/ 2 1/ 2, 1/

2

P S

3/ 2, 1/ 2 1/ 2, 1/

2

P

3/ 2, 1/ 2 1/ 2, 1/

2

P

2 5


3

3

2

5

S


3

3

2

1

S


3

3

3/ 2, 1 / 2 1/ 2, 1 / 2

P

S

forbidden, m j

2

3/ 2, 3/ 2 1/ 2, 1/

2

P

S

Ep 2BB Es BB Ep Es BB

3/ 2, 3/ 2 1/ 2, 1 / 2

There are six different photon energies emitted (two transitions are forbidden); their

spectrum looks as below:

anomolous

normal

7-72. (a) Substituting r,

yields (eventually)

into Equation 7-9 and carrying out the indicated operations

2 2

2 2 2

r, 2/ r 1/ 4a r, 2/ r V r, E r,

0

2

2

183



Canceling r,

and recalling that

r

4a

(because given is for n = 2) we

2 2

0

2

2

have 14

0

/ a v E

2

The circumference of the n = 2 orbit is:

C 2 4a 2 a / 4

1/ 2k.

0 0

Thus,

2 2 2

1

k

V E V E

2

2

4 / 4k

2

(b) or

2

p

v

E and Equation 7-9 is satisfied.

2m

2

2 2

r

r/

a0

2 2

dx A e cos r sindrd d

0

a

0

2

2

2

r

r/

a0

2 2

A e r dr cos sin d d

a

1

0 0

0 0

Integrating (see Problem 7-23),

0 /

2 3

A a

6 2 3 2 1

A 1/ 8a A 1/

8a

2 3 3

0 0

1

7-73. gLBL

(Equation 7-43)

(a) The 1s state has 0, so it is unaffected by the external B.

The 2p state has 1, so it is split into three levels by the external B.

(b) The 2p1sspectral line will be split into three lines by the external B.

(c) In Equation 7-43 we replace with e / 2 m , so

B k k

11 e / 2m m / m

(From Equation 7-45)

kz k B e k

Then /

E m m B

B e k

5 79 10

. 5 eV / T 0 . 511 10 6 MeV / c 2 / 497 . 7 MeV / c 2

1 . 0 T

5 94

10

8

. eV

184



E (From Problem 7-62) where λ for the (unsplit) 2p → 1s transition

hc

is given by

k k 2 1

k e

3

hc E and E E E 13. 6eV m / m 11/ 4 9.

93

10 eV

3

and 1240eV nm 9. 9310 eV 0.

125nm

and

8

0. 125nm

5.

9410

eV

5.

98 10

1240eV nm

12

7-74.

2

ke

E

B S L where, for n 3,

r a n 9a

3 2

r m mc

2

0 0

For 3P states SL

2

8 9 16

3 2

6

90. 053nm 0.

51110

eV

2 2

1. 440eV nm 3. 00 10 m / s 10 nm / m 6.

58 10 eV s

E

1.

6010

4

eV

For 3D states

SL

2

/ 3

4 4

E 1. 60 10

eV / 3 0.

53 10 eV

7-75. (a) J = L+ S = L S

B

L 2 S / L + S

2 (Equation 7-71)

J B

B

J

L L 2S S 3S L

J J J

J

B 2 2

L 2S

3SL

(b) J 2 J J L + S L + S L L + S S + 2S L S L J 2 L 2 S

2

J

3

2

2 J

B

(c) 2 2 2 2 2 B

2 3

2 2 2

J

L S J L S J S L

J

J

J 2 J J 2 J

Z B 2 2 2 Z B 2 2 2

(d) Z

J

3J S L 3J S L

1

2

185



B 1

J S L

2

2J

2 2 2

J

Z

(e) E

B (Equation 7-69)

Z

j j

BB1

gm B

j

B

1 ss

1 1

2j

j1

(Equation 7-72)

m

j j 1 s s 1 1

where g 1

(Equation 7-73)

2j

j1

j

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)

e

N

5 05

10

2m

p

27

. /

J T

2 k m

2k 2. 8 2k

2.

8

B

r r a

m N m N

3 3 3

0

7 27

H / m . . J / T

3

10

0.

529 10

m

2 10 2 8 5 05 10

0.

0191T

(b)

4

E 2 2 5. 79 10 / 0. 0191 2.

21 10

6

BB eV T T eV

(c)

6

hc 1.

2410

eV m

0. 561m

56.

1cm

6

E

2.

2110

eV

186

Chapter 8 – Statistical Physics

8-1. (a)

(b)

v

rms

. J / mole K K

1/

2

3RT

3 8 31 300

1930m / s

3

M

2. 0079 10

kg / mole

3 3

. kg / mole . m / s

2

Mv 2 1 0079 10 11 2 10

rms

4

T 1.

01

10 K

3R

3 8. 31J / mole K

2

8-2. (a)

3 2E

2 13.

6eV

2 3k

3 8. 617 10

eV / K

k

5

Ek

kT T 1.

0510

K

5

3 3

E kT 8 67 10 10 1 29

2 2 . eV / K K . keV

5 7

(b)

k

8-3.

v

rms

3RT

M

(a) For O 2 :

v

rms

. J / K mol K

3 8 31 273

3210

3

kg / mol

461m / s

(b) For H 2 :

v

rms

. J / K mol K

3 8 31 273

210

3

kg / mol

1840m / s

8-4.

1 / 2 1 /

J / mole K K

/

2

1/

2 2 2

3RT kg m s

m/

s

M

kg / mole kg

3 3

E 1 8 31 273 3400

2 2 . /

K

n RT mole J mole K J

8-5. (a)

(b) One mole of any gas has the same translational energy at the same temperature.

187


8-6.

3/

2

2 1 2 m

4

v v n v dv 4 v e dv where m / 2kT

N

2 kT

0 0

2

v

3/

2

2 m

v I4 I4

4 where is given in Table B1-1.

2 kT

3 3

8 8

1/ 2 5 / 2 1/

2

I4

m/

2kT

5/

2

v

3/ 2 5 / 2

2 1/

2

4

m 3 2kT 3kT 3RT 3RT

2 kT 8 m m mN M

A

v

rms

v

2 3

RT

M

8-7.

v

8kT

23

8 1. 381 10 J / K 300

27

m 1. 009u 1. 66 10 kg / u

K

1/

2

2510m / s

v

m

2kT

23

2 1. 381 10 J / K 300

27

m 1. 009u 1. 66 10 kg / u

K

1/

2

2220m / s

3 / 2 2

2 mv / kT

n v 4 N m / 2 kT v e (Equation 8-28)

At the maximum:

dn

dv

3 / 2 2

2 mv / 2kT

0 4 N m / 2 kT 2v v mv / kT e

2

mv / 2kT

2

0 ve 2 mv / kT

The maximum corresponds to the vanishing of the last factor. (The other two factors give

minima at v = 0 and v = ∞.) So

2

2 0 and

m

2

1/

2

mv / kT v kT / m .

188


8-8.

8-9.

3/

2

m

2

2 mv / 2kT

n v dv 4 N v e dv (Equation 8-8)

2 kT

dn

dv

2

2 2vm

mv / 2kT

A v 2 v e The v for which dn / dv 0 is vm.

2kT

A

2mv

2kT

3

2

mv / 2kT

2v e 0

Because A = constant and the exponential term is only zero for v → ∞, only the quantity

in [] can be zero, so

2mv

2kT

3

2v

0

or

2 2kT

2kT

v vm

m

m

(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 molecules / g mol

A

Each molecule has, on the average, 3kT/2 kinetic energy, so the total translational kinetic

189



energy in one liter is:

KE

23

6.

02 10 23 3 1. 381 10 J / K 293

22.

4 2

K

163J

8-11.

n g e g e

n

E2

/ kT

2 2 2

E1

/ kT

1

ge

1

g1

E2 E1

/ kT

T

E E kT g n g n

e E E kT

2 1 / 2 1 2 1

2 1

/ ln

g1 n2 g1 n2

E E 10.

2eV

2 1

k ln g / g n / n 8. 617 10 eV / K ln 4 10

2 1 1 2

5 6

7790K

8-12.

n2 g2

E 3

2 E1

/ kT

e e

n

g

1 1

1

3

4 10 eV

5

8. 617 10 eV / K 300K

2.

57

8-13. There are two degrees of freedom, therefore,

C 2 R / 2 R, C R R 2 R, and 2R / R 2.

v

p

8-14. c 3 R / M

v

(a)

(b)

(c)

3 1. 99cal / mole K

Al: cv

27. 0g / mole

0. 221 cal / g K 0. 215cal / g K

3 1. 99cal / mole K

Cu: cv

62. 5g / mole

0. 0955 cal / g K 0. 0920cal / g K

3 1. 99cal / mole K

Pb: cv

207g / mole

0. 0288 cal / 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


8-15.

2 N

n( E)

E e

3/

2

kT

1/ 2 E / kT

(Equation 8-13)

At the maximum:

dn

dE

0

2 N 1 1

E E e

3/

2

kT 2

kT

1/ 2 1/ 2

E / kT

E e E / kT

2

1/ 2 E / kT 1

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

.

8-16. (a)

3/

2

m

2

2 mv / 2kT

n v 4 N v e (Equation 8-8)

2 kT

4 N

3/

2

m

2kT

3/

2

2

ve

2

v m / 2kT

4

2

Nv 2 2

v / v m

where 2 /

3 e v

m kT m

vm

4N

1 v v/ vm

e

v v

m

m

2

2

4N

2

A

1 v v/

v

N n v v e

m

0.

01v

v v

m

m

2

m

22 2 v/

vm

. v vm

e

1 36 10

2

(b)

(c)

(d)

(e)

22 2 0

N 1.

36 10 0 e 0

22 2 1 21

N 1. 36 10 1 e 5.

00 10

2

22 2 2

20

N 1. 36 10 2 e 9.

96 10

2

22 2 8

4

N 1. 36 10 8 e 1.

369 10 (or no molecules most of the time)

191


8-17. For hydrogen:

E

to six decimal places.

E1 13.

605687eV

n

mk e

2

2 4

1 13.

605687

eV

n n

2 2 2

using values of the constants accurate

E 3. 401422eV E E 10.

204265eV

2 2 1

E 1. 511743eV E E 12.

093944eV

3 3 1

n g 8

e e 4 e 4 10 0

n g

2

2 2 E2 E1 / kT 10. 20427 / 0.

02586 395 172

(a)

1 1

n3 g3

E3 E1 / kT 18

e e

12. 09394 / 0.

02586 e

468 203

n

g

1 1

2

9 9 10 0

(b)

n

n

2

1

10. 20427 / kT

10. 20427 / kT

0. 01 4e

e 0.

0025

10. 20427 / kT ln 0. 0025 5.

99146

T

10.

20427eV

5. 99146 8.

61734 10

5

eV K

19,

760K

(c)

n

n

3

1

5

12. 09394 / 8. 61734 10 19,

760

9e

0. 00742 0. 7%

8-18.

no field

E

hf

B field

m

m

m

0

1

1

E

E

ground state

Neglecting the spin, the 3p state is doubly degenerate: 01 , hence, there are two m = 0

levels equally populated.

E hf hc / 1. 8509eV 670.

79nm

192



E

eB

2m

e

2.

315 10

4

eV

(a) The fraction of atoms in each m-state relative to the ground state is: (Example 8-2)

n

1 e

e 10 8 .

n

18 10

1. 8511/ 0. 02586 71. 58 31.

09 32

n

0 2 e

2 e 2 10 1 .

n

64 10

n

n

0

1

e

1. 8509 / 0. 02586 71. 57 31.

08 31

e

10 8.

30 10

1 . 8507 / 0 . 02586 71 . 56 31 . 08 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.

intensities will be: 8. 30 / 16. 4/ 8. 18 0. 51/ 1. 00 / 0.

50

With that as the “standard”, the relative

8-19. (a)

e

N

V

22

h

3

m kT

e

3/

2

(Equation 8-44)

3/

2 2

N 2 2 m 2 2

ekT

mec kT

e

3 e

3

V h hc

3/

2

1/

2

5 2 3

7

2 5. 11 10 eV 2.

585 10 eV 10 nm

1 2 2. 51 10 / cm

3

1240eV nm

1cm

19 3

8-20. (a)

e

O

N

V

2 3/

2

2

h

3

MkT

(Equation 8-44)

N

V

A

M

2

hc

3

2

Mc kT

3/

2

6. 022 10 / mole

1.

24 10

23 4

eV cm

3 3 2 6 5

22. 4 10 / 2 32 931. 5 10 / 8. 617 10 / 273

cm mole uc eV u eV K K

3

1.

75 10

7

193



(b) At temperature T,

7 3 2 3 2

2

1 75 10 273 / /

e O . / T

7 3 / 2 3 / 2 3 / 2 7

3 / 2

1 1. 75 10 273K / T T 1.

75 10 273K

2/

3

7

T 1. 75 10 273K 8.

5mK

8-21. Assuming the gasses are ideal gases, the pressure is given by:

P

2

3

N E

V

for classical,

FD, and BE particles. P FD 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, P BE 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.

8-22. (a)

1

fBE

For = 0 and f 1 , at 5800

E / kT

BE

T K

ee 1

e

1

E/

5800k

1 e 2

E/

5800k 1

E K eV K

5

5800 8. 62 10 / ln2

E 0.

347eV

(b) For E 0. 35 V, = 0 and f 0.

5

e

1

BE

0. 35 / kT

0.

5 e 3

0. 35 / kT

1

5

0. 35eV 8. 62 10 eV / K T ln3

T

0.

35eV

5

ln 3 8. 62 10 /

eV K

3700K

h h h h

8-23.

1/

2

p 2mE

2m

3kT

/ 2 3mkT

The distance between molecules in an ideal gas

1/

3

V / N is found from

194



1/ 3 1/

3

PV nRT nRT N N NkT V / N kT / P

A

A

and equating this to

above,

kT / P

1/

3

h

3mkT

1/

2

kT h P h

P 3mkT

k 3mk

3 3

5/

2

and solving for T, yields: T

3 / 2 3 / 2

T

Ph

3

25 /

3

34

101kPa

6.

63 10

J s

3/ 2 5/

2

27 23

k 3mk 3 2 1. 67 10 kg 1. 38 10 J / K

25 /

44 . K

8-24.

3/

2

N0 T

1 (Equation 8-52)

N T

C

(a)

(b)

(c)

(d)

3/

2

N0 3T

For T 3TC

/ 4 1 0.

351

N 4T

N

C

3/

2

0

For T TC

/ 2 1 0.

646

N 2TC

N

T

3/

2

0

For T TC

/ 4 1 0.

875

N 4TC

N

T

3/

2

0

For T TC

/ 8 1 0.

956

N 8TC

T

8-25. For small values of

1

, e / ! N N e

e 1

2

1 2 and

0 0

1 1

1

which for small values becomes: N0 1 1 N0 1 or N

0

8-26.

T

C

2

h N

2mk

2 2.

315

V

2/

3

(Equation 8-48)

The density of liquid Ne is 1.207 g/cm 3 , so

195



N

V

1. 207g / cm 6. 022 10 molecules / mol 10 cm / m

3 23 6 2 3

20. 18g / mol

3. 601 10 / m

28 3

T

2

34 2/

3

28 3

6. 626 10 Js

3.

601 10 m

27 23

2 20u 1. 66 10 kg / u . 381 10 J / K 2 2.

315

0.

895K

Thus,

TC

at which

20

temperature of 24.5K.

Ne would become a superfluid is much lower than its freezing

8-27. Power per unit area R arriving at Earth is given by the Stefan-Boltzmann law:

where

is Stefan’s constant. For a 5% decrease in the Sun’s temperature,

R

T

4

4

R T R 0. 95T T 0.

95T

R T

T

4

4

4

1 0. 95 0.

186 , or a decrease of 18.6%.

hf

8-28. E

(Equation 8-60)

hf / kT

e 1

hf kT / 10

(a) For T 10hf / k; hf kT / 10 E 0.

951kT

1/

10

e 1 0.

1051

hf kT

(b) For T hf / k; hf kT E 0.

582kT

1

e 1 1.

718

(c) For

hf 10kT

e 1 2.

20 10

4

T 0. 1hf / k; hf 10kT E 4.

54 10 kT

10 4

According to equipartition E kT in each case.

8-29.

2

hf / kT

hf e

CV

3 N

Ak As T , hf / kT gets small and

2

kT hf / kT

e 1

hf / kT

e 1 hf / kT

2

hf 1 hf / kT

CV 3 N

Ak 3N 3 / 3

2

Ak N

A

R N

A

R

kT hf / kT

The rule of Dulong and Petit.

196


8-30.

2

hf / kT

hf e

CV

3 R (Equation 8-62)

kT

2

hf / kT

e 1

Writing

hf kT Af A h kT T K

13

/ where / 2. 40 10 when 200 ,

Af

2 e

2 1

C 3 R Af eR Af

e 2e

1 e 2 1/

e

V 2

Af

Af

Af Af

Because Af is “large”, 1/ Af

Af

e 0 and e dominates Af , so

Af Af

C 3R / e e 3R / C f ln 3R / C 1/

A

V V V

For Al, C 200K 20. 1 J / K mol (From Figure 8-13)

V

2

f

3 8.

31

ln 1 / 2. 40 10 8.

97 10

20.

1

13 11

Hz

For Si, C 200K 13. 8 J / K mol (From Figure 8-13)

V

f

3 8.

31

ln 1 / 2. 40 10 2.

46 10

13.

8

13 12

Hz

8-31.

2

hf / kT

hf e

CV

3 R (Equation 8-62)

kT

2

hf / kT

e 1

At the Einstein temperature TE

hf / k ,

1

2 e

CV

3R 1 3R 0. 9207 3 8. 31J / K mol 0.

9207

2

1

e 1

22. 95K / K mol 5. 48cal / K mol

8-32. Rewriting Equation 8-69 as

3/

2

2 1/

2

n E 8mc E

V hc

2

2 E E F / kT

e 1

Set up the equation on a spreadsheet whose initial and final columns are E(eV) and

n(E)/V (eV•nm 3 ) -1 , respectively.

197



E eV

n E / V eV nm

3

1

4.5 14.4

4.6 14.6

4.7 14.5

4.8 (=E F ) 7.46

4.9 0.306

5 0.0065

5.1 0.00014

The graph of these values is below.

From the graph, about 0.37 electrons/nm 3 or

E F have been excited to levels above E F .

26 3

3. 7 10 electrons / m within 0.1eV below

198


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

2 2 8 62 10 1 1 72 10

5 4

kT . eV K K . 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.

T Au 136K T Al 243K

E

E

T Be 575 K T Diamond off the graph (well over 1000K)

E

E

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

2 2 2

En

n h 8mL and six levels will be occupied in 22 Ne , five levels with 10

protons and six levels with 12 neutrons.

E

F

2 2

5 1240MeV fm

protons 516MeV

2

8 1. 0078u 931. 5MeV / u 3.

15 fm

199



E

F

2 2

6 1240MeV fm

neutrons 742MeV

2

8 1. 0087u 931. 5MeV / u 3.

15 fm

E protons 3/ 5 EF

310MeV

E neutrons 3/ 5 EF

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.

E h / 8 mL . All 10 bosons can be in this level, so E total 10h / 8mL

.

2 2 2 2

1 1

8-37. (a)

f

E

1

FD E EF

/ kT

e

e

1

(Equation 8-68)

1 1

E EF / 0. 1EF 10 E EF / EF

1 e 1

1 1

(b) fFD E

E EF / 0. 5EF 2 E EF / EF

e

1 e 1

200


8-38.

N

N

O

3/

2

T

1 (Equation 8-52)

T

c

(a)

(b)

N

N

O

N

N

O

3/ 2 3/

2

Tc

/ 2 1

1 1 0.

646

T

2

c

3/ 2 3/

2

Tc

/ 4 1

1 1 0.

875

T

4

c

8-39. For a one-dimensional well approximation,

2 2 2

En

n h 8 mL . At the Fermi level E F ,

n=N/2, where N = number of electrons.

2 2 2 2

N / 2 h h N

EF

where N / L = number of electrons/unit length,

2

8mL

32m

L

i.e., the density of electrons. Assuming 1 free electron/Au atom,

1/

3

23 3 2

N N 6. 02 10 electrons / mol 19. 32g / cm 10 cm / m

A

L M 197g / mol

3

1/

3

3.

81 10 m

9 1

E

F

2 2

34 9 1

6. 63 10 J s 3.

81 10 m

31 19

32 9. 11 10 kg 1. 602 10 J / eV

1.

37eV

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 6.

2 eV deep.

8-40. (a)

/ 2

23

2kT

2 1. 3807 10 J / K 850K

At T 850K vm

183. 9m / s

25

m

6.

94 10 kg

1/

2

1/ 2 1/

2

8kT

4

v vm

207. 5m / s

m

1/ 2 1/

2

3kT

3

vrms

vm

225. 2m / s

m 2

201



The times for molecules with each of these speeds to travel across the 10cm diameter

of the rotating drum is:

t v

m

0.

10m

183. 9m/

s

0.

10m

207. 5m/

s

5.

44 10

4

t v 4.

82 10 s

t v

rms

0.

10m

225. 2m/

s

4

4.

44 10

s

4

s

3

The drum is rotating at 6250rev/min = 104.2rev/s or 9. 600 10 s / rev . The fraction

of a revolution made by the drum while molecules with each of these three speeds are

crossing the diameter is:

4

5.

44 10 s

for vm

: 0.

05667rev

3

9. 600 10 s / rev

4

4.

82 10 s

for v : 0.

05021rev

3

9. 600 10 s / rev

4

4.

44 10 s

for vrms

: 0.

04625rev

3

9. 600 10 s / rev

Assuming that point A is directly opposite the slit s 2 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 0.

10

v : 0. 05667rev 0. 314159m / rev 0. 01780m 1.

780cm

m

v : 0. 05021rev 0. 314159m / rev 0. 01577m 1.

577cm

v : 0. 04625rev 0. 314159m / rev 0. 01453m 1.

453cm

rms

(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 N 2 molecules at 850K is 710.5m/s, since they have a smaller

mass than Bi 2 molecules. Repeating for them the calculations in part (a), N 2

molecules moving at v m would strike the plate only 0.4cm from A. Molecules moving

at

v and v rms

would be even closer to A.

m.)

202


8-41.

1 1

2

EK(

escape)

m vescape

mv F v dv

2 2

0

1

2

2 3 mv / 2kT

mv v e dv

2 1 I

m

2 I

v e dv

0 5

0

2

3 mv / 2kT

3

3

1 2

m m 2 kT where

2

2 / 2 m

2

m kT m

kT

8-42. (a)

2

E / kT Au / kT

f u du Ce du Ce du

(from Equation 8-5)

2 2

Au / kT Au / kT

1 f u du Ce du 2C e du

1/

2

2

0

2 2 where

CI C / A/

kT

C kT / A C A/

kT

(b)

2

2 2 2 Au / kT

E Au Au f u du Au A/

kT e du

A A/ kT I A A/ kT / A/

kT

3/

2

2

2

2 4 where

1 3/

2 1

A A/ kT kT / A kT

2 2

8-43.

1 / 2 2

mvx

/ 2kT

/

f v m 2 kT e (from Equation 8-6)

x

vx vx f vx dv

x

0

1/ 2 2 1 2 2

/ 2 / / 2

/ 2 mv x kT x

/ 2

mv kT

x x x x

0

v m kT e dv v m kT e dv

0

1/ 2 2

mvx

/ 2kT

/

x

2 m 2 kT v e dv

1/

2

2 m / 2 kT I with m / 2kT

1

x

203



1/

2 2kT

2kT

2 m / 2 kT 1/

2

m m

8-44.

f

1 1

e e

/

1 e

F /

1

FD E kT E E kT

where

EF

kT

For

f

FD

E E e

E EF

/ kT

F, 1 and

1 1 1

=

E EF

/ kT EF

/ kT E / kT e

/

e e e e

E kT

f

B

8-45.

4 2m

3/

2

e

1/ 2 E / kT

N e E e dE

3

h

0

V

(Equation 8-43)

Considering the integral, we change the variable:

E / kT

2

u , then

2 1/

2

1/

2

E kTu E kT u dE kT u du

, , and 2 . So,

1/ 2 E / kT

3/

2 2

0 0

2

u

E e dE 2 kT u e du

The value of the integral (from tables) is

/ 4 , so

3/ 2 3/ 2 3/

2

4 2me

V 2 kT 2 2me

kT V

N e or e

3 3

h

4

Nh

, which is Equation 8-44.

8-46. (a) N ni

f0 E0 f1 E1 (with g0 g

1

1)

i

0 / kT

/ kT

Ce Ce C 1 e

N

So, C

1 e

/ kT

0 n0 n1

Ce N e e

(b) E

/ kT

N N 1 e N 1 e

/ kT / kT / kT

/ kT

204



/ /

As 0, kT

kT

T e 1/ e 0 , so E 0

/ kT

/ kT

As T , e 1/ e 0 , so E / 2

(c)

C

V

dE d N E d N e

dT dT dT 1 e

/ kT

/ kT

N

kT

2

2

1

e

e

2

/ kT

/ kT

2

/ kT

e

1 e

/ kT

Nk kT 1 e

2

e

/ kT

/ kT

2

(d)

1

T / k 0.1 0.25 0.5 1.0 2.0 3.0

CV

Nk 0.005 0.28 0.42 0.20 0.06 0.03

8-47. (a)

2 2

1/

2

2 2 2 n1 n2

n3

x y z

p k k k

L L L

2

1/

2

L

n n n

2 2 2

1 2 3

1/ 2 N

L

205



E

pc

c N

L

(b) Considering the space whose axes are n1, n2,

and n

3

. The points in space correspond

to all possible integer values of n1, n2,

and n

3

, 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

2

/ .

N dN Because photons can have two

polarizations (spin directions), the number of possible state is

2 1 8 4

2 2

/ N dN N 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

2

N dN

2

EL c L c dE

8 L 8

2

L

3 3

2 2

E dE

3 3

c

hc

E dE

The probability that a photon exists in a state is given by:

f

E

1 1

e e

/ 1 e

/ 1

BE E kT E kT

(Equation 8-24)

The number of photons with energy between E and E+dE is then:

n E dE f E g E dE

BE

8

L / hc

e

E / kT

3 2

E dE

(d) The number of photons per unit volume within this energy range is

Because each photon has energy E, the energy density for photons is:

1

3

n E dE / L .

u E dE

E n E dE / L

3

hc

8

3

3

E dE

e

E / kT

1

which is also the density of photons with wavelength between λ and λ+dλ, where

206



hc / E E hc /

. So,

2

d hc hc

d dE dE dE dE d

2 2

dE E hc

u d u E dE

3 2 3

8 hc / hc / d 8 hc d

3 hc / kT

hc / kT

hc e 1 e 1

207

Chapter 9 – Molecular Structure and Spectra

9-1. (a)

19 23

eV eV 1. 609 10 J 6.

022 10

molecules

1 1

molecule

molecule

eV mole

(b)

(c)

E

E

d

d

1

96472 J cal

23057 cal 23.

06

kcal

mole

4.

184J

mole mole

eV 23. 06kcal / mole

4. 27 98. 5kcal / mole

molecule

1eV / molecule

eV 1eV / molecule

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 Cl 2 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.

ke 2

E U E

d C ion

r0

2

ke 1.

440eV nm

CsI : Eion

3. 89eV 3. 06eV Ed

3.

44eV

r

0.

337nm

0

2

ke 1.

440eV nm

NaF : Eion

5. 14eV 3. 40eV Ed

5.

72eV

r

0.

193nm

0

0

2

ke 1.

440eV nm

LiI : Eion

5. 39eV 3. 06eV Ed

3.

72eV

r

0.

238nm

While E d for CsI is very close to the experimental value, the other two are both high.

Exclusion principle repulsion was ignored.

209


9-5. (a) Total potential energy:

2

ke

U r Eex

E

r

ion

(Equation 9-1)

attractive part of U r

(b) The net ionization energy is:

ion

0

2

ke 1.

44eV nm

5.

39eV

r

0.

267nm

0

ionization energy of electron affinity of

E Rb

Cl

4. 18eV 3. 62eV 0.

56eV

Neglecting the exclusion principle repulsion energy E ex ,

0

dissociation energy U r 5. 39eV 0. 56eV 4.

83eV

(c) Including exclusion principle repulsion,

dissociation energy 4 37 5 39 0 56

E 5. 39eV 4. 37eV 0. 56eV 0.

46eV

ex

. eV U r0

. eV . eV Eex

2

ke 1.

440eV nm

Uc

Eion

4. 34eV 3. 36eV 4.

13eV

r

0.

282nm

9-6.

0

The dissociation energy is 3.94eV.

E U E 3. 94eV 4.

13eV E

d c ex ex

E 0. 19 eV at r 0.

282nm

ex

0

9-7.

E

ex

A

(Equation 9-2) 0.

19eV

n

r

A

0.

282nm

n

At r 0

the net force on each ion is zero, so we have (from Example 9-2)

2

Uc

r0

ke nA n A n

eV nm

2 n1

n

r0 r0 r0 r0 r0 r0

18. 11 / 0.

19eV

18. 11eV / nm0.

282nm

n 26.

9 27

0.

19eV

27 16 27

. . .

n

A Eexr0 0 19eV 0 282nm 2 73

10 eV nm

210


9-8. E 3.

81 eV per molecule of NaBr (from Table 9-2)

d

19

1eV / molecule 1eV / molecule 1. 609 10

J / eV

23

6. 0210 molecules / mol / 1cal / 4. 186J 23. 0kcal / mol

E NaBr 3. 81eV / molecule 23. 0kcal / mol / 1eV / molecule 87. 6kcal / mol

d

1.

440eV nm

UC

4. 34eV 3. 36eV 4.

13eV

0.

282nm

9-9. For KBr :

E 3. 94eV U E 4.

13eV E

d C ex ex

Eex

0.

19eV

1.

440eV nm

UC

4. 18eV 3. 62eV 4.

60eV

0.

279nm

For RbCl :

E 4. 37eV U E 4.

60eV E

d C ex ex

Eex

0.

23eV

9-10. H2S, H2Te, H3P,

H3Sb

9-11. (a) KCl should exhibit ionic bonding.

(b) O 2 should exhibit covalent bonding.

(c) CH 4 should exhibit covalent bonding.

9-12. Dipole moment pionic

er0 (Equation 9-3)

1 609 10

. 19 C 0 . 0917 nm

1 47 10 10

20 9

. C nm /

1 47 10

29

. Cm

m nm

if the HF molecule were a pure ionic bond. The measured value is

30 29

the HF bond is C m C m

6 64 10

6. 4010 1. 47 10 0. 44 or 44 % ionic.

29

. Cm

, so

211


19 9

9-13. pionic

er0

C m

1.609 10 0.2345 10 (Equation 9-3)

29

3. 757 10 Cm, if purely ionic.

The measured value should be:

ionic

29

p 0. 70 0. 70 3. 757 10 2.

630 10

29

ionic

measured p C m C m

19 9

9-14. pionic

er0

C m

3 09

10

1.609 10 0.193 10 (Equation 9-3)

29

. Cm

The measured values is

29

2. 67 10 C m, so the BaO bond is

29 29

C m C m

2. 67 10 3. 0910 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 = p p

30

1 2

and p 6. 46 10 C m and p p1cos 52. 25 p2cos 52.

25

If bonding were ionic,

ionic 0

19

1.609 10 0.0956 10 9

p er C m 1.532 10

29 C m

30

p actual p / cos . . C m/ cos . .

30 C m

1

2 52 25 6 46 10 2 52 25 5 276 10

Ionic fraction = fraction of charge transferred =

5.

276 10

1.

532 10

30

29

Cm

0. 34 or 34%

Cm

9-17.

2 2 2

U k p1 / r (Equation 9-10)

(a) Kinetic energy of N2 0. 026 eV , so when U 0.

026eV

the bond will be broken.

0.

026eV

r

37 2 9 2 2 30

1. 110 m C / N 910 N m / C 6.

4610

C m

r

6

2 2

37 2 9 2 2 30

1. 110 m C / N 910 N m / C 6.

4610

C m

19

0. 026eV

1. 6010

J / eV

2 2

8.

9410

6 56 6

10

r 6. 710 m 0.

67nm

m

212



2

ke

1.

440eV nm

(b) U U 0.

026eV r 55nm

r

r

(c) H 2 O-Ne bonds in the atmosphere would be very unlikely. The individual molecules

will, on average, be about 4nm apart, but if a H 2 O-Ne molecule should form, its

U 0. 003 eV at r 0.

95nm

, a typical (large) separation. Thus, a N 2 molecule with

the average kinetic energy could easily dissociate the H 2 O-Ne bond.

9-18. (a)

3

E 0. 3eV hc / 1240eV nm / 1240eV nm / 0. 3eV 4.

13 10 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) O 2 is nonpolar. The covalent bond involves no separation of charges, hence no

polarization of the molecule.

4 2 1 2

9-20. For N2 E0r 2. 4810 eV / 2 I where I mr0

and m 14.

0067u

2

2. 4810 eV 2I

r

4 2

2

0 4

r

2. 4810 eV 14.

0067u

2

2

34

1.

05510

Js

2. 4810 eV 1. 6010 J / eV 14. 0067u 1. 6610

kg / u

0 4 19 27

10

1. 61 10 m 0.

161

nm

1/

2

213


9-21.

1

E I mr

2I

2

E

2

2

0r

(Equation 9-14) where

0

2

for a symmetric molecule.

2 2

c 197.

3eV nm

16 931. 510 / 0.

121

1.

7810

0r

2 2 2 2 6 2

2

mr0 mc r0

uc eV uc nm

4

eV

9-22. For Co:

f

13

6.

42 10 Hz (See Example 9-6)

E v 1/

2 hf (Equation 9-20)

V

(a) E1 E0 3hf / 2 hf / 2 hf

(b)

n

n

1

0

e

4 14 10

. 15 eV s 6 . 42 10 13 Hz

0. 27eV

E1E 0 / kT

5

0. 27 / 8.

6210

0.

01 e

(from Equation 8-2)

T

5

0 01 0 27eV

8 62

10

ln . . . eV / K T

0.

27eV

T

ln . . /

5

0 018 62 10

eV K

T

680K

9-23. For LiH: f

13

4.

22 10 Hz (from Table 9-7)

15 13

(a) EV

v hf E0

hf eV s Hz

E0 0.

087eV

mm

1 2

(b)

m m

1 2

1/ 2 / 2 4. 14 10 4. 22 10 / 2

(Equation 9-17)

7. 0160u1.

0078u

7. 0160u 1.

0078u

0.

8812u

(c)

1

f

2

K

(Equation 9-21)

214



2 2 13

2 2

4. 22 10 0. 88121. 66 10

27 /

K f Hz kg u

(d)

K 117 N / m

E n h 8mr r n h 8mE

n

2 2 2 2 2 2

0 0

n

1 /

8

2

r h mE

0 0

34

6.

6310

Js

r

8 0 8812 1 66 10 0 087 1 60 10

0 1/

2

27 19

. u . kg / u . eV . J / eV

r m nm

11

0

5. 1910 0.

052

mm 1.

0078u

1 2

9-24. (a) For H 2 : 0.

504u

m1

m2

21.

0078u

2

14.

0067u

(b) For N 2 : 7.

0034u

214.

0067u

12. 0111u 15.

9994u

(c) For CO:

6.

8607u

12. 0111u

15.

9994u

(d) For HCl:

1. 0078u35.

453u

1. 0078u

35.

453u

2

0.

980u

9-25. (a)

(b)

39. 1u 35.

45u

mm

m m 39. 1u 35.

45u

1 2

1 2

2

2

0r

(Equation 9-14)

0

E I r

2I

E

18.

6u

c c

2

2 2

2

0r

r

2 2 2 0

2

2r0 2c r0

2c EV

c

r

197.

3eV nm

2 2 6 2 5

2cE

V 210. 6uc 931. 510 eV / uc 1.

4310

eV

0 1 / 2 1 / 2

r0 0.

280nm

215


9-26.

f

1

2

K

(Equation 9-21)

(a) For H 35 Cl: μ = 0.980u and f

13

8.

97 10 Hz .

2 2 13 2

27

K 2 f 2 8. 97 10 Hz 0. 980u 1. 6610 kg / u 517N / m .

(b) For K 79 Br:

39. 102u78.

918u

mm

m m 39. 102u 78.

918u

1 2

1 2

26.

147u

and

2 2 12 2

27

12

f 6.

93

10 Hz

K 2 f 2 6. 9310 Hz 26. 147u 1. 6610 kg / u 82. 3N / m

9-27. 1.

f

1 K

2

(Equation 9-21) Solving for the force constant,

2

K (2 f)

2. The reduced mass of the NO molecule is

m m

m m

N O

N

O

(14.01u)(16.00u)

14.01u 16.00u

7.47 u

3.

K

13 2 27 3

(2

5.63 10 Hz) 7.47u 1.66 10 kg/u 1.55 10 N/m

( Note: This is equivalent to about 8.8 lbs/ft, the force constant of a moderately strong

spring.)

9-28.

E

I Treating the Br atom as fixed,

2

0r

2

. . / . 2

I m r u kg u nm

E

2 27

H 0

1 0078 1 66 10 0 141

2

34

1.

05510

Js

27 2 9

. u . kg / u . nm m / nm

2 1 0078 1 66 10 0 141 10

0r

2

1. 67 10 J 1.

04

10

0

22 3

eV

E 1 E r

for 0, 1, 2 , (Equation 9-13)

216



14

The four lowest states have energies:

E0 0

3

E1 2E0r

2.

08

10 eV

3

E2 6E0r

6.

27 10 eV

3

E3 12E0r

12.

5

10 eV

E

10

3

eV

12

10

8

6

4

3

2

2

0

1

0

9-29.

13

E hf where f 1. 05 10 Hz for Li.

Approximating the potential (near the bottom)

2 2

hf

2

2 mr0

2

with a square well, E 2 1 2 1

For Li 2 :

r

2

0

2

3

1 3

2 2 f 4 f

34

3

1.

05510

Js

r

4

1 0510 6 939 1 66 10

13 27

. Hz . u . kg / u

11

4. 53 10 m 0.

045

nm

1/

2

9-30.

2

2

0r

0

E where I r (Equation 9-14)

2I

For K 35 Cl:

For K 37 Cl:

r0 0. 267 nm for KCl.

39. 102u34.

969u

39. 102u

34.

969u

39. 102u34.

966u

39. 102u

34.

966u

18.

46u

19.

00u

217



E

E

K Cl

2

34

1.

05510

Js

27 9

. u . kg / u . m

2 18 46 1 66 10 0 267 10

35

0r

2

K Cl

2. 5510 J 1.

59

10

24 5

eV

2

34

1.

05510

Js

27 9

. u . kg / u . m

2 19 00 1 66 10 0 267 10

37

0r

2

E0 r

0.

04

10

2. 4810 J 1.

55

10

5

eV

24 5

eV

9-31. (a) NaF – ionic (b) KBr – ionic

(c) N 2 – covalent

(d) Ne – dipole-dipole

9-32.

2 2

1

E

01 ,

2

I r0

(a) For NaCl: r 0

0.

251nm

(from Table 9-7)

35

Cl

35

22. 989834.

9689

22. 9898 34.

9689

m Na m

13.

8707u

m Na m Cl

E

2

34

1.

05510

Js

27 9

13. 8707u 1. 6610 kg / u0.

25110

m

01 ,

2

01 ,

7. 67 10 4.

80

10

24 5

E J eV

(b)

7.

67 10

E hf f E / h

6.

6310

24

0, 1 0,

1 34

10

f 1.

16

10 Hz

J

Js

8 10

c / f 3. 0010 m/ s 1. 1610 Hz 0. 0259nm 2.

59cm

218


9-33. (a)

1240eV nm

2400nm E hc / 2400nm 0.

517eV

2400nm

E E 3. 80eV E E 0. 500eV E E 2. 9eV E E 0.

30eV

2 1 3 2 4 3 5 4

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 E 1 is occupied, states up to E 3 can be

reached, so the E2 E1 and E3 E2

transitions will occur, as well as E3 E1.

(d) E4 E3 2. 9 eV hc or 1240eV nm 2.

9eV 428nm

E4 E2 3. 4 eV hc or 1240eV nm 3.

4eV 365nm

E4 E1 7. 2 eV hc or 1240eV nm 7.

2eV 172nm

9-34.

A21

B u f

21

hf / kT

e

1 (Equation 9-42)

For the Hα line λ=656.1nm

At T = 300K,

hf hc 1240eV nm

73.

1

kT kT nm eV K K

656. 1 8. 6210 5 / 300

e

1 e 1 5.

5

10

hf / kT 73.

1 31

Spontaneous emission is more probable by a very large factor!

9-35.

n E

n E

1

0

E1

/ kT

e

i.e., the ratio of the Boltzmann factors.

E0

/ kT

e

For O 2 : f

13

4.

74 10 Hz and

15 13

E0 hf 2 4. 1410 eV s 4. 7410 Hz 2 0.

0981eV

E1 3hf 2 0.

294eV

At 273K, 5

kT 8. 6210 eV / K 273K 0.

0235eV

219



n E e e

n E e e

0. 0294 / 0. 0235 12.

5

1 4

2.

4

10

0. 0981/ 0. 0235 4. 17

0

Thus, about 2 of every 10,000 molecules are in the E 1 state.

Similarly, at 77K,

n E

n E

1 13

0

1.

4

10

E 1 E r

for 0, 1, 2 , (Equation 9-13)

9-36. 0

Where

2

2

0r

0

E and I r with m / 2

2I

2

34

1.

05510

Js

27 9

. u . kg / u . m

23 4

E0r

1. 8010 J 1.

1210

eV

2

(a) E0 0

2 18 99 1 66 10 0 14 10

E 2E 2. 2410 eV E E 2.

24

10 eV

4 4

1 0r

1 0

E 6E 6. 7210 eV E E 4.

48

10 eV

4 4

2 0r

2 1

E 12E 13. 410 eV E E 6.

72

10 eV

4 4

3 0r

3 2

E

10

4

eV

14

12

10

3

8

6

2

4

2

0

1

0

220



(b) 1 E hc hc E

For

For

For

1240eV nm

2.

2410

eV

6

E1 E0 : 5. 5410 nm 5.

54nm

4

1240eV nm

4.

4810

eV

6

E2 E1: 2. 77 10 nm 2.

77nm

4

1240eV nm

6.

7210

eV

6

E3 E2 : 1. 8510 nm 1.

85nm

4

9-37. (a)

10 10 10 1 5 10 1 5 10

7 7 9

MW J / s E J / s . s .

2 J

(b) For ruby laser: 694. 3nm,

so the energy/photon is:

E hc 1240eV nm 694. 3nm 1.

786eV

Number of photons =

2

1.

510

J

19

1. 786eV

1. 6010

J / eV

5.

2310

6

9-38.

3

4mW

4

10 /

J s

1240eV nm

E hc 1.

960eV per photon

632.

8nm

410

Number of photons =

19

1. 960eV

1. 6010

J / eV

3

J / s

1 28 10

16

. /

s

9-39. (a) D m m

sin 1. 22 / 1. 22 600 10 1010 7.

32

10

7 32

10

6

. radians

9 2 6

S / R where S diameter of the beam on the moon and R distance to moon.

8 6 3

S R

3. 8410 m 7. 3210 radians 2. 8110 m 2.

81km

(b)

9

sin 1. 22 600 10 m 1m 7.

32

10

7 radians

8

. .

7

S R

3 8410 m 7 3210 radians 281m

221


9-40. (a)

n E

n E

2

1

E2

/ kT

e

E2E1

/ kT

e E2 E1

hc 1240eV nm 420nm 2.

95eV

E1

/ kT

e

At 5

T 297K, kT 8. 6110 eV / K 297K 0.

0256eV

2. 95 / 0.

0256 21 115 29

.

n E2 n E1 e 2 510 e 210 0

(b) Energy emitted =

1. 810 21 2. 95 / 5.

3110 21 850

eV photon eV J

9-41. (a) Total potential energy:

2

ke

U r Eex

E

r

2

ke 1.

44eV nm

the electrostatic part of U r

at r0

is 6.

00eV

r 0.

24nm

(b) The net ionization energy is:

ion

E 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 27 6 00 1 52 4 67

. eV U r0

. eV . eV . eV Eex

E 6. 00eV 4. 27eV 1. 52eV 0.

21eV

ex

A

(c) Eex

(Equation 9-2)

n

r

At r 0

0. 24nm, E 0.

21eV

At

At

At

ex

2

ke

r0 0. 14nm, U r

0 and Eex

Eion

8.

77eV

r

A

r0 : Eex

0. 21eV A 0. 21eV 0.

24nm

n

0.

24nm

0

ion

A

r 0. 14nm : Eex

8. 77eV A 8. 77eV 0.

14nm

n

0.

14nm

Setting the two equations for A equal to each other:

2 2

0. 24nm

0. 24 8.

77eV

n

1. 71

41.

76

n

0.

14nm

0. 14

0.

21eV

n

n

222



nlog 1. 71 log 41.

76

n log 41. 76 log 1. 71 6.

96

n

6 . 96 5 6.

96

A 0. 21eV 0. 24nm 0. 21eV 0. 24nm 1.

02

10 eV nm

9-42. (a) D m m

9-43. (a) U

sin 1. 22 / 1. 22 694. 310 0. 01 8.

47 10

8 47 10

5

. radians

9 5

(b) E hc / 1240eV nm / 694. 3nm 1. 786eV / photon

For

att

photon

18

10 photons / s :

total

1. 786 / 1. 602 10 19 / 10

18 /

E eV photon J eV photons s

0. 286J / s 0.

286W

2

Area of spot A is: A d cm 2

/ 4 8. 47 / 4

E Etotal

/ A 0. 286W / 8. 47cm / 4

5. 0810

W / cm

and 2 3 2

2

ke 1 440eV nm

5 39eV

r

.

0 267nm

.

.

(b) To form K and Cl requires E 4. 34eV 3. 61eV 0.

73eV

2

ke

Ed U C

Eion

5. 39eV 0. 73eV 4.

66eV

r

(c) Eex

4. 66eV 4. 43eV 0.

23 eV at r0

ion

9-44.

2

2

0r

0 0

E where I r with r 0.

267 nm and

2I

39. 102u35.

453u

mm

m m 39. 102u 35.

453u

1 2

1 2

18.

594u

2

34

1.

05510

Js

27 9

. u . kg / u . m

24 5

E0r

2. 5310 J 1.

5810

eV

2

2 18 594 1 66 10 0 267 10

223


kp1

9-45. (a) E where ,

d

p

3 1

qa being the separation of the charges q and q of the dipole

x

(b) U p E and p E p = E

So the individual dipole moment of a nonpolar molecule in the field produced by p 1 is

2

p E kp x and U p E kp x

3 6

2 d 1 2 d

1

dU d

Fx

k p x 6 k p x

dx dx

2 2 6 2 2 7

1 1

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.

Similarly,

mm

(1.01u)(1.01u)

H H

H2

mH

m

H

1.01u 1.01u

HD

D2

0.67 u

1.01u

0.51u

2. The vibrational frequencies for the molecules are then:

f

1 K 1 580 N/m

2 2 (0.51u)(1.66 10 kg/u)

H2

27

H

2

14

1.32 10 Hz

Similarly,

f

f

HD

D2

14

1.1510 Hz

13

9.3610 Hz

3. The energies of the four lowest vibrational levels are then:

For H

2

:

1 1 34 14 20

0 H2

(6.63 10

hf

J s)(1.32 10 Hz) 4.38 10

J

E

E

2 2

20

4.3810 J

0

19

0.27eV

1.6010

J/eV

224



Similarly,

E 0.82eV

E

E

1

2

3

1.37eV

1.91eV

For HD:

E

1 1 34 14 20

0 HD (6.63 10

hf

J s)(1.15 10 Hz) 3.81 10

J

2 2

20

3.8110 J

0

19

E

0.24eV

1.6010 J/eV

And again similarly,

E 0.72eV

E

E

1

2

3

1.19eV

1.67eV

For D

2

:

1 1 34 13 20

0 D2

(6.63 10

J s)(9.36 10 Hz) 3.10 10

E hf J

2 2

20

3.1010 J

E0

0.19eV

19

1.6010 J/eV

And once again similarly,

E 0.58eV

1

2

0.97eV

E3

1.36eV

4. There are three transitions for each molecule:

E

3 2; 2 1; 1

0

For H

2

: E hf hc / hc / E 1240eV nm / E eV

Similarly,

E E E (1.91 1.37)eV 0.54eV

32

3 2

1240eV nm / 0.54eV

21

10

3

2.30 10 nm

3

2.2510 nm

3

2.2510 nm

For HD:

E E E (1.67 1.19)eV 0.48eV

32

3 2

1240eV nm / 0.48eV

3

2.58 10 nm

225



And again similarly,

21

10

3

2.6410 nm

3

2.5810 nm

For D

2

:

E E E (1.36 0.97)eV 0.39eV

3 2

3 2

1240eV nm / 0.39eV

And once again similarly,

21

10

3

3.1810 nm

3

3.1810 nm

3

3.18 10 nm

9-47. (a)

6

E hc eV nm . mm nm / mm .

3 eV

3

1240 0 86 10 1 44 10

6

E eV nm . mm nm / mm .

4 eV

2

1240 1 29 10 9 61 10

6

E eV nm . mm nm / mm .

4 eV

1

1240 2 59 10 4 79 10

These are vibrational states, because they are equally spaced. Note the v 0 state

at the ½ level spacing.

18

16

v 3

14

Ev

10

4

eV

12

10

8

v 2

v 1

6

4

2

v 0

0

226



(b) Approximating the potential with a square well (at the bottom),

2 2

4 2

1

4.

7910

2

2 mr0

E eV n

r

2

2 2 2 34

2 1

1.

05510

Js

. u . kg / u . eV . J / eV

2 28 01

1 66 10 4 79 10 1 60 10

0 27 4 19

10

2. 15 10 m 0.

215

nm

1/

2

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):

13

The vibrational frequency for NaCl is 1. 14 10 Hz (from Table 9-7) and two vibrational

levels, for example v = 0 and v = 10 yield (from Equation 9-20)

E 1/ 2hf 0. 0236eV E 11/ 2hf 0.

496eV

0 10

above the bottom of the well. Clearly, the average separation for v 10 > v 0 .

9-49. (a)

2

E where r 0.

128 nm for HCl and

0r

2

0

2r0

1. 0079u35.

453u

mm

m m 1. 0079u 35.

453u

1 2

1 2

0.

980u

2

34

1.

05510

Js

27 9

. u . kg / u . m

22 3

E0r

2. 089 10 J 1.

30310

eV

2

2 0 980 1 66 10 0 128 10

E

E0

1

r

E 0 E 2E 2. 606 10 eV E 6E 7.

82

10 eV

3 3

0 1 0r

2 0r

E E E 2. 606 10 eV E E E 5.

214 10 eV

3 3

01 1 0 12 2 1

2.

606 10

eV

4.

136 10

eV s

3

12

f01 E01 h 0.

630 10 Hz

15

227



5.

214 10

eV

4.

136 10

eV s

3

12

f12 E12 h 1.

26 10 Hz

15

01

01

6 884 10 0 6310 6 890 10 6 878

10

14 12 14 14

f f f . Hz . . Hz; . Hz

01

c f

01

435. 5nm; 436.

2nm

02

02

6 884 10 1 2610 6 897 10 6 871

10

14 12 14 14

f f f . Hz . . Hz; . Hz

02

c f

02

435. 0nm; 436.

6nm

(b) From Figure 9-29:

The agreement is very good!

f 0. 610 Hz and f 1.

2

10 Hz

12 12

01 12

9-50. (a) : /

Li2 Ev

v 12hf

15 13

E / . eV s . Hz . eV .

2 eV

E

1

3 2 4 14 10 1 05 10 0 0652 6 52 10

E0

1

r

5 4

E .

eV . eV

1

2 8 39 10 1 68 10

79

(b) : 12 /

K Br E v hf

v

15 12

E / . eV s . Hz .

2 eV

E

1

3 2 4 14 10 6 93 10 4 30 10

E0

1

r

6 5

E .

eV . eV

1

2 9 1 10 1 8 10

9-51. HCl

0.

980 u (See solution to Problem 9-49)

From Figure 9-29, the center of the gap is the characteristic oscillation frequency f :

13 1 K

f 8. 65 10 Hz E 0.

36 eV Thus, f or K 2 f

2

2 13 2

27

K 2 8. 6510 Hz 0. 980u 1. 6610 kg / u 480N / m

2

228


9-52.

n E g E e

n

n

En

/ kT

1240eV nm

694. 3nm E E hc 1.

7860eV

694.

3nm

21 2 1 21

E2 E1 1. 7860eV 0. 0036eV 1.

7896eV

Where E 2 is the lower energy level of the doublet and E

2

is the upper.

Let T = 300K, so kT = 0.0259eV.

n E g E E 2 1

2 E1

/ kT

. / .

(a) e e e

n E g E

4 2

n E

n E

2 2 1 7896 0 0259 69 31

1 1

2 1. 7896 / 0.

0259 31

1

1

e

2

5.

64 10

4.

91 10

(b) If only E2 E

1

transitions produce lasing, but E2 and E

2

are essentially equally

populated, in order for population inversion between levels E2 and E

1, at least 2/3

rather than 1/2) of the atoms must be pumped. The required power density (see

Example 9-8) is:

p

19 3 34 14

2N

hf 2 2 10 cm 6. 63 10 J s 4.

32 10 Hz

3

3 t

3 3 10 s

s

1273W / cm

3

9-53. (a) E v 1/

2 hf (Equation 9-20)

v

For v = 0,

(b) For

E hf J s Hz eV

34 13

0

/ 2 6. 63 10 8. 66 10 / 2 0.

179

2

1, E / I h f

so

6.

63 10 Js

I h / 4 f 2.

8 10 kg m

2 11

4 6 10 Hz

34

2 47 2

(c)

I r

2

,

where is given by Equation 9-17.

mH

mCl

m m

H

Cl

0. 973u r 0.

132nm

229


9-54. (a)

dU U

12 13 6 7

0

12 a r 2 6 a r

dr

12 6 6 6 6

For U , min

dU / dr 0 , so 12a r 12a 0 r a r a

12 6

a a

(b) For U Umin,

r a then Umin

U 2 1 2 U U

a a

0 0 0

(c) From Figure 9-8(b): r0 0. 074nm ( a) U0

32.

8eV

(d)

r/

r

0

r / r 12

6

0

2

0

/

r r U

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

230


9-55. (a)

2

ke

U r Eex

E

r

ion

(Equation 9-1)

For NaCl, Ed

4. 27 eV and r0

0.

236 nm (Table 9-1).

E E Na E Cl 5. 14 3. 62 1. 52 eV and U r E 4.

27eV

ion ion aff 0 d

E

ex

2

ke

4. 27 1. 52 0.

31eV

0.

236

n

(b) E Ar 0.

31 eV (Equation 9-2)

ex

Following Example 9-2,

2

ke n A n

25. 85eV / nm

0.

31eV

r r r r

2

n

0 0 0 0

Solving for n: n 25. 85eV / nm 0. 236nm / 0. 31eV

19.

7 20

20 14 20

A 0. 31eV 0. 236nm 8.

9 10 eV nm

9-56. For

2

ke

H H system, U r E

r

ion

There is no

Eex

term, the two electrons of H are in the n = 1 shell with opposite spins.

E ionization energy for H electron affinity for H = 13. 6eV 0. 75eV 12. 85eV

.

ion

U r

1. 440eV nm

dU r 1.

440

12.

85eV

2

r dr r

For U r to have a minimum and the ionic H H molecule to be bound, dU / dr 0.

As we see from the derivative, there is no non-zero or finite value of r for which this

occurs.

9-57. (a)

u

u

35

1. 007825u

34.

968851u

1. 007825u

34.

968851u

0.

979593u

37

1. 007825u

36.

965898u

1. 007825u

36.

965898u

0.

981077u

35

1.

52 10

3

(b) The energy of a transition from one rotational state to another is:

231



E

, 1

1 / I hf (Equation 9-15)

f

2 2

1 1 h 1

hI 4 h r 4 r

h

2 2 2 2

0 0

f

df

d

4

1 h 1

r

2 2 2

0

f 1 h

1 h

f 4 r 4 r

2 2 2 2

0 0

1

(c)

f

f

1.

53 10

3

from part (b). In Figure 9-29 the

f

between the

35

Cl

lines (the taller ones) and the 37 Cl lines is of the order of

13

0 01 10 so 0 0012

. Hz, f / f . , about 20% smaller than / .

mm 12. 0112u

15.

9994u

1 2

9-58. (a) For CO : r0

0. 113 6.

861u

m m 12. 0112u 15.

9994u

1 2

2

2 27 9 46 2

I r0 6. 861u 1. 66 10 kg / u 0. 113 10 1.

454 10 kg m

2

2

34

1.

055 10 Js

23 4

E0r

3. 827 10 J 2.

39 10 eV

46 2

2I

2 1.

454 10

(b) E E

0

E

0

0

1

r

4

E1 2E0r

4.

78 10 eV

3

E2 6E0r

1.

43 10 eV

3

E3 12E0r

2.

87 10 eV

3

E4 20E0r

4.

78 10 eV

3

E5 30E0r

7.

17 10 eV

kg m

232



(c) (See diagram)

3 3

E . . eV . eV

54

7 17 4 78 10 2 39 10

3 3

E . . eV . eV

43

4 78 2 87 10 1 91 10

3 3

E . . eV . eV

32

2 87 1 43 10 1 44 10

3 3

E . . eV . eV

21

1 43 0 48 10 0 95 10

E10 4.

78 10

4

eV

8

7

5

6

Ev

10

3

eV

5

4

4

3

3

2

1

0

2

1

0

(d) hc / E

1240eV nm

2.

39 10 eV

54 3

1240eV nm

1.

91 10 eV

43 3

1240eV nm

1.

44 10 eV

32 3

1240eV nm

0.

95 10 eV

21 3

1240eV nm

4.

78 10 eV

10 4

5

5. 19 10 nm 0.

519

5

6. 49 10 nm 0.

649

5

8. 61 10 nm 0.

861

5

13. 05 10 nm 1.

31

5

25. 9 10 nm 2.

59

All of these are in the microwave region of the electromagnetic spectrum.

mm

mm

mm

mm

mm

233


9-59.

# 2

n 1, v 1,

1

(a)

# 1 n 1, v 0,

0

1 1

E 1 hfH 1 E

2 0r hf

H

since 0

2

2 2

3

E 2 hfH

2 E

2 0r

since 1

2

1

E 3 hfH

6 E

2 0r

since 2

2

3 1

A 2 1 2 1 356 10

2 2

14

E E hfH E

2 0r hfH

h . Hz

2

# 3 n 1, v 0,

2

3 1

B 2 3 2 6 1 246 10

2 2

14

E E hfH E

2 0r hfH E

2 0r

h . Hz

Re-writing [A] and [B] with

E

/ I :

2

0r

2

A1 1 356 10

2 14

hfH

I h . Hz

2

B1 2 1 246 10

2 14

hfH

I h . Hz

2

Subtracting [B1] from [A1] and cancelling an h from each term gives:

3 4 0 110 10

2 14

h I . Hz

I

3h

4 0.

110 10

2 14

Hz

4.

58 10

kg m

48 2

(b)

2

I r

0

For

H

2

1.

007825

: 0.

503912u

2 1.

007825

2

1/

2 48 2 27

r0 I / 4. 58 10 kg m 0. 5039u 1. 66 10 kg / u

1/

2

r m nm in agreement with Table 9-7.

11

0

7. 40 10 0.

0740

Canceling an h from [B1] and substituting the value of I from (a) gives:

2 14

fH

2h 4 I 1.

246 10 Hz

2

14

fH

1.

32 10 Hz also in agreement with Table 9-7.

2

234

Chapter 10 – Solid State Physics

10-1. Ur

0

E

ke

r

0

2

1

1 (Equation 10-6)

n

2

ke 1

U r0

1

r

n

0

741 0 257

1

0

1

1 Er kJ / mol . nm

d

eV / ion pair

0.

7844

2

n ke 1. 7476 1. 44eV nm 96. 47kJ / mol

1

n 4.

64

1

0.

7844

10-2. The molar volume is

M

3

2Nr

A 0

M 74. 55g / mole

. / . /

8

r0 3. 1510 cm 0.

315nm

23 3

2N A

2 6 022 10 mole 1 984g cm

1/

3

10-3. The molar volume is

M

3

2Nr

A 0

M

42. 4g / mole

2. 07g / cm

3

3

2Nr A 0 2 6 022 10 mole 0 257 10 cm

23 7

. / .

3

10-4. (a)

U

att

ke

r

0

2

(Equation 10-1)

2

ke 1

0

1 (Equation 10-6)

r

n

(b) E U r

d

0

1

8. 01eV

1 7. 12eV / ion pair

9

96. 47kJ / mol 1cal

7. 12eV / ion pair 164kcal / mole

1eV / ion pair

4/

186J

235



(c)

165 5 0 314

1 Er . kcal / mol . nm

d 0

4. 186J 1eV / ion pair

1 0.

8960

2

n ke 1. 7476 1. 44eV nm 1cal

96. 47kJ / mol

1

Therefore, n 9.

62

1

0.

8960

10-5. Cohesive energy (LiBr)

=

1mol

1eV

/ . ev / ion pair

6. 0210 ion pairs

1.

6010

3

78810 J mol

8 182

23 19

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/cm 3 , then

mol of Na / cm 0. 3934 2. 16g / cm 2. 990g / mol 0. 03696 mol / cm

3 3 3

mol Cl cm mol cm

3 3

of / 0. 03696 / , also

since there is one ion of each element per molecule.

3

Number of ions 0 03696

Na / cm . NA

3

Number of ions 0 03696

Cl / cm . NA

Total number of

ions / cm 0. 07392 6. 022 10 4.

45

10

3 23 22

Nearest neighbor distance = equilibrium separation r 0

.

1

10

r0 2. 8810 m 0.

288nm

3

22 3 2

4. 4510 ions / cm 10

cm / m

1/

3

236


r0 KCl 0. 315nm 3.

15

10 m

10

10-7.

N ions / m 1/ r 3. 2010 / m 3. 20

10 / cm

Half of the ions in

element.

3 3 28 3 22 3

0

This number of ions equals:

1. 6010 ions 1.

60 10

N 6 022 10

A

22 22

3

1cm are K and half are Cl , so there are

23

. /

ions

mol

This is the moles of each ion in 1 cm 3 .

Molecular weight of K = 39.102 g/mol.

Molecular weight of Cl = 35.453 g/mol.

0.

02657 mol

Weight of K/cm 3 =

39. 102g / mol 0. 02657mol / cm 1. 039 g / cm

Weight of Cl/cm 3 =

3 3

35. 453g / mol 0. 02657mol / cm 0. 942 g / cm

3 3

. . / . /

density of KCl 1 039 0 942 g cm 1 98 g cm

3 3

1. 60 10 /cm

22 3

of each

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


10-9.

Uatt

2 2 2 2 2 2 2

ke

a 2a 3a 4a 5a 6a

2 2 1 2 1

Uatt

ke 21

3 3 5 3

The quantity in parentheses is the Madelung constant α. The 35 th term of the series (2/35)

is approximately 1% of the total, where α = 4.18.

me

v

10-10. (a) (Equation 10-13)

2

ne

31 5

9. 1110 kg 1. 17 10

m / s

2

8 47 10 28 3

. / 1. 60 10 19

electrons m C 0.

410

9 m

7

1.

2310

/

(b) v kT m 1 2 (from Equation 10-9)

/

e

m

v

100

1/

2

100K

1

300K

3

100 300 3 7 0010

8

. m

10-11. (a)

3

I I 4 10 A

j

2

A d / 4 3

1.

6310

m

2

3

479 A/

m (from Equation 10-10)

(b)

v

d

2

I d 479A/

m

3 5310

Ane ne m C

3 53

10

6

. /

28 3 19

8. 47 10 / 1.

602 10

cm s

8

. /

m s

10-12. (a) There are n

a

conduction electrons per unit volume, each occupying a sphere of

volume

3

4 r s

3:

3

na rs

/

4 3 1

1 /

3/

4

3

r r n

3 3

s s a

4

n

a

238



(b)

3

4

8. 47 10

/ m

1/

3

10

rs

1. 4110 m 0.

141nm

28 3

10-13. (a) n N / M for 1 electron/atom

A

10. 5g / cm

3 6. 022 10

23 / mole

n 5. 86

10 / cm

107. 9g / mole

19. 3g / cm

3 6. 022 10

23 / mole

(b) n 5. 90

10 / cm

196. 97g / mole

Both agree with the values given in Table 10-3.

22 3

22 3

10-14. (a) n 2 N / M for two free electrons/atom

(b)

A

3 23

. g / cm . / mole

2 1 74 6 022 10

n 8. 6210 / cm 8. 62

10 / m

24. 31g / mole

3 23

. g / cm . / mole

22 3 28 3

2 7 1 6 022 10

n 13. 110 / cm 13. 1

10 / m

65. 37g / mole

22 3 28 3

Both are in good agreement with the values in Table 10-3,

8. 61 10 /m

28 3

for Mg and

13. 2 10 /m

28 3

for Zn.

10-15. (a)

2

me

v 1 ne

(Equation 10-13)

(Equation 10-13)

2

ne

m v

31 5

9. 1110 kg 1. 0810

m / s

2

8 47 10 28 3

. 1. 602 10 19

m C 0.

37 10

9 m

7

1.

2210

1 1

1.

2210

6

8.

17 10

7

m

m 1

e

m

239



(b)

v

8kT

m

e

(Equation 10-9)

200K 300K v200K v300K

1 /

300K 200K / 300K

2

1 / 2

7 8

1. 2210 200 / 300 9.

9610

m K K m

1 1

200K

1 00 10

200K

9.

9610

m

7

.

8

m 1

(c) 100K 300K v100K v300K

1 / 2

7 8

1. 2210 100 / 300 7.

0410

m K K m

1 1

100K

1 42 10

100K

7.

0410

m

7

.

8

m 1

10-16.

E

3

E

5 F

(Equation 10-22)

E 3 7 . 06 eV 4 . 24 eV

5

(a) for Cu:

E 3 4 . 77 eV 2 . 86 eV

5

(b) for Li:

E

F

2 2/

3

hc 3 N

mc

V V

2

2 8

2 28 3 3

9

1240eV nm 3 5.

9010

m

10

m

5

25.

1110

eV 8

1nm

1/

3

5.

53eV

240


10-17. A long, thin wire can be considered one-dimensional.

E

F

hc 2

2

h N 2 N

2

(Equation 10-15)

2

32m L

32mc

L

For Mg: 1 /

28 2

3

N / L 8. 61

10 / m

E

F

9 28 3

1240eV nm 10 m / nm 8. 6110

/ m

6

320.

51110

eV

2 2 / 3

1.

87eV

10-18. (a) For Ag:

E

F

2

2 2 3

9

28 3

/

h 3N 1240eV nm 10

m / nm 35.

86 10

m

5.

50eV

6

2m

8V

2 0.

51110

eV 8

For Fe: Similarly, E 11.

2eV

(b) For Ag: E kT (Equation 10-23)

T

F

F

F

F

EF

5.

50eV

4

6.

3810

K

5

k 8. 617 10

eV / K

For Fe: Similarly,

TF

4

13.

0

10 K

Both results are in close agreement with the values given in Table 10-3.

2/

3

10-19. Note from Figure 10-11 that most of the excited electrons are within about 2kT above the

Fermi energy E , F

i.e., E 2kT

. Note, too, that kT E

F

, so the number N of excited

electrons is: 1/

22

N N E n E E N E kT N E kT and

F F F F

N

8V

2m

3 h

2

3/

2

E

3/

2

F


Differentiating Equation 10-19 gives: N E

V

8m

2 h

F

2

32 /

E

12 /

F

Then,

2

N

2 h

32 /

V

8m

12 /

EF

kT

32 /

N 8V 2m

32 /

E

2 F

3

h

3

kT

2

E

F

241



N

3

5

EF

for Ag 5. 35 eV, so 8. 617 10 eV / K 300K 5. 53eV

0. 0070 0. 1%

N 2

10-20.

u

F

1/ 2 1/

2

2E

F

2E

F

c 2

me

mec

(a) for Na:

F

(Equation 10-24)

1/

2

2 3.

26eV

8 6

u 3. 0010 m / s

1. 07 10 m / s

5

5.

1110

eV

(b) for Au:

F

1/

2

2 5.

55eV

8 6

u 3. 0010 m / s

1. 40 10 m / s

5

5.

1110

eV

(c) for Sn:

F

1/

2

2 10.

3eV

8 6

u 3. 0010 m / s

1. 90 10 m / s

5

5.

1110

eV

meuF meuF

10-21. (Equation 10-25)

2 2

ne

ne

(a) for Na:

(b) for Au:

31 6

9. 1110 kg 1. 07 10

m / s

2

2 65 10 28 3

. 1. 609 10 19

m C 4.

210

8 m

8

3. 42 10 m 34.

2

nm

31 6

9. 1110 kg 1. 4010

m / s

2

5 90 10 28 3

. 1. 609 10 19

m C 2.

0410

8 m

(c) for Sn:

8

4. 14 10 m 41.

4

nm

31 6

9. 1110 kg 1. 90 10

m / s

2

14 8 10 28 3

. 1. 609 10 19

m C 10.

610

8 m

8

4. 31 10 m 43.

1

nm

242


2

T

Cv

electrons R

2 T

10-22.

F

(Equation 10-30)

2

kT

Cv

electrons

R because EF kTF. C

v

due to the lattice vibrations is 3R,

2 E

assuming

E

F

2

2

kT

E

T T (rule of Dulong and Petit): R 0.

103R

E eV

0. 10 3 2

F

0. 60 7.

06

3

T 4.

9810

K

2 2 5

k 8. 617 10

eV / K

This temperature is much higher than the Einstein temperature for a metal such as copper.

F

10-23.

3

kT

U NEF

N

kT

5 EF

(Equation 10-29)

2

3 kT 3

Average energy/electron = U N EF

kT EF

5 EF

5 4

For copper: E 7. 06 eV,

so

kT

2

E

F

U N 3 7 . 06 eV 4 . 236 eV

5

At T = 0K:

At T = 300K:

5

. eV / K K

2 2

3

2

U N 8 61 10 300

7 . 06 eV 4 . 236 eV

5 4 7.

06

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 T

F

for Cu (81,600K). The classical value

U N 3/ 2 kT 0. 039 eV,

is far too small.

of

2

T

Cv

electrons R

2 T

10-24.

F

(Equation 10-30)

Melting temperature of Fe = 1811K (from Table 10-1)

T Fe K

4

F

for 13

10 (from Table 10-3)

The maximum Cv

for the Fe electrons, which is just before Fe melts, is:

243



2

1811K

4

Cv

electrons R 0.

0219

2

R

1310

K

The heat capacity of solids, including Fe, is 3R (rule of Dulong and Petit, see Section 8-1).

0.

0291

Cv

electrons R

0.

0073

C 3R

v

M B

10-25. P (Equation 10-35)

kT

24

9. 28510 J / T 2.

0T

23

1. 3810 J / K 200K

P 6.

7 10

3

10-26.

2

0M

0

B kT

units

N 1 J 1

2 3

A

m

T

J

2

NJ NJ NJ NJA m

A m T J

A m Wb m A m N Am

A m N

2 2 2

/ /

2 3 2 2 3 2 2 3

2 2 3 2

J Nm

1 dimensionless

Nm

Nm

10-27. E hc

3 6 3

(a) For Si: hc E 1240eV nm 1. 14eV 1. 08810 nm 1. 0910 m 1.

09

10 nm

(b) For Ge:

3 6 3

Similarly, 1. 72210 nm 1. 7210 m 1.

72

10 nm

(c) For diamond:

Similarly, 1. 77 10 nm 1.

72

10

2 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



(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 33 . eV.

3

10-29. (a)

E hc 1240eV nm 3. 35 m10 nm / m 0.

37eV

(b)

5

E kT 0. 37eV T 0. 37eV / k 0. 37eV / 8. 617 10 eV / K 4300K

10-30. (a)

N

3 3

2. 33g / cm 100nm 10 7 cm / nm 6. 0210

23 / mol

mN

A

VN

A

M M 28g / mol

7

5.

01 10 Si atoms

(b)

7

E 13eV 4 5. 01 10 6.

5 10

8 eV

10-31. (a)

E

2

2

1ke

m 1

1

2 2

2 1

(Equation 10-43)

9 2 2 19

34

1.

05510

Js

2

910 N m / C 1. 6010 C

1

0. 2 9.

1110

kg

E1

2 2

2 11.

8

21

3. 12 10 J 0.

0195

Ionization energy = 0.0195eV

2

(b)

1

0 e

eV

r a 1 m m (Equation 10-44)

0. 0529nm

1/ 0. 2 11. 8 3.

12nm

(c) E ( g

Si ) 1 . 11 eV at 293K

2

31

245



E1 / Eg

0. 0195 / 1. 11 0. 0176 or about 2%.

10-32. (a)

E

2

2

1ke

m 1

1

2 2

2 1

(Equation 10-43)

9 2 2 19

34

1.

05510

Js

2

910 N m / C 1. 6010 C

1

0. 34 9.

1110

kg

E1

2 2

2 15.

9

2

31

21

2. 92 10 J 0.

0182

2

(b)

1

0 e

eV

r a 1 m m (Equation 10-44)

0. 0529nm

1/ 0. 34 15. 9 2.

48nm

10-33. Electron configuration of Si:

2 2 6 2 2

1s 2s 2p 3s 3p

(a) Al has a

2

3s

3

p configuration outside the closed n = 2 shell (3 electrons), so a p-type

semiconductor will result.

(b) P has a

2 3

3s

3p configuration outside the closed n = 2 shell (5 electrons), so an n-type

semiconductor results.

10-34.

5

E kT 0. 01eV T 0. 01eV / 8. 617 10 eV / K 116K

dBw iB

10-35. (a) VH

vdBw

(Equation 10-45 and Example 10-9)

nq qnt

The density of charge carriers n is:

20A0.

25T

iB

n 7.

1010 carriers/m

qtV C m V

H

19 3 6

1. 6010 0. 210 2.

210

N

5. 7510 3 kg / m

3 6. 0210

26 / mol

A

(b) N 2.

92

10

M

118. 7kg / mol

28

28 3

28 28

Each Sn atom contributes n/ N 7. 1010 / 2. 9210 2.

4 charge carriers

246

eV /

10-36.

b kT

Inet

I e

0

1 (Equation 10-49)

(a) eV b / kT

e eV kT

10 , so / ln 10 . Therefore,

b

1 60 10 19 1 381 10

. C V . 23 / 300 ln 10

b

J K K

23 19


V 1. 3810 J / K 300K ln 10 1. 6010 C 0. 0596V 59.

6mV

b

(b) eV b / kT

e 01 .

V 0. 0596V ln 0. 1/ln 10 0. 0596 59.

6mV

b

eV /

10-37.

b kT

Inet

I e

(a)

0

1 (Equation 10-49)

eV b / kT

e 5 , so eV / kT ln 5 . Therefore,

b

23

5 1. 3810 J / K 200K

ln5

kT ln

Vb

0. 0278V 27.

8mV

19

e

1.

6010

C

(b) eV b / kT

e 05 .

eV / kT ln 0.

5

b

23

05 1. 3810 J / K 200K

ln 0.

5

kT ln .

Vb

0. 0120V 12.

0mV

19

e

1.

6010

C

eV /

10-38.

b kT

Inet

I e

0

1 (Equation 10-49)

Assuming T = 300K,

0. 2 0.

1

.

0 2

0 1

e . V / kT e . V / kT

0 0

1 1 e0. 2V / kT e0. 1V / kT

I V I V I e I e e e

47.

7

I V

I e e

e 0. 1V / kT e 0. 1V / kT

01

0

1 1

10-39. (a) From Equation 10-49, exp eV

/ kT 10

Taking ln of both sides and solving for V b ,

b

23

1. 3810 J / K 77K

ln10

Vb

kT / e ln 10 0. 0153 volts 15.

3mV

19

1.

6010

C

247



(b) Similarly, for exp eV / kT 1;

V 0

b

(c) For (a): eV /

b kT

Inet

I0 e 1 Inet

=1mA 10 1 9 mA

For (b): Inet

0

b

10-40. E hc For 484nm E E

/

gap

E hc / 484nm 1240eV nm / 484nm 2.

56eV

gap

10-41. M T c

constant (Equation 10-55)

First, we find the constant for Pb using the mass of natural Pb from Appendix A,

T c for Pb from Table 10-6, and α for Pb from Table 10-7.

0.

49

u K

constant 207. 19 7. 196 98.

20

206

For 0 . 49

Pb : T constant / M 98. 20 205. 974u 7.

217K

c

207

For 0 . 49

Pb : T constant / M 98. 20 206. 976u 7.

200K

c

208

For 0 . 49

Pb : T constant / M 98. 20 207. 977u 7.

183K

c

10-42. (a) E 3.

5 kT (Equation 10-56)

g

c

5 3

Tc

for I is 3. 408 K, so, Eg

3. 5 8. 617 10 eV / K 3. 408K 1.

028

10 eV

(b) E hc /

g

3 6

hc / Eg

1240eV nm / 1. 028 10 eV 1.

206 10 nm

3

1. 206 10 m 1.

206

mm

10-43. (a) E 3. 5 kT For Sn : T 3.

722K

g c c

5

E 3. 5 8. 617 10 eV / K 3. 722K 0.

0011eV

g

248



(b) E hc /

g

4 6 3

hc / Eg

1240eV nm / 6 10 eV 2. 07 10 nm 2.

07 10 m

T / T 0. 5 E T / E 0 0. 95 where E 0 3.

5 kT (Equation 10-56)

10-44. At

c g g g c

E T 0. 95 3. 5 kT 3.

325kT

So,

g c c

(a)

For 3 325 8 617 10 3 722 1 07 10

5

Sn : E . . / . .

3

g

T eV K K eV

(b)

For 3 325 8 617 10 9 25 2 65 10

5

Nb : E . . / . .

3

g

T eV K K eV

(c)

For 3 325 8 617 10 1 175 3 37 10

5

Al : E . . / . .

4

g

T eV K K eV

(d)

For 3 325 8 617 10 0 85 2 44 10

5

Zn : E . . / . .

4

g

T eV K K eV

10-45. B T B 0 1 T / T 2

C C C

(a) B T B 0 0. 1 1

T / T 2

C C C

T/ T 2 1 0. 1 0.

9

C

T/ T 0.

95

(b) Similarly, for

C

B T B 0 0.

5

C

T/ T 0.

71

(c) Similarly, for

c

C

B T B 0 0.

9

C

T/ T 0.

32

c

C

10-46. 1. Referring to Figure 10-56, notice that the length of the diagonal of a face of the cube is

2 2 2

r; therefore, a a (4 r) a 8 r

2. Volume of the unit cube V a 8 r

cube

3 3/2 3

249



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

3

4 r / 3

. The total volume

occupied by atoms is then

Vatoms

3

4 (4 r

/ 3) .

4. The fraction of the cube’s volume occupied by atoms is then:

V

V

atoms

cube

3

4 (4 r )

0.74

3/2 3

8 r

10-47. T F 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 E F is approximately:

f kT / E T / T

F

F

(a)

(b)

f 300K / 81, 700K

3.

7

10

3

f 1000K / 81, 700K

12.

2

10

3

10-48.

−e +e −e +e −e +e −e +e −e +e −e +e −e

● ● ● ● ● ● ● ● ● ● ● ● ●

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

r

0

(a) For the negative ion at the origin (position 0) the attractive potential energy is:

V

2

2ke

1 1 1 1 1

1

r

2 3 4 5 6

0

(b) V

2

ke

, so the Madelung constant is:

r

0

1 1 1 1 1

21

2 3 4 5 6

250



2 3 4

x x x

1 1 1

1 2 1

2 3 4 2 3 4

Noting that ln x x

, ln

and 2ln 2 1.

386

10-49.

C

v

2

2

T

R T

F

(Equation 10-30)

T

F

2

R

2 3 74 10

T

4

. /

J mol K T

and

2

kT

EF

kTF

R

4

2 3. 7410

J / mol K T

8. 314J / mol K 1. 3810

J / K

4

23. 7410

J / mol K

2 23

18 19

1. 5110 J 1/ 1. 6010 J / eV 9.

45eV

EF

EF

10-50. (a) 2 / 3 2 / 3

N g E dE AE dE A E A E F

1 / 2 3 / 2 3 / 2

0 0 0

EF

1/ 2 3/

2

(b) 2 3 3 /

N AE dE A/

E 2

F

EF

kT

EF

kT

3/ 2 3/

2

2 3 1

3 /

A/ E E kT / E

2

Because kT

F F F

E for most metals,

1 kT / E 1

3/

2

(c) For Cu

F

F

3/ 2 1/

2

kTEF

2 / 3

3/

2

N A

E E kTE

AkTE

EF

3/ 2 3/ 2 1/ 2 1/

2

F F F F

The fraction within kT of E F is then f

N

AkTE 3kT

N 2A/

3 E 2E

1/

2

F

3/

2

F

3 0.

02585eV

EF

7. 04 eV; at 300K, f 0.

0055

2 7.

04eV

F

251


10-51. (a)

6 6

N . eV / photon . eV / pair . pairs / photon

1

1 17 10 0 72 1 63 10

6 6

N . eV / photon . eV / pair . pairs / photon

2

1 33 10 0 72 1 85 10

(b)

N N 1. 2310 N / N 7.

8

10

3 4

1 1 1 1

N N 1. 3610 N / N 7.

4

10

3 4

2 2 2 2

E

N

(c) Energy resolution

E N

E1/ E1 0. 078%

E2/ E2 0. 074%

meff

v

10-52. (Equation 10-13)

2

ne

23

J K K

1/ 2

3 1. 38 10 / 300

5

v 3kT / meff

2. 6110

m / s

31

0. 29.

1110

kg

Substituting into λ:

31 5

0. 29. 1110 kg 2. 6110

m / s

22 3 3 19

10 m 510 m1.

6010

C

2

1/

2

For Cu:

8

3.

7 10 m 37

1/

2

1/ 2 2 7.

06eV

6

uF 2EF / me

1. 57 10 m / s

31

9.

1110

kg

n m m

28 3 8

8. 47 10 and 1.

7 10 (Example 10-5)

8

Substituting as above, 3. 910 m 39nm

The mean free paths are approximately equal.

nm

252


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

3 3/2 3

and the volume of the cube is V a 8 r . (See problem 10-46 and

cube

Appendix B3.) The density of fcc Cu would be:

mCu

4( M / NA)

d

3/2 3

V 8 r

cube

where M = molar mass (atomic weight) and

d

4(63.55g/mol)

3/2 8 3 23

8 (1.28 10 cm) (6.02210 / mol)

N

A

= Avagadro’s number.

8.90g/cm

3

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

2 2 2 2

a f b (4 r)

and

2 2 2

a a f . Combing these yields:

2 2 2 2 4r

a f 3 a (4 r)

a

3

The unit cube has 8 corners with ⅛ of a Cu atom at each corner plus 1 Cu atom at the

center, or 8×⅛ + 1 = 2 atoms. The density of bcc Cu is then:

3/2

m 2( M / NA) 2(63.55g/mol) 3

d 8.17g/cm

3 3 8 3 23

V (4 r / 3) 4 (1.28 10 cm) (6.02210 /mol)

cube

3

Based on the above density calculations, metallic Cu is most likely a face-centered

cubic crystal.

253


10-54. (a) For small V b (from Equation 10-49)

eV b / kT

e 1 eV / kT, so I I eV / kT V / R

b 0 b b

9

b

R Vb

I0eV / kT kT eI0 0. 025eV e10 A 25.

0M

9

(b) For V 0. 5V ; R V I 0.

5 10 A 500M

b

9 0 5 0 025

(c) For . / .

b

b

V 0. 5V ; I 10 A e 1 0.

485A

Thus, R V I 0. 5/ 0. 485 1.

03

b

(d)

dI eI

dV kT

dV kT dI eI

0 eVb

/ kT b

eVb

/ kT

20

e Rac

e 25Me

0.

0515

b

0

10-55.

a

h h

2 2 2

0

o

0 2 2 2

mee me e

me

ke

eff

eff

silicon:

a

2

34

121.

05510

Js

31 9 2 2 19

kg N m C C

0. 2 9. 11 10 9 10 / 1.

602 10

0 2

germanium: :

9

3. 17 10 m 3.

17

nm

This is about 14 times the lattice spacing in silicon (0.235nm)

a

2

34

161.

05510

Js

31 9 2 2 19

kg N m C C

0. 10 9. 11 10 9 10 / 1.

602 10

0 2

9

8. 46 10 m 8.

46

nm

This is nearly 35 times the lattice spacing in germanium (0.243nm)

10-56. (a)

E

n

2

2

1ke

m 1

2

n

2 2

(Equation 10-44)

2

6 2

1 1.

440eV nm

0. 015 0. 51110

eV / c 1

2

6.

582 10 eV s

18 n

or

16 2 2

254



2

2

9 2 2 19 31

N m / C . C . . kg

1 2

9 10 1 60 10 0 015 9 11 10 1

2 6.

63 10 J s 18 n

22

1.

0110

J 6. 28

10 eV / n

2

n

donor ionization energy =

34 2 2

4 2

6 28 10

4

. eV

m e

(b) r1 a0 (Equation 10-45)

m

0. 0529nm

1/ 0.

015 18

r1 63.

5nm

(c) Donor atom ground states will begin to overlap when atoms are 2r1

127nm

apart.

9

10 nm / m

donor atom concentration

4.

8810

127nm / atom

mu

e F

10-57. (a) (Equation 10-25)

2

ne

3

m

20 3

So the equation in the problem can be written as m i. Because the impurity

increases

m

8 8

by 1. 110 m, i

1.

110 mand

mu

e F

i

2 8

ne 1.

1810

m

and 1 2 6

i

/

28 3

where n 8. 47 10 electrons / m (from Table 10-3)

u 2E / m 1. 57 10 m/ s.

Therefore,

F F e

31 6

9. 1110 kg 1. 57 10

m / s

2

8 47 10 28 3

. / 1. 60 10 19

m C 1.

110

8 m

8

6. 00 10 m 60.

0

nm

(b)

2

1 / nar and d 2 r (Equation 10-12)

So we have

d 4 / n

where n 1 % of n 8. 47 10 / m

2 26 3

i i i

2 28 3 8 22 2

d 4/ 8. 47 10 / m 6. 6010 m 2. 2810 m d 0.

0151nm

255


10-58. (a) The modified Schrödinger equation is:

2

ke 1

2 2

d dR r

2

r R

2 2

r

ER r

2m r dr dr r

2m r

The solution of this equation, as indicated following Equation 7-24, leads to solutions

r / a

of the form:

0 n

R 1 2 2 2

n

r a 0e r L r / a 0

, where a

n

0

/ ke m

(b) By substitution into Equation 7-25, the allowed energies are:

2

2

1 ke m E1

1 ke

En

where E

2 2

1

m

2 n n

2

2

(c) For As electrons in Si:

m

. (see Problem 10-31) and Si 11 8

02m

e

.,

9 2 2 19

34

1.

05510

Js

2

2

N m / C . C

31

. . kg

910 1 6010 1

0 29 1110

E1

2 2

2 11.

8

21

3. 12 10 J 0.

0195

eV

Energy 10 2 eV

0

1.0

3

4

n

5

2

2.0

3.0

4.0

1

10-59.

U

2

n

ke r0 1 r0

r0

r n

r

(Equation 10-5)

dU

F Kr yields K

dr

2

n

1 ke

(a) For NaCl: 1. 7476 , n 9. 35 , and r0

0.

282 nm and

r

3

0

256



22. 99 35.

45

22. 99 35.

45

m Na m Cl u u

13.

95u

m Na m Cl u u

2

1 K 1

n 1

ke

f 3

2

M 2

13.

95u r0

1/

2

2

1

1.

2810

2

13 95 1 66 10 0 282 10

9 2 2 19

1. 74768. 35910 N m / C 1.

6010

C

3

27 9

. u . kg / u . m

(b)

8 13

c / f . 00 10 m/ s / 1. 28 10 Hz 23.

4m

This is of the same order of magnitude as the wavelength of the infrared absorption

bands in NaCl.

1/

2

13

Hz

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 e ne m ne m

2

E

/ E / , which satisfies Ohm’s law because .

2 2

ne / m and 1/ m/ ne

.

j E Thus,

10-61. (a) For r, s, and t all even

rst

1 1 6010 1 60 10

19 19

. C . C.

Similarly, for any permutation of

r, s even; t odd:

rst

1 1 and the ion’s charge at that location is:

1 1, ion charge =

rst

r even; s, t odd: 1

1, ion charge =

r, s, and t all odd:

rst

1 60

10

19

. C.

1 60

10

19

. C.

19

1 1, ion charge = 1. 60

10 C.

257



(b)

2

ke

U

r

If the interatomic distance r = a, then a cube 2a on each side

U

2 4 4 2 4 4 4 4

ke

a

2a a 2a 2a 3a 3a

2

ke

U 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

B / kT B / kT

M e e

tanh

/ /

B / kT

B kT B kT

e e

and M tanh B / kT

(b) For 0 and

B kT, T tanh B / kT B / kT

2

0M

0B

0

B BkT kT

258

Chapter 11 – Nuclear Physics

11-1.

Isotope Protons Neutrons

18 F 9 9

25 Na 11 14

51 V 23 28

84 Kr 36 48

120 Te 52 68

148 Dy 66 82

175 W 74 101

222 Rn 86 136

11-2. The momentum of an electron confined within the nucleus is:

34 14

p / x 1. 055 10

J s / 10 m

1 055 10 1 1 602 10

20 13

. J s / m / . /

6 59

10

8

. /

MeV s m

The momentum must be at least as large as

electron’s kinetic energy is:

J MeV

8 8

p, so p 6 59 10 MeV s m and the

8

min

. /

Emin pmin c 6. 5910 MeV s / m 3. 0010 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

6

the lowest available energy states is 1 2 so 0

/ , Li S cannot have such a structure.

259


11-4. A proton-electron model of 14 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 28 . , 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

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

Isotones

(a)

18 F

17 F

19 F

16 N

17 O

(b)

208 Pb

206 Pb

210 Pb

207 Tl

209 Bi

(c)

120 Sn

119 Sn

118 Sn

121 Sb

122 Te

11-7.

Nuclide Isobars Isotopes

(a)

14

8O

6

14

6C

8

14

7N

7

16

8O

8

(b)

63

28

Ni

35

63

29Cu

34

63

30

Zn

33

60

28

Ni

32

(c)

236

93

Np

143

236

92U

144

236

94

Pu

142

235

93

Np

142

260


11-8. mass Au A1. 66 10 27 kg / u

3 1 /

volume 4 / 3 R

4 / 3 R 3

3

0A

where

15

R . fm . m

0

1 2 1 2 10

27

A1. 6610 kg / u

3

15

/

. m

mass

density = 2. 29 10

kg / m

volume 4 3 1 2 10 A

17 3

11-9.

B ZM c Nm c M c

(a)

2 2 2

H N A

(Equation 11-11)

. . .

2 2 2

uc uc MeV uc

Be B 4 1 007825uc 5 1 008665uc 9 012182uc

9 2 2 2

4 5

0. 062443 0. 062443 931. 5 /

58.

2MeV

B / A 58. 2MeV / 9 nucleons 6. 46MeV / nucleon

(b)

(c)

. . .

2 2 2

. uc . uc . MeV / uc

C B 6 1 007825uc 7 1 008665uc 13 003355uc

13 2 2 2

6 7

0 104250 0 104250 931 5

91.

1MeV

B / A 91. 1MeV / 13 nucleons 7. 47 MeV / nucleon

. . .

2 2 2

uc uc MeV uc

Fe B 26 1 007825uc 31 1 008665uc 56 935396uc

57 2 2 2

26 31

0. 536669 0. 536669 931. 5 /

499.

9MeV

B / A 499. 9MeV / 57 nucleons 8. 77 MeV / nucleon

11-10.

R R A where R 1.

2 fm (Equation 11-3)

13 /

0 0

16

(a) 1 / 3

O R 1. 2 fm 16 3.

02 fm

56

(b) 1 / 3

Fe R 1. 2 fm 56 4.

58 fm

197

(c) 1 / 3

Au R 1. 2 fm 197 6.

97 fm

238

(d) 1 / 3

U R 1. 2 fm 238 7.

42 fm

261


11-11. (a)

n

B M He c m c M He c

3 2 2 4 2

3. 016029 uc 1. 008665 uc 4.

002602 uc

2 2 2

0. 022092uc 2 931. 5MeV / uc 2 20.

6MeV

(b)

n

B M Li c m c M Li c

6 2 2 7 2

6. 015121 uc 1. 008665 uc 7.

016003 uc

2 2 2

0. 007783uc 2 931. 5MeV / uc 2 7.

25MeV

(c)

n

B M N c m c M N c

13 2 2 14 2

13. 005738 uc 1. 008665 uc 14.

003074 uc

2 2 2

0. 011329uc 2 931. 5MeV / uc 2 10.

6MeV

B a A a A a Z A a A 2Z A a A c

11-12. 2

2 / 3 2 1/ 3 1 1/

2 2

1 2 3 4 5

(This is Equation 11-13

on the Web page www.whfreeman.com/tiplermodernphysics6e.) The values of the a

i

in

MeV/c 2 are given in Table 11-3 (also on the Web page).

For 23 Na:

B 15. 67 23 17. 23 23

0. 75 11 23 93.

2 23 211 23 0 23

184. 9MeV

23 2 2 2

p

n

c

2 / 3 2 1/ 3 2 1 1/

2 2

M Na c 11m c 12 m c B (Equation 11-14 on the Web page)

2 2

11 1. 007825uc 12 1. 008665uc

184.

9MeV

23 23 . 190055 184 . 9 / 2 1 / 931 . 5 /

2

M Na u MeV c MeV c u

23. 190055 u 0. 198499 u 22.

991156 u

This result differs from the measured value of 22.989767u by only 0.008%.

R 1. 07 0. 02 A fm (Equation 11-5) R 1.

4 A fm (Equation 11-7)

1/ 3 1/

3

11-13.

(a) 16 1/ 3 1/

3

O : R 1. 07A 2. 70 fm and R 1. 4A 3.

53 fm

(b) 63 1/ 3 1/

3

Cu : R 1. 07A 4. 26 fm and R 1. 4A 5.

57 fm

262



(c) 208 1/ 3 1/

3

Pb : R 1. 07A 6. 34 fm and R 1. 4A 8.

30 fm

2

3 1 e

54

R

11-14. 2

U Z Z

0

2

1 (Equation 11-2)

where Z = 20 for Ca and U

5.

49MeV

from a table of isotopes (e.g., Table of Isotopes

8 th ed., Firestone, et al, Wiley 1998).

3 1 e

1

54

RU

2

2

R Z Z

0

2

9 2 2 19 2 2 6

N m C C eV

0. 6 8. 9910 / 1. 60 10 2. 0 19 / 5.

49 10

15

6. 13 10 m 6.

13

fm

11-15. (a)

ln 2t/

t1/

2

t

R R0e R0 e (Equation 11-19)

at t 0 : R R0

4000counts / s

at

0

ln 210

s / t1 2

/

t 10 s : R R e

ln 210s / t1 2

/

1000 4000

e

ln 210s / t1 2

/

1/

4

e

1/ 2 1/

2

ln 1 / 4 ln 2 10s / t t 5.

0s

(b) at

s

ln 2 20 / 5s

t 20s : R 4000counts / s e 200counts / s

11-16.

ln 2 t / 2min

0 0

R R e at t 0 : R R 2000counts / s

(a)

min

ln 2 4 / 2 min

at t 4min : R 2000counts / s e 500counts / s

(b)

min

ln 2 6 / 2 min


(c)

min

ln 2 8 / 2 min


263


11-17.

ln 2t/

t1/

2

t

R R0e R0 e (Equation 11-19)

(a) at t 0 : R R0

115. 0 decays / s

at t 2. 25h : R 85. 2decays / s

85. 2decays / s 115. 0decays / s e

85. 2 / 115.

0

2.

25h

e

85 2 115 0 2 25

ln . / . . h

2.

25h

1

85 2 115 0 2 25h

0 133h

ln . / . / . .

t h h

1

1/ 2 ln 2/ ln 2/ 0. 133 5.

21

dN

dN0

(b) N R N

dt

dt

0 0


1

N0 R0/ 15. 0atoms / s / 0. 133h 1h / 3600s

3 11

10

6

. atoms

11-18. (a) 226 Ra t1 2

1620y

/

dN ln2 ln2

Nm

A

R N N

dt t t M

1Ci

3 7 10

1/ 2 1/

2

23

ln 26. 022 10 / mole1g

7

1620 y3. 1610 s / y26. 025g / mole

3.

6110

10 1

. s , or nearly the same.

(b)

Q M 226 Ra c 2 M 222 Rn c 2 M 4 He c

2

2 2 2

226. 025402uc

222. 017571uc 4.

002602uc

0. 005229uc 0. 005229uc 931. 5MeV / uc

4. 87MeV

2 2 2

s

10 1

264


11-19. (a)

dN ln 2t/

t1/

2

R R0 e (from Equation 11-19)

dt

when t 0, R R0

8000counts / s

when

ln / t1 2

10 2 /

t 10min, R 1000 counts / s 8000 counts / s e

ln / t1/

2

10 2

e 1000 / 8000 1/

8

10ln( 2)/ t ln 1/

8

t

1/

2

1/

2

10ln

2

3.

33min

ln 18 /

Notice that this time interval equals three half-lives.

ln 2 / t ln 2 / 3.

33min = 0.208 min

-1

(b)

1/

2

(c)

t ln 2 / t1/

2 t

0.

208 1

R R0e R0 e Thus, R 8000 counts / s e 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


11-21. 62 Cu is produced at a constant rate R , so the number of 62 0 Cu atoms present is:

0

/ 1

N R e t (from Equation 11-26). Assuming there were no 62 Cu atoms initially

present. The maximum value N can have is R0 / N0,

t

ln 2 / t1/

2

N N 1e

0.

90N N 1e

e

0

0 0

2 t1/

2

0 10

t

tln / ln .

t

t

ln 2 / t1/

2

1 0. 90 0.

10

10ln 0. 10 / ln 2 33.

2min

t1/ 2 ln 2 / ln 2 / 9. 810 y 7.

07 10 y (Equation 11-22)

10 1 8

11-22. (a)

(b) Number of 235 U atoms present is:

10 6 g6. 02 10

23 atoms / mol

10 . gN

A

N 2.

56

10

M

235g / mol

dN

N y s y atoms

dt

15

atoms

10 1 7 15

9. 8 10 1/ 3. 16 10 / 2.

56 10 (Equation 11-17)

0. 079 decays / s

(c) N N0 e t

(Equation 11-18)

9 8 10 10 1 6

. y 10 y

N 2. 5610 e

2.

558 10

15 15

t1/ 2 ln 2 / ln 2 / 0. 266y 2.

61 y (Equation 11-22)

1

11-23. (a)

(b) Number of N atoms in 1 g is:

23

1g

6. 0210

atoms / mol

10 . gN

A

N 2.

74

10

M

22g / mol

22

atoms

dN

N y s y atoms

dt

0 266

. 1 1 / 3 . 16 10 7 / 2 . 74 10 22

2. 310 decays / s 2.

3

10

14 14

Bq

266



dN

t

(c) Ne

0

(Equation 11-19)

dt

0 266 1 7 22 0 266 3 5

1 3 16 10 2 74 10

. y . y

. y / . s / y . e

9. 110 decays / s 9.

1

10

13 13

(d) N N0 e t

(Equation 11-18)

0 266 1

. y 3 . 5 y

Bq

22

22

N 2. 7410 e 1.

08

10

11-24. (a) 22 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.

logt AE B

1/

2

1/

2

(Equation 11-30)

10

for

1/

210 ,

5.

4

t s E MeV

for t 1s, E 7.

0MeV

1/

2

from Figure 11-16

10 10

5 4

1 /

A

2

log . B

(i) 10 0. 4303 A

B

1 7 0 1 /

A

2

log . B

(ii) 0 0. 3780 A B B 0.

3780 A

Substituting (ii) into (i),

10 0. 4303 A 0. 3780 A 0. 0523 A, A 191, B 0. 3780 A 72.

2

267


11-26.

144

237 Np

142

233 P

a

140

229 A

c

229 T

138

225 R

h

136

a

225 Ac

N

134

132

217 A

c

221 F

r

130

213 B

i

217 R

n

128

209 Tl

213 P

o

126

209 P

b

11-27.

Th Ra

232 228

90 88

80 82 84 86 88

Q M Th c M Ra c M He c

232 2 228 2 4 2

Z

92

232. 038051 uc 228. 031064 uc 4.

002602 uc

2 2 2

0. 004385uc 2 931. 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 2 m E where E E 4.

085MeV

Ra Ra Ra Ra

2m E 2M E 2M 4.

085 E

Ra Ra Ra

M

228. 031064 4.

085MeV

Ra

E

4. 085MeV 0. 9834. 085MeV 4.

01MeV

M m

228. 031064 4.

002602

Ra

268


11-28. 7 Be 7 Li v

4 3 3 4 e

(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.

EC . .

67 67

Ga Zn v e

Q M Ga c M Zn c

67 2 67 2

66. 9282uc

66.

972129 uc

2 2

0. 001075uc 2 931. 50MeV / uc 2 1.

00MeV

11-30. 72 Zn 72 Ga ve

Q M Zn c M Ga c

72 2 72 2

71. 926858 uc 71.

926367 uc

2 2

0. 000491uc 2 931. 50MeV / uc 2 0.

457MeV 457keV

11-31.

Np Np n and Np U p

233 232 233 232

For n emission:

Q M Np c M Np c m c

233 2 232 2 2

n

233. 040805 uc 232. 040022 uc 1.

008665 uc

0. 007882uc

2

2 2 2

233

Q 0 means M products M Np ; prohibited by conservation of energy.

For p emission:

Q M Np c M U c m c

233 2 232 2 2

n

233. 040805 uc 232. 037131 uc 1.

008665 uc

0. 004991uc

2

2 2 2

269



233

Q 0 means M products M Np ; prohibited by conservation of energy.

11-32.

286 280 247 235 174 124 80 61 30 0

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 −

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 M He c

8 2 4 2

8. 005304uc

2 4.

002602 uc

2 2

0. 000100uc 2 931. 50MeV / uc 2 0.

093MeV 93keV

11-34.

EC . .

80 80

80 80

80 80

Br Kr ve and Br Se ve and Br Se ve

For decay:

Q M Br c M Kr c

80 2 80 2

79. 918528 uc 79.

916377 uc

2 2

0. 002151uc 2 931. 50MeV / uc 2 2.

00MeV

270



For decay:

Q M Br c M Se c m c

80 2 80 2 2 2

e

2 2

79. 918528uc 79. 916519uc 2 0.

511MeV

0. 002009uc 2 931. 50MeV / uc 2 1. 022MeV 0.

85MeV

For E.C.:

Q M Br c M Se c

80 2 80 2

79. 918528 uc 79.

916519 uc

2 2

0. 002009uc 2 931. 50MeV / uc 2 1.

87MeV

11-35.

R R A where R 1.

2 fm (Equation 11-3)

13 /

0 0

For 13 /

C : R 1. 2 12 2. 745 fm 2. 745 10 m and the diameter = 5.

490 10 m

12 15 15

Coulomb force:

F

Gravitational force:

C

9 19

9. 0010 1.

610

C

2

15

5.

490 10

m

2

ke

7.

65N

2

r

2

2

11 27

m 6. 67 10 1.

67 10

kg

p

36

FG

G 6.

1810

N

2

2

15

r

5.

490 10

15

The corresponding Coulomb potential is: UC

FC

r 7. 65N 5.

490 10 m

The corresponding gravitational potential is:

G

G

2

m

4 2010 1 60

10

14 13

. J . /

0. 26MeV

6 . 18 10 36

5 490 10

.

15

U F r N m

J MeV

50

3 39 10 1 1 60 10 13

. J / . 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


11-36. The range R of a force mediated by an exchange particle of mass m is:

R / mc (Equation 11-50)

2

mc c / R 197. 3MeV fm/ 5 fm 39.

5MeV

m 39. 5MeV / c

2

11-37. The range R of a force mediated by an exchange particle of mass m is:

R / mc (Equation 11-50)

2

mc c / R 197. 3MeV fm / 0.

25 fm 789MeV

m 789 MeV / c

2

11-38.

Nuclide Last proton(s) Last neutron(s) j

Si d 6

29

14 15

37

17 20

1

5 / 2

2

1 / 2

Cl d 4

1

3 / 2

s 0 1/2

1

3 / 2

d 2 3/2

Ga p 3

2

71

31 40

59

27 32

2

3 / 2

2

1 / 2

Co f 7

4

73

32 41

1

7 / 2

p 1 3/2

2

3 / 2

Ge p 4

33

16 17

2

3 / 2

p 3 7/2

1

9 / 2

S s 2

81

38 49

2

1 / 2

g 4 9/2

1

3 / 2

Sr p 6

9

2

3 / 2

d 2 3/2

1

9 / 2

g 4 9/2

The nucleon configurations are taken directly from Figure 11-35, and the

are those of the unpaired nucleon.

and j values

272


11-39.

Isotope Odd nucleon Predicted μ μ

N

29

14

Si neutron −1.91

27

17Cl proton +2.29

71

31Ga proton +2.29

59

27Co proton +2.29

73

32Ge neutron −1.91

33

16

S neutron −1.91

87

38Sr neutron −1.91

11-40.

14

Nuclear spin of must be 1 because there are 3 states, +1, 0, and 1.

N m I

11-41.

36 53 82 88 94 131 145

S, Mn, Ge, Sr, Ru, In,

Eu

11-42.

3 H

3 He

14 N

14 C

273



n

15 N

p

n

15 O

p

n

16 O

p

11-43.

He, Ca, Ni, Sn,

Pb

3 40 60 124 204

2 20 28 50 82

11-44. (a)

1p 1/2

1p 3/2

1s 1/2

n

13 N

p

(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



(d)

1p 1/2

1p 3/2

1s 1/2

n

p

First excited state. There are several diagrams possible.

11-45.

30

14Si j 0

37

17Cl j 3/

2

55

27Co j 7/

2

90

40Zr j 0

49

In j 9/

2

107

11-46 (a)

Q M H c M H c M H c M H c

2 2 2 2 3 2 1 2

2 2. 014102uc 3. 016049uc 1.

007825uc

2 2 2

0. 004330uc 2 931. 5MeV / uc 2 4.

03MeV

(b)

Q M He c M H c M He c M H c

3 2 2 2 4 2 1 2

3. 016029 uc 2. 014012 uc 4. 002602 uc 1.

007825 uc

2 2 2 2

0. 019704uc 2 931. 5MeV / uc 2 18.

35MeV

275



(c)

n

Q M Li c m c M H c M He c

6 2 2 3 2 4 2

6. 01512uc 1. 008665 uc 3. 016049 uc 4.

002602 uc

2 2 2 2

0. 005135uc 2 931. 5MeV / uc 2 4.

78MeV

11-47. (a)

Q M H c M H c M He c m c

3 2 1 2 3 2 2

N

3. 016049 uc 1. 007825 uc 3. 016029 uc 1.

0078665 uc

2 2 2 2

0. 000820uc 2 931. 5MeV / uc 2 0.

764MeV

(b) The threshold for this endothermic reaction is:

E

th

m

M

M

Q (Equation 11-61)

2 2

3. 016049uc

1.

007825uc

0. 764MeV

3.

05MeV

2

1.

007825uc

(c)

2 2

1. 007825uc

3.

016049uc

Eth

0. 764MeV 1.

02MeV

2

3.

016049uc

11-48.

14 2 16

N H O

Possible products:

16 14 2

O N H

16 16

O O

16 15

O O n

11-49. (a)

,

C p N

12 15

16 15

O N p

16 12

O C

Q M C c M He c M N c m c

12 2 4 2 15 2 2

p

12. 000000 uc 4. 002602 uc 15. 000108 uc 1.

007825 uc

2 2 2 2

0. 005331uc 2 931. 5MeV / uc 2 4.

97MeV

(b) Od,

p

16 17

O

Q M O c M H c M O c m c

16 2 2 2 17 2 2

p

15. 994915 uc 2. 014102 uc 16. 999132 uc 1.

007825 uc

2 2 2 2

0. 002060uc 2 931. 5MeV / uc 2 1.

92MeV

276


11-50. The number of 75 As atoms in sample N is:

N

VN

M

A

1 2 30 10

cm cm m 4 cm / m 5 . 73 g / cm 3 6 . 02 10 23 atoms / mol

20 75

2. 76 10 As atoms

The reaction rate R per second per 75 As atoms is:

R

I

(Equation 11-62

74. 9216g / mol

4 5 10

. 24 cm 2 75 As 0 . 95 10 13 neutrons / cm 2 s

4. 28

10 s

11 1

Reaction rate = NR

20 11 10

. 7610 atoms 4. 2810 / s atom 1. 18

10 / s

23 23

22 23

11-51. (a) Ne p,

n

Na Ned,

n

Na

,

F n Na

20 23

11 14

14 14

(b) B , p

C

N n,

p

C C d,

p

13 14

29 31

32 31

(c) Si , d P

P p,

d P Si n,

d

32 31

C

P

11-52. (a) 14 C (b) n (c) 58 Ni (d) α

(e) 14 N (f) 160 Er (g) 3 H (i) p

Q

11-53.

2 m

p m

n m

d

c

1. 007276 u 1. 008665 u 2.

013553 u

0. 002388 u (See Table 11-1.)

. . /

2 2

Q 0 002388u 931 5MeV uc c 224MeV

277


11-54.

dW dN

P E

dt dt

6

dN P 500 10 J / s 1eV

6 19

dt E 200 10 eV / fission

1.

6010

J

1 5610

19

. /

fissions s

11-55. The fission reaction rate is:

N

R N R 0 k (see Example 11-22 in More section)

/

0

N

k R N R

0

N logk log R N / R

R0

log R N /

N

logk

(a) For the reaction rate to double

log10

R N 10R 0 : N 24.

2

log 11 .

(b) For

log100

R N 100R 0 : N 48.

3

log 11 .

(c) For

(d) Total time

(e) Total time

log2

R N 2R 0 : N 7.

27

log 11 .

t N 1 ms N ms : (a) 7.27 ms (b) 24.2 ms (c) 48.3ms

t N 100ms 100 N ms : (a) 0.727 ms (b) 2.42 ms (c) 4.83ms

11-56. 101 134

40 61 52 82

U

235

92 143

Zr Te n

101 133

41Nb60

51Sb82

2n

n

Tc In 3n

101 132

43 58 49 83

Rh Ag 4n

102 130

45 57 47 83

278


11-57.

J 1MeV 1fusion

500MW

500 1 78 10

13

s

1. 6010 J

17.

6MeV

14

. /

fusions s

Each fusion requires one 2 H atom (and one 3 H atom; see Equation 11-67) so 2 H must be

14

provided at the rate of 1. 78

10 atoms / s.

11-58. The reactions per 238 U atom is:

R

I

(Equation 11-62)

2

24 2 11 2

1m

18

0. 0210 cm atom5. 010 n / m

1. 00 10 / atom

4 2

10

cm

The number N of 238 U atoms is:

23

5. 0g6. 0210

atoms / mol

N

238. 051g / mol

Total 239 U atoms produced = RN

22 238

1.

26 10 U atoms

18 22 4 239

1. 00 10 / atom 1. 26 10 atoms 1.

26 10 U atoms

11-59.

1

Q M H c M H c M H c

2

1 2 1 2 2 2

1. 007825 uc 1. 007825 uc 2.

014102 uc

2 2 2

0. 001548uc 2 931. 50MeV / uc 2 1.

4420MeV

Q M He c M H c M He c

2 2 1 2 3 2

2. 014102 uc 1. 007825 uc 3.

016029 uc

2 2 2

0. 005898uc 2 931. 50MeV / uc 2 5.

4940MeV

Q M He c M He c M He c m H c

3 2 3 2 4 2 1 2

3

2

. uc 2 . uc 2 . uc

2

2 3 016029 4 002602 2 1 007825

0. 013806uc 2 931. 50MeV / uc 2 12.

8603MeV

Q Q1 Q2 Q3 1. 4420MeV 5. 4940MeV 12. 8603MeV 19.

80MeV

279


11-60. Total power = 1000MWe

/ 0.

30 3333MW

4

(a) 1 day 8. 64

10 s

9

3. 3310 J / s 1MeV / 1. 6010 13 J 2. 08

10

22 MeV / s

Energy/day 2. 0810 22 MeV / s 8. 64 10 4 s / day 1. 80 10

27 MeV / day

The fission of 1kg of

235 26

U provides 4.

95

10 MeV (from Example 11-19)

235 27 26

1 1 8010 4 9510 3 64

kg U / day . MeV . MeV / kg . kg / day

1 3 64 365 1 33 10

235 235 3

kg U / year . kg U / day days / year . kg / year

(b)

(c) Burning coal produces

7

3. 1510 J / kg 1MeV / 1. 6010 13 J 1. 97 10

20 MeV / kg coal

26 235

4. 9510

MeV / kg U

Ratio of kg coal needed per kg of 235 U is:

2.

5110

20

1. 97 10

MeV / kg coal

For 1 day:

3. 64kg 235 U 2. 5110 6 9.

1 10

6 kg This is about 10,000 tons/day,

the approximate capacity of 100 railroad coal hopper cars.

6 9

9. 1210 kg / day 365days / year 3. 33

10 kg / year

For 1 year:

6

H2O

1000 kg / m , so

11-61.

3

(a) 1000kg

:

.

2

/ / .

6 23 2

10 g 6 02 10 molecules H O mol 2H molecule 0 00015 H

1 00

10

25 2

. H atoms

Each fusion releases 5.49MeV.

Energy release = MeV

18. 02g / mol

1. 0010 5. 49 5.

49

10

25 25

MeV

25 13 12

5. 49 10 MeV 1. 60 10 J / MeV 8.

78 10

J

(b) Energy used/person (in 1999) =

20 9

3. 5810 J 5.

9

10 people

6 07 10

10

. J /

person y

280



10 1y

Energy used per person per hour = 6. 07 10

J / person y

8760h

6 93

10

6

. J /

person h

At that rate the deuterium fusion in 1m 3 of water would last the “typical” person

12

8.

8010

J

6

6. 9310

J / person h

6

1.

27 10 h 145

y

11-62. (a) n

Q M U c m c M Cd c M Ru c 5m c

235 2 2 120 2 110 2 2

n

235. 043924 uc 1. 008665 uc 119. 909851 uc 109.

913859 uc

5 1. 008665uc

2 2 2 2

1. 186 931. 5 / 1.

10

10

2

uc 2 MeV uc 2 3 MeV

(b) This reaction is not likely to occur. Both product nuclei are neutron-rich and highly

unstable.

11-63. The original number

N of 0

14

C nuclei in the sample is:

N0 15g 6 78 10 nuclei g 1 017 10

. 10 / . 12 where the number of 14 C nuclei per gram

of C was computed in Example 11-27. The number N of 14 C present after 10,000y is:

N N e N e

ln 2t/

t

t

1/

2

0 0

(Equation 11-18)

ln 210000 / 5730

e

1. 017 10 3.

034 10

12 11

1/

2

R N ln2 t N (Equation 11-19)

ln 2 5730 y1y 3. 16 10 7 s3.

034 10

11

1. 16decays / s

11-64. If from a living organism, the decay rate would be:

15. 6decays / g min 175g 20. 230 decays / min (from Example 11-27)

8. 1decays / s 60s / min 486decays / min

The actual decay rate is:

281



2

1 486decays / min

2

20, 230decays / min

(from Example 11-27)

2 n 20, 230 / 486

2 20 230 486

nln ln , /

n ln 20, 230 / 486 ln( 2) 5.

379 half lives

5. 379 half lives 5730 y / half life 30,

800 y

Age of bone =

11-65.

t

1 /

t 2 ln 1 N /

ln

(Equation 11-92)

D

NP

2

t Rb y N N

87 10

1/ 2

4. 8810 and

P/ D

36.

5

10

4.

8810

y

9

t ln 1 1/ 36. 5

1.

90 10

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

intensity, where

250nA eC / proton 1. 56

10 protons / s

I = proton intensity = 1 12

x

650b 650 10 cm

24 2

m mass 0. 35mg / cm 0. 00001 3. 5

10 mg / cm

2 6 2

n number of atoms of interest mN / A

t exposure time

detector efficiency 0.

0035

overall efficiency 0.

60 detector efficiency

mN

A

A

N I

x

t

A

x

, times the particle beam

N

250 10

1 60 10

9

19

. C /

C/

s

650 10

proton

cm

24 2

282



0 35 10 3 2 0 00001 6 02 10 23

. g / cm . . mol

1

80g / mol

15 min 60s/min 0. 60 0.

0035

4

5.

06 10 counts in 15 minutes

11-67.

t

1 /

t 2 ln 1 N /

ln

(Equation 11-92)

D

NP

2

232 10

t1/ 2

Th 1.

40

10 y

23

4. 11g

6. 0210

atoms / mol

232 22

NP

Th 1.

066 10 atoms

232. 04g / mol

23

0. 88g

6. 0210

atoms / mol

208 21

ND

Pb 2.

547 10 atoms

208g / mol

N N

D

P

21 22

2. 547 10 1. 066 10 0.

2389

10

1.

4010

y

9

t ln 1 0. 2389

4.

33

10 y

ln

2

11-68.

E

2 B z p

f

h h

8 4

T

2 2. 79

N

3. 1510 eV / T 1

N

0.

510

(a) For Earth’s field: f

15

4.

1410

eV s

3

2. 13 10 Hz 2.

12

kHz

(b) For B = 0.25T:

3 4 7

f 2. 1210 Hz 0. 25T 0. 510 T 1. 0610 Hz 10.

6MHz

(c) For B= 0.5T:

3 4 7

f 2. 1210 Hz 0. 5T 0. 510 T 2. 1210 Hz 21.

2MHz

283


11-69. (a) N C

6

12 10 C / s10 min60s / min

1.

5010

12 3 16

19

3 1.

6010

C

(b)

C

C ratio 1500 1.

5010 10

14 12 16 13

mass

12

C

15

1. 5010 atoms / min75 min121.

6610

27 kg

0.

015

7 4

1. 4910 kg 1. 4910 g 0.

149mg

(c) The

C C ratio in living C is 1. 35

10 .

14 12 12

14 12 13

n

sample C C 10 0.

10 1

14 12 12

living C C 1. 35 10 1.

35

2

where n = # of half-lives elapsed. Rewriting as (see Example 11-28)

1.

35

2 n 13.

5

0.

10

n

nln 2 ln 13. 5 ln 13. 5 /ln 2 3.

75

3. 75t 3. 75 5730 y 2.

15

10 y

age of sample =

4

1/

2

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

1 2. 05decays / g min

2

15. 6decays / g min

(from Example 12-13)

2 n 15. 6/ 2.

05

2 15 6 2 05

nln ln . / .

14

15 6 2 05 2 2 928 half lives of

nln . / . ln .

nt1/ 2 2. 928 5730 y 16,

800 y

Age of spear thrower =

C

284


11-71. Writing Equation 11-14 as:

2 2 2 2 3 1 3 2 2 1 1 2 2

, / / 1 2 3 4

2

/

p n

5

M Z A c Zm A Z m c a A a A a A z a A Z A a A c

and differentiating,

M m

p m

n

2 a A Z 2 a A 2 Z 2 A

Z

1/

3 1

3 4

0 m m 2a A Z 4a 8a A Z

p

n

1/

3 1

3 4 4

0 m m 4a 2a A 8a A Z

p

n

1/

3 1

4 3 4

m m 4a

Z where a 0 75 and a 93 2

2a A 8

p n 4

1 3 1

3

.

/ 4

.

3

a4A

(a) For A = 27:

Z

2

1. 008665 1. 007825 931. 5MeV / uc 4 93.

2

1/

3 1

2 0. 75 27 8 93.

2 27

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)

(b)

(c)

3/

2

3/

2

0. 31 0. 31 5 3.

47

R E MeV cm

R g / cm R cm 3. 47cm 1. 29 10 g / cm 4. 47 10 g / cm

2 3 3 3 2

2 3 2 3 3

R cm R g / cm 4. 47 10 g / cm 2. 70g / cm 1.

66 10 cm

11-73. For one proton, consider the nucleus as a sphere of charge e and charge density

3

c

3e 4 R . The work done in assembling the sphere, i.e., bringing charged shell dq

up to r, is:

U

c

3

4 r

2 1

c c c

4

dU k r dr and integrating from 0 to R yields:

3

r

k 16 R 3 ke

15 5 R

2 2 5 2

c

For two protons, the coulomb repulsive energy is twice U c , or

2

6ke

5 R .

285


11-74. The number N of 144 Nd atoms is:

N

53 94 6 02 10

23

. g . /

144g / mol

atoms mol

2.

25 10

23

atoms

dN

dt

N dN dt N

2. 36s 2. 25 10 1.

05 10 s

1 23 23 1

23 1 23 15

t s s y

1/ 2

ln 2 ln 2 1. 05 10 6. 61 10 2.

09 10

t ln 2 / t

1/

2

11-75. R R0 e (from Equation 11-19)

(a) At t = 0: R = R 0 = 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. 0 decays / min e

ln 2 4. 21d

/ t1 2

/

73. 5/ 115.

0 e

ln 73. 5/ 115. 0 ln 2 4. 21 d / t

/

1 2

t1/ 2

ln 2 2. 41d /ln 73. 5/ 115. 0 6.

52d

(b)

R 10decays / min = 115. 0decays / min e

ln 2 t/ 6.

52d

(c)

t ln 10 / 115. 0 6. 52d ln 2 23.

0d

R 2. 5decays / min = 115. 0decays / min e

ln 2 t/ 6.

52d

t ln 2. 5/ 115. 0 6. 52d ln 2 36.

0 d (because t = 0)

This time is 13 days (= 2t 1/2 ) after the time in (b).

11-76. For 227 Th : t1 2

18.

72 d (nucleus A)

/

For 223 Ra : t1 2

11.

43 d (nucleus B)

/

At t = 0 there are 10 6 Th atoms and 0 Ra atoms

A

(a) N N t

0

e (Equation 11-18)

A

N

A

0 A A A t B t B t

NB

e e N

Be

B A

0

(Equation 11-26 on the Web page)

286



6 ln 2 15d

/ 18.

72d

5

N 10 e

5.

74 10

A

6

10 ln 2 18.

72d

NB

e e

ln 2 1/ 11. 43d

1/ 18.

72d

N

A 0 A A A B

(b) N N means N e e e

A B 0 A

Cancelling N 0A and rearranging,

t t t

B A

ln 2 15 / 18. 72 ln 2 15 / 11.

43 5

0 2.

68 10

B

A

A

A B

1 e

t

A

ln 2 ln 2

0. 0370d

0.

0606d

18. 82d

11.

43d

1 1

B

ln

B A

A

1

B

A

t

t

ln

B A

A

1

A

B

0. 0606 0.

0370

ln 1 0 . 0370 0 . 0606

0.

0370

43. 0d

11-77. (a)

16 9 6

/ 6. 582 10 eV s 0. 13 10 s 5.

06 10 eV

(b)

E

r

hf

2 2

0.

12939MeV

2Mc 2M I c

2 191 2

(Equation 11-47)

4. 71 10 MeV 4.

71 10

8 2

(c) (See Section 1-5)

The relativistic Doppler shift Δf for either receding or approaching is:

f v

f c

0

eV

h f hf E

6 8

v

5. 06 10 eV 3. 00 10 /

c

6

E c E 0. 12939MeV 10 eV / MeV

0

m s

v 0. 0117m / s 1. 17cm / s

287


11-78.

B ZM H c Nm c M c

1 2 2 2

n A

For 3 2 2 2

He : B 2 1. 007825uc 1. 008665uc 3.

016029uc

2 2

0. 008826 931. 5 / 7.

72

uc MeV uc MeV

For 3 2 2 2

H : B 1. 007825uc 2 1. 008665uc 3.

016029uc

2 2

0. 009106 931. 5 / 8.

48

uc MeV uc MeV

1/ 3 1/

3 15

R R A . fm . fm . m

0

1 2 3 1 730 1 730 10

2 5

Uc

ke / R ke / R eV 8. 32 10 eV 0.

832MeV or about 1/10 of the binding

energy.

11-79. For 47 Ca :

46 2 2 47 2

B M Ca c mnc M Ca c

45. 953687 uc 1. 008665 uc 46.

954541 uc

2 2 2

2 2

0. 007811 931. 5 / 7.

28

For 48 Ca :

uc MeV uc MeV

47 2 2 48 2

B M Ca c mnc M Ca c

45. 954541 uc 1. 008665 uc 47.

952534 uc

2 2 2

2 2

0. 010672 931. 5 / 9.

94

uc MeV uc MeV

Assuming the even-odd 47 Ca to be the “no correction” nuclide, the average magnitude of

the correction needed to go to either of the even-even nuclides 46 Ca or

approximately B – average binding energy of the odd neutron,

48 Ca is

9. 94MeV 7. 28MeV / 2 8. 61MeV . So the correction for 46 Ca is 8.16 – 7.28 = 0.88

MeV and for 48 Ca is 9.94 – 8.16 = 1.78 MeV, an “average” of about 1.33 MeV. The

estimate for a 5 is then:

a A 1. 33MeV a 1. 33/ 48 9.

2 . This value is about

1 / 2 1 / 2

5 5

30% below the accepted empirical value of a 5 = 12.

288


11-80. For a nucleus with I > 0 the α feels a centripetal force

2

Fc

mv / r dV / dr where r distance of the from the nuclear center. The

corresponding potential energy V

ln

r and becomes larger (i.e., more negative) as r

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 R A where R 1.

2 f (Equation 11-3)

13 /

0 0

141 15 1/

3

15

R Ba 1. 2 fm 10 m / fm 141 6.

24 10 m

92 15 1/

3

15

R Kr 1. 2 fm 10 m / fm 92 5.

42 10 m

(b)

V kq q / r

9 2 2 19 19

8. 998 10 N m / C 56 1. 60 10 C 36 1.

60 10 C

1 2 15 15 19

6. 24 10 m 5. 42 10 m 1. 60 10 J / eV

8

2. 49 10 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 v L is at x.

M 0 mx mx

x dx m x / dt

m

xCM

vL

and V

M m M m dt dt M m

mv

V

L

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 1/

2 1 2mvL

1 2 4mM

M 2V M mvL

2 2 M m 2 M m

2

289



Before collision:

E

i

1

mv

2

2

L

After collision:

1 1 4mM

4mM

E mv mv E 1

2 2 M m M m

2 2

L L 2 i

2

11-83. At the end of the two hour irradiation the number of

32 56

P and Mn atoms are given by

R

0

t

N e R0

I

1 from Equation 11-26 where (Equation 11-62)

For 32 P :

24 2 12 2 13 32 31

R . cm neutrons / cm s . P atoms / s per P

0

0 180 10 10 1 80 10

1. 80 10 / s 3600s / h 342.

2h

13

Rt

0 1/

2 ln 2 t / t1/

2

ln 2 2h / 342.

2h

N0

1 e 1 e

For 56 Mn :

ln 2 ln 2

1. 29 10 P atoms /

9 32 31

P atom

24 2 12 2 11 56 55

R . cm neutrons / cm s . Mn atoms / s per Mn

0

13 3 10 10 1 33 10

11

1. 33 10 / s 3600s / h 2.

58h

N0

1 e

ln 2

ln 2 2h/ 2.

58h

7. 42 10 Mn atoms /

8 56 55

Mn atom

(a) Two hours after the irradiation stops, the activities are:

dN

dt

N e

0

t

N

0

t

ln

1/

2

2 ln 2 t/

t1/

2

e

For 32 P :

dN

9

1. 29 10 ln 2

4

dt 14. 26d 8. 64 10 s / d

ln 2 2h/ 342.

2h

16 31

e 7. 23 10 decays / P atom

For 56 Mn :

290



8

7. 42 10 ln 2

dN

dt 2. 58h 3600s / h

ln 2 48h/ 2.

58h

17 56

e 1. 39 10 decays / Mn atom

The total activity is the sum of these, each multiplied by the number of parent atoms

initially present.

11-84. Q 200 MeV / fission .

19 13

E NQ 7. 0 10 J N 200MeV / fission 1. 60 10 J / MeV

N

7.

0 10

13

MeV / fission . J / MeV

19

200 1 60 10

J

2.

19 10

30

fissions

Number of moles of 235 U needed =

30 23 6

N / N 2. 19 10 6. 02 10 3.

63 10 moles

A

Fissioned mass/y =

6 8 5

3. 63 10 moles 235g / mole 8. 54 10 g 8.

54 10 kg

That is 3% of the mass of the 235 U atoms needed to produce the energy consumed.

Mass needed to produce

7 0 10 8 54 10 0 03 2 85 10

19 5 7

. J . kg / . . kg .

Since the energy conversion system is 25% efficient:

Total mass of 235 U needed =

8

1. 14 10 kg .

11-85. The number of 87 Sr atoms present at any time is equal to the number of 87 Rb nuclei that

have decayed, because 87 Sr is stable.

N Sr N0 Rb N Rb N Sr N Rb N0 Rb N Rb 1N

N Sr N Rb

0.

010

N0 Rb N Rb N Sr N Rb 1 1.

010

and also

ln 2 t/

t1/

2

0

1/ 1.

010

N Rb N Rb e

ln2

t

1/

2

t

ln 1/ 1.

010

10 8

t t y y

1/ 2ln 1/ 1. 010 /ln( 2) 4. 9 10 ln 1/ 1. 010 /ln( 2) 7.

03 10

291


11-86. (a) Average energy released/reaction is: 3. 27MeV 4. 03MeV / 2 3.

65MeV

E

t

13

P 4W 4J / s N 3. 65MeV 1. 60 10 J / MeV

N

4J / s

3 65 1 60 10

13

. MeV / reaction . /

Half of the reactions produce neutrons, so

J MeV

6 85 10

12

. /

reactions s

12

3. 42 10 neutrons / s will be released.

(b) Neutron absorption rate

12 11

0. 10 3. 42 10 3. 42 10 neutrons / s

Energy absorption rate =

0. 5MeV / neutron

11

3. 42 10 neutrons / s

13

1. 60 10 J / MeV

2

2. 74 10 J / s

Radiation dose rate =

2

2. 74 10 J / s / 80kg 100rad / J / kg

3

3. 42 10 rad / s

2

3 42 10 4 0 137 493

. rad / s . rem / s rem / h

(c) 500 rem, lethal to half of those exposed, would be received in:

500rem / 492rem / h 1.

02h

t

11-87. R t N0 I 1 e (Equation 11-86)

35Bq

For Co: N0

6

24 2 12 2

.

19 10 cm 3. 5 10 / s cm 1 e

1 319 10 2

17

2.

00 10 atoms

For Ti:

N

115Bq

0

24 2 12 2

0.

120 2

0. 15 10 cm 3. 5 10 / s cm 1 e

15

1.

03 10 atoms

11-88. The net reaction is:

2 3 4 1

5 H He He H n 25MeV

Energy release / 2 H 5MeV (assumes equal probabilities)

4 water 4000g 2 1. 007825 15. 994915 g / mol 222.

1 moles

4 of water thus contains 2(222.1) moles of hydrogen, of which

Number of 2 H atoms =

4 2

1. 5 10 is H , or

292



2 222. 1 moles 6. 02 10 atoms / mol 1. 5 10 4.

01 10

23 4 22

Total energy release =

22 23 10

4. 01 10 5MeV 2. 01 10 MeV 3.

22 10 J

20 2

Because the U.S. consumes about 1. 0 10 J / y, the complete fusion of the H in 4

2

of water would supply the nation for about 1. 01 10 s 10.

1ms

11-89. (a)

2 hc / Mc

2

E

hc hc

2 2

2

hc

2

E

hc

p

E 2hc 2E

2 2

2 2

E E hc Mc Mc

1 2

2 2 2

E Mc E / 2 E Mc E / 2 /

p

p

4mM

4mM

E E E E 1

E E

M m M m

f i i 2 i i

2

E 4mM 4m / M

E M m 1 m / M

i

2 2

which is Equation 11-82 in More section.

(b)

1/

2

E 5. 7MeV 938. 28MeV / 2 51.

7MeV

(c)

vL

neutron (m)

x

CM

O 14 N (M)

The neutron moves at v L in the lab, so the CM moves at v vLmN / mN

M toward

the right and the 14 N 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

2

2 mv

2 2 2

L

2

M v Mv M m M

2

293



2

2Mm mvL

4Mm

1

2 2

m M m M

mv

2

2

L

4 14. 003074u

1.

008665u

1. 008665u

14.

003074u

2

57 . MeV

= 1.43 MeV

(d)

1/

2

2 2

14. 003074 931. 5 / 1. 43 / 2 96.

5

E uc MeV uc MeV MeV

11-90. In the lab frame:

photon

E hv pc

p hv / c E / c

deuteron,

M at rest

In CM frame:

photon

E pc

p E / c

deuteron, M

2 2

EK

1/ 2Mv p / 2M

p

2ME

K

For E pc in CM system means that a negligible amount of photon energy goes to

recoil energy of the deuteron, i.e.,

2

p

pc

pc E or pc pc 2Mc

2

2M

2Mc

2

2

E pc 2Mc 2 1875. 6MeV 3751.

2MeV (see Table 11-1)

In the lab, that incident photon energy must supply the binding energy B = 2.22 MeV plus

the recoil energy E K given by:

2 2 2 2 2

E p / 2M pc / 2Mc B / 2Mc

K

2

2.

22MeV

2 1875.

6MeV

2

0.

0013MeV

294



So the photon energy must be E 2. 22MeV 0. 001MeV 2. 221MeV , which is much

less than 3751 MeV.

11-91. (a)

B ZM H c Nm c M c

1 2 2 2

n A

(Equation 11-11)

For Li : B 3 1. 007825uc 4 1. 008665uc 7.

016003uc

7 2 2 2

2 2

0. 042132 931. 5 / 39.

25

uc MeV uc MeV

For Be : B 4 1. 007825uc 3 1. 008665uc 7.

016928uc

7 2 2 2

2 2

0. 040367 931. 5 / 37.

60

uc MeV uc MeV

B

1.

65MeV

For B : B 5 1. 007825uc 6 1. 008665uc 11.

009305uc

11 2 2 2

2 2

0. 0081810 931. 5 / 76.

21

uc MeV uc MeV

For C : B 6 1. 007825uc 5 1. 008665uc 11.

011433uc

11 2 2 2

2 2

0. 078842 931. 5 / 73.

44

uc MeV uc MeV

B

2.

77MeV

For N : B 7 1. 007825uc 8 1. 008665uc 15.

000108uc

15 2 2 2

2 2

0. 123987 931. 5 / 115.

5

uc MeV uc MeV

For O : B 8 1. 007825uc 7 1. 008665uc 15.

003065uc

15 2 2 2

2 2

0. 120190 931. 5 / 112.

0

uc MeV uc MeV

B

3.

54MeV

(b)

B a Z A a B Z A

2 1/ 3 2 1/

3

3 3

3

1/

3

For A 7; Z 4 : a 1. 65MeV 7 7 0.

45MeV

3

1/

3

For A 11; Z 6 : a 2. 77 MeV 11 11 0.

56MeV

295



3

1/

3

For A 15; Z 8 : a 3. 54 MeV 15 15 0.

58MeV

a3 0.

53MeV

These values differ significantly from the empirical value of a3 0. 75MeV

.

11-92. (a) Using M Z 0 from Problem 11-71.

m m 4

Z a MeV c a MeV c

2a A 8

n p 4 2 2

where

1 3 1

3

0. 75 / ,

4

93. 2 /

/

3

a4A

1 m 4

4

n

m

n p p

a4

1 m m 4a A

2A a A 2a

2 1 a A 4a

1 1/

3

3 4

2/

3

3 4

29 1 1. 008665 1. 007276 931. 5 / 4 93.

2

(b) & (c) For A 29 : Z

14

2/

3

2 1 0. 75 29 / 4 93.

2

The only stable isotope with A = 29 is 29

14 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. 78

38Sr is not

stable. Stable isotopes with A = 78 are 78

78

34

Se and Kr . 38

The only stable isotope with A = 119 is 119

50 Sn .

For A = 140: Computing as above with A = 140 yields Z = 69. 140

69Tm is not

stable. The only stable isotope with A = 140 is 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


11-93. (a) (b)

j = 3/2

j = 1/2

1d 5/2

1d 5/2

1p 1/2

1p 1/2

1p 3/2

1p 3/2

1s 1/2

1s 1/2

n

Ground state 11 B

p

n

1 st excited state 11 B

p

(c)

j = 5/2

1d 5/2

1p 1/2

1p 3/2

1s 1/2

n

2 nd excited state 11 B

p

297



(d)

1d 5/2

1d 5/2

1p 1/2

1p 1/2

1p 3/2

1p 3/2

1s 1/2

1s 1/2

n

17 O p

n

17 O

p

Ground state (j = 5/2)

1 st excited state (j = 1/2)

(e)

2s 1/2

1d 5/2

1p 1/2

1p 3/2

1s 1/2

n

17 O

2 nd excited state (j = 1/2)

p

298


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] M Z , 151 [adjusted]

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



(b) The drip lines occur for:

protons: M Z, 151 M Z 1, 150 m

p

0

neutrons: M Z, 151 M Z, 150 m

n

0

Write a calculator or computer program for each using Equation 11-14 (on Web page)

and solve for Z.

11-95. (a)

,

p n

2 / 3 1/ 3 2 1 1/

2

1 2 3 4

2

5

M Z A Zm Nm a A a A a A a A Z A a A

from Equation 11-14 on the Web page.

M 126,

310 126m 184m

p

n

15. 67 310 17.

23 310

2 1/

3

0.

75 126 310

2/

3

93.

2 310 2 126 310

1 1/

2

12 310 12 310

2

M

126, 310 313.

969022u

(b) For β − decay: (126, 310) → (127, 310) + β − + v

e

Computing M(127, 310) as in (a) yields 314.011614u.

300



Q M 126, 310 c M 127,

310 c

2 2

313. 969022 uc 314.

011614 uc

2 2

2

0. 042592 931. 5 / 39.

7

u MeV uc MeV

Q < 0, so β − decay is forbidden by energy conservation.

For β + decay: (126, 310) → (125, 310) + β + + v e


Q M 126, 310 c M 125, 310 c 2mec

2 2 2

2 2

313. 969022 313. 923610 1.

022

41. 3MeV

uc uc MeV

β + decay and electron capture are possible decay modes.

For α decay: (126, 310) → (124, 310) + α


Q 313. 969022uc 309. 913540uc 4.

002602uc

49. 3MeV

α decay is also a possible decay mode.

2 2 2

11-96. (a) If the electron’s kinetic energy is 0.782MeV, then its total energy is:

E MeV m c MeV MeV MeV

2

0. 782

e

0. 782 0. 511 1.

293

2 2 2

e

E pc m c

2

(Equation 2-32)

2 2

e

2

1/

2

p E m c c

1/

2

2 2

1. 293MeV 0.

511MeV c

1. 189MeV / c

(b) For the proton p 1. 189MeV / c also, so

2 2 2

E p / 2m pc 2mc

kin

2 4

1. 189MeV 2 938. 28MeV 7. 53 10 MeV 0.

753keV

301



(c)

4

7.

53 10 MeV

0.

782MeV

100 0. 0963%

dN

11-97. R

p N (Equation 11-17)

dt

(a) N R N0e R R e R 1 e

t t t

p p p p

t

N R / 1 e (from Equation 11-17)

p

At t = 0, N(0) = 0. For large t, N t R

p

/ , its maximum value

(b) For dN / dt 0

R N N R / R / ln / t

p p p

2

1/

2

1 1

N 100s / ln 2/ 10min 100s 60s / min / ln 2/

10min

4 62

8. 66 10 Cu nuclei

11-98. (a) 4n + 3 decays chain

below.)

235 207

92U143 82

Pb

125

There are 12 α decays in the chain. (See graph

302


(b) There are 9 β − decays in the chain. (See graph below.)


(c)

235 2 207 2 4 2

Q M U c M Pb c 7M He c

235. 043924uc 206. 975871uc 7 4.

002602 uc

0. 049839 931. 50 /

2 2 2

2 2

uc MeV uc 46 43

. MeV

(d) The number of decays in one year is:

dN

dt

N e where ln 2 / t ln 2 / 7. 04 10 y 9.

85 10 y

t

0 1/

2

8 10 1

N

0

1 1000 6 02 10

23

kg g / kg . atoms / mol

235g / mol

2.

56 10

24

atoms

dN

dt

10 1 24 1y

15

9. 85 10 y 2. 56 10 e 2. 52 10 decays / y

Each decay results in the eventual release of 46.43 MeV, so the energy release

per year Q is:

15

Q 2. 52 10 decays / y 46. 43MeV / decay

17 13

1. 17 10 MeV / y 1. 60 10 J / MeV

The temperature change ΔT is given by:

4 3

1. 87 10 J / y 1cal / 4. 186J 4. 48 10 cal / y

Q cm T or T Q/

cm where m 1kg 1000g

and the specific heat of U is c 0. 0276cal / g C .

3

T 4. 48 10 cal / y 0. 0276cal / g C 1000g 162 C

303



144

α

235 U

231 Th

142

140

237 Ac

β −

α

231 Pa

N

138

136

134

132

α β −

β − α

α

β − α

Rn

223 Ra

223 Fr

219 At

α

215 Bi

215 Po

α

227 Th

130

128

α

211 Pb

β −

β −

211 Bi

α

215 At

126

124

α

207 Tl

β −

207 Pb

β −

α

211 Po

80

82 84 86 88 90

Z

92

11-99. The reactions are:

(1) 1 H 1 H 2 H v

e

(2) 1 H 2 H 3 He

followed by

(3) 3 He 3 He 4 He 1 H 1 H or

304



(4) 1 H 3 He 4 He v

e

(a) From 2(1) + 2(2) + (3):

1 2 3 2 3 4 1

6 H 2 H 2 He 2 H 2 He He 2 H 2 2v

e

2

Cancelling

1 2 3

2 H, 2 H,

and 2 He on both sides of the sum,

(b)

1 4

4 H He 2 2v

e

2

From (1) + (2) + (4):

1 2 3 2 3 4

4 H H He H He He 2 2v

e

Cancelling 2 H,

and

3 He on both sides of the sum,

1 4

4 H He 2 2v

e

1 2 4 2 2

Q 4m H c M He c 2m e

c

4 938. 280MeV 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. 7MeV 2 mec

2

24. 7MeV 2 1.

022MeV 26. 7MeV

Each cycle uses 4 protons, thus produces 26. 7MeV / 4 6. 68MeV / proton .

Therefore, 1 H (protons) are consumed at the rate:

dN P

26

4 10 J / s 1eV

dt E

6

6. 68 10 eV

19

1.

60 10 J

The number N of 1 H nuclei in the Sun is:

N

M

M

1

H

1/

2 2 10

1.

673 10

30

27

kg

kg

5.

98 10

3 75 10

38

. /

56

protons

which will last at the present consumption rate for

protons s

t

N

dN dt

56

5.

98 10 protons

protons s

38

/ 3. 75 10 /

1.

60 10

18

s

1y

1. 60 10 s

5.

05 10

3.

16 10 s

18 10

7

y

305


11-100. At this energy, neither particle is relativistic, so

p

p

E E p p E E 17.

7MeV

2 2

He

n

He n He n He n

2mHe

2mn

2m E 2m E 2m 17.

7MeV E

He He n n n He

m

m 17. 7 Therefore, n

He

mn EHe MeV mn EHe

17.

7MeV

m m

E

He

1. 008665u

17.

7MeV

3.

56MeV

2. 002602u

1.

008665u

E 17. 7MeV E 17. 7 3. 56 MeV 14.

1MeV

n

He

He

n

11-101. (a) The number N of generations is:

N

5s

0. 08s / gen

62.

5 generations

Percentage increase in energy production =

R N

R

0

R

0

100

R N

R

0

N

1 100 where R N / R 0 k (from Example 11-22 in More section)

N

k

62.

5

1 100 1 005 1 100 137

. %

(b) Because k neutron flux,

the fractional change in flux necessary is equal to the

fractional change in k:

k 1 1.

005 1

k 1.

005

0.

00498

11-102. (a) For 5% enrichment:

0. 05 U 0.

95

235 238

f f f

U

0. 05 584b

0. 95 0 29.

2b

0. 05 U 0.

95

235 238

a a a

U

0. 05 97b 0. 95 2. 75b 7.

46b

306



k

f

2. 4 29.

2b

2. 4 1.

91 (Equation 11-68 in More section)

29. 2b

2.

46b

f

a

(b) For 95% enrichment:

0. 95 U 0.

05

235 238

f f f

U

0. 95 584b

0. 05 0 554.

8b

0. 95 U 0.

05

235 238

a a a

U

0. 95 97b 0. 05 2. 75b 92.

3b

N

The reaction rate after N generations is R N R 0 k .

N

For the rate to double R N 2R 0 and 2 k N ln 2 /ln k .

N

N

5% ln 2 /ln 1. 91 1.

07 generations

95% ln 2 /ln 2. 06 0.

96 generations

Assuming an average time per generation of 0.01s

t

5% 1.

07 10

2

s

t

95% 0.

96 10

2

s

Number of generations/second = 1/seconds/generation

In 1s: N 5% 93. 5 and N 95%

104

One second after the first fission:

93.

5 26

N

R 5% R 0 k 1 1. 91 1.

9 10

Energy rate =

26

1 9 10 200

. fissions / s MeV / fission

3 8 10 1 60 10

28 13

. MeV / s . /

15 15

6. 1 10 J / s 6.

1 10 W

104 32

R 95% R 0 k

N 1 2. 06 4.

4 10

J MeV

Energy rate =

32

4 4 10 200

. fissions / s MeV / fission

8 8 10 1 60 10

34 13

. MeV / s . /

22 22

1. 4 10 J / s 1.

4 10 W

J MeV

307

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 .

2

E m

c MeV

139.

6 (from Table 12-3)

(c)

hc 1240MeV fm

E hf hc / Thus, 8.

88 fm

E 139.

6MeV

12-2. (a)

2 2

E

mc mc 2285MeV 139. 6MeV 2424.

6MeV

E 2mpc 2 938. 28MeV

1876.

56MeV

2

(b)

E 2m c 2 105. 66MeV 211.

32MeV

2

(c)

12-3. (a)

e

e

e

e

γ

Z

e

e

e

e

(b)

e

γ

e

γ

γ

e

e

γ

e +

e

309


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 32 S e assuming no neutrino

Q M P c M S c

32 2 32 2

(electron’s mass is included in that of 32 S )

31. 973908uc

31.

972071uc

2 2

2 2

0. 001837uc 931. 5MeV / uc 1.

711MeV

To a good approximation, the electron has all of the kinetic energy

E Q 1.

711MeV

k

(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 m c

2

2

e

2 2

Ek mec mec

(Equation 2-32)

2 2

2 2

Q m c m c Q 2Qm c

2 2 2 2

e e e

310



The kinetic energy of the 32 S is then:

E

k

2

2 2

2

e

2 32 2

2

p pc Q Qm c

2M 2Mc 2M S c

2

1. 711MeV 21. 711MeV 0.

511MeV

2 2

. uc . MeV / uc

2 31 972071 931 5

5

7. 85 10 MeV 78.

5

eV

(c) As noted above, the momenta of the electron and 32 S are equal in magnitude and

opposite direction.

pc 2 Q 2 2Qm c 2 1. 711MeV 2

21. 711MeV 0.

511MeV

e

1 /

2

1. 711 2 1. 711 0.

511

2

p MeV MeV MeV 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:

2

E mpc

938.

28MeV

(c)

(d)

hc 1240MeV fm

E

hv hc / 1.

32 fm

E 938.

28MeV

8

c 3. 0010

m / s

v 2.

27 10

15

1.

3210

m

23

Hz

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


12-8. (a) 2/ 5 2/

5

a b

m m c

Since the units of must be seconds s, we have

5 2

5

5

1 Nm

2 1 a b

2

s kg e c

C

c

having substituted the internal units of ke

2 / c.

Re-writing the units, noting that

J N m, gives:

s

kg N m C s

1 10 5 5 5

C 1 J s m a b

5 10 10

c

5

Noting that

J kg m s

2 2

/ and cancelling yields,

1 2

a ab ab

s kg m s

Since a 1 must be 0, a 1 , and since a b 1, b 2.

2 5

2 / mc

34

2 1.

05510 Js 137

10

(b) 1. 2410 s 0.

124ns

2

31 8

9. 1110 kg 3. 0010

m / s

5

12-9. (a) Weak interaction

(b) Electromagnetic interaction

(c) Strong interaction

(d) Weak interaction

0

12-10. 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


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.

d 1 d 1

t t

t 1

c c

2

1

2

1

(Equation 1-19)

1 1

2

1 1 1

1 1

1 since 1

2 1

2

2

Substituting into t,

d 1

t E mc E mc

2

c 2

/ 2

2 2 2

(Equation 2-10)

2

d mc

t

2c

E

2

1/ 2

2 1/

2

2

t 2cE

2

2tE

mc

d

d / c

2

6

212.

5s10 10

eV

7

170, 000c y / c3. 16 10

s / y

1/

2

21.

6eV

m 22 eV / c

2

313


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)

ν ν

μ −

μ e

e + π−

ν τ

μ −

W +

W −

W −

μ +

ν μ

τ −

ν μ

12-16. (See Table 12-8 in More Section.)

(a) 30 MeV

(b) 175 MeV

(c) 120 MeV

m c m m c Conservation of energy and lepton number are violated.

12-17. (a)

2 2

p n e

(b)

n p

2 2

m c m m c 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, pp 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


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

1 1

T3

1 2 2

T 1

(b) n p

1 1

T3

0

2 2

T 1 or 0

315



(c) p

1 3 3

T3

1 T

2 2 2

(d) n

1 3 3

T3

1 T

2 2 2

(e) n

1 1 1 3

T3

1 T or

2 2 2 2

12-22. (a) e

Electron lepton number changes from 0 to 1; violates conservation of

electron lepton number.

0

(b) e e v v Allowed by conservation laws, but decay into two photons via

e

e

electromagnetic interaction is more likely.

(c) e e v

Allowed by conservation laws but decay without the electrons

is more likely.

(d)

(e)

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.

n p e v Allowed by conservation laws. This is the way the neutron decays.

e

12-23. (a)

(b)

(c)

0 0

p

n

0

p

n

0 0 0

0 0

(d) e e e e

(e)

0

v

12-24.

p

0

Because s have B = 0 and p has B = 1, conservation of B requires the to have B = 1.

0

0

The has B = 0, so conservation of B requires that the have B = 1.

316


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 ++

0

(d) uss +1 0 −2

(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

0

(e) ds 0 0 −1

* forms and along with uu and dd

317


12-28.

u

d

0

K

d

c

g

g

u

b

0

K

d

b

12-29. (a) T 3 = 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 C = 2/3 – 1/3 – 1/3 = 0

S = 0 + 0 + −1 = −1 The T = 1 state is the Σ 0 . The T = 0 state is the Λ 0 .

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.)

2

12-31. The range R is R c / mc (Equation 11-50). Substituting the mass of W + (from Table

12-4),

34 8

1. 05510 J s3. 0010

m / s

2 10

81GeV / c 1. 6010

J / GeV

18 3

R 2. 4410 m 2.

4410

fm

12-32.

d

u

0

ds du

v

W

v

d

0

s

318


12-33.

p

u d u d

0

p

uds uud ud

Weak decay

0

Z

u d s

0

12-34. n p

Q m c m c m c

2 2 2

n p

939. 6 938. 3

139. 6 MeV

138.

3MeV

This decay does not conserve energy.

12-35.

p

u d u u d

0

p

uds uud ud

Strong decay

0

Z

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 (quark-

antiquark) structure. us produces the correct set of quantum numbers. (From Table

12-5.)

319



(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 ds. (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 cd.

12-39. The

0

decays via the electromagnetic interaction whose characteristic time is

20

10 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 v would have 1 = 0 + 0 = 0 and B is not conserved. The

e

lepton numbers would not be conserved either; a “leptoquark” number would be

conserved.

2

12-41. V H2O 0. 75 V 0.

754

R R

, where R(Earth) =

6

6. 37 10 m and

R km m

3

1 10 .

V H2O 0. 75 4 6. 37 10 m 10 3.

8210

m

2

6 3 17 3

M H O V H O . 17 m 3 kg / m 3 .

20 kg

2 2

3 82 10 1000 3 82 10

23 22

Number of moles (H 2 O) = 3. 8210 g / 18. 02g / mole 2.

12

10 moles

Number of H 2 O molecules = N # of moles

A

6 . 02 10 23 molecules / mole 2 . 12 10 22 moles

46

1. 28 10 molecules H2O

Each molecule contains 10 protons (i.e., 2 in H atoms and 8 in the oxygen atom), so the

47

number of protons in the world’s oceans is N 1. 28

10 .

320



The decay rate is

dN

dt

N where 1/ 1/

10 y 10 y

10 32

y

1 1 . 28 10 47 protons

32 32 1

15 7

1. 2810 proton decays / y 4

10 decays / s

12-42. (a)

0

p e v e

Q m c M c m c MeV

2 0 2 2

p

e

938. 3 1116 0. 511 MeV 178MeV

Energy is not conserved.

(b) p

(c)

Spin (angular momentum) 1 0 1 1.

Angular momentum is not conserved.

2

0

p

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

S 0 +0 +0 = 0

n , B = −1, Q = 0, spin = 1/2, S = 0


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

S 0 +0 +0 = 0

321



(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

S 0 −1 −1 = −2

(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

S 0 +0 −1 = −1

(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

S −1 −1 −1 = −3

(e)

, B = 1, Q = −1, spin = 1/2, S = −2


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

S 0 −1 −1 = −2

322


12-44. (a)

(b)

(c)

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

S 0 +0 +0 = 0

Quark strucure u c

B 1/3 −1/3 = 0

Q 2/3 −2/3 = 0

spin 1/2 1/2 = 1, 0

S 0 +0 = 0

Quark strucure u b

(d)

B 1/3 −1/3 = 0

Q 2/3 +1/3 = 1

spin 1/2 1/2 = 1, 0

S 0 +0 = 0

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, 1/2

S 1 +1 +1 = 3

12-45. The

0

Z 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

e

p e v v v

(c) Conservation of charge: 0 11 0 0 0 0

Conservation of baryon number: 11 0 0 0 0 1

323



Conservation of lepton number:

(i) for electrons: 0 0 11 0 0 0

(ii) for muons: 0 0 0 0 11 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

cuu

0, 1, 2, 1, 0 c s s

0, 0, 1, 0, 1

bs

0, 1, 1, 0, 0 s d u

0, 1, 1, 1, 0 c s d

1, 1, 3, 0, 0 sss

12-48. (a)

x x

t1 x / u1 t2 x / u2 t t2 t1

u u

u1

u

2

xu

t

x

2

u1u

2 c

2 1

(b)

E

mc

2

1 u / c

2 2

(Equation 2-10)

2 2 2 2 2

2

2 ( mc ) 2 2 ( )

1 / mc u

1

mc

E u c

2 2 2

1 u / c E c E

2 2

Expanding the right side of the equation in powers of ( mc / E)

and keeping only

1/2

the first term yields

324



u

c

2

1 mc

1 2 E

2

(c)

2 2

2 2

1 mc 1 mc

u1 u2

c 1 1

2 E1 2 E

2

2 2 2 2 2 2

c( mc ) 1 1 c( mc ) E1 E

2

2 2 2 2

2 E2 E1 2 E1 E2

c(2.2eV / c c

) (20MeV) (5MeV)

2 (20MeV) (5MeV)

2 2 2 2 2

u u1 u2 2 2

2 (20) (5) (MeV)

2 6 2

c(2.2eV) (10 MeV/eV) 375

2 2 2

2 12

(2.2) (375) 10

c

2 2

2(20) (5)

5

2.7 10 m/s

And therefore,

t

5 15 5

1.710 c y9.4610 m / c y2.710 m/s

(3.010 m/s)

8 2

0.48s

12-49. (a) e ve

v

Electron lepton number: 0 11 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 11 0 0

Tau lepton number: 1 0 0 1 1

(c)

v

Electron lepton number: 0 0 0 0

Muon lepton number: 0 11 0

Tau lepton number: 0 0 0 0

325


0

12-50.

hf

2

2

E pc m c

2 2

In lab:

0

hf

Conservation of momentum requires that each carry half of the initial momentum, hence

1/

2

2

2 2

the total energy: hf / ccos p hf E / pc m

c

cos

2

p

hf /

c

2 2 2

p

1/

2

2 2

2

2 pc m

c

2c

pc

850MeV

0.

9876

1/ 2 1/ 2

2 2 2 2

2

pc m c

850MeV

135MeV

cos 1 0 . 9876 9 . 02

12-51. (a)

0

p

1116MeV 938 140 MeV 38MeV

conserved.

Energy:

Electric charge: 0 11 0 conserved.

Baryon number: 11

0 1 conserved.

Lepton number: 0 0 0 0 conserved.

(b) n

p

1197MeV 940 938 MeV 681MeV

not conserved.

Energy:

Electric charge: 1 0 1 1 conserved.

Baryon number: 111 0 not conserved.

Lepton number: 0 0 0 0 conserved.

This reaction is not allowed (energy and baryon conservation violated).

326



(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 11 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 v e

Only the e + and e − are stable.

(b) The net effect of the chain reaction is: p 3e e 3v 2v 2v

(c) Charge: 1 1 31 1 conserved

Baryon number: 11 0 0 0 0 0 0 1 conserved.

Lepton number:

(i) electrons: 0 0 31 31 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.

e

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/c 2 ,

about 1/3 of the proton’s mass.

12-54. (a)

0

p

Ekin

M m m

c

0 2

p

1116MeV / c 938. 3MeV / c 139. 6MeV / c

c

38. 1MeV

2 2 2 2

327



(b) Because the

opposite direction.

0

decayed at rest, the p and

m v m v m / m v / v

p p p p

1 2 2

Ekin

mv

2 m mp

m

p 938.

3

6.

72

1

Ekin

p 2

mv

mp

m

m

139.

6

p p

2

(c)

have momenta of equal magnitudes and

E E p E E p 6. 72E p 7. 72E p 38.

1MeV

kin kin kin kin kin kin

kin

38. 1 / 7. 71 4.

94

E p MeV MeV MeV

E 6. 72E p 33.

2MeV

kin

kin

12-55.

0 0

(a)

E

T

for decay products is the rest energy of the

0

, 1193MeV.

(b) The rest energy of

0

,

116 MeV so E 1193MeV 1116MeV 77MeV

(c) The

and p E / c 77MeV / c .

0

decays at rest, so the momentum of the

photon.

0 2 / 2 77 / 2

/ 21116

/

2

Ekin

p M MeV c

MeV c

2. 66MeV

small compared to E

0

equals in magnitude that of the

(d) A better estimate of E

and p

are then E 77MeV 2. 66MeV

74.

3MeV

and

p

74. 3MeV / c.

12-56. (a)

x x x u u

t t t Note that u u c

1 2 2

2 1 1 2

u2 u1 u1u

2

x u1

u2

xu

t

2 2

c c

328



(b)

(c)

1/

2

2

2 2

2

2

mc

u 1 mc

o

1

o

(Equation 2-10). Thus,

mc

1

2 2

2

u / c

c E 2 E

E

1

2 2

2 2

1 moc

1 moc

u1 u2

c 1 1

2 E1

2 E

2

o

2 2

2 2 2 2 2

o

c c m c

o

1

2

2 2

2 1 1 2

c m c m c E E

2 E 2 E 2 E E

c

2 6 6

20eV

2010 eV 510

eV

2 2

2

6 6

2010 eV 510

eV

2 2

2 2 2

c20eV

20 5

7.

510

2 2 2

2

6

20 5 10

eV

12

170, 000c y7.

510

C

12

xu

c

c

6

T 1. 28 10 y 40.

3s

2 2

(d) If the neutrino rest energy is 40eV, then

c

11

u 3. 0010 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


12-58. () i dp ev B

dt

( ii) dE

dt

0 which follows from the fact that the Lorentz force is v ; therefore, v v =

constant and thus

v

constant.

Equation (i) then becomes: d p m

dv ev B

dt dt

For circular orbits

mv

R

2

evB or, re-writing a bit,

1GeV

m v p eBR and pc ceBR 0.

35BR GeV

19

1.

6010

J

and finally, p 0. 35BR GeV / c

330

Chapter 13 – Astrophysics and Cosmology

v W

13-1. v v 4 km / s.

Assuming Sun’s rotation to be

W

E

r

uniform, so that v v , then v v 2km / s.

W E W E

v E

ω

Because v 2

/ T, v 2 r / T or

E

5

2

r

6

T 2. 1910 s 25.

3days

v

E

2

6.

9610

km

2km / s

13-2.

2GM

11 30

. . 2

2 6 67 10 1 99 10

2

41

U J 7.

5910

J

8

R

6.

9610

The Sun’s luminosity

L

3.

85

10

26

W

U 7.

5910

J

L 3. 8510

J / s

41

15 7

tL

1. 97 10 s 6.

26 10 years

26

13-3. The fusion of 1 H 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 4 12

Be He C to proceed. This 3- 4 He fusion to 12 C 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 yfrom Galactic center = radius of orbit

2 r 2

28 000 9 45 10

time for 1 orbit

15

, c y . /

5

v 2. 510

m / s

m c y

15 8

6. 65 10 s 2.

1110

yr

331


13-5. Observed mass (average)

3 27 3

1 / 1. 67 10 / 10% of total mass

H atom m kg m (a)

missing mass 9 1. 67 10 kg / m 1. 56 10 kg / m

500 photons cm 500 10 photons m

27 3 26 3

3 6 3

/ / , so the mass of each photon would be

26 3

1. 50 10 kg / m

500 10 photons / m

6 3

3.

01 10

35

kg

or

3.

01 10 kg m

mv

c eV c

1. 60 10 J / eV s

35 2

2 2

16. 9 /

19 2

13-6.

5

4

T

4

10 K

3

2

1

0

0 1

2

3

4

38

M / M /

6

7

8

13-7.

1 1 3 00 10 1 3 00 10 3 00 10

8 8 5

c s c s . m/ s s . m . km

1 min 1min 60 min 3 00 10 60 1 80 10

5 7

c c s / . km s . km

1 1 3600 1 08 10

9

c h c h s / h . km

1 24 3600 2 59 10

10

c day c h s / h . km

13-8. (a) See Figure 13-16.

1 1 496 10

11

AU . m . R 1 pc when 1 ", so

R

1AU

1"

332



or

1AU

3600"

180

1"

1 rad

16

R 3.

086 10 m 1pc

16

3.

086 10 m

1pc

3.

26c y

15

9. 45 10 m / c y

(b) When 0. 01", R 100 pc and the volume of a sphere with that radius is

4

3

3 6 3

V R 4. 19 10 pc . If the density of stars is

of stars in the sphere is equal to

0 08

3

. / ,

0. 08/ 4. 19 10 3.

4 10

pc then the number

3 6 3 5

pc pc stars.

13-9.

L r f m m f f

2

4

1 2

2. 5log 1

/

2

Thus,

L r f L r f L L

2 2

p

4

p p

and

B

4

B B

and

p B

r f r f r r f / f

2 2 2 2

p p B B B p p B

1. 16 0.

41

log f

p

/ fB 0. 30 f

p

/ f

B

2.

00

25 .

Because

1/

2

r 12 pc, r r f / f 12 2 17.

0 pc

p B p p B

13-10. (a)

(b)

2 25

M 0. 3M Te

3300K L 5 10 L 1.

93 10 W

2 28

M 3. 0M Te

13, 500K L 10 L 3.

85 10 W

(c) R M R M R / M

R

03 .

0. 3 0. 3 0. 3 2.

09 10

M

8

R M M R m

Similarly,

9

R R m

30 .

3. 0 2.

09 10

t M t M t / M t M

3 3 3 3

L L L L

(d)

3 3

3 3

L

0. 3 0. 3

L

0. 3 0.

3

L

t M t M M t

or t 0. 3 37 t . Similarly, t 3. 0 0.

04t

L L L L

333


13-11. Angular separation

S

R

distance between binaries

distance Earth

6

100 10 km

11

10 m

100 100 3 15 10

7

cy c y . s / y

1.

057 10

7

rad

6 9

6. 06 10 degrees 1.

68 10 arcseconds

13-12. Equation 13-18:

Fe 13 He 4 n . m 56 55. 939395 u , m 4 4. 002603 u , m 1. 008665 u .

56 4

26 2

Fe

Energy required: 13 m 4 4 m m 56 0. 129104 u .

He

2

1 931. 49432 / 0.

129104 120

u MeV c u MeV

4 1

Equation 13-19:

2

He 2 H 2n m 1 1.

007825

n

Energy required: 2 m 1 2 m m 4 0. 020277 u 28.

3 MeV

H

n

He

Fe

H

He

n

13-13. 24 km / s (a)

15 . cy

15 . cy

r ; assuming constant expansion rate,

2

Age of Shell

1. 5cy/

2

4

2. 4 10 m/

s

(b) L 12L T 1.

4T

star e star e

11

2.

95 10 s 9400

y

R M R M T M T M L M L M

e

R / M , T / M , L / M

e

1/ 2 1/

2 4 4

e

1/

2 4

R T L

R m , T M , L M

M M M

e 1/

2 4

star star e star 1/

2 star star 4 star

Using either the

M T

star

e star

2

Te

or L relations, Rstar

R R 1. 4 R 1.

96R

M T

e

2

or

R

star

L

L

star

1/

2

1.

86R

334


13-14.

R GM c

S

2

2 / (Equation 13-24)

(a) Sun

11 30 2 3

RS

2 6. 67 10 1. 99 10 / c 2.

9 10 m 3km

(b) Jupiter m 318m R 2.

8m

J E S

(c) Earth

3

RS

8 86 10 m 9mm

. !

13-15. M 2M

(a) (Equation 13-22)

14 1/

3 14 1/

3

4

R 1. 6 10 M 1. 6 10 2M 1.

01 10 m

(b) 05 . rev / s rad / s

1 2

where for a sphere

2 I 2

2 1 2 2 1 01 10 8 0 10

5 2 5

2 4 38

I MR M . . J

(c)

d

1

10 /

2

d I d I where day

8

d

38

d 2 8.

0 10 J

25 25

2 1. 85 10 J / s L 1.

85 10 W

8 5

10 d 8. 64 10 s / d

13-16. Milky Way contains

11

10 stars of average mass M , therefore the visible mass =

30 11 41

1. 99 10 10 1. 99 10 kg 10% of total

(a) Mass of a central black hole =

41 42

9 1. 99 10 1. 8 10 kg

(b) Its radius would be

2

RS

2 GM / c (Equation 13-24).

11 42 2 15

RS

2 6. 67 10 1. 8 10 / c 2. 6 10 m 17,

000AU

13-17. v 72, 000 km / s .

(a)

v 72, 000km / s

H 21. 2km / s 10 c y

9

v Hr r 3.

40 10 c y

6

(b) From Equation 13-29 the maximum age of the galaxy is:

1 / 4. 41 10 1.

4 10

17 10

H s y

335



1/

H r

1/ H r / v 1/ H r / v

10%

1/

H r

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

7 8 16

of c, thus the core can be at most 1. 5y 3. 15 10 s / y 3. 0 10 m / s 1.

42 10 m

4

9. 45 10 AU in diameter. The Milky Way diameter is

60 000 3 8 10

9

, c y . AU .

13-19. Combining Hubble’s law (Equation 13-28) and the definition of the redshift (Equation 13-

27) yields

z

Hor

c

0

0

Hor

c

1

0

(a)

r

6

5 10 c y

6

km 5 10 c y

21.7 1 656.3nm

6 8

s 10 c y 3 10 m/s

656.5nm

(b)

r

6

50 10 c y

Similarly,

658.7nm

(c)

r

6

500 10 c y

Similarly,

680.0nm

(d)

r

9

5 10 c y

Similarly,

893.7nm

336


13-20. Equation 13-33:

c

2

3H

3

2

8 G 8 1/ H G

c

3

2

10 7 11 2 2

8 1. 5 10 y 3. 15 10 s / y 6. 67 10 Nm / kg

8. 02 10 kg / m

27 3

(This is about 5 hydrogen atoms/m 3 !)

13-21. Present size

10 1 1

10

10 c y S

p

S

p

with T 2. 7K 2.

7 10 c yK

T T

(a) 2000 years ago, S = S p

(b) 10 6 years ago, S = S p

(c) 10 seconds after the Big Bang,

10 9 9

S 2 7 10 c yK 10 K 2 7 10 S 25c y

. / .

p

(d) 1 second after the Big Bang,

10 9 10

S 2 7 10 c yK 5 10 K 5 4 10 S 5c y

. / .

p

(e) 10 -6 seconds after the Big Bang,

S 2. 7 10 c yK / 5 10 K 5. 4 10 S 0. 005c y 6.

4 10 AU

10 12 13 4

p

13-22.

8

mpl

5.

5 10 kg

Planck time 5. 5 10 kg / m

3

3

35 3

pl 10 m

97 3

27

1.

67 10 kg

proton 1. 67 10 kg / m

3

15 3

10 m

osmium 2. 45 10 kg / m

4 3

18 3

13-23. Wien’s law (Equation 3-11):

max

2. 898mm K 2.

898mm K

T 2.

728K

(this is in the microwave region of the EM spectrum)

1.

062mm

13-24. Muon rest energy

2

208m

e

106 / .

MeV c The universe cooled to this energy (average)

at about 10 -3 s (see Figure 13-34). 2.728K corresponds to average energy = 10 -3 eV.

Therefore,

m

3 19

10 eV 1. 6 10 J / eV

2

c

1.

8 10

39

kg

337


13-25.

M / 4 / 3 r t t M / 4 / 3 r t

3 3

0 0

r t R t r t

0

(Equation 13-37) \

M

M

R t

4/ 3 r t / R t

4/

3 r

3

0 3 2

t

3

0

R t t

13-26.

B

If Hubble’s law applies in A, then

FBC

vBA HrBA, vCA Hr

CA.

FBA

From mechanics,

C

vBC vBA vCA H rBA rCA Hr

BC

FCA

A (Milky Way)

and Hubble’s lab applies in C, as well, and

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

article of radius a is: GM m

F

where

grav 2

r

the particle due to the Sun’s radiation pressure at r is given by: (See Equation RP-9.)

3

m (4 / 3) a . The force acting on

2 2 U

Frad

a Prad

a where

3

2

a 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 4 L

R

c c 4 r

2

Therefore,

F

rad

a

2

1

3 4

4L

2

rc

The minimum value of a is obtained from the condition that Fgrav

F

rad

:

338



GM m

2 1 4L

a

2 2

r 3 4 r c

GM a L

3

(4 / 3)

2 1 4

a

2 2

r

3 4

r c

Simplifying this expression yields:

a

4

L

cGM

26

3.84 10 W

a

8 11 2 2 30 3

4 (3.00 10 m/s)(6.67 10 N m / kg )(1.99 10 kg)(5500kg/m )

7 5

a 1.40 10 m or 1.40 10 cm

Note that (i) a is very small and (ii) the magnitude of a is independent of r.

13-28. (Equation 13-31) Robs

Remit

1 Z r0

r 1 Z

Z M / 4 / 3 r M / 4 / 3 r

3 3

0 0

Substituting for r 0

in the 0

equation:

3

3 3

0

M / 4/ 3 r 1 Z M / 4/

3 r 1 Z

0

Z / 1 Z or

3

Z 1 Z 0

3

13-29. (a) H available for fusion =

30 29

M 0. 75 0. 13 2. 0 10 kg 0. 75 0. 13 2.

0 10 kg

(b) Lifetime of H fuel =

29

2.

0 10 kg

1

6. 00 10 kg / s

3.

3 10

17

s

17 7 10

3. 3 10 s / 3. 15 10 s / y 1.

03 10 y

10 10 9

(c) Start being concerned in 1. 03 10 y 0. 46 10 y 5.

7 10 y

13-30. SN1987A is the Large Magellanic cloud, which is 170,000c•y away; therefore

(a) supernova occurred 170,000 years BP.

339



(b)

E K m c

o

2

mc

1

o

2

v

c

2

2

9.

38 10

eV m eV v c

v

1

2

c

8

9 2 8 9 8

10 ,

o

9. 28 10 10 9. 38 10 or 0.

875

2

Therefore, the distance protons have traveled in 170,000y

= v 170, 000 y 149, 000c y . No, they are not here yet.

13-31.

M

30

1. 99 10 kg .

(a) When first formed, mass of H =

1 27

0. 7M , m H 1. 007825u 1. 66 10 kg / u , thus

07 . M

number of H atoms 8.

33 10

27

1. 007825u

1. 66 10 kg / u

56

(b) If all

1 4

H He; 4 H He 26. 72eV . The number of He atoms

produced =

8.

33 10

4

Total energy produced =

(c) 23% of max possible =

0. 23 8.

89 10

L

6

.

8.

33 10

4

44

0. 23 8. 89 10 J

6

26. 72 5. 56 10 8.

89 10

57 44

MeV MeV J

44

17 10 26

tL

5. 53 10 s 1. 7 10 y L 3.

85 10 W

13-32. (a)

F Gm m / r a m v / r m

2 2

1 2 c 2 2

2 2

v / r Gm / r f v / r

1

and orbital frequency 2

Substituting for f and noting that the period

T 1/ f , 4 f Gm / r

2 2 3

1

2 2 3

or, T 4 r / Gm , which is Kepler’s third law.

(b) Rearranging Kepler’s third law in part (a),

1

340



m 4 r / GT

E

2 3 2

moon

2 8

4 3.

84 10 m

11 2 2 4

6. 67 10 Nm / kg 27. 3d 8. 64 10 s / d

3

2

24

6. 02 10 kg

(c)

1/

2

1/

2 3

3

6

r

6.

67 10

sh

3

2 2 5. 44 10 1.

5

11 24

GmE

6. 67 10 6.

02 10

T s h

(d)

mcomb

4 1.

97 10

2 7

6 67 10 6 46 3 1 10

11 4

. . d . /

3

s d

2

1.

48 10

22

kg

2 2

6

13-33. (a) T 12d 6. 06 10 / s 2

T 12 24 3600

(b) For

2

m1m

2

v m1m

2

m1 m2

: reduced mass , then G and

m m r r

1 2

m m r Gm m

r

m m r r G

2 2 2 3

1 2 1 2

or m

2

1

m2

1 2

(c)

1 1 1 2 2 2 1 2 1 2

v r , v r , and from the graph v 200 km / s and v 100km / s

3

200 10 m/

s

10 10

r1 3. 3 10 m and, similarly, r

6

2

1.

6 10 m

6. 06 10 / s

10

r r r . m

1 2

4 9 10

Assuming circular orbits,

m v m v rv

2 2 2

1 1 2 2 1 2

and m1 m

2 2

r1 r2 r2v

1

Substituting yields,

m 6. 63 10 kg and m 1.

37 10 kg

30 31

1 2

341


13-34.

1 2 2 2

E mv GmM / r FG

GM m / r mv / r

2

2 1 2 1

or GM m / r mv mv GM m / r

2 2

1 GM m GM m 1 GM m

E

2 r r 2 r

13-35.

universe V

dV

20km / s

H

6

10 cy

Current average density = 1H atom / m

3

dR

4

3

3 2

V R dV 4 R dR

The current expansion rate at R is:

20km / s

10 cy

10 4 7

v HR 10 c y 20 10 km / s 20 10 m / s

6

7 7

dR 20 10 m / s 3. 16 10 s / y

6

10 y

6

10 y

2 2

2 10 15 7 7

dV 4 R dR 4 10 9. 45 10 m / c y 20 10 m / s 3. 16 10 s / y

6

10 y

6

10 y

74 3

7. 07 10 m # of H atoms

6 6 to be added

10 c y 10 c y

4

Current volume V

3

3

10 77 3

10 8.

4 10

m

74 6

7. 07 10 atoms / 10 c y

3 6

"new" H atoms = 0. 001 "new" H atoms / m 10 c y ; no

77 3

8.

4 10 m

342


13-36. (a) Equation 8-12: v 3 RT / M is used to compute v rms vs T for each gas ® = gas

constant.

Gas

(

M

3

10 kg )

rms

3 R/

M

v

rms

(m/s) at T = :

50K 200K 500K 750K 1000K

H 2 O 18 37.2 263 526 832 1020 1180

CO 2 44 23.8 168 337 532 652 753

O 2 32 27.9 197 395 624 764 883

CH 4 16 39.5 279 558 883 1080 1250

H 2 2 111.6 789 1580 2500 3060 3530

He 4 78.9 558 1770 1770 2160 2500

The escape velocities vsc

2gR 2 GM / R , where the planet masses M and

radii R, are given in table below.

Planet Earth Venus Mercury Jupiter Neptune Mars

v esc (km/s) 11.2 10.3 4.5 60.2 23.4 5.1

v esc /6 (m/s) 1870 1720 750 10,000 3900 850

On the graph of v rms vs T the v esc /6 points are shown for each planet.

343



(b) v 2GM / R v 2GM / R v 2GM / R

esc Pl Pl Pl E E E

v M / R ME

/ RE

M / M

vPl

vE

v M / R M / R R / R

Pl Pl Pl Pl E

E E E E E Pl E

(c) All six gases will still be in Jupiter’s atmosphere and Netune’s atmosphere, because

v esc for these is so high. H 2 will be gone from Earth; H 2 and probably He will be gone

from Venus; H 2 and He are gone from Mars. Only CO 2 and probably O 2 remain in

Mercury’s atmosphere.

13-37. (a) α Centauri

d

in pc

Earth's orbit radius (in AU)

sin p

1AU

sin 0. 742"

5 5

d 2. 78 10 pc 9.

06 10 c y

(b) Procyon

1AU

sin 0. 0286"

5 6

d 7. 21 10 pc 2.

35 10 c y

13-38. Earth is currently in thermal equilibrium with surface temperature 300 K . Assuming

Earth radiates as a blackbody

4 4

2

I T and I 300 459 W / m . The solar constant

is

3 2

f 1. 36 10 W / m currently, so Earth absorbs 459/1360 = 0.338 of incident solar

energy. When

2 2

L 10 L then f 10 f . If the Earth remains in equilibrium.

3 2 4

I 0. 338 1.

36 10 10 T or T 994K 676 C sufficient to boil the oceans

away. However, the v rms for H 2 O molecules at 994K is

3 8.

31 949

vrms

3RT / M 1146m / s 1. 15km / s . The v

3

esc = 11.2km/s (see

18 10

solution to problem 14-26).

atmosphere.

Because v rms ≈ 0.1 v esc , the H 2 O will remain in the

344


13-39. (a)

a

n = grains/cm 3

R

photons

N/mrs

R

dx

a

dust grain

total scattering area =

which is

2 2

R a ndx

a

2

2 2

R na dx

2

R ndx

of the

total area = fraction scattered = dN/N

N

N0

dN

N

d

2

n R dx or N N e

0

0

2

n R d

From those photons that scatter at x = 0 (N 0 ), those that have not scattered again after

traveling some distance x = L is

given by:

L

2

n R L

0

.

N N e The average value of L (= d 0 ) is

d

dNL

L dL

dL

0

L

2

0 2

0

n R

dL

L

0

dN dL

dL

1

Note:

dN

n R N e

2

n R L

d/

d0 5

(b) I I e near the Sun d 3000c y R 10 cm

0 0

1

3000c y 9. 45 10 cm / c y n 1. 1 10 / cm

n 10

17 12 3

2

5

(c)

grains

2 gm / cm

3

cm

3

m grains

4

2 10 1. 1 10 / cm 9. 41 10 gm / cm

of space 3

3

5 12 3 27 3

9. 41 10 gm / cm

M

27 3

3

17

mass in 300c y

9. 45 10 cm / c y 300

0. 0012 0. 1% M

345


13-40.

56 H 14 He Fe 2 2e 2 2e 2. 04MeV / c

1 14 56 2

1 2 26

14 m 14 4. 002603 m 55.

939395 u

4He

56Fe

56.

036442u

Net energy difference (release) =

14 26.

72MeV

2.

04MeV

90.

40MeV

466.

5MeV

2 Fe Cd 4 4e 4 4e 4. 08MeV / c

56 112 2

26 48

56 112

2 2 55. 939395 111.

902762

m Fe u m Cd u

Net energy required = 2 m m 56 112

0. 023972 u 4. 08 MeV 18.

25 MeV

Fe

Cd

13-41. (a)

dt

1 024 10

hc

4 2 2

. G M dM

4

rearranging, the mass rate of change is

dM

dt

hc

1.

024 10

4

GM

24 2 2

Clearly, the larger the mass M, the lower the rate at which the black hole loses mass.

(b)

t

2 2

4 2 11 30

1. 024 10 6. 62 10 2.

0 10

hc

4

44 37

t 3. 35 10 s 1.

06 10 y far larger than the present age of the universe.

(c)

t

2 2

4 4 11 30 12

1. 024 10 6. 67 10 2.

0 10 10

6. 63 10 3.

00 10

34 8

4

68 61

t 3. 35 10 s 1.

06 10 y

346

Tipler - Modern Physics 6th ed 2013 solutions

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