Física para Ciencias e Ingeniería. Volumen II - Douglas C. Giancoli - 4ta edición [Solucionario]

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

52. (a) Since the quality factor for protons is 1, the effective dose (in rem) is the same as the absorbed
dose (in rad). Thus the absorbed dose is 1.0 rad or 0.010 Gy.

(b) A Gy is a J per kg.

 0.010 Gy  1J kg   0.25 kg  1eV J  1p   1.31010 p
 1Gy  1.60 1019  1.2 106eV 

53. The counting rate will be 85% of 25% of the activity.

 0.035106Ci  3.7 1010 decays s   1   0.25 0.85  275.2 counts s 280 counts s
 1Ci   1decay 

54. The two definitions of roentgen are 1.6 1012 ion pairs g produced by the radiation, and the newer

definition of 0.878102 J kg deposited by the radiation. Start with the current definition, and
relate them by the value of 35 eV per ion pair.

   1R  0.878 102 J kg 1kg 1000 g 1eV 1.60 1019 J 1ion pair 35eV

 1.567 1012 ion pairs g
The two values of ion pairs per gram are within about 2% of each other.

55. We approximate the decay rate as constant, and find the time to administer 36 Gy. If that calculated

time is significantly shorter than the half-life of the isotope, then the approximation is reasonable. If
1.0 mCi delivers about 10 mGy/min, then 1.6 mCi would deliver 16 mGy/min.

dose  rate  time  time  dose 36 Gy min  1day   1.56 day  1.6 day
rate  16 103 Gy  1440 min 
 

This is only about 11% of a half life, so our approximation is reasonable.

56. Since the half-life is long (5730 yr) we will consider the activity as constant over a short period of
time. Use the definition of the curie from Section 42-6.

 2.00106 Ci  3.701010 decays s   7.40 104 decays s dN  N  ln 2 N 
 1Ci  dt T1/ 2


dN T1/2   5730 y 
dt ln 2  ln 2 
   N 
7.4104 decays s 3.156107 s y  1.9311016 nuclei

1.9311016 nuclei  0.014 kg   4.491010 kg
 6.021023 nuclei 

57. (a) According to Appendix F, I131 decays by beta decay.

53

I131  131 Xe    + v
54
53

(b) The number of nuclei present is given by Eq. 41-6.

ln N T1/ 2 ln N 8.0d ln 0.070
t   N0 N0
N  N0et      30.69d  31d
 ln 2 ln 2

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601

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(c) The activity is given by dN dt   N. This can be used to find the number of nuclei, and then

the mass can be found. Note that the numeric value of dN dt is negative since the number of

undecayed nuclei is decreasing.

dN   N 
dt

  N
dN dt  T1/ 2   dN dt  8.0 d 86400s d 1 103Ci 3.70 1010 decays s

 ln 2  ln 2

 3.69 1013 nuclei

3.69  1013 nuclei  0.131 kg   8 1012 kg  8 ng
 6.02 1023nuclei 

58. The activity is converted to decays per day, then to energy per year, and finally to a dose per year.
The potassium decays by gammas and betas, according to Appendix F. Gammas and betas have a
quality factor of 1, so the number of Sv is the same as the number of Gy, and the number of rem is
the same as the number of rad.

 2000 1012 Ci   3.70 1010 decays s  12 hr   3600s   0.5 L   1.598 106 decays
 L   1Ci   hr   day  day

1.598 106 decays  365 day  0.10 1.5 MeV   1.60 1013J   1.40 105 J
day   yr  decay   MeV  yr

(a) For the adult, use a mass of 60 kg.

Effective dose  1.40 105 J  1  1Gy  1Sv 
yr   60 kg  1J kg  1Gy 

 2.33107 Sv yr  105 mrem   2.33  102 mrem yr
 Sv 
 

 2 107 Sv yr or 2 102 mrem yr

2.33102 mrem
year
fraction of allowed dose  100 mrem  2 104 times the allowed dose

year

(b) For the baby, the only difference is that the mass is 10 times smaller, so the effective dose is 10
times bigger. The results are as follows.

 2 106 Sv yr , 0.2 mrem yr , and 2 103 times the allowed dose

59. Each decay releases one gamma ray of energy 122 keV. Half of that energy is deposited in the body.
The activity tells at what rate the gamma rays are released into the body. We assume the activity is
constant.

   1.55106 Ci s  1
 3.70 1010 Ci  86400 s day0.50122 keV  1.601016 J keV  58kg 

 8.338107 J kg  8.3107 Gy
day day

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602

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

60. We use the dose, the mass of the beef, and the energy per electron to find the number of electrons.

4.5 103 Gy  1J kg   5 kg   1MeV J  1e   11017 e
 1Gy   1.60 1013 1.2 MeV 

61. (a) The reaction has Z = 86 and A = 222 for the parent nucleus. The alpha has Z = 2 and A = 4, so

the daughter nucleus must have Z = 84 and A = 218. That makes the daughter nucleus Po.218

84

(b) From Figure 41-12, polonium-218 is radioactive . It decays via both alpha and beta decay ,

each with a half-life of 3.1 minutes .

(c) The daughter nucleus is not a noble gas, so it is chemically reacting . It is in the same group as

oxygen, so it might react with many other elements chemically.

(d) The activity is given by Eq. 41-7a, R  N  ln 2 N .
T1/ 2

ln 2 1.6109 g 6.021023 nuclei
 R  ln 2 N 
T1/2 3.8235d 86400s d 222 g

 9.104106 decays s  9.1106 Bq  0.25 mCi

To find the activity after 1 month, use Eq. 41-7d.

 ln2t  ln 2 30d
R  R0e T1/2  9.104106 decays s e 3.8235d  3.956104 decays s

 4.0104 Bq  1.1Ci

62. (a) For parallel rays, the object and the image will be the same size, and so the magnification is 1.

(b) When the film is pressed against the back, the image of the back on

the film will be the same size as the back, since there is no appreciable d1 d2
spreading of the rays from the back to the film. So mback = 1. But,
from the diagram, we see that the rays which define the boundary of

the area on the chest will have a much larger image with the film at the

back. The height of the image is proportional to the distance from the

point source, since the rays travel in straight lines. h1 h2

mfront  h2  d1  d2 1 d2  1 25  2.67
h1 d1 d1 15

So the range of magnifications is 1  m  2.67, depending on which

part of the body is being imaged.

63. The frequency is given in Example 42-14 to be 42.58 MHz. Use that to find the wavelength.

c f    c  2.998 108 m s  7.041m
f 42.58 106 Hz

This lies in the radio wave portion of the spectrum.

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603

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

64. We use Eq. 42-12, but with the neutron’s magnetic moment.

hf  2μn BT 

BT  hf  hf  hf eh  2 fmp e
2μ n
20.7023 μN 2  0.7023  0.7023

4 mp

   2 42.58106 Hz 1.6731027 kg  3.978 T
  0.7023 1.602 1019 C

65. (a) The reaction is 9 Be  4 He  n?. There are 6 protons and 13 nucleons in the reactants, and
4 2

so there must be 6 protons and 13 nucleons in the products. The neutron is 1 nucleon, so the

other product must have 6 protons ad 12 nucleons. Thus it is C12 .

6

(b) Q  m c2 9  m c2 4  mnc2  m c212
4 Be 2 He 6 C

 9.012182 u  4.002603 u  1.008665 u  12.000000 u  931.5 MeV c2  c 2  5.701MeV
u 


66. The energy and temperature are related by the Boltzmann constant, which has units of

energy/temperature.

k  1.381  1023 J 1eV 1 keV 8.620 108 keV K
K  1.602 1019 J  1000 eV 

67. From Eq. 18-5, the average speed of a gas molecule (root mean square speed) is inversely
proportional to the square root of the mass of the molecule, if the temperature is constant. We

assume that the two gases are in the same environment and so at the same temperature. We use UF6
molecules for the calculations.

v 235 UF6  m 238 UF6  238  619  1.0043 :1
235  619
92 92

v 238 UF6 m 235 UF6

92 92

68. (a) We assume that the energy produced by the fission was 200 MeV per fission, as in Eq. 42-6.

 20 kilotons TNT   5 1012J   1 MeV   1fission atom   0.235 kg 
 1 kiloton   1.60 1013J   200 MeV   6.02 1023 atom 

 1.220 kg  1kg

(b) Use E  mc2 .

E  20 kilotons TNT   5 1012J 
c2  1kiloton 
 E  mc2  m    1.11103 kg  1g
3.0 108 m s 2

This is consistent with the result found in Problem 26.

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604

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

69. The effective dose (in rem) is equal to the actual dose (in rad) times the quality factor, from Eq. 42-
11a.

dose rem  29 mrad yr X-ray,  -ray 1  3.6 mrad yr 10  65mrem yr

70. The oceans cover about 70% of the Earth, to an average depth of approximately 4 km. The density

of the water is approximately 1000 kg m3 . Find the volume of water using the surface area of the

Earth. Then convert that volume of water to mass, to the number of water molecules, to the number
of hydrogen atoms, and then finally to the number of deuterium atoms using the natural abundance
of deuterium from Appendix F.

   Mass of water  surface area depth density  4 6.38 106 m 2 4000m 1000 kg m3

 2.05 1021kg water

 2.051021kg water  6.02 1023molecules  2 H atoms  0.000115 d atoms 
 0.018 kg water  1 molecule  1 H atom 

 1.58 1043d  2 1043d atoms

 1.58 1043 d  2 103 kg   5 1016 kg d
 6.02 1023d atoms 

From Eqs. 42-9a and 42-9b, if the two reactions are carried out at the same rate then 4 deuterons

would produce 7.23 MeV of energy. Use that relationship to convert the number of deuterons in the

oceans to energy.

 1.58 1043d  7.23 MeV 1.60 1013J  4.57 1030 J  5 1030 J

4 d 1MeV

71. Because the quality factor for gamma rays is 1, the dose in rem is equal in number to the dose in rad.
Since the intensity falls off as r2, the square of the distance, the exposure rate times r2 is constant.

Allowed dose  5 rem  1rad   1 year  1week   2.747 103 rad
year  1rem   52 weeks   35 hours  hour
   

 2.747 103 rad  r2   5.2 102 rad  1m2 
hour hour

 5.2 102 rad  1m2
hour
r  4.351m  4.4 m
 rad 
2.747 103 hour

72. (a) The reaction is of the form ?  4 He  222 Rn . There are 88 protons and 226 nucleons as
2 86

products, so there must be 88 protons and 226 nucleons as reactants. Thus the parent nucleus is

226 Ra .
88

226 Ra  4 He  222 Rn
88 2 86

(b) If we ignore the kinetic energy of the daughter nucleus, then the kinetic energy of the alpha

particle is the Q-value of the reaction.

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605

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

K  m c  m c  m c2 2 2

 226 Ra 4 He 222 Rn
88 2 86

 226.025410 u  4.002603 u  222.017578 u 931.5 MeV c2  c2  4.871 MeV
u 

(c) From momentum conservation, the momentum of the alpha particle will be equal in magnitude
to the momentum of the daughter particle. At the energy above, the alpha particle is not

relativistic, and so K  p2  p  2m K .
2m

p  2maKa  2 4.002603u  931.5 MeV c2  4.871MeV  191MeV c
 1u 

(d) Since p  p ,daughter K daughter 2  p2 .
2mdaughter
 pdaughter
2mdaughter

K daughter  p2  191MeV c2  8.82 102 MeV
2mdaughter
2 222 u   931.5 MeV c2 
 1u 

Thus we see that our original assumption of ignoring the kinetic energy of the daughter nucleus

is valid. The kinetic energy of the daughter is less than 2% of the Q-value.

73. (a) The mass of fuel can be found by converting the power to energy to number of nuclei to mass.

 2400 106 J s 1 y   3.156  107s   1 MeV   1 fission atom   0.235 kg 
 1 y   1.60 10 J13   200 MeV   6.02 1023atom 

 9.240 104 kg  920 kg

(b) The product of the first 5 factors above gives the number of U atoms that fission.

 #Sr atoms  0.06  3.156  107s   1MeV   1fission atom 
2400 106 J s 1 y  1y   1.60 1013J   200 MeV 

 1.42 1026Sr atoms
The activity is given by Eq. 41-7a.

  N  ln 2 N  ln 2
3.156 107 s yr
T1/ 2 29 yr 
dN

   dt
1.42 1026  1.076 1017 decays s

   1Ci  106 3 106 Ci
1.076 1017 decays s  3.70 1010 decays s   2.91  Ci 

74. This “heat of combustion” is 26.2 MeV / 4 hydrogen atoms.

26.2 MeV  1.60 1013 J   1H atom   1u   6.26 1014 J kg
4 H atoms  1 MeV   1.0078 u    1027 
    1.66 kg

This is about 2 107 times the heat of combustion of coal.

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606

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

75. (a) The energy is radiated uniformly over a sphere with a radius equal to the orbit radius of the
Earth.

   1300 W m2 4 1.496 1011m 2  3.656 1026 W  3.7 1026 W

(b) The reaction of Eq. 42-8 releases 26.2 MeV for every 4 protons consumed, assuming we ignore
the energy carried away with the neutrinos.

 3.656  1026 J   4 protons   1 MeV J   3.489  1038 protons s
 s   26.2 MeV   1.60 1013 
 

 3.5 1038 protons s

(c) Convert the Sun’s mass to a number of protons, and then use the above result to estimate the
Sun’s lifetime.

2.0  1030 kg  1 proton kg  1s  1 yr   1.1 1011 yr
 1.673 1027   3.489 1038 protons   3.156 107 

76. For the net proton cycle, Eq. 42-8, we see that there are two neutrinos produced for every four
protons consumed. Thus the net number of neutrinos generated per second from the sun is just half
the value of protons consumed per second. That proton consumption rate is calculated in Problem
75b.

 3.489 1038 protons s  2   1.745  1038  s
 4p 

We assume the neutrinos are spread out uniformly over a sphere centered at the Sun. So the fraction

that would pass through the area of the ceiling can be found by a ratio of areas, assuming the ceiling

is perpendicular to the neutrino flux. But since the window is not perpendicular, a cosine factor is

included to account for the angle difference, as discussed in Eq. 22-1a. Finally, we adjust for the

one-hour duration, assuming the relative angle is constant over that hour.

180 m2
1.496 1011 m
   1.7451038s 4 2 cos 38 3600 s  3.2 1020 v

77. We use the common value of 200 MeV of energy released per fission. We then multiply that by the
number of fissions, which we take as 5.0% of the number of U-238 atoms.

 200 MeV  1.602 1013 J   U235 nuclei  
1  1MeV   nuclei  
 Total 235 U  0.05 92  2.0 kg 238 U 
nucleus of 92 92 
U238 
energy  
92

  6.022  1023 nuclei of 238 U nuclei 
  0.238 kg 92 
  238 
92 U

 8.107 1012 J  8 1012 J

78. (a) The energy released is given by the Q-value.

 Q 2  MeV c2  2
 2m c2  m c24  2 12.000000 u  23.985042 u  931.5 u  c  13.93 MeV
12
6 12
C Mg

(b) The total kinetic energy of the two nuclei must equal their potential energy when separated by

6.0 fm.

2K  1 q1q2 
40 r

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607

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

6 1.60 1019 C 2  1MeV 
   K 1 q1q2
 1 4 0 r  1 8.99 109 Nm2 C2 6.0 1015m  1.60 1013 J 
2 2

 4.315 MeV  4.3 MeV

(c) The kinetic energy and temperature are related by Eq. 18-4.

 4.315 MeV  1.60 1013 J 
 1 MeV 
K  3 kT  T  2 K  2  3.3 1010 K
2 3 k
3 1.38 1023 J K

79. (a) A Curie is 3.7 1010 decays s.

  0.10 106 Ci 3.7 1010 decays s  3700 decays s

(b) The beta particles have a quality factor of 1. We calculate the dose in gray and then convert to
sieverts. The half life is over a billion years, so we assume the activity is constant.

   3700 decays s 1.4 MeV decay 1.60 1013 J MeV 1
3.156 107 s y  55 kg 

 4.756 104 J kg y  4.756 104 Gy y  4.8 104 Sv y

This is about 4.756 104 Sv y  0.13 or 13% of the background rate.
3.6 103 Sv y

80. The surface area of a sphere is 4 r2 .

  Activity
 m2
 2.0 107 Ci  2.0 107 Ci 3.7 1010 decays s  1400 decays s
4 r2 4 6.38 106 m 2 m2

Earth

 81.  MeV c2 
Q  3m4 c2  m c2  34.002603 u  12.000000 u  931.5 u  c 2  7.274 MeV

2 He 12 C 
6

82. Since the half-life is 30 years, we assume that the activity does not change during the 2.0 hours of
exposure. We calculate the total energy absorbed, and then calculate the effective dose. The two
energies can be added directly since the quality factor for both gammas and betas is about 1.

    1010 decays   3600 s  
s  1 hr  

Energy  
1.2 106 Ci 3.7 1.6 hr 

 eV J   3.478 105 J
 decay eV 
850  103   1.60×10-19  


dose  3.478 105 J  100rad  5.351105 rad  5.4 105 rem
65 kg 1J kg

83. The half life of the strontium isotope is 28.79 years. Use that with Eq. 41-7c to find the time for the

activity to be reduced to 15% of its initial value.

R  R0et  t   1 ln R   T1/2 ln R  28.79 y ln 0.15  79 y
 R0 ln 2 R0

ln 2

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608

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

84. Source B is more dangerous than source A because of its higher energy. Since both sources have the
same activity, they both emit the same number of gammas. Source B can deposit twice as much
energy per gamma and therefore cause more biological damage.

Source C is more dangerous than source B because the alphas have a quality factor up to 20 times
larger than the gammas. Thus a number of alphas may have an effective dose up to 20 times higher
than the effective dose of the same number of like-energy gammas.

So from most dangerous to least dangerous, the ranking of the sources is C > B > A .

We might say that source B is twice as dangerous as source A, and source C is 20 times more
dangerous than source B.

85. The whole-body dose can be converted into a number of decays, which would be the maximum
number of nuclei that could be in the Tc sample. The quality factor of gammas is 1.

 50 mrem  50 mrad   1J kg   1eV 
50 103 rad  100 rad   60 kg   1.60 1019 J   1.875  1017 eV
 

 1.875 1017 eV  1 effective   2 decays   1 nucleus   2.679  1012 nuclei
 140 103eV   1 effective    1 decay 
 

This is the total number of decays that will occur. The activity for this number of nuclei can be

calculated from Eq. 41-7a.

 R
 N  ln 2 N  ln 2 2.6791012 decays  1Ci   2.32 103 Ci  2 mCi
T1/ 2  3.70 1010 decays 
6 h3600 s h s

86. The number of 60 Co nuclei  N60  can be calculated from the activity and the half-life of 60 Co. The
27 27

number of 60 Co nuclei can also be calculated from the cross section and the number of 59 Co nuclei
27 27

 N59  present in the paint. By combining these two calculations, we can find the number of 59 Co in
27

the paint. We assume that, since the half life is relatively long, that all 60 Co are made initially,
27

without any of them decaying.

dN 60   N60  ln 2 N 60  N 60  dN 60 T1/ 2
dt T1/ 2 dt ln 2

We assume the paint is thin and so use Eq. 42-3 for the cross section. Let R represent the rate at

which 60 Co nuclei are made (i.e., the collision rate), and t the elapsed time of neutron bombardment.
27

R0 is the rate at which the neutrons hit the painting, and so is the given neutron flux times the area of
the painting.

N 60  Rt  R0nl t  R0  N59 paint  volume of paint   t  R0 N59 t
 volume of   surface area of paint  A

  neutron flux  AN59 t   neutron flux  N59 t 

A

N59  N 60  dN 60 T1/ 2
dt
neutron flux  t ln 2 neutron flux  t

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609

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

    55decays
s  5.27 yr  3.156 107 s yr

 ln 2   5.0 1012   104 cm2  19 1028 m 2 300 s
 cm2 s   m2 

  4.6311017 atoms  58.933 g 
 m59  4.6311017 atoms  6.022 1023 atoms   4.5 105 g

87. Since all speeds are relativistic, we may use p  2mK to relate momentum and kinetic energy. If

we assume the target is at rest, then the total momentum of the products must equal the momentum
of the bombarding particle. The total (kinetic) energy of the products comes from the kinetic energy
of the bombarding particle and the Q-value of the reaction.

ppr  pb  2mpr Kpr  2mb Kbr  K pr  mb Kb  Q  Kb  Kb   Qmpr
mpr mpr  mb

88. (a) We assume a thin target, and use Eq. 42-3.

   R  R 1  1.6 105
  R0nl R0 nl
1 4.0 107 m  1bn   6.8 bn
5.9 1028 m3  1028 m2 

(b) We assume that the cross section is the area presented by the gold nucleus.

  r2  1  d 2  d 46.8 bn    1bn   2.9 1014 m
4  1028 m 
 2

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610

CHAPTER 43: Elementary Particles

Responses to Questions

1. p + n → p + p + – .

2. No. In the rest frame of the proton, this decay is energetically impossible, and so it is impossible in
every other frame as well. In a frame in which the proton is moving very fast, the decay products
must be moving very fast as well to conserve momentum. With this constraint, there will not be
enough energy to make the decay possible.

3. “Antiatoms” would be made up of antiprotons, antineutrons, and positrons. If antimatter came into
contact with matter, the corresponding pairs of particles would annihilate, producing energy in the
form of gamma rays and other particles.

4. The photon signals the electromagnetic interaction.

5. (a) Yes, if a neutrino is produced during a decay, the weak interaction is responsible.
(b) No, for example, a weak interaction decay could produce a Z0 instead of a neutrino.

6. The neutron decay process also produces an electron and an antineutrino; these two particles only
interact via the weak force. In addition, the strong force dominates at extremely small distances and
at other distances the weak force dominates.

7. An electron takes part in electromagnetic, weak, and gravitational interactions. A neutrino takes part

in weak and gravitational interactions. A proton takes part in all four interactions: strong,
electromagnetic, weak, and gravitational.

8. All of the gauge bosons, leptons, and mesons have baryon number equal to zero and there are no
baryons produced in the decays, so baryon number is conserved for these groups. The chart below
shows charge conservation for a few examples of decays in these groups.

Particle name Decay Charge conservation check
W +1 = +1 + 0
muon W   e  e –1 = –1 + 0 + 0
tau    e  e   –1 = –1 + 0 + 0

pion         –1 = –1 + 0 + 0
eta    e  e  +1 = +1 + 0
       0=0+0+0
 0  3 o
0   0 0 = +1 –1 + 0

The next chart shows charge conservation and baryon number conservation for a sample of the
baryon decays shown in Table 43-2.

Particle name Decay Charge conservation check Baryon number (B)
neutron 0 = +1 –1 + 0 conservation check
n  p  e e
+1 = +1 + 0 + 0

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611

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

sigma   p  o +1 = +1 + 0 +1 = +1 + 0
  n   +1 = 0 + 1 +1 = +1 + 0

9. Decays via the electromagnetic interaction are indicated by the production of photons. In Table 43-2,
the decays of 0, Σ0, and η0 proceed via the electromagnetic interaction.

10. All of the decays listed in Table 43-2 in which a neutrino or antineutrino is one of the decay products
occur via the weak interaction. These include the W, muon, tau, pion, kaon, and neutron. In addition,
the Z particle decays via the weak interaction.

11. The Δ baryon has a baryon number of one and is therefore made of three quarks. The u, c, and t
quarks have a charge of +2/3 e each and the d, s, and b quarks have a charge of -1/3 e each. There is
no way to combine three quarks for a total charge of -2e.

12. As evidenced by their shorter lifetimes, the J/ψ and Υ particles decay via the electromagnetic
interaction.

13. Based on the lifetimes listed, all of the particles in Table 43-4, except the J/ψ and the Υ, decay via
the weak interaction.

14. Baryons are formed from three quarks or antiquarks, each of spin 1 or  1 , respectively. Any
2 2

combination of quarks and antiquarks will yield a spin magnitude of either 1 or 3 . Mesons are
2 2

formed from two quarks or antiquarks. Any combination of two quarks or antiquarks will yield a

spin magnitude of either 0 or 1.

15. The “neutrinolet” would not interact via the gravitational force (no mass), the strong force (no color
charge), or the electromagnetic force (no electrical charge). In addition, it does not feel the weak
force. However, it could possibly exist. The photon, for example, also has no rest mass, color charge,
or electric charge and does not feel the weak force.

16. (a) No. Leptons are fundamental particles with no known internal structure. Baryons are made up
of three quarks.

(b) Yes. All baryons are hadrons.
(c) No. A meson is a quark–antiquark pair.
(d) No. Hadrons are made up of quarks and leptons are fundamental particles.

17. No. A particle made up of two quarks would have a particular color. Three quarks or a quark–
antiquark pair are necessary for the particle to be white or colorless. A combination of two quarks
and two antiquarks is possible, as the resulting particle could be white or colorless.

18. Inside the nucleus, a neutron will not decay because the dominant interaction is the strong interaction
with the other nucleons. A free neutron will decay through the weak interaction.

19. The reaction is not possible, because it does not conserve lepton number. L = 1 on the left-hand side
of the reaction equation, and L = –1 on the right-hand side of the reaction equation.

20. The reaction proceeds by the weak force. We know this because an electron anti-neutrino is
produced in the reaction, which only happens in reactions governed by the weak interaction.

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612

Chapter 43 Elementary Particles

Solutions to Problems

1. The total energy is given by Eq. 36-11a.
E  m0c2  K  0.938 GeV  4.65GeV  5.59 GeV

2. Because the energy of the electrons is much greater than their rest mass, we have K  E  pc.

Combine that with Eq. 43-1 for the de Broglie wavelength.

  E  pc ; p  h
  
 E  hc    hc  6.631034 Js 3.00 108 m s  4.4 1017 m
 E 28 109 eV 1.60 1019 J eV

3. The frequency is related to the magnetic field in Eq. 43-2.

  f  qB
2 m
 B  2mf 2 1.67 1027 kg 3.1107 Hz  2.0T
q  1.60 1019 C

4. The time for one revolution is the period of revolution, which is the circumference of the orbit

divided by the speed of the protons. Since the protons have very high energy, their speed is

essentially the speed of light.

 T
 2 r  2 1.0103 m  2.1105 s
v 3.0108 m s

5. Use Eq. 43-2 to calculate the frequency. The alpha particle has a charge of +2e and a mass of 4

times the proton mass.

 f 
qB 2 1.601019 C 1.7 T  1.3107 Hz  13MHz

 2 m 2 4 1.67 1027 kg 

6. (a) The maximum kinetic energy is K  q2B2R2  1 mv2 . Compared to Example 43-2, the charge
2m 2

has been doubled and the mass has been multiplied by 4. These two effects cancel each other in

the equation, and so the maximum kinetic energy is unchanged. The kinetic energy from that

example was 8.653 MeV.

K  8.7 MeV

   v 
2K 28.653MeV 1.60 1013 J MeV  2.042 107 m s  2.0 107 m s

4 1.66 1027 kg
m

(b) The maximum kinetic energy is K  q2B2R2  1 mv2 . Compared to Example 43-2, the charge
2m 2

is unchanged and the mass has been multiplied by 2. Thus the kinetic energy will be half of what

it was in Example 43-2 (8.653 MeV).

K  4.3MeV

   v  2 
2K 1  8.653 MeV 1.60 1013 J MeV  2.042 107 m s  2.0 107 m s
m 2

2 1.66 1027 kg

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613

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

The alpha and the deuteron have the same charge to mass ratio, and so move at the same speed.
(c) The frequency is given by f  qB . Since the charge to mass ratio of both the alpha and the

2 m
deuteron is half that of the proton, the frequency for the alpha and the deuteron will both be half
the frequency found in Example 43-2 for the proton.

f  13MHz

7. From Eq. 41-1, the diameter of a nucleon is about dnucleon  2.41015 m. The 25-MeV alpha particles

and protons are not relativistic, so their momentum is given by p  mv  2mK . The wavelength is

given by Eq. 43-1,   h  h.
p 2mK

h 6.631034 Js  2.88 1015 m
 1.6 1019 J eV
24 1.66 1027 kg 25 106eV
2 mKE
    

h 6.63 1034 Js  5.75 1015


2 mpKE
   p 
2 1.67 1027 kg 25 106eV 1.6 1019 J eV

We see that   dnucleon and p  2dnucleon . Thus the alpha particle will be better for picking out

details in the nucleus.

8. Because the energy of the protons is much greater than their rest mass, we have K  E  pc.

Combine this with the expression (given above Example 43-2) relating the momentum and radius of
curvature for a particle in a magnetic field.

v  qBr  mv  qBr  p  qBr  E  qBr 
mc

  B  E 
   qrc
1.01012 eV 1.601019 J eV  3.3T
1.601019 C 1.0103 m 3.00108 m s

9. Because the energy of the protons is much greater than their rest mass, we have K  E  pc. A

relationship for the magnetic field is given right before Eq. 43-2.

v  qBr  mv  qBr  p  qBr  E  qBr 
mc

  B  E 
   qrc
7.01015 eV 1.601019 J eV  5.5T
1.601019 C 4.25103 m 3.00108 m s

10. (a) The magnetic field is found from the maximum kinetic energy as derived in Example 43-2.

K  q2B2R2  B  2mK 

2m qR

   22.014 1.661027 kg 12106 eV 1.601019 J eV  0.7082T  0.71T
 B 
1.601019 C 1.0 m

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614

Chapter 43 Elementary Particles

(b) The cyclotron frequency is given by Eq. 43-2.

   f
 qB 1.60 1019C 0.7082T  5.394 106 Hz  5.4 MHz
2 m  2 2.014 1.66 1027 kg

(c) The deuteron will be accelerated twice per revolution, and so will gain energy equal to twice its
charge times the voltage on each revolution.

12106 eV
1.601019 C 22103 V
    number
of revolutions  n  2 1.601019 J eV

 273 revolutions  270 revolutions

(d) The time is the number of revolutions divided by the frequency (which is revolutions per
second).

t  n  273 revolutions  5.1  106 s  51s
f 5.394 106 rev s

(e) If we use an average radius of half the radius of the cyclotron, then the distance traveled is the
average circumference times the number of revolutions.

distance  1 2 rn  1.0 m273  860 m
2

11. Because the energy of the protons is much greater than their rest mass, we have K  E  pc.

Combine that with Eq. 43-1 for the de Broglie wavelength. That is the minimum size that protons of

that energy could resolve.

  E  pc ; p  h
  
 E  hc    hc  6.631034 Js 3.0108 m s  1.81019 m
 E 7.01012 eV 1.601019 J eV

12. If the speed of the protons is c, then the time for one revolution is found from uniform circular

motion. The number of revolutions is the total time divided by the time for one revolution. The

energy per revolution is the total energy gained divided by the number of revolutions.

v  2 r  T  2 r  2 r n  t  ct
T vc T 2 r

     Energy
revolution  E  E 2 r  1.0 106 MeV 150 103 MeV 2 1.0 103m

n ct 3.00 108 m s 20s

 0.89 MeV rev  0.9 MeV rev

13. Start with an expression from Section 42-1, relating the momentum and radius of curvature for a

particle in a magnetic field, with q replaced by e.

v  eBr  mv  eBr  p  eBr
m

In the relativistic limit, pE c and so E  eBr . To put the energy in electron volts, divide the
c

energy by the charge of the object.

E  eBr  E  Brc
ce

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

14. The energy released is the difference in the mass energy between the products and the reactant.

E  m0 c2  mnc2  m c 2  1115.7 MeV  939.6 MeV 135.0 MeV  41.1MeV

0

15. The energy released is the difference in the mass energy between the products and the reactant.

E  m c2  m c 2  m c2  139.6 MeV 105.7 MeV  0  33.9 MeV
 
 

16. Use Eq. 43-3 to estimate the range of the force based on the mass of the mediating particle.

  mc2  hc
  2d
 d hc  2 6.631034 Js 3.00 108 m s  3.98 1016 m
 2 mc2 497.7 106eV 1.60 1019 J eV

17. The energy required is the mass energy of the two particles.

E  2mnc2  2939.6 MeV  1879.2 MeV

18. The reaction is multi-step, and can be written as shown here: 0  0 +

The energy released is the initial rest energy minus the  p   
final rest energy of the proton and pion, using Table 43-2.

 E  c2  1192.6 Mev  938.3MeV 139.6 MeV  114.7 MeV
m0  mp  m 


19. Because the two protons are heading towards each other with the same speed, the total momentum of
the system is 0. The minimum kinetic energy for the collision would result in all three particles at
rest, and so the minimum kinetic energy of the collision must be equal to the mass energy of the  0 .

Each proton will have half of that kinetic energy. From Table 43-2, the mass of the  0 is

135.0 MeV c2 .

2Kproton  m c 2  135.0 MeV  Kproton  67.5 MeV

0

20. Because the two neutrons are heading towards each other with the same speed, the total momentum
of the system is 0. The minimum kinetic energy for the collision would result in all four particles at
rest, and so the minimum kinetic energy of the collision must be equal to the mass energy of the

KK pair. Each neutron will have half of that kinetic energy. From Table 43-2, the mass of each
of the K and the K is 493.7 MeV c2 .

2Kneutron  2mKc2  Kneutron  mKc2  493.7 MeV

21. We treat the neutrino as massless, but it still has momentum and energy. We use conservation of
momentum and conservation of energy, along with Eqs. 36-11 and 36-13.
(a) To find the maximum kinetic energy of the positron, we assume that the pion has no kinetic
energy, and so the magnitude of the momenta of the positron and the neutrino are the same.

pe  pv ;

 mKc2  m c2  Ee  Ev  mKc2  m c2  Ee 2  Ev2  pv2c2  pe2c2  Ee2  me2c4 

mKc2  m c2 2  me2c4
2 mKc2  m c2
   Ee   Ke  mec2 

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Chapter 43 Elementary Particles

mKc2  m c2 2  me2c4
2 mKc2  m c2
   Ke   mec2

 497.7MeV 139.6 MeV2  0.511MeV2  0.511MeV  178.5 MeV
2497.7MeV 139.6 MeV

(b) To find the maximum kinetic energy of the pion, we assume that the positron has no kinetic
energy, and so the magnitude of the momenta of the positron and the neutrino are the same.
The result is found by just interchanging the pion and positron.

mKc2  mec2 2  m2c4
2 mKc2  mec2
   Ke   m c2

 497.7MeV  0.511MeV2  139.6 MeV2  139.6 MeV  128.6 MeV
2497.7MeV  0.511MeV

22. The energy of the two photons (assumed to be equal so that momentum is conserved) must be the

combined rest mass energy of the proton and antiproton.

  2m0c2
 2hf  2h c  hc 6.631034 Js 3.00 108 m s  1.32 1015 m
   m0c2  938.3106eV 1.60 1019 J eV

23. (a) 0  n    Charge conservation is violated, since 0  0 1
(b) 0  p  K Strangeness is violated, since 1  0  0
(c) 0   +   
Energy conservation is violated, since

1115.7 MeV c2  938.3MeV c2  493.7 MeV c2  1432.0 MeV c2

Baryon number conservation is violated, since 1  0  0

Strangeness is violated, since 1  0  0

Spin is violated, since 1 00
2

24. (a) The Q-value is the mass energy of the reactants minus the mass energy of the products.

 Q  m0 c2 
mpc2  m c2  1115.7 MeV  938.3MeV 139.6 MeV  37.8 MeV


(b) Energy conservation for the decay gives the following.

m0 c2  Ep  E  E  m0 c2  Ep
 

Momentum conservation says that the magnitudes of the momenta of the two products are

equal. Then convert that relationship to energy using E2  p2c2  m02c4 , with energy
conservation.

ppc 2  p c2

   ppp  


 Ep2 E2 m2 2
 mp2c4     c 4  m0 c2  Ep  m2 c4


 Ep2  mp2c4  Ep2 4
m c2 4  2Epm0 c2   m2 c 
0 

Ep  m c2 4  mp2c4  m2 c4  1115.7 MeV2  938.3MeV2  139.6 MeV2  943.7 MeV
0  21115.7 MeV

2m0 c2

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617

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

E  m0 c2  Ep  1115.7 MeV  943.7 MeV  172.0 MeV


Kp  Ep  mpc2  943.7 MeV  938.3MeV  5.4 MeV

K  E  m c 2  172.0 MeV 139.6 MeV  32.4 MeV
  


25. (a) We work in the rest frame of the isolated electron, so that it is initially at rest. Energy
conservation gives the following.

mec2  Ke  mec2  E  Ke  E  Ke  E  0

Since the photon has no energy, it does not exist, and so has not been emitted.
(b) For the photon exchange in Figure 43-8, the photon exists for such a short time that the

uncertainty principle allows energy to not be conserved during the exchange.

26. The total momentum of the electron and positron is 0, and so the total momentum of the two photons

must be 0. Thus each photon has the same momentum, and so each photon also has the same energy.

The total energy of the photons must be the total energy of the electron / positron pair.

 E  Ee+/e pair  2hf  2h c
photons  2 m0c2  KE  

  hc
    m0c2  KE 
6.631034 Js 3.00 108 m s  1.335 1012 m  1.3 1012 m
0.511106eV  420 103eV 1.60 1019 J eV

27. Since the pion decays from rest, the momentum before the decay is zero. Thus the momentum after
the decay is also zero, and so the magnitudes of the momenta of the positron and the neutrino are
equal. We also treat the neutrino as massless. Use energy and momentum conservation along with
the relativistic relationship between energy and momentum.

   m c2 4 E2

 E  E ; p  p  p2 c2  p2c2  E2  m2 c 
e e e e e

 E2 2 E2 2E m c2
e e e  
 m2 c4  m c2  E  m2 c4  2 E m  c2    m2 c4  m2 c4
e  e  e   e

E  m c1 2  m2 c2  K  m c 2  m c1 2  m2 c2 
e e e e e
2  2 
2m 2m
 

  K
 e
 1 m c2  m c2  m2 c2  1 139.6 MeV  0.511MeV  0.511MeV c2 0.511MeV
2  e e 2
2 139.6 MeV c2
2m


 69.3MeV

28. (a) For the reaction    p  n 0 , the conservation laws are as follows.

Charge: 11  0  0 Charge is conserved.
Baryon number: 0 1 1 0 Baryon number is conserved.
Lepton number: 0  0  0  0 Lepton number is conserved.
Strangeness: 0  0  0  0 Strangeness is conserved.

The reaction is possible.

(b) For the reaction    p  n   0 , the conservation laws are as follows.

Charge: 11  0  0 Charge is NOT conserved.

The reaction is forbidden, because charge is not conserved.

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618

Chapter 43 Elementary Particles

(c) For the reaction    p  p  e , the conservation laws are as follows.

Charge: 11  11 Charge is conserved.
Baryon number: 0 1 1 0 Baryon number is conserved.

Lepton number: 0  0  0 1 Lepton number is NOT conserved.

The reaction is forbidden, because lepton number is not conserved.

(d) For the reaction p  e  ve , the conservation laws are as follows.

Charge: 1 1 0 Charge is conserved.

Baryon number: 1  0  0 Baryon number NOT conserved.

Mass energy is fine, because mp  me  mv.

The reaction is forbidden, because baryon number is not conserved.

(e) For the reaction   e   , the conservation laws are as follows.

Charge: 1  1 0 Charge is conserved.

Baryon number: 0  0  0 Baryon number is conserved.

Electron lepton number: 0  1 0 Lepton number is NOT conserved.

Mass energy is fine, because m  me  mv.

The reaction is forbidden, because lepton number is not conserved.

(f) For the reaction p  n  e  ve , the conservation laws are as follows.
Mass energy: 938.3MeV c2  939.6 MeV c2  0.511MeV c2
Mass energy is NOT conserved.

The reaction is forbidden, because energy is not conserved.

29. Since the  decays from rest, the momentum before the decay is zero. Thus the momentum after

the decay is also zero, and so the momenta of the 0 and   are equal in magnitude. Use energy

and momentum conservation along with the relativistic relationship between energy and momentum.

m c2  E0  E   E  m c2  E0
   

p0 c 2  p c2

   p0p  


 E2 c4 m2 2 m2 c4
0  
 m2  E2  c 4  m c2  E0 
0  

 E2 m2
0 0
 c4  m2 c4  2 E0 m c 2  E2  m2 c4 
  0 

E0  m2 c4  m c2 4  m2 c4  1321.3 MeV2  1115.7 MeV2  139.6 MeV2  1124.3MeV
 0  2 1321.3 MeV

2m c2


E  m c2  E0  1321.3MeV 1124.3MeV  197.0 MeV
 

K0  E0  m0c2  1124.3MeV 1115.7 MeV  8.6 MeV

K  E  m  c2  197.0 MeV 139.6 MeV  57.4 MeV
  

30. p  p  p  p : This reaction will not happen because charge is not conserved 2  0 ,
and baryon number is not conserved 2  0 .

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619

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

pp  ppp: This reaction will not happen because charge is not conserved 2  1 ,
and baryon number is not conserved 2  1 .
pp  pppp:
p  p  p  e  e  p : This reaction is possible. All conservation laws are satisfied.

This reaction will not happen. Baryon number is not conserved

2  0 , and lepton number is not conserved 0  2 .

31. The two neutrinos must move together, in the opposite direction of the electron, in order for the

electron to have the maximum kinetic energy, and thus the total momentum of the neutrinos will be
equal in magnitude to the momentum of the electron. Since a neutrino is (essentially) massless, we

have Ev  pvc. We assume that the muon is at rest when it decays. Use conservation of energy and

momentum, along with their relativistic relationship.

p  p  p 
e e

 m c2  E  E  E  E  p c  p c  E  p  p c  E  pc 
 e e  e e  e e  e e

   m c2 2 2 E2 m2
 e e
 E  pc  m c2  E  pc   c4 
e e  e e

m2 c4  2m c2 E  E2  E2  m2 c4  E  m2 c4  m2 c4  K  m c 2 
  e e e e e  e e e

 2m c2



K  m2 c4  m2 c4  m c 2  105.7 MeV2  0.511MeV2  0.511MeV  52.3 MeV
e  e e 2105.7 MeV

2m c2


32. A   could NOT be produced by p  p  p  n    . The pion has a mass energy of 139.6 MeV,

and so the extra 100 MeV of energy could not create it. The Q-value for the reaction is

 Q  2mpc2  c2
2mpc2  m  139.6 MeV , and so more than 139.6 MeV of kinetic energy is


needed. The minimum initial kinetic energy would produce the particles all moving together at the

same speed, having the same total momentum as the incoming proton. We consider the products to

be one mass M  mp  mn  m since they all move together with the velocity. We use energy and


momentum conservation, along with their relativistic relationship, E2  p2c2  m02c4 .

ppc 2  pMc 2
   Ep  mpc2  EM Ep2 m2 c4 EM2 M 2c4
; pp  pM       

 Ep2  mp2c4  Ep  mpc2 2  M 2c4  Ep2  2Epmpc2  mp2c4  M 2c4 

Ep  M 2c4  2mp2c4  Kp  mpc2 
2mpc2

Kp  M 2c4  2mp2c4  mpc2  M 2c4  2mpc2
2mpc2 2mpc2

 938.3MeV  939.6 MeV 139.6 MeV2  2938.3MeV  292.4 MeV
2  938.3 MeV

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620

Chapter 43 Elementary Particles

33. We use the uncertainty principle to estimate the uncertainty in rest energy.

 E  h 
  2t 2
6.631034 Js  9420eV  9 keV
7 1020s 1.60 1016 J eV

34. We estimate the lifetime from the energy width and the uncertainty principle.

E  h t  h  6.631034 Js  2 1021s
2t 2E 2
   1.601019 J 
300 103 eV  1eV 

Apply the uncertainty principle, which says that E  h .
2t

E h t  h  6.631034 Js 7.5 1021 s
2t 2E
    1.601019 J  
 1eV 
2 88 103 eV

36. (a) For B  b u , we have

Charge: 1   1  2 Spin: 0  1  1
3 3 Strangeness: 2 2
Baryon number: Bottomness:
Charm: 0  1  1 000
Topness: 3 3

000 1  1 0

000

(b) Because B is the antiparticle of B , B  b u . The B0 still must have a bottom quark, but

must be neutral. Therefore B0  b d . Because B0 is the antiparticle to B0 , we must have

B0  b d .

37. We find the energy width from the lifetime in Table 42-2 and the uncertainty principle.

E  h  6.631034 Js
2t 2 1018s 1.60 1019 J eV
  (a) t  1018s  659eV  700eV

E  h  6.631034 Js  6.59 107 eV  70 MeV
2t 2 1023s 1.60 1019 J eV
  (b) t  1023s

38. (a) Charge: 0  1  1 Charge is conserved.

Baryon number: 1  1  0 Baryon number is conserved.

Lepton number: 0  0  0 Lepton number is conserved.

Strangeness: 2  1  0 Strangeness is NOT conserved.

Energy: 1314.9 Mev c2  1189.4 Mev c2  139.6 Mev c2  1329 Mev c2
Energy is NOT conserved.

The decay is not possible, because energy is not conserved.

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621

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(b) Charge: 1  0  1  0 Charge is conserved.
Baryon number: 1  1  0  0 Baryon number is conserved.
Lepton number: 0  0  0  1 Lepton number is NOT conserved.
Strangeness: 3  1  0  0 Strangeness is NOT conserved.

Energy: 1672.5 Mev c2  1192.6 Mev c2  139.6 Mev c2  0 Mev c2  1332.2 Mev c2
Energy is conserved.

The decay is not possible, because lepton number is not conserved.

(c) Charge: 0  0 0 0 Charge is conserved.
Baryon number: 0  0 0 0 Baryon number is conserved.
Lepton number: 0  0 0 0 Lepton number is conserved.
Strangeness: 1  1  0  0 Strangeness is conserved.

Energy: 1192.6 Mev c2  1115.7 Mev c2  0 Mev c2  0 Mev c2  1115.7 Mev c2
Energy is conserved.

The decay is possible.

39. (a) The 0 has a strangeness of –2, and so must contain two strange quarks. In order to make a
neutral particle, the third quark must be an up quark. So 0  u s s .

(b) The  has a strangeness of –2, and so must contain two strange quarks. In order to make a
particle with a total charge of –1, the third quark must be a down quark. So   d s s .

40. (a) The neutron has a baryon number of 1, so there must be three quarks. The charge must be 0, as
must be the strangeness, the charm, the bottomness, and the topness. Thus n  u d d .

(b) The antineutron is the anti particle of the neutron, and so n  u d d .
(c) The 0 has a strangeness of -1, so it must contain an “s” quark. It is a baryon, so it

must contain three quarks. And it must have charge, charm, bottomness, and topness equal to 0.
Thus 0  u d s .
(d) The 0 has a strangeness of +1, so it must contain an s quark. It is a baryon, so it must
contain three quarks. And it must have charge, charm, bottomness, and topness equal to 0. Thus
0  u d s .

41. (a) The combination u u d has charge = +1, baryon number = +1, and strangeness, charm,
bottomness, and topness all equal to 0. Thus u u d  p .

(b) The combination u u s has charge = -1, baryon number = -1, strangeness = +1, and
charm, bottomness, and topness all equal to 0. Thus u u s   .

(c) The combination u s has charge = -1, baryon number = 0, strangeness = -1, and charm,
bottomness, and topness all equal to 0. Thus u s  K .

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622

Chapter 43 Elementary Particles

(d) The combination d u has charge = -1, baryon number = 0, and strangeness, charm,

bottomness, and topness all equal to 0. Thus d u   
(e) The combination c s has charge = -1, baryon number = 0, strangeness = -1, charm = -1, and

bottomness and topness of 0. Thus c s  D 
S

42. To form the D0 meson, we must have a total charge of 0, a baryon number of 0, a strangeness of 0,

and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we must

have a “c” quark, with a charge of  2 e . To have a neutral meson, there must be another quark with
3

a charge of  2 e . To have a baryon number of 0, that second quark must be an antiquark. The only
3

candidate with those properties is an anti-up quark. Thus D0  c u .

43. To form the DS meson, we must have a total charge of +1, a baryon number of 0, a strangeness of

+1, and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we

must have a “c” quark, with a charge of  2 e . To have a total charge of +1, there must be another
3

quark with a charge of  1 e . To have a baryon number of 0, that second quark must be an
3

antiquark. To have a strangeness of +1, the other quark must be an anti-strange. Thus DS  c s .

44. Here is a Feynman diagram for the reaction    p   0  n. u_ 0 n

u d ud

u_  d u u d

 p

45. Since leptons are involved, the reaction n  v  p    is a weak
interaction. Since there is a charge change in the lepton, a W boson must
be involved in the interaction. If we consider the neutron as having
emitted the boson, then it is a W , which interacts with the neutrino. If

we consider the neutrino as having emitted the boson, then it is a W ,
which interacts with the neutron.

46. To find the length in the lab, we need to know the speed of the particle which is moving
relativistically. Start with Eq. 36-10a.

 1 c2   vc 1  1 12 c 1  1 12  0.7585c
K  m0c2  1 v2 1 
 K   950 MeV 
 m0c2  1777 MeV


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623

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

tlab  t0  2.911013s  4.4651013s
1 v2 c2
1 0.75852

   xlab  vtlab  0.7585 3.00108 m s 4.4651013s  1.02104 m

47. (a) At an energy of 1.0 TeV, the protons are moving at practically the speed of light. From uniform
circular motion we find the time for the protons to complete one revolution of the ring. Then
the total charge that passes any point in the ring during that time is the charge of the entire
group of stored protons. The current is then the total charge divided by the period.

v  2 R  T  2 R  2 R
T vc

   I  Ne  Nec  5.01013protons
 T 2 R
1.601019 C proton 3.0108 m s  0.38A
2 1.0103 m

(b) The 1.0 TeV is equal to the kinetic energy of the proton beam.

Kbeam  Kcar  Kbeam  1 mv2 
2

   v 
2Kbeam  2 1.0 1012 eV proton 5.0 1013protons 1.60 1019 J eV
 103m s
m 1500 kg

 1.0 102 m s

48. By assuming that the kinetic energy is approximately 0, the total energy released is the rest mass
energy of the annihilating pair of particles.

(a) Etotal  2m0c2  2 0.511MeV  1.022 MeV

(b) Etotal  2m0c2  2 938.3 MeV  1876.6 MeV

49. These protons will be moving at essentially the speed of light for the entire time of acceleration. The

number of revolutions is the total gain in energy divided by the energy gain per revolution. Then the

distance is the number of revolutions times the circumference of the ring, and the time is the distance

of travel divided by the speed of the protons.

 N
 E  1.01012 eV 150109 eV  3.4105 rev
E rev 2.5106 eV rev

   d  N 2 R  3.4105 2 1.0103 m  2.136109 m  2.1109 m

t  d  2.136109 m  7.1s
c 3.00108 m s

50. (a) For the reaction    p  K0  p   0 , the conservation laws are as follows.

Charge: 11  0 1 0 Charge is NOT conserved.

The reaction is not possible, because charge is not conserved.

Also we note that the reactants would have to have significant kinetic energy to be able to

“create” the K0.

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624

Chapter 43 Elementary Particles

(b) For the reaction K  p  0   0 , the conservation laws are as follows.

Charge: 11  0  0 Charge is conserved.

Spin: 0 1  1 0 Spin is conserved.
2 2
Baryon number is conserved.
Baryon number: 0 1  1 0 Lepton number is conserved.
Strangeness is conserved.
Lepton number: 0  0  0  0

Strangeness: 1 0  1 0

The reaction is possible, via the strong interaction.

(c) For the reaction K  n     0   , the conservation laws are as follows.

Charge: 1 0  1 0  0 Charge is conserved.

Spin: 0 1   1  0 1 Spin is conserved.
2 2

Baryon number: 0 1  1 0  0 Baryon number is conserved.

Lepton number: 0  0  0  0  0 Lepton number is conserved.
Strangeness: 1 0  1 0  0 Strangeness is NOT conserved.

The reaction is not possible via the strong interaction because strangeness is not

conserved. It is possible via the weak interaction.

(d) For the reaction K   0  0    , the conservation laws are as follows.

Charge: 1  0  0 1 Charge is conserved.

Spin: 0  0  0  0 Spin is conserved.

Baryon number: 0  0  0  0 Baryon number is conserved.

Lepton number: 0  0  0  0 Lepton number is conserved.

Strangeness: 1  0  0  0 Strangeness is NOT conserved.

The reaction is not possible via the strong interaction because strangeness is not

conserved. It is possible via the weak interaction.

(e) For the reaction    e e , the conservation laws are as follows.

Charge: 1  1 0 Charge is conserved.

Spin: 0   1  1 Spin is conserved.
2 2 Baryon number is conserved.

Baryon number: 0  0  0

Lepton number: 0  11 Lepton number is conserved.

Strangeness: 0  0  0  0  0 Strangeness is conserved.

The reaction is possible, via the weak interaction.

51. (a) For the reaction    p  K   , the conservation laws are as follows.

Charge: 11  11 Charge is conserved.
Baryon number: 0 1  0 1 Baryon number is conserved.
Lepton number: 0  0  0  0 Lepton number is conserved.
Strangeness: 0  0 11 Strangeness is conserved.

The reaction is possible, via the strong interaction.

(b) For the reaction    p  K   , the conservation laws are as follows.

Charge: 1  1  1  1 Charge is conserved.
Baryon number: 0 1  0 1 Baryon number is conserved.

Lepton number: 0  0  0  0 Lepton number is conserved.
Strangeness: 0  0 11 Strangeness is conserved.

The reaction is possible, via the strong interaction.

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625

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(c) For the reaction    p  0  K0   0 , the conservation laws are as follows.

Charge: 11  0  0  0 Charge is conserved.
Baryon number: 0 1  1 0  0 Baryon number is conserved.
Lepton number: 0  0  0  0  0 Lepton number is conserved.
Strangeness: 0  0  11 0 Strangeness is conserved.

The reaction is possible, via the strong interaction.

(d) For the reaction    p  0   0 , the conservation laws are as follows.

Charge: 11  0  0 Charge is NOT conserved.

The reaction is not possible, because charge is not conserved.

(e) For the reaction    p  p  e e , the conservation laws are as follows.

Charge: 11 11 0 Charge is conserved.

Baryon number: 0 1 1 0  0 Baryon number is conserved.

Lepton number: 0  0  0 11 Lepton number is conserved.

Strangeness: 0  0  0  0  0 Strangeness is conserved.

The reaction is possible, via the weak interaction.

Note that we did not check mass conservation, because in a collision, there is always some kinetic
energy brought into the reaction. Thus the products can be heavier than the reactants.

52. The   is the anti-particle of the   , so the reaction is        . The conservation rules are

as follows.

Charge: 1  1 0 Charge is conserved.

Baryon number: 0  0  0 Baryon number is conserved.

Lepton number: 0  11 Lepton number is conserved.

Strangeness: 0  0  0 Strangeness is conserved.

Spin: 0  1  1 Spin is conserved
2 2

53. Use Eq. 43-3 to estimate the mass of the particle based on the given distance.

  mc2  hc 
 2d
6.631034 Js 3.0108 m s 1 J eV   1.98 1011 eV  200 GeV
2 1018 m  1.601019 

This value is of the same order of magnitude as the mass of the W .

54. The Q-value is the mass energy of the reactants minus the mass energy of the products.
For the first reaction, p  p  p  p   0 :

 Q  2mpc2 
2mpc2  m c 2   m c2  135.0 MeV
 
0 0

For the second reaction, p  p  p  n    :

 Q  2mpc2 
mpc2  mnc2  m c2  mpc2  mnc2  m c2a
 

 938.3MeV  939.6 MeV 139.6 MeV= 140.9 MeV

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626

Chapter 43 Elementary Particles

55. The fundamental fermions are the quarks and electrons. In a water molecule there are 2 hydrogen
atoms consisting of one electron and one proton each, and 1 oxygen atom, consisting of 8 electrons,
8 protons, and 8 neutrons. Thus there are 18 nucleons, consisting of 3 quarks each, and 10

electrons. The total number of fermions is thus 183 10  64 fermions .

56. We assume that the interaction happens essentially at rest, so that there is no initial kinetic energy or

momentum. Thus the momentum of the neutron and the momentum of the  0 will have the same

magnitude. From energy conservation we find the total energy of the  0 .

m c2  mpc2  E 0  mnc2  Kn 


 E0
 m c2  mpc2  mnc2  Kn  139.6 MeV  938.3MeV  939.6 MeV  0.60 MeV


 137.7 MeV

From momentum conservation, we can find the mass energy of the  0 . We utilize Eq. 36-13 to
relate momentum and energy.

   pn  p0  2
pnc 2  p c  En2  mn2c4  E2  m2 c 4  m2 c 4  E2  En2  mn2c4 
 0 0 0 0
0

m c 2  E2  En2  mn2c4  137.7 MeV2  939.6 MeV  0.60 MeV2  939.6 MeV2 1/2
 0
0

 133.5 MeV  m0  133.5 MeV c2
The value from Table 43-2 is 135.0 MeV.

57. (a) First we use the uncertainty principle, Eq. 38-1. The energy is so high that we assume E  pc,

and so p  E .

c

xp  h  x E  h 
2 c 2

    E  hc 
   2x
6.63 1034 Js 3.00 108 m s 1GeV 109 eV
2 1032 m 1.60 1019 J eV  2 1016 GeV

Next, we use de Broglie’s wavelength formula. We take the de Broglie wavelength as the
unification distance.

h h 
p Ec

    E  hc 
   
6.63 1034 Js 3.00 108 m s 1GeV 109 eV  11017 GeV
1032 m 1.60 1019 J eV

Both energies are reasonably close to 1016 GeV . This energy is the amount that could be

violated in conservation of energy if the universe were the size of the unification distance.

(b) From Eq. 18-4, we have E  3 kT .
2

    E
 3 kT  T  2E  2 1025 eV 1.6 1019 J eV  7.7 1028 K  1029K
2 3k 3 1.38 1023 J K

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627

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

58. The Q-value is the mass energy of the reactants minus the mass energy of the products.

 Q
 m c2  mpc2  m0 c2  mK0 c2  139.6 MeV  938.3MeV  1115.7 MeV  497.7 MeV


 535.5MeV

We consider the products to be one mass M  m0  mK0  1613.4 MeV c2 since they both have the

same velocity. Energy conservation gives the following: E  mpc2  EM . Momentum


conservation says that the incoming momentum is equal to the outgoing momentum. Then convert

that relationship to energy using the relativistic relationship that E2  p2c2  m2c4.

   p 2

 pM  pc  pMc 2  E2  m2 c4  EM2  M 2c4 
  

 E2 m2 c4 2 E2 2  mp2c4 M 2c4
  
  E  mpc2  M 2c4   2 E  mpc  
 

E  M 2c4  m2  c4  mp2c4  K  m c2 
   

2mpc2 

K  M 2c4  m2 c4  mp2c4  m c 2
  

2mpc2 

 1613.4 MeV2  139.6 MeV2  938.3MeV2  139.6 MeV  768.0 MeV
2  938.3 MeV 

59. Since there is no initial momentum, the final momentum must add to zero. Thus each of the pions
must have the same magnitude of momentum, and therefore the same kinetic energy. Use energy
conservation to find the kinetic energy of each pion.

2mpc2  2K  2m c2  K  mpc2  m c2  938.3 MeV  139.6 MeV  798.7 MeV

60. The Q-value is the energy of the reactants minus the energy of the products. We assume that one of
the initial protons is at rest, and that all four final particles have the same velocity and therefore the
same kinetic energy, since they all have the same mass. We consider the products to be one mass

M  4mp since they all have the same velocity.

Q  2mpc2  4mpc2  2mpc2  Mc2  2mpc2

Energy conservation gives the following, where Kth is the threshold energy.

 Kth  mpc2  mpc2  EM  KM  Mc2

Momentum conservation says that the incoming momentum is equal to the outgoing momentum.

Then convert that relationship to energy using the relativistic relationship that E2  p2c2  m02c4 .

      pp  pM  ppc 2  pMc 2  Kth  mpc2 2  mp2c4  KM  Mc2 2  M 2c4 

 K2 K2 2 c4
th th
 2Kthmpc2  mp2c4  mp2c4   4Kthmpc2  4mp2c4  4mp 

2Kthmpc2  4Kthmpc2  4mp2c4  16mp2c4  2Kthmpc2  12mp2c4 

Kth  6mpc2  3 Q

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628

Chapter 43 Elementary Particles

61. The total energy is the sum of the kinetic energy and the mass energy. The wavelength is found
from the relativistic momentum.

E  K  mc2  15109 eV  938106 eV  1.5941010 eV  16 GeV

hh hc
  
   p E2  mc2 2
E2  mc2 2

c

     
6.631034 Js 3.00108 m s 1  7.81017 m
1.601019 J eV
1.5941010 eV 2 938106 eV 2



62. We use 0 to represent the actual wavelength, and  to define the approximate wavelength. The

approximation is to ignore the rest mass in the expression for the total energy, E  K  mc2. We also
use Eqs. 36-10, 36-11, 36-13, and 43-1.

     p2c2  E2 mc2 2 2 mc2 2 K2 1 mc2 
K  mc2   2 K 
 

0 h hc ;  hc  0 ;   1.010 
   K
p 2mc2 1/ 2
K 1 K 

 
hc  1.01 hc K 2mc2 2 9.38108 eV  9.3331010 eV  9.31010 eV
K 2mc2 0.0201
K 1 1/ 2  1.012 1 
K 

    hc 
  K
6.631034 Js 3.00108 m s  1.3321017 m  1.31017 m
9.3331010 eV 1.601019 J eV

 1 c2  
K   1 v2 1 mc2

vc 1   K  12 c 1   9.3331010 eV  12  0.99995c
 mc2  9.38108 eV 

63. As mentioned in Example 43-9, the  0 can be considered as either u u or d d. There are various
models to describe this reaction. Four are shown here.

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629

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

64. (a) To conserve charge, the missing particle must be neutral. To conserve baryon number, the
missing particle must be a meson. To conserve strangeness, charm, topness, and bottomness,
the missing particle must be made of up and down quarks and antiquarks only. With all this

information, the missing particle is  0 .
(b) This is a weak interaction since one product is a lepton. To conserve charge, the missing

particle must be neutral. To conserve the muon lepton number, the missing particle must be an

antiparticle in the muon family. With this information, the missing particle is   .

65. A relationship between total energy and speed is given by Eq. 36-11b.

E  mc2 
1 v2 c2

v 1  mc2 2  1   9.38108 eV 2  1  1  9.38108 eV 2  1 9.0109
c  E   7.0 1012eV  2  7.0 1012eV 
     

66. We write equations for both conservation of energy and conservation of momentum. The
magnitudes of the momenta of the products are equal. We also use Eqs. 36-11 and 36-13.
p1  p2 ; E0  mc2  E1  E2 

 mc2  E1 2  E22  p22c2  m22c4  p12c2  m22c4  E12  m12c4  m22c4 

m2c4  2mc2E1  E12  E12  m12c4  m22c4  E1  m2c4  m12c4  m22c4
2mc2

K1  E1  m1c2  m2c4  m12c4  m22c4  m1c2  m2c4  m12c4  m22c4  2mc2m1c2
2mc2 2mc2

mc2  m1c2 2  m22c4
2mc2
 

67. The value of R0 is not known until we draw the graph. We note the following:

R  R0et/  R R0  et/  ln R  ln R0   t   ln R   t   ln R0

A graph of ln R vs. t should give a

straight line with a slope of 1/ 5.0

and a y-intercept of ln R0. The ln R (t ) 4.0

determination of the mean life does 3.0
not depend on R0, and so to find
2.0 y = -0.4413x + 4.7124
the mean life, we may simply plot 1.0 R2 = 0.9639
ln R vs. t. That graph is shown,

along with the slope and y- 0.0

intercept. The spreadsheet used for 1.0 2.0 3.0 4.0 5.0 6.0 7.0

this problem can be found on the t (s)

Media Manager, with filename

“PSE4_ISM_CH43.XLS,” on tab “Problem 43.67.”

1  2.266s  2.3s % diff  2.266  2.197 100  3.1%
 2.197
0.4413

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630

CHAPTER 44: Astrophysics and Cosmology

Responses to Questions

1. The Milky Way appears “murky” or “milky” to the naked eye, and so before telescopes were used it

was thought to be cloud-like. When viewed with a telescope, much of the “murkiness” is resolved
into stars and star clusters, so we no longer consider the Milky Way to be milky.

2. If a star generates more energy in its interior than it radiates away, its temperature will increase.
Consequently, there will be greater outward pressure opposing the gravitational force directed
inward. To regain equilibrium, the star will expand. If a star generates less energy than it radiates
away, then its temperature will decrease. There will be a smaller outward pressure opposing the
gravitational force directed inward, and, in order to regain equilibrium, the star will contract.

3. Red giants are extremely large stars with relatively cool surface temperatures, resulting in their
reddish colors. These stars are very luminous because they are so large. When the Sun becomes a
red giant, for instance, its radius will be on the order of the distance from the Earth to the Sun. A red
giant has run out of hydrogen in its inner core and is fusing hydrogen to helium in a shell
surrounding the core. Red giants have left their main sequence positions on the H–R diagram and
moved up (more luminous) and to the right (cooler).

4. A star moving along arrow #1 would increase in luminosity

while maintaining the same surface temperature. It would

therefore also have to increase in size, since each square L 1
meter of its surface would have the same color and therefore 2
same energy output as before. A star moving along arrow #2
would increase in luminosity and decrease in temperature. It 34

would also increase in size, since it would need to produce a 5
greater luminosity even though each unit area of its surface

would now be producing less energy. A star moving along

arrow #3 maintains the same luminosity while increasing its

surface temperature. It will become smaller, since a unit area

of this star will increase its energy and therefore a smaller T

overall area will be needed to maintain the same luminosity.

A star moving along arrow #4 decreases in both surface temperature and luminosity. Finally, a star

moving along arrow #5 will decrease in luminosity while maintaining the same surface temperature

and decreasing in size. Note that these arrows do not necessarily represent “natural” paths for stars

on the H–R diagram.

5. The H–R diagram is a plot of luminosity versus surface temperature of a star and therefore does not
directly tell us anything about the core of a star. However, when considered in conjunction with
theories of stellar evolution, the H–R diagram does relate to the interior of a star. For instance, all
main sequence stars are fusing hydrogen to helium in their cores, so the location of a particular star
on the main sequence does give us that information.

6. The fate of a star depends on the mass of the star remaining after the red giant phase. If the mass is
less than about 1.4 solar masses, the star will become a white dwarf. If the mass is greater than this
limit, than the exclusion principle applied to electrons is not enough to hold the star up against its
own gravity and it continues to contract, eventually becoming a neutron star or, if its mass at this
stage is more than two or three solar masses, a black hole.

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631

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

7. Yes. Hotter stars are found on the main sequence above and to the left of cooler stars. If H–R
diagrams of clusters of stars are compared, it is found that older clusters are missing the upper left
portions of their main sequences. All the stars in a given cluster are formed at about the same time,
and the absence of the hotter main sequence stars in a cluster indicates that they have shorter lives
and have already used up their core hydrogen and become red giants. In fact, the “turn-off” point, or
point at which the upper end of the main sequence stops, can be used to determine the ages of
clusters.

8. The baseline used in measuring parallaxes from the Earth is the distance from the Earth to the Sun.
(See Figure 44-11.) If you were measuring parallaxes from the Moon instead, you would need to
make a slight correction based on the position of the Moon with respect to the Earth–Sun line at the
time of the measurement. If you were measuring parallaxes from Mars, you would need to use the
distance from Mars to the Sun as the baseline. In addition, you would need to wait half a Martian
year between measurements instead of half an Earth year.

9. Watch the star over a period of several days and determine its period through observation. Use the
known relationship between period and luminosity to find its absolute luminosity. Compare its
absolute luminosity to its apparent luminosity (observed) to determine the distance to the galaxy in
which it is located.

10. A geodesic is the shortest distance between two points. For instance, on a flat plane the shortest
distance between two points is a straight line, and on the surface of a sphere the shortest distance is
an arc of a great circle. According to general relativity, space–time is curved. Determining the nature
of a geodesic, for instance by observing the motion of a body or light near a large mass, will help
determine the nature of the curvature of space–time there.

11. If the redshift of spectral lines of galaxies were discovered to be due to something other than
expansion of the universe, then the Big Bang theory and the idea that the universe is expanding
would be called into question. However, the evidence of the cosmic background microwave
radiation would conflict with this view, unless it too was determined to result from some cause other
than expansion.

12. No. In an expanding universe, all galaxies are moving away from all other galaxies on a large scale.
(On a small scale, neighboring galaxies may be gravitationally bound to each other.) Therefore, the
view from any galaxy would be the same. Our observations do not indicate that we are at the center.
(See Figure 44-23.)

13. They would appear to be receding. In an expanding universe, the distances between galaxies are
increasing, and so the view from any galaxy is that all other galaxies are moving away.

14. An explosion on Earth would be affected by the Earth’s gravity and air resistance. Each piece of
debris would act like a projectile, with its individual initial velocity. More distant particles would not
spread at a higher speed. This corresponds somewhat to a closed universe, in which the galaxies
eventually stop and then all come back together again. In the case of the explosion on Earth, most of
the particles would eventually stop. Most would land on the ground. Some might escape into space.
The particles would not all reassemble, as in the “big crunch.”

15. Black holes have tremendous gravity, so we can detect them by the gravitational deflection of other
objects in their vicinity. Also, matter accelerating toward a black hole gives off x-rays, which can be
detected. In addition, gravitational lensing, the bending of light coming from stars and galaxies
located behind the black hole, can indicate that the black hole is present.

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632

Chapter 44 Astrophysics and Cosmology

16. R = 2GM/c², so M = Rc²/2G .

M = (5.29 x 10-11)(3.00 x 108)²/[2(6.67 x 10-11)] = 3.57 x 1016 kg

17. Both the formation of the Earth and the time during which people have lived on Earth are on the far
right edge of Figure 44-30, in the era of dark energy.

18. The 2.7 K cosmic microwave background radiation is the remnant radiation of the Big Bang. As the
universe expanded, the wavelengths of the Big Bang radiation lengthened and became redshifted.
The 2.7 K blackbody curve peaks at a wavelength of about 7.35 cm, in the microwave region. The
temperature of this radiation is low because the energy spread out over an increasingly large volume
as the universe expanded.

19. The early universe was too hot for atoms to exist. The average kinetic energies of particles were high
and frequent collisions prevented electrons from remaining with nuclei.

20. (a) Type Ia supernovae have a range of luminosities that can be extracted from their observable
characteristics and can be derived from the rate at which they brighten and fade away.

(b) The distance to a supernova can be determined by comparing the relative intensity to the
luminosity.

21. The initial Big Bang was not perfectly symmetric. Deviations in the symmetry enabled the
development of galaxies and other structures.

22. If the average mass density of the universe is above the critical density, then the universe will
eventually stop its expansion and contract, collapsing on itself and ending finally in a “big crunch.”
This scenario corresponds to a closed universe, or one with positive curvature.

23. If there were 7 protons for every neutron, and it takes two protons and two neutrons to create a
single helium nucleus, then for every helium nucleus there would be 12 hydrogen nuclei. Since the
mass of helium is four times the mass of hydrogen, the ratio of the total mass of hydrogen to the
total mass of helium should be 12:4, or 3:1.

24. (a) Gravity between galaxies should be pulling the galaxies back together, slowing the expansion of
the universe.

(b) Astronomers could measure the redshift of light from distant supernovae and deduce the
recession velocities of the galaxies in which they lie. By obtaining data from a large number of
supernovae, they could establish a history of the recessional velocity of the universe, and
perhaps tell whether the expansion of the universe is slowing down.

Solutions to Problems

1. Convert the angle to seconds of arc, reciprocate to find the distance in parsecs, and then convert to

light years.

   o  3600 
2.9 104  1o   1.044

d pc  1  1  0.958 pc  3.26 ly   3.1ly
 1.044  1 pc 

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633

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

2. Use the angle to calculate the distance in parsecs, and then convert to light years.

d  pc   1  1  3.704 pc  3.704 pc  3.26 ly   12 ly
  0.27  1pc 


3. Convert the light years to parsecs, and then take the reciprocal of the number of parsecs to find the
parallax angle in seconds of arc.

65 ly  1 pc   19.94 pc  20 pc 2 sig. fig.   1  0.050
 3.26 ly  19.94 pc

4. The reciprocal of the distance in parsecs is the angle in seconds of arc.

(a)   d 1  1  0.01786  0.018
56 pc
 pc 

 1o  o 5.0 106 o
 3600 
   (b) 
0.01786  4.961 106

5. The parallax angle is smaller for the further star. Since tan  d D , as the distance D to the star
increases, the tangent decreases, so the angle decreases. And since for small angles, tan    , we
have that   d D . Thus if the distance D is doubled, the angle  will be smaller by a factor of 2 .

6. Find the distance in light years. That value is also the time for light to reach us.

85 pc  3.26 ly   277 ly  280 ly  It takes light 280 years to reach us.
 1 pc 

7. The apparent brightness of an object is inversely proportional to the square of the observer’s distance

from the object, given by Eq. 44-1. To find the relative brightness at one location as compared to

another, take a ratio of the apparent brightness at each location.

L

bJupiter  4 d 2 2   d Earth 2   1 2  0.037
bEarth Jupiter  d Jupiter   5.2 
L d Earth  
2
4 d 2
Earth d Jupiter

8. (a) The apparent brightness is the solar constant, 1.3103 W m2 .

(b) Use Eq. 44-1 to find the intrinsic luminosity.

L  L  4 d 2b  4 1.496 1011 m2 2 1.3 103 W m2  3.7 1026 W
   b4 d 2

9. The density is the mass divided by the volume.

M M 1.99 1030 kg 2 103 kg m3
  V   
4  r 3 6 1010 m 3
3 4 
3

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634

Chapter 44 Astrophysics and Cosmology

10. The angular width is the inverse tangent of the diameter of our Galaxy divided by the distance to the

nearest galaxy. According to Figure 44-2, our Galaxy is about 100,000 ly in diameter.

  tan1 Galaxy diameter  tan 1 1.0 105ly  0.042 rad  2.4o
Distance to nearest galaxy 2.4 106ly

Moon  tan1 Moon diameter  tan 1 3.48 106 m  9.1 103 rad  0.52o
Distance to Moon 3.84  108 m

The Galaxy width is about 4.5 times the Moon width.

11. The Q-value is the mass energy of the reactants minus the mass energy of the products. The masses

are found in Appendix F.

4 He  4 He  8 Be
2 2 4

   Q  2mHec2  mBec2  2 4.002603 u   8.005305 u c2 931.5 Mev c2  0.092 MeV

4 He  8 Be  C12
2 4 6

 Q  mBec2  mHec2  mCc2  4.002603 u  8.005305 u 12.000000c2 931.5 Mev c2

 7.366 MeV

12. The angular width is the inverse tangent of the diameter of the Moon divided by the distance to the
Sun.

3.48 106 m
1.496 1011m
   tan1 Moon diameter tan 1 o
Distance to Sun
  2.33105 rad  1.33  103  4.79

13. The density is the mass divided by the volume.

M Sun 1.99 1030 kg
 R4 3

3 Earth
 M   3  1.83109 kg m3


V
4  6.38106 m
3

Since the volumes are the same, the ratio of the densities is the same as the ratio of the masses.

  M 1.99 1030 kg  3.33105 times larger
Earth M Earth  5.98 1024 kg

14. The density of the neutron star is its mass divided by its volume. Use the proton to calculate the

density of nuclear matter. The radius of the proton is taken from Eq. 41-1.

 M 1.5 1.99 1030 kg  5.354 1017 kg m3  5.4 1017 kg m3
  neutron
3
 Vstar
4  11103 m
3

neutron 5.354 1017 kg m3  2.9 108  neutron 5.354 1017 kg m3  2.3
 1.83109 kg m3  1.6731027 kg
star star

 white  nuclear

dwarf  matter 3
4  1.2 1015 m
3

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635

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

15. Wien’s law (Eq. 37-1) says that the PT   , where  is a constant, and so P1T1  P2T2. The

Stefan–Boltzmann equation (Eq. 19-17) says that the power output of a star is given by P   AT 4 ,

where  is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the

same as the luminosity L in this chapter. The luminosity L is related to the apparent brightness b by

Eq. 44-1. It is given that b1 b2  0.091 , d1  d2 , P1  470 nm, and P2  720 nm.

P1T1  P2T2  T2  P1 ; b1  0.091b2  L1  0.091 L2 
T1 P2 4 d12
4 d 2
2

 1  2 0.091 4 r22T24 T24 r22
d 2  0.091L2  0.091P2  0.091A2T24  4 r12T14  0.091 T14 r12 
L1 P1 A1T14
d12

r1  0.091  T2 2  0.091  P2 2  0.091  470 nm 2  0.1285
r2  T1   P1   720 nm 

The ratio of the diameters is the same as the ratio of radii, so D1  0.13 .
D2

16. Wien’s law (Eq. 37-1) says that the PT   , where  is a constant, and so P1T1  P2T2. The

Stefan–Boltzmann equation (Eq. 19-17) says that the power output of a star is given by P   AT 4 ,

where  is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the

same as the luminosity L in this chapter. The luminosity L is related to the apparent brightness b by

Eq. 44-1. It is given that b1  b2 , r1  r2 , P1  750 nm, and P2  450 nm.

P1T1  P2T2  T2  P1
T1 P2

b1  b2  L1  L2  d 2  L2  P2   A2T24  4 r22T24  T24   T2 4 
4 d12 2 L1 P1  A1T14 4 r12T14 T14  T1 
 
4 d 2 d12
2

d2   T2 2   P1 2   750 2  2.8
d1  T1   P2   450 
   

The star with the peak at 450 nm is 2.8 times further away than the star with the peak at 750 nm.

17. The Schwarzschild radius is 2GM .
c2

   REarth
 2GM Earth 2 6.67 1011 Nm2 kg2 5.98 1024 kg  8.86 103 m  8.9 mm
c2  3.00 108 m s2

18. The Schwarzschild radius is given by R  2GM . An approximate mass for our Galaxy is
c2

calculated in Example 44-1.

   R
 2GM 2 6.67 1011 Nm2 kg2 2 1041 kg  3 1014 m
c2  3.00 108 m s
2

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636

Chapter 44 Astrophysics and Cosmology

19. The limiting value for the angles in a triangle on a sphere is 540o . Imagine drawing an equilateral

triangle near the north pole, enclosing the north pole. If that triangle were small, the surface would
be approximately flat, and each angle in the triangle would be 60o . Then imagine “stretching” each
side of that triangle down towards the equator, while keeping sure that the north pole stayed inside
the triangle. The angle at each vertex of the triangle would expand, with a limiting value of 180o .

The three 180o angles in the triangle would sum to 540o .

20. To just escape from an object, the kinetic energy of the body at the surface of the body must be equal
to the magnitude of the gravitational potential energy at the surface. Use Eq. 8-19.

vesc  2GM 2GM  c
 2GM c2

RSchwarzchild

21. We find the time for the light to cross the elevator, and then find how far the elevator moves during

that time due to its acceleration.

 t  x
 c
; y  1 g t 2  g x2  9.80 m s2 2.4 m 1.3 1016 m
2
2c2 2 3.00 108 m s 2 

Note that this is smaller than the size of a proton.

22. Use Eq. 44-4, Hubble’s law.
v  Hd  d  v  1850 km s  84 Mly  8.4 107 ly
H 22 km s Mly

23. Use Eq. 44-4, Hubble’s law.

 v  Hd
 d  v  0.015 3.00 108 m s  204.5 Mly  2.0 102 Mly  2.0 108 ly
H
2.2 104 m s Mly

24. (a) Use Eq. 44-6 to solve for the speed of the galaxy.

z  obs  rest  v  v  c  455 nm  434 nm   0.04839c  0.048 c
rest c  
 434 nm 

(b) Use Hubble’s law, Eq. 44-4, to solve for the distance.

 v  Hd
 d v 0.04839 3.00 108 m s  660 Mly  6.6 108 ly

H 22000 m s Mly

25. We find the velocity from Hubble’s law, Eq. 44-4, and the observed wavelength from the Doppler

shift, Eq. 44-3.

(a) v  Hd  22000 m s Mly 7.0 Mly  5.133 104
c c
3.00 108 m s

  0 1 v c  656 nm 1  5.133 104  656.34 nm  656 nm
1 v c 1  5.133 104

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(b) v  Hd  22000 m s Mly 70 Mly  5.133 103
c c
3.00 108 m s

  0 1 v c  656 nm 1  5.133 103  659.38 nm  659 nm
1 v c 1  5.133 103

26. Use Eqs. 44-3 and 44-4 to solve for the distance to the galaxy.

 
 obs rest 1v c vc  2 2
1v c obs rest

 2 2
obs rest

     d 423.4 nm2  393.4 nm2 
 v  c  2 2  3.00 108 m s 423.4 nm2  393.4 nm2 
H H obs rest 2.2 104 m s Mly

 2 2
obs rest

 1.0 103 Mly  1.0 109 ly

27. Use Eqs. 44-3 and 44-5a to solve for the speed of the galaxy.

z  obs  rest  obs 1  1  v c 1 
rest rest 1v c

v  z  12 1  1.0602 1  0.05820  v  0.058 c
c z  12 1 1.0602 1

The approximation of Eq. 44-6 gives v  zc  0.060 c .

28. Use Eqs. 44-3 and 44-5a to solve for the redshift parameter.

z   obs rest  obs  1  1  v c  1  1  0.075  1  0.078
rest rest 1v c 1  0.075

Or, we use the approximation given in Eq. 44-6.

z  v  0.075
c

29. Eq. 44-3 states   rest 1v c .
1v c

1 v c  rest 1  v 1/ 2 1 v 1/ 2  rest 1  v  1 v  1 v 2
1 v c c  c  c  c  c 
  rest  1    1   rest  1
2 2 2

  rest 1  2  1 v   rest  1  v   rest  rest v v   v
 2 c    c  c    rest    rest c 

rest c

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638

Chapter 44 Astrophysics and Cosmology

30. For small relative wavelength shifts, we may use Eq. 44-6 to find the speed. We use Eq. 44-4 to find

the distance.

v    v  c  ; v  Hd 
c rest rest

d  v  c    3.00108 m s   0.10 cm   65 Mly
H H rest  22, 000 m s Mly   21cm 
   

31. Wien’s law is given in Eq. 37-1.

PT  2.90 103 mK  P  2.90 103 mK  2.90 103 mK  1.1103 m
T 2.7 K

32. We use Wien’s law, Eq. 37-1. From Figure 44-30, the temperature is about 1010 K.

PT  2.90 103 mK  P  2.90 103 mK  2.90 103 mK  3 1013 m
T 1010 K

From Figure 31-12, that wavelength is in the gamma ray region of the EM spectrum.

33. We use the proton as typical nuclear matter.

 1026 kg   1nucleon   6 nucleons m3
 m3   1.67 1027 kg 

34. If the universe’s scale is inversely proportional to the temperature, the scale times the temperature
should be constant. If we call the current scale “1,” and knowing the current temperature to be about
3 K, then the product of scale and temperature should be about 3. Use Figure 44-30 to estimate the
temperature at various times. For purposes of illustration, we assume the universe has a current size
of about 1010 ly. There will be some variation in the answer due to reading the figure.

(a) At t  106 yr , the temperature is about 1000 K. Thus the scale is found as follows.

ScaleTemperature  3  Scale  3  3  3103 

Temperature 1000

   Size  3103 1010 ly  3107 ly

(b) At t  1 s , the temperature is about 1010 K .

Scale  3 3  3  1010    Size  31010 1010 ly  3ly
Temperature  1010

(c) At t  106 s , the temperature is about 1013 K .

Scale  3 3  3 1013 
Temperature  1013

   Size  31013 1010 ly  3103 ly  31013m

(d) At t  1035 s , the temperature is about 1027 K .

Scale  3 3  3  1027
Temperature  1027

   Size  31027 1010 ly  31017 ly  0.3 m

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639

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35. We approximate the temperature–energy relationship by kT  E  mc2 as suggested on page 1217.

kT  mc2  T  mc2 .
k

   (a)
T  mc2  500 MeV c2 c2 1.60 1013 J MeV  6 1012 K
k 1.38 1023 J K

From Figure 44-30, this corresponds to a time of  105 s .

   (b)
T  mc2  9500 MeV c2 c2 1.60 1013 J MeV  1 1014 K
k 1.38 1023 J K

From Figure 44-30, this corresponds to a time of  107 s .

   (c)
T  mc2  100 MeV c2 c2 1.60 1013 J MeV  1 1012 K
k 1.38 1023 J K

From Figure 44-30, this corresponds to a time of  104 s .
There will be some variation in the answers due to reading the figure.

36. (a) According to the text, near Figure 44-33, the visible matter makes up about one-tenth of the

total baryonic matter. The average baryonic density is therefore 10 times the density of visible

matter.

 10 M Rvisible visible
  baryon  10
43
3

     1011galaxies 1011 stars galaxy 2.0 1030 kg star
   10
4   14 109 ly 9.46 1015 m ly 3
3

 2.0551026 kg m3  2.11026 kg m3

(b) Again, according to the text, dark matter is about 4 times more plentiful than normal matter.

 dark  4 baryon  4 2.055 1026 kg m3  8.2 1026 kg m3

37. (a) From page 1201, a white dwarf with a mass equal to that of the Sun has a radius about the size
of the Earth’s radius, 6380 km . From page 1202, a neutron star with a mass equal to 1.5 solar

masses has a radius of about 20 km . For the black hole, we use the Schwarzschild radius

formula.

    R
 2GM 2 6.67 1011 Nm2 kg2 3 1.99 1030 kg  8849m  8.85 km
c2  3.00 108 m s 2

(b) The ratio is 6380 : 20 : 8.85  721 : 2.26 : 1  700 : 2 : 1 .

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640

Chapter 44 Astrophysics and Cosmology

38. The angular momentum is the product of the rotational inertia and the angular velocity.

    I  Iinitial final 

I initial MR2 2 Rinitial 2  7 108 m 2
I final Rfinal  8 103 m 
5 initial 

MR2 2

5 final
        final initial 1rev
 initial  initial month

 7.66 109 rev month  7.66 109 rev 1month 1d 1h  2953rev s
 
month 30 d 24 h 3600s

 3000 rev s

39. The rotational kinetic energy is given by 1 I2 . The final angular velocity, from problem 43, is
2

7.66 109 rev month .

2 22 2
5 final final
12

K I  MR   R  final 2 final final
final final
K  I   MR    R  initial
12 2 22 initial initial
2 initial initial 5 initial initial

   2
    
8 103 m 7.66 109 rev month 8 109

7 108 m 1rev month

40. The apparent luminosity is given by Eq. 44-1. Use that relationship to derive an expression for the
absolute luminosity, and equate that for two stars.

b  L  L  4 d 2b
4 d 2

2 2
distant distant Sun Sun
 b  bL  L  4 d  4 d distant
Sun

star star star

d  ddistant Sun  lSun  1.5 1011m 1  1ly   5 ly
1011  9.461 1015m 
star ldistant

star

41. A: The temperature increases, the luminosity stays the same, and the size decreases.
B: The temperature stays the same, the luminosity decreases, and the size decreases.
C: The temperature decreases, the luminosity increases, and the size increases.

42. The power output is the energy loss divided by the elapsed time.

P  K  Kinitial fraction lost   1 I2 fraction lost   1 2 MR 2 2 fraction lost 
2 2 5

t t t t

     1 1.5 1.99 1030 kg 8.0 103 m 2 2 rad s2 1109  1.7461025 W  1.7 1025 W

 5 1d 24 h d 3600s h

43. Use Newton’s law of universal gravitation.

31041 kg 2
2 106 ly 9.46 1015 m
      Fm1m2  1.68 1028 N
G r2  6.67 1011 Nm2 kg2  2

 ly

 2 1028 N

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44. (a) Assume that the nucleons make up only 2% of the critical mass density.

 nucleon mass density  0.02 1026 kg m3

 0.02 1026 kg m3

nucleon number density  1.67 1027 kg nucleon  0.12 nucleon m3

neutrino number density  109  nucleon number density  1.2 108 neutrino m3

 0.98 1026 kg m3  8.17 1035 kg 9.315 108 eV c2  46 eV c2
neutrino  1.66 1027 kg
1.2 108 neutrino m3

(b) Assume that the nucleons make up only 5% of the critical mass density.

 nucleon mass density  0.05 1026 kg m3

 0.05 1026 kg m3

nucleon number density  1.67 1027 kg nucleon  0.30 nucleon m3

neutrino number density  109  nucleon number density  3.0 108 neutrino m3

 0.95 1026 kg m3  3.17 1035 kg 9.315 108 eV c2  18 eV c2

3.0 108 neutrino m3 neutrino 1.66 1027 kg

45. The temperature of each star can be found from Wien’s law.

PT  2.90 103 mK 

T660  2.90 103 mK  4390 K T480  2.90 103 mK  6040 K
660 109 m 480 109 m

The luminosity of each star can be found from the H–R diagram.

L660  31025 W L480  3 1026 W

The Stefan–Boltzmann equation says that the power output of a star is given by P   AT 4 , where

 is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the same as

the luminosity L given in Eq. 44-1. Form the ratio of the two luminosities.

  L480 A T4 r T2 4 2
  L660 480 480 480 480 2
 4  r480  2 31026 W 4390 K  1.67
A T4 4 r T2 4 r660 31025 W 6040 K
660 660 660 660 L T480 660
2

L T660 480

The diameters are in the same ratio as the radii.

d480  1.67  1.7
d 660

The luminosities are fairly subjective, since they are read from the H–R diagram. Different answers
may arise from different readings of the H–R diagram.

46. (a) The number of parsecs is the reciprocal of the angular resolution in seconds of arc.

   100 parsec  1  1  1o 
  60   60 
    0.01  2.78 106  3 106 o

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642

Chapter 44 Astrophysics and Cosmology

(b) We use the Rayleigh criterion, Eq. 35-10, which relates the angular resolution to the diameter of

the optical element. We choose a wavelength of 550 nm, in the middle of the visible range.

   1.22  D  1.22 =
1.22 550 109 m =13.83m  14 m

 D   2.78106   rad 180

The largest optical telescopes currently in use are about 10 m in diameter.

47. We approximate the temperature–kinetic energy relationship by kT  K as given on page 1217.

  kT  K
 TK 1.96 1012eV 1.60 1019 J eV  2 1016 K
k 1.38 1023 J K

From Figure 44-30, this is in the hadron era .

48. We assume that gravity causes a centripetal force on the gas. Solve for the speed of the rotating gas,

and use Eq. 44-6.

m mgas black m v2
gas gas
F  Fgravity centripetal  G hole  r 

r2

mblack      6.67 1011 Nm2 kg2
vgas  G hole 2 109 1.99 1030 kg  6.42 105 m s

r  68 ly   9.46 1015 m 
 1ly 

z v  6.42 105 m s  2.14 103  2 103
c 3.00 108 m s

49. (a) To find the energy released in the reaction, we calculate the Q-value for this reaction. From Eq.

42-2a, the Q-value is the mass energy of the reactants minus the mass energy of the products.
The masses are found in Appendix F.

 Q  2mCc2  mMgc2  2 12.000000 u  23.985042 uc2 931.5 Mev c2  13.93 MeV

(b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when
they are just touching. The distance between the two nuclei will be twice the nuclear radius,
from Eq. 41-1. Each nucleus will have half the total kinetic energy.

   r  1.2 1015 m  A 1/3  1.2 1015 m 121/3 U 1 q2
nucleus

40 2r

K  1 U  1 1 q2
2 2 4 0 nucleus
2r

1.60 1019 C 2

 1015 m 12 1/ 3
    1 62  1 MeV 
 2  1.60 1013J 
8.988 109 Nn2 C2 2 1.2  4.711 MeV

 4.7 MeV

(c) We approximate the temperature–kinetic energy relationship by kT  K as given on page

1217.

 kT  K
 T  K  4.711MeV 1.60 1013 J MeV  5.5 1010 K
k
1.38 1023 J K

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643

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

50. (a) Find the Q-value for this reaction. From Eq. 42-2a, the Q-value is the mass energy of the
reactants minus the mass energy of the products.

16 O  16 O  28 Si  4 He
8 8 14 2

   Q  2mOc2  mSic2  mHec2  215.994915 u   27.976927 u  4.002603 c2 931.5 Mev c2

 9.594 MeV

(b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when
they are just touching. The distance between the two nuclei will be twice the nuclear radius,
from Eq. 41-1. Each nucleus will have half the total kinetic energy.

      r  1.2 1015 m A 1/3  1.2 1015 m 16 1/3 U 1 q2
nucleus

40 2r

K nucleus  1 U  1 1 q2
2 2 4 0 nucleus
2r

82 1.60 1019 C 2
 2 1.2 1015 m 16 1/3
      C2 1eV
1 8.988 109 Nn2  1.60 1019J  7.609 MeV
2

 7.6 MeV

(c) We approximate the temperature–kinetic energy relationship by kT  K as given on page
1217.

 kT  K  1.60 1019 J 
TK 7.609 106eV  1eV 
k
 1.38 1023 J K  8.8 1010 K

51. We treat the energy of the photon as a “rest mass,” and so "mphoton  " Ephoton c2 . To just escape

from a spherical mass M of radius R, the energy of the photon must be equal to the magnitude of the

gravitational potential energy at the surface.

  c2
Ephoton  GMmphoton R  GMmphoton GM Ephoton GM
R Ephoton  Ephoton  c2

52. We use the Sun’s mass and given density to calculate the size of the Sun.

MM
  4 3 
V r3 Sun

  rSun 1/ 3 1/ 3
  3M    3 1.99 1030 kg   3.62 1018 m  1ly   382 ly  400 ly
 4   9.46 1015 m
 4 1026 kg m3

rSun 3.62 1018 m  2 107 ; rSun  382 ly  4 103
d Earth-Sun  1.50 1011m dgalaxy 100,000 ly

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644

Chapter 44 Astrophysics and Cosmology

53. We approximate the temperature–kinetic energy relationship by kT  K as given on page 1217.

  kT  K
 T  K  14 1012eV 1.60 1019 J eV  1.6 1017 K
k 1.38 1023 J K

From Figure 44-30, this might correspond to a time around 1015s . Note that this is just a very
rough estimate due to the qualitative nature of Figure 44-30.

54. (a) We consider the photon as p  E  mc
entering from the left, grazing the c
Sun, and moving off in a new x
direction. The deflection is
assumed to be very small. In 
particular, we consider a small part
of the motion in which the photon r yR
moves a horizontal distance R
dx  cdt while located at (x,y)
relative to the center of the Sun.  F  GMm
r2
Note that y  R and r2  x2  y2.

If the photon has energy E, it will

have a “mass” of m  E c2 , and a

momentum of magnitude p  E c  mc. To find the change of momentum in the y-direction,

we use the impulse produced by the y-component of the gravitational force.

Fy dt GMm dx GMm R dx GMmR dx
r2 c r2 r c
c x2  R2 3/2
 dpy   cos    

To find the total change in the y-momentum, we integrate over all x (the entire path of the
photon). We use an integral from Appendix B-4.



 GMmR dx GMmR x 2GMm 2GMp
c x2  R2 3/2   c R2 x2  R2 1/2   cR   c2R
py    



The total magnitude of deflection is the change in momentum divided by the original

momentum.

py 2GMp

  p  c2R 2GM

p  c2R

(b) We use data for the Sun.

 Nm2 
 kg 2 
     
2GM 2 6.67 1011 1.99 1030 kg
 c2R
 

3.00 108 m s 2 6.96 108 m

 4.238 106 rad  180   3600   0.87
  rad   1 

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645

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

55. Because Venus has a more negative apparent magnitude, Venus is brighter. We write the
logarithmic relationship as follows, letting m represent the magnitude and b the brightness.

m  k log b ; m2  m1  k log b2  log b1   k log b2 b1  

k  m2  m1  5  2.5

log b2 b1  log 0.01

mVenus mSirius
2.5
 m2  m1  k log b2 b1 m2 m1 Venus 4.41.4
Sirius 2.5
b b2 k
 b  10  b  10  10  161

56. If there are N nucleons, we assume that there are approximately 1 N neutrons and 1 N protons.
2 2

Thus, for the star to be neutral, there would also be 1 N electrons.
2

(a) From Eq. 40-12 and 40-13, we find that if all electron levels are filled up to the Fermi energy

EF , the average electron energy is 3 EF .
5

   Ee  Ne3EF  1 N 3 h2  3 Ne 2/3  3  1 N  h2  3 N 2/3
5 2 5 8me   V  5  2  8me   2V 

(b) The Fermi energy for nucleons would be a similar expression, but the mass would be the mass
of a nucleon instead of the mass of the electron. Nucleons are about 2000 times heavier than
electrons, so the Fermi energy for the nucleons would be on the order of 1/1000 the Fermi
energy for the electrons. We will ignore that small correction.

To calculate the potential energy of the star, think about the R dr
mass in terms of shells. Consider the inner portion of the r
star with radius r < R and mass m, surrounded by a shell of
thickness dr and mass dM. See the diagram. From Gauss’s
law applied to gravity, the gravitational effects of the inner
portion of the star on the shell are the same as if all of its
mass were at the geometric center. Likewise, the
spherically-symmetric outer portion of the star has no
gravitational effect on the shell. Thus the gravitational
energy of the inner portion–shel) combination is given by a

form of Eq. 8-17, dU  G mdM . The density of the star
r

is given by  4 M . We use that density to calculate the masses, and then integrate over the
3  R3

full radius of the star to find the total gravitational energy of the star.

M r3
   m  4r 3  4  r 3  M R3
3 3
4  R 3
3

M 3Mr 2
   dM   R3
4 r2dr  4  R3 4 r 2dr  dr
3

mdM M r3 3Mr 2 dr 3GM 2
r  G R3   R6
dU  G R3 r 4dr
r

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646

Chapter 44 Astrophysics and Cosmology

R  3GM 2  3GM 2 R 3GM 2 R5 3 GM 2
  R6 0   R6 5 
  U r4  dr r4 dr
  R6   5R
0

(c) The total energy is the sum of the two terms calculated above. The mass of the star is primarily

due to the nucleons, and so M  Nmnucleon .

Etotal  Ee U  3 1 N  h2 3 N 2/3  3 GM 2
5  2  8me   2V  5 R

3h2  M   3 M 2/3 3 GM 2  92/33h2 M 5/3  3 GM 2
     
80me  mnucleon    2mnucleon 4  R3  5 R 320 m m R4/3 5/3 2 5 R
3 e nucleon

Let a  92/33h2 M 5/3 and b  3 GM 2 , so Etotal  a  b. We set dEtotal dR  0 to find
320 m m4/3 5/3 5 R2 R

e nucleon

the equilibrium radius.

92/ 33h2 M 5/3
320 m m4/3 5/3
dEtotal 2a  b 0  Req  2a  2  92/3 h2
dR   R3 R2 b e nucleon

3 GM 2 32 GM m m4/3 1/3 5/3

e nucleon

5

We evaluate the equilibrium radius using the Sun’s mass.

Req  32 92/3 h2

GM m m4/3 1/3 5/3

e nucleon

 92/3 6.631034 Js 2
    
32 4/3  6.67 1011 Nm2  2.0 1030 kg 1/3 9.111031 kg 1.67 1027 kg 5/3
 kg 2 


 7.178 106 m  7.2 103 km

57. There are N neutrons. The mass of the star is due only to neutrons, and so M  Nmn . From Eqs.

40-12 and 40-13, we find that if all energy levels are filled up to the Fermi energy EF , the average

energy is 3 EF . We follow the same procedure as in Problem 56. The expression for the
5

gravitational energy does not change.

N 2/3 2/3 3 18 h M2/3 2 5/3
     En  Nn V  
E3  N 3 h2 3  M  3h2 3 M  
5 8mn     
5F  mn  40mn   4  R 3mn 160 4 / m3 8/ 3 R 2
3 n

3 18 2/3 h2M 5/3
 Etotal 3 GM 2
5 R
 En U  160 4 /3 m8/3 R 2 
n

Let a   3 18 2/3 h2M 5/3 and b 3 GM 2 , so Etotal  a  b. We set dEtotal dR  0 to find the
5 R2 R
160 m4/3 8/ 3
n

equilibrium radius.

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647

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

2 3 18 h M2/3 2 5/3
   dEtotal 160 4 /3 m8/3 18 h2/3 2
n
dR
2a  b 0  Req 2a 3 GM 2  16 4/3GM m1/3 8/3
  R3 R2  n

b

5

We evaluate the equilibrium radius for a mass of 1.5 solar masses.

18 h2/3 2
 Req 
16 4/3GM m1/3 8/3
n

 182/3 6.631034 Js 2
   
16 4/3  6.67 1011 Nm2  1.5 2.0 1030 kg 1/3 1.67 1027 kg 8/3
 kg 2 


 1.086 104 m  11km

58. We must find a combination of c, G, and  that has the dimensions of time. The dimensions of c

are  L  , the dimensions of G are  L3  , and the dimensions of  are  ML2  .
 T     
 MT 2   T 

 L   L3   ML2  L  3 2 M   T  2 
T   MT 2   T 
       tP  cG   T  

  3  2  0 ;     0 ;    2    1    5  0 ;   1  3 

5  1 3    1 ;  1 ;   5
2 2 2

tP  c G 5/2 1/2 1/2  G    6.67 1011 Nm2 kg2 1 6.631034 Js  5.38 1044 s
c5  2
 3.00 108 m s 5

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648