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

Chapter 40 Molecules and Solids

61. The arrow at the emitter terminal, E, indicates the direction of current IE. The current into the
transistor must equal the current out of the transistor.

IB  IC  IE

62. For an electron confined in 1 dimension, we find the uncertainty in the momentum from Eq. 38-1,

p   . The momentum of the electron must be at least as big as the uncertainty in the
x

momentum, so we approximate p   . Finally, we calculate the kinetic energy by K  p2 . Find
x 2m

the difference in the two kinetic energies based on the two position uncertainties.

K  p2  2
2m
2m x2

K molecule 2  1 1
   K 2m 
 Kin atoms    x 2  x 2 
in atoms molecule 

       2  
1.055 1034 Js  1 1  1
 0.074 109 m   1019 
  
2 9.111031 kg  0.053 109 m 2 2  1.60  J eV 
 molecule
in atoms

 6.62eV

There are two electrons, and each one has this kinetic energy difference, so the total kinetic energy

difference is 26.62eV  13.2eV  13eV .

63. We find the temperature from the given relationship.

   (a)
K  3 kT  T  2K  24.0eV 1.60 1019 J eV  3.1104 K
2 3k
3 1.38 1023 J K

   (b)
K  3 kT  T  2K  2 0.12 eV 1.60 1019 J eV  930 K
2 3k
3 1.38 1023 J K

64. (a) The potential energy for the point charges is found as from Eq. 23-10.

    U
  1 e2   2.30 1028 Jm  5.32 eV  5.3eV
4 r 0.27 109 m 1.60 1019 J eV
0

(b) Because the potential energy of the ions is negative, 5.32 eV is released when the ions are
brought together. The other energies quoted involve the transfer of the electron from the K
atom to the F atom. 3.41 eV is released and 4.34 eV is absorbed in the individual electron
transfer processes. Thus the total binding energy is as follows.

Binding energy  5.32 eV  3.41eV  4.34 eV  4.39eV  4.4eV

65. The diagram here is similar to Figure 40-9 and Figure 40-11. U
The activation energy is the energy needed to get the (initially)
stable system over the barrier in the potential energy. The
activation energy is 1.4 eV for this molecule. The dissociation
energy is the energy that is released when the bond is broken.
The dissociation energy is 1.6 eV for this molecule.

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551

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

66. (a) The equilibrium position is the location where the potential energy is a minimum. We find that

location by setting the derivative of the potential energy equal to 0.

 U  U0 1  earr0  2 dU 1  earr0   0
; dr  2U0 1  earr0   aearr0  0 

 earr0   1  a r  r0  0  r  r0

The dissociation energy is the energy difference between the two states of equilibrium

separation and infinite separation.

ear0  2 ear0 r0  2 2 2
   U
 Ur  U rr0  U0 1   U0 1    U0 10 U0 11  U0

(b) See the included graph. The U (eV) 8
spreadsheet used for this problem can 6
be found on the Media Manager, with 4
filename “PSE4_ISM_CH40.XLS,”
on tab “Problem 40.66b.”

2

0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

r/r 0

67. (a) The reduced mass is defined in Eq. 40-4.

  mH mCl  1.008 u   35.453 u   0.9801u
mH  mCl 1.008u  35.453u

(b) We find the effective spring constant from Eq. 40-5.

f  1 k 
2 

   k  4 f 2  4 2 8.66 1013 Hz 2 0.9801u 1.6605 1027 kg u  482 N m

The spring constant for H2 is estimated in Example 40-6 as 550 N/m.

kCO  482 N m  0.88
kH2 550 N m

68. Vibrational states have a constant energy difference of Evib  0.54eV, as found in Example 40-7.

Rotational states have a varying energy difference, depending on the l value, of Erot  2l , where l
I

represents the upper energy state, as given in Eq. 40-3.

1.055 1034 Js 2
1.66 1027 kg u 0.074 109 m
      Erot2l2l
 I  r02 l 2 1.60 1019 J eV

 0.5 1.008 u 

 0.0152eV l

Each gap, as represented in Figure 40-17, is larger. We add those gaps until we reach 0.54eV.

l  1  l  0 : Erot  0.0152 eV  Erot  0.0152 eV
Erot  30.0152eV  0.0456eV
l  2  l  1: Erot  20.0152 eV  Erot  60.0152 eV  0.0912 eV
l  3  l  2 : Erot  30.0152eV

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552

Chapter 40 Molecules and Solids

l  4  l  3 : Erot  4 0.0152 eV Erot  100.0152eV  0.152eV

l  5  l  4 : Erot  50.0152eV Erot  150.0152eV  0.228eV

l  6  l  5 : Erot  60.0152eV Erot  210.0152eV  0.3192eV

l  7  l  8 : Erot  70.0152 eV Erot  280.0152 eV  0.4256eV

l  8  l  7 : Erot  80.0152eV Erot  360.0152eV  0.5472eV

So we see that rotational states from l  0 to l  7 can be “between” vibrational states, or a total of
8 rotational states.

69. The Boltzmann indicates that the population of a state decreases as the energy of the state increases,
according to Nn  e .En /kT The rotation energy of states increases with higher l values, according to
Eq. 40-2. Thus states with higher values of l have higher energies, and so there are fewer molecules
in those states. Since the higher states are less likely to be populated, they are less likely to absorb a
photon. As an example, the probability of absorption between l  1 and l  2 is more likely than
between l  2 and l  3, and so the peak representing the transition between l  1 and l  2 is

higher than the peak representing the transition between l  2 and l  3.

The molecule is not rigid, and so the distance between the two ions is not constant. The moment of
inertia depends on the bond length, and the energy levels depend on the moment of inertia. Thus the
energy levels are not exactly equally spaced.

70. From Figure 40-17, a rotational absorption spectrum would show peaks at energies of 2 I , 22 I ,

32 I , etc. Adjacent peaks are separated by an energy of 2 I . We use the photon frequency at

that energy to determine the rotational inertia.

  E
 2  I 2 2 h 6.631034 Js  2.0 1047 kgm2
I  E  hf  4 2 f  4 2 8.4 1011 Hz

71. To use silicon to filter the wavelengths, wavelengths below the IR should cause the electron to be
raised to the conduction band, so the photon is absorbed in the silicon. Let us find the shortest
wavelength that will cause the electron to jump.

    
 c  hc  hc  6.631034 Js 3.00 108 m/s  1.09 106 m  1.09 m
f hf Eg
1.60 1019 J/eV 1.14 eV

Because this is in the IR region of the spectrum, the shorter wavelengths of visible light will excite

the electron and the photon would be absorbed. So silicon could be used as a window.

72. The kinetic energy of the baton is 1 I2 , and the quantum number can be found from Eq. 40-2. Let
2

the length of the baton be d. We assume the quantum number will be very large.

1 I2  l l 12  l22 
2 2I
2I

I 2mend 2 2  2 f
 l   1 d  1 mbar d   
2 12

   
2  0.38 kg   0.16 m2  1  0.26 kg   0.32 m2  2 1.6s1  2.07 1033
12  1.055 1034 Js

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553

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

The spacing between rotational energy levels is given by Eq. 40-3. We compare that value to the

rotational kinetic energy.

l2

 E

E
 I  1  2 1  2.4 1034
l22 2l 2.07 1033

2I

This is such a small difference that it would not be detectable, so no, we do not need to consider

quantum effects.

73. From the diagram of the cubic lattice, we see that an atom inside the cube

is bonded to the six nearest neighbors. Because each bond is shared by
two atoms, the number of bonds per atom is 3 (as long as the sample is

large enough that most atoms are in the interior, and not on the boundary

surface). We find the heat of fusion from the energy required to break the

bonds:

Lfusion   number of bonds  number of atoms  Ebond
 atom   mol 

    3 6.02  1023 atoms mol 3.9  103eV 1.60  1019 J eV

 1127 J mol  1100J mol

74. The longest wavelength will be the photon with the minimum energy.

    Emin
 hc  max  hc  6.63 1034 Js 3.00 108 m/s  3.5 107 m
max Emin
3.6eV 1.60 1019 J/eV

So the photon must have   3.5 107 m .

75. The photon with the minimum frequency for conduction must have an energy equal to the energy

gap.

  Eg
 hf  hc  6.631034 Js 3.00 108 m/s  5.50eV
 1.60 1019 J/eV 226 109 m

76. (a) We calculate the Fermi temperature, for a Fermi energy of 7.0 eV.

 TF
 EF  7.0eV 1.60 1019 J/eV  8.1104 K
k
1.38 1023 J K

(b) We are given that T  TF, and we assume that eE / kT  1.

 fE  1  1  1  1  e E / kT
eEEF  / kT 1 1
e E  kTF  E E
 kT kT  e kT   1
e kT 

This is not useful for conductors like copper, because the Fermi temperature is higher than the

melting point, and we would no longer have a solid conductor.

77. We use Eq. 40-11 with the limits given in order to determine the number of states.

E2 dE  8 2 m3/ 2 E2 8 2 m3/ 2 2 E3/2
h3 E1 h3 3 1
     N  V g
E1
E V E1/ 2dE  V E3/2 
2

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554

Chapter 40 Molecules and Solids

3/ 2
 16
   
2 9.111031kg 0.1m3 6.2 eV 3/ 2  4.0 eV 3/ 2  1.60 1019 J eV 3/2
3 6.63 1034 Js 3 

 3.365 1025  3 1025

78. (a) For the glass to be transparent to the photon, the photon must have less energy than 1.14 eV,

and so the wavelength of the photon must be longer than the wavelength corresponding to 1.14

eV.

Eband gap  hc 
min

    min
 hc  6.631034 Js 3.00 108 m s  1.09 106 m    1.09 106 m
Eband gap
1.14eV 1.60 1019 J eV

The minimum wavelength for transparency is in the infrared region of the spectrum. Since IR

has longer wavelengths than visible light, the silicon would not be transparent for visible light.
The silicon would be opaque, as in Example 40-14.
(b) The minimum possible band gap energy for light to be transparent would mean that the band
gap energy would have to be larger than the most energetic visible photon. The most energetic
photon corresponds to the shortest wavelength, which is 450 nm in this problem.

   Eband gap
 Emin  hc  6.63 1034 Js 3.00 108 m s  2.7625eV  2.8eV
min 450 109 m 1.60 1019 J eV

79. The photon with the maximum wavelength for absorption must have an energy equal to the energy

gap.

  Eg
 hf  hc  6.63 1034 Js 3.00 108 m/s  6.47 104 eV
 1.60 1019 J/eV 1.92 103 m

80. (a) The electrons will not be moving fast enough at this low temperature to use relativistic

expressions, so the momentum is just the mass times the speed. The kinetic energy of the

electrons can be found from the temperature, by Eq. 18-4. The kinetic energy is used to

calculate the momentum, and the momentum is used to calculate the wavelength.

K  3 kT  p2  p 3mkT
2 2m

    
 h  h 6.63 1034 Js  6.27 109 m  6 nm
p 3mkT 
3 9 1031kg 1.38 1023 J K 300 K

(b) The wavelength is much longer than the opening, and so electrons at this temperature would
experience diffraction when passing through the lattice.

81. The photon with the longest wavelength has the minimum energy.

  Eg
 hc  6.631034 Js 3.00 108 m/s  1.130eV  1.1eV
 1.60 1019 J/eV 1100 109m

If the energy gap is any larger than this, some solar photons will not have enough energy to cause an
electron to jump levels, and so will not be absorbed.

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555

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

82. The energy gap is related to photon wavelength by Eg  hf  hc  . Use this for both colors of

LED.

  6.631034 Js 3.00 108 m/s
  Green: Eg  1.60 1019 J/eV 525 109 m  2.37eV

Blue:   6.631034 Js 3.00 108 m/s
  Eg  1.60 1019 J/eV 465 109 m  2.67eV

83. The arsenic ion has a charge of +1, since we consider the ion as having been formed by removing
one electron from the arsenic atom. Thus the effective Z will be 1, and we can use the Bohr theory

results for hydrogen. We also substitute KÅ0 in place of Å0.

(a) The Bohr energy is given in Eq. 37-14a and b. The binding energy is the opposite of the ground
state energy, so we use n = 1.

Ebinding  Z 2e4m 1  1  Z 2e4m  1 13.6eV  0.094 eV
n2 K2   
8KÅ0 2 h2  8Å02h2  122

(b) The Bohr radius is given in Eq. 37-11.

h2 KÅ0  h2Å0
 me2
 mZe2
 r1
  K  12 0.529 1010 m  6.31010 m

84. From Eq. 25-13, the number of charge carriers per unit volume in a current is given by n  I,
evdrift A

where vdrift is the drift velocity of the charge carriers, and A is the cross-sectional area through which

the carriers move. From Eq. 27-14, the drift velocity is given by vdrift  eH , where eH is the Hall-
Bd

effect voltage and d is the width of the strip carrying the current (see Figure 27-32). The distance d

is the shorter dimension on the “top” of Figure 40-47. We combine these equations to find the

density of charge carriers. We define the thickness of the current-carrying strip by t  A d .

     n
 I  IBd  IB  0.28 103 A 1.3T  1.264 1020 electrons m3
evdrift A eeH A eeHt 1.60 1019 C 0.018V 1.0 103 m

The actual density of atoms per unit volume in the silicon is found from the density and the atomic
weight. We let that be represented by N.

 N  m3  1mole   6.02  1023 atoms  4.994 1028 m3
2330 kg  28.0855 103   1mole    atoms
 kg   

n  1.264  1020 electrons m3  2.5 109 electrons atom
N 4.994 1028 atoms m3

85. We assume the 130 V value is given to the nearest volt.

(a) The current through the load resistor must be maintained at a constant value.

I load  Voutput  130 V   7.22 mA
Rload 18.0 k 

At the minimum supply voltage, there will be no current through the diode, so the current

through R is also 7.22 mA. The supply voltage is equal to the voltage across R plus the output

voltage.

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556

Chapter 40 Molecules and Solids

VR  IloadR  7.22 mA 2.80 k  20.2 V ; Vsupply  VR  Voutput  150 V
min

At the maximum supply voltage, the current through the diode will be 120 mA, and so the
current through R is 127 mA.

VR  IloadR  127.22 mA 2.80 k  356 V ; Vsupply  VR  Voutput  486 V
max

(b) The supply voltage is fixed at 245 V, and the output voltage is still to be 130 V. The voltage
across R is fixed at 245V  130 V  115V. We calculate the current through R.

IR  VR  115 V   41.1mA
R 2.80 k

If there is no current through the diode, this current will be in the load resistor.

Rload  Vload  130 V  3.16 k
I load 41.1mA

If Rload is less than this, there will be a greater current through R, meaning a greater voltage
drop across R, and a smaller voltage across the load. Thus regulation would be lost, so

3.16 k is the minimum load resistance for regulation.

If Rload is greater than 3.16 k, the current through Rload will have to decrease in order for the
voltage to be regulated, which means there must be current through the diode. The current
through the diode is 41.1 mA when Rload is infinite, which is less than the diode maximum of

120 mA. Thus the range for load resistance is 3.16 k  Rload  .

86. The voltage as graphed in Figure 40-40c decays exponentially, according to Eq. 26-9b. As
suggested in the problem, we use a linear approximation for the decay, using an expansion from
Appendix A-3. From Figure 40-40c, we see that the decay lasts for approximately one-half of a
cycle, before it increases back to the peak value.

   Vmin  Vpeaket /
 Vmin  et / 1 t 1 t 1 s1 1  0.97
Vpeak  RC
2 60

7.8 103 36 106 F

The voltage will decrease 3% from its maximum, or 1.5% above and below its average.

87. The spreadsheet used for this problem can be found on the Media Manager, with filenamef (E )
“PSE4_ISM_CH40.XLS,” on tab “Problem 40.87.”

(a) 1.00

0.75

T  500K

0.50

0.25

0.00
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

E /E F

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557

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

(b)f (E ) f (E )

1.00

0.75

T  1000K

0.50

0.25

0.00
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

E /E F

(c)

1.00

0.75

T  5000K

0.50

0.25

0.00
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

E /E F

(d) 1.00f (E )

0.75

T  10, 000K

0.50

0.25

0.00
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

E /E F

We see that for temperatures even as high as 1000 K, there is very little probability of electrons
being above the Fermi level. Only when the temperature gets quite high do we see a significant
increase in the probability of electrons to have an energy higher than the Fermi energy.

88. (a) The total potential energy is due to the electron-electron interaction, the proton-proton
interaction, and 4 electron-proton interactions.

U  Ue-e  Up-p  4U p-e  e2 1  e2 1  4   e2  1
4 0 d 4 0 r0  4 0 
   1 d 2   1 r0 2
2 2

 e2  1  1  8 
4 0  d r0 
 r02  d 2 

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558

Chapter 40 Molecules and Solids

(b) U has a minimum at U (eV) 200
150
d  0.043nm . U  0 for 100
50
the approximate range
0.011nm  d  0.51nm. 0
-50
The spreadsheet used for -100
this problem can be found
on the Media Manager, 0
with filename
“PSE4_ISM_CH40.XLS,”
on tab “Problem 40.88b.”

0.2 0.4 0.6 0.8

d (nm)

(c) To find the point of greatest stability, set the derivative of U with respect to d (indicated by U  )
equal to 0 and solve for d.

U  e2  1  1  8  
4 0  d r0 
r02  d 2 

e2  1  1  82d   e2  8d 1  0
4 0  d2  2   4 0  r02  d 2 d2 
 r02  d 2 3/ 2   
   U       3/ 2   



8d 1

     r02  d 2 3/2  d 2
 8d 3  r02  d 2 3/ 2  2d  r02  d 2 

4d 2  r02  d 2  d  r0  0.074  0.0427 nm
33

89. We first find the wavelength for a photon that has 1.14 eV of energy. This is the maximum

wavelength that will be able to make electron-hole pairs.

    Egap
 hc    hc  6.631034 Js 3.00 108 m s  1.09 106 m  1090 nm
 Egap
1.14eV 1.60 1019 J eV

Any wavelength shorter than 1100 nm will be effective. The 1000 W m2 value includes all

wavelengths of solar photons reaching the Earth, so we need to find what fraction of solar photons
have wavelengths below 1100 nm. We do this using the Planck formula, from Section 37-1. We
find the following using numeric integration, for a temperature of 6000 K.

1100 nm 2 hc2 5
0 ehc / kT  1
fraction d 

of effective photons   2 hc2 5
0 ehc / kT  1
d 

The Planck function is shown in the figure for T = 6000 K. We approximated the upper limit for the
full integration as 4000 nm, and obtained a ratio of 0.79. Thus we use an effective solar energy input
of 790 W m2 . To estimate the number of incoming photons, we use an average photon wavelength

of 500 nm, estimated simply by looking at the Planck function graph. We also assume that each

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559

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photon produces only one electron-
hole pair, even though many of
them would have enough energy to
create more than one.

The spreadsheet used for this I(,T)
problem can be found on the
Media Manager, with filename 0 500 1000 1500 2000 2500 3000 3500 4000
“PSE4_ISM_CH40.XLS,” on tab
“Problem 40.89.”  (nm)

I   Solar energy   1 "average" photon  1 electron produced  Charge 
A  sm2   Energy for 500 nm photon  1 solar photon   electron 
  

          
790 W m2  1.60 1019 C  790 W m2 500 109 m 1.60 1019 C
 hc  6.631034 Js 3.00 108 m s

 318C sm2  32 mA cm2

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560

CHAPTER 41: Nuclear Physics and Radioactivity

Responses to Questions

1. All isotopes of the same element have the same number of protons in their nuclei (and electrons in

the atom) and will have very similar chemical properties. Isotopes of the same element have
different numbers of neutrons in their nuclei and therefore different atomic masses.

2. (a) uranium; (b) nitrogen; (c) hydrogen; (d) strontium; (e) berkelium. The element is determined by
the atomic number.

3. Z is the number of protons and N is the number of neutrons.
(a) Z = 92, N = 140; (b) Z = 7, N = 11; (c) Z = 1, N = 0; (d) Z = 38, N = 44; (e) Z = 97, N = 150.

4. If there are 88 nucleons and 50 neutrons, then there are 38 protons. Strontium is the element with 38
protons.

5. The atomic masses given in the periodic table are averaged over the isotopes of the element in the
percentages in which they occur in nature. For instance, the most common form of hydrogen has one
proton and no neutrons, but other naturally occurring isotopes include deuterium (one proton and
one neutron) and tritium (one proton and 2 neutrons). These latter two are much less common, and
the resulting “weighted average” for the mass of hydrogen is 1.0079.

6. A force other than the gravitational force or the electromagnetic force is necessary to explain the
stability of nuclei. In most nuclei, several protons and neutrons are confined to a very small space.
The gravitational attractive force between the nucleons is very small compared with electromagnetic
repulsion between the protons. The strong nuclear force overcomes the electromagnetic repulsion
and holds the nucleus together.

7. The strong force and the electromagnetic force are two of the four fundamental forces in nature.

They are both involved in holding atoms together: the strong force binds quarks into nucleons and
nucleons together in the nucleus; the electromagnetic force is responsible for binding negatively-
charged electrons to positively-charged nuclei and atoms into molecules. The strong force is the
strongest of the four fundamental forces; the electromagnetic force is about 100 times weaker at
distances on the order of 10-17 m. The strong force operates at short range and is negligible for
distances greater than about the size of the nucleus. The electromagnetic force is a long range force
that decreases as the inverse square of the distance between the two interacting charged particles.
The electromagnetic force operates only between charged particles. The strong force is always
attractive; the electromagnetic force can be attractive or repulsive. Both these forces have mediating
field particles associated with them. The gluon is the particle for the strong force and the photon is
the particle for the electromagnetic force.

8. Chemical processes are the result of interactions between electrons. Radioactivity is not affected by
the external conditions that normally affect chemical processes, such as temperature, pressure, or
strong chemical reagents. Therefore, radioactivity is not a chemical process, but a nuclear one. In
addition, the energies associated with radioactivity are generally larger than energies corresponding
to electron orbital transitions, indicating that radioactivity is a nuclear process.

9. The resulting nuclide for gamma decay is the same isotope in a lower energy state:

64 Cu* 64 Cu  γ .
29 29

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561

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

The resulting nuclide for beta-minus decay is an isotope of zinc, 64 Zn :
30

64 Cu 64 Zn  e  .
29 30

The resulting nuclide for beta-plus decay is an isotope of nickel, 64 Ni :
28

64 Cu 64 Ni  e  .
29 28

10. 238 U decays by alpha emission into 23940Th, which has 144 neutrons.
92

11. Alpha (α) particles are helium nuclei. Each α particle consists of 2 protons and 2 neutrons, and
therefore has a charge of +2e and an atomic mass value of 4 u. Beta (β) particles are electrons (beta-
minus) or positrons (beta-plus). Electrons have a charge of –e and positrons have a charge of +e. In
terms of mass, beta particles are much lighter than protons or neutrons, by a factor of about 2000, so
are lighter than alpha particles by a factor of about 8000. Gamma (γ) particle are photons. They have
no rest mass and no charge.

12. (a) Magnesium is formed: 24 Na  1224Mg  e  .
11

(b) Neon is formed: 22 Na  22 Ne  e  .
11 10

(c) Lead is formed: 210 Po  206 Pb  4 He .
84 82 2

13. (a) Sulfur is formed: 32 Pb  S32  e  .
15
16

(b) Chlorine is formed: S35  1375Cl  e  .

16

(c) Thallium is formed: 211 Bi  82107Tl  4 He .
83 2

14. (a) 45 Sc ; (b) 58 Cu ; (c) e  ; (d) 230 U ; (e) e 
21 29 92

15. The two extra electrons held by the newly formed thorium will be very loosely held, as the number
of protons in the nucleus will have been reduced from 92 to 90, reducing the nuclear charge. It will
be easy for these extra two electrons to escape from the thorium atom through a variety of
mechanisms.

16. When a nucleus undergoes either β– or β+ decay it becomes a different element, since it has either
converted a neutron to a proton or a proton to a neutron and therefore its atomic number has
changed. The energy levels of the atomic electrons will adjust to become the energy levels of the
new element. Photons are likely to be emitted from the atom as electrons change energies to occupy
the new levels.

17. Alpha particles from an alpha-emitting nuclide are part of a two-body decay. The energy carried off
by the decay fragments is determined by the principles of conservation of energy and of momentum.
With only two decay fragments, these two constraints require the alpha particles to be
monoenergetic. Beta particles from a beta-emitting nucleus are part of a three-body decay. Again,
the energy carried off by all of the decay fragments is determined by the principles of conservation
of energy and of momentum. However, with three decay fragments, the energy distribution between
the fragments is not determined by these two constraints. The beta particles will therefore have a
range of energies.

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562

Chapter 41 Nuclear Physics and Radioactivity

18. Below. During electron capture, a proton in the nucleus becomes a neutron. Therefore, the isotopes
that undergo electron capture will most likely be those that have too few neutrons in the nucleus
(compared to their number of protons) and will lie below the line of stability in Figure 41-2.

19. No. Hydrogen has only one proton. Deuterium has one proton and one neutron. Neither has the two
protons and two neutrons required to form an alpha particle.

20. Many artificially produced radioactive isotopes have very short half-lives, and so are rare in nature
because they do not last long if they are produced naturally. Many of these isotopes also have a very
high energy of formation, which is generally not available in nature.

21. No. At the end of one month, ½ the sample will remain. At the end of two months, ¼ of the original
sample will remain.

22. For Z > 92, the short range of the attractive strong nuclear force means that no number of neutrons is
able to overcome the electrostatic repulsion of the large concentration of protons.

23. Helium-3, 3 He , will be the other particle released. There are a total of four protons and three
2

neutrons in the reactant particles. The alpha particle carries off two protons and two neutrons,

leaving two protons and one neutron.

24. No. Carbon-14 dating can only be used to date objects that were once living. The stone used to build
walls was never alive.

25. In β decay, a neutrino and a β particle (electron or positron) will be emitted from the nucleus, and the

number of protons in the nucleus changes. Because there are three decay products (the neutrino, the
β particle, and the nucleus), the momentum of the β particle can have a range of values. In internal
conversion, only an electron is emitted from the atom, and the number of protons in the nucleus
stays the same. Because there are only two decay products (the electron and the nucleus), the
electron will have a unique momentum and, therefore, a unique energy.

26. Figure 41-6 shows the potential energy curve for an alpha particle and daughter nucleus for the case
of radioactive nuclei. The alpha particle tunnels through the barrier from point A to point B in the
figure. In the case of stable nuclei, the probability of this happening must be essentially zero. The
maximum height of the Coulomb potential energy curve must be larger and/or the Q-value of the
reaction must be smaller so that the probability of tunneling is extremely low.

27. The decay series of Figure 41-12 begins with a nucleus that has many more neutrons than protons
and lies far above the line of stability in Figure 41-2. In a β+ decay, a proton is converted to a
neutron, which would take the nuclei in this decay series farther from the line of stability and is not
energetically preferred.

28. There are four alpha particles and four β– particles (electrons) emitted, no matter which decay path is

chosen. The nucleon number drops by 16 as 222 Rn decays into 206 Pb 206Pb, indicating the presence
86 82

of four alpha decays. The proton number only drops by four, from Z = 86 to Z = 82, but four alpha
decays would result in a decrease of eight protons. Four β– decays will convert four neutrons into

protons, making the decrease in the number of protons only four, as required. (See Figure 41-12.)

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563

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

Solutions to Problems

1. Convert the units from MeV c2 to atomic mass units.

 m  c2  1u 
139 MeV    0.149 u
 931.49 MeV c2 

2. The  particle is a helium nucleus and has A = 4. Use Eq. 41-1.

   1 1

r  1.2 1015 m A3  1.2 1015 m 43  1.9 1015 m  1.9 fm

3. The radii of the two nuclei can be calculated with Eq. 41-1. Take the ratio of the two radii.

 r238  2381 / 3
 r232  232 1 / 3
 1.2 1015 m   238 1/ 3  1.0085
1.2 1015 m  232 

So the radius of 238 U is 0.85% larger than the radius of 232 U.
92 92

4. Use Eq. 41-1 for both parts.

   (a) r  1.2 1015 m A1/3  1.2 1015 m 1121/3  5.8 1015 m  5.8fm

 (b)  r 3  3.7 1015 m 3
r  1.2 1015 m A1/ 3  A   1.2  1015 m   1.2 1015 m   29.3  29
  


5. To find the rest mass of an  particle, we subtract the rest mass of the two electrons from the rest
mass of a helium atom:
m  mHe  2me

    4.002603u 931.5MeV uc2  2 0.511MeV c2  3727 MeV c2

This is less than the sum of the masses of two protons and two neutrons because of the binding
energy.

6. Each particle would exert a force on the other through the Coulomb electrostatic force. The distance

between the particles is twice the raiius of one of the particles. The Coulomb force is given by Eq.

21-2.

      F  2
 1 q q  8.988 109 N  m2 C2 2 1.60 1019 C  63.41N  63 N
4 0
2r 2 2 41/3 1.2 1015 m 2

The acceleration is found from Newton’s second law. We use the mass of a “bare” alpha calculated

in Problem 5.

F  ma  a  F  63.41N  9.5 1027 m s2
m
3727 MeV c2  1.6605 1027 kg 
 931.49 MeV c2 
 

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564

Chapter 41 Nuclear Physics and Radioactivity

7. (a) The mass of a nucleus with mass number A is approximately (A u) and its radius is

 r  1.2  1015 m A1/3. Calculate the density.

     
 m  A 1.66  1027 kg u  A 1.66  1027 kg u  2.293 1017 kg m3 
V  1015 m 3A
4  r 3 4  1.2
3 3

 2.31017 kg m3

We see that this is independent of A.

(b) We find the radius from the mass and the density.

   M 1/ 3 1/ 3
 3M  3 5.98 1024 kg   184 m  180 m
  4  R 3  R   4    2.293 1017 kg m3 
3   
 4

(c) We set the density of the Earth equal to the density of the uranium nucleus. We approximate

the mass of the uranium nucleus as 238 u.

Earth  U   M Earth  mU 
R4 3 r4 3

3 Earth 3U

   rU 1/ 3 1/ 3
 REarth  mU   6.38 106 m  238 1.66 1027 kg 2.58 1010 m
 M Earth   
  5.98 1024 kg 


8. Use Eq. 41-1 to find the value for A. We use uranium-238 since it is the most common isotope.

  runknown
rU
 r  1.2  1015 m A1/ 3  A  2380.53  29.75  30
1.2  1015 m 238 1/ 3  0.5

From Appendix F, a stable nucleus with A  30 is 31 P .
15

9. The basic principle to use is that of conservation of energy. We assume that the centers of the two

particles are located a distance from each other equal to the sum of their radii. That distance is used

to calculate the initial electrical potential energy. Then we also assume that, since the nucleus is

much heavier than the alpha, that the alpha has all of the final kinetic energy when the particles are

far apart from each other (and so have no potential energy).

Ki Ui  Kf Uf  0 1 q qFm  K  0 
4 0
r  rFm 

2100 1.60 1019 C 2

41/ 3  2571/ 3 1.2 1015 m 1.60 1019 J eV
       K  8.988 109 N  m2 C2  3.017 107 eV

 30 MeV

10. (a) The hydrogen atom is made of a proton and an electron.

 mp
   mp  me
1.67 1027 kg  0.99945
 1.67 1027 kg  9.111031kg

(b) Compare the volume of the nucleus to the volume of the atom. The nuclear radius is given by

Eq. 41-1. For the atomic radius we use the Bohr radius, given in Eq. 37-12.

  3 3

  

  
rnucleus   1.2 1015 m 1.2 1014
ratom  0.53 1010 m


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

11. Electron mass is negligible compared to nucleon mass, and one nucleon weighs about 1.0 atomic
mass unit. Therefore, in a 1.0-kg object,

 (1.0 kg) 6.02  1026 u kg  6  1026 nucleons.

N  1.0 u nucleon

No, it does not matter what the element is, because the mass of one nucleon is essentially the same
for all elements.

12. The initial kinetic energy of the alpha must be equal to the electrical potential energy when the alpha

just touches the uranium. The distance between the two particles is the sum of their radii.

Ki Ui  Kf Uf  K  0  0 1 q qU  
4 0
r  rU

292 1.60 1019 C 2

41/ 3  2381/ 3 1.2 1015 m 1.60 1019 J eV
       K  8.988 109 N  m2 C2  2.832 107 eV

 28 MeV

13. From Figure 41–1, we see that the average binding energy per nucleon at A = 63 is about 8.7 MeV.
Multiply this by the number of nucleons in the nucleus.

638.7 MeV  548.1MeV  550 MeV

14. Deuterium consists of one proton, one neutron, and one electron. Ordinary hydrogen consists of one

proton and one electron. We use the atomic masses from Appendix F, and the electron masses

cancel.

     Binding energy  m1Hm 1 n m 2 H  c2
1 0 1

  1.007825u  1.008665u  2.014082 u c2 931.5MeV uc2

 2.243MeV

15. We find the binding energy of the last neutron from the masses of the isotopes.

     Binding energy  mP31m 1 n m P32  c2
0
15 15

  30.973762 u  1.008665u  31.973907 u c2 931.5MeV / c2

 7.94 MeV

16. (a) 7 Li consists of three protons and three neutrons. We find the binding energy from the masses,
3

using hydrogen atoms in place of protons so that we account for the mass of the electrons.

     Binding energy  3m1H 1 n 7 Li  c2
1  4m 0 m 3

  31.007825u  41.008665u  7.016005u c2 931.5MeV / c2

 39.24 MeV

Binding energy  39.24 MeV  5.61MeV nucleon
nucleon 7 nucleons

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566

Chapter 41 Nuclear Physics and Radioactivity

(b) 197 Au consists of 79 protons and 118 neutrons. We find the binding energy as in part (a).
79

     Binding energy  79m 1 H 1 197 Au  c2
1  118m 0 n m 79

  791.007825u 1181.008665u  196.966569 u c2 931.5MeV / c2

 1559 MeV

Binding energy  1559 MeV  7.91MeV nucleon
nucleon 197 nucleons

17. 23 Na consists of 11 protons and 12 neutrons. We find the binding energy from the masses:
11

     Binding energy  11m 1 1 n 23  c2
1 H  12m 0 m 11 Na

  111.007825u  121.008665u  22.989769 u c2 931.5MeV uc2

 186.6 MeV

Binding energy  186.6 MeV  8.113 MeV nucleon
nucleon 23

We do a similar calculation for 24 Na, consisting of 11 protons and 13 neutrons.
11

     Binding energy  11m 1 H 1 24 Na  c2
1  13m 0 n m 11

  111.007825u  131.008665u  23.990963u c2 931.5MeV uc2

 193.5MeV

Binding energy  193.5 MeV  8.063 MeV nucleon
nucleon 24

By this measure, the nucleons in 23 Na are more tightly bound than those in 24 Na.
11 11

18. We find the required energy by calculating the difference in the masses.

(a) Removal of a proton creates an isotope of carbon. To balance electrons, the proton is included

as a hydrogen atom: 15 N  1 H  14 C.
7 1 6

     Energy needed  m14 C m 1 H m 15 N  c2
6 1 7

  14.003242 u  1.007825u  15.000109 u 931.5MeV uc2

 10.21MeV

(b) Removal of a neutron creates another isotope of nitrogen: 15 N  1 n  14 N.
7 0 7

     Energy needed  m14 N m 1 n m 15 N  c2
7 0 7

  14.003074 u  1.008665u  15.000109 u 931.5MeV uc2

 10.83MeV

The nucleons are held by the attractive strong nuclear force. It takes less energy to remove the
proton because there is also the repulsive electric force from the other protons.

19. (a) We find the binding energy from the masses.

   Binding Energy  2m 4 He m 8 Be  c2
2 4

  24.002603u  8.005305u c2 931.5MeV uc2   0.092 MeV

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567

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

Because the binding energy is negative, the nucleus is unstable. It will be in a lower energy
state as two alphas instead of a beryllium.
(b) We find the binding energy from the masses.

   Binding Energy  3m 4 He m 12 C  c2
2 6

  34.002603u  12.000000 u c2 931.5MeV uc2   7.3MeV

Because the binding energy is positive, the nucleus is stable.

20. The decay is 3 H  3 He  0 e  v. When we add one electron to both sides to use atomic masses,
1 2 1

we see that the mass of the emitted  particle is included in the atomic mass of 3 He. The energy
2

released is the difference in the masses.

   Energy released  m 3 H m 3 He  c2
1 2

  3.016049 u  3.016029 u c2 931.5MeV uc2  0.019 MeV

21. The decay is 1 n  1 p  0 e  v. The electron mass is accounted for if we use the atomic mass of
0 1 1

1 H. If we ignore the recoil of the proton and the neutrino, and any possible mass of the neutrino, we
1

get the maximum kinetic energy.

     Kmax  m
1 n m 1 H  c2  1.008665u  1.007825u c2 931.5MeV uc2
0 1

 0.782 MeV

22. For the decay 11 C  10 B  1 p, we find the difference of the initial and the final masses:
6 5 1

     m  m161 C 10 B 1 H
m 5 m 1

 11.011434 u  10.012937 u  1.007825u   0.009328u

 11.011433u  10.012936 u  1.007825u   0.0099318u.

Since the final masses are more than the original mass, energy would not be conserved.

23. The wavelength is determined from the energy change between the states.

  E  hf  h c
 
   hc  6.63  1034 Js 3.00 108 m s  2.6 1012 m
E 1.60 1013 J MeV
0.48 MeV

24. For each decay, we find the difference of the initial and the final masses. If the final mass is more
than the initial mass, then the decay is not possible.

     (a) 1
m  m U232 m 0 n m U233  232.037156 u  1.008665u  233.039635u  0.006816 u

92 92

Because an increase in mass is required, the decay is not possible.

     (b) 13 N 1 n 14 N  13.005739 u  1.008665u 14.003074 u  0.011330 u
m  m 7 m 0 m 7

Because an increase in mass is required, the decay is not possible.

     (c) K39 1 n K40  38.963707 u  1.008665u  39.963998u  0.008374 u
m  m m 0 m
19 19

Because an increase in mass is required, the decay is not possible.

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568

Chapter 41 Nuclear Physics and Radioactivity

25. (a) From Appendix F, 24 Na is a   emitter .
11

(b) The decay reaction is 24 Na  24 Mg   v . We add 11 electrons to both sides in order to
11 12

use atomic masses. Then the mass of the beta is accounted for in the mass of the magnesium.

The maximum kinetic energy of the   corresponds to the neutrino having no kinetic energy (a

limiting case). We also ignore the recoil of the magnesium.

   K  m 24 Na m 24 Mg  c2
11 12

  23.990963u  23.985042 u c2 931.5MeV uc2  5.52 MeV

26. The kinetic energy of the electron will be maximum if the (essentially) massless neutrino has no
kinetic energy. We also ignore the recoil energy of the sodium. The maximum kinetic energy of the
reaction is then the Q-value of the reaction. Note that the “new” electron mass is accounted for by
using atomic masses.

     K  Q  m23 23  c2
10 Ne m 11 Na  22.9945u  22.9898u c2 931.5MeV uc2

 4.4 MeV

If the neutrino were to have all of the kinetic energy, then the minimum kinetic energy of the
electron is 0. The sum of the kinetic energy of the electron and the energy of the neutrino must be
the Q-value, and so the neutrino energies are 0 and 4.4 MeV, respectively.

27. (a) We find the final nucleus by balancing the mass and charge numbers.

Z  X   Z U  Z He  92  2  90

A X   AU  AHe  238  4  234

Thus the final nucleus is 234 Th .
90

(b) If we ignore the recoil of the thorium, the kinetic energy of the  particle is equal to the Q-
value of the reaction. The electrons are balanced.

     K  Q  m  c2
U238 m 23940Th m 4 He 
2
92

     m K
23940Th m U238 m 4 He  c2
2
92

 238.050788u  4.002603u  4.20 MeV  1u 
  
c2  931.5 MeV c2 

 234.04368u

This answer assumes that the 4.20 MeV value does not limit the sig. fig. of the answer.

28. The reaction is 60 Co  60 Ni    . The kinetic energy of the   will be maximum if the
27 28

(essentially) massless neutrino has no kinetic energy. We also ignore the recoil of the nickel.

   K  m 60Co  m 60 Ni  c2

  59.933822 u  59.930791u c2 931.5MeV / uc2  2.82 MeV.

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569

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

29. We use conservation of momentum – the momenta of the two particles must be equal and opposite if

there are only two products. The energies are small enough that we may use non-relativistic

relationships.

p  pX  2m K  2mX KX  KX  m K  4 5.0 MeV   0.078 MeV
mX 256

30. For alpha decay we have 218 Po  214 Pb  4 He. We find the Q value.
84 82 2

     Q  m218Pom 214 Pb m 4 He  c2
84 82 2

  218.008965u  213.999805u  4.002603uc2 931.5MeV uc2

 6.108 MeV

For beta decay we have 218 Po  218 At  0 e  v . We assume the neutrino is massless, and find the Q
84 85 1

value.

   Q  m218 Po m 218 At  c2
84 85

  218.008965u  218.008694 uc2 931.5MeV uc2  0.252 MeV

31. (a) We find the final nucleus by balancing the mass and charge numbers.

Z  X   Z P  Z e  15  1  16

A X   AP  Ae  32  0  32

Thus the final nucleus is 32 S .
16

(b) If we ignore the recoil of the sulfur and the energy of the neutrino, the maximum kinetic energy

of the electron is the Q-value of the reaction. The reaction is P32  S32   v. We add 15

15 16

electrons to each side of the reaction, and then we may use atomic masses. The mass of the

emitted beta is accounted for in the mass of the sulfur.

   K  Q  m P32 m S32  c2 

15 16

   mS32 P32 K 31.973907 1.71MeV  1u 
m  c2   u  c2  931.5 MeV c2   31.972071u
16 15  

32. We find the energy from the wavelength.

  E  hc 
  
6.63  1034 Js 3.00  108 m s  12.4 MeV
1.00  1013 m 1.602  1019 J eV

This has to be a  ray from the nucleus rather than a photon from the atom. Electron transitions do
not involve this much energy. Electron transitions involve energies on the order of a few eV.

33. We add three electron masses to each side of the reaction 7 Be  0 e  7 Li  v. Then for the mass
4 1 3

of the product side, we may use the atomic mass of 7 Li. For the reactant side, including the three
3

electron masses and the mass of the emitted electron, we may use the atomic mass of 7 Be. The
4

energy released is the Q-value.

   Q  m7 Be m 7 Li  c2
4 3

  7.016930 u  7.016005u c2 931.5MeV uc2  0.862 MeV

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570

Chapter 41 Nuclear Physics and Radioactivity

34. The emitted photon and the recoiling nucleus have the same magnitude of momentum. We find the
recoil energy from the momentum. We assume the energy is small enough that we can use classical
relationships.

p  E  pK  2mK KK 
c

E2 1.46 MeV2  2.86  105 MeV  28.6 eV
2mKc2 u 931.5MeV
 KK  uc2 c2
2  39.96

35. The kinetic energy of the   particle will be maximum if the (almost massless) neutrino has no

kinetic energy. We ignore the recoil of the boron. Note that if the mass of one electron is added to

the mass of the boron, then we may use atomic masses. We also must include the mass of the  .

(See Problem 38 for details.)

 11C 11 B  0 e  10   v
5 1
6

             K  m1111 0 0  11 11 0  c2
6 C m 5 B m 1 e m  c2  m 6 C m 5 B  2m 1 e
1

  11.011434 u  11.009305  20.00054858u c2 931.5MeV uc2  0.9612 MeV

If the   has no kinetic energy, then the maximum kinetic energy of the neutrino is also

0.9612 MeV . The minimum energy of each is 0, when the other has the maximum.

36. We assume that the energies are low enough that we may use classical kinematics. In particular, we

will use p  2mK . The decay is 238 U  23940Th  4 He. If the uranium nucleus is at rest when it
92 2

decays, the magnitude of the momentum of the two daughter particles must be the same.

p  pTh ; KTh  pTh 2  p 2  2m K  m K   4u   4.20 MeV   0.0718 MeV
2mTh 2mTh 2mTh mTh  234 u 
 

The Q-value is the total kinetic energy produced.

Q  K  KTh  4.20 MeV  0.0718MeV  4.27 MeV

37. Both energy and momentum are conserved. Therefore, the momenta of the product particles are
equal in magnitude. We assume that the energies involved are low enough that we may use classical

kinematics; in particular, p  2mK .

p  pPb ; K Pb  pP2b  p2  2m K   m  K  4.0026 K
2mPb 2mPb 2mPb  mPb  205.97
 

The sum of the kinetic energies of the product particles must be equal to the Q-value for the reaction.

     KPb  K  m 210 Po m 206 Pb m 4 He  c2  4.0026 K  K 
84 82 2 205.97

     K  m210Po m 206 Pb m 4 He  c2
84 82 2

 4.0026  1
 205.97

209.982874 u   205.974465 u    4.002603 u  c2
 
 4.0026 1 931.5MeV uc2  5.31MeV
 205.97


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571

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

38. For the positron-emission process, A N  Z A N  e  v. We must add Z electrons to the nuclear
Z 1

mass of N to be able to use the atomic mass, and so we must also add Z electrons to the reactant

side. On the reactant side, we use Z 1 electrons to be able to use the atomic mass of N. Thus we

have 1 “extra” electron mass and the -particle mass, which means that we must include 2 electron
masses on the right-hand side. We find the Q-value given this constraint.

Q  M P   M D  2me  c2   M P  M D  2me c2.

39. (a) The decay constant can be found from the half-life, using Eq. 41-8.

 ln 2  ln 2  1.5 1010 yr1  4.9 1018s1
 T1/ 2 4.5 109 yr

(b) The half-life can be found from the decay constant, using Eq. 41-8.

T1/ 2  ln 2  ln 2  21660s  6.0 h
 3.2 105 s1

40. We find the half-life from Eq. 41-7d and Eq. 41-8.

R R0et R e ln 2 t T1/ 2 ln 2 t ln 2 3.6h  1.8 h
T1/ 2 ln R ln 320
  0     

R0 1280

We can see this also from the fact that the rate dropped by a factor of 4, which takes 2 half-lives.

41. We use Eq. 41.6 to find the fraction remaining.

N  e  ln 22.0 yr12mo yr 
N0 
N  N0et   et  9 mo   0.158  0.16


42. The activity at a given time is given by Eq. 41-7b. The half-life is found in Appendix F.

ln 2 ln 2
T1/ 2 3.16 107 s yr
dN

   dt
 N  N  5730 yr 8.11020 nuclei  3.1109 decays s

43. Every half-life, the sample is multiplied by one-half.

N   1 n   1 6  0.015625
N0 2 2

44. The activity of a sample is given by Eq. 41-7a. There are two different decay constants involved.

Note that Appendix F gives half-lives, not activities.

 I NI  Co NCo  I N0eIt  N eCot   I  eI Co t 
 Co
Co 0

 TCo 
 1/2 
ln   ln   5.2710 y365.25d y
TI  8.0233 d 
 ln I  Co  1/2     63.703d

t
 I  Co ln 2 ln 2  1 1 
TI  TCo ln 2   
 8.0233d  5.2710 y   365.25 d y  
1/2 1/2

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572

Chapter 41 Nuclear Physics and Radioactivity

45. We find the number of nuclei from the activity of the sample and the half-life.

dN  N  ln 2 N 
dt T1/ 2

  N
 T1/ 2 dN  4.468 109 yr 3.156 107 s yr 340decays s  6.9 1019 nuclei
ln 2 dt ln 2

46. Each  emission decreases the mass number by 4 and the atomic number by 2. The mass number
changes from 235 to 207, which is a change of 28. Thus there must be 7 particles emitted. With

the 7  emissions, the atomic number would have changed from 92 to 78. Each   emission
increases the atomic number by 1, so to have a final atomic number of 82, there must be
4   particles emitted.

47. We need both the decay constant and the initial number of nuclei.

  ln 2  ln 2 day   3600 s h  9.99905 107 s1
T1/ 2
8.0233days24 h

  782 106 g 
 N0
  130.906 g mol   6.02 1023 atoms mol  3.596 1018 nuclei.
 

(a) We Eq. 41-7b to evaluate the initial activity.

  dN
dt  9.99905 107 s1 3.596 1018  3.5957 1012 decays s  3.60 1012 decays s

0

(b) We evaluate Eq. 41-7c at t  1.0 h.

 dN dN et e   9.99905107 s1 3600s
dt
dt 0
  3.5957 1012 decays s

 3.58 1012 decays s

(c) We evaluate Eq. 41-7c at t = 4 months. We use a time of 1/3 year for the 4 months.

 dN dN e     9.99905107 s1 0.333yr  3.156107 s yr
dt
dt
 et  3.5957 1012 decays s

0

 9.72 107 decays s

48. We will use the decay constant frequently, so we calculate it here.

  ln 2  ln 2  0.022505s1
T1/ 2 30.8s

(a) We find the initial number of nuclei from the atomic mass.

 7.8  106 g  3.787 1016  3.8  1016 nuclei.
 N0  124 g mol
6.02  1023 atoms mol

(b) Evaluate Eq. 41-6 at t  2.6 min.

 N  N0et  3.787  1016 e   0.022505s1 2.6min60s min  1.1311015  1.1  1015 nuclei

(c) The activity is found by Eq. 41-7a.

   N  0.022505s1 1.131  1015  2.545 1013  2.5  1013 decays s

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573

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

(d) We find the time from Eq. 41-7a.
N   N0et 

ln  N  ln  1decay/s 
  N0   0.022505s1 3.787 1016 
  decay/s   1528s  25.46 min 
0.022505s1
  t    25 min



49. We find the mass from the initial decay rate and Eq. 41-7b.

dN  N0   m 6.02 1023 nuclei mole 
dt mole
0 atomic weightg

dN 1 atomic weight dN T1/ 2 atomic weight
dt 0 dt
6.02 1023 0 ln 2 6.02 1023
   m  

       2.0 105s1
1.265 109 yr 3.156 107 s yr  39.963998 g   0.76 g

ln 2 6.02 1023

50. The number of nuclei is found from the mass and the atomic weight. The activity is then found from

number of nuclei and the half-life, using Eq. 41-7b.

 dN
 dt
 N  ln 2 N  ln 2  8.7 106 g  6.02 1023 atoms mol  9.2 1010 decays s
T1/ 2 1.23  106 s  
 31.974 g mol 

51. We find the mass from the initial decay rate and Eq. 41-7b.

dN  N0   m 6.02 1023 nuclei mole 
dt mole
0 atomic weightg

dN 1 atomic weight dN T1/2 atomic weight 
dt 0 dt
6.02 1023 0 ln 2 6.02 1023
   m  

87.32d86,400s d 34.969032g
    2.31  1011 g
3.65 104 s1 ln 2 6.02 1023 

52. (a) The decay constant is found from Eq. 41-8.

ln 2 ln 2  1.38 1013 s1
T1/ 2 1.59 105 yr 3.156 107 s yr
   

(b) The activity is the decay constant times the number of nuclei.

   N  1.38 1013 s1  7.59 105 decays  60 s  4.55 107 decays
5.50 1018 s  1min   min .
 

53. We use Eq. 41-7c.

R R0et R e ln 2 t T1/ 2 ln 2 t ln 2 8.6 min  4.3 min
T1/2 ln
  0    R   1 
R0 4
ln

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574

Chapter 41 Nuclear Physics and Radioactivity

54. Because the fraction of atoms that are 14 C is so small, we use the atomic weight of 12 C to find the
6 6

number of carbon atoms in the sample. The activity is found from Eq. 41-7a.

 385g 
 N  1.931025 atoms
  12 g mol  6.02 1023 atoms mol
 

 N14  1.3 
  1012  1.93  1025  2.511013 nuclei

 ln 2 
 3.156 107 s 

5730 yr
   N
yr 2.511013  96decays s

55. We find the mass from the activity. Note that NA is used to represent Avogadro’s number.

R  N  ln 2 mNA 
T1/ 2 A

    m 
RT1/2 A   370 decays s 4.468 109 yr 3.156 107 s yr  238.05 g mole 2.98 102 g
NA ln 2 6.02 1023 nuclei mole ln 2


56. We assume that the elapsed time is much smaller than the half-life, so that can approixmate the

decay rate as being constant. We also assume that the 87 Sr is stable, and there was none present
38

when the rocks were formed. Thus every atom of 87 Rb that decayed is now an atom of 87 Sr.
37 38

NSr   NRb   NRb t  t  NSr T1/ 2  0.0260 4.75 1010 yr  1.78 109 yr
N Rb ln 2 ln 2

This is  4% of the half-life, so our original assumption is valid.

57. The activity is given by Eq. 41-7a.

0.975 N0   N0et  ln 0.975  t   ln 2 t 
T1/ 2

T1/ 2   ln ln 2 31.0 h   848.71h  1d   35.4 d
0.975  24 h 
 

58. The activity is given by Eq. 41-7a.
(a) We use Eq. 41-7c.We find the number of half-lives from

R  R0et  t   1 ln R   T1/ 2 ln R   53d  ln 15 decays s   240.85 d  240 d
 R0 ln 2 R0 ln 2  350 decays s 
 

(b) We find the mass from the activity. Note that NA is used to represent Avogadro’s number.

R0   N0  ln 2 m0 NA 
T1/ 2 A

R0T1/ 2 A 350 decays s53d86, 400s d 7.017 g mole 2.7 1014 g
NA ln 2 ln 2
6.02 1023 nuclei mole
 m   

59. 232 Th  228 Ra  24 ; 228 Ra  228 Ac   ; 228 Ac  228 Th   ; 228 Th  224 Ra  42 ;
90 88 88 89 89 90 90 88

224 Ra  220 Rn  42
88 86

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575

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

235 U  231 Th  24 ; 231 Th  231 Pa   ; 231 Pa  227 Ac  42 ; 227 Ac  227 Th   ;
92 90 90 91 91 89 89 90

22970Th  223 Ra  24
88

60. Because the fraction of atoms that are 14 C is so small, we use the atomic weight of 12 C to find the
6 6

number of carbon atoms in 85 g. We then use the ratio to find the number of 14 C atoms present
6

when the club was made. Finally, we use the activity as given in Eq. 41-7c to find the age of the

club.

 85g 
 N12
6
C   12 g mol   6.02 1023 atoms mol  4.264 1024 atoms
 

  N 14
6
C 1.3  1012 4.264 1024  5.5431012 nuclei

    N 14
6
C N e t14
today 6 C 0

   t   1 ln N 14 C today   T1/ 2 ln  N 14 C today
  6 ln 2 6

 N 14 C  ln 2 N 
6  T1/ 2 14
0  C
6
0

  5730 yr ln 7.0decays s  9178 yr  9200 yr
ln 2
 ln 2 
3.156 107 s yr 
  0
 5730 yr
5.5431012 nuclei

61. The number of radioactive nuclei decreases exponentially, and every radioactive nucleus that decays
becomes a daughter nucleus.
N  N0et

 ND  N0  N  N0 1  et

62. The activity is given by Eq. 41-7d.

ln R ln 2 t ln 2 4.00 hln 2
 T1/2 ln R
R  R0et     R0  T1/ 2     ln 0.01050  0.6085h  36.5min
t

R0

From Appendix F we see that the isotope is 211 Pb .
82

63. Because the carbon is being replenished in living trees, we assume that the amount of 14 C is constant
6

until the wood is cut, and then it decays. We use Eq. 41-6.

ln N T1/ 2  5730 yr  ln 0.060
N0 ln 2 ln 2 N
N  N0et   t  T1/2  t   ln N0   ln 2  23,000 yr

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576

Chapter 41 Nuclear Physics and Radioactivity

64. (a) The mass number is found from the radius, using Eq. 41-1.
(b)
 r  1.2 1015 m A1/3  A   r 3   5000 m 3  7.231055  7 1055
 1015   
 1.2  m   1.2 1015 m 

The mass of the neutron star is the mass number times the atomic mass unit conversion in kg.

    m  A 1.66 1027 kg u  7.231055 u 1.66 1027 kg u  1.20 1029 kg  11029 kg

Note that this is about 6% of the mass of the Sun.

(c) The acceleration of gravity on the surface of the neutron star is found from Eq. 6-4 applied to

the neutron star.

  g
 Gm  6.67 1011 Nm2 kg2 1.20  1029 kg  3.20 1011 m s2  31011 m s2
r2
5000 m2

65. Because the tritium in water is being replenished, we assume that the amount is constant until the

wine is made, and then it decays. We use Eq. 41-6.

ln N ln 2 T1/ 2 N 12.3 yrln 0.10
 T1/2 ln 2 N0
N  N0et     N0  t   ln   ln 2  41yr
t

66. We assume a mass of 70 kg of water, and find the number of protons, given that there are 10 protons

in a water molecule.

   Nprotons
 70 103 g water  6.02 1023 molecules water mol water  10 protons 
    water molecule 
 18 g water mol water  

 2.34 1028 protons

We assume that the time is much less than the half-life so that the rate of decay is constant.

N  N  ln 2  N  t  N  T1/ 2   1 proton  1033 yr   60,000 yr
t  T1/ 2  N  ln 2   
 2.34 1028 protons  ln 2 


This is almost 1000 times a normal life expectancy.

67. Consider the reaction n  p  e  v. The neutron, proton, and electron are all spin 1 particles. If
2

the proton and neutron spins are aligned (both are 1 , for example), then the electron and neutrino
2

spins must cancel. Since the electron is spin 1 , the neutrino must also be spin 1 in this case.
2 2

The other possibility is if the proton and neutron spins are opposite of each other. Consider the case

of the neutron having spin 1 and the proton having spin  1 . If the electron has spin 1 , then the
2 2 2

spins of the electron and proton cancel, and the neutrino must have spin 1 for angular momentum to
2

be conserved. If the electron has spin  1 , then the spin of the neutrino must be 3 for angular
2 2

momentum to be conserved.

A similar argument could be made for positron emission, with p  n  e  v.

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577

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

68. We assume that all of the kinetic energy of the alpha particle becomes electrostatic potential energy

at the distance of closest approach. Note that the distance found is the distance from the center of the

alpha to the center of the gold nucleus.

Ki Ui  Kf Uf  K 00 1 q qAu 
4 0 r

279 1.60 1019 C 2
7.7MeV 1.60 1013 J MeV
    r1 q qAu  2.9511014 m
 4 0 K  8.988 109 N  m2 C2

 3.0 1014 m

We use Eq. 41-1 to compare to the size of the gold nucleus.

 rapproach
rAu
 2.9511014 m  4.2
1971/3 1.2 1015 m

So the distance of approach is about 4.2  the radius of the gold nucleus.

69. We find the number of half-lives from the change in activity.

dN

dt   1 n  0.0100  n ln 0.0100  6.64 half-lives
dN 2
ln 1
2

dt 0

70. We find the mass from the activity. Note that NA is used to represent Avogadro’s number.

R  N  ln 2 m40 NA 
T1/ 2 A

    m40
 RT1/2 A  45 decays s 1.265109 yr 3.156 107 s yr 39.964g mole  1.721104 g
NA ln 2
 6.02 1023 nuclei mole ln 2

 1.7 104 g

We find the number of K39 atoms from the number of K40 atoms and the abundance given in

19 19

Appendix F. That is then used to find the mass of 39 K.
19

R   N40  N40  R  RT1/ 2 ; N40  0.000117NK ; N39  0.93258NK  0.93258 N 40
 ln 2 0.000117

m39  N39 A  0.93258 N 40 A  0.93258 RT1/ 2 A
NA 0.000117 NA 0.000117 ln 2 NA

    
 0.93258   45 decays s 1.265 109 yr 3.156 107 s yr 38.964g mole
 0.000117  ln 2
6.02 1023 nuclei mole

 1.3g

71. (a) If the initial nucleus is at rest when it decays, momentum conservation says that the magnitude
of the momentum of the alpha particle will be equal to the magnitude of the momentum of the
daughter particle. We use that to calculate the (non-relativistic) kinetic energy of the daughter
particle. The mass of each particle is essentially equal to its atomic mass number, in atomic
mass units.

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578

Chapter 41 Nuclear Physics and Radioactivity

p  pD ; KD  pD2  p2  2m K  m K  A K  4 K
2mD 2mD 2mD mD AD AD

KD 4 K K 1
K  KD AD K 
  
 4   AD  1 1 AD
 K  AD K   4 K  4



(b) We specifically consider the decay of 226 Ra. The daughter has AD  222.
88

KD  1  1 1  0.017699  1.8%
K  KD
1 1 AD 1  222
4 4

Thus the alpha particle carries away 1  0.018  0.982  98.2% .

72. We see from the periodic chart that Sr is in the same column as calcium. If strontium is ingested,

the body may treat it chemically as if it were calcium, which means it might be stored by the body

in bones. We use Eq. 41-6 to find the time to reach a 1% level.

ln N
N0 ln 2
N  N0et   t  T1/2 

t   T1/ 2 ln N    29 yrln 0.010  192.67 yr  190 yr
ln 2 N0
ln 2

The decay reactions are as follows. We assume the daughter undergoes beta decay.

90 Sr  Y90  0 e  v ; Y 90 90 Zr  0 e  v
38 1 1
39 39 40

73. We take the momentum of the nucleon to be equal to the uncertainty in the momentum of the

nucleon, as given by the uncertainty principle. The uncertainty in position is estimate as the radius

of the nucleus. With that momentum, we calculate the kinetic energy, using a classical formula.

px    p  p    
x r

       K 2
 p2  2  1.67 1027 kg 1.055 1034 Js 1.60 1013 J
2m 2mr 2 2 561/3 1.2 1015 m  2
 MeV

 0.988MeV  1MeV

74. (a) The reaction is 236 U  23920Th  4 He. If we assume the uranium nucleus is initially at rest, then
92 2

the magnitude of the momenta of the two products must be the same. The kinetic energy

available to the products is the Q-value of the reaction. We use the non-relativistic relationship

that p2  2mK.

pHe  pTh ; KTh  pT2h  pH2 e  2mHe KHe  mHe K He
2mTh 2mTh 2mTh mTh

Q  KTh  K He   mHe  1 KHe 
 mTh 


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579

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

 mTh   mTh  c2
 mHe  mTh   mHe  mTh 
 KHe  Q  
  mU  mTh  mHe



  232.038055 u u  236.045568 u  232.038055 u  4.002603 u  c2
 4.002603u  232.038055 
 

  0.004827u 931.5MeV uc2  4.496 MeV

(b) We use Eq. 41-1 to estimate the radii.

 rHe  1.2 1015 m  4 1/3  1.905 1015 m  1.9 1015 m

 rTh  1.2 1015 m 2321/3  7.374 1015 m  7.4 1015 m

(c) The maximum height of the Coulomb barrier will correspond to the alpha particle and the

thorium nucleus being separated by the sum of their radii. We use Eq. 23-10.

1 qHe qTh 1 qHe qTh
4 0 rA 4 0 rHe  rTh
 U 

 290 1.60 1019 C 2
      8.988 109 N  m2 C2 41/3  2321/3 1.2 1015 m 1.60 1013 J MeV

 27.898MeV  28MeV

(d) At position “A”, the product particles are separated by the sum of their radii, about 9.3 fm. At

position “B”, the alpha particle will have a potential energy equal to its final kinetic energy,

4.496 MeV. Use Eq. 23-10 to solve for the separation distance at position “B”.

UB  1 qHe qTh 
4 0 rB

290 1.60 1019 C 2
4.496 MeV 1.60 1013 J MeV
    rB1 qHe qTh C2
 4 0 UB  8.988 109 N  m2

 57.57 1015 m

rB  rA  57.6fm  9.3fm  48.3fm
Note that this is a center-to-center distance.

75. (a) We find the daughter nucleus by balancing the mass and charge numbers:

Z  X   Z Os  Z e   76  1  77 17961Os

A X   AOs  Ae   191  0  191  – (0.14 MeV)

The daughter nucleus is 191 Ir . 191 Ir*
77 77
 (0.042 MeV)
(b) See the included diagram.  (0.129 MeV) 191 Ir*
77
(c) Because there is only one  energy, the 
decay must be to the higher excited state. 191 Ir
77

76. The activity is the decay constant times the number of nuclei, as given by Eq. 41-7a.

(a) We calculate the activity for 131 I.
53

ln 2 ln 2  1.0g 
T1/ 2
d  86,
 R  N
 N  8.02 400 s d   130.906 g mol  6.02  1023 nuclei mol
 

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580

Chapter 41 Nuclear Physics and Radioactivity

 4.6 1015 decays s

(b) We calculate the activity for 238 U.
92

ln 2 ln 2  1.0g 
T1/ 2 4.47 109 yr 3.156 107 s yr
    R
 N   238.051g mol  6.02  1023 nuclei mol
 

 1.2 104 decays s

77. From Figure 30–1, we see that the average binding energy per nucleon at A  63 is ~8.6 MeV. We
use the mass average atomic weight as the average number of nucleons for the two stable isotopes of

copper. That gives a binding energy of 63.5468.6 MeV  546.5MeV  550 MeV .

The number of atoms in a penny is found from the atomic weight.

 3.0 g 
 63.546 g mol
 N
 6.02 1023 atoms mol  2.842 1022 atoms

Thus the total energy needed is the product of the number of atoms times the binding energy.

   2.842 1022 atoms 546.5MeV atom 1.60 1013 J MeV  2.5  1012 J

     78. (a)4He 4 He 4 He  4.002603u  4  0.002603u
 2 m 2 A 2

  0.002603u 931.5MeV uc2  2.425Mev c2

     (b)12C 12 C A 12 C  12.000000 u  12  0
 6 m 6  6

     (c)86Sr 86 Sr 86 Sr  85.909260 u  86   0.090740 u
 38 m 38 A 38

   0.090740 u 931.5MeV uc2  84.52 MeV c2

     (d)
 U235 m U235 A U235  235.043930 u  235  0.043930 u

92 92 92

  0.043930 u 931.5MeV uc2  40.92 MeV c2

(e) From the Appendix we see that   0 for 0  A  15 and A  214;
  0 for 0  Z  8 and Z  85;   0 for 16  A  214.
  0 for 9  Z  84.

79. The reaction is 1 H  1 n  2 H. If we assume the initial kinetic energies are small, then the energy of
1 0 1

the gamma is the Q-value of the reaction.

     Q  m1He m 1 n m 2 He  c2
1 0 1

  1.007825u  1.008665u  2.014082 u c2 931.5MeV uc2  2.243MeV

80. (a) We use the definition of the mean life given in the problem. We use a definite integral formula
from Appendix B-5.

   1 1
2 2
tN tN 0e  t dt te  t dt  1 et 1
    t dt  1
 0 t dt  0  0      
   0

N N 0e  t dt e  t dt

0 0 0

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581

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

(b) We evaluate at time t    1 .


N t    N 0e    e    e1  0.368
N t  0 N 0e 0 

81. (a) The usual fraction of 14 C is 1.3  1012. Because the fraction of atoms that are 14 C is so small,
6 6

we use the atomic weight of 12 C to find the number of carbon atoms in 72 g. We use Eq. 41-6
6

to find the time.

 72g 
 N12   3.612 1024 atoms
  12 g mol  6.02 1023 atoms mol


  N14  3.612 1024 atoms 1.31012  4.6956 1012 atoms

N  N0et  t   1 ln N   T1/ 2 ln N   5730 yr  ln 1  2.4 105 yr
 N0 ln 2 N0 4.6956 1012
ln 2

(b) We do a similar calculation for an initial mass of 270 grams.

 270 g 
 12g mol 
  N14 
 6.02 1023 atoms mol 1.3  1012  1.7611013 atoms

N  N0et  t   1 ln N   T1/ 2 ln N    5730 yr  ln 1  2.5 105 yr
 N0 ln 2 N0 ln 2 1.7611013

This shows that, for times on the order of 105 yr, the sample amount has fairly little effect on

the age determined. Thus, times of this magnitude are not accurately measured by carbon
dating.

82. (a) This reaction would turn the protons and electrons in atoms into neutrons. This would eliminate
chemical reactions, and thus eliminate life as we know it.

(b) We assume that there is no kinetic energy brought into the reaction, and solve for the increase of
mass necessary to make the reaction energetically possible. For calculating energies, we write

the reaction as 1 H  1 n  v, and we assume the neutrino has no mass or kinetic energy.
1 0

     Q  m
1 H m 1 n  c2  1.007825 u   1.008665u c2 931.5MeV uc2
1 0

  0.782 MeV

This is the amount that the proton would have to increase in order to make this energetically
possible. We find the percentage change.

 
 
m  100   0.782 MeV c2  0.083%
m   938.27 MeV c2
  100 



83. We assume the particles are not relativistic, so that p  2mK . The radius is given in Example 27-7

as r  mv . Set the radii of the two particles equal. Note that the charge of the alpha particle is twice
qB

that of the electron (in absolute value). We also use the “bare” alpha particle mass, subtracting the
two electrons from the helium atomic mass.

m v  m v  m v  2m v  p  2 p
2eB eB

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582

Chapter 41 Nuclear Physics and Radioactivity

p2 4 p2

K  2m  2m  4m  40.000549 u u   5.48  104
K p2 p2 m 4.002603u  20.000549

2m 2m

84. Natural samarium has an atomic mass of 150.36 grams per mole. We find the number of nuclei in

the natural sample, and then take 15% of that to find the number of 147 Sm nuclei. We first find the
62

number of 147Sm nuclei from the mass and proportion information.

 N147
62
  0.15 Nnatural  0.151.00g 6.02 1023 nuclei / mol  6.006 1020 nuclei of 147 Sm
62
Sm 150.36g mol

The activity level is used to calculate the half-life.

Activity  R  N  ln 2 N 
T1/ 2

ln 2 ln 2  1yr 
R  120decays  3.156 107 s 
 T1/2
 N s 6.006 1020  3.469 1018 s  1.11011 yr

85. Since amounts are not specified, we will assume that “today” there is 0.720 g of U235 and

92

100.000  0.720  99.280g of 238 U. We use Eq. 41-6.
92

(a) Relate the amounts today to the amounts 1.0 109 years ago.

t ln 2

N  N0et  N0  Net  NeT1/2

t ln 2  1.0109 yr
e  ln 2
e T1/2 7.04108 yr
     N0
235  N 235 0.720 g  1.927 g

e t ln 2  1.0109
T1/2 e 4.468109 ln 2  115.94 g
     N0
238  N 238 99.280 g

0.693 t  0.693 1.0109

T1/ 2 99.28 g e  4.468109
238
 N  N e0,238   115.937 g.

The percentage of U235 was 1.927  100%  1.63%
1.927  115.94
92

(b) Relate the amounts today to the amounts 100 106 years from now.

 t ln 2  100106 yr
T1/2 0.720 g e   7.04108 yr ln 2  0.6525g
     N  N0et 
N 235  N e 0 235

 t ln 2  100106 yr
T1/2 99.280 g e   4.468109 yr ln 2  97.752 g
     N238
 N e 0 238

The percentage of U235 will be 0.6525  100%  0.663%
0.6525  97.752
92

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583

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

86. We determine the number of K40 nuclei in the sample, and then use the half-life to determine the

19

activity.

  N 40
19
K   0.000117 N 39 K  0.000117 400 103 g 6.02  1023 atoms mol  7.2311017
19 38.9637 g mol

ln 2 ln 2  1yr 
T1/ 2 1.265 109  3.156 107 
 R
 N  N  yr 7.2311017 s  12.55 decay s  13decay s

87. We use Eq. 41-7a to relate the activity to the half-life.

dN  R  N  ln 2 N 
dt T1/ 2

ln 2 ln 2  6.02 1023 nuclei   1yr 
R 1decay  152 g   3.156 107 
 T1/2   1.31021 yr
 N  s 1.5 107 g s 

88. The mass number changes only with  decay, and changes by  4. If the mass number is 4n, then

the new number is 4n  4  4n  1  4n. There is a similar result for each family, as shown here.
4n  4n  4  4n 1  4n
4n  1  4n  4  1  4n 1 1  4n  1
4n  2  4n  4  2  4n 1  2  4n  2
4n  3  4n  4  3  4n 1  3  4n  3

Thus the daughter nuclides are always in the same family.

89. We calculate the initial number of nuclei from the initial mass and the atomic mass.

1 atom

13.005739 u  1.6605 1027 kg
   N  1.80 109 kg  8.3349 1016 nuclei  8.331016 nuclei

See the adjacent graph. From the graph, the half-life is approximately 600 seconds.

The spreadsheet used for this N (1016 nuclei) 10.0
problem can be found on the 8.0
Media Manager, with filename 6.0
“PSE4_ISM_CH41.XLS,” on 4.0
tab “Problem 41.89.” 2.0
0.0
0 300 600 900 1200 1500 1800
t (s)

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584

Chapter 41 Nuclear Physics and Radioactivity

90. See the following graph. The spreadsheet used for this problem can be found on the Media Manager,
with filename “PSE4_ISM_CH41.XLS,” on tab “Problem 41.90.”

Binding energy per nucleon, MeV 10
9
8 30 60 90 120 150
7 Mass Number, A
6
5
4
3
2
1
0
0

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585

CHAPTER 42: Nuclear Energy: Effects and Uses of Radiation

Responses to Questions

1. (a) 138 Ba ; (b) p or 1 H ; (c) γ ; (d) 199 Hg
56 1 80

2. S32

16

3. 20 Ne
10

4. Neutrons have no net charge, and therefore do not have to overcome the Coulomb barrier (Coulomb
repulsion) to get into the nucleus. Neutrons are also massive and can carry more energy than a
lighter particle.

5. 17 F is the residual nucleus. The reaction equation is: 20 Ne  p    17 F.
9 10 9

6. Fission fragments have more neutrons than are required for nuclear stability and will decay by β–
emission in order to convert a neutron to a proton.

7. The energy from nuclear fission appears in the thermal (kinetic) energy of the fission fragments and
the neutrons that are emitted, and in the thermal energy of nearby atoms with which they collide.

8. 239 Pu has a smaller critical mass than 235 U . Since there are more neutrons released per decay in Pu-
94 92

239, fewer nuclei are needed to release sufficient neutrons to create and sustain a chain reaction.

9. Yes, a chain reaction would be possible, since the multiplication factor f is greater than 1. However,
the chain reaction would progress slowly, and care would have to be taken to prevent neutron loss.

10. When uranium is enriched, the percentage of U-235 nuclei in a given mass of uranium is increased.
This process involves the nuclei of the atoms. Chemical processes typically involve the electrons in
atoms, and not the nuclei. The chemical behavior of all the isotopes of uranium is nearly identical,
therefore chemical means could not be used to separate isotopes and enrich uranium.

11. Neutrons are neutral; they are not repelled by the electrons surrounding the atom or by the positively
charged protons in the nucleus. They are able to penetrate easily into the nucleus, and, once there,
are held in place by the strong nuclear force. A neutron brings to the nucleus its kinetic energy and
the binding energy given up as the neutron is bound to the rest of the nucleus. This binding energy
can be very large, and is enough to move the nucleus into an excited state, from which it will fission.

12. The hydrogen atoms in water act as a moderator to slow down the neutrons released during fission
reactions. The slower neutrons are more likely to be absorbed by other uranium nuclei to produce
further fission reactions, creating a chain reaction that could lead to an explosion. A porous block of
uranium in air would be less likely to undergo a chain reaction due to the absence of an effective
moderator.

13. Ordinary water does not moderate, or slow down, neutrons as well as heavy water; more neutrons
will also be lost to absorption in ordinary water. However, if the uranium in a reactor is highly
enriched, there will be many fissionable nuclei available in the fuel rods. It will be likely that the few

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586

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

moderated neutrons will be absorbed by a fissionable nucleus, and it will be possible for a chain
reaction to occur.

14. A useful fission reaction is one that is self-sustaining. The neutrons released from an initial fission
process can go on to initiate further fission reactions, creating a self-sustaining reaction. If no
neutrons were released, then the process would end after a single reaction and not be very useful.

15. Heavy nuclei decay because they are neutron-rich, especially after neutron capture. After fission, the
smaller daughter nuclei will still be neutron-rich and relatively unstable, and will emit neutrons in
order to move to a more stable configuration. Lighter nuclei are generally more stable with
approximately equal numbers of protons and neutrons; heavier nuclei need additional neutrons in
order to overcome Coulomb repulsion between the protons.

16. The water in the primary system flows through the core of the reactor and therefore could contain
radioactive materials, including deuterium, tritium, and radioactive oxygen isotopes. The use of a
secondary system provides for isolation of these potentially hazardous materials from the external
environment.

17. Fission is the process in which a larger nucleus splits into two or more fragments, roughly equal in
size. Fusion is the process in which smaller nuclei combine to form larger nuclei.

18. Fossil fuel power plants are less expensive to construct and the technology is well known. However,
the mining of coal is dangerous and can be environmentally destructive, the transportation of oil can
be damaging to the environment through spills, the production of power from both coal and oil
contributes to air pollution and the release of greenhouse gases into the environment, and there is a
limited supply of both coal and oil. Fission power plants produce no greenhouse gases and virtually
no air pollution, and the technology is well known. However, they are expensive to build, produce
thermal pollution and radioactive waste, and when accidents occur they tend to be very destructive.
Uranium is also dangerous to mine. Fusion power plants produce very little radioactive waste and
virtually no air pollution or greenhouse gases. Unfortunately, the technology for large-scale
sustainable power production is not yet known, and the pilot plants are very expensive to build.

19. To ignite a fusion reaction, the two nuclei must have enough kinetic energy to overcome electrostatic
repulsion and approach each other very closely in a collision. Electrostatic repulsion is proportional
to charge and inversely proportional to the square of the distance between the centers of the charge
distributions. Both deuterium and tritium have one positive charge, so the charge effect is the same
for d-d and d-t ignition. Tritium has one more neutron than deuterium and thus has a larger nucleus.
In the d-t ignition, the distance between the centers of the nuclei will be greater than in d-d ignition,
reducing the electrostatic repulsion and requiring a lower temperature for fusion ignition.

20. The interiors of stars contain ionized atoms (a plasma) at very high temperature and with a high
density of nuclei. The nuclei have high enough kinetic energy and a great enough likelihood of
colliding with other nuclei to allow fusion to occur.

21. Stars maintain fusion confinement with gravity. The large amount of mass in a star creates a
tremendous gravitational attraction on the gas particles which is able to overcome the repulsive
Coulomb force and radiation pressure.

22. Younger women may suffer damage to reproductive cells as well as somatic damage. Genetic
damage caused by radiation can cause mutations and be passed on to future generations. A fetus is
particularly susceptible to radiation damage due to its small mass and rapid cell development. Since
it is quite possible, especially early in a pregnancy, for a woman to be pregnant and not know it, it is

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587

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

reasonable to have lower recommended dose levels for women of childbearing years. Beyond the
reproductive years, the acceptable exposure dosage for women can be increased.

23. Alpha particles are relatively large and are generally emitted with relatively low kinetic energies.
They are not able to penetrate the skin, and so are not very destructive or dangerous as long as they
stay outside the body. If alpha emitters are ingested or breathed, however, the protective layer of
skin is bypassed, and the alpha particles, which are charged, can do tremendous amounts of damage
to lung or other delicate internal tissue due to ionizing effects. Thus, there are strong rules against
eating and drinking around alpha-emitters, and the machining of such materials, which would
produce fine dust particles that could be inhaled, can be done only in sealed conditions.

24. The absorbed dose measures the amount of energy deposited per unit mass of absorbing material and
is measured in Grays (SI) (1 Gy = 1J/kg) or rads. The effective dose takes into account the type of
radiation depositing the energy and is used to determine the biological damage done by the radiation.
The effective dose is the absorbed dose multiplied by a quality factor, QF. The effective dose is
measured in rem or siverts (SI). 1 Sv of any type of radiation does approximately the same amount
of biological damage.

25. Appropriate levels of radiation can kill possibly harmful bacteria and viruses on medical supplies or
in food.

26. Allow a radioactive tracer to be introduced into the liquid that flows through the pipe. Then check
the pipe with a Geiger counter. When you find the tracer on the outside of the pipe (radiation levels
will be higher at that point), you will have found the leak.

Solutions to Problems

1. By absorbing a neutron, the mass number increases by one and the atomic number is unchanged.

The product nucleus is 28 Al . Since the nucleus now has an “extra” neutron, it will decay by  ,
13

according to this reaction: 28 Al  28 Si     ve . Thus the product is 28 Si .
13 14 14

2. If the Q-value is positive, then no threshold energy is needed.

 Q  MeV c2 
 2m2 c2  m3 c2  mnc2  22.014082 u  3.016029 u 1.008665u  931.5 u  c2

1 H 2 He

 3.232 MeV

Thus no threshold energy is required .

3. A “slow” neutron means that it has negligible kinetic energy. If the Q-value is positive, then the
reaction is possible.

 Q  m c  m c  m c 23822 2  MeV c2  c2
92 n  u 
U 239 U 238.050788u 1.008665u  239.054293u 931.5
92

 4.807 MeV

Thus the reaction is possible .

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588

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

4. The Q-value tells whether the reaction requires or releases energy.

 Q  MeV c2 
 mpc2  m c2  m c2  m c2  1.007825u  7.016005u  24.002603 u  931.5 u  c2
7 Li 4 He
3 2

 17.35MeV

The reaction releases 17.35 MeV .

5. The Q-value tells whether the reaction requires or releases energy.

Q  m c2  m c29  m c212  mnc2
4 Be 6 C

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

The reaction releases 5.701MeV .

6. (a) If the Q-value is positive, then no threshold energy is needed.

Q  mnc2  m c224  m c223  mdc2
12 Mg 11 Na

 1.008665 u  23.985042 u  22.989769 u  2.014082 u  931.5 MeV c2  c2  9.449 MeV
 u 

Thus more energy is required if this reaction is to occur. The 16.00 MeV of kinetic energy is

more than sufficient, and so the reaction can occur .

(b) 16.00 MeV  9.449 MeV= 6.55 MeV of energy is released

7. (a) If the Q-value is positive, then no threshold energy is needed.

Q  mpc2  m c27  4m c2  m c2
3 Li 2 He

 1.007825 u  7.016005 u  2 4.002603 u  931.5 MeV c2  c2  17.348 MeV
 u 

Since the Q-value is positive, the reaction can occur .

(b) The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy.

Ktotal  Kreactants  Q  3.5 MeV  17.348 MeV= 20.8 MeV

8. (a) If the Q-value is positive, then no threshold energy is needed.

Q  m c2  m c214  17m c2  mpc2
7 N 8 O

 4.002603 u  14.003074 u  16.999132 u  1.007825 u  931.5 MeV c2  c2  1.192 MeV
 u 

Thus more energy is required if this reaction is to occur. The 9.68 MeV of kinetic energy is

more than sufficient, and so the reaction can occur .

(b) The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy.

Ktotal  Kreactants  Q  9.68 MeV  1.192 MeV= 8.49 MeV

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589

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

9. Q  m c2  m c2 16  m 20 Ne
8 O
10

  4.002603 u 15.994915 u 19.992440 u 931.5 MeV c2  c 2  4.730 MeV
u 


10. The Q-value tells whether the reaction requires or releases energy.

Q  mdc2  m c213  m c214  mnc2
6 C 7 N

 2.014082 u  13.003355 u  14.003074 u  1.008665 u  931.5 MeV c2  c2  5.308 MeV
 u 

The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy.

Ktotal  Kreactants  Q  44.4 MeV  5.308 MeV= 49.7 MeV

11. The nitrogen-14 absorbs a neutron. Carbon-12 is a product. Thus the reaction is

n  14 N  C14  ?. The reactants have 7 protons and 15 nucleons, which means 8 neutrons. Thus
7
6

the products have 7 protons and 15 nucleons. The unknown product must be a proton. Thus the

reaction is n  14 N  C14  p .
7
6

Q  mn c 2  m c214  m c214  mpc2
7 N 6 C

 1.008665 u  14.003074 u 14.003242 u 1.007825 u 931.5 MeV c2  c2  0.626 MeV
u 


12. (a) The deuteron is 2 H , and so the reactants have 4 protons and 8 nucleons. Therefore the
1

reactants have 4 neutrons. Thus the products must have 4 protons and 4 neutrons. That means

that X must have 3 protons and 4 neutrons, and so X is 7 Li .
3

(b) This is called a “stripping” reaction because the lithium nucleus has “stripped” a neutron from
the deuteron.

(c) The Q-value tells whether the reaction requires or releases energy.

Q  mdc2  m c2 6  m c2 7  mpc2
3 Li 3 Li

 2.014082 u  6.015123 u  7.016005 u  1.007825 u  931.5 MeV c2  c2  5.007 MeV
 u 

Since the Q-value is positive, the reaction is exothermic .

13. (a) This is called a “pickup” reaction because the helium has “picked up” a neutron from the carbon

nucleus.

(b) The alpha is 4 He . The reactants have 8 protons and 15 nucleons, and so have 7 neutrons. Thus
2

the products must also have 8 protons and 7 neutrons. The alpha has 2 protons and 2 neutrons,

and so X must have 6 protons and 5 neutrons. Thus X is C11 .

6

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590

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

(c) The Q-value tells whether the reaction requires or releases energy.

Q  m c23  m c2 12  m c211  m c2
2 He 6 C 6 C

 3.016029 u  12.000000 u 11.011434 u  4.002603 u  931.5 MeV c2  c2  1.856 MeV
u 


Since the Q-value is positive, the reaction is exothermic .

14. (a) The product has 16 protons and 16 neutrons. Thus the reactants must have 16 protons and 16

neutrons. Thus the missing nucleus has 15 protons and 16 neutrons, and so is .P31

15

(b) The Q-value tells whether the reaction requires or releases energy.

Q  mp c 2  m c2 31  m c2 32
15 P 16 S

 1.007825 u  30.973762 u  31.972071 u  931.5 MeV c2  c 2  8.864 MeV
u 


15. We assume that all of the particles are essentially at rest, and so ignore conservation of momentum.
To just make the fluorine nucleus, the Q-value plus the incoming kinetic energy should add to 0.

K Q  K  mp c 2  m c218  m c2 18  mnc2  0 
8 O 9 F

m c218  K  mpc2  m c2 18  mnc2 
9 F 8 O

 2.438 MeV  1.007825 u  17.999161 u  1.008665 u 931.5 MeV c2  c 2
u 

 1.6767874 104 MeV

 m 18  1u 
9  931.5 MeV c2  
1.6767874 104 MeV c 2 18.000938 u

F

16. We assume that the energies are small enough that classical mechanics is applicable, particularly

p 2mK and K  1 mv2 . The least proton kinetic energy is required when the product particles
2

move together and so have the same speed. We write equations for 1-D momentum conservation

and for energy conservation, and then combine those to find the required proton energy.

pp  pn  pN   mn  mN  v

 Kp pp2  mn 2
2mp  mN  v mn  mN   mn  mN 
    m1  mN  v2  mp  K  KN  
2mp mp n
2n

mp Kp
mn  mN
 Kn  KN 

Energy conservation : Kp Q  Kn  KN  mp Kp  Kp  Q mn  mN
mn  mN  mp
 mn  mN 

Note that this result is also derived in Problem 87. We now substitute in the values for this specific

problem.

Kp  Q mn  mN   m c2  mpc2  m c13 2  mn c 2  Q mn  mN
mn  mN  mp mn  mN  mp
13 C 7N

6

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591

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

13.003355 u  1.007825 u  13.005739 u  1.008665 u  931.5 MeV c2  c 2  
 u   
   

  1.008665 13.005739  
1.008665 13.005739 1.007825  

 3.24 MeV

If some intermediate values are used, such as Q  3.003 MeV , and mn  mN  1.077, the mass
mn  mN  mp

factor as 1.077, then the value of 3.23 MeV will be obtained.

17. We use the same derivation algebra as in Problem 16, and so jump to the final expression.

Kp  Q mn  mN   m c2  mp c 2  m c14 2  mn c 2  Q mn  mN
mn  mN  mp mn  mN  mp
14 C 7N

6

14.003242 u  1.007825 u  14.003074 u  1.008665 u  931.5 MeV c2  c 2  
 u   
     

  1.008665 14.003074  
 1.008665 14.003074 1.007825  

 0.671MeV

18. We use Eq, 42-3.

   R  nl  1.7 1021 nuclei m3 0.120 m 40 1028 m2  8107

R0

19. From the figure we see that a collision will occur if R1 v
d  R1  R2. We calculate the area of the effective circle d
presented by R2 to the center of R1.

   d 2    R1  R2 2

R2

20. At a distance x into the target material, particles are R0 Rx Rx  dRx
arriving at a rate Rx. Due to interactions between the
particles and the target material, which remove particles x0 x x  dx l
from the stream, particles are arriving at a distance of
x  dx at a lower rate of Rx  dRx , where dRx  0. Thus

the collision rate is dRx. The cross section, given in Eq.

42-3, gives the relationship between the two rates. Also
see the diagram.

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592

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

  dRx  dRx  Rxn dx
Rxndx

Integrate the above differential relationship.

Rx 1 n x ln  Rx  Rx  R0en x
 dRx  Rxn dx  R0 Rx dRx  0 dx   R0   n x 
 

R0en l represents the rate at which particles leave the target material, unaffected by the target
material.

21. We use the result from Problem 20, where Rl  R0enl . We use the data for the 1.0-cm-thick target
to get an expression for n .

Rl  R0en l  n   1 ln Rl   1 ln 0.25  138.63 m1
l R0 0.010 m

Rx  R0en x  x 1 ln Rx  1  ln 1  9.966 102 m  10 cm
n R0   138.63 m1  10

Note that the answer is correct to 2 significant figures.

22. We assume a 2.0% reaction rate allows us to treat the target as thin. We use Eq. 42-3. We need the
volume density of the cadmium atoms.

 n  m3  6.02 1023 atoms  1028 nuclei m3
8650 kg  0.1139 kg   4.572
 

(a) The cross section for 0.1-eV neutrons is ~ 3000 bn.

R  nl 
R0

R 1  0.020 1  1.458106 m
R0 n 4.5721028 nuclei m3
  l
3000 1028 m2

 1.5μm

(b) The cross section for 5.0-eV neutrons is ~ 2 bn.

R 1  0.020 1  2.187 103 m
R0 n 4.5721028 nuclei m3
  l 
2 1028 m2

 2.2 mm

23. The Q-value gives the energy released in the reaction, assuming the initial kinetic energy of the
neutron is very small.

Q  m c  m c  m c  m c  3m c22 2 2 2
n n
235 U 141 Ba 92 Kr
92 56 36

 1.008665u  235.043930 u  140.914411 u  91.926156u  31.008665 u 931.5 MeV c2  c2
u 

 173.3MeV

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593

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

24. The Q-value gives the energy released in the reaction, assuming the initial kinetic energy of the

neutron is very small.

Q  mnc2  m c235 2  m c2  m c136 2 12mnc2

92U 88 Sr 54Xe
38

 1.008665u  235.043930 u 87.905612 u 135.907219 u 12 1.008665 u   931.5 MeV c2  c2
u 


 126.5MeV

25. The power released is the energy released per reaction times the number of reactions per second.

P energy # reactions 

reaction s

200106 W
200106 eV reaction 1.601019 J eV
  # reactions P   61018 reactions s
s energy

reaction

26. Compare the energy per fission with the rest mass energy.

200 MeV
931.5 MeV c2
 energy per fission c2  9.1104  1
1100
rest mass energy mc2
 235u

27. We convert the 880 watts over a year’s time to a mass of uranium.

 880 J   3.156 107 s   1MeV   1fission   0.235 kg U235   3.388 104 kg 235 U
 1 s   1y   1.60 1013   200 MeV   6.02 1023  92
J 92

atoms

 0.34 g 235 U
92

28. (a) The total number of nucleons for the reactants is 236, and so the total number of nucleons for

the products must also be 236. The two daughter nuclei have a total of 231 nucleons, so

5 neutrons must be produced in the reaction: 235 U  n  133 Sb  98 Nb  5n .
92 51 41

(b) Q m c m c m c m c m c     523522 2 2 2
n n
92U 133 Sb 98 Nb
51 41

 235.043930u  1.008665u 132.915250 u  97.910328u  51.008665 u 931.5 MeV c2  c2
u 

 171.1MeV

29. We assume as stated in problems 26 and 27 that an average of 200 MeV is released per fission of a

uranium nucleus. Also, note that the problem asks for the mass of 238 U, but it is the 235 U nucleus
92 92

that undergoes the fission. Since 238 U is almost 100% of the natural abundance, we can use the
92

abundance of 235 U from Appendix F as a ratio of 235 U to 238 U.
92 92 92

 3 107 J  1 MeV   1 nucleus 235 U   1 atom 238 U   0.238 kg 
1.60  1013  92   92   1023 nuclei 
   
J 200 MeV 0.0072 atoms 235 U 6.02 238 U
92 92

 5.15 105 kg U238  5  105 kg U238

92 92

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594

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

30. Since the reaction is 38% efficient, the fission needs to generate 950 0.38 MW of power. Convert

the power rating to a mass of uranium using the factor-label method. We assume 200 MeV is
released per fission, as in other problems.

950 106 J 1 atom 1eV 0.235 kg U 3.156107 s  962 kg U235
0.38 s  200106 eV  1.601019 J  6.021023 atoms 
1y 92

 960 kg 235 U
92

31. We find the number of collisions from the relationship En  E0  1 n , where n is the number of
2

collisions.

1n ln En ln 0.040 eV
2 n  E0 1.0 106eV
 En  E0    24.58  25collisions
ln 1 ln 1
2 2

32. If the uranium splits into equal fragments, each will have an atomic mass number of half of 236, or
118. Each will have a nuclear charge of half of 92, or 46. Calculate the electrical potential energy
using Eq. 23-10. The distance between the nuclei will be twice the radius of a nucleus, and the
radius is given in Eq. 41-1.

1019 C 2

m 1181/
   U  1 Q1Q2 
 40 r
8.99 109 Nm2 C2  462 1.60 3  1MeV J   260 MeV
 1.60 1013 
2 1.2 1015

This is about 30% larger than the nuclear fission energy released.

33. The height of the Coulomb barrier is given by the electrostatic potential energy, Eq. 23-10. The

distance to use is the sum of the radii of the two particles involved. For the alpha decay, the

daughter nucleus is 23920Th. We assume the fission results in two equal fragments, each with Z = 46

and A = 128. These are palladium nuclei.

U  1 Q QTh ; Ufission  1 QPdQPd
40 r  rTh 40 rPd  rPd

      U
  Ufission
 1 QQTh  2 90 290 2 1281/3  0.11
1.21015 m 41/3  2321/3   46 2 41/3  2321/3
40 r  rTh
1 QPdQPd 46 2

40 rPd  rPd    1.21015m 2 1281/3

34. The reaction rate is proportional to the number of neutrons causing the reactions. For each fission
the number of neutrons will increase by a factor of 1.0004, so in 1000 milliseconds the number of

neutrons will increase by a factor of 1.00041000  1.5 .

  35.
K  3 kT  3 1.38 1023 J K 2 107 K  4 1016 J
2 2

    1eV 
3 1.38 1023 J K 2 107 K 1.60 1019 J  2588 eV  3000 eV
2

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595

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

36. The Q-value gives the energy released in the reaction.

Q  m c22  m c23  m c24  mn c 2
1 H 1 H 2 He

 2.014082 u  3.016049 u  4.002603 u 1.008665 u  931.5 MeV c2  c2  17.57 MeV
 u 
 

37. Calculate the Q-value for the reaction 2 H  2 H  3 He  n
1 1 2

Q  2m2 c 2  m c2  mnc2

1 H 3 He
2

 2  2.014082 u  3.016029 u  1.008665 u 931.5 MeV c2  c2  3.23 MeV
u 

38. For the reaction in Eq. 42-7a, if atomic masses are to be used, then one more electron needs to be
added to the products side of the equation. Notice that charge is not balanced in the equation as

written. The balanced reaction is 1 H  1 H  2 H  e  v  e
1 1 1

Q  2m1 c 2  m c2  me+ c2  m c2
e
1 H 2 H
1

 2 1.007825 u   2.014082 u  2 0.000549 u  931.5 MeV c2  c2  0.4378 MeV  0.44 MeV
 u 

For the reaction in Eq. 42-7b, use atomic masses since there would be two electrons on each side.

Q  m c21  m c22  m c23
1 H 1 H 2 He

 1.007825 u  2.014082 u  3.016029 u 931.5 MeV c2  c 2  5.4753MeV  5.48 MeV
u 

For the reaction in Eq. 42-7c, use atomic masses since there would be two electrons on each side.

Q  2m c 2  m c2  2m1 c 2

3 He 4 He 1 H
2 2

  2  3.016029 u  4.002603  2 1.007825 u  931.5 MeV c2  c2  12.86 MeV
u 


39. (a) Reaction 42-9a: 4.00 MeV 1u 1 kg 5.98 1023 MeV g
2 2.014082 u  1.66 1027 kg  1000 g 

Reaction 42-9b: 3.23 MeV 1u 1kg  4.83 1023 MeV g
2 2.014082 u  1.66 1027 kg  1000 g

Reaction 42-9c: 17.57 MeV 1u 1kg  2.10 1024 MeV g
2.014082 u  3.016049 u  1.66 1027 kg  1000 g

(b) Uranium fission (200 MeV per nucleus):

200 MeV 1u 1 kg 5.131023 MeV g
 235 u   1.66 1027 kg  1000 g 

Reaction 42-9a gives about 17% more energy per gram than uranium fission. Reaction 42-9b
gives about 6% less energy per gram than uranium fission. Reaction 42-9c gives about 4 times
as much energy per gram than uranium fission.

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596

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

40. Calculate the Q-value for the reaction 238 U  n  U239
92 92

Q  m c  m c  m c238 2 2 2
n
92U 239 U
92

 238.050788 u 1.008665 u  239.054293 u 931.5 MeV c2  c2  4.807 MeV
u 


41. The reaction of Eq. 42-9b consumes 2 deuterons and releases 3.23 MeV of energy. The amount of
energy needed is the power times the elapsed time, and the energy can be related to the mass of
deuterium by the reaction.

 850 J  1 yr   3.156 107 s  1MeV  2d  2.014 103kg 
 s   yr  1.601013 J   3.23 MeV   6.021023 d 

 3.473104 kg  0.35g

42. (a) The reactants have a total of 3 protons and 7 neutrons, and so the products should have the
same. After accounting for the helium, there are 3 neutrons and 1 proton in the other product,

and so it must be tritium, 3 He. The reaction is 6 Li  1 n  4 He  3 H.
1 3 0 2 1

(b) The Q-value gives the energy released.

Q  m c26  m c21  m c24  m c23
3 Li 0 n 2 He 1 He

  6.015123 u  1.008665 u  4.002603 u  3.016049 u 931.5 MeV c2  c 2  4.784 MeV
u 


43. Assume that the two reactions take place at equal rates, so they are both equally likely. Then from
the reaction of 4 deuterons, there would be a total of 7.228 MeV of energy released, or 1.807 MeV

per deuteron on the average. A total power of 1250 MW  3788 MW must be obtained from the
0.33

fusion reactions to provide the required 1250 MW output, because of the 330% efficiency. We
convert the power to a number of deuterons based on the energy released per reacting deuteron, and
then convert that to an amount of water using the natural abundance of deuterium.

 3788 106 J  3600 s   1MeV  1d  1 H atom 
  s   1h      0.000115 d's  
  1.60  1013 J  1.807 MeV   
3788 MW 
 1 H2 O molecule 0.018 kg H2O 
 2 H atoms   6.02 1023 molecules  
   

 6131kg h  6100 kg h

44. We assume that the reactants are at rest when they react, and so the total momentum of the system is
0. As a result, the momenta of the two products are equal in magnitude. The available energy of
17.57 MeV is much smaller than the masses involved, and so we use the non-relativistic relationship

between momentum and kinetic energy, K  p2  p  2mK .
2m

K4 He  Kn  Ktotal  17.57 MeV p4 He  pn  2m K 4 He 4 He 2mn Kn 
2
2 2 2

 m K  m K  m K  m K  K4 
2
He 4 He nn 4 He 4 He n total 4 He
2 2 2 2

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597

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

K4  mn Ktotal   1.008665 17.57 MeV  3.536 MeV  3.5 MeV
 4.002603 1.008665
2 He m4  mn

2 He

Kn  Ktotal  K4 He  17.57 MeV  3.54 MeV  14.03MeV  14 MeV

2

If the plasma temperature were significantly higher, then the approximation of 0 kinetic energy being

brought into the reaction would not be reasonable. Thus the results would depend on plasma

temperature. A higher plasma temperature would result in higher values for the energies.

45. In Eq. 42-9a, 4.00 MeV of energy is released for every 2 deuterium atoms. The mass of water can be
converted to a number of deuterium atoms.

1.00 kg H2O  6.02 1023H2O   2H   1.15 104 d   7.692 1021 d nuclei 
 0.018kg H2O   1H2O   1H 

 7.692 1021d nuclei  4.00 106eV   1.60 1019 J  2.46 109 J
 2d atoms  1eV  

As compared to gasoline: 2.46 109 J  50 times more than gasoline
5 107 J

46. (a) We follow the method of Example 42-10. The reaction is C12  1 H  13 N   . We calculate
6 1 7

the potential energy of the particles when they are separated by the sum of their radii. The radii

are calculated from Eq. 41-1.

6 1 1.60 1019 C 2  1MeV 

1.2  1015m 11/ 3  121/ 3  1.60  1013 J 
      Ktotal1 qC qH
 4 0 rC  rH  8.99 109 Nm2 C2

 2.19 MeV

For the d-t reaction, Example 42-10 shows Ktotal  0.45 MeV . Find the ratio of the two
energies.

KC  2.19 MeV  4.9
Kdt 0.45 MeV
The carbon reaction requires about 5 times more energy than the d-t reaction.

(b) Since the kinetic energy is proportional to the temperature by K  3 kT , since the kinetic
2

energy has to increase by a factor of 5, so does the temperature.

 T  4.9 3 108 K  1.5 109 K .

47. (a) No carbon is consumed in this cycle because one C12 nucleus is required in the first step of the
6

cycle, and one C12 nucleus is produced in the last step of the cycle. The net effect of the cycle
6

can be found by adding all the reactants and all the products together, and canceling what

appears on both sides of the reaction.

12 C  1 H  13 N  
6 1 7

N13  C13  e  v

7 6

13 C  1 H  14 N  
6 1 7

14 N  1 H  15 O  
7 1 8

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598

Chapter 42 Nuclear Energy; Effects and Uses of Radiation

O15  N15  e  v

8 7

15 N  1 H  C12  4 He
7 1 6 2

____________________________________

C12  1 H  13 N  C13  1 H  14 N  1 H  O15  15 N  1 H
6 1 7 1 7 1 8 7 1
6

 13 N    13 C  e v 14 N    15 O    15 N  e  v 12 C  4 He
7 6 7 8 7 6 2

____________________________________

4 1 H  4 He  2 e  2v  3
1 2

There is a difference of one gamma ray in the process, as mentioned in the text.

(b) To use the values from Appendix F, we must be sure that number of electrons is balanced as

well as the number of protons and neutrons. The above “net” equation does not consider the

electrons that neutral nuclei would have, because it does not conserve charge. What the above

reaction really represents (ignoring the gammas and neutrinos) is the following.

4 1 H  4 He  2 e   4 1 p  2 1 p  2 1 n  2e
1 2 1 1 0

To use the values from Appendix F, we must add 4 electrons to each side of the reaction.

   41 p  4e   2 1 p  2e  2 1 n  2e  2e  4 1 H  4 He  2e  2e
1 1 0 1 2

The energy produced in the reaction is the Q-value.

Q  4m1 c 2  m c2  4me

1 H 4 He
2

 4 1.007825 u  4.002603 u  4  0.000549 u  931.5 MeV c2  c 2  24.69 MeV
 u 

As mentioned at the top of page 1143, the positrons and the electrons annihilate to produce

another 2.04 MeV, so the total energy released is 24.69 MeV  2 1.02 MeV  26.73 MeV .

(c) In some reactions extra electrons must be added in order to use the values from Appendix F.

The first equation is electron-balanced, and so Appendix F can be used.

Q  m c2  m c2  m c13 2

12 C 1 H 7N
6 1

 12.000000 u  1.007825 u  13.005739 u  931.5 MeV c2  c 2  1.943 MeV
u 


The second equation needs to have another electron, so that N13  C13  e  e  v .

7 6

Q  m c13 2  m c2  2mec2

7N 13 C
6

 13.005739 u  13.003355 u  2 0.000549 u   931.5 MeV c2  c 2  1.198 MeV
 u 

We must include an electron-positron annihilation in this reaction.

1.198 MeV  1.02 MeV  2.218 MeV

The third equation is electron-balanced.

Q  m c2  m c2  m c14 2

13 C 1 H 7N
6 1

 13.003355 u  1.007825 u  14.003074 u  931.5 MeV c2  c 2  7.551MeV
u 


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599

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

The fourth equation is electron-balanced.

Q  m c14 2  m c2  m c2

7N 1 H 15 O
1 8

 14.003074 u  1.007825 u  15.003066 u  931.5 MeV c2  c 2  7.296 MeV
u 
 

The fifth equation needs to have another electron, so that O15  15 N  e  e  v .
8 7

Q  m c2  m c15 2  2me c 2

15 O 7N
8

 15.003066 u  15.000109 u  2 0.000549 u   931.5 MeV c2  c 2  1.732 MeV
u 

We must include an electron-positron annihilation in this reaction.

1.732 MeV  1.02 MeV  2.752 MeV

The sixth equation is electron-balanced.

Q  m c15 2  m c2  m c2  m c2

7N 1 H 12 C 4 He
1 6 2

 15.000109 u  1.007825 u  12.000000 u  4.002603 u  931.5 MeV c2  c 2
 u 


 4.966 MeV

The total is found as follows.
1.943 MeV  2.218 MeV  7.551MeV  7.296 MeV  2.752 MeV  4.966 MeV

 26.73 MeV

(d) It takes a higher temperature for this reaction than for a proton-proton reaction because the
reactants have to have more initial kinetic energy to overcome the Coulomb repulsion of one
nucleus to another. In particular, the carbon and nitrogen nuclei have higher Z values leading to
the requirement of a high temperature in order for the protons to get close enough to fuse with
them.

48. Because the quality factor of alpha particles is 20 and the quality factor of X-rays is 1, it takes 20
times as many rads of X-rays to cause the same biological damage as compared to alpha particles.

Thus the 250 rads of alpha particles is equivalent to 250 rad  20  5000 rad of X-rays.

49. Use Eq. 42-11b to relate Sv to Gy. From Table 42.1, the quality factor of gamma rays is 1, and so
the number of Sv is equal to the number of Gy. Thus 4.0Sv  4.0 Gy .

50. A gray is 1 Joule per kg, according to Eq. 42-10.

3.0 J  65 kg  195 J  200 J (2 sig. fig.)
kg

51. The biological damage is measured by the effective dose, Eq. 42-11b.
65 rad fast neutrons 10  x rad slow neutrons  3 

x  65 rad 10  220 rad slow neutrons
3

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600