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Review problem. An alpha particle (Z = 2, mass 6.64 × 10–27 kg)
approaches to within 1.00 × 10–14 m of a carbon nucleus (Z = 6).
What are (a) the maximum Coulomb force on the alpha particle,
(b) the acceleration of the alpha particle at this point, and (c) the
potential energy of the alpha particle–nucleus system when the
alpha particle is at this point?
P44.3 (a)
(b)
F  ke
a
Q 1Q 2
r2

9
 8.99  10 N  m
2
C
2

 2 6 1.60  1019 C 

1.00  1014 m
F
27.6 N

 4.17  1027 m s2
m 6.64  1027 kg

2
 27.6 N
2
away from the
nucleus.
(c)

Q Q
U  ke 1 2  8.99  109 N  m
r
2
C
2

 2 6 1.60  1019 C 
1.00  10
14
m

2
 2.76  1013 J 1.73 M eV
How much energy (in MeV units) must an alpha particle have to
reach the surface of a gold nucleus (Z = 79, A = 197)? Assume the
gold nucleus remains stationary.
P44.6 It must start with kinetic energy equal to
stands for the sum of the radii of the
4
2H e
f

keqQ
rf
. Here
rf
197
79A u
nuclei, computed as
13
13
rf  r0A 1  r0A 2  1.20  1015 m   41 3  1971 3   8.89  1015 m .
Thus,
8.99 10

9
Ki  U
f
N m
2


and
Ki  U
C 2  2 79 1.60  1019 C
8.89  1015 m

2
 4.09  1012 J 25.6 M eV
.
7.
238
92
Find the radius of (a) a nucleus of 42 He and (b) a nucleus of
U.

  4

  238
P44.7 (a)
r  r0A 1 3  1.20  1015 m
(b)
r  r0A 1 3  1.20  1015 m
13
 1.90  1015 m
13
 7.44  1015 m
8.
Find the nucleus that has a radius approximately equal to
half the radius of uranium 238
92 U .
P44.8 From
r  r0A 1 3 ,
Thus, if
from which
rU  r0  238 .
1
13
r0A 1 3  r0  238
2
13
the radius of uranium is
A  30
r
1
rU
2
then
.
9.
A star ending its life with a mass of two times the mass of
the Sun is expected to collapse, combining its protons and
electrons to form a neutron star. Such a star could be thought of as
a gigantic atomic nucleus. If a star of mass 2 × 1.99 × 1030 kg
collapsed into neutrons (mn = 1.67 × 10–27 kg), what would its
radius be? (Assume that r = r0A1/3.)
P44.9 The number of nucleons in a star of two solar masses is
A

2 1.99  1030 kg
27
1.67  10
kg nucleon
Therefore


r  r0A 1 3  1.20  1015 m
 2.38  1057 nucleons.
 2.38 10 
57 1 3
 16.0 km
.
13. Nucleus 1 has eight times as many protons as nucleus 2, five
times as many neutrons, and six times as many nucleons as
nucleus 2. Nucleus 1 has four more neutrons than protons. (a)
What are the two nuclei? (b) Is each nucleus stable? If not, what is
the minimum number of neutrons that must be added to, or
removed from, each unstable nucleus to make it stable?
*P44.13 (a)
N 1  5N 2
Z1  8Z 2
N 1  Z1  6 N 1  Z2 
and
N 1  Z1  4
1 
1
N 1  Z1  6 N 1  Z1
5
8 
Thus:
5
Z1
4
5
 Z1  4  Z1
4
 Z 1  16
N 1 
N 1  Z 1  4  20, A 1  Z 1  N 1  36
N2 
1
1
N 1  4, Z 2  Z 1  2, A 2  Z 2  N 2  6
5
8
Hence: 36
16 S and
(b)
6
2H e
.
is unstable. Two neutrons must be removed to
make it stable  42H e .
6
2H e
14. Calculate the binding energy per nucleon for (a) 2H, (b) 4He,
(c) 56Fe, and (d) 238U.
P44.14
Using atomic masses as given in Table A.3,
(a)
For
2
1H
:
2.014102  11.008 665  11.007 825
2
Eb
 931.5 M eV 
  0.001194 u 
  1.11 M eV nucleon

A
u
.
(b)
(c)
For
For
4
2H e
56
26Fe
:
:
21.008 665  21.007 825  4.002 603
4
Eb
 0.007 59 uc2  7.07 M eV nucleon
A
.
301.008 665  261.007 825  55.934 942  0.528 u
Eb 0.528

 0.009 44 uc2  8.79 M eV nucleon
A
56
(d)
For
238
92U
:
.
1461.008 665  921.007 825  238.050783  1.934 2 u
Eb 1.934 2

 0.00813 uc2  7.57 M eV nucleon
A
238
.
15. The iron isotope 56Fe is near the peak of the stability curve.
This is why iron is generally prominent in the spectrum of the Sun
and stars. Show that 56Fe has a higher binding energy per nucleon
than its neighbors 55Mn and 59Co. Compare your results with
Figure 44.5.
P44.15
BE M  931.5

A
A
Nuclei
Z
25
Mn
56
26
Fe
59
27
Co
56
 Fe has a greater
55
M  Zm H  N m n  M
BE
A
in MeV
N
M in u
M in u
30
54.938 050 0.517 5
8.765
30
55.934 942 0.528 46
8.790
32
58.933 200 0.555 35
8.768
BE
than its neighbors. This tells us finer detail
A
than is shown in Figure 44.5.
16. Two nuclei having the same mass number are known as
isobars. Calculate the difference in binding energy per nucleon for
23
the isobars 23
11 Na and 12 Mg . How do you account for the difference?
P44.16
Use Equation 44.2.
Eb
The 23
 8.11 M eV
11N a ,
A
nucleon
and for
23
12M
g,
Eb
 7.90 M eV nucleon .
A
The binding energy per nucleon is greater for
(There is less proton repulsion in N a23 .)
23
11N a
by
0.210 M eV
.
17. Nuclei having the same mass numbers are called isobars. The
139
isotope 139
57 La is stable. A radioactive isobar, 59 Pr , is located below
the line of stable nuclei in Figure 44.4 and decays by e+ emission.
139
–
Another radioactive isobar of 139
57 La , 55 Cs , decays by e emission
and is located above the line of stable nuclei in Figure 44.4. (a)
Which of these three isobars has the highest neutron-to-proton
ratio? (b) Which has the greatest binding energy per nucleon? (c)
139
Which do you expect to be heavier, 139
59 Pr or 55 Cs ?
P44.17 (a)
The neutron-to-proton ratio
A Z
Z
is greatest for
139
55 Cs
and is equal to 1.53.
(b)
139
La
has the largest binding energy per nucleon of
8.378 MeV.
(c)
with a mass of 138.913 u. We locate the nuclei
carefully on Figure 44.4, the neutron–proton plot of
stable nuclei. Cesium appears to be farther from the
139
Cs
center of the zone of stability. Its instability means
extra energy and extra mass.
18. The energy required to construct a uniformly charged sphere
of total charge Q and radius R is U = 3keQ2/5R, where ke is the
Coulomb constant (see Problem 71). Assume that a 40Ca nucleus
contains 20 protons uniformly distributed in a spherical volume.
(a) How much energy is required to counter their electrical
repulsion according to the above equation? (Suggestion: First
calculate the radius of a 40Ca nucleus.) (b) Calculate the binding
energy of 40Ca. (c) Explain what you can conclude from comparing
the result of part (b) and that of part (a).
P44.18 (a)
The radius of the

40
Ca
nucleus is:
R  r0A 1 3  1.20  1015 m
  40
13
 4.10  1015 m
.
The energy required to overcome electrostatic
repulsion is
U 
3keQ
5R
2


 

2
C 2 20 1.60  1019 C 

  1.35  1011 J 84.1 M eV
15
5 4.10  10 m
3 8.99  109 N  m
2


.
(b)
The binding energy of
40
20 Ca
is
Eb  201.007 825 u   201.008 665 u   39.962 591 u   931.5 M eV u  342 M eV
.
(c)
The nuclear force is so strong that the binding energy
greatly exceeds the minimum energy needed to
overcome electrostatic repulsion.
19. A pair of nuclei for which Z1 = N2 and Z2 = N1 are called
mirror isobars (the atomic and neutron numbers are interchanged).
Binding-energy measurements on these nuclei can be used to
obtain evidence of the charge independence of nuclear forces (that
is, proton–proton, proton–neutron, and neutron–neutron nuclear
forces are equal). Calculate the difference in binding energy for the
two mirror isobars 158 O and 157 N . The electric repulsion among eight
protons rather than seven accounts for the difference.
P44.19
The binding energy of a nucleus is
Eb  M eV   ZM  H   N m n  M

For
15
8O
 X  931.494 M eV u .
A
Z
:
Eb  81.007 825 u  71.008665 u  15.003065 u  931.494 M eV u  111.96 M eV
For
15
7N
.
:
Eb  71.007 825 u  81.008665 u  15.000109 u  931.494 M eV u  115.49 M eV
Therefore,
the binding energy of 157N is largerby 3.54 M eV
.
.
20. Calculate the minimum energy required to remove a neutron
from the 2043 Ca nucleus.
43
P44.20
Removal of a neutron from 20
Ca would result in the residual
42
nucleus, 20
Ca . If the required separation energy is Sn , the overall process
can be described by
mass

S
43
20 Ca
n

 m ass
  m ass n
42
20 Ca
Sn   41.958618  1.008665 42.958767 u  0.008 516 u931.5 M eV u  7.93 M eV
.
23. (a) Use the semiempirical binding-energy formula to
compute the binding energy for 56
26 Fe . (b) What percentage is
contributed to the binding energy by each of the four terms
P44.23 (a)
E1  C 1A   15.7 M eV  56  879 M eV .
“Volume” term:
“Surface” term: E2  C 2A 2 3   17.8 M eV  56 2 3  260 M eV .
“Coulomb” term:
E3  C 3
Z  Z  1
A
13
   0.71 M eV 
 26 25
 121 M eV .
 561 3
“Asymmetry” term:
E4  C 4
 A  2Z  2
A
   23.6 M eV 
 56  52 2
56
 6.74 M eV
Eb  491 M eV
(b)
E1
 179%
Eb
;
E2
 53.0%
Eb
,
E3
 24.6%
Eb
;
E4
 1.37%
Eb
.