Chapter 42
1. Our calculation is similar to that shown in Sample Problem 421. We set
0
Cu
min
5.30MeV=
1/ 4
/
K
U
q q
r
and solve for the closest separation,
r
min
:
19
9
Cu
Cu
min
6
0
0
14
2
29
1.60
10
C
8.99
10 V m/C
4
4
5.30
10 eV
1.58
10
m
15.8 fm.
e
q q
kq q
r
K
K


ﾴ
ﾴ
ﾴ
ﾴ
ﾴ
We note that the factor of
e
in
q
= 2
e
was not set equal to 1.60
10
– 19
C, but was
instead allowed to cancel the “e” in the nonSI energy unit, electronvolt.
2. Kinetic energy (we use the classical formula since
v
is much less than
c
) is converted
into potential energy (see Eq. 2443). From Appendix F or G, we find
Z
= 3 for Lithium
and
Z
= 90 for Thorium; the charges on those nuclei are therefore 3
e
and 90
e
,
respectively. We manipulate the terms so that one of the factors of
e
cancels the “e” in the
kinetic energy unit MeV, and the other factor of
e
is set equal to its SI value 1.6
10
–19
C. We note that
k
1 4
0
can be written as 8.99
10
9
V·m/C. Thus, from energy
conservation, we have
K
U
r
k
K
e
q q

1 2
9
19
6
8 99
10
3
16
10
90
300
10
.
.
.
V m
C
C
eV
c
hc
hb g
which yields
r
= 1.3
10
– 13
m (or about 130 fm).
3. The conservation laws of (classical kinetic) energy and (linear) momentum determine
the outcome of the collision (see Chapter 9). The final speed of the
particle is
v
m
m
m
m
v
f
i

Au
Au
,
and that of the recoiling gold nucleus is
v
m
m
m
v
f
i
Au,
Au
2
.
(a) Therefore, the kinetic energy of the recoiling nucleus is
1629
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CHAPTER 42
2
2
2
Au
Au,
Au
Au,
Au
2
Au
Au
2
2
4
1
1
2
2
4 197u
4.00u
5.00MeV
4.00u+197u
0.390MeV.
f
f
i
i
m
m
m
K
m
v
m
v
K
m
m
m
m
�
�
�
�
�
�
(b) The final kinetic energy of the alpha particle is
2
2
2
2
Au
Au
Au
Au
2
1
1
2
2
4.00u
197u
5.00MeV
4.00u
197u
4.61MeV.
f
f
i
i
m
m
m
m
K
m v
m
v
K
m
m
m
m
�
�
�
�


�
�
�
�
�
�
�
�
�
�

�
�
�
�
We note that
K
K
K
af
f
i
Au,
is indeed satisfied.
4. (a) 6 protons, since
Z
= 6 for carbon (see Appendix F).
(b) 8 neutrons, since
A – Z
= 14 – 6 = 8 (see Eq. 421).
5. (a) Table 421 gives the atomic mass of
1
H as
m
= 1.007825 u. Therefore, the
mass
excess
for
1
H is
= (1.007825 u –
1.000000 u)= 0.007825 u.
(b) In the unit MeV/
c
2
,
= (1.007825 u – 1.000000 u)(931.5 MeV/
c
2
·u) = +7.290
MeV/
c
2
.
(c) The mass of the neutron is given in Sample Problem 423. Thus, for the neutron,
= (1.008665 u –
1.000000 u) = 0.008665 u.
(d) In the unit MeV/
c
2
,
= (1.008665 u – 1.000000 u)(931.5 MeV/
c
2
·u) = +8.071
MeV/
c
2
.
(e) Appealing again to Table 421, we obtain, for
120
Sn,
= (119.902199 u – 120.000000 u) = – 0.09780 u.
(f) In the unit MeV/
c
2
,
= (119.902199 u – 120.000000 u) (931.5 MeV/
c
2
·u) = – 91.10 MeV/
c
2
.
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 Fall '08
 Many
 Physics, Atom, Mass, Nuclear Fission, Neutron, Binding energy

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