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249
Chapter 11
!
Nuclear Physics
111.
Isotope
Protons
Neutrons
99
11
14
23
28
36
48
52
68
66
82
74
101
86
136
112. The momentum of an electron confined within the nucleus is:
The momentum must be at least as large as
)
p, so
and the
electron’s kinetic energy is
.
This is twenty times the observed maximum beta decay energy, precluding the existence of
electrons in the nucleus.
113. A protonelectron model of
6
Li would consist of 6 protons and 3 electrons.
Protons and
electrons are spin1/2 (FermiDirac) particles.
The minimum spin for these particles in the
lowest available energy states is
, so
6
Li (S=0) cannot have such a structure.
114. A protonelectron model of
14
N would have 14 protons and 7 electrons.
All are FermiDirac
spin1/2 particles.
In the ground state the proton magnetic moments would add to a small
fraction of the proton magnetic moment of 2.8
:
N
, but the unpaired electron would give the
system a magnetic moment of the order of that of an electron, about 1
:
B
.
Because
:
B
is
approximately 2000 times larger than
:
N
, the
14
N magnetic moment would be about 1000
times the observed value, arguing against the existence of electrons in the nucleus.
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115. The two proton spins would be antiparallel in the ground state with
.
So
the deuteron spin would be due to the electron and equal to 1/2
S
.
Similarly, the proton
magnetic moments would add to zero and the deuteron’s magnetic moment would be 1
:
B
.
From Table 111, the observed deuteron spin is 1
S
(rather than 1/2
S
found above) and the
magnetic moment is 0.857
:
N
, about 2000 times smaller than the value predicted by the
protonelectron model.
116.
Isotopes
Isotones
(a)
18
F
17
F
19
F
16
N
17
O
(b)
208
Pb
206
Pb
210
Pb
207
Tl
209
Bi
(c)
120
Sn
119
Sn
118
Sn
121
Sb
122
Te
117.
Nuclide
Isobars
Isotopes
(a)
(b)
(c)
118.
where R
0
= 1.2 fm = 1.2 × 10
!
15
m
Chapter 11
!
Nuclear Physics
251
119.
(a)
(b)
(c)
1110.
(a)
(b)
(c)
(d)
1111. (a)
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(Problem 1111 continued)
(b)
(c)
1112.
(This is Equation 1113
on the Web page
www.whfreeman.com/modphysics4e.) The values of the
a
i
in MeV/c
2
are
given in Table 113 (also on the Web page).
For
23
Na:
This result differs from the measured value of 31.972071u by only 0.009%.
1113.
(a)
:
(b)
:
(c)
:
Chapter 11
!
Nuclear Physics
253
1114.
(Equation 112)
where Z = 20 for Ca and
)
U = 5.49 MeV from a table of isotopes (e.g., Table of Isotopes 8th
ed., Firestone, et al., Wiley 1998).
1115. (a)
(Equation 1119)
at
(b)
1116.
(a)
(b)
(c)
1117.
(Equation 1119)
(a) At
At
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(Problem 1117 continued)
(b)
1118. (a)
or nearly the same.
(b)
Chapter 11
!
Nuclear Physics
255
1119. (a)
when
when
Notice that this time interval equals three halflives.
(b)
(c)
Thus,
1120. (a) and (b)
t
1/2
.
3.6 min
counts
'
s
1/2R
0
t, min
(c) Estimating from the graph, the next count (at 8 min) will be approximately 230 counts/s.
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This homework help was uploaded on 04/03/2008 for the course PHYSICS 7C taught by Professor Lin during the Spring '08 term at University of California, Berkeley.
 Spring '08
 LIN
 Energy, Kinetic Energy, Momentum, Neutron

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