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Chapter 45
607
P45.37
(a)
Starting with
N
=
0 radioactive atoms at
t
=
0, the rate of increase is (production – decay)
dN
dt
RN
=−
λ
so
dN
R
N dt
bg
.
The variables are separable.
dN
dt
Nt
−
=
zz
00
:
−
−
F
H
G
I
K
J
=
1
ln
R
t
so
ln
R
t
−
F
H
G
I
K
J
and
R
e
t
−
F
H
G
I
K
J
=
−
.
Therefore,
1
−=
−
R
Ne
t
N
R
e
t
−
1
ej
.
(b)
The maximum number of radioactive nuclei would be
R
.
Additional Problems
P45.38
(a)
Suppose each
235
U fission releases 208 MeV of energy. Then, the number of nuclei that
must have undergone fission is
N
==
×
×
=×
−
total release
energy per nuclei
J
208 MeV
J MeV
nuclei
51
0
160 10
15 10
13
13
24
af
.
..
(b)
mass
nuclei
6.02 10 nuclei mol
g mol
kg
23
=
×
×
F
H
G
I
K
J
≈
235
0 6
24
.
.
P45.39
(a)
At 6 10
8
×
K, the average kinetic energy of a carbon atom is
3
2
15 862 10
6 10
8 10
58
4
kT
B
eV K
K
eV
×
=
×
−
a
f
e
j
Note that 6 10
8
×
K is about 6
36
2
=
times larger than 1 5 10
7
.
×
K , the core temperature of
the Sun. This factor corresponds to the higher potentialenergy barrier to carbon fusion
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This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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