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71. (a) Adapting Eq. 4320, we fnd
N
0
=
(2
.
5
×
10
−
3
g)(6
.
02
×
10
23
/
mol)
239 g
/
mol
=6
.
3
×
10
18
.
(b) From Eq. 4314 and Eq. 4317,

∆
N

=
N
0
h
1
−
e
−
t
ln 2
/T
1
/
2
i
=(
6
.
3
×
10
18
)
h
1
−
e
−
(12 h) ln 2
/
(24
,
100 y)(8760 h
/
y)
i
=2
.
5
×
10
11
.
(c) The energy absorbed by the body is
(0
.
95)
E
α

∆
N

=(0
.
95) (5
.
2MeV)
(
2
.
5
×
10
11
)(
1
.
6
×
10
−
13
J
/
MeV
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy

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