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(0.3466/s)(173)
60/s
60 Bq
RN
λ
==
≈
=
.
36. The number of atoms present initially at
0
t
=
is
6
0
2.00 10
N
=×
. From Fig. 4219,
we see that the number is halved at
2.00 s.
t
=
Thus, using Eq. 4215, we find the decay
constant to be
1
00
0
11
1
ln
ln
ln 2
0.3466 s
2.00 s
/ 2
2.00 s
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 Spring '08
 Any
 Physics

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