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1
Radioactive Decay
We saw that:
N = N
0
exp(
λ
t)
23.3
Substituting:
N/N
0
= exp((ln 2)t/t
1/2
)
and:
λ
=
(ln 2)/t
1/2
Rearranging:
N/N
0
= exp((t/t
1/2
)ln 2)
Rearranging:
N/N
0
= exp(ln 2
)
(t/t
1/2
)
exp(ln x) = x, so:
N/N
0
= 2
(t/t
1/2
)
Radioactive Decay
23.3
N/N
0
= 2
(t/t
1/2
)
Example: How much of a radioisotope remains after:
(a) 1 half life? (b) 2 half lives? (c) 5 half lives?
(a) after one half life, t = t
1/2
N/N
0
= 2
(t/t
1/2
)
(b) after two half lives, t =
2
t
1/2
N/N
0
= 2
(t/t
1/2
)
(c) after five half lives, t =
5
t
1/2
Note that abundance goes to zero at
infinity
!
Radioactive Decay
23.3
N/N
0
= 2
(t/t
1/2
)
Example: How much
14
C is left after 15,000 years if the half
life is 5730 years?
N/N
0
N/N
0
N/N
0
Therefore
16.3
% of the sample of
14
C remains after
15,000 years of decay.
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Dating Based on Radioactive Decay
23.3
bring t’s down:
ln(N/N
0
) =

(t/t
1/2
)ln 2
take ln:
ln(N/N
0
) = ln 2

(t/t
1/2
)
We saw that:
N/N
0
= 2

(t/t
1/2
)
But what if we know the amount remaining and want
to calculate the age of the specimen? We need to
rearrange the equation to calculate t.
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 Spring '08
 ANDERSON
 Radioactive Decay

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