Radioactive half life for phosphorus-32 is 14.3 days , calculate activity of 1.6 mg sample phosphorus
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Answer:
26.8 days
Explanation:
The equation for 1st order decay is:
N
t
=
N
0
e
−
λ
t
We can get the value of the decay constant
λ
from the 1/2 life:
λ
=
0.693
t
1
2
=
0.693
14.3
=
0.0484
x
d
−
1
Taking natural logs of both sides:
ln
N
t
=
ln
N
0
−
λ
t
Putting in the numbers and converting to milligrams:
ln
0.6
=
ln
2.2
−
0.0484
t
∴
−
0.5108
=
0.7884
−
0.0484
t
∴
t
=
1.30
0.0484
=
26.8
x
days
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