The half-life of 158O is 122 s.
How long does it take for the number of 158O nuclei in a given sample to decrease to a factor of 5×10−4 of the initial value?
lambda = 0.693 / T1/2 = 0.693 / 122 = 5.68 * 10-3
N = N0 * e-lambda * t
5 * 10-4 N0 = N0 * e-lambda * t
-lambda * t = ln ( 5 * 10-4 )
- 5.68 * 10-3 * t = ln ( 5 * 10-4 )
t = 1338 s
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