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Question

A sample of radioactive material decays simultaneously by two processes A and B with half lives 12 h and 14h, respectively. For the first half hour it decays with the process A, next one hour with the process B, and for further half an hour with both A and B. If, originally there were N0 nuclei, find the number of nuclei after 2 h of such decay.

A
N0(2)8
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B
N0(2)4
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C
N0(2)6
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D
N0(2)5
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Solution

The correct option is A N0(2)8
After first half hours, N = N012
For t = 12 h to t = 112 h, 1 h = four half lives
Hence, N = (N012)(12]4 = N0(12)5
For t = 112 to t = 2 h
[For both A and B, 1t1/2 = 1t1/2+ 1t1/4 = 2+ 4 = 6 t1/2 = 16]
12 h = three half-lives
N = (N0(12)5](12)3 = N0(12)8

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