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Question

At some instant, a radioactive sample S1 having an activity 5 μCi has twice the number of nuclei as another sample S2 which has an activity of 10 μCi. The half lives of S1 and S2 are
  1. 20 years and 5 years, respectively
  2. 20 years and 10 years, respectively
  3. 10 year each
  4. 5 year each

A
10 year each
B
20 years and 5 years, respectively
C
20 years and 10 years, respectively
D
5 year each
Solution
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For sample 1 we can write,
N1=No1eλ1t
N2=No2eλ2t
Given that at a particular time t N1=2N2
Also given that
For sample 1,
A1=dN1dt=No1λ1eλ1t=5μCi
A2=dN2dt=No2λ2eλ2t=10μCi
Solvong above equations we get,
λ1λ2=14
half life1half life2=ln2λ1×λ2ln2=4

TS1=4TS2
TS1=20 years and TS2=5 years is the possible answer.

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