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# 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 are20 years and 5 years, respectively20 years and 10 years, respectively10 year each5 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=No1e−λ1tN2=No2e−λ2tGiven that at a particular time t N1=2N2Also given thatFor sample 1, A1=−dN1dt=No1λ1e−λ1t=5μCiA2=−dN2dt=No2λ2e−λ2t=10μCiSolvong above equations we get,λ1λ2=14half life1half life2=ln2λ1×λ2ln2=4TS1=4TS2⇒TS1=20 years and TS2=5 years is the possible answer.

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