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(B) The activity of a radioactive sample is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.Given: $A_1

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$20\, years$ and $10\, years$,respectively

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(B) The activity of a radioactive sample is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.Given: $A_1 = 5\,\mu Ci$ and $A_2 = 10\,\mu Ci$.Also,$N_1 = 2N_2$.Using the activity formula: $A_1 = \lambda_1 N_1$ and $A_2 = \lambda_2 N_2$.Substituting the values: $5 = \lambda_1 N_1$ and $10 = \lambda_2 N_2$.Since $N_1 = 2N_2$,we have $5 = \lambda_1 (2N_2) \Rightarrow 10 = 2\lambda_1 N_2$.Comparing with $A_2 = 10 = \lambda_2 N_2$,we get $\lambda_2 N_2 = 2\lambda_1 N_2$,which implies $\lambda_2 = 2\lambda_1$.The half-life $T_{1/2}$ is inversely proportional to the decay constant $\lambda$ $(T_{1/2} = \ln(2) / \lambda)$.Therefore,$T_2 = T_1 / 2$,or $T_1 = 2T_2$.Checking the options,if $T_2 = 5\, years$,then $T_1 = 10\, years$. If $T_2 = 10\, years$,then $T_1 = 20\, years$.Option $A$ gives $T_1 = 20\, years$ and $T_2 = 5\, years$,which does not satisfy $T_1 = 2T_2$. Option $B$ gives $T_1 = 20\, years$ and $T_2 = 10\, years$,which satisfies $T_1 = 2T_2$.

$A$ radioactive sample $S_1$ having an activity of $5\,\mu Ci$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10\,\mu Ci$. The half-lives of $S_1$ and $S_2$ can be

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