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(D) Given: $N_1 = 2N_2$.Activity of a radioactive substance is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of

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$20$ years and $5$ years,respectively

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(D) Given: $N_1 = 2N_2$.Activity of a radioactive substance is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.Half-life $T_{1/2} = \frac{\ln 2}{\lambda}$,which implies $\lambda = \frac{\ln 2}{T_{1/2}}$.For sample $S_1$: $A_1 = \lambda_1 N_1 = \frac{\ln 2}{T_1} N_1 = 5\,\mu Ci$  ...... $(i)$For sample $S_2$: $A_2 = \lambda_2 N_2 = \frac{\ln 2}{T_2} N_2 = 10\,\mu Ci$  ...... $(ii)$Dividing equation $(ii)$ by $(i)$:$\frac{A_2}{A_1} = \frac{\lambda_2 N_2}{\lambda_1 N_1} = \frac{T_1}{T_2} 	imes \frac{N_2}{N_1} = \frac{10}{5} = 2$Since $N_1 = 2N_2$,we have $\frac{N_2}{N_1} = \frac{1}{2}$.Substituting this into the ratio:$\frac{T_1}{T_2} 	imes \frac{1}{2} = 2 \Rightarrow \frac{T_1}{T_2} = 4 \Rightarrow T_1 = 4T_2$.This means the half-life of $S_1$ is four times that of $S_2$. Looking at the options,if $T_2 = 5$ years,then $T_1 = 20$ years. Thus,the half-lives are $20$ years and $5$ years,respectively.

At some instant,a radioactive sample $S_1$ having an activity $5\,\mu Ci$ 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 $S_2$ are:

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