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

The correct option is A $20$ years and $5$ years, respectively
Given that,
$A_1={\lambda }_1{N}_1=5\mu \text{Ci}$

$A_2={\lambda }_2{N}_2=10\mu \text{Ci}$

From given data, we get

${\lambda }_2{N}_2=2{\lambda }_1{N}_1$

Also, given that, $N_1=2N_2$

$\Rightarrow {\lambda }_2{N}_2=2{\lambda }_1\left(2N_2\right)$

$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=4$

As, we know, $\lambda \propto \frac{1}{T_{1/2}}$

If $T_1$ and $T_2$ represents the half life of the sample, $S_1$ and $S_2$

$\Rightarrow \frac{T_1}{T_2}=4$

Of the given options, option $\left(A\right)$ satisfies the above condition.

Hence, option $\left(A\right)$ is correct.