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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 halflives of $S_1$ and $S_2$ are
$10$ years and $20$ years, respectively
$5$ years and $20$ years, respectively
$20$ years and $10$ years, respectively
$20$ years and $5$ years, respectively
Solution
Given $: \mathrm{N}_{1}=2 \mathrm{N}_{2}$
Activity of radioactive substance $=\lambda \mathrm{N}$
Half life period $\mathrm{t}=\frac{\ln 2}{\lambda}$ or, $\lambda=\frac{\ln 2}{\mathrm{t}}$
$\lambda_{1} \mathrm{N}_{1}=\frac{\ln 2}{\mathrm{t}_{1}} \times \mathrm{N}_{1}=5 \mu \mathrm{C}_{\mathrm{i}}$ …… $(i)$
$\lambda_{2} \mathrm{N}_{2}=\frac{\ln 2}{\mathrm{t}_{2}} \times \mathrm{N}_{2}=10 \mu \mathrm{C}_{\mathrm{i}}$ …… $(ii)$
Dividing equation $(ii)$ by $(i)$
$\frac{t_{2}}{t_{1}} \times \frac{N_{1}}{N_{2}}=\frac{1}{2}$
$\frac{t_{2}}{t_{1}}=\frac{1}{4} \Rightarrow t_{1}=4 t_{2}$
i.e., Half life of $S_{1}$ is four times of sample $S_{2}$ Hence $5$ years and $20$ years.
