Q169 / 180 · KCET 2010 · Physics
A radioactive sample $S_{1}$ having the activity $\mathrm{A}_{1}$ has twice the number of nuclei as another sample $S_{2}$ of activity $A_{2}$. If $A_{2}=2 A_{1}$, then the ratio of half-life of $S_{1}$ to the half-life of $S_{2}$ is
- Correct answerA4
- B2
- C$0.25$
- D$0.75$
Solution
Activity, $\mathrm{A}=\lambda \mathrm{N}=\frac{0.693}{\mathrm{~T}_{1 / 2}} \mathrm{~N}$
where $\mathrm{T}_{1 / 2}$ is the half-life of a radioactive sample.
$\therefore$
$$
\begin{aligned}
\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}} &=\frac{\mathrm{N}_{1}}{\mathrm{~T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~N}_{2}} \\
\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}} &=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}} \times \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \\
&=\frac{2 \mathrm{~A}_{1}}{\mathrm{~A}_{1}} \times \frac{2 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{4}{1}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.