KCET · Physics · Nuclear 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
- A 4
- B 2
- C \(0.25\)
- D \(0.75\)
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
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.
\(\therefore\)
\[
\begin{aligned}
\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}} &=\frac{\mathrm{N}_{1}}{\mathrm{~T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~N}_{2}} \\
\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}} &=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}} \times \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \\
&=\frac{2 \mathrm{~A}_{1}}{\mathrm{~A}_{1}} \times \frac{2 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{4}{1}
\end{aligned}
\]
where \(\mathrm{T}_{1 / 2}\) is the half-life of a radioactive sample.
\(\therefore\)
\[
\begin{aligned}
\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}} &=\frac{\mathrm{N}_{1}}{\mathrm{~T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~N}_{2}} \\
\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}} &=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}} \times \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \\
&=\frac{2 \mathrm{~A}_{1}}{\mathrm{~A}_{1}} \times \frac{2 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{4}{1}
\end{aligned}
\]
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