KCET · Physics · Nuclear Physics
The masses of two radioactive substances are same and their half-lives are \(1 \mathrm{yr}\) and \(2 \mathrm{yr}\) respectively. The ratio of their activities after 4 yr will be
- A \(1: 4\)
- B \(1: 2\)
- C \(1: 3\)
- D \(1: 6\)
Answer & Solution
Correct Answer
(A) \(1: 4\)
Step-by-step Solution
Detailed explanation
Let initial activity of both substances are same.
\[
\begin{aligned}
\therefore \quad & \mathrm{R}=\mathrm{R}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}=\mathrm{R}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{t}_{1 / 2}} \\
\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} &=\frac{\left(\frac{1}{2}\right)^{4 / 1}}{\left(\frac{1}{2}\right)^{4 / 2}}=\frac{\left(\frac{1}{2}\right)^{4}}{\left(\frac{1}{2}\right)^{2}}=\left(\frac{1}{2}\right)^{2} \\
\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} &=\frac{1}{4}
\end{aligned}
\]
\[
\begin{aligned}
\therefore \quad & \mathrm{R}=\mathrm{R}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}=\mathrm{R}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{t}_{1 / 2}} \\
\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} &=\frac{\left(\frac{1}{2}\right)^{4 / 1}}{\left(\frac{1}{2}\right)^{4 / 2}}=\frac{\left(\frac{1}{2}\right)^{4}}{\left(\frac{1}{2}\right)^{2}}=\left(\frac{1}{2}\right)^{2} \\
\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} &=\frac{1}{4}
\end{aligned}
\]
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