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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
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The correct answer is:
$1: 4$
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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