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Two radioactive materials $\mathrm{X}_1$ and $\mathrm{X}_2$ have decay constants ' $5 \lambda$ ' and ' $\lambda$ ' respectively. Initially, they have the same number of nuclei. After time ' $t$ ', the ratio of number of nuclei of $X_1$ to that of $X_2$ is $\frac{1}{e}$. Then $t$ is equal to
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The correct answer is:
$\frac{1}{4 \lambda}$
Let $N_1$ be the number of $X_1$ and $N_2$ be the number of nuclei of $X_2$ after time t.
$$
\begin{aligned}
& \therefore \mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}} \text { and } \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \therefore \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{e}^{-\lambda, \mathrm{t}}}=\frac{1}{\mathrm{e}^{4 \lambda \cdot \mathrm{t}}}=\frac{1}{\mathrm{e}} \\
& \therefore 4 \lambda \mathrm{t}=1 \\
& \therefore \mathrm{t}=\frac{1}{4 \lambda}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}} \text { and } \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \therefore \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{e}^{-\lambda, \mathrm{t}}}=\frac{1}{\mathrm{e}^{4 \lambda \cdot \mathrm{t}}}=\frac{1}{\mathrm{e}} \\
& \therefore 4 \lambda \mathrm{t}=1 \\
& \therefore \mathrm{t}=\frac{1}{4 \lambda}
\end{aligned}
$$
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