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In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at time $t=\frac{1}{2 \lambda}$ is $[\lambda=$ decay constant $]$
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Verified Answer
The correct answer is:
$\sqrt{\mathrm{e}}$
According to radioactive disintegration law,
$$
\begin{aligned}
& \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda t} \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda \times \frac{1}{2 \lambda}} \\
& \left(\because t=\frac{1}{2 \lambda}\right) \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\frac{1}{2}} \\
& \frac{\mathrm{N}_0}{\mathrm{~N}}=\mathrm{e}^{\frac{1}{2}} \\
& \frac{\mathrm{N}_0}{\mathrm{~N}}=\sqrt{\mathrm{e}} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda t} \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda \times \frac{1}{2 \lambda}} \\
& \left(\because t=\frac{1}{2 \lambda}\right) \\
& \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\frac{1}{2}} \\
& \frac{\mathrm{N}_0}{\mathrm{~N}}=\mathrm{e}^{\frac{1}{2}} \\
& \frac{\mathrm{N}_0}{\mathrm{~N}}=\sqrt{\mathrm{e}} \\
&
\end{aligned}
$$
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