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In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at an instant of time equal to its mean life is
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Verified Answer
The correct answer is:
$\mathrm{e}$
Let the initial number of atoms at time $t=0$ be $\mathrm{N}_{0}$.
Let $N$ be the number of atoms at any instant $t$.
Mean life $\tau=\frac{1}{\lambda}$, where $\lambda$ is disintegration constant.
Given, $\mathrm{t}=\tau$
According to radioactive disintegration law, or
$$
\begin{gathered}
\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \\
\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}=\frac{\mathrm{N}_{0}}{\mathrm{e}}
\end{gathered}
$$
Or $$\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$$
Let $N$ be the number of atoms at any instant $t$.
Mean life $\tau=\frac{1}{\lambda}$, where $\lambda$ is disintegration constant.
Given, $\mathrm{t}=\tau$
According to radioactive disintegration law, or
$$
\begin{gathered}
\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \\
\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}=\frac{\mathrm{N}_{0}}{\mathrm{e}}
\end{gathered}
$$
Or $$\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$$
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