Q127 / 180 · KCET 2009 · Physics
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
- A$\frac{1}{\mathrm{e}^{2}}$
- B$\frac{1}{\mathrm{e}}$
- Correct answerC$\mathrm{e}$
- D$e^{2}$
Solution
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 $$\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$$
Hence, the correct answer is option: 'C'.