KCET · Physics · Nuclear 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 \(\frac{1}{\mathrm{e}}\)
- C \(\mathrm{e}\)
- D \(e^{2}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{e}\)
Step-by-step Solution
Detailed explanation
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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