MHT CET · Physics · Nuclear Physics
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 \(]\)
- A \(\frac{1}{\mathrm{e}}\)
- B \(\sqrt{\mathrm{e}}\)
- C \(\mathrm{e}\)
- D \(2 \mathrm{e}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
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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