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MHT CET · Physics · Nuclear Physics

Two radioactive materials \(\mathrm{X}_1\) and \(\mathrm{X}_2\) have decay constants ' \(5 \lambda\) ' and ' \(\lambda\) ' respectively. Initially, they have the same number of nuclei. After time ' \(t\) ', the ratio of number of nuclei of \(X_1\) to that of \(X_2\) is \(\frac{1}{e}\). Then \(t\) is equal to

  1. A \(\frac{\lambda}{2}\)
  2. B \(\frac{\mathrm{e}}{\lambda}\)
  3. C \(\lambda\)
  4. D \(\frac{1}{4 \lambda}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{4 \lambda}\)

Step-by-step Solution

Detailed explanation

Let \(N_1\) be the number of \(X_1\) and \(N_2\) be the number of nuclei of \(X_2\) after time t.
\(
\begin{aligned}
& \therefore \mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}} \text { and } \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \therefore \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{e}^{-\lambda, \mathrm{t}}}=\frac{1}{\mathrm{e}^{4 \lambda \cdot \mathrm{t}}}=\frac{1}{\mathrm{e}} \\
& \therefore 4 \lambda \mathrm{t}=1 \\
& \therefore \mathrm{t}=\frac{1}{4 \lambda}
\end{aligned}
\)