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
- A \(\frac{\lambda}{2}\)
- B \(\frac{\mathrm{e}}{\lambda}\)
- C \(\lambda\)
- D \(\frac{1}{4 \lambda}\)
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.
\(
\begin{aligned}
& \therefore \mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}} \text { and } \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \therefore \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{e}^{-\lambda, \mathrm{t}}}=\frac{1}{\mathrm{e}^{4 \lambda \cdot \mathrm{t}}}=\frac{1}{\mathrm{e}} \\
& \therefore 4 \lambda \mathrm{t}=1 \\
& \therefore \mathrm{t}=\frac{1}{4 \lambda}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}} \text { and } \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\
& \therefore \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{e}^{-\lambda, \mathrm{t}}}=\frac{1}{\mathrm{e}^{4 \lambda \cdot \mathrm{t}}}=\frac{1}{\mathrm{e}} \\
& \therefore 4 \lambda \mathrm{t}=1 \\
& \therefore \mathrm{t}=\frac{1}{4 \lambda}
\end{aligned}
\)
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