WBJEE · Physics · Nuclear Physics
Two radionctive substances \(A\) and \(B\) have decay constants \(5 \lambda\) and \(\lambda\) respectively. At \(t=0,\) they have the same number of nuclei. The ratio of number of nuclei of \(A\) to that of \(B\) will be \((1 / e)^{2}\)
after a time interval of
- A \(\frac{1}{\lambda}\)
- B \(\frac{1}{2 \lambda}\)
- C \(\frac{1}{3 \lambda}\)
- D \(\frac{1}{4 \lambda}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2 \lambda}\)
Step-by-step Solution
Detailed explanation
We know that \(N=N_{0} e^{-\lambda t}\) For \(A\), \(N_{A}=N_{0} e^{-5 \lambda_{L}}\) For \(B\), \(N_{B}=N_{0} e^{-\lambda t}\) Given, \(\begin{aligned} \frac{N_{A}}{N_{B}} &=\frac{1}{e^{2}} \\ \frac{N}{N_{B}} &=\frac{1}{e^{4 \lambda t}} \\ t &=\frac{1}{2 \lambda} \end{aligned}\)
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