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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2022
  • Medium
  • Numerical

Two radioactive materials \(A\) and \(B\) have decay constants \(25 \lambda\) and \(16 \lambda\) respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of \(B\) to that of \(A\) will be "\(e\)" after a time \(\frac{1}{a \lambda}\). The value of \(a\) is \(......\)

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Answer & Solution

Correct Answer

(A) \(9\)

Step-by-step Solution

Detailed explanation

\(N = N _{0} e ^{-\lambda t }\)

\(\frac{ N _{ B }}{ N _{ A }}=\frac{ e ^{-\lambda_{2} t }}{ e ^{-\lambda_{1} t }}= e ^{-\lambda_{2} t } \cdot e ^{\lambda_{1} t }\)

\(e ^{1}= e ^{\left(\lambda_{1}-\lambda_{2}\right) t }\)

\(\left(\lambda_{1}-\lambda_{2}\right) t =1\)

\(t =\frac{1}{\lambda_{1}-\lambda_{2}}=\frac{1}{25 \lambda-16 \lambda}=\frac{1}{9 \lambda}\)

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