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 \(......\)
Pick an option to check your answer.
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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