JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2022
- Medium
- Numerical
The activity of a radioactive material is \(6.4 \times 10^{-4}\) curie. Its half life is \(5\; days\). The activity will become \(5 \times 10^{-6}\) curie after \(.......day\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(35\)
Step-by-step Solution
Detailed explanation
\(A_{0}=6.4 \times 10^{-4}\) Curie
\(T _{1 / 2}=5 \text { days }=\frac{\ln 2}{\lambda}\)
\(A = A _{0} e ^{-\lambda t }\)
\(5 \times 10^{-6}=6.4 \times 10^{-4} e ^{-\lambda t }\)
\(\frac{5}{6.4} \times 10^{-2}=e^{-\lambda t}\)
\(7.8 \times 10^{-3}= e ^{-\lambda t}\)
\(\log \left(7.8 \times 10^{-3}\right)=-\lambda t \ln e\)
\(\ln \left(7.8 \times 10^{-3}\right)=-\frac{\lambda n 2}{5} \cdot t\)
\(\therefore \frac{5 \times 4.853}{0.693}= t =35 \text { days }\)
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