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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 \(.......day\)

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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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