JEE Mains · Physics · STD 12 - 13. Nuclei
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 \(.......day\)
- A \(7\)
- B \(15\)
- C \(25\)
- D \(35\)
Answer & Solution
Correct Answer
(D) \(35\)
Step-by-step Solution
Detailed explanation
\(A_{0}=6.4 \times 10^{-4}\) Curie \(T _{1 / 2}=5 \text { days }=\frac{\ln 2}{\lambda}\) \(A = A _{0} e ^{-\lambda t }\) \(5 \times 10^{-6}=6.4 \times 10^{-4} e ^{-\lambda t }\) \(\frac{5}{6.4} \times 10^{-2}=e^{-\lambda t}\) \(7.8 \times 10^{-3}= e ^{-\lambda t}\)…
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