JEE Mains · Physics · STD 12 - 13. Nuclei
The disintegration rate of a certain radioactive sample at any instant is \(4250\) disintegrations per minute.\(10\) minutes later, the rate becomes \(2250\) disintegrations per minute. The approximate decay cons \(.........\min^{-1}\)
- A \(0.02\)
- B \(2.7\)
- C \(0.063\)
- D \(6.3\)
Answer & Solution
Correct Answer
(C) \(0.063\)
Step-by-step Solution
Detailed explanation
At \(t=0\) disintegration rate \(=4250\,dpm\) At \(t=10\) disintegration rate \(=2250\,dpm\) \(A=A_{0} e^{-\lambda t}\) \(2250=4250\,e ^{-\lambda .(10)}\) \(\Rightarrow \lambda(10)=\ln \left(\frac{4250}{2250}\right)\) \(\Rightarrow \lambda=0.063 min ^{-1}\)
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