JEE Mains · Physics · STD 12 - 13. Nuclei
A radioactive nucleus decays by two different process. The half life of the first process is \(5\) minutes and that of the second process is \(30\,s\). The effective half-life of the nucleus is calculated to be \(\frac{\alpha}{11}\,s\). The value of \(\alpha\) is \(..............\)
- A \(301\)
- B \(302\)
- C \(300\)
- D \(303\)
Answer & Solution
Correct Answer
(C) \(300\)
Step-by-step Solution
Detailed explanation
\(\frac{ dN _1}{ dt }=-\lambda_1 N \quad \frac{ dN _2}{ dt }=-\lambda_2 N\) \(\frac{ dN }{ dt }=-\left(\lambda_1+\lambda_2\right) N\) \(\Rightarrow \lambda_{\text {eq }}=\lambda_1+\lambda_2\) \(\Rightarrow \frac{1}{t_{1 / 2}}=\frac{1}{300}+\frac{1}{30}=\frac{11}{300}\)…
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