JEE Mains · Physics · STD 12 - 13. Nuclei
A radioactive nuclei with decay constant \(0.5/s\) is being produced at a constant rate of \(100\, nuclei/s\). If at \(t\, = 0\) there were no nuclei, the time when there are \(50\, nuclei\) is
- A \(1\,s\)
- B \(2\ln \left( {\frac{4}{3}} \right)s\)
- C \(ln\, 2\, s\)
- D \(\ln \left( {\frac{4}{3}} \right)s\)
Answer & Solution
Correct Answer
(B) \(2\ln \left( {\frac{4}{3}} \right)s\)
Step-by-step Solution
Detailed explanation
Let \(N\) be the number of nucleiat any time \(t\) then, \(\frac{d N}{d t}=100-\lambda N\) or \(\int_0^N {\frac{{dN}}{{(100 - \lambda N)}}} = \int_0^t d t\) \(-\frac{1}{\lambda}[\log (100-\lambda N)]_{0}^{N}=t\) \(\log (100-\lambda N)-\log 100=-\lambda t\)…
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