A) 1s
B) \[2\ln \left( \frac{4}{3} \right)s\]
C) \[\ln 2s\]
D) \[\ln \left( \frac{4}{3} \right)s\]
Correct Answer: B
Solution :
Let N be the number of nuclei at any time t then,\[\frac{dN}{dt}=100-\lambda N\]or\[\int\limits_{0}^{N}{\frac{dN}{(100-\lambda N)}}=\int\limits_{0}^{t}{dt}\] \[-\frac{1}{\lambda }[log(100-\lambda N)]_{0}^{N}=t\] \[[log(100-\lambda N-log100=-\lambda t\] \[\log \frac{100-\lambda N}{100}=-\lambda t\] \[\frac{100-\lambda N}{100}={{e}^{-\lambda t}}\] \[1-\frac{\lambda N}{100}={{e}^{-\lambda t}}\] \[N=\frac{100}{\lambda }(1-{{e}^{-\lambda }}t)\] As, N = 50 and \[\lambda =0.5/\sec \] \[\therefore \]\[50=\frac{100}{0.5}(1-{{e}^{-0.5t}})\] Solving we get,\[t=2\ln \left( \frac{4}{3} \right)\sec \]
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