In a radioactive disintegration,the ratio of | vedclass

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(C) Let the initial number of atoms at time $t=0$ be $N_{0}$.Let the number of atoms at any instant of time $t$ be $N$.According to the radioactive de

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$e$

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(C) Let the initial number of atoms at time $t=0$ be $N_{0}$.Let the number of atoms at any instant of time $t$ be $N$.According to the radioactive decay law,$N = N_{0} e^{-\lambda t}$.Mean life is defined as $	au = 1/\lambda$,where $\lambda$ is the decay constant.Given that the time $t = 	au$,we substitute this into the decay equation:$N = N_{0} e^{-\lambda 	au} = N_{0} e^{-\lambda (1/\lambda)} = N_{0} e^{-1}$.Therefore,$N = N_{0}/e$.The ratio of the initial number of atoms to the number of atoms present at time $t = 	au$ is:$N_{0}/N = N_{0} / (N_{0}/e) = e$.

In a radioactive disintegration,the ratio of the initial number of atoms to the number of atoms present at an instant of time equal to its mean life is

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