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 $N$ be the number of atoms at any instant $t$.The mean life $	au$ is defined as $	au

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$e$

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(C) Let the initial number of atoms at time $t=0$ be $N_0$.Let $N$ be the number of atoms at any instant $t$.The mean life $	au$ is defined as $	au = 1/\lambda$,where $\lambda$ is the disintegration constant.According to the law of radioactive decay,the number of atoms remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.Given that the time $t$ is equal to the mean life,i.e.,$t = 	au = 1/\lambda$.Substituting $t = 1/\lambda$ into the decay equation:$N = N_0 e^{-\lambda (1/\lambda)} = N_0 e^{-1} = N_0/e$.We need to find the ratio of the initial number of atoms $(N_0)$ to the number of atoms present at time $t$ $(N)$:Ratio $= N_0 / N = N_0 / (N_0 / e) = e$.Therefore,the correct option is $C$.

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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