In a mean life of a radioactive sample,

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(B) The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$.The mean life $	au$ is defined as $	au = 1/\lambda$.At time $t = 	au$,the amo

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About $2/3$ of the substance disintegrates

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(B) The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$.The mean life $	au$ is defined as $	au = 1/\lambda$.At time $t = 	au$,the amount of substance remaining is $N = N_0 e^{-\lambda(1/\lambda)} = N_0 e^{-1} = N_0/e$.Since $e \approx 2.718$,we have $N \approx N_0 / 2.718 \approx 0.368 N_0 \approx N_0 / 3$.The amount of substance that has disintegrated is $N_{dis} = N_0 - N = N_0 - N_0/e = N_0(1 - 1/e)$.$N_{dis} \approx N_0(1 - 0.368) = 0.632 N_0 \approx 2/3 N_0$.Thus,about $2/3$ of the substance disintegrates.

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