For a substance,the fraction of its initial q | vedclass

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(C) For a radioactive substance,the number of nuclei remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.Given the mean life $	au = 1/\lambda$

Vedclass · Accepted Answer

$(2/3) N_0$

Vedclass · Answer

(C) For a radioactive substance,the number of nuclei remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.Given the mean life $	au = 1/\lambda$,so at $t = 	au$,we have $t = 1/\lambda$.Substituting this into the equation: $N = N_0 e^{-\lambda 	imes (1/\lambda)} = N_0 e^{-1} = N_0 / e$.Given $e = 2.71$,the amount remaining is $N = N_0 / 2.71 \approx 0.37 N_0$.The amount that has disintegrated is the initial amount minus the remaining amount.Disintegrated amount $= N_0 - 0.37 N_0 = 0.63 N_0$.Since $2/3 \approx 0.66$ and $1 - 1/e \approx 0.632$,the fraction is approximately $0.63 N_0$,which is closest to $(2/3) N_0$.

For a substance,the fraction of its initial quantity $(N_0)$ which will disintegrate in its average lifetime is about $(e = 2.71)$.

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