The half-life of a radioactive nucleus is 50 | vedclass

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Question

(D) Let $N_0$ be the initial number of nuclei.At time $t_1$,the amount decayed is $\frac{1}{3}N_0$,so the remaining amount is $N(t_1) = N_0 - \frac{1}

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Let $N_0$ be the initial number of nuclei.At time $t_1$,the amount decayed is $\frac{1}{3}N_0$,so the remaining amount is $N(t_1) = N_0 - \frac{1}{3}N_0 = \frac{2}{3}N_0$.Using the decay law $N(t) = N_0 (\frac{1}{2})^{t/T_{1/2}}$,we have $\frac{2}{3}N_0 = N_0 (\frac{1}{2})^{t_1/50}$,which implies $(\frac{1}{2})^{t_1/50} = \frac{2}{3}$.At time $t_2$,the amount decayed is $\frac{2}{3}N_0$,so the remaining amount is $N(t_2) = N_0 - \frac{2}{3}N_0 = \frac{1}{3}N_0$.Using the decay law,$\frac{1}{3}N_0 = N_0 (\frac{1}{2})^{t_2/50}$,which implies $(\frac{1}{2})^{t_2/50} = \frac{1}{3}$.Dividing the two equations: $\frac{(\frac{1}{2})^{t_1/50}}{(\frac{1}{2})^{t_2/50}} = \frac{2/3}{1/3} = 2$.This simplifies to $(\frac{1}{2})^{(t_1-t_2)/50} = 2^1$,or $2^{(t_2-t_1)/50} = 2^1$.Therefore,$\frac{t_2 - t_1}{50} = 1$,which gives $t_2 - t_1 = 50 \ days$.

The half-life of a radioactive nucleus is $50 \ days$. The time interval $(t_2 - t_1)$ between the time $t_2$ when $\frac{2}{3}$ of it has decayed and the time $t_1$ when $\frac{1}{3}$ of it has decayed is ......... $days$.

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