The half-life of a radioactive substance is 2 | vedclass

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(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.At time $t_1$,the amount decayed is $\frac{1}{3} N_0$,so the remaining amount is

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$20$

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(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.At time $t_1$,the amount decayed is $\frac{1}{3} N_0$,so the remaining amount is $N_1 = N_0 - \frac{1}{3} N_0 = \frac{2}{3} N_0$.At time $t_2$,the amount decayed is $\frac{2}{3} N_0$,so the remaining amount is $N_2 = N_0 - \frac{2}{3} N_0 = \frac{1}{3} N_0$.We observe that $N_2 = \frac{1}{2} N_1$.Since the amount of substance reduces to half of its value in one half-life period $(T_{1/2})$,the time interval required for the substance to decay from $N_1$ to $N_2$ is exactly one half-life.Therefore,$t_2 - t_1 = T_{1/2} = 20 \, min$.

The half-life of a radioactive substance is $20 \, min$. The approximate 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 ......... $min$.

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