Half-life of a radioactive substance is 20 , | vedclass

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(B) Let the initial amount of the radioactive substance be $N_0$.When $33 \%$ is disintegrated,the remaining amount is $N_1 = N_0 - 0.33 N_0 = 0.67 N_

Vedclass · Accepted Answer

$20$

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(B) Let the initial amount of the radioactive substance be $N_0$.When $33 \%$ is disintegrated,the remaining amount is $N_1 = N_0 - 0.33 N_0 = 0.67 N_0 \approx \frac{2}{3} N_0$.When $67 \%$ is disintegrated,the remaining amount is $N_2 = N_0 - 0.67 N_0 = 0.33 N_0 \approx \frac{1}{3} N_0$.We know that the amount remaining at time $t$ is given by $N(t) = N_0 (\frac{1}{2})^{t/T}$,where $T = 20 \, min$ is the half-life.For $N_1$: $\frac{2}{3} N_0 = N_0 (\frac{1}{2})^{t_1/20} \implies \frac{2}{3} = (\frac{1}{2})^{t_1/20}$.For $N_2$: $\frac{1}{3} N_0 = N_0 (\frac{1}{2})^{t_2/20} \implies \frac{1}{3} = (\frac{1}{2})^{t_2/20}$.Dividing the two equations: $\frac{1/3}{2/3} = \frac{(1/2)^{t_2/20}}{(1/2)^{t_1/20}} \implies \frac{1}{2} = (\frac{1}{2})^{(t_2-t_1)/20}$.Comparing the exponents: $1 = \frac{t_2-t_1}{20} \implies t_2 - t_1 = 20 \, min$.

Half-life of a radioactive substance is $20 \, min$. The difference between the points of time when it is $33 \%$ disintegrated and $67 \%$ disintegrated is approximately .......... $min$.

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