The half-life of a substance is 20 min. The | vedclass

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(A) Given,half-life $T_{1/2} = 20 \ min$.Let $N_0$ be the initial amount of the substance.At $33\%$ decay,the remaining amount is $N_1 = N_0(1 - 0.33)

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$20$

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(A) Given,half-life $T_{1/2} = 20 \ min$.Let $N_0$ be the initial amount of the substance.At $33\%$ decay,the remaining amount is $N_1 = N_0(1 - 0.33) = 0.67 N_0$.At $67\%$ decay,the remaining amount is $N_2 = N_0(1 - 0.67) = 0.33 N_0$.Using the radioactive decay law,$N = N_0(1/2)^{t/T_{1/2}}$,we have:For $N_1$: $0.67 N_0 = N_0(1/2)^{t_1/20} \Rightarrow 0.67 = (1/2)^{t_1/20} \dots (1)$For $N_2$: $0.33 N_0 = N_0(1/2)^{t_2/20} \Rightarrow 0.33 = (1/2)^{t_2/20} \dots (2)$Dividing equation $(1)$ by $(2)$:$\frac{0.67}{0.33} = \frac{(1/2)^{t_1/20}}{(1/2)^{t_2/20}} = (1/2)^{(t_1 - t_2)/20}$Since $0.67/0.33 \approx 2$,we have $2^1 = 2^{(t_2 - t_1)/20}$.Therefore,$(t_2 - t_1)/20 = 1$,which gives $t_2 - t_1 = 20 \ min$.

The half-life of a substance is $20 \ min$. The time taken between $33\%$ decay and $67\%$ decay will be ....... $min$.

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