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

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(B) Given,half-life $T_{1/2} = 20 	ext{ minutes}$.The law of radioactive decay is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$,where $N(t)$ is the amount

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$40$

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(B) Given,half-life $T_{1/2} = 20 	ext{ minutes}$.The law of radioactive decay is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$,where $N(t)$ is the amount remaining.For $20\%$ decay,the amount remaining is $N_1 = 80\% 	ext{ of } N_0 = 0.8 N_0$.$0.8 N_0 = N_0 (1/2)^{t_1/20} \implies (1/2)^{t_1/20} = 0.8$.For $80\%$ decay,the amount remaining is $N_2 = 20\% 	ext{ of } N_0 = 0.2 N_0$.$0.2 N_0 = N_0 (1/2)^{t_2/20} \implies (1/2)^{t_2/20} = 0.2$.Dividing the two equations:$\frac{(1/2)^{t_2/20}}{(1/2)^{t_1/20}} = \frac{0.2}{0.8} = \frac{1}{4}$.$(1/2)^{(t_2 - t_1)/20} = (1/2)^2$.Equating the exponents:$\frac{t_2 - t_1}{20} = 2$.$t_2 - t_1 = 40 	ext{ minutes}$.

The half-life of a radioactive substance is $20 \text{ minutes}$. The time interval between $20\%$ and $80\%$ decay is ........ $\text{minutes}$.

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