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:$\frac{t_2 - t_1}{20} = 2$.$t_2 - t_1 = 40 	ext{ minutes}$.

List-$I$	List-$II$
$(i) \ a \bar{a}$	$(A) -\frac{\pi}{3}$
$(ii) \ \arg \left(\frac{1}{\bar{a}}\right)$	$(B) -i \sqrt{3}$
$(iii) \ a-\bar{a}$	$(C) \frac{2i}{\sqrt{3}}$
$(iv) \ \operatorname{Im}\left(\frac{4}{3a}\right)$	$(D) 1$
	$(E) \frac{\pi}{3}$
	$(F) \frac{2}{\sqrt{3}}$

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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