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

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(B) Given,half-life $T_{1/2} = 20 \ min$. The decay constant is $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20} \ min^{-1}$.Let $N_0$ be the initi

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$40$

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(B) Given,half-life $T_{1/2} = 20 \ min$. The decay constant is $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20} \ min^{-1}$.Let $N_0$ be the initial amount of radioactive substance.At time $t_1$,$20\%$ has decayed,so the remaining amount is $N(t_1) = 0.8 N_0$. Using $N(t) = N_0 e^{-\lambda t}$,we have $0.8 N_0 = N_0 e^{-\lambda t_1} \implies e^{-\lambda t_1} = 0.8$.At time $t_2$,$80\%$ has decayed,so the remaining amount is $N(t_2) = 0.2 N_0$. Thus,$0.2 N_0 = N_0 e^{-\lambda t_2} \implies e^{-\lambda t_2} = 0.2$.Dividing the two equations: $\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = \frac{0.8}{0.2} = 4$.$e^{\lambda(t_2 - t_1)} = 4$.Taking the natural logarithm on both sides: $\lambda(t_2 - t_1) = \ln 4 = 2 \ln 2$.Substituting $\lambda = \frac{\ln 2}{20}$:$\frac{\ln 2}{20} (t_2 - t_1) = 2 \ln 2$.$t_2 - t_1 = 2 	imes 20 = 40 \ min$.

The half-life of a radioactive substance is $20 \ min$. The time taken for the decay to increase from $20\%$ to $80\%$ is ........ $min$.

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