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 $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20} \ min^{-1}$.Let $N_0$ be the initial

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$40$

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(B) Given: Half-life $T_{1/2} = 20 \ min$. The decay constant $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20} \ min^{-1}$.Let $N_0$ be the initial amount of the substance.At $20\%$ decay,the remaining amount is $N_1 = N_0 - 0.20 N_0 = 0.80 N_0$. The time taken is $t_1$.Using the decay law $N_1 = N_0 e^{-\lambda t_1}$,we have $0.80 N_0 = N_0 e^{-\lambda t_1}$,so $e^{-\lambda t_1} = 0.8$.At $80\%$ decay,the remaining amount is $N_2 = N_0 - 0.80 N_0 = 0.20 N_0$. The time taken is $t_2$.Using the decay law $N_2 = N_0 e^{-\lambda t_2}$,we have $0.20 N_0 = N_0 e^{-\lambda t_2}$,so $e^{-\lambda t_2} = 0.2$.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 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 interval between $20\%$ decay and $80\%$ decay is ......... $min$.

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