Half-life of a radioactive substance is 20 , | vedclass

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

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$40$

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(B) The half-life of the substance is $T_{1/2} = 20 \, min$. The decay constant is $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20}$.Let $N_0$ be the initial amount of the substance.For $20\%$ decay,the amount remaining is $N_1 = N_0 - 0.20 N_0 = 0.80 N_0$. The time taken is $t_1$,where $0.80 N_0 = N_0 e^{-\lambda t_1}$.For $80\%$ decay,the amount remaining is $N_2 = N_0 - 0.80 N_0 = 0.20 N_0$. The time taken is $t_2$,where $0.20 N_0 = N_0 e^{-\lambda t_2}$.Dividing the two equations:$\frac{0.80 N_0}{0.20 N_0} = \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}$$4 = e^{\lambda(t_2 - t_1)}$Taking the natural logarithm on both sides:$\ln 4 = \lambda(t_2 - t_1)$$2 \ln 2 = \left( \frac{\ln 2}{20} \right) (t_2 - t_1)$$2 = \frac{t_2 - t_1}{20}$$t_2 - t_1 = 40 \, min$.

Half-life of a radioactive substance is $20 \, min$. The time between $20\%$ and $80\%$ decay will be ......... $min$.

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