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(B) The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.At $t_1$,the decay fraction is $\frac{1}{5}$,so the remaining f

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$\frac{t_2 - t_1}{\ln 4}$

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(B) The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.At $t_1$,the decay fraction is $\frac{1}{5}$,so the remaining fraction is $1 - \frac{1}{5} = \frac{4}{5}$.Thus,$\frac{N(t_1)}{N_0} = e^{-\lambda t_1} = \frac{4}{5}$.Taking the natural logarithm,$-\lambda t_1 = \ln(4/5) = \ln 4 - \ln 5$.At $t_2$,the decay fraction is $\frac{4}{5}$,so the remaining fraction is $1 - \frac{4}{5} = \frac{1}{5}$.Thus,$\frac{N(t_2)}{N_0} = e^{-\lambda t_2} = \frac{1}{5}$.Taking the natural logarithm,$-\lambda t_2 = \ln(1/5) = -\ln 5$.Subtracting the two equations: $(-\lambda t_1) - (-\lambda t_2) = (\ln 4 - \ln 5) - (-\ln 5) = \ln 4$.$\lambda(t_2 - t_1) = \ln 4 = 2 \ln 2$.$\lambda = \frac{2 \ln 2}{t_2 - t_1}$.The mean life $	au$ is given by $	au = \frac{1}{\lambda} = \frac{t_2 - t_1}{2 \ln 2} = \frac{t_2 - t_1}{\ln 4}$.

$A$ fresh radioactive sample is given at $t = 0$. Its decay fraction is $\frac{1}{5}$ at $t_1$ instant and $\frac{4}{5}$ at $t_2$ instant. Its mean life is

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