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(B) Let $N_0$ be the initial number of nuclei. The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.At $t_1$,the decay f

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

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(B) Let $N_0$ be the initial number of nuclei. The number of nuclei remaining 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,$N(t_1) = N_0 \frac{4}{5} = N_0 e^{-\lambda t_1}$.At $t_2$,the decay fraction is $\frac{4}{5}$,so the remaining fraction is $1 - \frac{4}{5} = \frac{1}{5}$. Thus,$N(t_2) = N_0 \frac{1}{5} = N_0 e^{-\lambda t_2}$.Dividing the two equations: $\frac{N(t_1)}{N(t_2)} = \frac{4/5}{1/5} = 4 = \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = e^{\lambda(t_2 - t_1)}$.Taking the natural logarithm on both sides: $\ln 4 = \lambda(t_2 - t_1)$.Since the mean life $	au = \frac{1}{\lambda}$,we have $\lambda = \frac{1}{	au}$.Substituting this into the equation: $\ln 4 = \frac{t_2 - t_1}{	au}$.Therefore,$	au = \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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