A fresh radioactive sample is given at t = 0. | vedclass

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(B) Let $N_0$ be the initial number of radioactive nuclei.At time $t_1$,the decay fraction is $\frac{1}{5}$,so the remaining fraction is $1 - \frac{1}

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

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(B) Let $N_0$ be the initial number of radioactive nuclei.At time $t_1$,the decay fraction is $\frac{1}{5}$,so the remaining fraction is $1 - \frac{1}{5} = \frac{4}{5}$. Thus,$N(t_1) = N_0 \frac{4}{5}$.At time $t_2$,the decay fraction is $\frac{4}{5}$,so the remaining fraction is $1 - \frac{4}{5} = \frac{1}{5}$. Thus,$N(t_2) = N_0 \frac{1}{5}$.The decay law is $N(t) = N_0 e^{-\lambda t}$.For $t_1$: $\frac{4}{5} = e^{-\lambda t_1} \implies \ln(\frac{4}{5}) = -\lambda t_1$.For $t_2$: $\frac{1}{5} = e^{-\lambda t_2} \implies \ln(\frac{1}{5}) = -\lambda t_2$.Subtracting the two equations: $\ln(\frac{1}{5}) - \ln(\frac{4}{5}) = -\lambda t_2 - (-\lambda t_1) = -\lambda(t_2 - t_1)$.$\ln(\frac{1/5}{4/5}) = -\lambda(t_2 - t_1) \implies \ln(\frac{1}{4}) = -\lambda(t_2 - t_1)$.$-\ln(4) = -\lambda(t_2 - t_1) \implies \lambda = \frac{\ln 4}{t_2 - t_1}$.Mean life $T_m = \frac{1}{\lambda} = \frac{t_2 - t_1}{\ln 4}$.

$A$ 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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