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(B) The number of radioactive nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lambda = \frac{1}{	au_{mean}}$.For the fir

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$N_0 \left( \frac{e^4 + 1}{e^5} \right)$

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(B) The number of radioactive nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lambda = \frac{1}{	au_{mean}}$.For the first species with mean-life $	au_1 = 	au$,the decay constant is $\lambda_1 = \frac{1}{	au}$.The number of nuclei remaining at $t = 5	au$ is $N_1 = N_0 e^{-\lambda_1 t} = N_0 e^{-(1/	au)(5	au)} = N_0 e^{-5} = \frac{N_0}{e^5}$.For the second species with mean-life $	au_2 = 5	au$,the decay constant is $\lambda_2 = \frac{1}{5	au}$.The number of nuclei remaining at $t = 5	au$ is $N_2 = N_0 e^{-\lambda_2 t} = N_0 e^{-(1/5	au)(5	au)} = N_0 e^{-1} = \frac{N_0}{e}$.The total number of radioactive nuclei is $N_{total} = N_1 + N_2 = \frac{N_0}{e^5} + \frac{N_0}{e}$.Taking $N_0$ as a common factor,$N_{total} = N_0 \left( \frac{1 + e^4}{e^5} ight)$.

$A$ radioactive sample consists of two distinct species having an equal number of $N_0$ atoms initially. The mean-life of one species is $\tau$ and of the other is $5\tau$. The decay products in both cases are stable. The total number of radioactive nuclei at $t = 5\tau$ is

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