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(B) The number of radioactive nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-t/	au_{mean}}$.For the first species with mean-life $	au_1

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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 after time $t$ is given by $N(t) = N_0 e^{-t/	au_{mean}}$.For the first species with mean-life $	au_1 = 	au$ at time $t = 5	au$:$N_1 = N_0 e^{-5	au/	au} = N_0 e^{-5}$.For the second species with mean-life $	au_2 = 5	au$ at time $t = 5	au$:$N_2 = N_0 e^{-5	au/5	au} = N_0 e^{-1}$.The total number of radioactive nuclei is $N_{total} = N_1 + N_2$.$N_{total} = N_0 e^{-5} + N_0 e^{-1} = N_0 \left( \frac{1}{e^5} + \frac{1}{e} ight)$.$N_{total} = N_0 \left( \frac{1 + e^4}{e^5} ight)$.

$A$ radioactive sample consists of two distinct species having an equal number of atoms $N_0$ 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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