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(D) Let $N_0$ be the initial number of atoms of each species. The number of radioactive nuclei of the first species at time $t$ is $N_1(t) = N_0 e^{-t

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(D) Let $N_0$ be the initial number of atoms of each species. The number of radioactive nuclei of the first species at time $t$ is $N_1(t) = N_0 e^{-t/	au}$.The number of radioactive nuclei of the second species at time $t$ is $N_2(t) = N_0 e^{-t/(5	au)}$.The total number of radioactive nuclei is $N(t) = N_1(t) + N_2(t) = N_0(e^{-t/	au} + e^{-t/(5	au)})$.At $t = 0$,$N(0) = 2N_0$. As $t 	o \infty$,$N(t) 	o 0$.Since both $e^{-t/	au}$ and $e^{-t/(5	au)}$ are monotonically decreasing functions of time,their sum $N(t)$ must also be a monotonically decreasing function of time.Therefore,the total number of radioactive nuclei will continuously decrease with time,which is represented by the graph in option $(d)$.

$t (h)$	$0$	$1$	$2$	$3$	$4$
$R (MBq)$	$100$	$35.36$	$12.51$	$4.42$	$1.56$

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