Consider a radioactive nucleus $A$ which decays to a stable nucleus $C$ through the following sequence: $A \to B \to C$. Here $B$ is an intermediate nucleus which is also radioactive. Considering that there are $N_0$ atoms of $A$ initially,plot the graph showing the variation of the number of atoms of $A$ and $B$ versus time.

(N/A) The decay process follows the sequence $A \xrightarrow{\lambda_A} B \xrightarrow{\lambda_B} C$. The number of atoms of $A$ at time $t$ is given by $N_A(t) = N_0 e^{-\lambda_A t}$.
The rate of change of atoms of $B$ is given by $\frac{dN_B}{dt} = \lambda_A N_A - \lambda_B N_B$.
Initially,$N_B = 0$ at $t = 0$. As $A$ decays,$N_B$ increases,reaches a maximum value $(N_B)_{\max}$,and then decreases as $B$ decays into $C$.
At the instant when $N_B = (N_B)_{\max}$,the rate of production of $B$ equals its rate of decay,i.e.,$\lambda_A N_A = \lambda_B N_B$. Before this time,the growth rate of $B$ is greater than its decay rate,and after this time,its decay rate is greater than its growth rate.