Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is

$\mathop {^{38}S}\limits_{sulpher} \xrightarrow[{ - 2.48\,h}]{{half\,year}}\mathop {^{38}Cl}\limits_{chloride} \xrightarrow[{ - 0.62\,h}]{{half\,year}}\mathop {^{38}Ar}\limits_{Argon} $ 

Assume that we start with $1000$ $^{38}S$ nuclei at time $t = 0$. The number of $^{38} Cl$ is of count zero at $ t=0$ an will again be zero at $t = \infty $. At what value of $t,$ would the number of counts be a maximum ?

Taking sulphur as first element,

$\lambda_{1}=\frac{0.693}{\left(\tau_{1 / 2}\right)_{1}}$

$\therefore \lambda_{1}=\frac{0.693}{2.48}=0.2794 h^{-1}$ $...(1)$

Taking $\mathrm{Cl}$ as second element,

$\lambda_{2}=\frac{0.693}{\left(\tau_{1 / 2}\right)_{2}}$

$\therefore \lambda_{2}=\frac{0.693}{0.62}=1.118 h^{-1}$$.....(2)$

Here decay rate of first element (in magnitude) $\frac{d \mathrm{~N}_{1}}{d t}=\lambda_{1} \mathrm{~N}_{1}$

... $(3)$

Decay rate of second element $=-\lambda_{2} \mathrm{~N}_{2}$

Here first element gets transformed into second element. Now if net time rate of formation of second element is $\frac{d \mathrm{~N}_{2}}{d t}$ then,

$\frac{d \mathrm{~N}_{2}}{d t}=\lambda_{1} \mathrm{~N}_{1}-\lambda_{2} \mathrm{~N}_{2}$$....(4)$

Suppose, for second element, radioactive equilibrium is established at the end of time $t_{e^{*}}$

It means that at the end of this time, time rate of growth of second element becomes equal to its time rate of decay.

Hence at the end of time $t_{e^{\prime}}$ no. of nuclei in the sample of second element would become maximum equal to $\mathrm{N}_{2}$ and this number would remain constant also. Thus, since $\mathrm{N}_{2}=$ maximum,

$\frac{d \mathrm{~N}_{2}}{d t}=0$

$\therefore \lambda_{1} \mathrm{~N}_{1}-\lambda_{2} \mathrm{~N}_{2}=0$

$\therefore \lambda_{1} \mathrm{~N}_{1}=\lambda_{2} \mathrm{~N}_{2}$$....(5)$