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)$