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Radioacitive nuclei $A$ and $B$ disintegrate into $C$ with half lives $T$ and $2T$. At $t = 0$, pumber of nuclei of each $A$ and $B$ is $x$. The number of nuclei of $C$ when rate of disintegration of $A$ and $B$ are equal is
$1.5x$
$125x$
$x$
$1.75x$
Solution
$\lambda_{A} \mathrm{N}_{A}=\lambda_{B} \mathrm{N}_{B}$
$ \Rightarrow \left( {\frac{{ln2}}{T}} \right){N_A} = \left( {\frac{{ln2}}{{2T}}} \right){N_B}$
$\Rightarrow \mathrm{N}_{\mathrm{B}}-2 \mathrm{N}_{\mathrm{A}}$
$\Rightarrow \mathrm{t}=2 \mathrm{T} \Rightarrow 2$ half lives of $\mathrm{A}$
$\Rightarrow 1$ half lives of $\mathrm{B}$
$\Rightarrow$ no. of $A$ disintegrated
$=\mathrm{N}_{o}-\frac{\mathrm{N}_{o}}{4}=\frac{3 \mathrm{X}}{4}$
$\Rightarrow$ no. of $\mathrm{B}$ disintegrated
$=\mathrm{N}_{\mathrm{o}}-\frac{\mathrm{N}_{\mathrm{o}}}{2}=\frac{\mathrm{X}}{2}$
Net $\mathrm{C}$ formed $=0.75 \mathrm{x}+0.5 \mathrm{x}$
$=1.25 \mathrm{x}$
