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Two radioactive elements $R$ and $S$ disintegrate as
$R \rightarrow P + \alpha; \lambda_R = 4.5 × 10^{-3} \,\, years^{-1}$
$S \rightarrow P + \beta; \lambda_S = 3 × 10^{-3} \,\, years^{-1}$
Starting with number of atoms of $R$ and $S$ in the ratio of $2 : 1,$ this ratio after the lapse of three half lives of $R$ will be :
$3:2$
$1:3$
$1:1$
$2:1$
Solution
$\mathrm{N}_{1}=\mathrm{N}_{01} \mathrm{e}^{-\lambda_{2} \mathrm{t}}$
$\mathrm{N}_{2}=\mathrm{N}_{02} \mathrm{e}^{-\lambda_{2} \mathrm{z}}$
$\frac{{{{\rm{N}}_1}}}{{{{\rm{N}}_2}}} = \frac{2}{1} \times {{\rm{e}}^{ – \left( {{\lambda _1} – {\lambda _2}} \right)t}}$
${\lambda _1}t = 3\ln 2$
${\lambda _2}{\rm{t}} = 3\ln 2 \times \frac{3}{{4.5}} = 2\ln 2$
$ \Rightarrow \frac{{{{\rm{N}}_1}}}{{{{\rm{N}}_2}}} = 2 \times {{\rm{e}}^{ – \ln 2}} = 1$
