Question

${}_{83}{Bi}^{212}$ decays with parallel radioactive reactions resulting in ${}_{81}{\text{Ti}}^{208}$ and ${}_{84}{\text{Po}}^{212}$ by emitting $\alpha \text{and}\beta$ particles respectively. The $\alpha \text{and}\beta$ decay processes have decay constants ${\lambda }_1\text{and}{\lambda }_2$. Mean lifetime of ${}_{83}{\text{Bi}}^{212}$ is $500$ seconds. If ${}_{81}{\text{Ti}}^{208}$ and ${}_{84}{\text{Po}}^{212}$ are produced in ratio of $9:16$ then

• A. ${\lambda }_1=7.2\times {10}^{-4}{\text{s}}^{-1},{\lambda }_2=12.8\times {10}^{-4}{\text{s}}^{-1}$
• B. ${\lambda }_1=12.8\times {10}^{-4}{\text{s}}^{-1},{\lambda }_2=7.2\times {10}^{-4}{\text{s}}^{-1}$
• C. ${\lambda }_1=5.6\times {10}^{-4}{\text{s}}^{-1},{\lambda }_2=14.4\times {10}^{-4}{\text{s}}^{-1}$
• D. ${\lambda }_1=14.4\times {10}^{-4}{\text{s}}^{-1},{\lambda }_2=5.6\times {10}^{-4}{\text{s}}^{-1}$

Answer 1

The correct option is A ${\lambda }_1=7.2\times {10}^{-4}{\text{s}}^{-1},{\lambda }_2=12.8\times {10}^{-4}{\text{s}}^{-1}$

Since reaction takes place in parallel,
${\lambda }_1+{\lambda }_2=\frac{1}{\tau }=\frac{1}{500}=2\times {10}^{-3}{\text{s}}^{-1}$

Also, given that , $\frac{{\lambda }_1}{{\lambda }_2}=\frac{9}{16}$

${\lambda }_1=\frac{9}{25}\times 2\times {10}^{-3}{\text{s}}^{-1}=7.2\times {10}^{-4}{\text{s}}^{-1}$

${\lambda }_2=\frac{16}{25}\times 2\times {10}^{-3}{\text{s}}^{-1}=12.8\times {10}^{-4}{\text{s}}^{-1}$

Hence, option $\left(a\right)$ is correct.