Question

Half lives of $\alpha$ and $\beta$ emission of a radioactive material are $30$ years and $60$ years respectively. When material decays giving $\alpha$ and $\beta$ emission simultaneously, then time in which $\frac{7}{8}th$ of the material decays is

• A. $20$ yeras
• B. $60$ yeras
• C. $50$ yeras
• D. $90$ yeras

Answer 1

The correct option is B $60$ yeras

$\begin{array}{c}\text{Given.}\\ \text{Half life}\left(T_{1/2}\right)\text{for}\alpha \text{-decay}=30\text{year}\\ \text{Half life}\left(T_{1/2}\right)\text{for}\beta -decay=60\text{year}\end{array}$

$\begin{array}{c}\text{and also,}\\ \text{decay constant}=\frac{0.693}{T_{1/2}}\end{array}$

${\lambda }_{\alpha }=\frac{0.693}{30}\text{and}{\lambda }_{\beta }=\frac{0.693}{60}$

$\begin{array}{c}\text{Now, for simultaneous decay}\\ {\lambda }_{\text{net}}={\lambda }_1+{\lambda }_2+\dots \\ \text{so ,}{\lambda }_{net}={\lambda }_{\alpha }+{\lambda }_{\beta }\end{array}$

$\begin{array}{c}\text{Given,}\frac{7}{8}\text{th decay in time t}\\ \text{So, left will be}(1-\frac{7}{8})=\frac{1}{8}\end{array}$

$\begin{array}{cc}\text{Now,}{A}_t& =A_0{e}^{-\lambda t}\\ \frac{1}{8}& =1e^{-{\lambda }_{net}t}\\ \frac{1}{8}& =e^{-{\lambda }_{net}t}\\ \ln 8& ={\lambda }_{net}t\end{array}$

${\lambda }_{\text{net}}={\lambda }_{\alpha }+{\lambda }_B$

$\begin{array}{c}\frac{0.693}{30}+\frac{0.693}{60}\\ \therefore 0.693=\ln 2\end{array}$

$\begin{array}{c}{\lambda }_{\text{net}}=\frac{\ln 2\left(3\right)}{60}\\ {\lambda }_{net}=\frac{\ln 2}{20}\end{array}$

$\begin{array}{cc}\text{Now,}& t=\frac{\ln 8}{\ln 2}\times 20\text{year}\\ & =\frac{3\ln 2}{\ln 2}\times 20\\ t& =60\text{year.}\end{array}$