Question

The average (mean)life of a ration nuclide which decays by parallel path is
$A\frac{{\lambda }_1}{⟶}B,{\lambda }_1=1.8\times {10}^{2}se{c}^{-1}$ $;$ $2A\frac{{\lambda }_2}{⟶}B.{\lambda }_2\times {10}^{-3}se{c}^{-1}$

• A. 52.63 sec
• B. 500 sec
• C. 50 sec
• D. none of these

Answer 1

The correct option is C 50 sec

$\begin{array}{c}\text{Given,}\\ A⟶B{\lambda }_1=1.8\times {10}^{-2}{sec}^{-1}\\ 2A⟶B{\lambda }_2={10}^{-3}{sec}^2\end{array}$

$\begin{array}{c}\text{given, it decay paralally}\\ {\lambda }_{\text{net}}={\lambda }_1+{\lambda }_2\end{array}$

$\begin{array}{c}\text{but in second equation}2\text{mole of}\\ \text{A is decayng so, "}\end{array}$

$\begin{array}{cc}{\lambda }_{\text{net}}& ={\lambda }_1+2{\lambda }_2\\ \lambda & =1.8\times {10}^{-2}+2\times {10}^{-3}\\ \lambda & =0.02\end{array}$

$\begin{array}{ccc}\because \text{Now}& \text{mean}\text{life}& =\frac{1}{\lambda }\\ & =\frac{1}{0.02}& \\ \tau & =50\text{sec}& \end{array}$