Question

A radioactive element decays by parallel path as given below:
$A\underset{}{\overset{{\lambda }_1}{\to }}B$ ${\lambda }_1=1.8\times {10}^{-2}{sec}^{-1}$
$2A\underset{}{\overset{{\lambda }_2}{\to }}B$ ${\lambda }_2={10}^{-3}{sec}^{-1}$
Average life of radio-nuclide $A$ will be:

• A. $52.63$ $sec$
• B. $500$ $sec$
• C. $50$ $sec$
• D. $120$ $sec$

Answer 1

The correct option is C $50$ $sec$

$\lambda ={\lambda }_1+2{\lambda }_2$
$=1.8\times {10}^{-2}+2\times {10}^{-3}$
$=2\times {10}^{-2}{sec}^{-1}$
$\tau =\frac{1}{\lambda }=\frac{1}{2\times {10}^{-2}}=50$ $sec$