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The mean lives of a radioactive sample are $30$ years and $60$ years for $\alpha$-emission and $\beta $ -emission respectively. If the sample decays both by $\alpha$- emission and $\beta $-emission simultaneously, the time after which, only one-fourth of the sample remain is :- ........... $years$
$10$
$20$
$40$
$45$
Solution
${\lambda _{(\alpha + \beta )}} = {\lambda _\alpha } + {\lambda _\beta }$
$ \Rightarrow \frac{1}{{{T_{\frac{1}{2}(\alpha – \beta )}}}} = \frac{1}{{{T_{\frac{1}{2}(\alpha )}}}} + \frac{1}{{{T_{\frac{1}{2}\left( \beta \right)}}}}$
$\Rightarrow \frac{1}{T_{\frac{1}{2}(x+\beta)}}=\frac{1}{30}+\frac{1}{60}=\frac{1}{20}$
$\therefore {{\rm{T}}_{\frac{1}{2}(\alpha + \beta )}} = 20$ years.
$\therefore $ One-fourth of sample will remain after $2$ half life $=40$ years.