Question

The mean lives of a radioactive substance are $1620$yr and $405$yr for $\alpha$-emission and $\beta$-emission, respectively. Find out the time during which three-fourth of a sample will decay, if it is decaying both by $\alpha$-emission and $\beta$-emission simultaneously.

• A. $643$yr
• B. $449$yr
• C. $528$yr
• D. $279$yr

Answer 1

Answer: (B)

$449$yr

$\lambda ={\lambda }_{\alpha }+{\lambda }_{\beta }$
$=\frac{1}{T_{\alpha }}+\frac{1}{T_{\beta }}$ $[\because \lambda =\frac{1}{T}]$
$=\frac{1}{1620}+\frac{1}{405}$ [given, $T_{\alpha }=1620yr$ and $T_{\beta }=405$yr]
$=\frac{5}{1620}y{r}^{-1}$
$\frac{3}{4}$th sample will decay, i.e., remaining $1/4$th,
$N=N_o{\left(\frac{1}{2}\right)}^n$
$\frac{N_o}{4}=N_o{\left(\frac{1}{2}\right)}^n$
$\Rightarrow n=2$
$\therefore t=2T_{1/2}=n\frac{In2}{\lambda }$
$=2\times \frac{0.693}{5/1620}=449$yr.