Question

The mean lives of a radioactive substance are $1620$ and $405$ years for $\alpha$-emission and $\beta$-emission respectively. Find out the time (in years) after which three fourth of a sample will decay, if it is decaying both by $\alpha$-emission and $\beta$-emission simultaneously. (Take ln 2 = 0.693)

Answer 1

$\lambda N={\lambda }_{\alpha }N+{\lambda }_{\beta }N$
${\lambda }_{\alpha }=\frac{1}{T_{\alpha }}=\frac{1}{1620}{\text{yr}}^{-1}$
${\lambda }_{\beta }=\frac{1}{T_{\beta }}=\frac{1}{40}{\text{yr}}^{-1}$
$\frac{1}{T}=\frac{1}{T_{\alpha }}+\frac{1}{T_{\beta }}$
= Total decay constant
$T=\frac{T_{\alpha }{T}_{\beta }}{T_{\alpha }+T_{\beta }}=324$ years
$\frac{N}{N_0}=e^{-\lambda t}$
$t=\frac{1}{\lambda }\ln \frac{N_0}{N}=T\ln \frac{N_0}{N}$
Given that three fourth of the sample decays, ie $N=N_0/4$
$t=324\times 2ln2$
$t=449.06$ years
$t\approx 449$ years