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

(Round off to nearest integer)

Let at some instant of time t, number of atoms of the radioactive substance as N. It may decay either by  $\alpha -$emission or by  $\alpha -$emission. So, we can write, ${\left(\frac{-dN}{dt}\right)}_{net}={\left(\frac{-dN}{dt}\right)}_{\alpha }+{\left(\frac{-dN}{dt}\right)}_{\beta }$

If the effective decay constant is $\lambda ,$ then $\lambda N={\lambda }_{\alpha }N+{\lambda }_{B}N$ or $\lambda ={\lambda }_{\alpha }+{\lambda }_{\beta }=\frac{1}{1620}+\frac{1}{405}=\frac{1}{324}yea{r}^{-1}$

Now, $\frac{N_0}{4}=N_0{e}^{-\lambda t}$

$\therefore -\lambda t=\ln \left(\frac{1}{4}\right)=-1.386 or \left(\frac{1}{324}\right)t=1.386$

$t=449yr$