Question

The mean lives of a radioactive substance are $1200\text{yr}$ and $600\text{yr}$ for $\alpha -\text{emission}$ and $\beta -\text{emission}$ respectively. In $56\times n\text{yr}$ three fourth of the sample decays by simultaneous $\alpha -\text{emission}$ and $\beta -\text{emission}$ respectively. The value of $n$ is (integer only)
[Use $\ln \left(4\right)\approx 1.4$]

Answer 1

Given, $T_{\alpha }=1200\text{yr};T_{\beta }=600\text{yr}$

When a substance decays by $\alpha \&\beta -$ emission simultaneously, the average disintegration constant ${\lambda }_{av}$ is given by,

${\lambda }_{av}={\lambda }_{\alpha }+{\lambda }_{\beta }$

where ${\lambda }_{\alpha }=$ disintegration constant for $\alpha -$ emission only

${\lambda }_{\beta }=$ disintegration constant for $\beta -$ emission only

Mean life is given by $T_m=\frac{1}{{\lambda }_{av}}$

$\because {\lambda }_{av}={\lambda }_{\alpha }+{\lambda }_{\beta }$

$\Rightarrow \frac{1}{T_m}=\frac{1}{T_{\alpha }}+\frac{1}{T_{\beta }}=\frac{1}{1200}+\frac{1}{600}$

$\Rightarrow T_m=400\text{yr}$

Time taken to decay $\frac{3}{4}$ of the sample is given by,

$t=\frac{1}{{\lambda }_{av}}\ln \left(\frac{N_0}{N}\right)=T_m\ln \left(\frac{N_0}{N}\right)$

$\Rightarrow t=400\ln \left(\frac{100}{25}\right)=400\ln \left(4\right)$

$\Rightarrow t=400\times 1.4=560\text{yr}$

$\therefore n=10$