Question

The mean lives of radioactive substances are $1620$ are $405$ years for $\alpha -\text{emission}$ and $\beta -\text{emission}$ respectively. Find out the time during which three fourth of a sample will decay if it is decaying both by $\alpha -\text{emission}$ and $\beta -\text{emission}$ simultaneously. Choose the approximate answer. Given $\ln 4=1.386$.

• A. $320\text{Years}$
• B. $449\text{Years}$
• C. $640\text{Years}$
• D. $729\text{Years}$

Answer 1

The correct option is B $449\text{Years}$

When a substance decays $\alpha$ and $\beta$ emission simultaneously,

the effective disintegration constant $\lambda$ is given by

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

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

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

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

$\Rightarrow \lambda =\frac{1}{T_{\alpha }}+\frac{1}{T_{\beta }}$

$\Rightarrow \lambda =\frac{1}{1620}+\frac{1}{405}=\frac{1}{324}$

By law of radioactivity,

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

Given that, ${\left(\frac{3}{4}\right)}^{th}$ of the sample is decayed, then the active sample present

$N=\frac{N_0}{4}$

$\Rightarrow t=\frac{\ln 4}{\lambda }=324\left(1.386\right)=449\text{years}$

Hence, option $\left(B\right)$ is correct

Why this Question? Mean or Average Life $\left(T_a\right)$: It is the average of age of all active nuclei, i.e. $T_a=\frac{\text{sum of times of existence of all nuclei in a sample}}{\text{initial number of active nuclei in that sample}}$ $T_a=\frac{1}{\lambda }$