Question

The mean lives of radioactive substances are $1620y$ and $405y$ 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. $449y$
• B. $399y$
• C. $549y$
• D. $579y$

Answer 1

Answer: (A)

Let at some instant of time $t$ , number of atoms of the radioactive substance are $N$ . It may decay either by α - emission on by β- 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
$\lambda N={\lambda }_{\alpha }N+{\lambda }_{\beta }N$
∴ $\lambda ={\lambda }_{\alpha }+{\lambda }_{\beta }=\frac{1}{1620}+\frac{1}{405}$
$=\frac{1}{324}{y}^{-1}$
Now, $\frac{N_0}{4}=N{}_{0}e{}^{-\lambda t}$
∴ $-\lambda t=\text{ln}\left(\frac{1}{4}\right)=-1\text{.}386$
or $\left(\frac{1}{324}\right)t=1\text{.}386$
∴ $t=449y$