Question

A sample of radioactive material decays simultaneously by two processes A and B with half live 1/2 and 1/4 hr. respectively. For first half hour it decays with the process A, next one hour with the process B and for further half an hour with both A and B. If originally there were $N_0$ nuclei, the number of nuclei after 2 hours of such decay is $N_0{\left(\frac{1}{2}\right)}^x$ then find the value of $x$

• A. 5
• B. 4
• C. 3
• D. 8

Answer 1

The correct option is D 8

Half life when only process A occurs = $t_1$ = $\frac{1}{2}$ hr.

Half life when only process B occurs = $t_2$ = $\frac{1}{4}$ hr.

Half life when both process A and B occur simultaneously = $\frac{1}{t_1}+\frac{1}{t_2}$

= $\frac{1}{6}$ hr.

Now , number of particles at the end of $n$ half lives = $N_0(\frac{1}{2}{)}^n$

Therefore, number of particles after half hour = $N_0(\frac{1}{2}{)}^1$

Number of particles left after further one hour = $N_0\left(\frac{1}{2}{)}^1\right(\frac{1}{2}{)}^4$

Number of particles left after decay with both process A and B = $N_0\left({\frac{1}{2}}^1\right)\left({\frac{1}{2}}^4\right)\left({\frac{1}{2}}^3\right)$

= $N_0(\frac{1}{2}{)}^8$

Therefore , $x=8$.