Question

A sample of radioactive material decays simultaneously by two processes $A$ and $B$ with half lives $\frac{1}{2}$ h and $\frac{1}{4}$h, respectively. For the 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, find the number of nuclei after 2 h of such decay.

• A. $\frac{N_0}{{\left(2\right)}^8}$
• B. $\frac{N_0}{{\left(2\right)}^4}$
• C. $\frac{N_0}{{\left(2\right)}^6}$
• D. $\frac{N_0}{{\left(2\right)}^5}$

Answer 1

The correct option is A $\frac{N_0}{{\left(2\right)}^8}$

After first half hours, $N=N_0\frac{1}{2}$
For $t=\frac{1}{2}h\mathrm{to}t=1\frac{1}{2}h,1h=\mathrm{four}\mathrm{half}\mathrm{lives}$
Hence, $N=\left(N_0\frac{1}{2}\right){\left(\frac{1}{2}\right]}^4=N_0{\left(\frac{1}{2}\right)}^5$
For $t=1\frac{1}{2}\mathrm{to}t=2h$
[For both $A$ and $B$, $\frac{1}{t_{1/2}}=\frac{1}{t_{1/2}}+\frac{1}{t_{1/4}}=2+4=6\Rightarrow t_{1/2}=\frac{1}{6}$]
$\frac{1}{2}h=$ three half-lives
$\therefore N=\left(N_0{\left(\frac{1}{2}\right)}^5\right]{\left(\frac{1}{2}\right)}^3=N_0{\left(\frac{1}{2}\right)}^8$