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 $2h$ of such decay

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

Answer 1

Answer: (C)

$\frac{N_0}{(2{)}^6}$

First we need to find the effective half life when both process happen.
$t_{\frac{1}{2}}\propto \frac{1}{\lambda }$

$t_{\frac{1}{2}}=\frac{k}{\lambda }$

$\lambda =\frac{k}{t_{\frac{1}{2}}}$

Hence for a combination of one and two.
${\lambda }_{eff}={\lambda }_1+{\lambda }_2$

$\frac{1}{t_{\frac{1}{2}effective}}=\frac{1}{t_{\frac{1}{2}1}}+\frac{1}{t_{\frac{1}{2}2}}$

$\frac{1}{t_{\frac{1}{2}}}=2+4=6$

$t_{\frac{1}{2}}=\frac{1}{6}hours$

Hence, the first hour is in first process.
Contains two half lives.
The second hour contains $4$ half lives,
Hence, in the first two hours there are a total of $6$ half lives.
If the number of nuclei initially was $N_0$.
After $2$ hours it would be:
$\Rightarrow \frac{N_0}{2^6}$ (six half lives).