Question

A sample of radioactive material decays by two processes $\text{A}$ and $\text{B}$ with half life of $1/2\text{hr}$ and $1/4\text{hr}$ respectively. For the first half hour it decays with the process $\text{A}$, next one hour with the process $\text{B}$ and for a further half hour simultaneously with both processes $\text{A}$ and $\text{B}$. If originally there were $N_o$ nuclei, find the number of remaining nuclei after $2\text{hr}$ of these decays.

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

Answer 1

The correct option is A $\frac{N_o}{(2{)}^8}$

We know that,

$N=N_o{\left(\frac{1}{2}\right)}^n$

Where,
$n=$ Number of half life
$N_o=$ Initial number of nuclei

$\left(1\right)$ For the first half hour, nuclei decays by process $\text{A}$ :

Given, $t_{1/2}=1/2\text{hr}$

$\therefore n=\frac{\Delta t}{t_{1/2}}=\frac{1/2}{1/2}=1$

After the first half hour, the number of nuclei present will be,

$N=N_o{\left(\frac{1}{2}\right)}^1=N_o/2$

$\left(2\right)$ For the next one hour, nuclei decays by process $\text{B}$ :

Given, $t_{1/2}=1/4\text{hr}$

$\therefore n=\frac{\Delta t}{t_{1/2}}=\frac{1}{1/4}=4$

After the next one hour, the number of nuclei present will be,

$N=\left(N_o/2\right){\left(\frac{1}{2}\right)}^4=N_o/2^5$

$\left(3\right)$ For the last half hour, nuclei decays simultaneously by both the processes $\text{A}$ and $\text{B}$:

So,

$1/t_{1/2}=1/t_{1/2\left(A\right)}+1/t_{1/2\left(B\right)}=2+4=6$

$\Rightarrow t_{1/2}=1/6$

$\therefore n=\frac{\Delta t}{t_{1/2}}=\frac{1/2}{1/6}=3$

After the last half hour, the number of nuclei present will be,

$N=\left(N_o/2^5\right){\left(\frac{1}{2}\right)}^3=\frac{N_o}{(2{)}^8}$