Question

The nuclei of two radioactive isotopes of same substance $A^{236}$ and $A^{234}$ are present in the ratio of $4:1$ in an ore obtained from some other planet. Their half lives are $30\text{min}$ and $60\text{min}$ respectively. Both isotopes are $\text{alpha}$ emitters and the activity of isotope with half life $30\text{min}$ is ${10}^6\text{dps}$. Calculate after how much time their activities will become identical.

• A. $90\text{min}$
• B. $360\text{min}$
• C. $180\text{min}$
• D. $210\text{min}$

Answer 1

The correct option is C $180\text{min}$

Use $1$ for $A^{236}$ and, $2$ for $A^{234}$.

Given: $\frac{(N_1{)}_0}{(N_2{)}_0}=\frac{4}{1}$

$(T_{1/2}{)}_1=30\text{min};(T_{1/2}{)}_2=60\text{min}$
$(R_1{)}_0={10}^6\text{dps}$

As we know that, activity $R=\lambda N$

$\therefore \frac{(R_1{)}_0}{(R_2{)}_0}=\frac{{\lambda }_1(N_1{)}_0}{{\lambda }_2(N_2{)}_0}=\frac{(T_{1/2}{)}_2}{(T_{1/2}{)}_1}{\left(\frac{N_1}{N_2}\right)}_0$

$\Rightarrow \frac{{10}^6}{(R_2{)}_0}=\frac{60}{30}\left(\frac{4}{1}\right)$

$\Rightarrow (R_2{)}_0=\frac{{10}^6}{8}\text{dps}$

Let their activities become equal after time $t$,

$R_1=R_2$

$\Rightarrow (R_1{)}_0(\frac{1}{2}{)}^{t/(T_{1/2}{)}_1}=(R_2{)}_0(\frac{1}{2}{)}^{t/(T_{1/2}{)}_2}$

$\Rightarrow {10}^6(\frac{1}{2}{)}^{t/30}=\frac{{10}^6}{8}(\frac{1}{2}{)}^{t/60}$

$\Rightarrow (\frac{1}{2}{)}^{t/30}=(\frac{1}{2}{)}^{\left(t/60\right)+3}$

$\therefore \frac{t}{30}=\frac{t}{60}+3$

$\Rightarrow t=180\text{min}$

Therefore, option $\left(\text{C}\right)$ will be correct.