Question

A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of $20\text{s}$ and $10\text{s}$ respectively. Initially the mixture has $40\text{g}$ of $A_2$ and $160\text{g}$ of $A_2$. The amount of the two in the mixture will become

• A. $20\text{s}$
• B. $40\text{s}$
• C. $80\text{s}$
• D. $60\text{s}$

Answer 1

The correct option is B $40\text{s}$

Initial amount of $A_1=40\text{g}$
Initial amount of $A_2=160\text{g}$

We know,
$N=N_o{\left(\frac{1}{2}\right)}^{\frac{t}{T}}$

For$A_1,N_{A_1}=160{\left(\frac{1}{2}\right)}^{\frac{t}{20}}$

For$A_2,N_{A_2}=40{\left(\frac{1}{2}\right)}^{\frac{t}{10}}$

When,$N_{A_1}=N_{A_2}$

$\Rightarrow 160{\left(\frac{1}{2}\right)}^{\frac{t}{20}}=40{\left(\frac{1}{2}\right)}^{\frac{t}{10}}$

$\Rightarrow 4={\left(\frac{1}{2}\right)}^{\frac{t}{10}-\frac{t}{20}}$

$\Rightarrow 4={\left(\frac{1}{2}\right)}^{\frac{t}{20}}$

Taking $\log$ on both sides,
$\Rightarrow \log 4=\frac{t}{20}\log \frac{1}{2}$

$\Rightarrow 2\log 2=\frac{t}{20}\log 1-\frac{t}{20}\log 2$

$\Rightarrow t=40\text{s}$

Hence, option $\left(C\right)$ is correct.