A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of 20 s and 10 s…
A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of $\mathrm{A}_1$ and 160 g of $\mathrm{A}_2$. The amount of the two in the mixture will become equal after
60 s
80 s
20 s
40 s
Solution
For radioactive material $\mathrm{A}_1$,
$\mathrm{N}_1=\mathrm{N}_{\mathrm{ol}}\left(\frac{1}{2}\right)^{\mathrm{n}}=40\left(\frac{1}{2}\right)^{\mathrm{t}_{1 / 20}}\ldots(i)$
For radioactive material
$\mathrm{N}_2=\mathrm{N}_{\mathrm{o} 2}\left(\frac{1}{2}\right)^{\mathrm{n}_2}=160\left(\frac{1}{2}\right)^{\mathrm{t}_2 / 10}\ldots(ii)$
As, $N_1=N_2 \Rightarrow 40\left(\frac{1}{2}\right)^{t_1 / 20}=160\left(\frac{1}{2}\right)^{t_2 / 10}$
$\Rightarrow \frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{10}-\frac{\mathrm{t}}{20}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{20}}$
$\begin{aligned} & \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{\frac{t}{20}} \\ & \Rightarrow \frac{\mathrm{t}}{20}=2 \therefore \mathrm{t}=40 \mathrm{~s}\end{aligned}$