- Home
- Standard 12
- Physics
A mixture consists of two radioactive material $A_1$ and $A_2$ with half lives of $20\,s$ and $10\,s$ respectively . Initially the mixture has $40\,g$ of $A_1$ and $160\,g$ of $A_2$ . The amount of the two in the mixture will become equal after..........$sec$
$60 $
$80 $
$20 $
$40$
Solution
Let after $t\,s$ amount of the $A_{1}$ and $A_{2}$ will become equal in the mixture.
As $N=N_{0}\left(\frac{1}{2}\right)^{n}$
where $n$ is the number of half-lives
For $A_{1}, N_{1}=N_{01}\left(\frac{1}{2}\right)^{t / 20}$
For $A_{2}, N_{2}=N_{02}\left(\frac{1}{2}\right)^{t / 10}$
According to question, $N_{1}=N_{2}$
${\frac{40}{2^{t / 20}}=\frac{160}{2^{t / 10}}}$
${2^{t/10}} = 4\left( {{2^{t/20}}} \right)$ or ${2^{t/10}} = {2^2}{2^{t/20}}$ $ \Rightarrow {2^{t/10}} = {2^{\left( {\frac{t}{{20}} + 2} \right)}}$
${\frac{t}{10}=\frac{t}{20}+2 \text { or } \frac{t}{10}-\frac{t}{20}=2}$
or $\quad \frac{t}{20}=2 \quad$ or $\quad t=40 \mathrm{\,s}$
