Question

Two radioactive compounds A and B have half lives equal to 40 hrs and 24 hrs respectively. If initial concentrations of A and B are equal then after X hours the concentration of B will be half of the concentration of A. The value of $\frac{X}{10}$ is

Answer 1

Radioactive decay follow 1st order kinetics
$$\ln \left(\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right)=-{\lambda }_{A}t$$ ...(i)
$$\ln \left(\frac{{\left[B\right]}_t}{{\left[B\right]}_0}\right)=-{\lambda }_{B}t$$ ...(ii)
$[A{]}_0=[B{]}_0\text{and}[A{]}_t=2[B{]}_t$
$\left(i\right)–\left(ii\right)$
$\Rightarrow l{n}_2=t({\lambda }_B–{\lambda }_A)=$$$t(\frac{\ln 2}{24}-\frac{\ln 2}{40})$$
$\Rightarrow t=\frac{24\times 40}{16}=60hr=X$