Question

$A$ and $B$ are two substances undergoing radioactive decay in a container. The half life of $A$ is $15\min$ and that of $B$ is $5\min$. If the initial concentration of $B$ is $4$ times that of $A$ and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? ______ min.

Answer 1

Answer: 15

$[A{]}_t=[A{]}_0{e}^{-kt}$
For $A$ : Let $[A{]}_t$ be $y$ and $[A{]}_0$ be $x;k=\frac{\ln 2}{t_{1/2}}=$
$\frac{\ln 2}{15\min }$
$y=x{e}^{-kt}$
$=x{e}^{-\left(\frac{\ln 2}{15}\right)t}$
For B: $[B{]}_t=[B{]}_0{e}^{-kt}$
Let $[B{]}_t=y;[B{]}_0=4x;k=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{5\min }$
$y=4x{e}^{-\left(\frac{\ln 2}{5}\right)t}$
$\Rightarrow x{e}^{-\left(\frac{\ln 2}{15}\right)t}=4x{e}^{-\left(\frac{\ln 2}{5}\right)t}$
$e^{t\left(\frac{\ln 2}{5}\frac{\ln 2}{15}\right)}=4$
$t\times \left[\frac{\ln 2}{5}-\frac{\ln 2}{15}\right]=\ln 4$
$t\times \ln 2\left[\frac{1}{5}-\frac{1}{15}\right]=2\ln 2$
$t=15\min$