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

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