### Chemical Kinetics Question 13

#### Question 13 - 01 February - Shift 1

$A$ 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? $\min$.

## Show Answer

#### Answer: (15)

#### Solution:

$[A] _t=[A]_0 e^{-kt}$

For $\mathbf{A}$ : Let $[A] _t$ be $y$ and $[A]_0$ be $x ; k=\frac{\ln 2}{t _{1 / 2}}=$

$\ln 2$

$15 min$

$y=x e^{-k t}$

$=x e^{-(\frac{\ln 2}{15}) t}$

For $\mathbf{B}:[B] _t=[B]_0 e^{-kt}$

Let $[B] _t=y ;[B]_0=4 x ; k=\frac{\ln 2}{t _{1 / 2}}=\frac{\ln 2}{5 \min }$

$y=4 x e^{-(\frac{\ln 2}{5}) t}$

$x e^{-(\frac{\ln 2}{15}) t} \quad 4 x e^{-(\frac{\ln 2}{5}) t}$

$e^{t(\frac{\ln 2}{5}-\frac{\ln 2}{15})}=4$

$t \times[\frac{\ln 2}{5}-\frac{\ln 2}{15}]=\ln 4$

$t \times \ln 2[\frac{1}{5}-\frac{1}{15}]=2 \ln 2$

$t=15 min$