Question

A radioactive substance disintegrates at a rate proportional to its mass. When its mass is $10$ mg, the rate of disintegration is $0.051$ mg per day. How long will it take for the mass to be reduced from $10$ mg to $5$ mg? $({\log }_{2}2=0.6931)$.

Answer 1

Let $A$ be amount of mass at the time $t$
$\frac{dA}{dt}\propto A$
$\frac{dA}{dt}=KA$ ........(*)
$\frac{dA}{A}=Kdt$
$\int \frac{dA}{A}=\int Kdt$
$\Rightarrow \log A=Kt+\log c$
$\Rightarrow \log \frac{A}{c}=Kt$
$\Rightarrow \frac{A}{c}=e^{Kt}$
$\Rightarrow A=c{e}^{Kt}$ ......(1)
when $t=0$, $A=10$
(1) $\Rightarrow 10=c{e}^0$
$c=10$
$\therefore$ the solution is $A=10e^{Kt}$.
Given when $A=10$, $\frac{dA}{dt}=-0.051$ ($\therefore$ disintegration)
$\Rightarrow -0.051=10k$ (by*)
$\Rightarrow K=-0.0051$
$\therefore A=10e^{-0.0051t}$
To find $t$ when $A=5$
$\therefore A=10e^{-0.0051t}$
To find $t$ when $A=5$
$\Rightarrow 5=10e^{-0.0051t}$
$\Rightarrow \frac{1}{2}=e^{-0.0051t}$
$\Rightarrow 2=e^{-0.0051t}$
$\Rightarrow \log 2=\log e^{-0.0051t}$
$\Rightarrow 0.0051t=\log 2$
$\Rightarrow t=\frac{\log 2}{0.0051}=\frac{0.6931}{0.0051}=136$ days.