Question

Rate of decay of a radioactive body is proportional to its mass pressure that time. After a decay of $1$ hour the mass of body is $90$ grams $2$ hours it is $60$ grams find the initial mass of the body.

Answer 1

Suppose initial mass of Radioactive substance is $m_0$.
Rate of decay of radioactive substance at time t is
$\frac{dm}{dt}\propto m$
$\therefore \frac{dm}{dt}=-k\left(m\right)$
(Here k is constant of variance and substance looses it mass hence sign is negative)
$\therefore \frac{dm}{m}=-k$ dt
$\therefore \int \frac{1}{m}dm=-k\int 1dt$
$\therefore logm=-kt+logc$
$\therefore logm-logc=-kt$
$\therefore log\left(\frac{m}{c}\right)=-kt$
$\therefore \frac{m}{x}=e^{-kt}$
$\therefore m=c\cdot e^{kt}$
Now at $t=0$(at the initial time) mass $m=m_0$
$\therefore m_0=c{e}^{-k\left(0\right)}$
$\therefore x=m_0$
$\therefore m=m_0{e}^{-kt}$ .$\left(1\right)$
Now at $t=1$ hour mass m$=90$ gram
$\therefore 90=m_0{e}^{-k\left(0\right)}$ ..$\left(2\right)$
Now at $t=2$ hours mass $m=60$ gram
$\therefore 60=m_0{e}^{-2k}$ .$\left(3\right)$
Taking result $\left(2\right)$ $÷$ result $\left(3\right)$
$\therefore \frac{90}{60}=\frac{m_0{e}^{-k}}{m_0{e}^{-2k}}$
$\therefore \frac{3}{2}=ek$
We find initial mass $m_0$ from equation $\left(2\right)$
$90=m_0{\left(\frac{3}{2}\right)}^{-1}$
$\therefore 90=m_0\left(\frac{2}{3}\right)$
$\frac{270}{2}=m_0$
$\therefore 135=m_0$
$\therefore$ Initial mass of the radioactive substance is $135$ gram.