The rate of disintegration of a radio active element at time $\mathrm{t}$ is proportional to its mass at…
The rate of disintegration of a radio active element at time $\mathrm{t}$ is proportional to its
mass at that time. Then the time during which the original mass of $1 \cdot 5 \mathrm{gm}$. will
disintegrate into its mass of $0.5 \mathrm{gm}$. is proportional to
$\log 4$
$\log 5$
$\log 3$
$\log 2$
Solution
Let $m$ be the mass of the radioactive element at time $t$.
Then the rate of disintegration is $\frac{\mathrm{dm}}{\mathrm{dt}}$ which is proportional to $\mathrm{m}$.
$\therefore \frac{\mathrm{dm}}{\mathrm{dt}} \propto \mathrm{m} \Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km}$, where $\mathrm{k}>0$ $\therefore \frac{\mathrm{dm}}{\mathrm{m}}=-\mathrm{kdt} \Rightarrow \int \frac{1}{\mathrm{~m}} \mathrm{dm}=-\mathrm{k} \int \mathrm{dt}+\mathrm{c}$
$\therefore \log \mathrm{m}=-\mathrm{kt}+\mathrm{c}$
Initially, i.e. when $\mathrm{t}=0, \mathrm{~m}=1.5$
$\therefore \log (1.5)=-\mathrm{k} \times 0+\mathrm{c} \Rightarrow \mathrm{c}=\log \left(\frac{3}{2}\right)$ $\therefore \log \mathrm{m}=-\mathrm{kt}+\log \left(\frac{3}{2}\right) \Rightarrow \log \mathrm{m}-\log \frac{3}{2}=-\mathrm{kt}$
$\therefore \log \left(\frac{2 m}{3}\right)=-k t$
When $\mathrm{m}=0.5=\frac{1}{2}$, then
$\begin{aligned}
& \log \left(\frac{2 \times \frac{1}{2}}{3}\right)=-\mathrm{kt} \Rightarrow \log \left(\frac{1}{3}\right)=-\mathrm{kt} \Rightarrow-\log 3=-\mathrm{kt} \\
\therefore & t=\frac{1}{k} \log 3
\end{aligned}$
$\therefore$ Thus required time is proportional to $\log 3$.