The rate of disintegration of a radio active element at time $t$ is proportional to its mass at that time.…
The rate of disintegration of a radio active element at time $t$ is proportional to its mass at that time. Then the time during which the original mass of $6 \mathrm{gm}$ will disintegrate into its mass of $3 \mathrm{gm}$ is proportional to
$\log 4$
$\log 3$
$\log 5$
$\log 2$
Solution
$\begin{aligned} & \frac{d(m)}{d t} \propto m \\ & \Rightarrow \frac{d m}{d t}=-k m \\ & \int \frac{d m}{m}=-\int k d t \\ & \Rightarrow \log _e m=-k t+c \\ & \Rightarrow m=e^{-k t+c}=e^c \cdot e^{-k t} \\ & \text { at } t=0, m=6 \Rightarrow e^c=6 \\ & \text { i.e., } m=6 \cdot e^{-k t}\end{aligned}$
at $t=t, m=3$
Hence $3=6 \cdot e^{-k t}$
$\begin{aligned} & \Rightarrow \frac{1}{2}=e^{-k t} \\ & \Rightarrow \log \left(\frac{1}{2}\right)=-k t \\ & \Rightarrow-\log 2=-k t \\ & \left.\Rightarrow t=\frac{\log 2}{k} \text { [Propotional to } \log 2\right]\end{aligned}$