A radio active substance has half-life of \(h\) days, then its initial decay rate is given by Note that at \(\mathrm{t}=0, \mathrm{M}=\mathrm{m}_{\mathrm{o}}\)

Let M be the mass of substance at time t . Then,
\(\begin{aligned}
& \frac{\mathrm{dM}}{\mathrm{dt}}=-\mathrm{kM}, \text { where } \mathrm{k}\gt0 \\
& \Rightarrow \frac{\mathrm{dM}}{\mathrm{M}}=-\mathrm{kdt}
\end{aligned}\)
Integrating on both sides, we get
\(\log M=-k t+c\)
When \(\mathrm{t}=0, \mathrm{M}=\mathrm{m}_0\)
\(\begin{array}{ll}
\therefore \quad & \log \mathrm{m}_0=0+c \\
& \Rightarrow \mathrm{c}=\log \mathrm{m}_0 \\
\therefore \quad & \log \mathrm{M}=-\mathrm{kt}+\log \mathrm{m}_0 \\
& \Rightarrow \log \frac{M}{\mathrm{~m}_0}=-\mathrm{kt}
\end{array}\)
\(\begin{gathered}
\text { When } \mathrm{t}=\mathrm{h}, \mathrm{M}=\frac{1}{2} \mathrm{~m}_0 \\
\therefore \quad \log \left(\frac{\frac{1}{2} \mathrm{~m}_0}{\mathrm{~m}_0}\right)=-\mathrm{kh} \\
\Rightarrow \log \frac{1}{2}=-\mathrm{kh} \\
\Rightarrow \log 2=\mathrm{kh} \\
\Rightarrow \mathrm{k}=\frac{\log 2}{\mathrm{~h}} ...(i)\\
\mathrm{Initial} \text { decay rate, } \\
\frac{\mathrm{dM}}{\mathrm{dt}}=-\mathrm{km} \mathrm{~m}_0 \\
=\frac{-\mathrm{m}_0}{\mathrm{~h}} \log 2...[From(i)]
\end{gathered}\)