A radioactive substance, with initial mass \(m_0\), has a half-life of \(\mathrm{h}\) days. Then its initial decay rate is given by

Let \(\mathrm{m}\) be the mass of substance at time \(\mathrm{t}\). Then,
\(\frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km} \text {, where } \mathrm{k}>0 \)
\( \Rightarrow \frac{\mathrm{dm}}{\mathrm{m}}=-\mathrm{kdt}\)
Integrating on both sides, we get
\(\log m=-k t+c\)
When \(\mathrm{t}=0, \mathrm{~m}=\mathrm{m}_0\)
\(\therefore \log \mathrm{m}_0=0+\mathrm{c}\)
\(\Rightarrow \mathrm{c}=\log \mathrm{m}_0\)
\(\therefore \log m=-k t+\log m_0\)
\(\Rightarrow \log \frac{\mathrm{m}}{\mathrm{m}_0}=-\mathrm{kt}\)
When \(\mathrm{t}=\mathrm{h}, \mathrm{m}=\frac{1}{2} \mathrm{~m}_0\)
\(\therefore \log \left(\frac{\frac{1}{2} m_0}{m_0}\right)=-k h\)
\(\Rightarrow \log \frac{1}{2}=-\mathrm{kh}\)
\(\Rightarrow \log 2=\mathrm{kh}\)
\(\Rightarrow \mathrm{k}=\frac{\log 2}{\mathrm{~h}}\) ...(i)
Initial decay rate,
\(\frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km}_0\)
\(=\frac{-\mathrm{m}_0}{\mathrm{~h}} \log 2\) ...[From (i)]