The decay rate of radium is proportional to the amount present at any time \(t\). If initially 60 gms was present and half life period of radium is 1600 years, then the amount of radium present after 3200 years is

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 \(\mathrm{t}=0, \mathrm{~m}=60 \mathrm{gms}\)
\(\begin{array}{ll}
\therefore \quad & \log 60=-k(0)+c \\
& \Rightarrow \mathrm{c}=\log 60
\end{array}\)
\(\begin{array}{ll}
& \Rightarrow \mathrm{c}=\log 60 \\
\therefore \quad & \log \mathrm{~m}=-\mathrm{kt}+\log 60 \\
\therefore \quad & \text { when } \mathrm{t}=1600, \mathrm{~m}=\frac{60}{2}=30 \mathrm{gms} \\
\therefore \quad & \log 30=-1600 \mathrm{k}+\log 60 \\
& \Rightarrow 1600 \mathrm{k}=\log 2 \\
& \Rightarrow \mathrm{k}=\frac{1}{1600} \log 2
\end{array}\)
Equation (i) becomes
\(\log m=\frac{-1}{1600}(\log 2) t+\log 60\)
When \(t=3200\) years
\(\begin{aligned}
& \log \mathrm{m}=\frac{-1}{1600} \times \log 2 \times 3200+\log 60 \\
& \log \mathrm{~m}=-\log 4+\log 60 \\
& \Rightarrow \log \mathrm{~m}=\log \frac{60}{4} \\
& \Rightarrow \mathrm{~m}=15 \text { grams }
\end{aligned}\)