Radium decomposes at a rate proportional to the amount present. If half the orignal amount disappears in 1600 yrs, then the percentage loss in 100 years is \(\left(\right.\) Given \(\left.\log 2=0.6912 \& \mathrm{e}^{-0 \cdot 04320}=0 \cdot 9576\right)\)

Let \(R=\) Amount of radium present at time \(t\).
We have \(\frac{d R}{d t} \displaystyle\propto R \Rightarrow \frac{d R}{t}=k R \quad \rightarrow \int \frac{d R}{k}=\int {kdt}\)
\(\therefore \log R=kt+C\)...(1)
when \(t=0\), let \(R=R_{0}\) so we get
\(\log R_{0}=0+c \Rightarrow c=\log R\)
\(\log \frac{R}{R_{0}}=kt\)
\((\therefore)\)
When \(t=1600 \mathrm{yrs}, \mathrm{R}=\frac{1}{2} \mathrm{R_{0}}\)
\(\therefore \log \frac{1}{2} \frac{R_{0}}{R_{0}}=1000 \Rightarrow \log \frac{1}{2}-1000 k\) \(\therefore k=\frac{1}{1000} \log \frac{1}{2}-\frac{1}{1000}\left(0-k g_{2}\right)=\frac{-00012}{1000}\)
\(\therefore k=-0.000432t\)
When t \(=100\), we get
\(\therefore \log \frac{R}{R_{0}}=-0.0432 \Rightarrow \frac{R}{R_{0}}=e^{-4043}\)
\(\therefore \frac{R}{R_{0}}=0.9576 \Rightarrow R=0.9576 R_{0}\)
\(\% \operatorname{loss}=\frac{R_{0}-0.9576 R_{0}}{R_{0}} \times 100 \%\)
\(=0.0424 \times 100 \%=4.24 \%\)