Radium decomposes at a rate proportional to the amount present. If half the orignal amount disappears in…
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 $\left.\log 2=0.6912 \& \mathrm{e}^{-0 \cdot 04320}=0 \cdot 9576\right)$
$3 \cdot 24 \%$
$5 \cdot 24 \%$
$2 \cdot 24 \%$
$4 \cdot 24 \%$
Solution
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
$\therefore \log \frac{R}{R_{0}}=-0.0432 \Rightarrow \frac{R}{R_{0}}=e^{-4043}$
$\therefore \frac{R}{R_{0}}=0.9576 \Rightarrow R=0.9576 R_{0}$
$\% \operatorname{loss}=\frac{R_{0}-0.9576 R_{0}}{R_{0}} \times 100 \%$
$=0.0424 \times 100 \%=4.24 \%$