Question

Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of radium to decompose?

Answer 1

Let the original amount of radium be N and the amount of radium at any time t be P.
Given: $\frac{dP}{dt}\alpha P$
$$\begin{gathered} \Rightarrow \frac{dP}{dt}=-aP \\ \Rightarrow \frac{dP}{P}=-adt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \log \left|P\right|=-at+C.....\left(1\right) \\ \mathrm{Now}, \\ P=N\mathrm{when}t=0 \\ \mathrm{Putting}P=N\mathrm{and}t=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|N\right|=C \\ \mathrm{Putting}C=\log \left|N\right|\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|P\right|=-at+\log \left|N\right| \\ \Rightarrow \log \left|\frac{P}{N}\right|=-at.....\left(2\right) \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ P=\frac{98.9}{100}N=0.989N\mathrm{a}\mathrm{t}\mathit{}t=25 \\ \therefore \log \left|\frac{0.989N}{N}\right|=-25a \\ \Rightarrow a=-\frac{1}{25}\log \left|0.989\right| \\ \mathrm{Putting}a=-\frac{1}{25}\log \left|0.989\right|\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \log \left|\frac{P}{N}\right|=\left(\frac{1}{25}\log \left|0.989\right|\right)t \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{time}\mathrm{when}\mathrm{the}\mathrm{radium}\mathrm{becomes}\mathrm{half}\mathrm{of}\mathrm{its}\mathrm{quantity},\mathrm{we}\mathrm{have} \\ N=\frac{1}{2}P \\ \therefore \log \left|\frac{N}{{\displaystyle \raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right|=\left(\frac{1}{25}\log \left|0.989\right|\right)t \\ \Rightarrow \log \left|2\right|=\left(\frac{1}{25}\log \left|0.989\right|\right)t \\ \Rightarrow t=\frac{25\log 2}{\log 0.989}=\frac{25\times 0.6931}{0.01106}=1566.68\approx 1567\left(\mathrm{approx}.\right) \end{gathered}$$