Question

Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?

Answer 1

Let the original amount of the radium be N and the amount of radium at any time t be P.
Given: $\frac{dP}{dt}\alpha P$
$$\begin{gathered} \Rightarrow \frac{dP}{dt}=-aP \\ \Rightarrow \frac{dP}{P}=-adt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \log \left|P\right|=-at+C.....\left(1\right) \\ \mathrm{Now}, \\ P=N\mathrm{at}t=0 \\ \mathrm{Putting}P=N\mathrm{and}t=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|N\right|=C \\ \mathrm{Putting}C=\log \left|N\right|\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|P\right|=-at+\log \left|N\right| \\ \Rightarrow \log \left|\frac{P}{N}\right|=-at.....\left(2\right) \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ P=\frac{1}{2}N\mathrm{a}\mathrm{t}t=1590 \\ \log \left|\frac{N}{2N}\right|=-1590a \\ \Rightarrow -\log 2=-1590a \\ \Rightarrow a=\frac{1}{1590}\log 2 \\ \mathrm{Putting}a=\frac{1}{1590}\log 2\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \log \left|\frac{P}{N}\right|=-\left(\frac{1}{1590}\log 2\right)t \\ \frac{P}{N}=e^{-\frac{\log 2}{1590}t}.....\left(3\right) \\ \mathrm{Putting}t=1\mathrm{in}\left(4\right)\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{bacteria}\mathrm{after}1\mathrm{year},\mathrm{we}\mathrm{get} \\ \frac{P}{N}=0.9996 \\ \Rightarrow P=0.9996N \\ \mathrm{Percentage}\mathrm{of}\mathrm{amount}\mathrm{disapeared}\mathrm{in}1\mathrm{year}=\left(\frac{N-P}{N}\right)\times 100\%=\frac{N-0.9996N}{N}\times 100\%=0.04\% \end{gathered}$$