Question

The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.

Answer 1

Let N be the initial amount of radium and P be the amount of radium present at any time t.
$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dP}{dt}\alpha P \\ \Rightarrow \frac{dP}{dt}=aP,\mathrm{where}a<0 \\ \Rightarrow \frac{dP}{P}=adt \\ \Rightarrow \log \left|P\right|=at+C.....\left(1\right) \\ \mathrm{Now},P=N\mathrm{at}t=0 \\ \mathrm{Putting}P=N\mathrm{and}t=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|N\right|=C\mathit{} \\ \mathrm{Putting}C=\log \left|N\right|\mathit{}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|P\right|=at+\log \left|N\right| \\ \Rightarrow \log \left|\frac{P}{N}\right|=at.....\left(2\right) \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \log \left|\frac{2N}{N}\right|=at \\ \Rightarrow \log \left|2\right|=at \\ \Rightarrow t=\frac{1}{a}\log \left|2\right|,\mathrm{where}a\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{of}\mathrm{proportionality} \end{gathered}$$