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 the initial amount of radium be N and the amount of radium present at any time t be P.
Given:- \[\frac{dP}{dt}\alpha P\]
\[\Rightarrow \frac{dP}{dt} = - aP,\text{ where }a > 0\]
\[ \Rightarrow \frac{dP}{P} = - adt\]
Integrating both sides, we get
\[ \Rightarrow \log\left| P \right| = -\text{ at }+ C . . . . . \left( 1 \right)\]
Now,
\[P = N\text{ at }t = 0\]
\[\text{ Putting }P = N\text{ and }t = 0\text{ in }\left( 1 \right),\text{ we get }\]
\[\log\left| N \right| = C\]
\[\text{ Putting }C = \log\left| N \right|\text{ in }\left( 1 \right), \text{ we get }\]
\[\log\left| P \right| = - \text{ at }+ \log\left| N \right|\]
\[ \Rightarrow \log\left| \frac{N}{P} \right| =\text{ at }\]
According to the question, 
\[\log\left| \frac{N}{\frac{N}{2}} \right| =\text{ at }\]
\[ \Rightarrow \log\left| 2 \right| = \text{ at }\]
\[ \Rightarrow t = \frac{1}{a}\log\left| 2 \right|\]
Here, a is the constant of proportionality .