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.
We have,
\[\frac{dP}{dt}\alpha P\]
\[ \Rightarrow \frac{dP}{dt} = aP,\text{ where a }< 0\]
\[ \Rightarrow \frac{dP}{P} = adt\]
\[ \Rightarrow \log \left| P \right| = at + C . . . . . \left( 1 \right)\]
Now, P = N at t = 0
Putting P = N and t = 0 in (1), we get
\[\log \left| N \right| = C \]
\[\text{ Putting C }= \log\left| N \right|\text{ in }\left( 1 \right),\text{ we get }\]
\[\log \left| P \right| = at + \log \left| N \right|\]
\[ \Rightarrow \log \left| \frac{P}{N} \right| = at . . . . . \left( 2 \right)\]
According to the question,
\[\log\left| \frac{2N}{N} \right| = \text{at}\]
\[ \Rightarrow \log\left| 2 \right| = \text{at}\]
\[ \Rightarrow t = \frac{1}{a}\log\left| 2 \right|,\text{ where a is a constant of proportionality}\]