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: \[\frac{dP}{dt}\alpha P\]
\[\Rightarrow \frac{dP}{dt} = - aP\]
\[ \Rightarrow \frac{dP}{P} = - a dt\]
Integrating both sides, we get
\[ \Rightarrow \log \left| P \right| = - at + C . . . . . \left( 1 \right)\]
Now, 
P = N when t = 0 
\[\text{ Putting }P = N\text{ and }t = 0\text{ in }\left( 1 \right), \text{ we get }\]
\[\log \left| N \right| = C\]
\[\text{ Putting }C = \log \left| N \right|\text{ in }\left( 1 \right), \text{ we get }\]
\[\log \left| P \right| = -\text{ at }+ \log \left| N \right|\]
\[ \Rightarrow \log \left| \frac{P}{N} \right| = - \text{ at }. . . . . \left( 2 \right)\]
According to the question,
\[P = \frac{98 . 9}{100}N = 0 . 989N\text{ at }t = 25\]
\[ \therefore \log \left| \frac{0 . 989N}{N} \right| = - 25a\]
\[ \Rightarrow a = - \frac{1}{25}\log \left| 0 . 989 \right|\]
\[\text{ Putting }a = - \frac{1}{25}\log \left| 0 . 989 \right| \text{ in }\left( 2 \right), \text{ we get }\]
\[\log\left| \frac{P}{N} \right| = \left( \frac{1}{25}\log \left| 0 . 989 \right| \right)t\]
To find the time when the radium becomes half of its quantity, we have
\[N = \frac{1}{2}P\]
\[ \therefore \log \left| \frac{N}{\frac{N}{2}} \right| = \left( \frac{1}{25}\log \left| 0 . 989 \right| \right)t\]
\[ \Rightarrow \log \left| 2 \right| = \left( \frac{1}{25}\log \left| 0 . 989 \right| \right)t \]
\[ \Rightarrow t = \frac{25\log 2}{\log 0 . 989} = \frac{25 \times 0 . 6931}{0 . 01106} = 1566 . 68 \approx 1567 \left( \text{ approx . }\right)\]