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:-\[\frac{dP}{dt}\alpha P\]
\[\Rightarrow \frac{dP}{dt} = - aP\]
\[ \Rightarrow \frac{dP}{P} = - adt\]
Integrating both sides, we get
\[ \Rightarrow \log\left| P \right| = - \text{ at }+ C . . . . . \left( 1 \right)\]
Now,
\[P = N\text{ at }t = 0\]
\[\text{ Putting }P = N\text{ and }t = 0\text{ in }\left( 1 \right), \text{ we get }\]
\[\log\left| N \right| = C\]
\[\text{ Putting }C = \log\left| N \right|\text{ in }\left( 1 \right), \text{ we get }\]
\[\log\left| P \right| = - \text{ at }+ \log\left| N \right|\]
\[ \Rightarrow \log\left| \frac{P}{N} \right| = - \text{ at }. . . . . \left( 2 \right)\]
According to the question, 
\[P = \frac{1}{2}N\text{ at }t = 1590\]
\[\log\left| \frac{N}{2N} \right| = - 1590a\]
\[ \Rightarrow - \log 2 = - 1590a\]
\[ \Rightarrow a = \frac{1}{1590}\log 2\]
\[\text{ Putting }a = \frac{1}{1590}\log 2\text{ in }\left( 2 \right), \text{ we get }\]
\[\log\left| \frac{P}{N} \right| = - \left( \frac{1}{1590}\log 2 \right)t \]
\[\frac{P}{N} = e^{- \frac{\log 2}{1590}t} . . . . . . . . \left( 3 \right)\]
\[\text{ Putting }t = 1\text{ in }\left( 4 \right) \text{ to find the bacteria after 1 year, we get }\]
\[\frac{P}{N} = 0 . 9996\]
\[ \Rightarrow P = 0 . 9996N\]
\[\text{Percentage of amount disapeared in 1 year }= \left( \frac{N - P}{N} \right) \times 100\% = \frac{N - 0 . 9996N}{N} \times 100 \% = 0 . 04 \%\]