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 radium be N and the amount of radium at any time t be P.

We have,

\[\frac{dP}{dt}\alpha P\]

\[ \Rightarrow \frac{dP}{dt} = - aP\]

\[ \Rightarrow \frac{dP}{P} = - a dt\]

\[ \Rightarrow \log \left| P \right| = - at + C . . . . . \left( 1 \right)\]

Now, P = N at t = 0

Putting P = N and t = 0 in ( 1 ), we get

\[\log \left| N \right| = C\]

\[\text{ Putting C }= \log \left| N \right| \text{ in }\left( 1 \right),\text{ we get }\]

\[\log \left| P \right| = - at + \log \left| N \right|\]

\[ \Rightarrow \log\left| \frac{P}{N} \right| = -\text{ at }. . . . . \left( 2 \right)\]

According to the question,

\[P = \frac{1}{2}\text{ N at t }= 1590\]

\[\log\left| \frac{N}{2N} \right| = - 1590a\]

\[ \Rightarrow a = \frac{1}{1590}\log \left| 2 \right|\]

\[\text{ Putting a }= \frac{1}{1590}\log\left| 2 \right|\text{ in }\left( 2 \right),\text{ we get }\]

\[\log\left| \frac{P}{N} \right| = - \left( \frac{1}{1590}\log\left| 2 \right| \right)t\]

\[\frac{P}{N} = e^{- \frac{\log 2}{1590}t} . . . . . \left( 3 \right)\]

\[\text{ Putting }t = 1 \text{ in }\left( 3 \right)\text{ to find the bacteria after 1 year, we get }\]

\[\frac{P}{N} = e^{- \frac{\log 2}{1590}} \]

\[ \Rightarrow \frac{P}{N} = 0 . 9996\]

\[ \Rightarrow P = 0 . 9996N\]

\[\text{ Percentage of amount disapeared in 1 year }= \left( \frac{N - P}{N} \right) \times 100 % = \frac{N - 0 . 9996N}{N} \times 100 % = 0 . 04 %\]