Question

The activity of a radioactive isotope falls to 12.5% in 90 days. Compute the half-life and decay constant of the isotope.

Answer 1

Hint: All nuclear reactions follow first-order kinetics. The half-life of the reaction is equivalent to the half-life of isotope and the decay constant of the reaction is equivalent to the rate constant of the reaction. Use the integrated rate law of first-order kinetics and first find out the decay constant and then find out the half-life of the isotope.

Complete step by step solution:
- Nuclear reactions follow first-order kinetics.
- Let us assume ‘x’ amount of radioactive isotope was present at t=0s. According to the question, the activity of a radioactive isotope falls to 12.5% in 90 days. Therefore, $\dfrac{12.5}{100}\times x=0.125x$ of isotope is left behind after t=90days.
- From the integrated rate law of first order kinetics, we have the formula,
\[\lambda =\dfrac{2.303}{t}{{\log }_{10}}\dfrac{\left[ {{N}_{0}} \right]}{\left[ N \right]}\] where $\lambda $ is the decay constant and t is the time. ${{N}_{0}}$ is the number of radioactive isotopes at t=0 and N is the number of radioactive isotopes at time t.
- Therefore, substituting the values in the above equation we obtain,
\[\lambda =\dfrac{2.303}{90}{{\log }_{10}}\dfrac{x}{0.125x}=0.0231\,\,day{{s}^{-1}}\]
- Now, we know the formula to calculate half-life for a first order reaction. That is,
\[{{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\lambda }\]
\[\therefore {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{0.0231}=30\,days\]

Therefore, the half-life of the radioactive isotope is 30 days and the decay constant is \[0.0231\,\,day{{s}^{-1}}\].

Note: Remember all the nuclear reactions are first-order reactions. So, to calculate the half-life and the decay constant, integrated rate law of first-order kinetics is used.