Question

How can I calculate the rate of decay of a radioactive element?

Answer 1

The rate of decay of a radioactive element can be estimated from the rate constant, and for that, we need to know the half life of the reaction. It can be also calculated if the total time to complete the reaction, as well as the initial concentration of the reactants are given.

Complete step-by-step answer: In order to answer our question, we need to learn about kinetics and half life of a chemical reaction. Now, every reaction takes a certain amount of time to get completed. Moreover, the rates of reaction are different, for different reactions. More the rate of the equation, more is the speed and less is the time taken to complete the reaction. Now, let us come to the half life of a chemical reaction. Half life is defined as the time during which the concentration of the reactants is reduced to half of the initial concentration or it is the time required for the completion of half of the reaction. It is denoted by ${{t}_{1/2}}$. Now, let us calculate the half life for a first order reaction, in this case, decay of an isotope is an example of a first order reaction. Now,

$$k=\dfrac{2.303}{t}\log \dfrac{{{[R]}_{0}}}{[R]} \\\ at\,t={{t}_{1/2}},[R]=\dfrac{{{[R]}_{0}}}{2} \\\ \Rightarrow k=\dfrac{2.303}{{{t}_{1/2}}}\log \dfrac{{{[R]}_{0}}}{\dfrac{{{[R]}_{0}}}{2}} \\\ \Rightarrow k=\dfrac{2.303}{{{t}_{1/2}}}\log 2 \\\ \Rightarrow {{t}_{1/2}}=\dfrac{2.303}{k}\times \log 2=\dfrac{0.693}{k} \\\ \\\$$

Now, the rate of decay of a radioactive decay can be found out by seeing the half life of the reaction. More than half the life of the reaction, means that more time is taken for the disintegration to take place, and this means that the rate of the reaction is very slow.

Note: It is to be noted that in case of radioactive disintegrations, the decay graph is exponentially decreasing. If ${{[R]}_{0}}$ is the initial concentration and $[R]$ is the concentration after time “t”, then we can write the general formula $[R]={{[R]}_{0}}{{e}^{-kt}}$.