Question

An isotope has a half life of 90 days. How many days will it take for a 5 mg sample to decay to 1 mg?

Answer 1

Use the formula of half life of a first order reaction, that is ${{t}_{1/2}}=\dfrac{0.693}{k}$, and find the rate constant. Then, place this value of rate constant in the equation $[R]={{[R]}_{0}}{{e}^{-kt}}$

Complete 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 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} \\\ & \\\ \end{aligned}$$ Now, let us find the value of the rate constant from the data given in the question, which is $k=\dfrac{0.693}{90}=7.7\times {{10}^{-3}}$. From the equation, we have: $$\begin{aligned} & [R]={{[R]}_{0}}{{e}^{-kt}} \\\ & \Rightarrow 1=5{{e}^{-7.7\times {{10}^{-3}}t}} \\\ & \Rightarrow -7.7\times {{10}^{-3}}t=\ln 0.2 \\\ & \Rightarrow t=\dfrac{\ln 0.2}{7.7\times {{10}^{-3}}}=209\,days \\\ \end{aligned}$$ So, it will take 209 days for a 5mg sample of isotope, to decay to 1mg. **Note:** In the expression of half life, it does not carry conc. term hence the half life period or half change time for first order reaction does not depend upon initial concentration of the reactants. Similarly, the time required to reduce the concentration of the reactant to any fraction of the initial concentration for the first order reaction is also independent of the initial concentration.