
If actinium 227 has a half life of 21.77 years, how many years have gone by if actinium-227 undergoes 3 half-lives?
Answer
474.6k+ views
Hint: If one half life takes 21.77 years in order to decay the actinium isotope, then for 3 half lives to complete, the time taken should be thrice.
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 ${{t}_{1/2}}$. For example, the half life of a first order reaction can be found out in the following way:
\[\begin{align}
& k=\dfrac{2.303}{t}\log \dfrac{{{[R]}_{0}}}{[R]} \\
& at\,t={{t}_{1/2}},[R]=\dfrac{{{[R]}_{0}}}{2} \\
& \Rightarrow k=\dfrac{2.303}{{{t}_{1/2}}}\log \dfrac{{{[R]}_{0}}}{\dfrac{{{[R]}_{0}}}{2}} \\
& \Rightarrow k=\dfrac{2.303}{{{t}_{1/2}}}\log 2 \\
& \Rightarrow {{t}_{1/2}}=\dfrac{2.303}{k}\times \log 2=\dfrac{0.693}{k} \\
& \\
\end{align}\]
In case of radioactive decay, in this case, the decay of actinium, it is considered a first order reaction. So following is the reaction sequence of the decay of actinium isotope:
\[[Ac]\xrightarrow{21.77\,years}\dfrac{[Ac]}{2}\xrightarrow{21.77\,years}\dfrac{[Ac]}{4}\xrightarrow{21.77\,years}\dfrac{[Ac]}{8}\]
There are 3 half lives which are taken, and after 3 half lives, we can see that the concentration of the actinium isotope has become one- eighth of the original amount. So, the years that have gone by in order to finish 3 half lives is clearly $(21.77+21.77+21.77)=65.31\,years$, which is the required answer to our question.
Note:
It is to be noted that in the expression of half life, 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.
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 ${{t}_{1/2}}$. For example, the half life of a first order reaction can be found out in the following way:
\[\begin{align}
& k=\dfrac{2.303}{t}\log \dfrac{{{[R]}_{0}}}{[R]} \\
& at\,t={{t}_{1/2}},[R]=\dfrac{{{[R]}_{0}}}{2} \\
& \Rightarrow k=\dfrac{2.303}{{{t}_{1/2}}}\log \dfrac{{{[R]}_{0}}}{\dfrac{{{[R]}_{0}}}{2}} \\
& \Rightarrow k=\dfrac{2.303}{{{t}_{1/2}}}\log 2 \\
& \Rightarrow {{t}_{1/2}}=\dfrac{2.303}{k}\times \log 2=\dfrac{0.693}{k} \\
& \\
\end{align}\]
In case of radioactive decay, in this case, the decay of actinium, it is considered a first order reaction. So following is the reaction sequence of the decay of actinium isotope:
\[[Ac]\xrightarrow{21.77\,years}\dfrac{[Ac]}{2}\xrightarrow{21.77\,years}\dfrac{[Ac]}{4}\xrightarrow{21.77\,years}\dfrac{[Ac]}{8}\]
There are 3 half lives which are taken, and after 3 half lives, we can see that the concentration of the actinium isotope has become one- eighth of the original amount. So, the years that have gone by in order to finish 3 half lives is clearly $(21.77+21.77+21.77)=65.31\,years$, which is the required answer to our question.
Note:
It is to be noted that in the expression of half life, 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.
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