
If you had a 100-gram sample of plutonium, how much would still remain in 43 years?
Answer
533.1k+ views
Hint: Use the concept of half life and the formula ${{m}_{F}}={{m}_{I}}\times {{2}^{n}}$, where n belongs to the number of periods, which has the formula $n=\dfrac{total\,time}{half\,life}$. If the half life of a reaction is more, it means more time is required for the reaction to get completed, which means that the reaction is a slow one.
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, if we denote “n” as the number of periods, then we have $n=\dfrac{total\,time}{half\,life}=\dfrac{43}{14}=3.07$. If ${{m}_{I}}$ is the initial mass, then after “n” periods, the mass after shall be ${{m}_{F}}$. So, we have the relation between the two as:${{m}_{I}}={{m}_{F}}\times {{2}^{n}}$. Now, on substituting the values in the given equation, we have:
\[
{{m}_{F}}=\dfrac{{{m}_{I}}}{{{2}^{n}}} \\
\Rightarrow {{m}_{F}}=\dfrac{100}{{{2}^{3.07}}}=11.9g \\
\]
So, we come to know that 11.9 grams approximately, of plutonium is to be remained after 43 years, which is the required answer for our question.
Note: 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 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, if we denote “n” as the number of periods, then we have $n=\dfrac{total\,time}{half\,life}=\dfrac{43}{14}=3.07$. If ${{m}_{I}}$ is the initial mass, then after “n” periods, the mass after shall be ${{m}_{F}}$. So, we have the relation between the two as:${{m}_{I}}={{m}_{F}}\times {{2}^{n}}$. Now, on substituting the values in the given equation, we have:
\[
{{m}_{F}}=\dfrac{{{m}_{I}}}{{{2}^{n}}} \\
\Rightarrow {{m}_{F}}=\dfrac{100}{{{2}^{3.07}}}=11.9g \\
\]
So, we come to know that 11.9 grams approximately, of plutonium is to be remained after 43 years, which is the required answer for our question.
Note: 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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

