
The half-life of radium is 1622 years. How long will it take for seventh-eighth of a given amount of radium to decay
A. 3244 years
B. 6488 years
C. 4866 years
D. 811 years
Answer
583.2k+ views
Hint: Radioactive Decay of any radioactive substance follows first-order chemical kinetics. The amount of time for the radioactive substance to get halved remains constant. The amount of Radium remaining can be found as \[1-\dfrac{7}{8}=\dfrac{1}{8}\]. And the number of nuclei after n half lives is given by the formula given below.
Formulae Used:
\[N=\dfrac{1}{{{2}^{n}}}{{N}_{o}}\]
Complete answer:
Radioactive Decay of any radioactive substance follows first-order chemical kinetics. This means that the amount of time for the radioactive substance to get halved remains constant. The half time of radium is given to be 1622 years.
And number of nuclei after n half-lives is given by the formula
\[N=\dfrac{1}{{{2}^{n}}}{{N}_{o}}\]……(1)
Now, Lets first find the amount of radium remaining when seventh-eighth of a given amount of radium decays
\[1-\dfrac{7}{8}=\dfrac{1}{8}\]
Let the initial amount of radium taken be \[{{N}_{o}}\]. Then remaining radium can be written as
\[N=\dfrac{1}{8}{{N}_{o}}\]
\[\Rightarrow N=\dfrac{1}{{{2}^{3}}}{{N}_{o}}\]….(2)
Comparing (2) to (1) we get
\[\begin{align}
& \dfrac{1}{{{2}^{3}}}{{N}_{o}}=\dfrac{1}{{{2}^{n}}}{{N}_{o}} \\
& \Rightarrow n=3 \\
\end{align}\]
Thus, the time it takes for seventh-eighth of a given amount of radium to is three half-lives. So time taken will be
\[3\times 1622years=4866\text{ }years\]
Therefore, the time taken for seventh-eighth of a given amount of radium to decay will be 4866 years.
Option C is correct.
Note:
In a radioactive decay whenever the remaining radioactive substance is of the form \[\dfrac{1}{{{\left( 2 \right)}^{n}}}\] of the original amount of radioactive material taken. Then the time taken would be \[n\times {{t}_{\dfrac{1}{2}}}\], where \[{{t}_{\dfrac{1}{2}}}\] is the half-life of the radioactive substance.
Formulae Used:
\[N=\dfrac{1}{{{2}^{n}}}{{N}_{o}}\]
Complete answer:
Radioactive Decay of any radioactive substance follows first-order chemical kinetics. This means that the amount of time for the radioactive substance to get halved remains constant. The half time of radium is given to be 1622 years.
And number of nuclei after n half-lives is given by the formula
\[N=\dfrac{1}{{{2}^{n}}}{{N}_{o}}\]……(1)
Now, Lets first find the amount of radium remaining when seventh-eighth of a given amount of radium decays
\[1-\dfrac{7}{8}=\dfrac{1}{8}\]
Let the initial amount of radium taken be \[{{N}_{o}}\]. Then remaining radium can be written as
\[N=\dfrac{1}{8}{{N}_{o}}\]
\[\Rightarrow N=\dfrac{1}{{{2}^{3}}}{{N}_{o}}\]….(2)
Comparing (2) to (1) we get
\[\begin{align}
& \dfrac{1}{{{2}^{3}}}{{N}_{o}}=\dfrac{1}{{{2}^{n}}}{{N}_{o}} \\
& \Rightarrow n=3 \\
\end{align}\]
Thus, the time it takes for seventh-eighth of a given amount of radium to is three half-lives. So time taken will be
\[3\times 1622years=4866\text{ }years\]
Therefore, the time taken for seventh-eighth of a given amount of radium to decay will be 4866 years.
Option C is correct.
Note:
In a radioactive decay whenever the remaining radioactive substance is of the form \[\dfrac{1}{{{\left( 2 \right)}^{n}}}\] of the original amount of radioactive material taken. Then the time taken would be \[n\times {{t}_{\dfrac{1}{2}}}\], where \[{{t}_{\dfrac{1}{2}}}\] is the half-life of the radioactive substance.
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