
The half-life of radium is 1600 years. The fraction of a sample of radium that would remain after 6400 years
(A) \[1/4\]
(B) \[1/2\]
(C) \[1/8\]
(D) \[1/16\]
Answer
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Hint:Half-life is the time required for a quantity to reduce to half of its initial value. Half life is used in nuclear physics to describe how quickly unstable atoms undergo, or how long stable atoms survive radioactive decay. It is also used to characterize any type of exponential or non-exponential decay. Now that we’ve discussed it, let’s move to the solution of the problem at hand.
Formula Used:\[n=\dfrac{T}{{{t}_{1/2}}}\]
Step by Step Solution:
We all know that radioactive decay is a first-order reaction.
We have been given the time for which the reaction occurs and the half-life of the radioactive sample.
The time for which the reaction is taking place \[(T)=6400years\]
Half-life of the radioactive sample \[({{t}_{1/2}})=1600years\]
Now, we can calculate the number of half-lives that were completed in the total time given to us
No. of half-lives \[n=\dfrac{T}{{{t}_{1/2}}}\] where the meanings of the symbols have been discussed above
Substituting the values, we get, no. of half-lives
\[\begin{align}
& n=\dfrac{6400}{1600} \\
& \Rightarrow n=4 \\
\end{align}\]
For a first order reaction, after every half-life, the fraction of elements that remain is \[\dfrac{1}{2}\]
For n half-lives, the fraction of elements remaining at the end will be \[\dfrac{1}{{{2}^{n}}}\]
Therefore, for 4 half-lives, the fraction of elements remaining would be equal to \[\dfrac{1}{{{2}^{4}}}\]
Hence we can say that at the end of the time period provided to us, that is \[6400years\] , \[\dfrac{1}{16}\] of the initial element would be remaining.
Therefore, option (D) is the correct answer.
Note: Students often make the error of finding the number of half-lives and considering that as the final answer. For example, if the no. of half-lives calculated is four, it does not mean that the fraction of remaining elements would be one-fourth. This is the wrong approach. After every half-life, the remaining elements become half of their previous value such that the decrease is in the form of a geometric progression.
Formula Used:\[n=\dfrac{T}{{{t}_{1/2}}}\]
Step by Step Solution:
We all know that radioactive decay is a first-order reaction.
We have been given the time for which the reaction occurs and the half-life of the radioactive sample.
The time for which the reaction is taking place \[(T)=6400years\]
Half-life of the radioactive sample \[({{t}_{1/2}})=1600years\]
Now, we can calculate the number of half-lives that were completed in the total time given to us
No. of half-lives \[n=\dfrac{T}{{{t}_{1/2}}}\] where the meanings of the symbols have been discussed above
Substituting the values, we get, no. of half-lives
\[\begin{align}
& n=\dfrac{6400}{1600} \\
& \Rightarrow n=4 \\
\end{align}\]
For a first order reaction, after every half-life, the fraction of elements that remain is \[\dfrac{1}{2}\]
For n half-lives, the fraction of elements remaining at the end will be \[\dfrac{1}{{{2}^{n}}}\]
Therefore, for 4 half-lives, the fraction of elements remaining would be equal to \[\dfrac{1}{{{2}^{4}}}\]
Hence we can say that at the end of the time period provided to us, that is \[6400years\] , \[\dfrac{1}{16}\] of the initial element would be remaining.
Therefore, option (D) is the correct answer.
Note: Students often make the error of finding the number of half-lives and considering that as the final answer. For example, if the no. of half-lives calculated is four, it does not mean that the fraction of remaining elements would be one-fourth. This is the wrong approach. After every half-life, the remaining elements become half of their previous value such that the decrease is in the form of a geometric progression.
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