Question
Question: Half-life period of \(_{53}{I^{125}}\) is \(60\) days. Percentage of radioactivity present after \(1...
Half-life period of 53I125 is 60 days. Percentage of radioactivity present after 180 days is?
A.50%
B.75%
C.36%
D.12⋅5%
Solution
The radioactive reaction follows the first order kinetics. Half-life period of a reaction is the actual time required for the reaction to be half completed. Here the half-life of 53I125 is given as 60 days.
Complete answer:
In the question above, in order to find the percentage of radioactivity present after 180 days we need to do the following,
Amount of substance that is left unreacted after one half of life is,
2[A]0, Where [A]0 is initial concentration
Now for example, the amount of substance left after two half-lives is
22[A0]
Similarly, the amount of substance left unreacted after n half-lives is
2n[A0]
About the actual question, now the half-life of 53I125=60 days, after 180 days
Therefore, number of half-lives =60180 which is, total time divided by the half-life of 53I125
Now, after solving it we get, number half-lives as =3
So now, the amount of reactant that is left unreacted after 3 half-lives is,
23[A]0=8[A]0
Now finally, percentage of radioactivity present after 180 days is
%Left=[A]08[A0]×100=81×100=12⋅5%
Hence, the correct option is (D) .
Additional information:
It is important to know that these radioactive reactions follow the first order of kinetics. This is the order of chemical reaction in which the rate of the reaction depends on the concentration of only one reactant, and is proportional to the amount of the reactant.
Note:
Radioactive sources are very useful in the study of living organisms and to detect the diseases, in the food and also in the medical instruments. It is also used in agriculture purposes, carbon dating, geology and space exploration. It is used in the treatment of cancer. Gamma rays are used here as these rays are capable of passing deep inside the body.