InterviewSolution
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InterviewSolution
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The decay constant for the radioactive nuclide `64Cu` is `1.516 xx 10^(-5) s^(-1)` . Find the activity of a sample containing `1mug` of `^64Cu`. Atomic weight of copper = `63.5 g (mol)^-1`. Neglect the mass difference between the given radioisotope and normal coper. |
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Answer» `63.5 g `of copper has `6 xx 10^(23)` atoms. Thus, the number of atoms in `1 mg` of `Cu` is `N=(6 xx 10^(23) xx 1 mu g)/(63.5g) =9.45 xx10^(15)` The activity = `lambda N` `=(1.516 xx10^(-5) s^(-1)) xx (9.45 xx 10^(15))` `=1.43 xx 10^(11)` disintegrations `s^(-1)` `=(1.43 xx 10^(11))/(3.7 xx 10^(10)) Ci =3.86 Ci` Activity after n half - lives is `A_(0)//2^(n)` . `T_(avg) =("Sum agees of all the nuclei")/(N_(0))=(underset(0)overset(infty)(int)N_(0)e^(-lambdat)dt)/(N_(0))=(1)/(lambda)`. |
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