Obtain the amount of `.^60Co` necessary to provide a radioactive source of `8.0 Ci` strength. The half-life of `.^60Co` is `5.3` years?

Explanation

Given, half-life, `T_(1//2)=5.3` years `=5.3 xx 365 xx 24 xx 3,600`s
The rate of decay of radioactive material is given by
`(dN)/(dt) =-lambdaN`
where `lambda` is the decay constant and
N is the number of atoms
So, `N =(-T_(1//2))/0.693 (dN)/(dt)`
Here, the decay rate `(dN)/(dt) = 8.0 mCi`
`=8 xx 3.7 xx 10^(7)` disintegrations/s
`therefore N=(5.4 xx 365 xx 24 xx 3,600 xx 8 xx 3.7 xx 10^(7))/0.693`
`=7.14 xx 10^(16)` atoms.
Now, 60g of Co contains `6.023 xx 10^(23)` atoms, i.e.
Mass of `6.023 xx 10^(23)` atoms of Co = 60g ,
`therefore` Mass of `7.14 xx 10^(16)` atoms of Co `=(60 xx 7.14 xx 10^(16))/(6.023 xx 10^(23))g`
`=7.113 xx 10^(-6)` g
This will be amount of `""_(27)Co^(60)` necessary to obtain a source of required strength.