Determine the mass of `Na^(22)` which has an activity of 5mCi. Half life of `NA^(22)` is 2.6 years. Avogadro number `=6.023xx10^(23)` atoms.

Explanation

Given `A=5xx10^(-3)xx3.7xx10^(10)` dis. `s^(-1)=18.5xx10^(7)` dis. `s^(-1)`.
Half life period `T=2.6xx365xx86400=8.2xx10^(7)s`.
We know that `lambda =(0.693)/(T)`
`=(0.693)/(8.2xx10^(7)s^(-1)`
i.e., `lambda =8.45xx10^(-9)s^(-1)`
but `N=(A)/(lambda)`.
`=(18.5xx10^(7))/(8.45xx10^(-9))` atoms
i.e., `N=2.189xx10^(16)` atoms.
`6.023xx10^(23)` atoms of Na-22 weigh `22xx10^(-3)kg`.
`therefore 2.189xx10^(16)` atoms of Na-22 weigh x kg .
i.e., `x=(2.189xx10^(16)xx22xx10^(-3))/(6.023xx10^(23))`
`x=7.995xx10^(-10)`
`therefore` Mass of `Na^(22)` (Whose activity =5 mCi) is `7.995xx10^(-10)`kg.