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

Activity of `Na^(22) , A= (dN)/( dt) =5 m Ci`
`A= 5 xx 10^(-3) xx 3.7 xx 10^(10)`
`A=1.85 xx 10^(+7)` disintegration per second
Disintegration constant
`lambda = (0.693)/(T) = (0.693)/( 2.6 xx 365 xx 24 xx 60 xx 60)`
`lambda = 8.453 xx 10^(-9) s^(-1)`
Activity, `(d N)/( dt) = lambda N`
`N= (1)/( lambda) (dN)/( dt)`
`N= 2.189 x 10^(16)` atoms
According to Avagadro.s hypothesis
`6.023 xx 10^(23)` atoms of `Na^(22)` weigh 22 gm
`2.189 xx 10^(16)` atoms weigh
`= (22 xx 10^(-3))/( 6.023 xx 10^(23)) xx 2.189 xx 10^(16)`
`=7.996 xx 10^(-10)` kg