The half-life of ` .^198Au` is `2.7 days`. Calculate (a) the decay constant, (b) the average-life and (c) the activity of `1.00 mg` of `.^198Au`. Take atomic weight of `.^198Au` to be `198 g mol^(-1)`.

Explanation

The half life and the decay constant are related as
`t_(1/2) = (1n 2)/(lambda) = (0.693)/(lambda)`
or, `(lambda) = (0.693)/(t_(1/2) = (0.693)/(2.7 days)`
`= (0.693)/(2.7 xx 24 xx 3600 s) = 2.9 xx 10^(-6 s^(-1)`
(b) The average lif is `t_(av) = 1/(lambda) = 3.9 days.
(c) The activity is `A= lambdaN`. Now, `198 g` of `^198 Au` has `6 xx 10^23` atoms. The number of atoms in `1.00 mg` of `^198 Au` is`
`N= 6 xx 10^23 xx (1.0 mg)/(198 g) = 3.03 xx 10^18`.
Thus, `A = lambdaN`
`= (2.9 xx 10^(-6) s^(-1) (3.03 xx 10^(18)`
`= 8.8 xx 10^(12)` disintegration `s^(-1)`
`(8.8 xx 10^(12))/(3.7 xx 10^(10)) Ci = 240 Ci`.