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

(a) The half-life and the decay constant are related as
`t_(1//2)=(ln2)/(lambda)=(0.693)/(lambda) or, lambda=(0.693)/(t_(1//2))=(0.693)/(2.7 "days")`
`=(0.693)/(2.7xx24xx3600s)=2.9xx10^(-6)s^(-1)`
(b) The average-life is `t_(av)=(1)/(lambda)=3.9` days.
(c ) The activity is `A=lambdaN`. Now , 198 g of `.^(198)Au` has `6xx10^(23)` atoms.
The number of atoms in `1.00 mg` of `.^(198)Au` is
`N=6xx10^(23)xx(1.0mg)/(198g)=3.-03xx10^(8)`
Thus, `A=lambdaN= (2.9xx10^(-6)s^(-1))(3.03xx10^(18))`
`8.8xx10^(12) "disintergration"//s=(8.8xx10^(12))/(3.7xx10^(10))Cl=240 Ci`