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

Options

A. decay constant will be `2.9 xx 10^(-6) s^(-1)`
B. average life is 3.9 days
C. the number of atoms of 1 mg of `.^(198)Au` is `6 xx 10^(23)`
D. the activity of 1.00 mg of `.^(198)Au` is 238 Ci

Answer

A, B, D

Explanation

The half-life and the decay constant are related as
`T_(1//2) = (1n2)/(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)`
The average-life is `t_(av) = (1)/(lambda) = 3.9` days
The activity is `A = lambda` N. 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 = lambda N = (2.9 xx 10^(-6) s^(-1)) (3.03 xx 10^(18))`
`= 8.8 xx 10^(12)` disintegrations `s^(-1)`
`= (8.8 xx 10^(12))/(3.7 xx 10^(10)) Ci = 238 Ci`