Question

The half-life of ¹⁹⁹Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of ¹⁹⁸Au. (b) What will be the activity after 7 days? Take the atomic weight of ¹⁹⁸Au to be 198 g mol⁻¹.

Answer 1

Given:-

Half-life of ¹⁹⁹Au, T_1/2= 2.7 days

Disintegration constant, `lambda = 0.693/T_(1"/"2) = 0.639/(2.7 xx 24 xx 60 xx 60) = 2.97 xx 10^-6  "s"^-1`

Number of atoms left undecayed, N = `(1 xx 10^-6 xx 6.023 xx 10^23)/198`

Now, activity, `A_0 = lambdaN`

`= (1 xx 10^-6 xx 6.023 xx 10^23)/198 xx 2.97 xx 10^-6`

= 0.244 Ci

(b) After 7 days Activity,

`A = A_0e^-"λt"`

Here, `A_0 = 0.244` Ci

`therefore A = 0.244 xx e^(-2.97 xx 10^-6 xx 7 xx 3600 xx 24)`

`= 0.244 xx e^(17962.56 xx 10^-4)`

= 0.040 Ci