Half life period of `""^(215)At` is `100 mu s`. If a sample initially constains 6 mg of the element, calculate its activity, (i) initially, (ii) after `200 mu s`.

Explanation

Data supplied,
i. No. of atoms initially present, `""_(92)^(235)U N_0 = (6 xx 10^(-3) "gram")/(215//6.023 xx 10^(23))`
`N_0 = (6 xx 6.023 xx 10^(20))/(215) = 1.683 xx 10^(19)`
`T_(1//2) = 100 xx 10^(-6) "sec"`
Decay constant, `lambda = (0.6931)/(T_(1//2)) = (0.6931)/(100 xx 10^(-6)) = 6931 "sec"^(-1)`
`:.` Initial activity `A_0 = lambda N_0 = 6931 xx 1.683 xx 10^(19) = 1.166 xx 10^(23) Bq`
ii. `t = 200 mu s`
Also `t "sec. Activity ", A_t = A_0 e^(-lambda t) = A_0 (1/2)^n = 1.166 xx 10^(23) xx (1/2)^2`
`=1.66 xx 10^(23) xx 1/4 = 0.2915 xx 10^(23)Bq`
In terms of Curies (Ci)
`A_0 (1.166 xx 10^(23))/(3.7 xx 10^(10)) = 3.151 xx 10^(12) Ci`
`A_t = (0.2915 xx 10^(23))/(3.7 xx 10^10) =0.788 xx 10^12 Ci` .