Saved Bookmarks
| 1. |
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. |
|
Answer» Solution :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` . |
|