InterviewSolution
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InterviewSolution
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In an experiment on two radioactive isotopes of an element (which do not decay into each other), their number of atoms ratio at a given instant was found to be `3`. The rapidly decaying isotope has large mass and an activity of `1.0muCi` initially. The half-lives of the two isotopes are known to be `12` hours and `16` hours. what would be the activity of each isotope and their number of atoms ration after two days? |
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Answer» `((N_(A))/(N_(B)))_(0) =3` `(A_(A))_(0) = 1muCi` `(T_(1//2)) = 12h, (T_(1//2))B = 16` `((A_(A))_(0))/((A_(B))_(0))=(lambda_(A)(N_(A))_(0))/(lambda_(B)(N_(B))_(0))=((T_(1//2))_(B))/((T_(1//2))_(A))((N_(A))/(N_(B)))_(0)` `(1)/(A_(B))_(0) = (16)/(12) xx 3 rArr (A_(B))_(0) = (1)/(4)muCi` `t = 2 days = 2 xx 24 = 48h` where `A:` number of half lives in `2` days `= 48//12 = 4`. `A_(A) = (A_(A))_(0)((1)/(2))^(n) =1((1)/(2))^(4) = (1)/(16)muCi` `B:` number of half lives in `2` days `= 48//16 = 3`. `A_(B) = (A_(B))_(0)((1)/(2))^(n) =(1)/(4)((1)/(2))^(3) = (1)/(32)muCi` `(A_(A))/(A_(B)) = ((T_(1//2))_(B))/((T_(1//2))_(A))(N_(A))/(N_(B))` `(1//16)/(1//32) = (16)/(12)(N_(A))/(N_(B)) = (4)/(3)(N_(A))/(N_(B))` `(N_(A))/(N_(B)) = (3)/(2)` Alternatively `N_(A) = (N_(A))_(0)((1)/(2))^(n_(1)) = (N_(A))_(0)((1)/(2))^((48)/(12)) = (N_(A))_(0)((1)/(2))^(4)` `N_(B) = (N_(B))_(0)((1)/(2))^(n_(2)) = (N_(B))_(0)((1)/(2))^((48)/(16)) = (N_(B))_(0)((1)/(2))^(3)` `(N_(A))/(N_(B)) = ((N_(A))/(N_(B)))_(0) ((1)/(2)) = (3)/(2)` |
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