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A radioactive sample has 2.6 mu g of pure ""_(7)^(13)"N" which has a half - life of 10 minutes. (a) How many nuclei are present initially? (b) What is the activity initially?(c) What is the activity after 2 hours? (d) Calculate mean life of this sample. |
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Answer» Solution :(a) To find `N_(0)` we have to find the number of `""_(7)^(13)"N"` atoms is `2.6 mu g`. The atomic mass of nitrogen is 13. Therefore, 13 g of `""_(7)^(13)"N"` contains Avogadro number `(6.02 xx 10^(23))` of atoms. In 1 g, the number of `""_(7)^(13)"N"` is equal to be `(6.02 xx 10^(23))/(13)` atoms. So the numberof `""_(7)^(13)"N"` atoms in `2.6 mug` is `N_(0) = (6.02 xx 10^(23))/(13) xx 2.6 xx 10^(-6) = 12.04 xx 10^(6)` atoms. (b) To find the initial activity `R_(0)`, we have to evaluate decay CONSTANT `lambda` `lambda = (0.6931)/(T_(1/2)) = (0.6931)/(10 xx 60) = 1.155 xx 10^(-3) s^(-1)` Therefore `R_(0) = lambda N_(0) = 1.155 xx 10^(-3) xx 12.04 xx 10^(16)` `= 13.90 xx 10^(13) (decays)/(s)` `= 13.90 xx 10^(13)Bq` In terms or a curie, `R_(0) = (13.90 xx 10^(13))/(3.7 xx 10^(10)) = 3.75 xx 10^(3)Ci` since 1Ci = `3.7 xx 10^(10)Bq` (C) Activity after 2 hours can be calculated in TWO different ways: Methid 1 : `R = R_(0)e^(-lambda t)` At t = 2 hr = 7200 s `R = 3.75 xx 10^(3) xx e^(-7200) xx 1.155 xx 10^(-3)` `R = 3.75 xx 10^(3) xx 2.4 xx 10^(-4) = 0.9 Ci` Method 2 : `R = ((1)/(2))^(n) R_(0)` Here `n = (120 min)/(10 min) = 12` ` R = ((1)/(2))^(12) xx 3.75 xx 10^(3) approx 0.9 Ci` (d) mean life `tau = (T_(1/2))/(0.6931) = (10 xx60)/(0.6931) = 865.67 s` |
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