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A radioactive sample contains 2.2 mg of pure " "_(6)^(11)C which has half-life period of 1224 seconds. Calculate : (i) the number of atoms present initially. (ii) the activity when 5 mug of the sample will be left. |
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Answer» Solution :Here half-life period `T_(1/2) = 1224s` (i) As 11 g of `" "_(6)^(11)C` CONTAINS `N_(A) = 6.023 xx 10^(23)` atoms, hence number of atoms present initially in `2.2 mg = 2.2 xx 10^(-3) g` of SAMPLE `N_(0)=(2.2xx10^(-3)xx6.023xx10^(23))/11=1.2046xx10^(20)` (ii) Number of atoms ACTUALLY present in `5mug =5xx 10^(-6) g` of sample `N=(5xx10^(-6)xx6.023xx10^(23))/11=(5xx6.023)/11xx10^(17)` and decay constant `lambda=0.6931/T_(1/2)=0.6931/1224s^(-1)` `THEREFORE` Activity of the sample `R= lambdaN= 0.6931/1224 xx (5xx6.023)/11 xx 10^(17) = 1.55 xx 10^(14) Bq`. |
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