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(a) Determine the number of carbon ._6^(14)C atoms present for every gram of carbon ._6^(12)C in a living organism. Find (b) The decay constatnt and (c ) the activity of this sample. |
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Answer» Solution :The total number of carbon `._6^(12)C` atoms in onegram of carbon `._6^(12)C` is equal to the corresponding number od moles times Avogadro's number. Since there is only one `._6^(14)C` atom for every `8.3 xx 10^(11)` atoms of `._6^(12)C`, the number of `._6^(14)C` atoms is equal to the number of `._6^(12)C` atoms divided by `8.3 xx 10^(11)` . The decay constant `lambda` for `._6^(14)C` is `lambda =0.693//T_(1)//2`, Where `T_(1)//2` is the half-life. The activity is equal to the magnitude of `DELTA N//Delta t`, which is the decay constant times the number of `._6^(14)C` atoms present. (a) One gram of carbon `._6^(12)C` (atomic mass` = 12 u`) is equivalent to `1.0//1.2 mol`. Since Avogardo's number is `6.02 xx 10^(23)` atoms `mol^(-1)`and sinceof `._6^(14)C` atom for every `8.3 xx 10^(11)` atoms of `._6^(12)C` the number of `._6^(14)C` atoms FRO every `1.0 g` of carbon `._6^(12)C` is `((1.0)/(2)mol)(6.02 xx 10^(23) (("atoms")/(mol))((1)/(8.3 xx 10^(11)))` `=6.0 xx 1-^(10)` atoms (B) Since the half - life of `_(6)^(14)C` is `7530 ` years `(1.81 xx 10^(11) s)`, the decay constant is `lambda =(0.693)/(T_(1)//2) =(0.693/(1.81 xx10^(11) s)) =-3.83 xx 10^(-12) s^(-1)` (c ) The activity is the magnitude of `Delta N//Delta t` or `lambda N`. Thus, we find activity of `._6^(14)C` for every `1.0 g` of carbon `._6^(12)C` in a living organism`=lambdaN=(3.83xx10^(12)s^(-1))(6.0 xx10^(10)"atoms" )=0.23 Bq`. An organism that lived thousands of years ago presumbly had an activity of about `0.23 Bq` per gram of carbon. When the organism died, the activity began decreasing. From a sample of the remains, the current activity per gram of carbon can be measured and COMPARED to the value of `0.23 Bq` to determine the timethat has transpired since death. |
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