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InterviewSolution
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Determine the average `.^(14)C` activity in decays per minute per gram of natural carbon found in living organisms if the concentration of `.^(14)C` relative to that of `.^(12)C` is `1.4 xx10^(-12)` and half -life of `.^(14)C` is `T_(1//2)=57.30` years. |
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Answer» The decay constant is `lambda =(0.693)/(T_(1//2)) =(0.693)/(5730 xx 5.26 xx10^(5))` As `1` year `=3.01 xx 10^(9)` min and for `1.0 g` of carbon, the number of moles is `n=1//2`, so `N=nN_(A) =((1)/(2))(6.02 xx 10^(23))=5.0 xx10.^(22) "nulei" g^(-1)` . The given concentration factor is `(.^(12)C)/(.^(12)C) =1.4 xx10^(-12)` Thus, the number of `((.^(14)C)/(.^(12)C)) =(5.0 xx10^(22))(1.4 xx10^(-12))=7.0 xx10^(10) .^(14)C "nuclei" g^(-1)` The activity,number of deacys per minute, is `(Delta N)/(Delta t) lambda N=(2.30 xx 10^(-10))(7.0 xx10^(10))` `16 .^(14)C "decays" g.^(-1) min.^(-1)` . |
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