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Radiocarbon Dating
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The practical limit to ages that can be determined by radiocarbon dating is about 4100 yr. in a 41000 yr old sample, what percentage of the original carbon-14 remains? Please teach me step by step thx!! 更新: Btw,if anyone can check my answer for the following question... Sodium-24 emits a gamma raythat has an energy of 0.423 MeV.Assuming that the nucleus is initially at rest,determine the speed with which the nucleus recoils. 更新 2: 0.423*10^6eV*1.60*10^-19 =6.768*10^-14 mv=(E/c) (6.768*10^-14)/(398*10-26kg*3.00*10^8m/s) =5.67*10^-3m/s Am i right? Thanks a lot! 更新 3: The mass of the nucleus is 3.98*10^-26kg 更新 4: Sorry, the question should be like this ... not 4100 yrs The practical limit to ages that can be determined by radiocarbon dating is about 41000 yr. in a 41000 yr old sample, what percentage of the original carbon-14 remains? 更新 5: I just asked the teacher. My teacher said i have to find the half-live of carbon-14 to solve it..., But thx for helping me
最佳解答:
(1) The decay equation: dN/dt = -kN where k is the decay constant. Then dN/N = -kdt ∫dN/N = -∫kdt ln N = -kt + C where C is a constant N = N0e-kt where N0 is initial no. of redioactive nucleus So with e-4100k = 1/2, we have k = (ln 2)/4100 Thus, when t = 41000: e-41000k = e-10ln 2 = 2-10 which is the fraction of original carbon-14 remains. In percentage, 9.77% (2) In fact you should use conservation of momentum to solve this question: Momentum of the gamma ray = E/c = 0.423 x 106 x 1.6 x 10-19/(3 x 108) = 2.256 x 10-22 kg m/s So recoiling speed of nucleus = 2.256 x 10-22/(3.98 x 10-26) = 5568 m/s 2009-12-14 08:58:57 補充： Anyway, pls. wait and be patient, will ans ASAP. 2009-12-14 14:37:37 補充： According to the official information, half-life of carbon-14 is 5730 years, so: With e^(-2730k) = 1/2, we have k = (ln 2)/5730 Thus, when t = 41000: e^(-41000k) = e^(-7.16ln 2) = 0.00702 So percentage of original carbon-14 remains is 7.02% 2009-12-15 08:55:45 補充： Ar, yes, 0.00702 should be 0.702%
其他解答:8758B59A7FA1EEA7
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