a) a sample is being dated by the radiocarbon technique. If the sample was conta
ID: 2019139 • Letter: A
Question
a) a sample is being dated by the radiocarbon technique. If the sample was contaminated its activity would be 0.011Bq per gram of carbon. Given that ^(14)C activity of living organic material is 0.23Bq perm gram of carbon and the half life of ^(14)C is 5730 years find the true age (in years) of the sampleb) Suppose the sample is contaminated so that only 98% of its carbon is ancient carbon. The remaining 2%is fresh carbon, in the sence that its contents have not had any time to decay. Assuming that the lab technition is unaware of the contamination what apparent age (in years) would be determined for the sample?
Explanation / Answer
Given: (a) Intial activity = Ao = 0.23 Bq / gm Activity after time t = At = 0.011 Bq / gm = disntegration constant = 0.693 / T1/2 = 0.693 / 5730 = 1.20 x10-4 It is known by the formula At = Ao e -/t e -/t = (At/ Ao) = 0.0478 log ( e -/t ) = log(0.0478) -t = -1.32 t = 1.32 / 1.20x10-4 ˜ 11004.7 years (a) Intial activity = Ao = 0.23 Bq / gm Activity after time t = At = 0.011 Bq / gm = disntegration constant = 0.693 / T1/2 = 0.693 / 5730 = 1.20 x10-4 It is known by the formula At = Ao e -/t e -/t = (At/ Ao) = 0.0478 log ( e -/t ) = log(0.0478) -t = -1.32 t = 1.32 / 1.20x10-4 ˜ 11004.7 years (b) second case 98 % of the carbon is ancient => that 2% is disintegrated so, by the formual N = N0 e-t where : N= Number of mols left after the time t. N 0 = Initial number. Given: N / No = % of carbon present = 98 % 0.98 = e-t - t = -0.00877 t = 73.116 years t = 73.116 yearsRelated Questions
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