A sample of wood from an Egyptian mummy case taken from an archaeological site is subjected to radiocarbon dating. The activity due to 14C has a specific activity of 9.41 d/min.g. The activity of a carbon sample of equal mass from fresh wood is 15.3 d/min.g. How old is the case?
+2440 Solution:
\(\dfrac{9.41 \text{ d/min/g}} {15.93 \text{ d/min/g}} = 0.590709 |\text{ Fraction of remaining } ^{14}C\)
\( n^\frac{1}{2} = 0.590709 | \text{ Fraction converted to half-lives (n).}\\ \frac{1}{2} log (n) = log (0.590709) \\ (n) = 0.7594796 \text{ half-lives.}\\ \)
\(\text{Half-life of } ^{14}C = 5730 \text{ years ± 40 years.}\\ (n)*(5730) =\\ (0.7594796) * (5730) = 4351.82 \text{ years ± 30.4 years.}\\\)
The age of the case: 4351.82 years ± 30.4 years.
Additional notes: Definitions: The use of "d/min.g" where “.” implies multiplication, –normally presented as (d/min·g), with a center dot rather than a decimal or period. This presentation is common in many physics texts; however, this is a misnomer, because it implies that the disintegrations per second are multiplied by the mass of the sample, when infact this rate is divided by this mass . The correct form is "d/min/g" or "d/(min·g),” which is the form usually presented in biology texts on the same subject.
GA
--. .-
