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?

Guest Oct 26, 2020

#1**0 **

**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

GA

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GingerAle Oct 26, 2020