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# Pls help :(

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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?

Oct 26, 2020

#1
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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

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Oct 26, 2020
#2
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9.41 = 15.3 * 1/2^(t/5730)

9.41 / 15.3 = 1/2^(t / 5730)

Log(0.6150326797) =(t / 5730) * Log(1/2)

-0.4860798748  = (t / 5730) *  -0.69314718056

- 2,785.237682604 = -0.69314718056t

t = 4018.25 years + or - 28.3 years.

Oct 26, 2020