Does Carbon-14 Dating Agree w/ Mt. Vesuvius Eruption?

[vorcil]

Homework Statement

Nitrogen in the upper atmostphere is convereted by radiation to carbon 14
the half-life of carbon, $$\tau = 5730$$ years
carbon 14 makes up a known proportion of living plants and animals, after they die, the proportion of carbon 14 decays.

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History records that an eruption on Mount Vesuvius buried the city Pompeli in 79 AD
excavations uncovered bones and these contain 79.17 % of the original carbon 14:

the question:
Does the radiocarbon dating agree with the historical record?------------------------

The Attempt at a Solution

I'm going to assume that the percentage of undecayed mass, 79.17% was measured in 2010
as the question doesn't say when

2010 - 79 = 1931, the number of years passed

$$\frac{1931}{\tau}$$ = $$\frac{1931}{5730}$$ = 0.3369 % $$\tau$$

now that means it should be 100% - 33.69% of the original mass
= 0.663%,

So I'm going to say that the radiocarbon dating doesn't agree with the historical record, and that the bones uncovered had been of people that had died before the eruption on Vesuvius,

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Can someone confirm that I've done this correctly?
thanks

 

[Raskolnikov]

No, sorry but you didn't do this correctly at all.

The way to approach half-life is:

$$x(t) = Ce^{-kt}$$

where C = x(0) and k is some constant you solve for. Well we know that

$$x(5730) = \frac{C}{2} = Ce^{-5730k}$$

So we can easily find k. Now that you've got your full equation, plug in t = 1931 to check if x(t) is about 0.7917*C.

 

[vorcil]

$$x(t) = Ce^{-kt}$$

$$x(5730) = Ce^{-k5730} = \frac{C}{2}$$

$$ln(Ce^{-k5730}) = ln(\frac{C}{2})$$

$$ln(C) + (-k5730) = ln(C) - ln(2)$$

$$-k*5730 = -ln(2)$$

$$k = \frac{ln(2)}{5730}$$ = 0.000120968094

putting into the equation, $$e^{-kt}$$
I get, e^(-0.000120968094 * 1931) =0.79168% != 0.7917% but close enough I reckon

so the radiocarbon dating does match up with the historical record

 

[Raskolnikov]

yep, that's perfect. any other questions?

 

[vorcil]

yep, that's perfect. any other questions?

Nope, Thank you. =]

 

[HallsofIvy]

Since you are talking about a "half life" of 5730 years, that can be done more simply with $x(t)= C(1/2)^{t/5730}$

Solve $x(t)= C(1/2)^{t/5730}= .7917 C$ so you need to solve $(1/2)^{t/5730}= .7917$ for t. If that is the "current year" (whenever the excavations were done) then excavations agree with the rule.