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Homework Help: Radioactive decay

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

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


    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?


    3. 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 dosen't say when

    2010 - 79 = 1931, the number of years passed

    [tex] \frac{1931}{\tau} [/tex] = [tex] \frac{1931}{5730} [/tex] = 0.3369 % [tex] \tau [/tex]

    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 dosen't agree with the historical record, and that the bones uncovered had been of people that had died before the eruption on Vesuvius,


    Can someone confirm that i've done this correctly?
  2. jcsd
  3. Aug 4, 2010 #2
    No, sorry but you didn't do this correctly at all.

    The way to approach half-life is:

    [tex] x(t) = Ce^{-kt} [/tex]

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

    [tex] x(5730) = \frac{C}{2} = Ce^{-5730k} [/tex]

    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.
  4. Aug 4, 2010 #3
    [tex] x(t) = Ce^{-kt} [/tex]

    [tex] x(5730) = Ce^{-k5730} = \frac{C}{2}[/tex]

    [tex] ln(Ce^{-k5730}) = ln(\frac{C}{2}) [/tex]

    [tex] ln(C) + (-k5730) = ln(C) - ln(2) [/tex]

    [tex] -k*5730 = -ln(2) [/tex]

    [tex] k = \frac{ln(2)}{5730} [/tex] = 0.000120968094

    putting into the equation, [tex] e^{-kt} [/tex]
    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
  5. Aug 4, 2010 #4
    yep, that's perfect. any other questions?
  6. Aug 4, 2010 #5
    Nope, Thank you. =]
  7. Aug 5, 2010 #6


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    Science Advisor

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

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