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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?
    thanks
     
  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

    HallsofIvy

    User Avatar
    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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