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Homework Help: Radiocarbon Dating

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data
    The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the original 6 12 C atoms remains?

    2. Relevant equations

    3. The attempt at a solution
    t = 41,000 yrs
    T1/2 of Carbon = 5730 yrs
    ln2 = .693

    I basically plugged in the numbers and solved because you are given all the variables.
    (N1/No) = e^-(ln2/T1/2)t
    (N1/No) = e^-(.693/5730 yrs)(41,000 yrs)
    ln(N1/No) = -(.693/5730 yrs)(41,000 yrs)
    ln(N1/No) = -(1.2094E-4)(41,000 yrs)
    ln(N1/No) = -4.95863
    (N1/No) = e^(-4.95863)
    (N1/No) = .0102548​

    I then calculated for the % because the answer .012548 is a fraction.
    .0102548/100 %

    However that answer is incorrect and I'm not exactly sure why. I made sure that I was using the (ln)-function instead of the (log)-function. I don't know if it is my math or if I'm putting it into the website incorrectly.

    Any help would be greatly appreciated!
  2. jcsd
  3. Apr 16, 2014 #2


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    I get that e(-4.95863) ≈ 0.00702 .
  4. Apr 16, 2014 #3
    You are correct, I must have copied it down wrong.

    So then, I would take .00702 and divide by 100 to get the Percentage.

    .00702/100 = 7.0225E-5%

    This is correct, right?
  5. Apr 16, 2014 #4


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    Isn't 1/10 equal to 10 % ?

    You don't get that by dividing by 100 .
  6. Apr 16, 2014 #5
    Where are you getting the 10% from?

    I was dividing by 100 because initially you start out with 100% of the material. Or am I incorrect with that idea?
  7. Apr 16, 2014 #6


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    SammyS was illustrating that diving by 100 does not give the percentage.
    What is 0.1 as a percentage? How is it calculated?
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