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Number of C-14 atoms in old piece of wood

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm working a homework problem that states:

    A very old specimen of wood contained 1012 atoms of 14C in 1986.
    • How many 14C atoms did it contain in the year 9474 B.C.?
    • How many 14C atoms did it contain in 1986 B.C.?



    2. Relevant equations

    N(t) = N(0)exp^(-lambda*t)


    3. The attempt at a solution

    For C-14, lambda = ln2/T1/2 = 0693/5730 yrs = .000121 yr-1

    Since we do not know N0, I took the ratio of C-14 atoms in 1986 (A.D.) to C-14 atoms in 9474 B.C.:

    Let t1 reflect the time from t0 to 1986.
    Let t2 reflect the time from t0 to 9474 B.C.
    then the ratio of N(t1)/N(t2) is:

    N0 * e-lambda*t1
    -----------------------------------------------------
    N0 * e-lambda*t2

    N0 cancels and we know that N(t1) = 1012 atoms.

    Solving for N(t2) gives us:

    1012
    ----------------------------------------------------
    e-lambda(t2-t1)

    We know that the difference t2 - t1 is equal to 1986 + 9474 = 11460 yrs. Plugging the numbers and solving for N(T2), gives us

    1012
    ----------------------------
    e(-0.693(11460/5730))

    N(t2) = 4 x 1012 ===> number of C-14 atoms in 9474 B.C.

    Is this the correct approach? I used the approach for the second part of the question for number of C-14 atoms in 1986 B.C.. I got 1.6 x 1012 atoms.

    Thanks for the help.
     
  2. jcsd
  3. Sep 10, 2009 #2
    Well....*technically* [tex]N(t1) = N_0 * e^{-\lambda*t1} [/tex], so you shouldn't have cancelled out the N0, although it does not affect your final answer.
     
    Last edited by a moderator: Sep 10, 2009
  4. Sep 10, 2009 #3

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Perhaps an easier approach would be to note that the time elapsed, 11,460 years, is about 2 half-lives, 2.01 to be precise. What fraction of the original sample is left after 2 half-lives have gone by? If you know the fraction remaining, you can easily find the original amount.

    In general, I find it easier to answer this kind of problem using the half-life as a unit of time because the fraction remaining after n half-lives is

    f = 2-n.
     
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