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Homework Help: Date An Object Using Carbon Dating, Beta Decay

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data
    An old wooden tool, containing 75.0 grams of carbon, is found in an ancient tomb. The tool emits 500 electrons/minute from the beta decay of 146C. How old is the wood from which the tool was constructed? Given: The half-life of 146C is 5730 years, the ratio of 146C to 126C in living plants is 1.30 x 10-12, 1 year = 3.15576 x 107s, NA= 6.0221415 x 1023atoms/gram-mole.

    2. Relevant equations
    R = [itex]\lambda[/itex]N = [itex]\lambda[/itex]N0e-[itex]\lambda[/itex]t = R0e-[itex]\lambda[/itex]t
    R = rate of decay, [itex]\lambda[/itex] = decay constant, N0 = # of radioactive nuclei at t=0, N = number of radioactive nuclei now, R0 = decay rate at t=0

    T1/2 = ln|2|/[itex]\lambda[/itex]
    T1/2 = half life

    # of atoms of AZX = N = m * NA / A
    m = mass of sample, A = atomic mass of element, Z is atmoic # of element, N = number of atoms, NA = Avogadro's Constant

    3. The attempt at a solution

    R = 500
    N0 = 75 * 1.3*10-12 * NA / 14 = 4.194*1012
    T1/2 = ln|2|/[itex]\lambda[/itex] - > [itex]\lambda[/itex]= ln|2|/T1/2 = 3.83*10-12
    R = 500 = [itex]\lambda[/itex]N --> N = 1.305*1014
    R = [itex]\lambda[/itex]N0e-[itex]\lambda[/itex]t --> Solve for t
    I get t=3.0863*104years

    The correct answer is: 6.7*103 years
    Last edited: May 16, 2013
  2. jcsd
  3. May 16, 2013 #2


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    If the ratio 1.3*10-12 is the ratio of the number of C14 to C12, then you would first want to find the number of C12 and then multiply by 1.3*10-12.

    It's a good idea to carry along your units in the calculation. I think you'll see that the units are mismatched here.
  4. May 16, 2013 #3
    The 10-13 was a typo. I'm still having issues with this problem. I tried it again doing this:

    N0 = 75g * 1.3*10-12 * (6.022*1023)gram-mol / (14gram-mol) = 4,19 * 1012

    T1/2 = ln|2| / [itex]\lambda[/itex], solving for [itex]\lambda[/itex] gives [itex]\lambda[/itex] = 3.833*10-12

    R=[itex]\lambda[/itex]N0e-[itex]\lambda[/itex]t solving for t.
    Plugging in R as 500minutes / (60seconds/minute) = 8.3333 decays/second.
    This is giving me 7.1455 * 103 years.

    I think my mistake might be that 500 electrons/minute doesn't equal 500 decays/minute?
  5. May 16, 2013 #4


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    I think that's ok.

    Try calculating No as

    N0 = 75g * 1.3*10-12 * (6.022*1023)gram-mol / (12gram-mol)

    Think about why you would want to calculate it this way.
  6. May 16, 2013 #5
    I did this, and got the correct answer, but I am still somewhat confused as to why.

    My thought was that the C14 is the radioactive isotope, I should be trying to determine how much C14 is currently still present. To find that I need to figure out how much has decayed into C12. So I know the ratio at t=0, which lets me solve for N0. Why would I solve for N0 as C12 instead of C14 if C14 is the radioactive one?

    Also, a huge thank you for the help!
  7. May 16, 2013 #6


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    (75g / 12g mol-1) * (6.022*1023 atoms mol-1) gives the number of C12 atoms in the 75 g sample. You then multiply this by 1.3*10-12 to get the number of C14 atoms that were originally present in the 75 g sample.
  8. May 16, 2013 #7
    Ah, OK that makes sense now. Thank you. I'm guessing that since there are so few C14 atoms, its fair to make the assumption that the entire 75g sample is C12 when finding the number of C12 atoms?
  9. May 16, 2013 #8


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    Yes, that's right. Good.
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