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Radioactive Decay

  1. Oct 22, 2005 #1
    Q: A 50-g chunk of charcoal is found in the buried remains of an ancient city destroyed by invaders. The carbon-14 activity of the sample is 200 decays/min. Roughly when was the city destroyed?

    A: I used R = Ro*e^(-lambda*t) and t1/2 = ln2/lambda

    t1/2 = 3.834*10^-12 s-1

    and 200 decays/min = 3.33 decays/sec or 3.33 Bq

    What is the initial acitivity/kg (Ro)?
    The final acitivty/kg = 3.33 Bq/0.05 Kg = 66.66 Bq/Kg

    Therefore R/Ro = 66.66/255

    I then found the time to be 11 088 years

    Can someone please check over this and see if this looks correct and advise of any corrections?

    Many thanks!
  2. jcsd
  3. Oct 22, 2005 #2


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    I didn't check the numbers (for instance, is the initial activity of C-14 255 Bq/kg ?) but the approach seems totally correct.

  4. Oct 22, 2005 #3


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    What half-life of C-14 is used?

    Be careful with units. t1/2 should have units of time.

    1 Bq = 1 decay/s is correct.

    The decay constant [itex]\lambda[/itex] = ln 2 / t1/2, and customarily it is often expressed as s-1, but it should be compatible with whatever units (s, hr, days) are used for time, t.

    I calculate a decay constant of 3.9455E-12 s-1.

    Activity A(t) = Ao exp(-[itex]\lambda[/itex]t)

    So inital activity is A(t) exp ([itex]\lambda[/itex]t).

    So to solve the problem, one needs to know the normal specific activity for C-14 in charcoal. Where did 255 Bq/kg originate?

    255/66.66 = 3.825 half-lives.



    Useful calculator - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/raddec.html#c1
  5. Oct 24, 2005 #4
    In the textbook in a previous example - it says that the C-14 activity in a living tree is 255 Bq - that is where i got that value from but i wasnt sure if i could use that value in this question - ie is it always 255 Bq in living trees?!?

    The half-life is in s^-1 - therefore the answer would be in seconds and i just converted it back to years in the answer.

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