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Age of radioactive sample from sample

  1. Jul 30, 2016 #1
    1. The problem statement, all variables and given/known data
    About a day after the Chernobyl disaster the radioactive fallout landed in Sweden. The relative activity of ##^{133}I (t_{1/2} = 20.8h )## and ##^{131}I (t_{1/2} = 8.02 d)## were measured at 28/4 17:00 at ##270## mBq and ##1000## mBq respectively.
    Determine the time of the reactor failure. The exchange for creating different isobars with fission using thermal neutrons of ##^{235}U## can be found in the chart of nuclides.

    (I'm afraid I made a rather poor translation but I'm sure the meaning comes across)

    2. Relevant equations
    Exponential law of radioactive decay
    ##N(t) = N_0e^{-\lambda t}##

    Half-life
    ##t_{1/2} =\ln (2) / \lambda##

    Nuclear activity
    ##A(t) = \lambda N(t)##
    3. The attempt at a solution
    Lets refer to ##^{133}I## as element 1 and ##^{131}I## as element 2.
    The ratio of the activities is by the law of radioactive decay is then
    ##R = \frac{\lambda_1 N_1(t)}{\lambda_2 N_2(t)} = \frac{\lambda_1 N_1(0)}{\lambda_2 N_2(0)} e^{t (\lambda_2-\lambda_1)}##
    Rearranging we have
    ##t = \frac{1}{\lambda_2 - \lambda_1}\ln \left( R\frac{\lambda_2 N_2(0)}{\lambda_1 N_1(0)} \right)##

    We know that ##R=0.27## and the decay constants can be computed too
    ##\lambda_1 = \log (2) /(20.8 \cdot 3600) \approx 9.2568\cdot 10^{-6}##
    ##\lambda_2 = \log(2)/(8.02 \cdot 3600 \cdot 24) \approx 1.0003\cdot 10^{-6}##.

    To find the initial ratio of the elements I looked at the ratio in the Thermal fission yield column here https://www-nds.iaea.org/sgnucdat/c3.htm
    ##\frac{N_2(0)}{N_1(0)} \approx \frac{2.878}{6.59} \approx 0.4367##.

    Inserting all the values we have
    ##t = \frac{1}{1.003 \cdot 10^{-6}-9.2568\cdot 10^{-6}} \ln \left( 0.27\cdot 0.4367 \frac{1.003}{9.2568}\right) \approx 5.2882\cdot 10^5## seconds or ##146.8951## hours.

    So the answer I get Is 14:00 22/4. The provided answer on the other hand claim I should get 17:00 25/4. Am I using the wrong method or the wrong chart or I did I make a mistake somewhere?
     
    Last edited: Jul 30, 2016
  2. jcsd
  3. Jul 30, 2016 #2

    haruspex

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    Surely the activity ratio at time t (your R) is ##\frac{\lambda_1N_1(t)}{\lambda_2N_2(t)}##.
     
  4. Jul 30, 2016 #3
    Nice catch! Edited that now. I remembered the ##\lambda## in the next step in the equation however so it doesn't change the calculation itself.
     
  5. Jul 31, 2016 #4

    haruspex

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    I cannot find any flaw in your working. Interesting that the official answer is just about half.
     
  6. Jul 31, 2016 #5

    haruspex

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    I guess you mean isotopes.
     
  7. Jul 31, 2016 #6
    In that case I feel that at least I understand something:D I actually think I knew how they got the different answer now. Seems that if you calculate ##t = \frac{1}{1.003 \cdot 10^{-6}-9.2568\cdot 10^{-6}} \ln \left( 0.27\cdot 0.4367\right)##m that is we actually take the quantities as the amount instead of the activity we get ##71.94## hours. So I'm guessing this may be an error in the answer since they say relative activity and mBq is a unit of activity. Or I guess it could just be a coincidence.

    It should be, weirdly enough the question actually had isobars in it.
     
  8. Jul 31, 2016 #7

    haruspex

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    Yes, that must be it.

    The perils of predictive text?
     
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