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Neutron activation question

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data\
    (From a past exam paper)
    Na-24 can be produced by irradiation of 27-Al with energetic neutrons. It decays by beta emission.

    An aluminium sample of mass m=0.2g was irradiated by energetic neutrons for a period of 1 hour. Two hours after the end of irradiation the total 24-Na activity of the sample was 295 Bq.
    (i) Calculate the total number of Na-24 nuclei that were present in the aluminium sample immediately at the end of the irradiation.
    (ii) What was the neutron flux (use a reaction cross-section of [tex]\sigma[/tex]=0.125 barns).


    2. Relevant equations
    Basically, when the sample is being irradiated the number of 24-Na atoms [tex]N^{*}[/tex] obeys
    [tex]\frac{dN^{*}}{dt} = R - \lambda N^{*}[/tex]
    where [tex]R = N\sigma \Phi[/tex] is the rate of neutron absorption, [tex]N[/tex] is the number of atoms in the sample, [tex]\sigma[/tex] is the microscopic cross-section, and [tex]\Phi[/tex] is the neutron flux. [tex]\lambda[/tex] is the decay constant. Once the irradiation is stopped, it decays according to
    [tex]\frac{dN^{*}}{dt} = -\lambda N^{*}[/tex].

    3. The attempt at a solution
    It seems to be just a simple plug-and-chug question based on the rudimentary treatment of neutron activation analysis that we did in class, but the thing is that I can't see how you can possibly do it without knowing the Na-24 half-life and the atomic mass of aluminium, and neither of these values were given anywhere in the exam paper. Am I missing something obvious here?
     
    Last edited: Jun 21, 2009
  2. jcsd
  3. Jun 21, 2009 #2

    Astronuc

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    Staff: Mentor

    The atomic mass of Al-27 is 27 amu. To go from 27Al to 22Na requires an (n,α) reaction.

    The half-life of Na-22 is 14.951 hrs. One reference is http://www.nndc.bnl.gov/chart/.

    However, since A = λ N, then is it possible to determine λ, from the information given?
     
  4. Jun 21, 2009 #3
    Thank you for your answer.
    You're right, I didn't realise that the mass defects are so small that you can just ignore them for these purposes.
    But the point is, this question was in a past exam paper. That reference would not have been available.
    But we don't know N either, since the neutron flux is unknown.
     
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