1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook