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Homework Help: A question on Beryllium as a neutron moderator

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data

    I have an assignment question that asks the me to look up the partial mass attenuation coefficients for Beryllium for Rayleigh Scattering, Compton Scattering, the Photoelectric Effect, and Pair Production in both the nuclear and electron fields on the NIST database for 10MeV incident photons. That is fine - easy.

    But the next part of the question asks me (in the exact words) "...considering these partial interaction coefficients, briefly explain why Beryllium is commonly used in reactor systems and radiation units as a moderator, used to lower the energies of high energy particles while minimising particle number loss".

    2. Relevant Equations

    The incoherent scattering (Compton Scattering) partial mass attenuation coefficient is 1.37x10^-2 cm^2/g

    The coherent scattering (Rayleigh Scattering) partial mass attenuation coefficient is 1.86x10^-7 cm^2/g

    The photoelectric absorption partial mass attenuation coefficient is 8.81x10^-9 cm^2/g

    and for pair production (nuclear + electron fields), we have 2.603x10^-3


    3. The attempt at a solution

    This seems a bit odd to me - all of the above interaction coefficients are for photon interactions. It seems to me that this has little to do with what I would have thought were the main reasons for the use of Beryllium as a moderator in reactors which obviously have to do with neutron interactions - namely the low cross-section for thermal neutron absorption, low atomic number to make for efficient thermalising of the neutrons, and propensity to increase neutron flux through the (n,2n) reaction path.

    Now I guess that one could deduce from the values of the coefficients listed above as seen on NIST that beryllium is likely low a low Z material (due to the coherent/incoherent coefficient ratios and whatnot), and therefore probably an efficient moderator of neutrons, but it is bleeding obvious that Beryllium is low Z anyway, so this doesn't seem to me to be the answer.

    It seems to me that by the wording of the question, they are gunning for me to answer with something along the lines of the incoherent coefficient dominating the other processes, so that high-energy particles are reduced in energy quite efficiently. But the again, the coefficients are for photons and photonic processes, not neutrons, and unless there is a direct correspondence between the photon partial attenuation coefficients and the neutron partial cross sections, I don't see the relevance of photonic interaction cross sections when the crux of the matter concerns neutrons.

    The other thought I had was that Beryllium is quite a good neutron moderator/reflector, but due to it's low density, has very low linear attenuation coefficients (as opposed to the mass attenuation coefficients listed above, the difference being the mass attenuation coefficients have the density of the material divided out) for photonic interactions, and so Beryllium lets photons pass with relatively few interactions, while it still interacts strongly with neutrons. But why would this property mean it is good for use as a moderator? Why would a material that interacted weakly with photons and strongly with neutrons be a better moderator than a material that interacted strongly with both?

    Any help or thoughts in this matter would be greatly appreciated!
     
  2. jcsd
  3. Apr 3, 2010 #2
    Hey astrofiend,

    I think we must be doing the same assigment. Like you, I got quite confused by this question and asked myself the same thing.

    "briefly explain why Beryllium is commonly used in reactor systems and radiation units as a moderator". I think what she meant by this question is that we should explain in a first part why beryllium is use in nuclear reactor, and in a second and non-related part, the use of beryllium in radiation.

    The use of beryllium for radiation purpose because it does not interact that must with X-rays, and it can be used in x-rays detectors.
     
  4. Apr 5, 2010 #3
    Ahhh - OK. It didn't even occur to me that the wording could be taken in that way - that makes a great deal more sense. Cheers for that!
     
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