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Relation of atomic and mass enrichment?

  1. Jun 10, 2012 #1
    Hi all,

    I am preparing for a job interview in some nuclear facility and have to refresh my knowledge about nuclear stuff. So I started reading the book of Lewis "Fundamentals of reactor physics". I got stuck at page 35, formula (2.24). Does anyone have a clue how to arrive at this equation? Where does this factor 0.0128 come from? I played with the formulas a lot but never arrive at this equation. Help would be really appreciated guys. Since I am not allowed to put a link, please put a http: inside the following and you will see the page.

    //i210.photobucket.com/albums/bb283/DidgeFrank/Grafik1.jpg
     
  2. jcsd
  3. Jun 10, 2012 #2

    Bill_K

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    Science Advisor

    Replace the N's in favor of the M's. Up to a constant factor, N ~ M/A, so Eq 2.20 is

    ea = (M25/235)/((M25/235) + (M28/238))

    Then get rid of the M's in favor of ew:

    M25 = ew(M25 + M28)
    M28 = (1 - ew)(M25 + (M28)

    This gives you

    ea = (ew/235)/((ew/235) + ((1 - ew)/238)))

    so now just multiply out.
     
  4. Jun 10, 2012 #3

    Astronuc

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    Staff: Mentor

    The calculation of macroscopic cross-section requires atomic density, and enrichment on an atomic basis would be necessary. For manufacturing and accountability, the mass-based enrichment is required, since it is much easier to measure mass, and accountability records are provided in terms of mass.

    Bill K provided the method to compare mass fraction with atomic fraction.

    Remember that for an element or isotope, N = ρA/M, where ρ = density, A = Avogadro's Number, and M = atomic mass (of the element, which is weight by isotopic fractions, or by isotopic mass, if ρ is the isotpic mass density).

    Try 238/235.
     
    Last edited: Jun 10, 2012
  5. Jun 10, 2012 #4
    Oh yes, thanks! I was fooling around with the densities because it is mentioned in the text but this leads to nowhere.
     
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