## relation of atomic and mass enrichment?

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.

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 Blog Entries: 2 Recognitions: 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.
 Admin 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.

## relation of atomic and mass enrichment?

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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