Relation of atomic and mass enrichment?

[OneMoreName]

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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[Bill_K]

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

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

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

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

This gives you

e_a = (e_w/235)/((e_w/235) + ((1 - e_w)/238)))

so now just multiply out.

 

[Astronuc]

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.

 

[OneMoreName]

Oh yes, thanks! I was fooling around with the densities because it is mentioned in the text but this leads to nowhere.