# Relation of atomic and mass enrichment?

1. Jun 10, 2012

### 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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2. Jun 10, 2012

### 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

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.

3. Jun 10, 2012

### Astronuc

Staff Emeritus
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
4. Jun 10, 2012

### OneMoreName

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