Hi John,
I am looking forward to going though your code for this particular issue.
In the interim, I found some readings with reference to the mass dependency of ZAF corrections, as you mention:
Now having said that maybe the next generation of matrix corrections will not be based on concentrations, and that would (I think) solve your problem. I have to agree that when I got into this business I could not understand why everything was in mass concentrations. After all, the microprobe is *not* a balance and it certainly cannot distinguish isotopes! It really doesn't measure mass, just numbers of atoms! But like mineral names, this may be here to stay although no one can think of a good reason for it.
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From Goldstein et al. (2003):
Scanning Electron Microscopy and X-ray Microanalysis, Third Edition, 2003 by Goldstein, J., Newbury, D.E., Joy, D.C., Lyman, C.E., Echlin, P., Lifshin, E., Sawyer, L., and Michael, J.R.
9.4. The Approach to X-Ray Quantitation: The Need for Matrix Corrections (p. 402).
“As was first noted by Castaing (1951), the primary generated intensities are roughly proportional to the respective mass fractions of the emitting element. If other contributions to x-ray generation are very small, the measured intensity ratios between specimen and standard are roughly equal to the ratios of the mass or weight fractions of the emitting element. ”
And from Reed (1993):
Electron Microprobe Analysis, Second Edition, 1993 by Reed, S.J.B.
1.9 Relationship between X-ray intensity and elemental concentration (pp. 10-11).
“The reason that characteristic X-ray intensities are, to a first approximation, proportional to mass concentration (whereas atomic concentration might appear more reasonable) is related to the fact that incident electrons penetrate an approximately constant mass in materials of different composition. This is because these electrons lose their kinetic energy mainly through interactions with orbital electrons of the target atoms, the number of which is approximately proportional to atomic mass.
The consequences of this can be demonstrated as follows. Consider two elements, A and B, the latter being 'heavier' than the former. To determine the concentration of A in a sample containing a mixture of A and B, one compares the intensity of ‘A’ radiation emitted by the sample with that from pure A. For the sake of simplicity we assume that the sample contains equal numbers of A and B atoms. Fig. 1.7 shows diagrammatically the volumes excited in pure A (
a) and the A-B compound (
b), given that the masses excited are equal, as noted above. The number of atoms excited in pure A is greater than in the compound because of the presence of heavy B atoms in the latter. Consequently the ratio of the numbers of excited A atoms in sample and standard, and hence the X-ray intensity ratio, is less than 0.5 (the atomic concentration of A in the compound).
The following argument shows that this ratio is, in fact, equal to the mass concentration (given the assumptions stated above). If the atomic concentration of A is
nA, then the mass concentration is given by
CA =
nAAA/[
nAAA + (l-
nA)
AB] where
AA and
AB are the atomic weights of A and B respectively. The number of atoms excited in the pure A standard is equal to
Nm/
AA, where
N is Avogadro's number and
m the mass penetrated by the incident electrons. In the case of the compound, the number of A atoms in the excited volume is
nANm /[
nAAA + (l-
nA)
AB]. The X-ray intensity ratio (which is proportional to the ratio of the numbers of excited atoms) is given by this expression divided by
Nm/
AA, which is equal to the expression given above for
CA. In reality the intensity ratio is not exactly equal to the mass concentration, firstly because
Z /A is not constant and secondly because the 'stopping power' of all bound electrons is not the same. This, together with other factors which affect the measured intensities, gives rise to the need for matrix corrections, as described below.”
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I found this explanation to be useful, and have attached the 2 pages from Reed's book as a PDF (which shows the diagram).
Thanks,
Andrew
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