[Ifeffit] Lattice parameters: EXAFS vs. XRD
newville at cars.uchicago.edu
Wed Dec 30 10:47:50 CST 2009
I believe we had a conversation about this last January.
XAFS is not sensitive to the crystallographic lattice constants. It
measures the spacing between atoms. Because of thermal vibrations and
other disorder terms, the average distance between atoms is larger
than the distance between the lattice points.
On Fri, Dec 25, 2009 at 12:04 PM, Scott Calvin <SCalvin at slc.edu> wrote:
> Merry Christmas, everyone!
> Yes, I'm pondering EXAFS on Christmas...
> Here's an issue that I bet has been worked out, and I bet someone on this
> list knows the result and where it's been published.
> It's well known that the MSRD ("sigma squared") for EXAFS differs
> substantially from the "Debye-Waller factor" in XRD, because the first is
> the variance in the interatomic distance, and the second is the variance in
> the atomic position relative to a lattice point.
> But what about the lattice parameter implied by the nearest-neighbor
> distance in EXAFS as compared to the lattice parameter found by XRD?
> It is certainly true that in most materials, particularly highly symmetric
> materials, the nearest-neighbor pair distribution function is not Gaussian,
> and generally has a long tail on the high-r side. (This is largely because
> the hard-core repulsion keeps the atoms from getting much closer than their
> equilibrium positions.) So imagine a set of atoms undergoing thermal
> vibrations around a set of lattice points. For concreteness, let's consider
> an fcc material like copper metal. The lattice points themselves are further
> apart than they would be without vibration, sure, but that's not the
> question. The question is whether the square root of two multiplied by the
> average nearest-neighbor distance is still equal to the spacing between
> lattice points.
> My hunch is that the answer is no, and that the EXAFS implied value will be
> slightly larger. While the average structure is still closed-packed, the
> local structure will not be. And in a local structure that is not
> closed-packed, the atoms will occasionally find positions quite far from
> each other, but will never be very close. In a limiting case where melting
> is approached, it's possible to imagine an atom migrating away from its
> lattice point altogether, leaving a distorted region around the defect.
> While XRD would suppress the defect, EXAFS would dutifully average in the
> slightly longer nearest-neighbor distances associated with it.
> Just to be clear, I am not talking about limitations in some particular
> EXAFS model used in curve-fitting. For example, constraining the third
> cumulant to be zero is known to yield fits with nearest-neighbor parameters
> that are systematically reduced. In fact, limitations like that mean the
> question can't be answered just by looking at a set of experimental results:
> I can make my fitted lattice parameter for copper metal go up or down a
> little bit by changing details of a fitting model or tinkering with
> parameters that themselves have some uncertainty associated with them, like
> the photoelectron's mean free path. (Fortunately, this kind of tinkering
> will affect standards and samples in similar ways, and thus don't affect my
> confidence in EXAFS analysis as a tool for investigating quantitatively
> differences between samples, or between samples and a standard.) My question
> is about the ACTUAL pair distribution function in a real fcc metal. To the
> degree it's a question about analysis, it's about XRD:
> "In an fcc metal should the expectation value of the nearest-neighbor
> separation, multiplied by the square root of two, equal the lattice spacing
> as determined by XRD?"
> --Scott Calvin
> Sarah Lawrence College
> Ifeffit mailing list
> Ifeffit at millenia.cars.aps.anl.gov
More information about the Ifeffit