[Ifeffit] lambda in parameter Ei
Bruce Ravel
bravel at anl.gov
Mon May 15 08:39:00 CDT 2006
On Saturday 13 May 2006 17:20, Arkadiusz Onyszko wrote:
> Dear All,
> I found in manual, that lamda ~ 1/sqrt(Ei)
> I want calculate lambda with fitting parameter Ei. How do it ?
> What is the exact dependence Ei and lambda ?
> During fitting I have example Ei=7,5, so I want calculate lambda.
> This dependence: lamda ~ 1/sqrt(Ei) is not a complitly.
When fitting with Ifeffit, Ei is a correction to the broadening terms in Feff,
one of which is the mean free path. The exact effect of Ei on the exafs
equation as used in Feff and Ifeffit is explained in chapter 6 of this
document:
http://cars.uchicago.edu/~newville/feffit/feffit.ps
Asyou can see from equation 6.2, Ei is actually a correction to the complex
momentum. The complex momentum is related to the mean free path, but also to
other things in Feff's energy model. This makes the precise interpretation of
Ei rather subjective.
Is Ei a correction to Feff's loss terms? Is it a correction to Feff's value
of the mean free path? Both? Apparently, you are hoping that you can
interpret Ei entirely as a correction to the mean free path. I'm not sure how
you know that, but if so, Eq. 6.2 tells you how to do so. (And don't forget
the constants -- 2m_e/hbar^2!)
Hope that helps,
B
--
Bruce Ravel ---------------------------------------------- bravel at anl.gov
Molecular Environmental Science Group, Building 203, Room E-165
MRCAT, Sector 10, Advanced Photon Source, Building 433, Room B007
Argonne National Laboratory phone and voice mail: (1) 630 252 5033
Argonne IL 60439, USA fax: (1) 630 252 9793
My homepage: http://cars9.uchicago.edu/~ravel
EXAFS software: http://cars9.uchicago.edu/~ravel/software/
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