[Ifeffit] my problem

[Frenkel, Anatoly]

You cannot resolve these sites from EXAFS analysis. Even if you had only 2 inequivalent sites of the same central atom with occupancies N1 and N2, you would not be able to resolve N1 and N2 in the fit because different Nb atoms would have similar environments (e.g., 4 and 6 oxygens), and they will have similar Nb-O distances around each site. The best you can do is to try to fit the first nearest neighbor peak as an average Nb-O contribution (if it is a well isolated peak) and try to make use of sigma2 that may turn out to be larger than in some well known Nb oxide reference that you should analyze first. 
 
Anatoly

-----Original Message-----
From: ifeffit-bounces at millenia.cars.aps.anl.gov [mailto:ifeffit-bounces at millenia.cars.aps.anl.gov]On Behalf Of Tadej Rojac
Sent: Thursday, December 15, 2005 7:44 AM
To: Ifeffit at millenia.cars.aps.anl.gov
Subject: [Ifeffit] my problem


Dear collegues,
 
my name is Tadej Rojac and I'm writing from the Institute Jozef Stefan in Ljubljana, Slovenia. I work in the Department of electronic ceramics. Actually, I'm working on a EXAFS spectra with Artemis. What I am trying to do is mainly to describe an EXAFS spectrum with a model concerning the structure of Nb2O5. I'm working togheter with prof. Iztok Arcon, who is a specialist in the field. My main problem is that I have to do some suppositions. In fact Nb2O5 structure is quite complicated. It is composed of seven different Nb positions which I took into account using 7 feffs in Artemis. I don't know the exact occupancies of the seven positions. In order to make  the search easier I did at first a supposition that all the occupancies must be positive, which is realistic. I set this with the command "abs" in each feff file. Secondly, I need to supose that the sum of all occupancies is 1. In that way the search for the result is much easier, otherwise I get the sum greater than one, which, for sure, is not the case. My question is how can I do that? For example, if the occupancies are O1, O2, O3, O4, O5, O6 and O7 and I define them with a starting value (let use say the same, so 1/7) than I can do O = O1+O2+O3+O4+O5+O6+O7. Finally I need to do O =1, but here Artemis doesn't want to define the same parameter (O in this case) twice. So, how can I insert the conditions, that the sum of all the occupancies is 1, into Artemis? I hope you can help me.
 
I'm looking forward t hearing from you...thank you...  
 
Tadej Rojac
Jozef Stefan Institute 
Jamova 39
1000 Ljubljana
Slovenia
Tel.: +386 1 477 38 34
Fax: +386 1 477 38 87
E-mail: tadej.rojac at ijs.si

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