[Ifeffit] my problem
Kelly, Shelly D.
SKelly at anl.gov
Thu Dec 15 11:00:46 CST 2005
From: ifeffit-bounces at millenia.cars.aps.anl.gov
[mailto:ifeffit-bounces at millenia.cars.aps.anl.gov] On Behalf Of Tadej
Sent: Thursday, December 15, 2005 6:44 AM
To: Ifeffit at millenia.cars.aps.anl.gov
Subject: [Ifeffit] my problem
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.
[Kelly, Shelly D.]
In the path page for each path of each feff calculation you need to have
a variable that represents the occupancy.
S02xN = abs(S02 * O1)
Then in the variable page (guess set def)
Guess O1 = 0.15
Guess O2 = 0.15
Guess O3 = 0.15
Def O7 = 1 - abs(O1) -abs(O2) -abs(O3) - abs(O4) - abs(O5) -abs(O6)
Hence you get one occupancy for free for knowing one more piece of
Sounds like a tough problem.
I'm looking forward t hearing from you...thank you...
Jozef Stefan Institute
Tel.: +386 1 477 38 34
Fax: +386 1 477 38 87
E-mail: tadej.rojac at ijs.si
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