Re: [Ifeffit] error on installing demeter on MacOSX 10.9.5
Hi Christophe, For some reason, macports sometimes does not install all of the dependencies for Demeter. You may need to load the dependencies separately. In this case use "sudo ports install libcxx" Sincerely, Wayne On Wed, Nov 23, 2016 at 8:54 AM, christophe den auwer < christophe.denau...@unice.fr> wrote: > Dear all, > > I am trying to install Demeter on OS 10.9.5 and I am stuck! > > I installed Xcode 6.2 fine with the licence agreement and everything. > Then I installed MacPort 2.3.5 fine as well. I also checked updated > version and everything worked fine. > Then I tried to install Demeter and got stuck with the following lines : > > Error: org.macports.archivefetch for port libcxx returned: archivefetch > failed for libcxx @3.9.0_0+universalError: Failed to install libcxx > Please see the log file for port libcxx for details: > Error: Processing of port demeter failed > > I attached to my message, if of any use, the full procedure and feedback > from terminal window. > > If someone could help that would be really great!!! > > Thanks a lot > > Christophe > > > > > > > > > - > Christophe Den Auwer > Professeur > Institut de Chimie de Nice > Université de Nice Sophia Antipolis > 06108 Nice, France > 0492076362 > christophe.denau...@unice.fr > -- > > > > > > ___ > Ifeffit mailing list > Ifeffit@millenia.cars.aps.anl.gov > http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit > Unsubscribe: http://millenia.cars.aps.anl.gov/mailman/options/ifeffit > > -- Wayne Lukens Staff Scientist Lawrence Berkeley National Lab ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit Unsubscribe: http://millenia.cars.aps.anl.gov/mailman/options/ifeffit
Re: [Ifeffit] trouble installing deter on a mac
Hi Justin, It should be "sudo port install libcxx." After libcxx is installed, try installing demeter again. I had to install almost all of the packages that demeter uses one-by-one. I have no idea why, but it worked. Sincerely, Wayne On Wed, Nov 2, 2016 at 12:49 PM, justin park <justin_ap...@icloud.com> wrote: > Hello, > > I have followed the steps to install macports on my mac and it seems to be > working. However when I went to port in demeter it start and fails rather > quickly giving this error message. > > Error: org.macports.archivefetch for port libcxx returned: archivefetch > failed for libcxx @3.7.1_0+universal > > What do I need to do in macports to get it to properly fetch libcxx? > > > Thanks, > Justin > > > ___ > Ifeffit mailing list > Ifeffit@millenia.cars.aps.anl.gov > http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit > Unsubscribe: http://millenia.cars.aps.anl.gov/mailman/options/ifeffit > > -- Wayne Lukens Staff Scientist Lawrence Berkeley National Lab ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit Unsubscribe: http://millenia.cars.aps.anl.gov/mailman/options/ifeffit
Re: [Ifeffit] non-standard space group P 21/n
Dear Andrea,
To convert the atomic positions, use the following equations (starred
positions are in the P21/c setting)
a* = c
b* = b
c* = a+b
To convert the unit cell,
A* = C
B* = B
C* = sqrt{[Ccos(90-beta)]^2+[A-Csin(90-beta)]^2}
beta* = 90 - cos-1[Ccos(90-beta)/C*] where cos-1 is the inverse cosine
function
If you have problems, please double check my geometry for converting the
unit cell parameters.
Sincerely,
Wayne
On Fri, Dec 5, 2014 at 9:04 AM, andrea.san...@unipd.it wrote:
Dear Gordon,
thank you for your reply.
I tried to use the space group P1,
but the resulting local structure around the absorber atoms
is completely different from that expected.
Andrea
HI Andrea,
My suggestion, if you know where all the atoms are in the unit cell, is to
use
space group P1. It may be tedious and brute force in nature, but you won't
have
to worry about origin choices or non-standard space groups. You will still
know
which atoms are equivalent and can pick an appropriate one as your target
atom.
regards,
Robert
On Fri, Dec 5, 2014 at 8:07 AM, Sanson Andrea andrea.san...@unipd.it
wrote:
Dear all
does anyone know if is it possible to create with ATOMS
the non-standard space group P 21/n ?
Space group keywords are listed at the atoms website
http://iffwww.iff.kfa-juelich.de/icp/atoms/atoms.sgml-7.html#ss7.4
but no keyword for P 21/n group is available.
Thanks for any suggestion.
Best regards,
AS
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Staff Scientist
Lawrence Berkeley National Lab
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Re: [Ifeffit] Split white line in TcO2 spectra and similar Tc(IV) octahedral environments
Hi Fred, The second resonance is due to scattering, mainly the multiple scattering path involving the trans oxygen ligands of the octahedral Tc coordination sphere, I believe. In other words, it is the beginning of the EXAFS spectrum. These features are particularly prominent for Tc. This is the characteristic XANES spectrum of Tc(IV) in an octahedral environment with O neighbors. Sincerely, Wayne ps If you look at the theoretical scattering curves from Feff, you can identify which scattering path is responsible for this peak. If I recall correctly, it is the path that I mentioned, but I may misremember. On Mon, Aug 5, 2013 at 11:40 PM, fred.mosselm...@diamond.ac.uk wrote: Hi, ** ** Tc(IV) which is a d3 configuration ion, when octahedrally coordinated by O ligands eg TcO2 has what I would call a split white line (see attached image) in the K-edge XANEs. Others might call it a very strong second resonance . The image is from TC in an iron oxide phase by the way. One of my collaborators asked me why does the edge have two peaks.? This is a simple question but is there a simple answer? In a literature search although the shape seems accepted I couldn’t find a straightforward explanation. Ce(IV) L3 spectra have a split white line due to some sort of crystal field effect but there the transition (L3-edge) is direct to d states so it’s a bit more obvious. Here it’s an s-p transition formally so you’d think it must be some sort of split p state explanation , but I am not sure why? I am sure you could XANES simulate this to reproduce it but that doesn’t answer the question why in simple terms. It may be it ‘s a complicated band structure explanation and so there is no simple answer but if anyone has read or knows one , I ‘d be interested,. There are some similar effects in the white lines of Zr K-edge spectra , when the Zr is also octahedral but I can only find reports of that not the reason for it. BTW This is definitely pure Tc(IV) so the answer is not multiple oxidation states., which has been used to explain different Tc split white lines in other situations. Thanks Fred ** ** ** ** Prof. J F W Mosselmans Principal Beamline Scientist I18 Diamond Light Source Diamond House Harwell Campus Didcot Oxon OX11 ODE UK ** ** T 00 44 1235 778568 M 00 44 7785510211 E fred.mosselm...@diamond.ac.uk F 00 44 1235 778448 ** ** Never mind the W it's the tax that counts ** ** -- This e-mail and any attachments may contain confidential, copyright and or privileged material, and are for the use of the intended addressee only. If you are not the intended addressee or an authorised recipient of the addressee please notify us of receipt by returning the e-mail and do not use, copy, retain, distribute or disclose the information in or attached to the e-mail. Any opinions expressed within this e-mail are those of the individual and not necessarily of Diamond Light Source Ltd. Diamond Light Source Ltd. cannot guarantee that this e-mail or any attachments are free from viruses and we cannot accept liability for any damage which you may sustain as a result of software viruses which may be transmitted in or with the message. Diamond Light Source Limited (company no. 4375679). Registered in England and Wales with its registered office at Diamond House, Harwell Science and Innovation Campus, Didcot, Oxfordshire, OX11 0DE, United Kingdom ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit -- Wayne Lukens Staff Scientist Lawrence Berkeley National Lab ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
Re: [Ifeffit] Au-foil (fcc) splitted peak
Hi Julian, This looks like an example of the Ramsauer-Townsend effect. Maybe someone who has a stronger physics background can explain it, but here is a reference. Phys Rev B, 1995, 52, 6332-6348. Sincerely, Wayne On Tue, Oct 4, 2011 at 9:08 AM, Kaiser, Julian julian.kai...@helmholtz-berlin.de wrote: Dear Ifeffit Community, I'm trying to understand a basic question about an ordinary gold-foil in the fcc-crystal structure. If the spectra is plotted in the R-space, one will find two main-peaks at approx. 2.497 and 2.964 Angström that both should be attributed to the first shell. (see attached plot Au_foil.pdf) Would this mean, one can find a peak-splitting in the 1st NN shell? And if yes, why? Maby I am compleately wrong and the second peak demonstrates already the second shell.(?) But I think the second shell should not start below 4.07825 Armströng. I think most of fcc metal crystals show this behavior. Of cause it also varys depending on the k-weight(kw=2 in the attachment). Thanks a lot in advanced, Julian -- Helmholtz-Zentrum Berlin für Materialien und Energie GmbH Mitglied der Hermann von Helmholtz-Gemeinschaft Deutscher Forschungszentren e.V. Aufsichtsrat: Vorsitzender Prof. Dr. Dr. h.c. mult. Joachim Treusch, stv. Vorsitzende Dr. Beatrix Vierkorn-Rudolph Geschäftsführer: Prof. Dr. Anke Rita Kaysser-Pyzalla, Dr. Ulrich Breuer Sitz Berlin, AG Charlottenburg, 89 HRB 5583 Postadresse: Hahn-Meitner-Platz 1 D-14109 Berlin http://www.helmholtz-berlin.de ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
Re: [Ifeffit] How to calculate F-value for XANES PCA results
Hi Andrew, I don't completely understand what your question is, but I will try to answer you. First, Sam Webb should be able to answer exactly how F was calculated in Six-Pack. However, I can tell you how F is calculated in general. First of all, the number in the Excel spread-sheet is probably not F, it is the probability of F. In general, if the probability of F is greater than 5%, that component is part of the noise. This is the same as saying that that component is within 2-sigma of the noise. F is easy to calculate for least-squares fitting, and is nicely explained by wikipedia in the regression problems section: http://en.wikipedia.org/wiki/F-test where RSS is chi-square and n is the number of independent data points, which is the lesser of the number of data points or the spectral range divided buy the resolution. In your example, you have 18 independent data points (36 eV range divided by a 2 eV resolution). You then need to calculate the probability of that value of F given the number of parameters and number of independent data points, which is not explained by the wiki article but can easily be done in Excel. In your example, you have two components with probability of F less than 5%, these would be the components that you would retain. Sincerely, Wayne -- Wayne Lukens Staff Scientist Lawrence Berkeley National Laboratory email: wwluk...@lbl.gov phone: (510) 486-4305 FAX: (510) 486-5596 Andrew wrote: Hi everyone, I was looking through the literature on how to handle PC analysis data and saw that there are several different methods you can use for determining how many components there are in the series of scans. Included in SixPack are the indicator function, scree test, and the ability to quickly do the reduced eigenvalue ratios. I’ve been digging through the literature as to how to calculate the F-values. The closest to an answer that I got was: “The above-mentioned reduction of the body of experimental data, that is, the decision of what components correspond to the noise and what are the principal components, is now made on the basis of an F test of the variance associated with eigenvalue k and the summed variance associated with noise eigenvalues (k+1, ..., c). The null hypothesis is that a given factor k*/ /*is a member of the pool of noise factors. The probability that an F*/ /*value would be higher than the current value is given by %SL (percentage of significance level). Thus, the kth*/ /*factor is accepted as a principal component if %SL is lower than some test level.” (Garcia 1995). We ran the PCA on the reduction of iron while scanning at increased temperatures. I checked the foil standard but did not see any shift in the max at 7112, we scanned at 0.5 eV intervals (2 eV resolution at the beam). I thought I understood what that statement was saying but I’m almost certain I’m doing something wrong. I have attached the .xlsx file that I was working on and hope someone can point me to the right direction. The file includes the components of the PCA and some of the variances that I was calculating. If there is a paper that someone shows an actual calculation of this in the supplemental materials that would have been exactly what I was looking for! Thanks for the help! Andrew Campos Fernandez-Garcia, M., C. Marquez Alvarez, and G.L. Haller, The Journal of Physical Chemistry, 1995, *99*(33), 12565-12569. ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
Re: [Ifeffit] How to calculate F-value for XANES PCA results
Hi Andrew, It is easier to use Fdist to calculate the probability that a given value of F corresponds is within the noise. The way to do this is to use FDIST(F, (v1-v2), v1), where F, v1, and v2 are explained below. If you have two models, model 1 and model 2, where model 1 has an additional component versus model 2, and the chi-squared, number of parameters, and degrees of freedom for model 1 are X1, p1, and v1, where v1=n-p1 and n is the number of independent parameters, and model 2 has similarly defined X2, p2, and v2, then F= [(X2-X1)*v2]/[(v1-v2)*X1] the probability of F is FDIST(F, (v1-v2), v1). I assume the same is true for the PCA analysis; however, I don't know what the values for v and p are in this case. The formula for F looks right. At any rate, once you have done the PCA analysis, you need to figure out which standard spectra span the components (there should be the same number of standards as components). Then you need to fit your experimental spectra using the standards. At that point you can apply the F-test to determine whether the contribution from that standard is greater than 2 sigma over the noise. I hope that makes sense? Sincerely, Wayne Andrew wrote: Hi everyone, Thank you Dr. Lukens for your help! Let me see if I understand the method that was described for the F-value for the variances using the Fernandez-Garcia definition (that was previously mentioned), and please correct me if I am mistaken. The Principal Component Analysis returns the eigenvectors. Then, to calculate the F-value using the Fernandez-Garcia definition: F-value for component 1 = (variance of eigenvector 1)/ summation[(variance eigenvector 2) + (variance eigenvector 3) + … (variance eigenvector c)] F-value for component 2 = (variance of eigenvector 2)/ summation[(variance eigenvector 3) + (variance eigenvector 4) + … (variance eigenvector c)] F-value for component k = (variance of eigenvector k)/ summation[(variance of eigenvector k+1) + … + (variance of eigenvector c)] Where c is the number of components in the set. Then to calculate the probability of F corresponds to noise, then the that Excel can calculate this using the function Fdist(alpha, degree of freedom 1, degree of freedom 2). Alpha = the confidence interval desired (where 0.05 is generally used) degrees of freedom 1 = # of independent data points – 1 ((this is dependent on the resolution of the beam and Dr. Lukens provided an example calc.)) degree of freedom 2 = number of components on the denominator for the F-value being tested – 1 (i.e. for component k it would equal c-k-1-1 or c-k-2) Then, “if the probability of F less than 5%, these would be the components that you would retain.” Are these equations correct? Am I using the correct equation based on the Garcia-Fernandez definition? My main misunderstanding of this was what equation to use for the F-value. Sorry for killing a dead horse, but is this definition of degree of freedom 2 correct? Thanks again for all the help and sorry if this is poorly worded, and if this is on the outer-bounds for an IFEFFIT-relevant question. Andrew ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit -- Wayne Lukens Staff Scientist Lawrence Berkeley National Laboratory email: wwluk...@lbl.gov phone: (510) 486-4305 FAX: (510) 486-5596 ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
Re: [Ifeffit] limits for second shell
Dear Eugenio, There is a standard statistical test to answer just this problem, the F-test. To use the test, you first do the refinement including the contribution from the second shell and record chi-squared, which we will call c1, and the number of parameters, p1. Then you redo the fit without the second shell but keeping everything else the same including the k-weighting and the fit range; record the chi-squared, which we will call c2, and the number of parameters, p2. Finally, you will need the number of independent data points, which is the number given by ifeffit plus 2 (Stern’s rule), idp. Then you need to calculate F, which is given by F=(c2-c1)*(idp-p2)/[(p1-p2)*c1]. Using Excel, you can calculate the probability of F using FDIST(F,(p1-p2),(idp-p1)). This gives the probability that the improvement in chi-squared due to adding the second shell to the fit is due to noise in the data. The usual criterion for the F-test is that FDIST()0.05, which means that the improvement in the fit due to including that shell is two sigma over the noise. I think it is OK to include the shell even if FDIST()0.05, but you should report the probability that the improvement in the fit is due to random noise. This is a standard test in crystallography, where it is known as the Hamilton test. In short, the F-test tells you the probability that the improvement in the fit due to including a given shell is due to random error, or can be considered “real.” The F-test has one advantage of chi-squared tests in that it is a ratio of chi-squared of two fits, so the standard deviation of the data cancels. As with everything else, the F-test is model dependent and can give the wrong answer due to non-random errors such as problems with Feff. If you have noisy data, the F-test is probably pretty good; if you have data with little noise that goes high-k, I would be more careful applying it. The wikipedia entry on the F-test is OK, but used to have the formula wrong. There is a paper on using this test in EXAFS analysis: “A Variation of the F-Test for Determining Statistical Relevance of Particular Parameters in EXAFS Fits” Downward, L.; Booth, C. H.; Lukens, W. W.; Bridges, F. X-RAY ABSORPTION FINE STRUCTURE - XAFS13: 13th International Conference. AIP Conference Proceedings, Volume 882, pp. 129-131 (2007). Sincerely, Wayne -- Wayne Lukens Staff Scientist Lawrence Berkeley National Laboratory email: wwluk...@lbl.gov phone: (510) 486-4305 FAX: (510) 486-5596 Eugenio Otal wrote: Hi, I have a sample of a pure Er2O3 (blue line in the attached graph) and a sample of doped ZnO with erbium that has segregated the same oxide (red line) by thermal treatmen. The signal for de segregated oxide gets noisy around k=9 because the sample is so diluted, but the radial distribution shows second shell signal, smaller than the pre oxide, but still a signal. My doubt is about how to know if the second shell is real and if that second shell can be useful to obtain information. Is there a criteria to know that? Some limit in k-space? Thanks, euG ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit -- Wayne Lukens Staff Scientist Lawrence Berkeley National Laboratory email: wwluk...@lbl.gov phone: (510) 486-4305 FAX: (510) 486-5596 ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
Re: [Ifeffit] Atoms Feff for Ferrihydrite Structure
Hi Andreas, The problem was that for the atoms in position 6c, the sum of x and y must be exactly 1. I have attached a corrected atoms.inp file that seems to work. Sincerely, Wayne Lukens Voegelin Andreas wrote: Dear all, Using Atoms (latest version of Artemis), I tried to calculate a Feff input file based on a recently published structure for 6-Line Ferrihydrite (Michel et al, Science, 2007). Attached: Atoms input file generated with Artemis. For all three Fe sites as core (Fe1 and Fe2 octahedral, Fe3 tetrahedral), running Atoms yields Feff files with overlapping atoms located close to each other. The Artemis error messages after running Feff suggests that a shift vector may be needed in order to obtain correct Feff input files. As a non-crystallographer, I would be grateful for advice on how this problem can be solved. Best regards, Andreas ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit ! This atoms input file was generated by TkAtoms 3.0beta10 ! Atoms written by and copyright (c) Bruce Ravel, 1998-2001 title = 6L-Ferrihydrite title = Michel et al, Science, 2007 space = 186 a = 5.9280b = 5.9280c = 9.1260 alpha = 90.0 beta = 90.0 gamma = 120.0 core = Fe1 edge = K rmax =4.0 shift 0.0 0.0 0.0 atoms ! elem x y z tag occ. Fe0.169500.830500.63650 Fe1 1.0 Fe0.30.70.33790 Fe2 1.0 Fe0.30.70.95950 Fe3 1.0 O 0.00.00.04460 O11.0 O 0.30.70.76340 O21.0 O 0.169700.830300.24670 O31.0 O 0.522700.477300.97960 O41.0 ___ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
Re: [Ifeffit] Artemis fitting using FEFF paths plus experimentalreferences
Hi Matthew, It sounds like you have a problem caused by the finite size of the nanoparticles, which results in lower than expected amplitude of the shells in comparison to the amplitude of the shells generated from the xtal structures of maghemite and magnetite. If this is the case and you have a good idea how large the nanoparticles are, you can use Scott Calvin's formula for the amplitude of a scattering shell in a nanoparticle: APPLIED PHYSICS LETTERS 87 (23): Art. No. 233102 DEC 5 2005. This works very well for metallic nanoparticles, but I have never tried it on oxide nanoparticles. Maybe you could try using Scott's formula for the amplitudes and the distances from the crystal structures. If I recall correctly, Scott uses 3 Debye-Waller parameters as variables: one each for the first two shells and one for all other shells. The other possibility is to fit the XANES and low k portion of the EXAFS, but I assume that you have already tried that. Sincerely, Wayne Matthew wrote: Of course I tried PCA! Over a certain range of temperatures, one component looks tolerably like magnetite. The other is sort of, but not exactly, maghemite. However, if I go below that restricted range, it's not pseudobinary anymore. Also, I still don't know what the maghemite-like end-member actually is. Still, for the purposes of the ALS user-meeting poster I have to hang in a week or so, I'll probably take the PCA analysis plus a FEFF fit to the one end-member, and try to do a more thorough job when I have time. No, it doesn't fit well to a combo of the references I have, which does include maghemite. To be fair, what I have is not an isothermal series but an isochronal series at different temperatures, so there's less reason for expecting simple kinetics than in an isothermal series. mam - Original Message - *From:* Anatoly Frenkel mailto:[EMAIL PROTECTED] *To:* [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] *Sent:* Monday, September 24, 2007 6:59 PM *Subject:* RE: [Ifeffit] Artemis fitting using FEFF paths plus experimentalreferences Matthew - please forward my reply to the list if you consider it appropriate. I am spammed out. Anatoly * Matthew, May be it is not relevant, since I do not know if you have a series of EXAFS data at different stages of the oxidation, or just the initial and the final stages. If you do have several stages, including the starting (unoxidized) and final (oxidized, or a sum of oxidized and unoxidized, depending where you stopped at) ones, the principal component analysis seems to match your needs well. I assume you also have access to EXAFS data in bulk magnetite and maghemite references. PCA analysis may tell you (if you are lucky and your system is similar to those where such information was previously obtained): 1) The volume fractions of intermediate species at each stage of the oxidation: 2) Identities of each intermediate species The first part is entirely model-independent. The second part is not, but a reasonable selection of standard compounds (e.g., magnetite or maghemite) may help you isolating the intermediates. The good thing is that you do not need to fit a Fe-Fe and Fe-O bonds to the spectrum of the entire heterogeneous sample, where these contributions may be present in different species, and you have a large number of fitting parameters that describe such a model. Instead, you may end up obtaining that your species are identical with their bulk reference compounds (and thus no fit is required since the only thing you need to know is the volume fraction of this species) or, you may have to fit a couple of FEFF paths to that unknown species, after you deconvolute it from the mixture knowing what the rest of the phases are (it is doable in the case of one intermediate, but can be done in the case of a larger number of intermediates as well). Anatoly -Original Message- *From:* [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of *Matthew *Sent:* Monday, September 24, 2007 7:05 PM *To:* XAFS Analysis using Ifeffit *Subject:* [Ifeffit] Artemis fitting using FEFF paths plus experimentalreferences I have a system in which Fe:C alloy nanoparticles are being oxidized. I have the EXAFS of the unoxidized nanoparticles, and I'd like to use that as a reference for fitting as in: Fe_EXAFS(oxidation_stage) = x*(unox_particles)+(1-x)*(Fe-O paths + Fe-Fe paths +...) where the Fe-O and Fe-Fe are based on magnetite and maghemite, which seem to be products of the oxidation. It's not just a linear fit to
Re: [Ifeffit] Node in chi(k) envelope for a single shell
Dear Silvio, I was not able to view the image, but what you are describing seems like the Ramsauer-Townsend effect (Phys Rev B, 1988, V38, 10919–10921). As far as I understand it, the physical interpretation is that the at the energy of the dip, the potential well of the scattering atom has cross- section that is very close to the wavelength of the photoelectron. Consequently, the photoelectron is not scattered back to the absorbing atom efficiently at this energy. Perhaps someone in the list can explain this more clearly (or correctly). Sincerely, Wayne Lukens On Nov 5, 2006, at 4:38 AM, [EMAIL PROTECTED] wrote: Dear List, I have noticed in several fits that Artemis -- i.e., feff -- reports for a SINGLE shell (with absorber = scatterer = Hg, edge = L3) an almost-node in the envelope of chi(k) around 6 inverse angstroms. See the image at http://pjm.math.berkeley.edu/users/levy/files/HgNode.gif and the FEFF output after my signature. I've done my best to account for this behavior, but I simply don't have a physical intuition for why the F(k) factor should dip to zero or almost zero. Note that this is not a beat caused by two similar frequencies being added; the Fourier transform does of course show two peaks -- which is how my curiosity was aroused -- but the example figure comes from a SINGLE path, with 12 identical scatterers all at the same distance from the absorber. Looking forward to your two cents... Silvio Metacinnabar from CSDWeb, HgS Niculescu et al. 1970Feff 6L.02 potph 4.12 Abs Z=80 Rmt= 1.327 Rnm= 1.667 LIII shell Pot 1 Z=16 Rmt= 1.218 Rnm= 1.515 Pot 2 Z=80 Rmt= 1.310 Rnm= 1.624 Gam_ch=5.000E+00 H-L exch Mu=-2.727E+00 kf=1.809E+00 Vint=-1.519E+01 Rs_int= 2.005 Path2 icalc 2 Feff 6L.02 genfmt 1.4\4 --- ---\- 2 12.000 4.13662.9707 -2.72701 nleg, deg, reff, rnrmav(bohr), edge x y z pot at# 0.0.0. 0 80 Hg absorbing atom 0. -2.9250 -2.9250 2 80 Hg k real[2*phc] mag[feff] phase[feff] red factor lambda [EMAIL PROTECTED] 0.000 5.8876E+00 0.E+00 -5.3100E+00 0.1085E+01 5.5410E+00 1.8179E+00 0.200 5.8828E+00 6.5887E-02 -6.9020E+00 0.1085E+01 5.5726E+00 1.8284E+00 0.400 5.8685E+00 1.3063E-01 -8.3679E+00 0.1084E+01 5.6628E+00 1.8594E+00 0.600 5.8452E+00 1.9388E-01 -9.7059E+00 0.1082E+01 5.7987E+00 1.9101E+00 0.800 5.8135E+00 2.5592E-01 -1.0913E+01 0.1079E+01 5.9607E+00 1.9793E+00 1.000 5.7742E+00 3.1673E-01 -1.1989E+01 0.1077E+01 6.1257E+00 2.0656E+00 1.200 5.7279E+00 3.7510E-01 -1.2932E+01 0.1075E+01 6.2715E+00 2.1679E+00 1.400 5.6750E+00 4.2855E-01 -1.3742E+01 0.1074E+01 6.3832E+00 2.2854E+00 1.600 5.6157E+00 4.7354E-01 -1.4416E+01 0.1075E+01 6.4589E+00 2.4174E+00 1.800 5.5491E+00 5.0618E-01 -1.4949E+01 0.1076E+01 6.5142E+00 2.5639E+00 2.000 5.4733E+00 5.2301E-01 -1.5329E+01 0.1077E+01 6.5837E+00 2.7250E+00 2.200 5.3844E+00 5.2240E-01 -1.5543E+01 0.1078E+01 6.7192E+00 2.9013E+00 2.400 5.2972E+00 5.4699E-01 -1.5605E+01 0.1190E+01 4.9983E+00 3.0942E+00 2.600 5.3179E+00 5.7289E-01 -1.5987E+01 0.1239E+01 4.3905E+00 3.2252E+00 2.800 5.3348E+00 5.8762E-01 -1.6217E+01 0.1258E+01 4.1537E+00 3.3638E+00 3.000 5.3246E+00 5.9586E-01 -1.6383E+01 0.1257E+01 4.0743E+00 3.5096E+00 3.200 5.3065E+00 6.1650E-01 -1.6474E+01 0.1250E+01 4.0802E+00 3.6618E+00 3.400 5.2733E+00 6.3374E-01 -1.6567E+01 0.1239E+01 4.1398E+00 3.8197E+00 3.600 5.2309E+00 6.4470E-01 -1.6663E+01 0.1226E+01 4.2364E+00 3.9827E+00 3.800 5.1752E+00 6.3603E-01 -1.6766E+01 0.1212E+01 4.3604E+00 4.1501E+00 4.000 5.1090E+00 6.0562E-01 -1.6883E+01 0.1198E+01 4.5058E+00 4.3214E+00 4.200 5.0331E+00 5.5941E-01 -1.6984E+01 0.1183E+01 4.6683E+00 4.4959E+00 4.400 4.9489E+00 4.9872E-01 -1.7115E+01 0.1168E+01 4.8452E+00 4.6734E+00 4.600 4.8588E+00 4.2625E-01 -1.7230E+01 0.1154E+01 5.0343E+00 4.8533E+00 4.800 4.7653E+00 3.5048E-01 -1.7327E+01 0.1139E+01 5.2341E+00 5.0355E+00 5.000 4.6691E+00 2.7395E-01 -1.7453E+01 0.1125E+01 5.4434E+00 5.2195E+00 5.200 4.5722E+00 1.9793E-01 -1.7590E+01 0.1110E+01 5.6610E+00 5.4051E+00 5.400 4.4760E+00 1.3243E-01 -1.7866E+01 0.1096E+01 5.8864E+00 5.5922E+00 5.600 4.3817E+00 8.0754E-02 -1.8387E+01 0.1083E+01 6.1187E+00 5.7805E+00 5.800 4.2890E+00 7.7722E-02 -1.9408E+01 0.1070E+01 6.3576E+00 5.9699E+00 6.000 4.1996E+00 1.2225E-01 -1.9952E+01 0.1058E+01 6.6024E+00 6.1603E+00 6.500 3.9822E+00 2.6906E-01 -2.0350E+01 0.1031E+01 7.2385E+00 6.6398E+00 7.000 3.7681E+00 3.8170E-01 -2.0473E+01 0.1010E+01 7.9052E+00 7.1233E+00 7.500 3.5495E+00 4.3899E-01 -2.0529E+01 0.9919E+00 8.5988E+00 7.6098E+00 8.000 3.3245E+00 4.4840E-01 -2.0549E+01 0.9763E+00 9.3169E+00 8.0987E+00 8.500
