phase shift correction
Could you tell me... Why the peak in the Fourier transform of the experimental spectrum is less than the actual one. Why it is required to include the scattering phase shift correction in the experimental one? As it is expected that the experimental one should show the actual bond length. Bindu Dr.Bindu R. Visiting Scientist DCMP&MS Tata Institute of Fundamental Research Homi Bhabha Road Colaba Mumbai-400 005 India Mobile-919892536830 --------------------------------- Forgot the famous last words? Access your message archive online. Click here.
Hi Bindu,
Why the peak in the Fourier transform of the experimental spectrum is less than the actual one.
The EXAFS equation is: N S02 F(K) chi(k) = ----------- exp(-2k^2 sigma^2) sin(2kR + phi(k)) 2kR^2 In the oscillatory term, there is a piece that is the phase shift associated with the scattering of the photoelectron. When you do a Fourier transform, the peak of the sin wave is not at R, rather it is shifted by an amount that depends on the size of phi(k).
Why it is required to include the scattering phase shift correction in the experimental one? As it is expected that the experimental one should show the actual bond length.
I think that what I wrote above answers this question as well. B -- Bruce Ravel ----------------------------------- bravel@bnl.gov National Institute of Standards and Technology Synchrotron Methods Group at Brookhaven National Laboratory Building 535A Upton NY, 11973 My homepage: http://cars9.uchicago.edu/~ravel EXAFS software: http://cars9.uchicago.edu/~ravel/software/exafs/
Hi Bindu: On Sun, 2 Dec 2007, Ravel, Bruce wrote:
The EXAFS equation is:
N S02 F(K) chi(k) = ----------- exp(-2k^2 sigma^2) sin(2kR + phi(k)) 2kR^2
In the oscillatory term, there is a piece that is the phase shift associated with the scattering of the photoelectron. When you do a Fourier transform, the peak of the sin wave is not at R, rather it is shifted by an amount that depends on the size of phi(k).
In addition, because this is a sum of sinusoids with different phases, will not necessarily result in a single, distinct peak. Interference effects are quite common. Carlo -- Carlo U. Segre -- Professor of Physics Associate Dean for Special Projects, Graduate College Illinois Institute of Technology Voice: 312.567.3498 Fax: 312.567.3494 segre@iit.edu http://www.iit.edu/~segre segre@debian.org
Hi Bindu,
Why the peak in the Fourier transform of the experimental spectrum is less than the actual one.
The EXAFS equation is:
N S02 F(K) chi(k) = ----------- exp(-2k^2 sigma^2) sin(2kR + phi(k)) 2kR^2
In the oscillatory term, there is a piece that is the phase shift associated with the scattering of the photoelectron. When you do a Fourier transform, the peak of the sin wave is not at R, rather it is shifted by an amount that depends on the size of phi(k).
That should be, of course, "...the peak of the *Fourier transform* of the sine wave..." B -- Bruce Ravel ----------------------------------- bravel@bnl.gov National Institute of Standards and Technology Synchrotron Methods Group at Brookhaven National Laboratory Building 535A Upton NY, 11973 My homepage: http://cars9.uchicago.edu/~ravel EXAFS software: http://cars9.uchicago.edu/~ravel/software/exafs/
Hi all, Could someone please answer my questions? I would really appreciate your help. 1. For linear combination fitting, there are three indicators for the goodness of fitting: R-factor, chi-square and reduced chi-square. Could anyone tell me how they work? 2. Since TEY is sensitive for the surface and FY for the bulk (and surface?), species detected by TEY should be also detected by FY, right? 3. How to calculate the maximum analysis depths for TEY and FY? Thank you in advance. Jenny Cai
Hi Jenny,
1. For linear combination fitting, there are three indicators for the goodness of fitting: R-factor, chi-square and reduced chi-square. Could anyone tell me how they work?
This is actually documented in Athena, and the Users Guide. They are also defined in the Feffit documentation. In short, they are all scaled measures of Sum [(data-fit)^2] R-factor scales this my the data values, and chi-square scales by an estimate of the noise in the data. Reduced chi-square relates to chi-square through the usual statistical definition, in that it is chi-square / (number of free variables in the fit). Of course, chi-square requires one to know the uncertainty in the data -- generally we don't have a great estimate of this. I mean no offense of this, but if you're asking about these then you almost certainly haven't put in an estimate of the uncertainty. So chi-square is probably scaled incorrectly. On top of that, reduced chi-square needs to know the number of independent measurements. Normally one assumes each datum to be independent. This is arguable, but it we can make that assumption for now. But if chi-square is scaled poorly, so is reduced chi-square. If that's too vague, or I misunderstood the question, please ask again.
2. Since TEY is sensitive for the surface and FY for the bulk (and surface?), species detected by TEY should be also detected by FY, right?
Yes, but TEY samples a much smaller volume of material than FY, so the signal from the volume seen by TEY (that is, the surface) may be insignificant compared to the signal from the volume seen by FY.
3. How to calculate the maximum analysis depths for TEY and FY?
Google/Wikipedia might help here. The sampling depth for TEY is typically dominated by the mean-free-path for the Auger electrons, which is in the range of 20 - 50 Angstroms. Sampling depths for FY are typically set by the absorption length of fluorescence x-rays, which is in the range of 2 to 50 microns (yes, a much more variable range, depending on sample composition). In both cases, you'd need to calculate the depth that the x-ray beam penetrates the sample (depends strongly on matrix) too. In my experience, it's unusual for this to dominate the sampling depth, but it can be significant for FY in high-Z matrices. --Matt
Hi Jenny, I just wanted to add a bit to Matt's answer on TEY vs FY and suggest a couple of reference for you. There are three contributions to the TEY: photoelectrons, Auger Electrons, and secondary electrons. The secondary electrons will typically determine the maximum depth that you can observe in a TEY measurement. There is a good description of this in Jo Stohr's book, NEXAFS Spectroscopy which can be seen at google books. http://tiny.cc/44VGq There is also a quick synopsis of the book at: http://www-ssrl.slac.stanford.edu/stohr/nexafs.htm At this site, there is a photoelectron spectrum showing the relative intensity of Auger and Photopeaks. The secondaries are roughly a factor of 100 more intense than the other contributions. It is possible to measure both TEY and Auger electron yield (AEY) and you will see that the AEY is more surface sensitive that TEY. Measuring AEY requires an analyzer capable of discriminating the electron energy. AEY is typically performed at lower photon energies as there are not too many electron energy analyzers at hard x-ray facilities. It is actually pretty difficult to try to calculate the maximum analysis depth for TEY because of the three different contributions. Here is an attempt to understand the probe depth of TEY but neglects secondary electrons: http://www.icdd.com/resources/axa/vol44/v44_058.pdf For the K-edges in the hard x-ray regime, with TEY it would not be unrealistic to expect probe depths on the order of 1000 angstroms (100 nm). Jeff On Feb 20, 2008, at 6:33 PM, Jenny Cai wrote:
Hi all,
Could someone please answer my questions? I would really appreciate your help.
1. For linear combination fitting, there are three indicators for the goodness of fitting: R-factor, chi-square and reduced chi- square. Could anyone tell me how they work?
2. Since TEY is sensitive for the surface and FY for the bulk (and surface?), species detected by TEY should be also detected by FY, right?
3. How to calculate the maximum analysis depths for TEY and FY?
Thank you in advance.
Jenny Cai
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Hi Jeff, Thanks -- you're right. I was only considering Auger electrons which are more surface sensitive than the secondary electrons that actually dominate most TEY measurements. --Matt
Hi Matt and Jeff,
Thank you for the answer.
-Jenny
Matt Newville
participants (6)
-
Bindu R. -
Carlo Segre -
Jeff Terry -
Jenny Cai -
Matt Newville -
Ravel, Bruce