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 ----------------------------------- [EMAIL PROTECTED]
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/
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