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/