Hi Herve,
I haven’t looked at your project file, but the 1.80 A peak is not that big a surprise, as peaks in the Fourier transform typically show up 0.3 to 0.5 angstroms lower than the corresponding half-path length, due to the effects of the potential of the absorbing and scattering atoms. In other words, the Fourier transform is not a radial distribution function, although it’s correlated to one.
The interaction between the two similar path lengths further complicates peak location, since it can establish a “beating” between the two signals.
Under the circumstances, a peak at 1.80 angstrom is not remarkable.
The peak at 1.28 angstrom, on the other hand, is well in to the background region for this compound, and does not need to correspond to a path.
The fact that the Fourier transform is not a radial distribution function is why some sort of comparison to a standard is necessary. For Ifeffit users, that typically means fitting to a so-called “theoretical standard,” that is, a computed spectrum, using Artemis. It could also mean comparison to a well-chosen series of measured materials (“empirical standards”). But it is not possible to just look at the Fourier transform and read off bond lengths.
—Scott Calvin
Sarah Lawrence College
On Aug 14, 2015, at 4:41 AM, herve muguerra