Dears, there are already few posts about the doping, simulation of the doping and showing how to prepare the feff.inp file for doped material, but I would like to ask about the opposite situation for low doped materials - how to calculate the doping level from feff.inp file (from the list of ions). If there is already such problem discussed, then please excuse me for the discrepancy and send me the link (I couldn't find it). Just few examples: I construct the feff input file of the radius of 8A (FMS 8.0) and with the SCF of 7. I simulate the spectra with Jfeff9. Let it be Al doped ZnO - simulation on Al:K edge. 1) I investigate the change of the Al:K edge XANES spectra for the situation where I have Al in the centre and vary the distance to the next Al. At the beginning I have only 1 Al in the centre. The 8A radius consist 177 atoms: 86 Zn, 90 O and of course 1 Al. Can I simple say that my doping level in this situation is 1/177=~0.56 at% doping? But this could be also 1/88 if I will reduce the FMS to 6.2A, and the calculation will give the same result. Is any physical limitation to this calculation of doping level? 2) OK, lets add now another Al in the nearest Zn sphere - the XANES spectra changes, as I expected. The doping level doubled from 1/177 to 2/177=~1.13 at%. However, with the shifting of the second Al far away from the 1st Al, I do not observe the change of the XANES spectra, up to 4th sphere. With the 5th sphere (which is at ~6.5A) I can observe the tendency of the change of XANES spectrum toward this in point 1 (probably due to the limited size of the cluster). This means that with the change of the position of the second Al, I change the free means path from ~3.22A for 1st Zn sphere to ~6.14A for the 4th, this could be recalculated as the different doping level from ~5 at% to ~2 at%. Of course following the calculation from the point no. 1 we have in both cases ~1.1 at%... My impression is that I should think in the way of the free means path, but maybe I'm wrong? - I would like to know your attitude. Thanks and best regards kicaj