"The difference in absorption between energies before the edge and

energies just after the edge is referred to as the edge jump (or edge step)

and is often written Δμx. In theory, the edge jump in transmission is

directly proportional to the thickness of the sample and also to the concentration

of the target element in the absorbing material. Because of

undesirable photons, this won’t be exactly true, particularly for thicker

samples." So far a small excerpt from "XAFS for everyone". According to Achkar (https://pdfs.semanticscholar.org/47ac/b9e4903c2cd27ce107e5c55d198861330e0e.pdf) charging and saturation effects can be responsible for a negative edge step.

Dear Kritika Shukla ,

or simply you may have accidently interchanged I (transmitted spectrum ) and Io (primary spectrum) when doing the normalization I/Io (=exp(-µ*d)) in the case of a transmission experiment.

For fluorescence and photo-electron yield you have I/Io ~ µ.

So pay attention, a negative edge jump mistankenly may easily show up.

But you may share what kind of XAFS you are dealing with.

Thanks in advance

and best regards

in addition to what Garhard Martens already has mentioned, take care that you do NOT logarithmize fluorescence (or electron yield) data I_F/I_0, but don't forget to do so in case of transmission mode data. This may also lead to negative edge jumps. I also agree with Gerhard - please provide your raw beamline data, so coming to a helpful conclusion is much easier! Kind regards, Dirk

Dear Dirk,

we should more clearly state that logarithmizing itself does not invert the step direction; but the (-) sign in the -ln operation will do this.

Dear Kritika Shukla,

next idea: you may have acquired your data in the preset count mode, i.e. you may have measured the time at each photon energy position for achieving for example 10000cts at each point.

Your data points in the spectrum then will have a unit in 'time/count'.

To convert into intensity ( i.e. to 'counts/time') you have to take the inverse of your data and then go into the equation(s) above. When omitting the inverse-operation you will get a 'negative' step at the edge jump...

@ Gerhard Martens , this is exactly what I meant ...

But the time/counts resp. counts per time argument may also apply - good idea! Best regards, Dirk