[Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient

[Carlo Segre]

Hi Damon:

On Thu, 25 Jul 2013, Damon Turney wrote:

> Thank you Matthew Marcus.  To answer your questions: Most of the
> transmission loss is due to the Mn particulates.  The 0.01
> transmission coefficient (a.k.a. ~4 absorption lengths) is in regards
> to xray energies above the absorption edge.   Below the absorption
> edge the transmission coefficient should be 0.1 to 0.4.  I don't
> understand the definition of "edge jump" or how to calculate it (but
> trust me I will be reading up on all of this ASAP).  Anyway, the
> transmission coefficient contains all the required information.
>

The edge jump is basically the change in the absorption from just below 
the edge to the back-extrapolated baseline after the edge.  in your case, 
if the absorption is 0.4 below the edge and 0.001 above the edge (worst 
case, then assumint the variation of the absorption is all due to the Mn, 
you would have an edge jump of approximately

       delta mu ~ ln(1/0.001) - ln(1/0.4) = 6.91 - 0.91 = 5.0

This is  very large and would not work well for transmission at all 
because of the huge variation in intensity through the edge.  If you have 
your best case it would be

       delta mu ~ ln(1/0.01) - ln(1/0.1) = 4.6 - 2.3 = 2.3

This is acceptable even if not ideal for transmission.

> I doubt that I have "pinhole effect" because the MnO2 is small
> particulates (10 micron diameter) evenly dispersed throughout the 300
> micron thick sample, so I doubt there are any pinholes.
>

The absorption length of 6.6keV photons in MnO2 is 6.9 microns so there is 
a good chance that your particles, even though they are relatively small 
will distort the spectrum because they are significantly bigger than one 
absorption length.  This could also cause some problems in fluorescence. 
however, if this is your particle size and it is constrained due to your 
experiment, you will have to work with it.

> Matthew are these "distortions" that you speak of the same thing as
> "harmonics" that I've read about in "XAFS for Everyone" that can
> complicate XAFS analysis?
>

The distortions come because you have to take a ratio between the incident 
beam and the transmitted beam or the fluorescence signal and the incident 
beam intensity.   There are specific assumptions made in taking this ratio 
that, if not true, will distort your spectrum from what it really is. For 
transmission, the assumption is that any photon coming through the Io 
chamber has an equa probability of being absorbed in the sample.  If oyu 
have harmonice in teh beam (that is in stead of just having E, you have a 
component of photons with 3E or 5E), then this assumption is not valid. 
normally, when you have a relativley optically thin sample (not too much 
absorption, this is a good assumption.  When you have a highly absprbing 
sample, the harmonics will be the perdominant signal in the transmission 
chanber and thus you will ahve a problem with the ratio.

For fluorescence, there are other assumptions which when violated will 
distort the spectrum.

Carlo

-- 
Carlo U. Segre -- Duchossois Leadership Professor of Physics
Director, Center for Synchrotron Radiation Research and Instrumentation
Illinois Institute of Technology
Voice: 312.567.3498            Fax: 312.567.3494
segre at iit.edu   http://phys.iit.edu/~segre   segre at debian.org