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@iit.edu http://phys.iit.edu/~segre segre@debian.org