Re: [Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient
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.
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.
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?
Thanks,
Damon
On Thu, Jul 25, 2013 at 5:09 PM,
Send Ifeffit mailing list submissions to ifeffit@millenia.cars.aps.anl.gov
To subscribe or unsubscribe via the World Wide Web, visit http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit or, via email, send a message with subject or body 'help' to ifeffit-request@millenia.cars.aps.anl.gov
You can reach the person managing the list at ifeffit-owner@millenia.cars.aps.anl.gov
When replying, please edit your Subject line so it is more specific than "Re: Contents of Ifeffit digest..."
Today's Topics:
1. Is good EXAFS data possible with 0.01 transmission coefficient (Damon Turney) 2. Re: Is good EXAFS data possible with 0.01 transmission coefficient (Matthew Marcus) 3. Re: Is good EXAFS data possible with 0.01 transmission coefficient (Scott Calvin) 4. Re: Is good EXAFS data possible with 0.01 transmission coefficient (Scott Calvin) 5. Re: Is good EXAFS data possible with 0.01 transmission coefficient (Matthew Marcus)
----------------------------------------------------------------------
Message: 1 Date: Thu, 25 Jul 2013 16:07:57 -0400 From: Damon Turney
To: ifeffit@millenia.cars.aps.anl.gov Subject: [Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient Message-ID: Content-Type: text/plain; charset=ISO-8859-1 Hi IFEFFIT community,
I will make measurements soon at Brookhaven's NSLS to do EXAFS, and my sample is a powder mixture of carbon powder, MnO2 powder, and potassium hydroxide liquid. The transmission coefficient of 6.5keV x-rays (the Mn K edge is at ~6.5keV) through the sample is ~0.001 to 0.01. I am told that the ionization chamber detectors can easily detect the x-ray beam after this ~0.01 transmission loss (by increasing the gain on the detector), but I would like to ask the IFEFFIT community if there will be other problems with the EXAFS technique when the transmission coefficient is so low. Does anybody have comments?
Much thanks -- if you have any info I greatly appreciate it! Damon Turney City College of New York
------------------------------
Message: 2 Date: Thu, 25 Jul 2013 13:33:12 -0700 From: Matthew Marcus
To: XAFS Analysis using Ifeffit Subject: Re: [Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient Message-ID: <51F18B88.5020600@lbl.gov> Content-Type: text/plain; charset=ISO-8859-1; format=flowed Is most of the transmission loss due to the Mn or the matrix (C, KOH, etc)? Is that 0.01 above or below the edge? If most of the absorption is due to the matrix with the Mn providing an edge jump of <1, then I'd say that you have a shot. However, you become very sensitive to harmonics and glitches. If the Mn provides an edge jump of much less than 1, them fluorescence would be the way to go, if possible. With thick samples in transmission, you have an issue with hole effect. For instance, it's possible that your 1% transmission comes entirely from 1% area of pinholes, with the actual material being almost completely opaque. If that happens, you get a reduced edge jump and reduced EXAFS amplitude, with distortions resembling overabsorption.
If you're in the thick-matrix scenario, a possible workaround would be to make a sample which contains only matrix and is as nearly as possible exactly as thick as the real sample, and measure its transmission, and subtract. mam
On 7/25/2013 1:07 PM, Damon Turney wrote:
Hi IFEFFIT community,
I will make measurements soon at Brookhaven's NSLS to do EXAFS, and my sample is a powder mixture of carbon powder, MnO2 powder, and potassium hydroxide liquid. The transmission coefficient of 6.5keV x-rays (the Mn K edge is at ~6.5keV) through the sample is ~0.001 to 0.01. I am told that the ionization chamber detectors can easily detect the x-ray beam after this ~0.01 transmission loss (by increasing the gain on the detector), but I would like to ask the IFEFFIT community if there will be other problems with the EXAFS technique when the transmission coefficient is so low. Does anybody have comments?
Much thanks -- if you have any info I greatly appreciate it! Damon Turney City College of New York _______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
------------------------------
Message: 3 Date: Thu, 25 Jul 2013 16:33:40 -0400 From: Scott Calvin
To: XAFS Analysis using Ifeffit Subject: Re: [Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient Message-ID: <6A1D5E8E-88FB-4A0B-9CCC-9AF0CDB6623F@slc.edu> Content-Type: text/plain; charset="us-ascii" Hi Damon,
So you're saying you expect the total absorption of the sample to be less than 0.01 absorption lengths? In other words, if the detectors and electronics were identical, you would expect It = I0 exp(-0.01)?
While possible to measure in transmission, such a sample would normally be measured in fluorescence. If done that way, the measurement is routine.
Or am I misunderstanding your description? (I am not entirely sure what you mean by "transmission coefficient.")
--Scott Calvin Sarah Lawrence College
On Jul 25, 2013, at 4:07 PM, Damon Turney
wrote: Hi IFEFFIT community,
I will make measurements soon at Brookhaven's NSLS to do EXAFS, and my sample is a powder mixture of carbon powder, MnO2 powder, and potassium hydroxide liquid. The transmission coefficient of 6.5keV x-rays (the Mn K edge is at ~6.5keV) through the sample is ~0.001 to 0.01. I am told that the ionization chamber detectors can easily detect the x-ray beam after this ~0.01 transmission loss (by increasing the gain on the detector), but I would like to ask the IFEFFIT community if there will be other problems with the EXAFS technique when the transmission coefficient is so low. Does anybody have comments?
Much thanks -- if you have any info I greatly appreciate it! Damon Turney City College of New York _______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
------------------------------
Message: 4 Date: Thu, 25 Jul 2013 17:03:53 -0400 From: Scott Calvin
To: XAFS Analysis using Ifeffit Subject: Re: [Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient Message-ID: <0830DFDB-C085-45F5-B434-B79DF12ABD72@slc.edu> Content-Type: text/plain; charset="us-ascii" Oops--Matthew's interpretation is doubtless what you actually meant--i.e., that It < 0.01 I0. Therefore disregard my previous comment!
And I concur with Matthew's comments, now that I understand the circumstances.
--Scott Calvin Sarah Lawrence College
On Jul 25, 2013, at 4:33 PM, Matthew Marcus
wrote: Is most of the transmission loss due to the Mn or the matrix (C, KOH, etc)? Is that 0.01 above or below the edge? If most of the absorption is due to the matrix with the Mn providing an edge jump of <1, then I'd say that you have a shot. However, you become very sensitive to harmonics and glitches. If the Mn provides an edge jump of much less than 1, them fluorescence would be the way to go, if possible. With thick samples in transmission, you have an issue with hole effect. For instance, it's possible that your 1% transmission comes entirely from 1% area of pinholes, with the actual material being almost completely opaque. If that happens, you get a reduced edge jump and reduced EXAFS amplitude, with distortions resembling overabsorption.
If you're in the thick-matrix scenario, a possible workaround would be to make a sample which contains only matrix and is as nearly as possible exactly as thick as the real sample, and measure its transmission, and subtract. mam
On 7/25/2013 1:07 PM, Damon Turney wrote:
Hi IFEFFIT community,
I will make measurements soon at Brookhaven's NSLS to do EXAFS, and my sample is a powder mixture of carbon powder, MnO2 powder, and potassium hydroxide liquid. The transmission coefficient of 6.5keV x-rays (the Mn K edge is at ~6.5keV) through the sample is ~0.001 to 0.01. I am told that the ionization chamber detectors can easily detect the x-ray beam after this ~0.01 transmission loss (by increasing the gain on the detector), but I would like to ask the IFEFFIT community if there will be other problems with the EXAFS technique when the transmission coefficient is so low. Does anybody have comments?
Much thanks -- if you have any info I greatly appreciate it! Damon Turney City College of New York _______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
_______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
------------------------------
Message: 5 Date: Thu, 25 Jul 2013 14:09:49 -0700 From: Matthew Marcus
To: XAFS Analysis using Ifeffit Subject: Re: [Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient Message-ID: <51F1941D.3010507@lbl.gov> Content-Type: text/plain; charset=ISO-8859-1; format=flowed I'm pretty sure he has the opposite problem - the transmission is <0.01. That's assuming that "transmission coefficient" == "transmission". mam
On 7/25/2013 1:33 PM, Scott Calvin wrote:
Hi Damon,
So you're saying you expect the total absorption of the sample to be less than 0.01 absorption lengths? In other words, if the detectors and electronics were identical, you would expect It = I0 exp(-0.01)?
While possible to measure in transmission, such a sample would normally be measured in fluorescence. If done that way, the measurement is routine.
Or am I misunderstanding your description? (I am not entirely sure what you mean by "transmission coefficient.")
--Scott Calvin Sarah Lawrence College
On Jul 25, 2013, at 4:07 PM, Damon Turney
wrote: Hi IFEFFIT community,
I will make measurements soon at Brookhaven's NSLS to do EXAFS, and my sample is a powder mixture of carbon powder, MnO2 powder, and potassium hydroxide liquid. The transmission coefficient of 6.5keV x-rays (the Mn K edge is at ~6.5keV) through the sample is ~0.001 to 0.01. I am told that the ionization chamber detectors can easily detect the x-ray beam after this ~0.01 transmission loss (by increasing the gain on the detector), but I would like to ask the IFEFFIT community if there will be other problems with the EXAFS technique when the transmission coefficient is so low. Does anybody have comments?
Much thanks -- if you have any info I greatly appreciate it! Damon Turney City College of New York _______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
_______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
------------------------------
_______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
End of Ifeffit Digest, Vol 125, Issue 14 ****************************************
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
Just to amplify a bit on Carlo's point: The transmitted signal is the sum of the transmissions through different areas of the sample, thus
T/I0 = < exp(-mu(E)*t) >
where <...> is the spatial average. However, what you measure is
tmu_eff(E) = ln(I0/T) = ln
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
On 07/25/2013 08:02 PM, Matthew Marcus wrote:
Thus, a spectrum which should look like this * * * * * * * * * * ********
may wind up looking like this:
* * * ** * * ********
I want to nominate Matthew for the Oscar in best ASCII Art Design! That was awesome! B -- Bruce Ravel ------------------------------------ bravel@bnl.gov National Institute of Standards and Technology Synchrotron Science Group at NSLS --- Beamlines U7A, X24A, X23A2 Building 535A Upton NY, 11973 Homepage: http://xafs.org/BruceRavel Software: https://github.com/bruceravel
participants (4)
-
Bruce Ravel
-
Carlo Segre
-
Damon Turney
-
Matthew Marcus