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When analyzing manganese-rich samples with X-ray powder diffraction, sample fluorescence becomes a significant effect. Is this because the incident X-rays' energy - which, if Cu K$\alpha_1$, is approximately 8.048 keV - is getting close to the K$\alpha_1$ (approximate binding energy) of the inner-most electrons of Mn (5.899 keV)?

Sample fluorescence from Fe and Co is even stronger, which fits with the trend mentioned above. So, is this indeed a trend? If not, how do I generalize the increase in probability for sample fluorescence?

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  • $\begingroup$ With organic molecules (which is what I'm familiar with), the fluorescence of the molecule is a property. The amount of fluorescence is used to determine the concentration as you would use UV-Vis absorbences. So, other than using a brighter source, I don't think you can increase fluorescence. $\endgroup$ – LDC3 Feb 1 '15 at 15:56
  • $\begingroup$ I'm voting to close this question as off-topic because it's more physics than chemistry. $\endgroup$ – M.A.R. ಠ_ಠ Feb 1 '15 at 16:27
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    $\begingroup$ Like @LDC3, I'm familiar with fluorescence of organic compounds, rather than with XRF. If I remember correctly, XRF is not described by excitation and radiative deactivation. Instead, it results from the ionization of the sample from low-lying orbitals and subsequent electron transfer, upon which excess energy is released as photons. Consequently, a good match of incoming radiation with the relevant orbitals (spectral overlap) should result in fluorescence and more ionization events yield stronger emission. $\endgroup$ – Klaus-Dieter Warzecha Feb 1 '15 at 17:23
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    $\begingroup$ MARamezani: The line between physics and chemistry is floating. I am a chemist, and the question is in an analytical chemical context. Anyway, I understand your reaction. @KlausWarzecha, yes, you are correct that fluorescence occurs when the inner-most electrons are ejected by EM irradiation. My understanding of the increase in probability is similar to yours, but I was not sure, and could not find any references on it earlier. I did find an article by M. Fransen (2004) which actually says the same thing (page 224, fourth bullet point). icdd.com/resources/axa/vol47/V47_32.pdf $\endgroup$ – Yoda Feb 1 '15 at 22:04
  • $\begingroup$ I did XRF chemical analysis for years, so this is a chemistry question. // What you really need to solve the problem is a different x-ray tube. The K-alpha line of Cr (5.4 KeV) would not excite K lines of Mn (6.5374 KeV) or Fe (7.1109 KeV). $\endgroup$ – MaxW Feb 22 '17 at 18:18
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Short answer, yes.

As you note, the fluorescence signal is strongest when the exciting photon has an energy close to the excitation energy for the element absorbing the photon. There are actually two factors coming into play:

1) The closer the resonance of the incident photon with the electronic orbital it is exciting, the higher the cross-section to absorb the photon. (See here.)

2) The higher-Z the absorbing atom, the more likely it is that once excited, the atom will emit a fluorescence photon (as opposed to an Auger electron). That is, it has a higher fluorescence yield. (See here.)

In your case, the Z of the atoms are pretty close, but also in portion of the periodic table where the fluorescence yield is rapidly changing. The chance of a fluorescence photon being emitted once excited is about 20% greater for Co than Mn (and you can look it up in the literature).

The cross-section for absorption as a function of energy for these three elements looks like this (code for the plot here):

enter image description here

I also drew a vertical line for the exciting Cu-K$\alpha$ photon at 8048 eV, and you can see that the Co curve has the highest cross-section at about 316 cm$^2$/g, and Mn has the lowest at about 270 cm$^2$/g, which means the Co cross-section is about 17% higher than the Mn. The primary reason for the reduction in the sensitivity for the Mn edge is that the absorption cross-section drops off as $\approx\Delta E^3$. These data were plotted using the data files described by B. L. Henke, E. M. Gullikson, and J. C. Davis, Atomic Data and Nuclear Data Tables Vol. 54 No. 2 (July 1993)

So both these effects are colluding to increase your fluorescence for Co vs Mn.

As a note, if you want to reduce this effect in your X-ray diffraction patterns, you can change your source target material with a heavier element such as Zr which has it's K$\alpha$ up at 15.7 keV, yet the L lines are still around 2 keV (below the transition metal K-edges) or you can use a lighter target, such as Mn. Mn, of course, would not be able to excite its K-edge using its own K$\alpha$ fluorescence photon.

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  • $\begingroup$ I note a duplicate with physics.stackexchange.com/questions/162224/… If closing in favor of the physics.SE question, then I can transfer the answer there. $\endgroup$ – ZSG Feb 3 '15 at 8:47
  • $\begingroup$ I run into this effect in my research and I get around it by putting a thin piece of nickel foil (~50um thick) in front of the detector. It doesn't eliminate it, but does reduce it enough to make the pattern usable. $\endgroup$ – m3wolf Jan 12 '17 at 7:19
  • $\begingroup$ Nice answer, but I'd point out a couple of additional things. (1) the question is out of whack. The "probability of fluorescence" is normally taken to mean the fluorescence yield not the resultant x-ray intensity. (2) Once an atom is excited, it is excited. So the energy of the incident x-ray which created the excited state does not change the fluorescence yield itself. $\endgroup$ – MaxW Feb 22 '17 at 18:28

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