Hello Everyone,
Given the energy of an X photon probe, I would like to have a rough idea of the intensity of the X fluorescence lines produced by a desired element. I have found numerous tables showing the relative intensity within the same shell (eg Ka: Kb; La: Lb: Lg, etc.), but I can't find a value for different shell lines (eg K : L: M). My best guess is that it is not reported because it depends on the energy of the incident photon itself. But is there a formula to have at least an indicative idea?
Example:
If I have a sample of silver and I illuminate it with monochromatic 50 keV photons I know that the Ka1 line is about 54% compared to the other K emissions and the La1 line is about 80% compared to the other L emissions (values obtained from Hephaestus software), but what is the relationship between Ka1 and La1, or between all K and all L? If the photon, on the other hand, has an energy of 100 keV, do the ratios vary? if so, how?
Thank you.
Dear Walter Giurlani ,
we know that the x-ray fluorescence phenomenon is the relaxation reaction of a core hole excitation.
Once a core hole is created it is filled up by electrons from upper shells. The energy, which is released, escapes from the atoms by x-ray fluorescence photons or by Auger electrons. The set of K fluorescence lines (or L lines) showing up is characteristic for the type of the core shell excitation such as K- or L-shell(s). The photon energies of these lines and their relative intensities are characteristic for each type of atom, thus depending on the atomic number.
These relative intensities originating from the excitation of the same shells (e.g. K-alpha/K-beta or L-alpha/L-beta) do not depend on
a) how the core hole is created (e.g. by x-rays, electrons or protons etc.) and
b) do not depend on the energy of the exciting x-ray photon; the only requirement here is, that the x-ray energy has to be larger than the K- or L-excitation threshold (known as K-edge or L-edge energies).
The absolute intensity of the lines depends (among other parameters) on the excitation propability or better named 'the core hole excitation cross-section' of the atoms. This cross-section increases as the photo electric contribution to the x-ray attenuation coefficient increases.
It turns out, that the (local) fluorescence intensity (or better: yield) is propotional to the K-or L shell photo electric absorption contribution part µK or µLof the x-ray attenuation coefficient µ.
In the appendix I have attached a sketch, showing up, for the case of Ag, the energy dependence of the x-ray mass-attenuation coefficient µ/rho.
In the low energy regime below (Ag) K-edge energy, the attenuation coefficient is mainly given by energy dependence of the L-shell photo electric absorption µL(E). The fluorescence yield (line intensity) will go down as the photon energy increases. But the relative L-line distribution will remain.
But when passing the K-edge energy (Ag K-edge at 25,5keV) the K-edge raises up due to an additional µK-contribution, which is the photo electric absorption contribution of the K-shell. µK(E) also shows a decreasing behavior with increasing photon energy.
In the case of excitations of Ag in this energy regime, L- as well as K- fluorescence lines will show up. Their K-line to K-line or L-line to L-line relative contributions still remain, but their absolute values (yields) will go down as µK and µL decrease for increasing photon energy. The decreasing behaviors of µK and µL are similar, but unfortunately via a ratio µK/µL (or vice versa) any energy dependence will not be removed in the ratio; so still an energy dependence will remain*). Thus any combined ratio of K and L intensities (yields), which is proportional to µK/µL (or vice versa), will be also still energy dependent.
Up to now I am not aware on any simple formula for the calculation of µK(E) and µL(E) nor any database, which provides such data (not even NIST/XCOM)). µ calculations are done via quantum mechanics approaches.
For energies around the edge one may extrapolate the µL(E) dependence from below the K-edge to a short section above the K-edge by, for example, a so-called Victoreen fit (µ = a*E-3+b*E-4) in the energy regime below the edge energy. Such extrapolations have been (and still are) successfully used in XANES and EXAFS applications, which extend up to a very few 100 eV above edge energy; but in your case 25 keV above Ag K-edge and 75 keV above, the extrapolations will be too extremely stressed.
Sorry for that conclusion...
*) you may divide for example two different Victoreen formulae, mentioned in the upcoming description, and you will see that the energy E is not cancelled. Only in the limit for E = infinity the ratio µK/µL will approach a constant value. However for E= 50keV or 100keV, as mentioned above, you are far from infinity. 50 keV is only twice the K-edge energy of Ag ...
Best regards
G.M.
Dear Walter Giurlani ,
I just found a review article dealing with the K, L and M x-ray fluorescence yields and fits on measured data:
https://www.nist.gov/system/files/documents/2018/03/26/jpcrd473.pdf
But it is tough to read...
Good luck and
best regards
G.M.
Dear Gerhard Martens ,
Thank you very much for your detailed answer. It was very useful, it clarified my doubts and confirmed some of the hypotheses I had. The article you have attached is very interesting even if a bit complex, I will read it carefully.
Thanks again
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