Probability of fluorescence: matching of binding energy and incoming radiation energy?

Assume an X-ray diffractometer equipped with a copper anode X-ray tube. When a sample containing cobalt, iron, or manganese is irradiated by copper's K$\alpha_1$ radiation, sample fluorescence becomes significant.

I believe I read somewhere that when the incoming radiation's energy comes closer to the binding energy of the electrons, then the probability of fluorescence increases. Is this correct?

And also, can an inner-core electron's binding energy be approximated by the element's K$\alpha_1$ energy?

• Always beware of "I [read,heard,saw] somewhere..." . Thats a recipe for disaster. So, hint: what causes fluorescence? If you learn about electron (or nuclear) absorption and emission of photons, you'll get an idea of why the binding energy matters. – Carl Witthoft Jan 29 '15 at 15:33
• Disregarding Stoke's shift (and possible interactions within the electron cloud) I intuitively understand that the fluorescence energy should equal the binding energy. But since I do not have any proper citations on that, I was second-guessing, I suppose. Obviously, if the incoming radiation is lower than the binding energy, ionization will not occur. But why the probability for ionization (and hence fluorescence) should be greater when the incoming radiation's energy is close to the binding energy, rather than much greater than the binding energy, I do not know. I hope someone can help me. – Yoda Jan 29 '15 at 15:38
• The binding energies are given by the "K edge", "L edge", etc. – Steve Byrnes Feb 4 '15 at 12:53
• Fluorescence is a two-step process. First a deep core (e.g., 1s) electron is kicked out by a high-energy incident x-ray. Second, a higher energy (e.g., 2p) electron "falls down" to fill the hold left in the deep core 1s. This process of "falling down" in energy requires emission of a photon and that photon is the fluorescence photon. See my answer below. – hft Feb 7 '15 at 21:03

I believe this relates to density of states and the shape of the bands.

When you try to excite an electron from one state to another, there needs to be an available state "upon arrival" - and the probability of the transition is therefore directly related to the density of states. If I'm not mistaken, this is the reason that the excitation is strongest when you hit the exact frequency of the gap.

As for your second question - the $K\alpha$ line is emitted when an electron drops from 2p to the 1s orbital. This is less than the binding energy of the electron, since the 2p state is still a bound state and therefore has a lot of associated binding energy.

• Just to clarify, I am not talking about exciting K electrons into the L-shell, but fluorescence after ejection of K-shell electrons. In bullet point four of this article (icdd.com/resources/axa/vol47/V47_32.pdf), it is said that fluorescence increases as the radiation's energy is close to the fluorescent energy; that is, close to K$\alpha$. Your last paragraph is kind of obvious when I think about it - why didn't I realize that! – Yoda Feb 3 '15 at 6:54

[...]when the incoming radiation's energy comes closer to the binding energy of the electrons, then the probability of fluorescence increases. Is this correct?

When the incident energy is close to the binding energy there is a sharp increase in the absorption. See, for example, Figure 1-5 in the following document: http://xdb.lbl.gov/Section1/Sec_1-6.pdf, especially the bottom panel that shows a nice sharp peak for Carbon around 300eV. Compare the above figure with the following table of electron binding energies: http://xdb.lbl.gov/Section1/Table_1-1a.htm. Because electrons must first be absorbed before fluorescence can occur, it is reasonable to say that the probability of fluorescence increases when the incident radiation is near the binding energy.

And also, can an inner-core electron's binding energy be approximated by the element's Kα1 energy?

Compare this table of $K\alpha_1$ energies: http://xdb.lbl.gov/Section1/Table_1-3.pdf, with the table of electron binding energies given above. In general you will find the X-ray Data Booklet (http://xdb.lbl.gov) very useful for this sort of work (you can even get a free hard copy). For example, this section (http://xdb.lbl.gov/Section1/Sec_1-2.html) explains emission lines. In the referenced section the symbols "$L_2$" and "$L_3$" mean the same thing as "$2p$" (respectively, $2p_{1/2}$ and $2p_{3/2}$, where the 1/2 and 3/2 refer to spin-orbit splitting), and the symbol "$K_1$" means the same thing as "1s".

So, the short answer is: No, they are not the same, but are similar for light atoms. The $K\alpha_1$ energy is the energy difference between the $2p_{3/2}$ and $1s$ levels.