# Characteristic X-Rays and Energy Bands in Solids

1. Aug 6, 2013

### Alexander83

Hi there,
I'm an instructor teaching a Physics course for Radiography students designed for students with little to no Physics background. In prepping the course, I've come up with a couple of questions that are bothering me and hope someone here can be of assistance.

My questions pertain to x-ray production in radiography which is generally accomplished by firing beams of electrons at a metal target, generally made of Tungsten or an alloy of Tungsten. The resulting spectrum of x-rays is then described in terms of Bremsstrahlung and Characteristic x-ray production mechanisms.

Any Radiography Physics texts I've come across describe Characteristic x-ray production by invoking essentially the Bohr model of the atom, wherein x-rays are produced by an electron in a higher shell dropping down to a lower shell in the atom. It is claimed that this model explains why characteristic x-rays show discrete emission lines in the spectrum, with energies defined by the differences in the electron binding energy between the shells.

My belief is that this simplified explanation would be strictly only true for isolated target atoms, not for x-ray emission from the solid anode consisting of a large number of atoms. I've always qualitatively thought from my (very limited) understanding of solid state physics, that electron energy levels in a solid metal were better understood in terms of nearly continuous energy bands. My question then is how to reconcile the simple Bohr model description of characteristic x-rays with the presumably more complicated situation in a solid metal. Here are my questions:

1. In an actual x-ray spectrum from a metal target, characteristic radiation production is visible as a set of distinct peaks. Does each peak correspond to a photon of a single wavelength (i.e. are the peaks monoenergetic?), or does each peak correspond to a narrow range of energies (corresponding to transitions from a range of energy levels between bands), so that the peak appears to be monoenergetic without truly being so? My feeling is that it must be the latter, but I can't find a good reference source that addresses this

2. In looking up x-ray spectra data, characteristic line data are often provided for elements. I assume that these must be for isolated atoms of elements and not a large collection of atoms in a crystal lattice?

3. Is there a good reference paper or text that addresses some of these points that doesn't require extensive knowledge of solid state physics to parse?

Chris.

2. Aug 7, 2013

### DrDu

You are right, the peaks aren't monoenergetic but have a finite width. The various electronic bands (valence and conduction) have typically widths in the range of some electron volts which is by a factor 1000 smaller than the energy of the lines in the X-ray spectrum. So the variation is quite negligible.

3. Aug 7, 2013

### daveyrocket

Well I'm no expert on x-ray experiments. I know a bit more about x-ray absorption spectroscopy, which is a bit different than emission, which is what you're talking about. But to the best of my knowledge:

1. Yes, you should see broadening due to band effects. X-ray energies are on the order of keV and relevant band widths are generally less than 10 eV, so naively this would give you a broadening of 0.1% to 1%.

2. Possibly so. Even though band effects are rather minor in x-ray emission, the presence of a crystal environment will produce a shift in the core energy levels. This shift is detectable in x-ray absorption experiments and can be a useful tool for probing the local environment of atoms. A good source should indicate how experiments were performed.

3. I'm not sure of any references. I think part of the reason x-ray emission has such sharp peaks is because of selection rules. Selection rules allow only certain kinds of transitions, usually ones where the angular momentum quantum number only changes by one ($\Delta l = \pm 1$). So, if there is a vacancy in a 1s orbital in, say, Ag, you need a 3p, 4p electron to transition into that vacancy. The 3p -> 1s transition can come from an orbital which is occupied, and 3p states will have a fairly narrow bandwidth so this will give a sharp peak. You'd then get a secondary emission from something like 4d to the 3p vacancy, but this is only in the UV spectrum. [Selection rules aren't obeyed exactly in solids, but transitions which satisfy selection rules tend to have the highest probability of transition.] So I might start looking at materials on selection rules. There's a great deal of literature on this topic targeted at chemistry students so I think that will be easy for you to find.

4. Aug 7, 2013

### ZapperZ

Staff Emeritus
Hum.... The "band" effects should only be present in the first few "eVs" from the Fermi level. After that, the emission is from core-level states, i.e. from the individual atoms of the solid with very little signals from the collective effects. For example, when one does x-ray photoemission spectroscopy, one now starts to deal with the individual atom's atomic states, not band effects. The only possible signal that the atoms have formed a bond with other atoms might be through a slight shift or extra peaks in the signals.

So the broadening of the x-ray peaks here should have more to do with thermal broadening than from the band effects, which by the time we get to the x-ray shells, are very far away from the Fermi energy.

Zz.

5. Aug 7, 2013

http://arxiv.org/pdf/1303.3458.pdf

In the above paper, some brave souls actually attempt to model the X-ray emission lineshape for some of the lighter elements.

6. Aug 7, 2013

### ZapperZ

Staff Emeritus
Did you also notice that they were looking at the first few eVs of the spectra, which was what I stated in my post? In the hundreds of eVs, i.e. the core level, the many-body effect is no longer apparent.

Zz.

7. Aug 7, 2013

### cgk

In addition to what has been said before (all correct), I'd like to add that the band description of crystal electrons is not mandatory. The electronic states can also be described in terms of localized orbitals, or Wannier functions as they are called in solids, which are an equivalent representation of the occupied states[1]. For atomic orbitals with very small overlap with other atomic orbitals (e.g., core levels, especially deeper ones, as relevant in the characteristic x-rays), the local picture is often a better mode of visualizing them. Effectively, the core levels in solids behave just like core levels in an isolated atom, apart from some energy uncertainty arising from localizing them, and some energy shift arising from the orbital's environment[2]. As a result, a local description of the processes involved in x-ray related issues is technically fine. There *are* quantum states (even in a non-mean field theory) which correspond to a solid in which a single specific atom is core-ionized, and these effectively correspond to the "naive" picture explained in the Bohr model.

[1] Bands are the description of the solid in an effective mean field picture, in which the N-electron total wave function is approximated as a determinant of one-particle spin-orbitals. Such a determinant is invariant to unitary transformations (in fact, even general non-singular linear transformations) amonst those spin-orbitals: Different sets of one-particle states lead to the same total wave function, and all observable physical properties derived from them are identical.
This may be easier to understand in finite systems: Bands correspond to so called "canonical orbitals" of molecules (orbitals which are eigenfunctions of some effective 1-particle mean-field operator), while localized orbitals are obtained by an unitary transformation within the canonical orbitals.

[2] (as already mentioned, this shift can can be probed with XPS/ESCA to get information about the states of the atoms in question and their environment. This proved very useful in chemistry; Siegbahn got a Nobel price for it)

8. Aug 7, 2013

### DrDu

The question is, I think, not so much about the initial core state of the electron but about the final state. As it is a continuum state, there is not so much difference between a free atom and a bound atom as might seem.

9. Aug 7, 2013

### Alexander83

Thank you all for your thoughtful answers. I'll digest these for a little bit and do a little more reading on some of the topics raised, but your responses have been immensely helpful.

Chris.