X-rays mostly consist of bremsstrahlung and characteristic X-rays. What is the process called that gives rise to the latter (but not the former)?

I'm trying to write a sentence like "X-rays are produced by bremsstrahlung and _______."


The wikipedia article covers what is happening pretty well:

While light atoms have binding energies on order of a few $\text{eV}$, the binding energies of inner shell electrons on heavy atoms can be hundreds of $\text{eV}$. If anything happens to remove a tightly bound inner-shell electron (say a interaction with a high energy electron or other ionizing radiation), there will be a hole left deep in the atom's electron configuration.

That hole can be filled in several ways, including a having a loosely bound electron drop many levels in one step. The photon emitted by such a process will have a energy given by the difference in energies of shells and can be tens or hundreds of $\text{eV}$, well into the x-ray range.

The energy levels of the low lying states depend on the charge of the atomic nucleus, so the maximum energy photon generated this way is characteristic of the material.

Now, we come to the Particle Data Book chapter "Passage of particles through matter (rev.)" (under Experimental Methods and Colliders). If you look at figure 27.4 you can see several edges in the spectrum under $1\text{ MeV}$, those are the reductions in energy loss when the incident particle drops under threshold for a particular atom shell. The caption labels them as

$\sigma_\text{p.e.}$ = Atomic photoelectric effect (electron ejection, photon absorpsion)

In truth I'm not sure that this is a term that will be correctly understood by your readers, and would suggest spending some time reading through that chapter of the PDB so see if you find a better phrase in the text. Your really problem here is that this isn't a simple processes.

  • $\begingroup$ The question is: What is the process called? $\endgroup$ – nadia Jan 28 '12 at 19:13
  • $\begingroup$ So the photoelectric effect is the cause both of characteristic X-rays (electron strikes electron in the anode of the X-ray tube) and photoelectric scattering (photon/X-ray strikes electron in the patient)? $\endgroup$ – nadia Jan 28 '12 at 19:20
  • $\begingroup$ I'm still confused: "electron ejection" seems to imply that an electron is ejected, whereas x-rays are just photons. @Qmechanic: Same goes with the Auger effect. $\endgroup$ – nadia Jan 28 '12 at 19:31
  • $\begingroup$ @nadia It's a two stage process. (1) Something kicks a electron out of a deep-shell. (2) The resulting hole is re-filled with the emission of one or more photons, and when it is just one photon the energy is in the x-ray band and characteristic of the element and shell where the hole was. $\endgroup$ – dmckee --- ex-moderator kitten Jan 28 '12 at 19:34
  • $\begingroup$ @Qmechanic I think the Auger Effect is a noise channel in this process (i.e. it will smear out the characteristic line and/or reduce the rate). $\endgroup$ – dmckee --- ex-moderator kitten Jan 28 '12 at 19:36

The word that finishes your sentence would be "fluorescence". The two-step mechanism described by @dmckee where an electron is ejected, leaving a core hole which is then re-filled by another electron accompanied by the emission of a photon is correct.


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