I'm studying the XRF method cause I'm about to use a WDXRF spectrometer at college so I'm getting all the information about the physics and chemistry behind it.

I'm just curious about one thing, as you know when a beam of x-ray hits the electron of an inner shell if the energy is just about lower than the electron binding energy, that electron is expelled and the hole is filled with an electron from an outer shell and a photon (fluorescence) is emitted with the same energy as the difference between the shells.

What about the electron that has been expelled from the atom, where does it go and what happens to it?

  • $\begingroup$ After being used as a target, most metal blocks are positively charged. Hence atleast a few electrons must be leaving the target block. $\endgroup$ Commented Dec 1, 2013 at 7:39
  • $\begingroup$ OK they leave, but where do they go, they must be attracted t something else right or do they become free electrons? $\endgroup$ Commented Dec 1, 2013 at 17:56
  • $\begingroup$ Once they are out of the target, the potential which is accelerating the projectile electrons also starts accelerating these ejected electrons and they all become indistinguishable. $\endgroup$ Commented Dec 2, 2013 at 3:52

1 Answer 1


There are a number of techniques that use excitation of inner shell electrons. Auger spectroscopy, and ESCA for example. But XRF is a "bulk" analysis method, not surface such as Auger. The excitation energy from the initial x-ray will be keV whereas the empty orbitals in the atoms of the sample will be a few eV. The excited electron rips trough the sample creating various ions until its energy is reduced enough so that the emitted electron can drop into some empty low energy orbital. (In general it would a "different" electron that drops into the vacancy and creates the fluorescent x-ray.)

In general the x-ray can penetrated deeper into the sample than an excited electron can escape. So the bulk of the excited electrons stay within the sample.


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