I have been reading some papers on XPS and EDS and I saw that there's a difference between the width of the peaks (of the same material) in XPS and EDS. Is the reason for this difference because of the better energy resolution of XPS (spectral resolution)? EDS uses the signal from emitted X-rays, whereas XPS uses the emitted electron. There's a limit to what how much energy of the X-rays can be detected (about 2 eV) and electrons are easier in that respect and allow for measurement of lower eV.

Or is there another reason for the difference in the peaks? EDS (a) XPS (b)

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    $\begingroup$ Bluntly EDS measures energy of core electrons, and XPS measures the energy of valence electrons. $\endgroup$ – MaxW Aug 6 '20 at 18:21
  • $\begingroup$ Recommend you start reading in a textbook, and perhaps before also here en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy $\endgroup$ – Karl Aug 6 '20 at 18:41
  • $\begingroup$ I am reading a textbook and I have also searched for the answer on the wikipedia page of course, but I can't find the answer to my question. Perhaps I am missing a point, that is why I am asking it here. $\endgroup$ – Paperreader Aug 6 '20 at 18:47

The two techniques are used for different purposes.

EDS - Is a "bulk" analysis technique.

  • Since the exciting electron is of high energy it penetrates relatively deeply into the sample.

  • The emitted x-rays have a characteristic energy which is unaffected by the chemical bonding. (There is some effect due to bonding of course, but the difference in energy is infinitesimally small and hence insignificant.)

  • High energy x-rays can escape through the sample to a "great" depth.

XPS is a surface analysis technique.

  • Exciting electrons of low energy and only penetrate the surface layers of the sample.

  • The bonding energies of the valence electrons are low in energy.

  • The emitted electrons have a low energy (few eV) and hence don't penetrate out of the sample except for the surface layers.


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