Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

While the fasted emitted electrons comes from the fermi level of the material in UPS measurement, where does the slowest, which is called secondary electron cut-off orginate from? I really can't figure it out. Also, why isn't the smallest kinetic energy zero? At least it should be nearly zero considering a rather continuous distribution of energy states of electrons.

(ultraviolet photoelectron spectroscopy(UPS) involves photons of a certain energy exciting electrons out of the solids and kinetic energies of the electrons are measured)

share|cite|improve this question
    
Welcome to physics.SE zhanhaohu. Perhaps a better description of what ultraviolet photoelectron spectroscopy (UPS) is could help the other users to help you. – FraSchelle Jun 25 '13 at 10:58
    
You have "kinetic energy=hv-binding energy relative to vacuum level" and has already identified the highest possible starting energy (i.e. least bound). What is the lowest possible starting electron energy (i.e. most bound)? – dmckee Jun 25 '13 at 17:07

Assuming your detector has been corrected for its own work function (i.e. if you look at a metallic material, the Fermi edge should occur at the photon energy), and in the absence of stray magnetic fields, the energy that you measure with the secondary electron cutoff is the work function of the sample.

If your detector is not correcting for its work function, then the secondary cut-off will be the difference between the work function of the sample and the work function of the detector.

share|cite|improve this answer

You have to take into account that there are two different main processes that are observed in a UPS spectrum. After excitation of valence electrons by the incident ultraviolet radiation there are:

  1. The primary (photo-)electrons: the ones that do not undergo inelastic scattering within the material, i.e., they do not lose their energy and hence they do not lose information about the initial state they existed.
    The primary electrons produce distinct spectral features, representing in first approximation the Density of States of the material. Thus, one information one can extract from these features is the binding energies of electronic states.
  2. The secondary (photo-)electrons: these electrons undergo inelastic scattering events within the material as they travel out of it. The majority of the secondary electrons undergo this inelastic collisions, hence they are ejected out of the material with reduced kinetic energy (these electrons lose information about their initial state).
    The secondary electrons give a continuous background that is superimposed with the distinct features given by the primary electrons. The onset of the signal stemming from the secondary electrons is used to calculate the work function of the material.

The reason that the kinetic energy is not zero, is that upon electrical contact between sample and analyser (they are at electrical equilibrium at the end), there is a contact potential formed, which is equal to:wf_sample - wf_analyser, where wf = work function.
If wf_analyser < wf_sample, then this contact potential will accelerate electrons toward the analyser, giving them non-zero kinetic energy.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.