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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)

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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

1 Answer 1

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.

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