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First question: depends on your definition of photoelectric effect. In a CCD a photon is absorbed, and the energy given to an electron which is then promoted out of the Fermi sea, just as happens in the traditional experiment of a metal in a vacuum. However, in a CCD, the electron has only enough energy to get to the conduction band (generates an ...


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Usually we think of the Fermi tail as being due to electrons that are thermally excited into states that are above the Fermi energy. This has the effect of slightly reducing the work function. There aren't too many electrons promoted in such a way, so the tail is evident only at low voltages w.r.t. the work function. Hence: a tail. I would expect this ...


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Your method outlined will work. You can get a good approximation by ignoring the photon because the photon has a lot less momentum than the electron ejected.


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The electrons and holes are quasiparticles which can be hardly represented by a free-particle model due to their exchange-correlation coupling. For instance, at the extremely low densities, when low-energy electron and hole arrive at very short distance to each other, they do not necessary recombine, they probably form exciton, a stable bound state of the ...



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