Why can an electrons initial kinetic energy be considered negligible in the photoelectric effect?

Question

It says in my text book that einsteins photoelectric equation assumes the kinetic energy of a conduction electron can be considered negligible - this is confusing as everywhere else seems to say it's not the conduction electrons that are emmitted; it's the electrons in the atoms, and if it is the electrons in the atoms, i thought they had very high speeds?

Answer 1

Firstly I don't see why the electrons only in the valence band should be emitted, if I use a monochrome of line equal to the work function of the material you are sampling, only conduction electrons would be expected to be emitted. The reason, I think, that conduction electrons donot exhibit peaks as sharp as the valence band electrons on the XPS spectrum is because of the degree of delocalization of the electrons. The conduction electrons, according to the well accepted electron gas model are delocalized throughout the material and hence the size of the potential well is so large that the difference in the energies of the quantized levels are too small and hence generally donot get resolved as individual energy peaks, but a band. Another factor is that conduction bands have generally much lower occupation probabilities at room temperature because of the Fermi-Dirac distribution, since electrons are fermions. Core level bands on the other hand have much higher occupation probabilities and hence show in the spectrum much higher presence than the conduction electrons

For your second question, the equation of the Kinetic energy, $KE_{elec}$, of the emitted photoelectron is given as. $$KE_{elec}=h\nu -\phi-E_B$$, where $\phi$ is the work function of the material being sampled and $E_B$ is the Binding energy of the electron emitted from the valence band. The Binding Energy, $E_B$, is given by the equation $$E_B=E_F -E$$ where $E_F$ is the Fermi Level of the material being sampled and $E$ is the Bohr orbital energy of the electron in question. This energy $E=T+U$, where $T$ is the kinetic energy of the electron in the orbit, while $U$ is the potential energy of the electron in the orbit, so the Kinetic energy of the electrons is not assumed to be zero in the photo-electric effect conclusions, or else the technique of photo-electron spectroscopy would not exist.

Answer 2

To start with, the system is not a bohr atom system, it is a quantum mechanical system where the electrons in the atoms and molecules do not have orbits, but orbitals.

Thus only an average kinetic energy can be assigned from the energy level occupied by the given extracted electron.

Secondly, the electrons are in bound states in the solids, not only in energy levels of atoms but also common energy levels of the whole lattice. In metals the conduction electrons are almost free, the energy levels with a very small difference. It is a quantum mechanical solution with the whole lattice of the solid..

The photons will at first meet the surface of the metal, where the first thing they interact with is the conduction band electrons that are very loosely bound, and thus have small average kinetic energy to contribute to the interaction.