Discussion
Which of the following is correct ?
Answer: C
The correct statement is K. E of photoelectrons decrease when the oxidation state of the surface is increased.
In XPS the kinetic energy of an ejected electron is KE = hν − BE − φ. A higher oxidation state leaves the atom more positive, holding its core electrons more tightly; the binding energy rises (a chemical shift of a few eV), so at fixed photon energy the kinetic energy falls.
XPS ejects mainly core electrons, not valence electrons (A). Intensity changes the number of electrons, not their kinetic energy (B). And with a monochromatic source, the number of photoelectrons depends on intensity, not on frequency (D).
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