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The photoelectric equation, which relates the energy of the incident photons to the maximum kinetic energy of the photoelectrons, is given as $ E=h\upsilon -\phi$, where $h\upsilon$ represents the energy of the incident photon and $\phi$ represents the work function of the substance from which the photoelectrons are to be emitted.

My question: Is it possible for the photoelectrons to exchange energy among themselves by the means of Coulombic force? Or is the distance between them too large for the repulsive force to have a tangible effect on their respective energies?

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It's possible, but it's extremely unlikely.

Emission of a photoelectron froma metal surface is a two step process. The incident photon produces a photoelectron with almost 100% probability but that photoelectron is travelling in the same direction as the original photon i.e. down into the metal. For a photoelectron to escape the surface it has to be backscattered, and that process has a very low probability.

The overall quantum yield is usually around $10^{-5}$ to $10^{-6}$ i.e. only one photon in one hundred thousand to a million ejects an electron from the metal, and that low yield is because the backscattering is so inefficient.

It's conceivable that two or more backscattered electrons could strike another electron and hence eject it with (up to) twice the expected energy, but the probability of this happened is so low as to be completely negligable.

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  • $\begingroup$ Could you please expand a bit on the backscattering process? If the electrons get emitted in the same direction as that of the photon, how will they then, in any case, be able to come out of the metal surface? $\endgroup$ – Anindya Mahajan Oct 21 '16 at 16:12
  • $\begingroup$ @AnindyaMahajan: see Kinetic energy of photoelectrons $\endgroup$ – John Rennie Oct 21 '16 at 16:15

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