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Is it correct that

each photon above threshold frequency is absorbed by an electron

What is the probability of a photon absorbtion by an electron? Can a quantitative example be given?

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There are two steps to photoemission:

  1. an electron absorbs a photon. This produces an electron moving in the same direction as the original photon i.e. down into the bulk of the metal

  2. the electron either backscatters off other electrons in the lattice or it transfers its energy to other electrons. Either way this produces an electron moving in the opposite direction i.e. towards the metal surface and ultimately out of the metal

The yield for step 1 is almost 100%. We can measure this experimentally because exactly this principle is used in a photomultiplier tube. In a PMT a very thin sheet of metal is used so when an electron absorbs a photon it travels right through the metal sheet and out the other side. This design means no scattering is necessary for the original electron to escape. PMT efficiency can approach 100% in ideal conditions.

The yield for step 2 is absolutely awful. Experiment shows that only one in $10^5$ to $10^6$ photons produces a photoelectron. This shouldn't be surprising since the scattering of an electron headed down into the bulk of a metal is a random process and the probability of it producing an electron headed in the opposite direction with enough energy to escape is very low.

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  • $\begingroup$ Thank you for the answer. . . . So I have two more questions: 1. Does it mean that part (most?) of the electrons that absorbed a photon are not the photo-electrons emitted from the cathode? 2. Is it correct to say that the relevant cross section for photoelectric emission is determined by electron-electron interaction rather than by photon-electron interaction ("The yield for step 1 is almost 100%")? $\endgroup$
    – Dan
    Commented Apr 8, 2016 at 15:19
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    $\begingroup$ @Dan: yes, 99.99 to 99.999% of electrons that absorb a photon just rattle around inside the metal and do not escape, and yes the photoelectron yield is determined by interactions in the metal not the original photon electron interaction. $\endgroup$
    – John Rennie
    Commented Apr 8, 2016 at 15:23
  • $\begingroup$ This was quick! So there are a lot of misconceptions about this subject in a lot of textbooks . . . $\endgroup$
    – Dan
    Commented Apr 8, 2016 at 15:30
  • $\begingroup$ @Dan: the exact nature of the process is usually glossed over when introducing it to students, but I'd be surprised if any of the well known textbooks were actively misleading. What I've described in my answer is standard stuff that's well known in the physics community. $\endgroup$
    – John Rennie
    Commented Apr 8, 2016 at 15:35
  • $\begingroup$ Sorry but I don't understand a thing: if only $10^{5}$ of photons produce a photoelectron, that means that a solar panel only have $0.00001%$ efficiency? So each $100\text{ KW}$ of light only will be produced $1\text{ W}$ of electricity? $\endgroup$
    – Ender Look
    Commented Jul 24, 2017 at 1:58

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