I was reading about the photoelectric effect and a few questions come to my mind.

For the context, I read that if you shine some light (of high energy) on a metal, this latter will emit electrons.

So I was wondering:

  • how can a metal lose an electron? Would electrons 'fly' in the air? Or are we just talking about the metal having more "free" electrons?
  • if electrons are freed 'in the air', would that mean the metal would become electropositive? I recall in my chemistry classes at high-school when making batteries, we needed a salt bridge between the oxydation and reduction half cells to maintain their electrical neutrality. Do metals need electrical neutrality?
  • again if the metal "loses" electrons in the air: if we radiate enough light would the metal end up being short of electrons? as in, would the photo electric effect stop working on a given piece of metal after e.g an hour?

(note: I have read many questions on this website, including Photoelectric effect vs. electronegativity? and the wikipedia pages)


Take a look at https://en.wikipedia.org/wiki/Photoelectric_effect.

The experiment is done in vacuum! In air electrons would still be released but they would not get far because of collisions with air molecules.

Indeed the metal would become positively charged over time. But in the experiment there is a wire connecting the receiving anode and the electron emitting cathode. Thus the emitted electrons flow back (over the wire) to the cathode and the metal is kept neutral.

My guess: Without the current of electrons flowing from anode to cathode, a cloud of electrons will established around the cathode. The metal will be positively charged because of the lack of electrons that are in the cloud. This causes constant emission of electrons (by the light) balanced by electrons being attracted by the positively charged cathode and absorbed again. The absorption will itself emit photons.

  • $\begingroup$ "Without the current of electrons flowing from anode to cathode, a cloud of electrons will established around the cathode" ─ not quite. Electrons will fly away, but as they do so, the cathode will start to develop a positive charge, and, depending on the geometry (specifically, the cathode's self capacitance), this will produce an additional voltage $V_{\rm c\infty}$ compared to infinity, which acts in the same way as the bias voltage used in normal experiments. (cont.) $\endgroup$ – Emilio Pisanty Nov 11 '19 at 11:06
  • $\begingroup$ The kinetic energy gained by electrons at infinity will then be $K = h\nu - \Phi - V_{rm c\infty}$, with $\Phi$ the work function of the metal. Once $V_{rm c\infty}$ becomes greater than $\Phi-h\nu$, electrons can no longer be emitted by the cathode: they will attempt to fly away, but they'll be trapped in the charged cathode's Coulombic electrostatic potential well, and they'll fall back to the cathode. $\endgroup$ – Emilio Pisanty Nov 11 '19 at 11:08
  • $\begingroup$ @EmilioPisanty Well, to me it seems that this is exactly what I stated (or at least this is what I meant), though you put it more quantitatively. $\endgroup$ – Hartmut Braun Nov 11 '19 at 11:24

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