I've read many posts and textbooks on modern physics and quantum mechanics but none seem to explain this very well. Some say that it doesn't depend on the frequency of the photons, for a reason similar to this post: Basics of photoelectric effect.

However, in the lab I got a result that is similar to this graph: enter image description here originally posted here.

In fact with a frequency of $8.21 \times 10^{14}$Hz I got a photocurrent of $2.57 \times 10^{-9}$A at $0$V, while with $6.88 \times 10^{14}$Hz and double the intensity I got a photocurrent of $1.39\times 10^{-9}$A at $0$V. This clearly shows that with a higher frequency the value of the photocurrent is larger.

But how can I explain this effect?


1 Answer 1


A photocathode doesn't produce one electron per photon. We parameterize this as quantum efficiency, a quantity that is less than one.

A photon incident on a photocathode may be reflected rather than absorbed. Absorption doesn't necessarily produce a free electron. If it does, the free electron may not escape the material. These factors depend on energy, so the quantum efficiency varies with the photon's energy.

  • $\begingroup$ In order to reproduce the experimental result of the OP, the quantum efficiency should (quite strongly) decrease with increasing wavelength (decreasing frequency), but the quantum efficiency in this wavelength region (around 400 nm) does in fact vary in the opposite sense (as is evident from the graphics in your linked Wikipedia article) $\endgroup$
    – Thomas
    Commented Oct 6, 2022 at 20:42
  • $\begingroup$ @Thomas The Wikipedia curves are for the bulk photoelectric effect in silicon. Silicon sensors and photocells usually have a "dead layer" on the surface, which absorbs short wavelengths. Metal surfaces in vacuum have different curves, depending on the metal. $\endgroup$
    – John Doty
    Commented Oct 6, 2022 at 22:17
  • $\begingroup$ You should then not link to a resource that shows quamtum efficeincy curves which invalidate your answer. In any case, your conclusion is pure speculation without knowing which kind of detector the OP used.. $\endgroup$
    – Thomas
    Commented Oct 7, 2022 at 18:31
  • $\begingroup$ @Thomas The OP was obviously doing a classic stopping potential experiment. The question was "Does the photocurrent depend on the frequency?". The answer is generally yes, because the QE depends on frequency. I cannot address further details without knowing what they are. $\endgroup$
    – John Doty
    Commented Oct 7, 2022 at 19:13

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