In a photoelectric experiment, if the frequency of incident light is slightly raised while holding intensity constant, I understand that the number of incident photons decreases. This in turn results in the emission of fewer photoelectrons and thus a smaller saturated photocurrent.

While this seems to be a logical flow of events, I am slightly confused by why exactly the photocurrent would fall.

When the frequency of light is raised, the photon energies also rise. This allows for the emission of photoelectrons further below the fermi level. The distribution of electrons in metals would suggest that it is a fair approximation that the fermi level is indeed the highest occupied level at 298K.

As a result, the ability of the incident photons to excite photoelectrons is increased significantly. This would translate to a higher absorption coefficient due to the increased number of electrons that can be excited. A higher absorption coefficient then means that a higher number of photoelectrons can be emitted near the surface of the metal, translating to a higher photocurrent.

The question then is, what are the relative impacts of these competing effects and why is it generally assumed that the photocurrent decreases at higher frequencies of light?

I would guess that the energy band of metals is very narrow. Having a very narrow band would mean that any photons with energy well above the work function would not have a higher probability of exciting electrons as no electrons will be present at levels significantly below the fermi level (until the next quantum shell is reached)


1 Answer 1


We define the quantum efficiency as the number of photoelectrons ejected per incident photon. The quantum efficiency is generally staggeringly low, at around $10^{-5}$ to $10^{-6}$. See Quantum efficiency of Photoelectric effect for more on this. Your question comes down to the change of quantum efficiency with frequency - you argue that higher frequencies should have greater quantum efficiency, generate more photoelectrons and therefore produce a higher current.

This doesn't happen because photoelectron emission is a multistage process. The first step is that incident photon generates a photoelectron. However that photoelectron is travelling in the same direction as the photon i.e. down into the bulk of the metal. The second step is for the electron to scatter off the atoms in the metal and either ricochet back and escape, or transfer its momentum to some other electron so the other electron can escape.

The initial photoelectron generation is actually very efficient with a quantum efficiency approaching unity. However the second step is exceedingly inefficient. You argue that more energetic electrons should increase the efficiency of the first step, but that efficiency is alreay close to one. The limitation is the random scattering process that follows generation of the photoelectron, and a (slightly) higher photon energy has little effect on this.

  • $\begingroup$ Oh, that makes sense! Would the number of photoelectrons emitted in one step increase as the number of electrons emitted near the surface of the metal rises? Or perhaps when a high potential is applied to help nudge the electrons in the right direction? Shouldn't these help increase the quantum efficiency? $\endgroup$
    – Dr Coconut
    Sep 28, 2015 at 7:48
  • $\begingroup$ John, What should be the argument to justify that -- "the more energetic electrons produced in the first step won't increase the efficiency of second step" ??? $\endgroup$
    – Virender
    Feb 8, 2021 at 10:17

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