Meaning of the first and second laws of the photoelectric effect

Question

I was reading the introduction to quantum mechanics in my physics book and it begins with a discussion of the photoelectric effect and energy quantas. The first law, the one that says that the intensity of the photoelectric saturation current is directly proportional to the flux of incident EM radiation, means that the more radiation will hit the plate the higher the intensity of the current produced (via displacement of electrons), right?

And the second one, the one that says the kinetic energy of the emitted photoelectrons varies linearly with the frequency of incident EM radiation, and does not depend on the flux, means that the further up the EM spectrum the radiation is, the higher the voltage (electric potential) produced?

Answer 1

In the photoelectric effect one photon displaces one electron, so the way to understand it is to consider the properties of the photons.

If you take light of a fixed frequency, $\nu$, then the energy of the photons is fixed at $h\nu$. That means the intensity of light is proportional to the number of photons, and because one photon = one displaced electron, the intensity of light is proportional to the number of photoelectrons.

When a photon ejects a photoelectron the energy of the electron is equal to the energy of the photon, $h\nu$, minus the work function, $\phi$. This explains why the energy of the electrons, $E$, increases with frequency of the light, $\nu$.

$$E = h\nu - \phi$$

In the last part of your question you say:

means that the further up the EM spectrum the radiation is, the higher the voltage (electric potential) produced?

The photoelectric effect ejects electrons, so whatever you are illuminating will get a positive charge and therefore a potential difference relative to ground. However the potential difference is just related to the total charge, i.e. the number of electrons ejected, so it isn't affected by the energy of the photoelectrons.