# How could you influence the number of photons emitted from a metal

#### JDiorio

Just wondering, how could you influence the number of photons emitted from a metal surface. I would assume that increasing the intensity of the light would cause an increase because the intensity is a measure of the amount of photons traveling in the light. But what about other factors such as frequency or even the work function of the metal. How would that affect this number of emitted photons?

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#### JDiorio

Re: photoelectrons..

so I've done a little more research and I am not sure about my original hypothesis now.. I have found that the frequency at which the light is emitted is much more important then the intensity. If a "critical frequency" is not met, then no electrons will be emitted despite how intense the light it.

However, what if the intensity is changed once this critical frequency is reached. Then would there be a direct relationship between the intensity and number of emitted electrons.. or no. I believe there is a connection between the energy.. but to conclude a larger number of emitted electrons is not vaild. Any information would be great..

#### sophiecentaur

Gold Member
Re: photoelectrons..

I think the following is true:
The Work Function refers only to the minimum binding energy of electrons to the surface - i.e. the minimum frequency of photons which will release a photoelectron. It's only the tip of the iceberg but proved A Einstein's point! There will be a distribution of electron energies near the surface. For very low intensity radiation with energy very near to the Work Function, some of the incident photons may not get to interact with an appropriately weakly bound electron. For a high intensity, the number of electrons produced would be proportional to intensity because there are more than enough incident photons to hit every available electron. But at low intensity, for the above reason, the current might not be proportional but related to the actual distribution of suitable electrons on the surface.

Also, as the frequency is increased above the threshold, the number of 'available' electrons will also increase non-proportionally, according to the statistics of the energy distribution of the electrons. For a better idea, you'd need to look into the actual energy distributions.

#### Galap

Re: photoelectrons..

I think the following is true:
The Work Function refers only to the minimum binding energy of electrons to the surface - i.e. the minimum frequency of photons which will release a photoelectron. It's only the tip of the iceberg but proved A Einstein's point! There will be a distribution of electron energies near the surface. For very low intensity radiation with energy very near to the Work Function, some of the incident photons may not get to interact with an appropriately weakly bound electron. For a high intensity, the number of electrons produced would be proportional to intensity because there are more than enough incident photons to hit every available electron. But at low intensity, for the above reason, the current might not be proportional but related to the actual distribution of suitable electrons on the surface.

Also, as the frequency is increased above the threshold, the number of 'available' electrons will also increase non-proportionally, according to the statistics of the energy distribution of the electrons. For a better idea, you'd need to look into the actual energy distributions.
Hmm, non-proportionally concave up or concave down? (just trying to get a general feel for what happens)

#### sophiecentaur

Gold Member
Re: photoelectrons..

Concave down, in the case of intensity - approaching 45degrees - one photon for one electron, eventually.
I would thing S shaped for the frequency / current - ending up horizontal (i.e. once saturation has occurred, no extra electrons will be available to come off. I pictures a sort of Fermi distribution but that's very speculative - verging on BS, I'm afraid.

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