# Photoelectric effect intensity

I understand the PE effect quite well but I'm failing to understand one thing. Intensity is the amount of energy per second incident to a given area.

So can you can increase the intensity by either increasing the electrical energy or making the area smaller? By increasing the electrical energy, what exactly is happening? I assume that by increasing intensity you are increasing the number of photons incident to an area of the metal thus increasing the rate of photoelectric emission. But why doesn't increasing intensity via more electrical power lead to greater energy of the electrons, if by increasing intensity you are increasing the energy of the incident waves?

I understand the results but I'm trying to understand why the intensity works the way it is thanks

• The crucial point is to realize that each photon is assumed to transfer its energy only to one electron. This prevents adding more photons per second (i.e. increasing intensity) without increasing the energy per photon (i.e. fixed frequency) from having any effect on the energy of individual electrons. – Danu Nov 19 '14 at 9:11

You can increase the intensity (Power/Area) in any one of three ways: increasing the rate of incident photons, decreasing the area, or increasing the energy of the photons. (In each case we take the other two as remaining constant.)

Increasing the energy per photon will increase the energy of the emitted electrons, but not the rate of emission.

Increasing the photon rate will increase the rate of electron emission, but not the energy of the electrons.

Decreasing the area will have no effect, provided that the area does not become so small that electrons become depleted from the exposed region. (see the comment by @Danu)

• "Increasing the energy per photon will increase the energy of the emitted electrons, but not the rate of emission." To increase the energy of each photon, E = hf, you would have to increase the frequency so use other EM light. How does changing power/area increase the ENERGY of an individual photon? – salman Nov 19 '14 at 14:55
• To increase the energy per photon, you have to increase the frequency. That's another way to increase intensity. Perhaps our misunderstanding is your phrase "electrical energy". I don't know what you mean by electrical energy, so I ignored the phrase. I should have asked what you mean. – garyp Nov 19 '14 at 15:43
• Im not sure where my confusion lies I know that frequency is increased by using light with a higher frequency like those on the other end of the EM spectrum. So KE max can be increased by using photons that carry more energy. But when we are increasing intensity, what exactly are we increasing, is it the number of photons? If so we are increasing the energy of the beam but NOT the individual photons right? – salman Nov 19 '14 at 15:48
• You defined intensity as power/area. There are three ways to do that. I think that you are thinking of a special case in which the light source has a fixed spectrum, which is fairly typical. In that case, you are right, and intensity can be considered to be photons/second/area. But be careful about phrases like "energy in the beam". That's meaningless. A beam can have intensity. You'll have an easier time making sense of things if you use words whose meanings are well-defined, and whose meanings you understand. – garyp Nov 19 '14 at 17:57

The intensity can only be increased by greater energy of the photon. When you increase the energy of the photon, the energy is absorbed into the electron, which consequently flies off with the photon's energy. The more energy the photon has, the more energy the electron has.