# Photoelectric Effect and intensity of light

1. Apr 11, 2012

### maxbur

OK, so i have a question about the nature of light, in the context of the photoelectric effect.

So when photons are incident on a metal, given that they have a high enough frequency they can cause electrons to be emitted from the metal. Now i know that intensity of light is to do with the amplitude, and energy is to do with the frequency (E=fh).

Apparantly, when the frequency (thus energy) of the incident light is increased, and intensity is kept constant, the current measured in the circuit connected to the photoelectric set up decreases. Why is this?

The only thing i can think of is that by increasing the energy of the photons, if the intensity is kept constant, the number of photons would have to decrease (which explains the lower current), but again, I dont know why this is?

Thanks for any help, the internet is dreadful for finding such answers :)

2. Apr 11, 2012

### mikeph

If the frequency of the light is increased, the photon energy is increased, so if the power per unit area incident on the metal is unchanged then the number of photons incident per second (photon flux) must be proportionally reduced. The quantum nature of the photon in this experiment means a single photon (with sufficient energy) releases a single electron, so fewer incident photons means fewer released electrons.

3. Apr 11, 2012

### maxbur

Thanks :) so just to double check that I've understood, youre saying that intensity=power per unit area. And since power=energy per second, the factors affecting power (thus intensity) are both the frequency of the photons and the number of photons, right?

4. Apr 11, 2012

### mikeph

Intensity has lots of meanings and can often depend on what field you're studying. When I say it that's what I mean - wattage of incident radiation per unit area. 2nd question: yes, that's it.