Photoelectric Effect and intensity of light

[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 don't know why this is?

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

 

[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.

 

[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?

 

[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.

 

[sardar]

The photoelectric effect is a phenomenon where electrons are emitted from a metal surface when it is exposed to light. This effect is dependent on the frequency of the incident light, as you have mentioned. The energy of a photon is directly proportional to its frequency, so when the frequency of the incident light is increased, the energy of the photons also increases.

Now, the intensity of light is related to the number of photons hitting a surface per unit time. This means that when the intensity is kept constant, the number of photons hitting the surface remains the same. However, as the frequency (and thus energy) of the photons increases, the energy of each individual photon is also increased. This means that the energy of the photons is now high enough to overcome the binding energy of the electrons in the metal, causing them to be emitted.

In other words, when the frequency of the incident light is increased, the energy of each individual photon is also increased, allowing them to eject more electrons from the metal surface. This leads to a decrease in the number of electrons available to flow in the circuit, resulting in a decrease in the measured current.

It is important to note that the intensity of light does not directly affect the energy of the photons, only the number of photons hitting the surface. So even though the intensity is kept constant, the energy of the photons can still vary based on their frequency.

I hope this helps to clarify the relationship between the photoelectric effect and the intensity of light. Keep exploring and asking questions – that's what science is all about!