I am reading about the photoelectric effect. Two classical results are:

1) Number of electrons emitted increases with intensity (above the threshold frequency)

2) Number of electrons emitted is independent of frequency

However if I want to increase frequency but keep intensity constant doesn't this mean that I have to send in fewer photons which would decrease the number of emitted electrons?

Note 1: Source (for assumptions and image)

Note 2: I found questions with the same title but they seemed to be asking something different!

  • $\begingroup$ Why do you think intensity and frequency are interrelated? $\endgroup$ – Hritik Narayan Jun 21 '17 at 9:10
  • $\begingroup$ If the energy of one photon is E = hf. Then I think that intensity should be I = Anhf where A is a constant and n is the number of photons per unit time. $\endgroup$ – Beetroot Jun 21 '17 at 9:13
  • 1
    $\begingroup$ I think intensity in this context is just the number of photons. The graphs make no sense otherwise. $\endgroup$ – Hritik Narayan Jun 21 '17 at 9:16
  • $\begingroup$ But is it not well known that photoelectric current is independent of frequency? (I thought current would be directly proportional to number of electrons) $\endgroup$ – Beetroot Jun 21 '17 at 9:23
  • $\begingroup$ Yes. Why are you confused here? That is exactly what the graphs show. Frequency here is the frequency of the photons and intensity is the number of photons. $\endgroup$ – Hritik Narayan Jun 21 '17 at 9:23

I would hazard that the source you have linked is incorrect. Most often when I read about the photoelectric effect the main result is that:

1) The kinetic energy of the emitted electrons depends only on the frequency of the incident light. In particular, there are no emitted electrons until the light is at least a threshold frequency, $\nu_0$ and then after that the kinetic energy of the emitted electrons increases linearly with frequency of the incident light.

$$\text{KE} = hf - \phi$$

($\text{KE}$ is the kinetic energy of the emitted electrons, $h$ and $f$ are Planck's constant and the frequency of the incident light and $\phi$ is the work function of the material under consideration, the minimum amount of energy needed to eject an electron.)

2) (For fixed incident frequency) the number of emitted electrons increases linearly with the intensity of the incident light.

$$N_{\text{electron}} = N_{\text{photon}} = \frac{IA}{h f} = \frac{P}{hf} $$

($P$ is the power in the beam, $A$ is the area of the beam so the intensity, $I$ times $A$ gives the power. It is in fact power $P$ which equals $N_{\text{photon}} hf$ where $N_{\text{photon}}$ is the number of photons impinging upon the surface per second)

These two facts are consistent with the particle or photon description of light. That is, light comes in discrete packets where each packet carries a quanta of energy. The size of this quanta of energy is linearly proportional to the frequency of light and the intensity of the light is proportional to the number flux.

I don't know if I have seen many sources that claim the photocurrent is independent of the frequency of the incident field. As you have correctly noted (and shown in my 2nd equation above) the flux of photoelectrons depends directly on the flux of photons but, because of the relationship between intensity and flux, this in turn depends on intensity and frequency.

In short, I would expect that, at constant intensity, increasing the frequency of incident light would decrease the photocurrent. Someone should please correct me if I'm mistaken.

I think it would be extremely misleading to claim the intensity, in the context of the photoelectric effect, refers only to the number of photons in the beam and not the actual energy flux per unit area of the incident beam. Particularly because the experimenters at the time of the discovery would have been quantifying the experimentally accessible quantity of real intensity.


I don't see the confusion here.

Light intensity is determined by photon flux ($\phi $), the number of photons per second per unit area. The number of electrons increases as the intensity increases, because more photons get to kick out electrons from the metal.

It is independent of frequency because a change in frequency only causes the change in the energy of photons ($E \propto f $), thereby, increasing/decreasing the kinetic energy of the photoelectrons that get kicked out of the metal, not the photon flux.

In your example, if you kept the intensity constant, then the number of photons that kicks out the electrons remains the same, ergo, it doesn't increase/decrease the number of electrons that are ejected out. But if you were to increase the frequency of the incoming radiation, the kinetic energy with which the photoelectrons get ejected from the metal increases. That's it.

  • $\begingroup$ Thank you for your answer. Perhaps you could explain why intensity is normally defined as power per unit area but in the photoelectric effect is only dependent on the photon flux (and not the energy). $\endgroup$ – Beetroot Jun 21 '17 at 9:47
  • $\begingroup$ It refers to a different meaning here. The intensity really depends on the number of photons, while explaining the photoelectric effect. $\endgroup$ – vs_292 Jun 21 '17 at 10:00

The photoelectric effect is one of the basic pylons of the need to introduce quantum mechanics. (An other is the black body radiation).

The data can be interpreted only as a particulate nature of light, one photon kicking out one electron, and the electron is bound with a minimum energy. This cannot be explained with classical electromagnetic theory, which assumes that the electric and magnetic fields adding up in intensity would be able to free an electron even for very low frequencies.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.