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Why does there need to be a particle representation of light? Doesn't light as a wave explained the observations of the photoelectric perfectly? When the frequency of light is increased, the speed of electrons also increasing. With greater frequency of light, the changing electric and corresponding magnetic fields apply a force on the electrons in a shorter amount of time. And the intensity of light waves, leading to an increase in the number of electrons being released from the metal can be explained by the number of electromagnetic waves reaching the metal surface. So why does there need to be a particle representation of light? Any ideas?

Qmechanic
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  • "When the frequency of light is increased, the speed of electrons also increasing." No. Electrons and light are different things. – John Doty Dec 31 '23 at 12:58
  • @JohnDoty : The OP is not confusing light with electrons. "The electrons" refers to electrons interacting with light via the photoelectric effect. I misread this exactly as you did at first, then recognized my error. – WillO Dec 31 '23 at 13:10
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    @JohnDoty I concur with WillO, I think what the OP has in mind is that when the incident light has a surplus of energy that surplus goes to kinetic energy of the electron that is emitted. The larger the surplus, the larger the subsequent kinetic energy. – Cleonis Dec 31 '23 at 13:12
  • It's a proven fact that light has dual nature, particle and waves, in some scenarios it behaves like only wave, double slit experiment with light, and in other scenarios it behaves like only particle, Compton scattering. One can't explain Compton scattering without considering particle nature. – Aman pawar Dec 31 '23 at 13:21
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    In your question you offer this expression: "the number of electromagnetic waves reaching the metal surface". Here's the thing: in terms of wave mechanics there is no such thing as 'the number of waves'. Example: a vibrating string, vibrating exclusively in the first harmonic. There is no such thing as counting the number of waves. Your question comes with an implicit assumption; an assumption of some form of quantization. – Cleonis Dec 31 '23 at 13:44

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Take for example light detection with a photomultiplier tube

A photomultiplier tube can be operated in a mode called 'Geiger mode'. In Geiger mode a single quantum of energy is sufficient to trigger an avalanche.

By placing filters a light source can be attenuated down to any level of luminosity. No matter how dim the luminosity, each detection event is triggered by transfer of an amount of energy corresponding to the frequency of the light.

At the same time: no matter how dim the luminosity, interference effect can still be obtained, for instance with a double slit setup.

For instance, let's say there is a row of photomultiplier tubes. Such a row of tubes acts as a very coarse grained photo-sensitive array, such that imaging is enabled.

When given sufficient time an interference pattern builds up.


For that reason, and many others, there is no way to avoid acknowledging that quantum excitations of the quantum electrodynamical field can act as particles.

Cleonis
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