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I am able to understand how light can be modeled to have wave characteristics from Young's double slit experiment.

But I am unable to comprehend how we can understand light to have particle characteristics from the photoelectric experiment. How is it wave character not able to explain the phenomena observed in this experiment? And how is it that particle nature defeats the wave theory?

Jyotishraj Thoudam
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There have been attempts to describe the photoelectric effect by taking the EM field as a classical wave. For a discussion see a previous question "Can the photoelectric effect be explained without photons?". One of the answers describes that the photoelectric effect can be well explained considering the EM field more or less as a classical wave. To explain other experimental data though a quantized version of EM waves is needed.

On the second part of your question "And how is it that particle nature defeats the wave theory?" The above does not mean that these "wave quanta" (or photons) are particles in the sense of being localized objects flying around in space until they "hit" an atom kicking out an electron. A photon as a quantum of the waves is not localized or trackable as what one would think of a particle. Some physicists refere to these photons as particles which could lead to confusion (e.g. through which slit did they fly in these double slit experiments), but the bottom line is that even if you call these quanta particles they certainly do not defeat the quantum version of the wave theory.

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Jan Bos
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Take it by the other way, for the emission as well for the absorption of EM radiation photons are a good description. The photoelectric phenomenon is a good example for the point that EM radiation is made of photons.

The description of EM radiation as a wave has some weaknesses. For a thermic source of EM radiation one will not be able to measure nor the amplitude nor the wavelength directly. For a monocromatic source (with small aperture) one can measure behind a double slit the distances between the fringes of the intensity distribution which is explained by the interference of outgoing from the two slits circular waves (Huygens principle). But since the phenomenon of fringes appears behind every single sharp edge and even for a stream of single emitted photons, the explanation with Huygens principle is not holdable for this case. Than more the evidence for wave nature is made from a pattern which intensity distribution has the equation of a wave (a sin equation).

But really there is EM radiation which directly measurable frequency. Radio waves are produced by accelerating electrons in an antenna rod by a wave generator fore and back. This induces the emission of photons. Such a EM radiation clearly has the properties of energy transfer with a wave characteristics.

The description of monochromatic EM radiation with a associated wavelength /frequency is helpful but not necessary. The description by the energy of the involved photons is enough.

More about photons, EM radiation and radio waves see this answer.

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HolgerFiedler
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Why don't you look a this video an A levels tutorial for the photoelectric effect.

Summary of the observations of the puzzling photoelectric effect:

  1. The electrons were emitted immediately - no time lag!

  2. Increasing the intensity of the light increased the number of photoelectrons, but not their maximum kinetic energy!

  3. Red light will not cause the ejection of electrons, no matter what the intensity!

  4. A weak violet light will eject only a few electrons, but their maximum kinetic energies are greater than those for intense light of longer wavelengths!

The basic point is that the effect disappears at a threshold frequency and is not dependent on the intensity of the light impinging on the metal.

photoelectric

A wave formalism cannot explain all of these, because the energy in waves is additive, the more intense the beam, the more energy it delivers, but the photoelectric experiments show that this is not true for "light wave" + "electrons in metal" scattering.

No matter how strong the incoming beam of light, if it is below a threshold in frequency (depending on the metal), the electrons will not budge.It shows a one to one correlation that only a particle model can explain.

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anna v
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  • @JanBos you just said it, in gases. There are still the metals. In addition understanding QFT is out of the scope of the amateurs in physics – anna v Oct 02 '16 at 12:47
  • @JanBos As an experimentalist I would rather talk of particles as measured, i.e. approximations to classical concepts. For me QFT is another good mathematical model of nature, not nature. Using it as an ether on which the "particle " concept rides as a wave is twice removed from the data as far as I am concerned. – anna v Oct 02 '16 at 13:33
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    There is nothing particle like about the photon and using it in this way does not help any understanding. It actually creates more confusion in my opinion. That's why I mentioned the example of the photoelectric effect in gasses in which the particle concept is in a very clear manner inappropriate to describe the data as compared to the photoelectric effect on metal. – Jan Bos Oct 02 '16 at 13:59
  • @JanBos Look at this single photon at a time double slit and tell me there is nothing like a particle http://www.sps.ch/en/articles/progresses/wave-particle-duality-of-light-for-the-classroom-13/ . dots are footprints of classical particles after all – anna v Oct 02 '16 at 14:22
  • These dots are a consequence of the interaction of the EM field with the matter used in the detectors. See Fermi's golden rule. No need for particles to describe them. – Jan Bos Oct 02 '16 at 14:39
  • @JanBos sure, if you are good in mathematics. But the same is true of bullet holes, and sand blast holes, they are footprints that a "particle " passed, interacting electromagnetically with matter, that is what we mean by "particle" in everyday life . – anna v Oct 02 '16 at 16:13
  • @Annav, this answer doesn't get to the "why" of the matter, because in principle there could be a classical wavelength threshold, due to the size scales involved. I.e. if the wavelengths get too large compared to the atomic scale, then polarizability becomes important, the EM fields are effectively lowered, the long wavelength only "sees" the neutral atom as a whole, etc. – user1247 Oct 02 '16 at 17:31
  • @user1247 What you describe will need numbers for the dimensions of the metal lattice etc, and also discontinuities are hard to model with classical waves. Do you have a link ? – anna v Oct 02 '16 at 17:44
  • @Annav, you gave a conceptual answer, ("A wave formalism cannot explain this because the energy in waves is additive, the more intense the beam, the more energy it delivers"), and I was merely pointing out that this conceptual answer is flawed, because it's easy to come up with plausible conceptual reasons why the wave formalism could easily accommodate such behavior. The point is that an answer that actually explains the "why this is evidence for photons" should address why the classical formalism doesn't work, despite perfectly plausible reasons why it could. – user1247 Oct 02 '16 at 19:01
  • @Anna The video is very elementary and doesn't cater my need. But I do like the answer you gave. And still I could use some more explanation of the failure of wave theory other than frequency. – Jyotishraj Thoudam Oct 03 '16 at 01:24
  • @Anna I accidentally deleted some of my comments on my smart phone but wave theory of the EM field can still explain the photoelectric effect as I pointed out in my version of an answer. That this effect leads to particles is kind of an urban myth. But even worse, it leads to confusion when talking about other experiments with EM waves like the double slit which uses something like the photoelectric effect for its detectors. – Jan Bos Oct 03 '16 at 02:02