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I Photoelectric Effect: quantum energy doubts

  1. Nov 7, 2016 #1
    Hey, I've being studying the Photoelectric effect, I think I understand it superficially. One thing that has been bugging me is this:
    "Given that it is possible to move electrons with light and given that the energy in a beam of light is related to its intensity, classical physics would predict that a more intense beam of light would eject electrons with greater energy than a less intense beam no matter what the frequency." -physics.info

    This is easy enough to understand, but what I can't find an explanation for is why is energy related to the frequency, or wavelength, and not the amplitude. From what I've read and seen I am certain that it's true but I don't know why it is the way it is, maybe I'm missing something simple or maybe I have to study planck's postulate in depth.

    thanks
     
  2. jcsd
  3. Nov 7, 2016 #2
    The answer here is quantization. A particle of light (a photon) can be thought as a fundamental unit of amplitude.
    A brighter beam of light has more amplitude (more photons) and therefore more energy in total.
    However, the individual photons might not have enough energy to do something like knock an electron out of its orbital, like in the PE effect.
     
  4. Nov 8, 2016 #3

    vanhees71

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  5. Nov 8, 2016 #4
    Simply put, I think it's so because a single photon can excite only a single electron by transferring its energy to it. So by conservation of energy we can say that the energy lost by the photon is gained by the electron. This energy is used to overcome the electrostatic force of attraction between an electron and its respective nucleus. Any remaining energy is converted to kinetic energy of the electron. This can be given numerically by-> Kinetic energy of electron= Energy of photon- work function of the metal, where work function is just a fancy way of saying energy required to overcome the attractive force of the nucleus. Hence the kinetic energy of the electron depends only on the energy of the photon hitting it, which given by planck's equation where E=h*frequency, where h is planck's constant. As you can see the energy of the photon and hence the electron it hits only depends on the frequency of the incident photon. Intensity of a beam of light on the other hand governs the "number of photons" present in the beam of light. Hence the kinetic energy of the electron only depends on the frequency of the incident radiation. Hope this answered you question :)
     
    Last edited: Nov 8, 2016
  6. Nov 8, 2016 #5

    vanhees71

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    Yes, but as detailed in my Insight article, this is an idea about photons that's not accurate.
     
  7. Nov 8, 2016 #6

    zonde

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    Certainly you are not claiming that the idea about energy of electromagnetic field being quantized is inaccurate.
    So what is inaccurate in the idea about photons that you detail in your article? It seemed to me that in the article you say photoelectric effect is not proof of photon, but now it seems I have misunderstood the point of your article.
     
  8. Nov 9, 2016 #7

    vanhees71

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    Why don't you read the article? The energy of the electromagnetic field is of course not quantized. The spectrum of the free Hamiltonian of the electromagnetic field is ##\mathbb{R}_{\geq 0}##. What's "quantized" in the photoelectric effect is the electromagnetic energy absorbed and emitted by the bound electron. The calculation is a standard exercise in QM1 as a simple but important example for the application of Dirac's time-dependent perturbation theory for a harmonic perturbation. For the details, see my Insights article ;-).
     
  9. Nov 9, 2016 #8

    zonde

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    I read the article of course. But in the article the argument seems to go like:
    Do not assume quantization of electromagnetic energy -> some math -> correct prediction for photoelectric effect.
    But in order to claim that the idea about photons is not accurate the argument has to be like:
    Assume quantization of electromagnetic energy -> some math -> wrong prediction for photoelectric effect.
     
  10. Nov 9, 2016 #9

    vanhees71

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    No, what's wrong is the claim that photons are necessary to understand the photoelectric effect. It's also wrong to claim that electromagnetic energy is "quantized", i.e., somehow discrete. The Hamiltonian of the free quantized electromagnetic field has ##\mathbb{R}_{\geq 0}## as its spectrum. Even single-photon states can have any positive energy, and it's not discrete (a photon of frequency ##\omega## has an energy ##\hbar \omega##, and ##\omega=|\vec{p}|## can take any positive real value). Of course the resonant absorption (and also of course the stimulated emission) of electromagnetic energy is quantized, as is demonstrated in the article. Of course the photoelectric effect also works with single photons. Then it's just scattering of a photon on a bound electron kicking the electron into an unbound (scattering) state. There's no contradiction between the semiclassical and the full quantum treatment of the photoeffect. Of course, there are higher-order quantum corrections that cannot be explained with the semiclassical theory.

    The quantization of the electromagnetic field is demonstrated not by the photoeffect but by, e.g., spontaneous emission, and that's how Dirac introduced this idea in ~1927, i.e., to explain spontaneous emission, which was discovered by Einstein in 1917 in explaining the Planck spectrum of black-body radiation statistically within "old quantum theory", in terms of "modern quantum theory".
     
  11. Nov 9, 2016 #10

    zonde

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    Can you point out such claims (with emphasis on the word "necessary") in this thread?
     
  12. Nov 9, 2016 #11

    vanhees71

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    See my Insights article. There I don't need photons to derive Einstein's results from quantum mechanics of a bound electron interacting with a classical electromagnetic (plane-wave) field.
     
  13. Nov 9, 2016 #12

    PeterDonis

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    Yes, he is. The spectrum of energies for the free electromagnetic field is continuous, as he has said. "Quantized" in this connection means "can only assume discrete values"; a continuous spectrum is not quantized.

    No, what is inaccurate is the idea about photons that werty021 described in post #4 of this thread.

    Yes, post #4. See above.
     
  14. Nov 9, 2016 #13

    PeterDonis

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    Quantization of electromagnetic energy is not an assumption. You don't assume an energy spectrum for a system; you find out what it is by figuring out the appropriate Hamiltonian and solving the appropriate equations. That's what vanhees71 does in his article.
     
  15. Nov 9, 2016 #14

    dextercioby

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    In case the user zonde has not noticed, the "Planck quantum hypothesis" of 1900 is made in the context of "blackbody radiation", that is of a box of arbitrary shape with a very small opening (so that radiation could get in, but it would be extremely unlikely to come out) filled with thermal (usually IR) radiation. For this model with these specifications, indeed, there's a need for the box inner walls to absorb and emit this radiation in granular energy amounts. In the photoelectric effect the incoming radiation is assumed free (plane wave solutions of the wave equations are used), thus its source is very far away from the metal. There's no box scenario anywhere.

    Remember the simple example (yes, highly unphysical) of a 1D box (notice the word box) with infinite walls containing a spinless massive particle. Its momentum is ill-defined as an observable, but the energy (Hamiltonian) is, and its spectrum is discrete (countable). You can say that this unphysical system is quantized.
     
    Last edited: Nov 9, 2016
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