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Photoelectric effect [was: Quantum Mechanics Questions]

  1. May 1, 2015 #1
    I'm currently self-studying quantum mechanics and instead of starting a new thread every time I have a new question I figure I'd just make one thread dedicated to all of them.

    I'm going over the Photoelectric Effect. The way I understand it is when light is shone on a metallic surface, the photons provide energy to free electrons from the surface. Since this is instantaneous and varies with frequency (not intensity), this shows evidence that light behaves as a stream of particles as opposed to a wave.

    First, correct me if my understanding is wrong. Second, here is my question: so if the electrons are freed from the energy provided by the photon, where exactly does this energy come from and what does this say about the photon? At first it appears to come from conservation of momentum (KE is exchanged from the photon to the e- with some of the energy absorbed into the surface), but clearly this must be impossible since photons are massless. Does this imply that photons are essentially just packets of energy? How does this reaction occur?
     
    Last edited: May 1, 2015
  2. jcsd
  3. May 1, 2015 #2
    I think your understanding is good. This energy is inherent to the photon by the Einstein Planck relation ##E=h\nu##. Note that you could identify ##\nu## with the frequency of the electromagnetic wave associated with the photon. Einsteins explanation of the photoelectric effect, according to my Modern Physics book is that the energy in the EM wave is distributed in discrete bundles (light quanta) of energy ##h\nu## and momentum ##p=E/c##. I've seen people contradict this assertion, but anyways thats what my book says.

    You are right that momentum coming from a massless particle is strange, but thats the way it is. You might understand it by looking at the total energy relativistic equation ##E^2=(pc)^2+(mc^2)^2## If you look closely the mass of the photon is zero, but then that reduces the equation to ##E=pc##. Therefore you can determine the particle's momentum by knowing its energy. You can know its energy by known its color, so to speak.

    That light has momentum has many applications, and is a well known phenomenon. Take for the techique used in biology that moves cells around by shining light on it, or that other technique that accelerates spaceships with sun rays.
     
  4. May 2, 2015 #3

    jtbell

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    I suggest you re-consider this. Many or most people don't look at all threads in a forum. People who know something about e.g. the photoelectric effect are more likely to look at a thread that contains "photoelectric effect" in the title. Likewise for other people who are looking for information about the photoelectric effect, and who might benefit from the answers to your question.
     
  5. May 2, 2015 #4

    ZapperZ

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    Photons do have momentum, per the full relativistic equation mentioned earlier. However, its momentum is minuscule when compared with other types of momentum involved in the interaction (such as the momentum of the ejected photoelectrons).

    But more importantly, take note that the conduction electrons, while "free", are not completely decoupled from the solid. So any momentum change or any needed momentum is taking up by the lattice ions of the solid. This is especially true with regards to the recoil momentum when a photoelectron is ejected from the solid. It is also the reason why you will not get a photoelectric effect in a bunch of free, floating, electron gas.

    Please also note jtbell's response. The PF Rules require that you make descriptive thread title, not something generic such as this. So mixing a bunch of different topics in one thread should not be done, not just because it will not be reflected in the thread title, but also because it makes for a jumbled thread that will be very difficult to follow.

    Zz.
     
  6. May 3, 2015 #5
    First off, thanks for all the replies

    So the energy still comes from the momentum of the photon, just as it would from a body of mass (even though massless)? I can understand it at face value, I'm just having trouble "picturing" the actual energy transfer, but maybe it's something the will become clear as I continue on into the subject.

    As for the thread - mods, you can change the title to "photoelectric effect" and I'll make separate posts for each individual question.
     
  7. May 3, 2015 #6

    jtbell

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    Done! (sort of).
     
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