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Is Photoelectric effect on ATOMIC ions possible?

  1. Jul 3, 2011 #1
    Hi,

    I am interested to know if photoelectric effect happens in a "single atom" of say Iron which is in a vacum... ???

    if yes, what is the workfunction of Fe2+ ?

    If no what is the ionization energy of Fe2+ in eV?

    Thanks in advance.
     
  2. jcsd
  3. Jul 3, 2011 #2

    mathman

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    The photoelectric effect is an absorption of a photon of sufficient energy to eject an electron from an atom. Where the atom is (vacuum or otherwise) is not relevant.

    Try Google or some other source to get the numbers you want.
     
    Last edited: Jul 3, 2011
  4. Jul 3, 2011 #3
    The work function and ionisation energy are equivalent in that case.
     
  5. Jul 3, 2011 #4

    ZapperZ

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    Photoelectric effect is a specific phenomena involving solids, not "atoms". What you are asking about is photoionization.

    Zz.
     
  6. Jul 4, 2011 #5

    mathman

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    From Wikipedia, the free encyclopedia


    In the photoelectric effect, electrons are emitted from matter (metals and non-metallic solids, liquids or gases) as a consequence of their absorption of energy from electromagnetic radiation of very short wavelength, such as visible or ultraviolet light. Electrons emitted in this manner may be referred to as "photoelectrons".
     
  7. Jul 4, 2011 #6

    ZapperZ

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    .. and as we know, Wikipedia is never wrong. But wait, it gets better. Look up "Photoionization" on Wikipedia:

    No, really?

    How much do you want to bet that the Einstein's photoelectric effect equation doesn't work as described for photoionization process? Any takers?

    As someone who worked in photoemission spectroscopy, I guess I know less than the person/s who wrote that wikipedia entry.

    Zz.
     
    Last edited: Jul 4, 2011
  8. Jul 5, 2011 #7

    mathman

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    This is a quibble. During my working days (I am a retired mathematician), I was involved in calculation of nuclear radiation transport. One of the processes involved for gamma rays is that what you call photo-ionization. The physicists, as well as all the technical material involved, used the term photo-electric effect, irrespective of what the medium was (air, water, lead, iron, etc.)
     
  9. Jul 5, 2011 #8

    ZapperZ

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    But this is a practice among those who KNEW what the physics is for each of the process! I would hate to teach someone who is ASKING about such a thing and trying to learn something that is essentially an inaccurate (at best) description of a phenomena. That's like trying to each someone relativity and setting c=1 right off the bat just because those who work with it use it!

    There are many things professionals do that shouldn't be used to teach students. Many of them are pedagogically unsound! The photoelectric effect, in the annals of physics, refers to a very specific phenomenon, with a specific theoretical description. There is no ambiguity with that, is there?

    Zz.
     
  10. Jul 5, 2011 #9
    Can you inform us why it doesn't (or wouldn't) work? Isn't the basic idea merely the conservation of energy, which will work equally well for atoms and solids?

    Is it because you have to take into account the recoil effect and the momentum of the ion, or something else?
     
  11. Jul 5, 2011 #10

    SpectraCat

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    Well, aside from the "work function" being ill-defined for an atom (i.e. there are multiple electronic eigenstates from which the electron can be ejected), Einstein's equation works pretty well ... as jostpuur says, it is just a restatement of conservation of energy. A very similar equation (where the work function is replaced by the binding energy) forms the basis for photoelectron spectroscopy after all, and that is one way we can experimentally measure the relative energies of electronic states of atoms and molecules.

    So, were you just referring to the fact that photoelectrons can come out at multiple energies when you ionize atoms, when you said Einstein's equation for the photoelectric effect wouldn't work for photoionization of atoms, or is there something else I am missing?
     
  12. Jul 5, 2011 #11
    There are multiple possible initial eigenstates for electrons in solids too. The possible eigenvalues form the energy band.
     
  13. Jul 6, 2011 #12

    ZapperZ

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    It isn't just a conservation of energy. There is a very explicit relationship between energy of the photoelectrons and the photon's energy. In other words, this is what was tested by Millikan when he decided to challenge the Einstein's photoelectric theory.

    We shouldn't be using the "photoelectric effect" to describe photoemission in semiconductors, multiphoton photoemission, Schottky-enabled photoemission, etc.. etc. Just because these phenomena produce photoelectrons due to light impinging on something, doesn't mean that they are all the same phenomenon, especially when there is a perfectly good "word or phrase" being given describe each of them. What's wrong with calling this phenomenon in atomic/molecular gases as photoionization?

    Zz.
     
  14. Jul 6, 2011 #13

    SpectraCat

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    The reason it is called a "band" is that the energy levels aren't discrete .. but rather a continuum (or at least quasi-continuum) of electronic states that you can think of as being made up of a superposition of degenerate (or near-degenerate) atomic states. Therefore I don't think it is correct to call them "eigenstates", but it has been a while since I took solid state physics.

    Anyway, I am pretty sure that you only eject electrons from the "top" of the conduction band in the photoelectric effect .. that is one reason why the work function makes sense for metals in a way that it doesn't for atoms or molecules.
     
  15. Jul 6, 2011 #14

    SpectraCat

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    Well, what I meant was that both Einstein's equation and the equation for PES amount to:

    (photon energy)-(electron binding energy)-(kinetic energy of ejected electron)=0

    The 3 phenomena represented in that equation are common to both processes, so that is probably why people (not just wikipedia) tend to draw parallels between the two.

    I tend to agree with you that it is better to keep the concepts of photoionization and the photoelectric effect distinct, however it can also be useful to point out their similarities, particularly in a pedagogical context.

    Also, the only reason I posted in this thread at all was that I was intrigued by your comment that Einstein's photoelectric effect equation "does not work as described for photoionization processes". If there is more to that statement than the distinction I already noted between the work function for metals, and the electron binding energy (which reflects the different atomic energy levels) for atoms, then I would like to learn more.
     
  16. Jul 6, 2011 #15

    ZapperZ

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    Er.. that isn't the complete photoelectric effect equation. That is some generic equation of the emission of electrons.

    Again, Einstein's photoelectric effect equation has an explicit relationship between the photoelectron energy and the photon energy. From that, one can obtain not only the "work function" but also the planck constant. This is what Millikan tried to verify in his infamous experiment (http://www.ffn.ub.es/luisnavarro/nuevo_maletin/Millikan_1916_1.pdf" [Broken]). My argument is, you won't get this with photoionization! If Millikan tried to check that equation using atomic gasses, he'll think that Einstein was wrong. Now, would that be a valid test?

    I'm all for pointing out similarities. But I'm against clouding the issue by calling them the same thing.

    I've pointed out a while back that I routinely violate Einstein's photoelectric effect equation. In fact, I can easily show you, even in a standard metal that's commonly used in photoelectric effect experiments, that I can get photoelectrons with photon energies below the work function!

    This is why I keep emphasizing that the photoelectric effect (as opposed to photoemission phenomenon in general) is a very specific phenomenon with a very specific theoretical description. It is done on a standard, polycrystaline metal with unpolarized light, and under the condition of single-photon photoemission and no Schottky effect. If you do that, then THAT is the photoelectric effect we know and love, and the Einstein's photoelectric effect equation WORKS! Deviate from that, then it may not and you'll get puzzling results. The general photoemission phenomena in solids require a more detailed description, which includes band strucuture, polarization, etc.. etc. It is also why when I see people coming up with non-photon theory (such as SED) to claim that they can also duplicate the result of photoelectric effect, I tell them that what they have just done is to match the most naive and simplest form of photoemission, i.e. the photoelectric effect. They have done nothing more than successfully approximating a cow at infinity to be a sphere.

    Zz.
     
    Last edited by a moderator: May 5, 2017
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