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Photoelectric effect and polarization

  1. Aug 2, 2005 #1

    vanesch

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    Hi,

    Can someone point me to a treatment of the preferential direction of emission of a photoelectron in the case of polarized incident light ?
    I think that there is a relationship a la cos^2 theta between the plane of polarization of the light (E-field) and the probability of emission of the photoelectron, but I'd like to know where it comes from and if it is related to the atom doing the emission or not.
    I think I know how to work out the (if I'm not mistaking, opposite) relationship in the Compton effect when we can consider the electron to be free (this is treated in Peskin and Schroeder), but I don't know how to do it for the photo-electric effect.

    thanks and cheers,
    Patrick.
     
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  3. Aug 2, 2005 #2

    ZapperZ

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    Er.. Unless I missed something, the preferential direction of emission only comes in when a single-crystal surface is used. This would work even for an unpolarized light. A polycrystal surface would not give such preferential direction.

    The main reason for the preferential direction is that the electrons in the solid has certain crystallographic direction in which it has a distinct momentum (crystal momentum). If you look in this paper http://arxiv.org/abs/cond-mat/0507653, you'll see the "polar angle" being used to designate the in-plane momentum. This polar angle is the angle of emission with respect to the normal of the surface.

    I would say most books on photoemission covering angle-resolved photoemission (such as Huffner's text) would have a treatment on this.

    Zz.
     
  4. Aug 2, 2005 #3

    vanesch

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    In fact, we're more thinking about nobel gasses: photons come in, and liberate a photo-electron: this photo-electron's direction is not isotropic, but correlated with the incident polarization, and we can reconstruct the direction of the photo-electron and hence try to learn something about the polarization of the incident photon. This technique is used, but I'd like to know more about the theory behind it (for instance, I don't know if the particular atom, say, Helium or Argon, plays a role, or whether this is simply something based upon conservation of angular momentum or so).

    Thanks, don't know much about it, or the literature involved. Now, I don't know if it makes a difference, but we're more involved with X-ray photons than UV or visible, and where Compton and photo-electric effect are in competition.

    cheers,
    Patrick.
     
  5. Aug 2, 2005 #4

    ZapperZ

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    Oh, then you are not asking about "photoelectric" effect, but rather "photoionization" effect. There's a difference.

    Zz.
     
  6. Aug 2, 2005 #5

    vanesch

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    I bow down to you and flaggelate myself for using the wrong term :blushing:

    But now that my back is bleeding, and that I have put small sharp stones in my running shoes :tongue:, do you know about a pointer to some readable theory behind the photoionization effect which gives me the relationship between the incident polarization and the outgoing photo-electron ?

    thanks :smile:
    Patrick
     
  7. Aug 2, 2005 #6
    What exactly is the difference between photo ionisation and photo electric effect ?

    Both of them have incident photons and outgoing electrons. The only difference i see is that in the case of photo electric effect, the electrons are coming from the conduction band, so they do not belong to a specific atom. In the case of ionization, the electrons belonged to a specific atom. Can't we say that photo electric effect is just a special case of ionization ? Or am i missing something ?

    marlon
     
  8. Aug 2, 2005 #7

    Gokul43201

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    You can talk of the photoelectric effect insulators as well, so it's not necessarily about conduction electrons. With the photoelectric effect, you have to worry about things like dipole selection rules (for insulators) which lead to a polarization dependent absorption cross-section. Anisotropy plays a big role.

    I don't imagine these are relevant to photoionization in a monoatomic gas. In fact, the only way I can think of a polarization dependence for such a case is if I think purely classically (actually, semi-classically). It would seem like the photoelectron density would want to be highest along the direction of the E-field (and be azimuthally symmetric about it - the [itex]cos^2 \theta [/itex] behavior is like the differential cross-section for a scattering event, which to me, is a nice way to picture it).
     
    Last edited: Aug 2, 2005
  9. Aug 2, 2005 #8
    Well, this can all be true but if you just look at how Einstein proved the photo electric effect, how is that different from ionization ?

    marlon
     
  10. Aug 2, 2005 #9

    Gokul43201

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    It isn't very different, except for the value of the work function. It also tells you nothing about the dependence on polarization or the angular distribution of the photoelectron density.
     
  11. Aug 2, 2005 #10

    ZapperZ

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    Well, first of all, I wasn't being nitpicky about the term. There are important but subtle differences between the two. You don't get the typical Millikan-type photocurrent versus photon energy distribution in a photoionization measurement, for example. And a standard photoelectric effect done on metals do not show any strong anisotropy of the photoelectrons either in terms of intensity or direction. And not only that, but photoemission/photoelectric is done in condensed matter physics, while photoionization tends to be more in atomic/molecular physics study.

    In any case, you may want to look at this paper regarding the anisotropy in photoionization:

    S.T. Manson and A.F. Starace, Rev. Mod. Phys. v.54, p.389 (1982).

    Zz.
     
  12. Aug 3, 2005 #11
    How is that ?

    Are you referring to the work function being dependent of the crystallographic direction along which a crystal has been cut and surface electron density relaxation?

    marlon
     
  13. Aug 3, 2005 #12

    vanesch

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    I just skimmed through it (needs careful reading to understand it) and I think indeed this is what I was asking for, so thank you very much, again, I bow down to your encyclopedic knowledge :approve: (and this time without irony :smile: )

    EDIT:
    Just to tell you where I might need it (or more, a collegue of mine):
    we make gas detectors for thermal neutron imaging, based upon the nuclear reaction of thermal neutrons with Helium-3. This reaction gives rise to a small track of ionisation in the detector. However, we also have contamination from gamma radiation: photons interacting with the gas produce fast electrons which also give small ionisation tracks in the detector. There are of course ways to discriminate between both, but they are not perfect, so we end up with a background of gamma radiation in the neutron picture. The technique we often use is simply to look at the projected charge density of the track, which is usually much lower than that of a neutron interaction. But not always.
    Now, we are develloping techniques which can reconstruct the track direction and we try to use that information to further reduce the gamma background. Normally we have two populations of produced electrons: from the photo-ionisation effect, and from Compton scattering. They have different angular properties, and in order to test our model of what's happening (that will be used to subtract the gamma contamination), a collegue suggested of using polarized X-rays to find out the populations and their angular distributions. You can of course do this correction entirely by empirical "calibration", but I wanted to have a more theoretical understanding of what was going on, hence my question. It would be fun to have a simple monte carlo correctly predicting the populations.
    I think I master the Compton effect (as long as I can consider the electron as "free"), but I didn't have a clue about the photoionisation effect.
    Before HEP people make a fool of us, we're talking about tracks that are less than 1 mm long, so it is not so easy, in a gas detector, to reconstruct that track !
     
    Last edited: Aug 3, 2005
  14. Sep 23, 2005 #13

    Hi Patrick,

    If you are still interested in this, I have a QM textbook with the theoretical treatment of the angular dependence for the photoelectric effect. I can photograph the corresponding pages and email them to you (unfortunately it is only in German, but you should be able to understand what is going on because of the formulae; basically it is a perturbation theory describing the effect of an electromagnetic wave on an atom; the cos^2 factor comes as usual from the dipole matrix element).

    For an experimental reference, see http://prola.aps.org/abstract/PR/v37/i10/p1233_1 .

    I just wonder why this topic is so little known among physicists. Maybe because it flatly contradicts the particle model for light in connection with the photoelectric effect (obviously for the latter you would expect the photoelectron ejection to be primarily in the direction of propagation of light).

    Thomas
     
  15. Sep 23, 2005 #14

    ZapperZ

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    Er.. Hello? Little known?

    There's a WHOLE field of study called angle-resolved photoemission spectroscopy (ARPES). Pick up any paper by Spicer, Z.X. Shen, P.D. Johnson, J.C. Campuzano, etc. etc. The physics is well known, especially in terms of why the photon momentum doesn't have any significant effect on the photoelectron momentum [Hint: photoelectric effect doesn't occur in free-electron gasses].

    So no, there's nothing in the photoelectric effect that would be "obvious" that one would expect the photoelectron momentum to be in the direction of propagation of light.

    Zz.
     
  16. Sep 23, 2005 #15
    As I can see from your posts above, you are referring to photoelectron emission from a solid surface. However, Patrick's question referred to the photoelectric effect in gases (i.e. for single atoms) and this was also what my link http://prola.aps.org/abstract/PR/v37/i10/p1233_1 was referring to.

    This is not correct, as a fully quantum mechanical perturbation theory also yields the cos^2 factor with regard the the electric field direction.

    Thomas
     
  17. Sep 23, 2005 #16

    ZapperZ

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    And if you have bothered to read the thread carefully, I clearly made the distinction between "photoionization" and "photoemission/photoelectric". I'm not the one confusing the two, you are, based on your treatement of the photoelectric effect.

    And I was addressing YOUR comment when you said

    "I just wonder why this topic is so little known among physicists. Maybe because it flatly contradicts the particle model for light in connection with the photoelectric effect (obviously for the latter you would expect the photoelectron ejection to be primarily in the direction of propagation of light)."

    This is bogus. I gave vanesh a reference about the photoionization phenomenon, so one can't say there's little known on this. And *I* have personally worked in photoemission and can testify to VOLUMES and volumes of stuff on this. So where is your evidence that these things are "little known among physicists"? Did you just made that up? What KIND of physicists do you hang around anyway?

    Zz.
     
  18. May 19, 2006 #17
    Hello:
    I need info about photoionisation.Which wawelenght effeckts on h2.Please send related literature.Thanks
     
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