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I Wavelength of a Stationary Particle

  1. Aug 26, 2016 #1

    Drakkith

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    We were going over the basics of the photoelectric effect today in my solid state chemistry class when my instructor gave us a question asking what the wavelength of an ejected electron was. We worked through the question and got the answer, but that got me thinking.

    If the wavelength is: λ=h/p, or λ=h/(mv), what happens when the velocity of the electron is zero? The equation seems to imply that the wavelength goes to infinity as v approaches zero. What's going on there?
     
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  3. Aug 26, 2016 #2

    DrClaude

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    Ask Heisenberg. He knows, but he is not certain :wink:
     
  4. Aug 26, 2016 #3

    DrClaude

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  5. Aug 26, 2016 #4

    Drakkith

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    Thanks DC!
     
  6. Aug 30, 2016 #5
    Always tag @ZapperZ for photoelectric questions :smile:
     
  7. Aug 30, 2016 #6

    ZapperZ

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    Thanks, @Greg Bernhardt , but DrClaude got it already. And besides, this is more of a deBroglie wavelength question rather than a photoelectric effect question.

    But to add, when we look at the spectrum of the photoelectrons, we often look at the Energy versus momentum spectrum (E vs k). So already, there is a description of the wavelength of the photoelectrons. The λ=∞ simply means that there's zero momentum. It is not a wave and it is not moving, which for an electron, is highly unusual in most cases.

    ... and that reminds me, I really should finish my photoelectric effect article. Argh!

    Zz.
     
  8. Sep 1, 2016 #7

    dlgoff

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    bold by me

    Thanks for this. Now I understand what @Demystifier was getting at in his reply.
     
  9. Sep 2, 2016 #8

    vanhees71

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    Again and again the same nonsense!

    Plane waves (and for 0 momentum it's a constant!) are not representing states but are generalized functions living in the dual of the nuclear space (where the self-adjoint operators representing position and momentum are defined) of the rigged Hilbert space used in quantum theory. A particle cannot have an exactly sharp momentum. That's reflected by the Heisenberg uncertainty principle, ##\Delta x \Delta p_x \geq \hbar/2##.
     
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