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I Photon Momentum/Energy

  1. May 23, 2016 #1
    Hi,

    Is the energy a photon carries with respect to its frequency the same as that of its momentum energy? My understanding is that it is by the energy relations,

    $$hf=E$$
    $$E^2=p^2c^2+m^2c^4=p^2c^2$$ for a photon with ##m=0##, frequency ##f## and plank constant ##h## . So we have from both,

    $$p=\frac{hf}{c}$$ Where ##p## and ##c## are the momentum of the photon and speed of light respectively.

    The issue I have is with photon absorption, as an atom absorbs a photon of frequency ##f## and excites an electron to the next highest state, that state being ##hf## higher than the previous, then how does the photon also transfer momentum to the atom if all its energy is taken by the excitation of the electron?
     
    Last edited: May 23, 2016
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  3. May 23, 2016 #2

    jbriggs444

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    A photon with energy exactly equal to the required delta will fail to excite the atom. The kinetic energy of the recoiling atom needs to be provided.

    However, if one has a body with atoms in random thermal motion, some of them will be moving toward the oncoming photons and some will be moving away. This has the effect of smearing out the absorption band -- some atoms will be able to absorb low energy photons and some will be able to absorb high energy photons.
     
  4. May 23, 2016 #3
    But if the atom is moving towards the photon, it see's a blue shift in the photon's frequency, if this blue shifted frequency matches that of delta then surely it can be absorbed even though the energy of the photon is actually less, the extra energy coming from the recoil of the atom in the opposite direction?
     
  5. May 23, 2016 #4

    jbriggs444

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    Yes.

    Note that the notion of the photon's "actual" energy is not well founded. The energy of a photon depends on your choice of reference frame. There is no one particular energy that is more actual than the others. If one chooses a center-of-momentum reference frame in which the combined momentum of the atom plus photon is zero then the photon's energy can all be used up in increasing the energy level of the electron.
     
  6. May 23, 2016 #5

    ZapperZ

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    The entire atom takes up the momentum of the photon. However, since the atom (with its nucleus) is so much more massive, and the momentum of a photon is extremely small, the change in momentum of the atom is negligible.

    Zz.
     
  7. May 23, 2016 #6
    The change in atomic momentum is equal to the photon momentum.
    The associated change in kinetic energy is much smaller than the photon's energy.
     
  8. May 23, 2016 #7

    ZapperZ

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    I'm not sure what you mean by "atomic momentum". The change in the orbital angular momentum quantum number is due to the photon's orbital momentum. But it also has a linear momentum k. This is what is absorbed by the atom and caused it to negligibly recoil. In photoabsoroption in solids, the lattice ions absorbed this linear momentum.

    Zz.
     
  9. May 23, 2016 #8
    Mv, where M is the atomic mass.
    I believe it is due to the photon's intrinsic momentum aka spin.
    Mv=hk
    The entire crystal moves with momentum hk after the absorption.
    Indeed, this involves negligible kinetic energy.
     
  10. May 23, 2016 #9

    ZapperZ

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    Then what are you disputing or arguing against with my post?

    This is very puzzling.

    Zz.
     
  11. May 23, 2016 #10
    Your question about the meaning of "atomic momentum" surprised me but your account is adequate.
    One thing, I would call the angular momentum of a photon "spin" and not "orbital" momentum.
     
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