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Homework Help: Absorption of photon by an electron?

  1. Dec 29, 2005 #1
    This question has really left me puzzled. I have no clue on how to go about. :grumpy:
    How do I show that a Photon can never be completely absorbed by an electron?
  2. jcsd
  3. Dec 29, 2005 #2


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    Is it true that a photon can never be completly absorbed by an electron?
    What about the photoelectric effect then?
  4. Dec 29, 2005 #3


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    perhaps this is in reference to the compton effect
  5. Dec 29, 2005 #4


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    Are you asking how to show the change in potential energy of the electron's new orbit doesn't account for all of the photon's energy? Since photons have momentum, conservation of momentum will result in a change in velocity of the object that absorbed the photon (the entire atom would have to move, not the just the electron). That means a change in kinetic energy of the object in addition to the electron moving to a higher orbital.

    I'm not positive that's exactly the right line of reasoning, but it has to be pretty close.
  6. Dec 29, 2005 #5


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    I think the original question is asking why a photon cannot be absorbed by a FREE electron. In a photoelectric effect, the photon is absorbed by the WHOLE SOLID, not by the electron. The absorbed energy causes a transition by a conduction electron. Similarly, a photon can also be absorbed by the WHOLE ATOM, not by the electron. This process causes an electron in an atomic orbital to make a transition to a higher state of the atom. A free electron doesn't have such "states".

    To answer the original question, one needs to look at the conservation of angular momentum. A photon has spin of 1, while an electron has a spin of 1/2. See if you can answer the question after being given that.

  7. Dec 30, 2005 #6
    Thanks for the responses. Perhaps my original question is a bit ambiguous.
    Yes, you have understood my question perfectly well!
    A truly free isolated electron cannot absorb a photon and remain an electron since this would violate the principle of conservation of energy or momentum.
    I am denoting a Photon by the symbol [itex]\gamma[/itex].

    [tex]\gamma + e^{-} \rightarrow e^{-}[/tex]
    This isn't possible since photons have spin 1 and electrons have spin 1/2, so angular momentum is not conserved.
  8. Dec 30, 2005 #7
    Conservation of energy

    I found that it can be proved that conservation of energy does not take place by choosing a convenient reference frame. By choosing the frame of reference as the rest frame of the final electron, the momentum of the final electron is zero in this frame.

    For the initial state if the electron approaches from right with momentum [itex]-p_e[/itex] and the photon approaches from left with momentum [itex]p_{\gamma}[/itex] then conservation of momentum gives:
    [tex]p_{\gamma} - p_e = 0[/tex]
    [tex]p_{\gamma} = p_e[/tex]

    Applying the conservation of energy law;
    Energy of initial photon is: [itex]E_{\gamma} = p_{\gamma}c[/itex]
    Energy of initial electron is: [itex]E_e =\sqrt{{p_e}^2c^2 + {m_e}^2c^4}[/itex]
    Energy of final electron is : [itex]m_ec^2[/itex]

    Conservation of energy of final electron reads as:
    [tex]E_{\gamma} + E_e = m_ec^2[/tex]

    [tex]E_{\gamma} + \sqrt{{p_e}^2c^2 + {m_e}^2c^4} = m_ec^2[/tex]

    [tex]E_{\gamma} + \sqrt{{E_{\gamma}}^2 + {m_e}^2c^4} = m_ec^2[/tex]

    The quantity on the left hand side of this equation exceeds the one on the right (since [itex]{E_{\gamma}[/itex] cannot be negative) hence this equation is absurd which proves that it is not possible for a free electron to completely absorb a photon.
    Last edited: Dec 30, 2005
  9. Dec 31, 2005 #8
    In case of a bound electron, electron in a particular state has a particular orbital angular momentum characteristic of that orbital , when collided with a photon which is a 1-spin particle with unit angular momentum, the electron absorbs the photon , and because electron is a fermion , it makes a transition so as to conserve angular momentum . Something like the change in angular momentum of electron is compensated by the momentum brought in by the photon.It wont absorb the photon as long as it does not contain the minimum amount of energy for electron to make a transition.
  10. Jan 2, 2006 #9


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    GCT is partly correct with respect to Compton scattering, in that if the photon's energy is much, much greater than the binding energy of the electron in the atom, the electron may be treated as a free electron in terms of scattering. And the key is scattering, based on the conservation of momentum and energy.

    Here is my approach to the problem.

    Take a free electron at rest, so the total energy is simply the rest energy ([itex] m_oc^2[/itex]), and a photon of energy [itex]E_{ph}[/itex].

    The photon strikes the electron and is to be completely absorbed.

    Applying conservation of momentum:

    [itex]E_{ph}/c\,\bar{u}\,=\,m_e v\,\bar{u}[/itex] => [itex]E_{ph}\,=\,m_e vc\,=\,\gamma m_o vc[/itex], where [itex]\bar{u}[/itex] is the unit vector in the direction of the photon and recoiled electron, [itex]m_e[/itex] is the relativistic electron mass, and [itex]m_o[/itex] is the rest mass.

    For energy, one has [itex]E_{ph}\,+\,m_o c^2\,=\,m_e c^2\,=\,\gamma m_o c^2[/itex]

    then [itex]E_{ph}\,=\,(\gamma - 1)m_o c^2[/itex]

    and substituting for [itex]E_{ph}[/itex] from the momentum equation, one obtains

    [itex]E_{ph}\,=\,\gamma m_o vc\,=\,(\gamma - 1)m_o c^2[/itex] which implies all of the photon energy goes into kinetic energy of the electron.

    This last expression simplies to [itex]\gamma v\,=\,(\gamma - 1)c[/itex], which after some manipulation require v = c, which is not possible.

    Rather the photon is 'scattered' by a process known as Thompson scattering for a free electron.



    For a neutral atom, the scattering process is known as Rayleigh scattering or, if the photon energy is great than the binding energy of an electron to the atom, Compton scattering may occur.

    Compton scattering - http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compeq.html#c1
  11. Jan 2, 2006 #10
    Thanks for the explanation, Astronuc!

    Just another question I found in my coursework book on similar grounds for which I need some clarification.
    The following is not an allowed reaction([itex]\gamma[/itex] denotes a photon).
    [tex]\gamma + e^{-} \rightarrow e^{-}[/tex]

    But the book says,
    [tex]e^{-} + p^{+} \rightarrow H + \gamma[/tex]
    is an allowed reaction called radiative capture of electrons(p stands for proton and H stands for Hydrogen).

    I don't see spin conservation here...so why is it an allowed reaction? :confused:
  12. Jan 2, 2006 #11


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    Well in the hydrogen atom, H, the electron and proton are preserved and so are their spins. They are simply in a bound state, the atom, and the photon is emitted as a consequence.

    Recombination is just the opposite of ionization.
  13. May 19, 2006 #12
    You can prove using momentum and energy considerations that a free cannot absorb a photon "completely":

    [tex]0 = \frac{hf}{c} - P_{e}[/tex]
    [tex]\frac{P_{e}^2}{2m_{e}} = hf[/tex]

    Solve for [itex]P_{e}[/itex] and then find the electron recoil speed. It equals [itex]2c[/itex] which is absurd :approve:

    EDIT: Now I see you already knew this...lol

    EDIT #2: This derivation is technically wrong in the relativistic domain, as point out above. You can resort to it only if you really want to avoid relativity (a bad idea).
    Last edited: May 20, 2006
  14. May 19, 2006 #13


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    This is not an acceptable proof in that context since you are using a nonrelativistic expression for the energy of the electron. The full proof involves the relativistic expressions for momenta and energies and has been presented in this thread.

  15. May 19, 2006 #14


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    I personally like to use 4-momenta in that type of problem.
    By conservation of four-momentum, we have

    [tex] P_{\gamma} + P_{i} = P_f [/tex]

    where P_i and P_f are the four-momenta of the electron before and after. Squaring both sides, we get

    [tex] P_{\gamma}^2 + 2 P_{\gamma} \cdot P_i + P_i^2 = P_f^2 [/tex]

    which means (using P^2 = m^2 c^4) that
    [tex] 2 P_{\gamma} \cdot P_i = 0 [/tex]
    Considering the frame in which the electron before the collision was at rest, we get
    [tex] 2 E_{\gamma} m c^2 = 0 [/tex]
    which is impossible.

    I find the use of four-momenta very powerful in those types of problems.


  16. May 19, 2006 #15


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    I'm afraid this is not the correct way of working things out. You're confusing Newtonian formalisms with Relativistic ones.

    The momentum formula is OK, but the energy one should be [tex]E_k = \Delta m c^2[/tex] rather than [tex]E_k = \frac{1}{2}mv^2[/tex]

    Working through the momentum conservation first :

    [tex]m'v = p[/tex] where m' is the relativistic mass of the electron, v is its recoil velocity and p is the momentum of the incident photon.

    [tex]\frac{m}{\sqrt{1 - \frac{v^2}{c^2}}}v = p[/tex] where m is the rest mass of the electron.

    Solving for [tex]\frac{v^2}{c^2}[/tex], we find

    [tex]\frac{v^2}{c^2} = \frac{p^2}{p^2 + m^2c^2}[/tex]

    which is always less than one (hence v < c).

    Now, let's assume conservation of energy and work through the equations like so :

    [tex]E_k = \Delta m c^2 = (m' - m)c^2 = pc[/tex]
    [tex]m(\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1) = \frac{p}{c}[/tex]

    Solving for [tex]\frac{v^2}{c^2}[/tex] we get

    [tex]\frac{v^2}{c^2} = \frac{p(p + 2mc)}{(p + mc)^2}[/tex]

    which still gives us v < c.

    But the expression we get from the cons. of energy approach is unequal to the one from cons. of momentum. This is a contradiction which shows that our original premise of the electron absorbing the photon is flawed. (QED)

    On a macro scale between large masses, this would be called a completely inelastic collision, and we wouldn't expect KE to be conserved, so it would be meaningless to work out the latter part like we did here. But at a subatomic level, I'm not aware of any other way the energy can be dissipated in a photon-free electron collision, so cons. of energy must be invoked (as it is in the usual Compton scattering formula derivation). This leads to a contradiction, proving that completely inelastic collisions are not possible.

    That's the explanation, not the superluminal recoil velocity of the electron, which doesn't happen when you use the correct equations. :smile:
  17. May 19, 2006 #16


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    Hi, you had submitted your post while I was typing mine. That's an elegant and sophisticated approach you're using, but I'm not familiar with it. Is my more long-winded proof OK ? :smile:
  18. May 19, 2006 #17


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    There is no requirement for spin conservation - only for the conservation of TOTAL angular momentum. This can indeed be ensured by specific values of angular momentum of the incident electron (in the proton rest frame, say).
  19. May 19, 2006 #18


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    Hi curious. Yes, your derivation is good.
    I simply used 4-vectors. The advantage is that it basically inclused conservation of energy *and* three-momenum at the same time. In addition, the intermediate steps of my derivation are valid in any frame (any equation expressed in terms of four-vectors is relativistically invariant) . This is a powerful approach for any problem involving conservation of energy/momentum is relativistic situations. For example, the Compton effect formula can be derived much more efficiently than the usual approach shown in elementary textbooks (maybe I should post that). One thing that simplifies the calculations is that the square of any four-momentum is equal to m^2c^4 where m is the rest mass of the particle, so zero for a photon and m^2 c^4 for any massive particle.

    Best regards, and keep the good work...I see that you have been posting high-quality posts.

  20. May 20, 2006 #19


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    Thank you. :smile:

    The 4-vector approach is something that I've heard of before, but never learned (basically all my knowledge of advanced Physics is self-taught). I must acquaint myself with this formalism, it looks very powerful. From my very cursory skimmming of references, I understand it's defined on Minkowski space and has its own definitions for the inner product and other operations. I also see that it can easily prove results like E = mc^2, E^2 = p^2c^2 + m^2c^4 and the Compton scattering formula, all of which I've proven myself using the hard and tedious way! :biggrin:
  21. May 20, 2006 #20
    I know but thats how you explain it you want to avoid relativity. Thats not meant to be a relativistic explanation. And I read somewhere that how would photelectric effect take place if electrons couldn't absorb photons. Well, those are bound electrons so its an altogether different situation. A free electron is free because its not under the influence of a core or nucleus. As for a bound electron, it is bound by lots of other cores in the material and it must be given enough energy to get out of their influence before it comes out of the surface.

    But yes, you're right about the relativistic "proper" explanation. I wanted to avoid it. Nobody's confusing anything.

    EDIT: As you say the important point is inconsistency of momentum and energy in the calculation.

    EDIT #2: Please see my comments at the end of the wrong derivation.
    Last edited: May 20, 2006
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