1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electron changing energy state

  1. May 4, 2013 #1
    When a photon is absorbed by an electron the electron moves from ground state to a higher energy lets say n=3. this electron can then drop back down and emit a photon of the exact energy as the photon that was absorbed.
    My question is why is there no work function or energy transfer as it where, when the electron moves up from ground to n=3? what made me think about this was the photo electric effect. the Kmax of an electron is hf-∅. I think i may be trying to think about it using classical physics instead of quantum.
  2. jcsd
  3. May 4, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    This is not really the photoelectric effect, because there is no "electric" part, i.e. no liberation of electron. The electron is still in a bound state.

    Secondly, the typical photoelectric effect is done on metals, not isolated atoms.

    Thirdly, the energy level, in some way, IS the "work function" in a loose sense, because this is the energy needed to promote the electron from ground state to your n=3 level. Without that energy, such a transition can't occur.

    BTW, you can't simply equate the KE of the electron as being such a difference. The photon energy went into the change in the electrons potential and kinetic energy, as dictated by the solution of the Schrodinger equation for that particular state.

  4. May 4, 2013 #3
    A nit pick for clarificaton: An electron cannot absorb a photon (at least not a real one). Think of it as the atom that is absorbing the energy,
  5. May 4, 2013 #4
    So when thinking about the photoelectric effect we have to consider the properties of the material rather than a single atom that makes sense.

    I am not advanced enough for the Shrodinger equation i'm afraid. but I get what your saying I think. lol

    Thanks for the clarification Popper I was unaware of this.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook