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Photoelectric effect and energy levels in atoms

  1. Jul 21, 2006 #1
    Hi guys,

    I've got a little conceptual question here. I became a little confused after my exam.

    Firstly, I understand that in the photoelectric effect, the electrons either accept all or none of the energy from the particular incident photon and the photon must have a minimum frequency before the electrons can be emmitted and this frequency is characteristic of the material irradiated.

    What's the difference in this case?:

    For example, in a hydrogen atom, there are different orbits for electrons and they have different energy levels, of which the lowest is closest to the nucleus at -13.6eV if I remember right. There cannot be any other orbits hence the fixed amount of energy at each level.

    Say when I put electrons with an amount of energy which is enough to remove the electron from nucleus (ionisation energy) through the atom, why is it that electrons can accept different amounts of energy from the incident electrons and transit between different energy levels instead of accepting all becoming a "free electron?

    An example for hydrogen, I inject electrons (also, must it always be electrons, can it be photons?) of 13.6eV passing through the hydrogen gas. By principle since the hydrogen electrons accept all the energy, wouldn't it remove the electrons from the attraction of the nucleus? Why is it possible still for it to accept a certain amount to excite it to say level 2, 3, 4.. etc? When it only accepts a certain amount of energy for excitation, where does the rest of the energy go to?

    Thanks apologise for the poor expression.
  2. jcsd
  3. Jul 22, 2006 #2
    I am not an expert in physics, but I am going to try to answer al's question as an exercise. If I am wrong or not quite right, could one of you experts tell me? Thanks.

    Any time you add energy to a system, whether it be through a stream of electrons or photons (and yes al, it can be either), you are going to end up with a statistical distribution of energies on the receiving end. It is just the way Quantum Mechanics works.

    The Uncertainty principle says that we can only know the product of the location and momentum to within a certain limit. We can't know them both exactly. That means that the electrons you are injecting will have a statistical distribution of energies because you know their location (moving toward your target) within a certain limit. If you didn't know their location, you wouldn't be able to hit the target. This distribution of energies means that the collisions are going to excite the H2 atoms to different levels.
  4. Jul 22, 2006 #3


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    Mostly, the difference is self-imposed. In the first case you are talking about a specific outcome of electrons interacting with photons in a solid, and in the second case, you have electrons interacting with electrons in isolated atoms.

    Why would you expect the two to be the same?

    Nevertheless, there's a good chance that a 13.6eV electron will cause an ionization event in atomic hydrogen gas. If it doesn't, the rest of the energy lives in the incident electron.
  5. Jul 22, 2006 #4
    Thanks guys.

    So to sum up my understanding,

    Actually in short the electrons in the photoelectric effect receive energy from a photon and it either receive the entire photon or nothing at all? And whether it will be irradiated depends on the frequency of the incident photons?

    For the atom, whether bombarded with electrons or photons the electrons that are bounded to the nucleus can transit between different energy levels that are characteristic of the atom. They don't have to receive all the energy and if bombarded with an amount of energy equivalent to the atom's ionisation energy there would be electrons being completely removed from the influence of the nucleus and at the same time some would receive only some of the energy and transit between different energy states and the remaining energy not received remains with the incident photos or electrons?

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