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Energy levels, de Broglie

  1. Mar 31, 2008 #1
    The Formula for computing E(n) = -(Z^2)/(n^2) * Ei

    is fairly straight forward. Am I right in thinking this formula refers to the energy level of the atom with Z number of protons? This formula, what it yields at least, doesn't depend on the number of electrons the atom has?

    Also, when trying to find the wavelength of the emitted/absorbed photon, should we use hc/E or h/p ? What makes a wave a de Broglie wave? I have deeper doubts but I don't even know where to start, I hope with more replies it'll become clearer.
  2. jcsd
  3. Mar 31, 2008 #2


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    Staff: Mentor

    This formula works only for hydrogen and for "hydrogen-like" ions, that is, ions with only one electron, e.g. He+, Li++, Be+++, etc. If there's more than one electron, the electrons interact with each other and this affects the energy levels.

    It doesn't make any difference, because for a photon, E = pc. This follows from the general formula for energy, momentum and rest mass:

    [tex]E^2 = (pc)^2 + (m_0 c^2)^2[/tex]

    with [itex]m_0 = 0[/itex] for a photon.
  4. Mar 31, 2008 #3
    I suppose for the purpose of my class (and my exam today) this general formula will be enough. Actually, they love to use the energy 13.6 eV in the book, I believe this is the same case that only applies for single electron-atoms.

    That's exactly right..I didn't think it through very well.

    What about when they ask for the energy of the photon that is emitted? The wavelength of the emitted/absorbed photon will always equal the Energy needed for the electron to move from one energy level to the next, correct?
  5. Apr 1, 2008 #4


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    The wavelength of an emitted/absorbed photon will be related to the energy decrease/increase of the electron that absorbed it by:

    [tex]\Delta E=hc/\lambda[/tex]
    Last edited: Apr 1, 2008
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