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Bohr Model and photon emission

  1. Apr 16, 2012 #1
    I am feeling a little stupid asking this considering I am about to graduate with my BS in chemistry. But I have never given this much thought, nor do I remember learning this and I cant figure out a proper explanation. Im sure I am overlooking a simple detail, but I can not figure it out.

    Ok, so I understand the bohr model and the principles behind photon absorption/emission perfectly fine. But one thing that is puzzling to me is why is it when the bohr model is applied to an atom such as hydrogen, only 4 emission lines observed? Can't it have more? I ask this because, the energy levels, n, go from n=0,1,2,3,4,...,infinity, right? So obviously there are more than 4 different energy levels that are capable of emitting a photon. Anybody have an answer?


    I think I figured it out...is it due to when n>4 that the only wavelength emitted are beyond the visible spectrum, thus meaning it is not visible to us?
    Last edited: Apr 16, 2012
  2. jcsd
  3. Apr 16, 2012 #2
    Essentially, you have answered your own question, I think. However, it will be easiest to restrict our attention to the Balmer series, especially to the spectral lines H-alpha to H-gamma. Where λ >400nm (i.e. (n=3) to (n=6)) the photon emitted lies within the visible spectrum.

    There are many series in addition to Balmer, even where n-prime is greater than 6.

    There are series constructed entirely within the UV portion of the spectrum, for example.

    If you are interested in physical fundamentals, you will have to go beyond the Bohr model, however for the purposes of the foundations of chemistry this could be unnecessary. Most applications of partially quantised classical theories and models are fine (i.e. work) within certain scales.
  4. Apr 16, 2012 #3


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