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Angular momentum of excited electron in hydrogen

  1. Feb 24, 2013 #1
    9qzbt5.png

    I am sure there must be a better way to do this, but I just used the method I thought would work. Essentially, I just want to find n and put that in p=nh2π, but my method to find n will give me a complex answer, which is clearly not what I am looking for. If someone could either tell me what I've done wrong, show me a different method, or both, that'd be great.

    Thanks :)
     
  2. jcsd
  3. Feb 24, 2013 #2
    There is a simple expression connecting the energy of the electron with the quantum number n.
    It should be in your textbook/notes.
     
  4. Feb 24, 2013 #3

    TSny

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    In using the Balmer (or Rydberg) formula, you have 1/22, which means you are considering a transition from the nth state to the n = 2 state. But your calculation of the wavelength assumed a transition all the way to the ground state.

    An easier way to find n would be to use the formula for the energy levels of the atom.
     
  5. Feb 24, 2013 #4
    What would that be?
     
  6. Feb 24, 2013 #5

    TSny

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    If you are using a textbook that discusses the Bohr model, the formula should be there. But you can find it here.

    You'll just need to find the value of n that corresponds to an energy of -1.51 eV.
     
  7. Feb 24, 2013 #6
    Ok, so going through my textbook, I recall there is a formula: [itex]E_n = \frac{-me^4}{8n^2h^2\epsilon_0^2}[/itex] which, upon putting the numbers in, gives me the formula in the text you have linked me to (Thank you for that). I can get the answer quite easily from here, but I have to ask: Would this formula be appropiate in general, or is this something specific to hydrogen?

    Edit: For what it's worth, I am pretty sure it'd be applicable for ions that have only one electron. But the derivation used Coloumbs law, and I know the position of an electron is a really shady concept, so I'd imagine things get more complicated when you have more than one electron.
     
    Last edited: Feb 24, 2013
  8. Feb 24, 2013 #7

    TSny

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    It's specific for the hydrogen atom. [EDIT: But there is a simple modification of the formula that you can use for "hydrogen-like atoms" which are ions that contain only one electron. Examples are singly ionized helium (He+) and doubly ionized lithium (Li++).]
     
  9. Feb 24, 2013 #8
    Ahh, so I'm wrong about ions with one electron, then?
     
  10. Feb 24, 2013 #9

    TSny

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    As mentioned in my edit above, you can slightly modify the formula to apply to one-electron ions. You just have to multiply the formula by the square of the number of protons in the nucleus.
     
  11. Feb 24, 2013 #10
    Ahh, ok. Thank you for your help :)
     
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