Angular momentum of excited electron in hydrogen

  • #1
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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 :)
 

Answers and Replies

  • #2
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There is a simple expression connecting the energy of the electron with the quantum number n.
It should be in your textbook/notes.
 
  • #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.
 
  • #4
An easier way to find n would be to use the formula for the energy levels of the atom.
What would that be?
 
  • #5
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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.
 
  • #6
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.
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.
 
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  • #7
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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++).]
 
  • #8
It's specific for the hydrogen atom.
Ahh, so I'm wrong about ions with one electron, then?
 
  • #9
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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.
 
  • #10
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.
Ahh, ok. Thank you for your help :)
 

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