Angular momentum of excited electron in hydrogen

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 :)

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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.

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.

An easier way to find n would be to use the formula for the energy levels of the atom.
What would that be?

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.

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: $E_n = \frac{-me^4}{8n^2h^2\epsilon_0^2}$ 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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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++).]

It's specific for the hydrogen atom.
Ahh, so I'm wrong about ions with one electron, then?

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.

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 :)