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In my studies, it is stated that an atomic orbital is usually described in terms of three quantum numbers: $n$, the principal quantum number, $l$, the orbital quantum number and $m$, the magnetic quantum number. My question is:

Given that the principal quantum number, $n$, has a value of $k \in \mathbb{N}$, is there a closed form equation for calculating the number of atomic orbitals in terms of $k$?

As an example, for $n=2$, the number of atomic orbitals is $4$. But can this value be calculated for any $n = k$?

Qmechanic
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Winter Soldier
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1 Answers1

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This very much depends on the potential. The example you quote is for hydrogen but for the 3d harmonic oscillator there are $\frac{1}{2}(n+1)(n+2)$ states.

In the nuclear shell model for instance, there is strictly speaking no degeneracy beyond the $2\ell+1$ states with angular momentum $\ell$: the energy depends on $n$ and $\ell$ and states with the same $n$ but different $\ell$ have different energies.

In the case of the infinite spherical well, there is no limit on the possible values of $\ell$ for a given $n$, and all these $E_{n,\ell}$ states have different energies.

ZeroTheHero
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  • OP didn't ask for the number of degenerate orbitals, but rather for the total number of orbitals with a given $n$, regardless of energy. – probably_someone Sep 25 '18 at 13:47
  • @probably_someone there is no such number and the question only makes sense for degenerate orbitals. In an infinite 3d square well the number of orbitals of different $\ell$ with given $n$ is unbounded, and they all have different energies, as these depend on $n$ and $\ell$. – ZeroTheHero Sep 25 '18 at 13:53
  • ^I think this should be added to your answer. – probably_someone Sep 25 '18 at 13:57
  • I think it is fairly obvious that the OP is talking about the hyrdogen atom. – BioPhysicist Sep 25 '18 at 14:16
  • @AaronStevens yeah the example is very likely from hydrogen but I'm not so sure about the question. anyways it would be poor form to leave the OP with the impression that there are $n^2$ orbitals for every potential. – ZeroTheHero Sep 25 '18 at 14:29
  • All: Thanks for the answers and comments; together they have answered my question. – Winter Soldier Sep 25 '18 at 15:18