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So when learning about the Bohr model of hydrogen and de Broglie waves, it was shown that treating the electron of hydrogen as a de Broglie wave results in the relationship $$L=n\hbar, \qquad n\in\mathbb{N}.$$ However, when learning about the azimuthal quantum number, it was stated that $$L=\sqrt{\ell(\ell+1)}\hbar.$$ So how come in the ground state ($n=1, \ell=0$), these two equations give different values for angular momentum? I feel like I'm missing something really important here. If it's the case that the Bohr model doesn't accurately describe the angular momentum of the electron in the ground state, why is the angular momentum zero?

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    $\begingroup$ You're missing nothing. The Bohr model is false, and doesn't correctly describe the hydrogen atom. $\endgroup$
    – ACuriousMind
    Oct 31 '14 at 13:50
  • $\begingroup$ Oh ok, the textbook didn't really make that clear. I knew the Bohr model wasn't complete, but I didn't expect it to be this inconsistent with quantum mechanics. $\endgroup$ Oct 31 '14 at 13:52
  • $\begingroup$ Hm...what do you mean "why is the angular momentum zero"? Solving the hydrogen atom quantum mechanically for the allowed states, it just turns out that there are states with $l=0$. What sort of reason would you expect? (Note that, quantumly, you should not think about electrons actually orbiting the nucleus) $\endgroup$
    – ACuriousMind
    Oct 31 '14 at 14:01
  • $\begingroup$ Well the reason why I didn't understand why there were states with L=0 was because the second equation was just presented to me without justification. However, I'll be sure to look up how it arises from the Schrodinger equation. $\endgroup$ Oct 31 '14 at 14:16
  • $\begingroup$ This is a well-known deficiency of the Bohr model, a pedagogical dilemma resolved by J Dahl and M Springborg, Mol Phys 47 (1982) 1001-1019, and especially their appendix. Indeed, the Wigner transform (inverse Weyl transform) of the square of the quantum angular momentum ${\mathfrak L}\cdot {\mathfrak L}$ turns out to be $l^2 - 3 \hbar^2/2$, Where l is the classical quantity, significantly for the ground-state Bohr orbit. $\endgroup$ Oct 26 at 21:33
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TLDR: $\sqrt{l(l+1)} \rightarrow l$ for large values of $l$, but the largest value it can take within an orbital is $l=n-1$, and $n-1 \approx n$ for large values of $n$. Thus $\sqrt{l(l+1)} \hbar \rightarrow n \hbar$ for $l, n \gg 1$.

Long answer:

The Bohr model was a bridge between the Rutherford's and the quantum mechanical atomic model we know today. The greatest achievement of the Bohr model is the prediction of the QM energy levels in hydrogen all the way to the ground state $n=1$, something that might qualify as a coincidence, but it is more like the Bohr quantization rules being educated postulates.

In this sense, Bohr model is less "wrong" than classical physics. Nevertheless, there is the correspondence principle (also by Bohr) from the "correct" quantum world to this "wrong" classical one. Furthermore, while not being a correspondence principle as such, it is not surprising that few aspects from the QM theory can be corresponded to Bohr's angular momentum quantization $L=n \hbar$ in some cases where Bohr's model did a good job. Ultimately, this viewpoint also helps with the intuition of QM as a student.

One example of this correspondence (QM $\rightarrow L=n \hbar$) applies for the rigid rotor of moment of inertia $I$. While the full QM solution gives $E_l = \frac{\hbar l(l+1)}{2 I}$, Bohr's model predicts $E_n = \frac{\hbar n^2}{2 I}$.

In conclusion, it is not surprising that Bohr's angular momentum rule coincides with the QM angular momentum for large quantum numbers. Although not correct in general, it serves as a way to reconcile the old Bohr model in particular cases with the larger QM. In this way, Bohr's model is not seen as being just so inconsistent with QM.

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In your formulas $n$ doesn't have the same meaning. The 1st formula means that the orbital angular momentum is an integer (or zero) multiple of $\hbar$. But for a level with principal quantum number $n$, the angular momentum varies from $(n-1)\hbar$ to $0$, not from $n\hbar$ to $0$. Thus, you don't have a contradiction.

See the page in Wikipedia https://en.wikipedia.org/wiki/Atomic_orbital, and go to the issue "Complex orbitals".

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For $l=0$ the potential minimum which physically means that our electron in the H-atom should fall in the nucleus. Actually, this doesn’t happen in QM due to the Heisenberg uncertainty principle.

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