# Quantised Angular Momentum?

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?

• You're missing nothing. The Bohr model is false, and doesn't correctly describe the hydrogen atom. – ACuriousMind Oct 31 '14 at 13:50
• 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. – Muster Mark Oct 31 '14 at 13:52
• 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) – ACuriousMind Oct 31 '14 at 14:01
• 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. – Muster Mark Oct 31 '14 at 14:16

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.