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Looking at an hydrogen atom with the bohr model we get

$$L=n\hbar, \qquad n\in\mathbb{N}\tag{1}$$

for the angular momentum.

But by solving the Schrödinger equation, we get

$$L=\hbar\sqrt{\ell(\ell+1)}.\tag{2}$$

How is it possible these two equations give different values in some cases?

I have already seen this post regarding this topic this post regarding this topic. But the answer didn't really satisfy me. Because if i have $\ell=1$, the the solution of the Schrödinger equation gives me $L=\sqrt 2\hbar$. Thus I have still a contradiction to the Bohr-solution.

What am I missing here?

Qmechanic
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Jonas Peter
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2 Answers2

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The angular momentum values predicted by the Bohr model are plain incorrect. For instance, the ground state of hydrogen would have $\ell=1$ as per Bohr but $\ell=0$ as per Schrodinger. Moreover, Bohr predicts a single value of $\ell$ per energy level, whereas Schrodinger predicts many.

Experimental evidence (from the Zeeman effect) contradicted the Bohr model and reconciliation between observed and predicted values of $\ell$ was a triumph of Schrodinger's approach.

ZeroTheHero
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The Bohr model is wrong in this context. The solution of the angular part of the Shrodinger equation for the hydrogen atom gives you the spherical hatmonics $Y^m_\ell(\theta,\phi)$ which are the eigenfunctions of the angular momentum poerator $\hat L^2$ (and $\hat L_z$). Those functions are described by the two quantum numbers $\ell$ and $m$ which are different from $n$. That's why the eigenvalues of the angular momentum are given in terms of $\ell$ not $n$.
$$\hat L^2 Y^m_\ell(\theta,\phi)=\hbar \ell(\ell+1)Y^m_\ell(\theta,\phi)$$

Samà
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