The angular momentum of every S-subshell of an atom is 0 by Azimuthal Quantum No. Relation. But if angular momentum of S-subshell is zero. Then by, Angular Momentum =Mass×Velocity×Radius; Radius of S-subshell shoul be 0 which is not the the case. Then why do we consider Angular Momentum of S-subshell zero.

  • $\begingroup$ Because electrons do not behave classically. $\vec{l} = \vec{r} \times \vec{p}$ is a classical relationship. This is one of the shortcomings of the Bohr model. $\endgroup$
    – orthocresol
    Mar 28 '16 at 14:41
  • $\begingroup$ @orthocresol I thought the relation $\vec{l} = \vec{r} \times \vec{p}$, where $\times$ denotes the cross product, carries over from classical to quantum mechanics simply by replacing the vectors with their respective quantum mechanical operators? I've always taken that for granted. Are there any exceptions I should know about? $\endgroup$
    – Philipp
    Mar 28 '16 at 16:27
  • $\begingroup$ @Philipp Yeah, the ang mom relations $\hat{l}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p}_y$ etc. are derived from the cross product. I think the problem with OP's question is that he's implicitly using the Bohr model by talking about the "radius" of the s orbital - the electron isn't going in uniform circular motion around the nucleus with a defined value of $r$ and so it's not appropriate to use the classical relation here (which applies to the movement of a particle). I'd say $l = 0$ for an s orbital simply because the s orbital is an eigenfunction of $\hat{l^2}$ with eigenvalue 0 $\endgroup$
    – orthocresol
    Mar 28 '16 at 16:56

Your confusion arises from the wrong choice of equation for the angular momentum. Your equation

\begin{align} \lVert \vec{l} \rVert = m \lVert \vec{v} \rVert \lVert \vec{r} \rVert \ , \end{align}

where $\vec{l}$ denotes the angular momentum vector ($\lVert \vec{l} \rVert$ is its length, i.e. the angular momentum), $\vec{v}$ the velocity vector, $\vec{r}$ the radius vector and $m$ the mass, does already entail some assumptions that are not necessarily valid. It derives from the more general equation

\begin{align} \vec{l} = \vec{r} \times \vec{p} , \end{align}

where $\vec{p}$ is the momentum vector which under certain circumstances can indeed be written as $\vec{p} = m \vec{v}$ (there are a few subtleties that I don't want to touch here, keyword: canonical momentum, since they don't relate to the problem at hand) and $\times$ denotes the cross product between two vectors. If you now take the absolute value of the angular momentum vector and use the properties of the cross product you arrive at

\begin{align} \lVert \vec{l} \rVert &= \lVert \vec{r} \times \vec{p} \rVert \\ &= \lVert \vec{r} \rVert \lVert \vec{p} \rVert \sin \theta \ , \end{align}

where $\theta$ is the angle between the vectors $\vec{r}$ and $\vec{p}$. So, the factor that is missing in your equation is the $\sin \theta$ term, whose absence means that your equation implicitly assumes that $\theta = 90°$ since $\sin 90° = 1$. And indeed if this would be the case, then for $ \lVert \vec{l} \rVert = 0$ to be true either $ \lVert \vec{r} \rVert$ or $ \lVert \vec{p} \rVert$ had to be equal to zero. However, since this assumption isn't made anywhere in the quantum mechanical description of an atom the $\sin \theta$ term has to be kept and there is another way to achieve $\lVert \vec{l} \rVert = 0$, namely $\vec{r}$ and $\vec{p}$ being either parallel ($\theta = 0°$) or antiparallel ($\theta = 180°$) to each other because then the $\sin \theta$ term is equal to zero.

Of course, for the quantum mechanical description you have to replace the vectors by the quantum mechanical operators and the absolute values of the vectors by the expectation values of the operators.


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