# Question b41bf

Aug 22, 2017

Zero, regardless of which $s$ orbital.

The angular momentum is given typically as the expectation value (or eigenvalue) of the squared operator, $\hat{{L}^{2}}$:

ul(hat(L^2))Y_(l)^(m_l)(theta,phi) = ul(ℏ^2l(l+1))Y_(l)^(m_l)(theta,phi)

where ℏ = h//2pi is the reduced Planck's constant, and $l$ is the angular momentum quantum number. ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ is the angular component of the wave function $\psi$.

The expectation value therefore corresponds to the angular momentum squared:

hat(L^2) harr ℏ^2l(l+1)#

And so, the magnitude of the angular momentum vector is:

$\textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "|vecL| = ℏsqrt(l(l+1))" }}{|}}}}$

An $s$ orbital has an angular momentum quantum number of $l = \boldsymbol{0}$, so an $s$ orbital has no angular momentum.

That should make sense, because a sphere has no angular deviations. That's why the $p$, $d$, $f$, . . . orbitals have more interesting shapes.