Question #b41bf

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Answer 1 · 2017-08-22T19:30:31

Zero, regardless of which $s$ $s$ orbital.

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

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

where $ℏ = h//2pi$ is the reduced Planck's constant, and $l$ is the angular momentum quantum number. $Y_(l)^(m_l)(theta,phi)$ 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:

$color(blue)(barul(|stackrel(" ")(" "|vecL| = ℏsqrt(l(l+1))" ")|))$

An $s$ orbital has an angular momentum quantum number of $l = bb(0)$ , 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.

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