Question

Question #b41bf

Answer 1

Zero, regardless of which `s` orbital.

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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)(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.