Question #b41bf

1 Answer
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)(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.