Why do we say that #s# orbitals have zero angular momentum?
1 Answer
The angular momentum of any
In other words, recall that angular momentum gives rise to irregular shapes of a given atomic orbital. Well, all
Consider some of the
#psi_(1s)(r,theta,phi) = 1/sqrtpi (Z/a_0)^(3//2)e^(-Zr//a_0)#
#psi_(2s)(r,theta,phi) = 1/(4sqrt(2pi)) (Z/a_0)^(3//2)(2 - (Zr)/a_0)e^(-r//2a_0)#
#psi_(3s)(r,theta,phi) = 1/(81sqrt(3pi)) (Z/a_0)^(3//2) (27 - 18 (Zr)/a_0 + 2((Zr)/(a_0))^2)e^(-Zr//3a_0)#
If you look closely, there is no