In a silver atom, the first 46 electrons are all paired and according to David McIntyre in Quantum Mechanics,
The electrons in the closed shells can be represented by a spherically symmetric cloud with no orbital or spin angular momentum.
How do we show that the total orbital angular momentum for a spherically symmetric distribution is zero – using geometry?
furthermore, does this configuration of angular momentum – orbital angular momentum – change if the atoms are subject to an inhomogeneous magnetic field, like in the Stern Gerlach experiment? If not then why?
Best Answer
For a full shell, the addition of the expectation values of any angular momentum $L_i$ is zero, and similarly for the spin operators $\sigma_i$. This is not hard to see - for $l(l+1)$ as the expectation value of $L^2$ for a s,p,d,f subshell, the basis of that subshell is spanned by states indexed by integers between $-l$ and $l$, and since that l is also the actual value for the angular momentum in a given direction, full occupancy of these states implies vanishing total angular momentum. Even easier for spin - there are only up and down states, and equally many of them in every shell, so if they're all occupied, the net spin is zero.