There are 47 electrons in a Silver atom, but talking about its orbital angular momentum we only take the outermost valence electron which occupies the 5s orbital. Why don't the remaining inner 46 electrons contribute to the total orbital angular momentum of a silver atom?

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    $\begingroup$ Are the contributions of all electrons equal in sign? $\endgroup$
    – DJohnM
    Jun 24 '19 at 20:24
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    $\begingroup$ Silver is perhaps not a great example, since the $s$-wave orbitals (with angular momentum quantum number $\ell=0$) don't carry any orbital angular momentum. $\endgroup$
    – rob
    Jun 24 '19 at 21:25

We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher orbitals would have the angular momentum of only the valence electrons. Now I will demonstrate why a full shell must have zero angular momentum.

An example using the simplest S-shell.

We have two states available, "up" $|\uparrow{} \rangle$, and "down" $|\downarrow{} \rangle$. We also have the constraint that these are fermions, meaning any combination has to be entirely antisymmetric when two particles are interchanged.

If we are placing a single electron into the S-shell we have 2 states available, either: $$|\uparrow{}\rangle \text{ & } |\downarrow{}\rangle$$ Each of these have angular momentum $\frac{1}{2}$. However if we want to add another electron we only have 1 possible state which satisfies antisymmetry, $$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow{}\downarrow{}\rangle - |\downarrow{}\uparrow{}\rangle\right)$$ This is the singlet configuration which has angular total angular momentum zero.

Importantly there is only one state with total angular momentum $J=0$, two states with $J=\frac{1}{2}$, three states with $J=1$ (triplet), and so on.

Here I will outline the logic of the general proof. Firstly ignoring spin, each single particle orbital $l$ has $2l +1$ states with angular momentum projections ranging from $-l \leq m_l \leq l$. Secondly not ignoring spin you can place 2 electrons in each orbital with spin up or down. This gives $2(2l + 1)$ single particle states. If we have completely filled this shell that means that we have placed an electron in each single particle orbital.

Now if we count all of the unique ways we can fill all of the orbitals, there is only one way to do this. That means that this is a singlet configuration and not a member of a higher multiplet (such as the triplet with 3 states mentioned above).

The total state is defined in terms of having a total angular momentum and total angular momentum projection. Clearly it has angular momentum projection of $0$ since $\sum_{m= -l}^{m = l} m = 0$. However since it is a singlet state it also has total angular momentum $0$, and we can treat it like a "core" with no angular momentum.

As a side note this is used extensively in atomic physics as well as nuclear physics. For nuclear physics we would not be talking about electrons but instead protons and neutrons. Therefore we not only have the choice of spin up or down for each orbital but also between proton and neutron. This gives us 4 particles in each orbital, and is made more rigorous with the idea of "isospin". So far as most nuclear interactions care their interactions are the same so we can treat them as 2 projections of one object, the "nucleon". The total wavefunction must be antisymmetric under interchange in the combined spatial-spin-isospin space. A filled orbital momentum shell would therefor have zero angular momentum as well as zero total isospin.

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    $\begingroup$ MathJax suggestion: hide a $\providecommand{\ket}[1]{\left|{#1}\right>}$ somewhere near the top of your answer. Then use $\ket\uparrow$, $\ket{\uparrow\downarrow}$, etc., which have nicer-looking kerning than just $|\uparrow\rangle$. $\endgroup$
    – rob
    Jun 24 '19 at 21:29
  • $\begingroup$ what about the orbital, Spin and total angular momentum of a Nitrogen atom $\endgroup$
    Jun 25 '19 at 5:00
  • $\begingroup$ My answer was generic so it can be applied to any atom (or really any multi-particle spherical system). Again only the valence electrons will contribute to the angular momentum. You need to actually work through the angular momentum coupling, but not for all of the electrons in the system, as the lowest filled S-shell will not contribute. $\endgroup$
    – TEH
    Jun 25 '19 at 14:49

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