I would like to ask the following question: I have read in more than one texts that the silver atom, in its ground state, posseses 47 electrons, 46 of which are in sphericaly symmetric distribution around the nucleus, and the 47th. posseses the outermost orbit in a 5s state (orbital angular momentum [math]l=0[/math]). Hence - the texts read - the ground state silver atom posseses a total orbital angular momentum [math]\vec{L}=0[/math]. The Stern - Gerlach experiment, which proved the existence of electron spin, was based on this very fact.

 

My question is: Why is it that the total (orbital) angular momentum of the silver atom is that of its 47th. outermost electron? Why is it that the other 46 electrons do not contribute to this angular momentum? Are there any references that deal with this topic?

It's because the other 46 electrons are spin-paired. Each one of the inner electrons that is spin up has a partner that is spin down. Vectorially the angular momenta add up to zero. The only one that's left is that lonely little 47th electron.

It's because the other 46 electrons are spin-paired. Each one of the inner electrons that is spin up has a partner that is spin down. Vectorially the angular momenta add up to zero. The only one that's left is that lonely little 47th electron.

 

My question was about ORBITAL angular momentum, not spin. Is it the same situation?

good point

My question was about ORBITAL angular momentum, not spin. Is it the same situation?

Yes.

good point

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