jefdryck

[jefdryck]

Lowest level:
J is given (see picture) by 3/2
We have only one unfilled (sub)shell (5d) with only one electron. Therefore S = 1/2.
If less than half of the subshell is filled, J is given by |L-S|, therefore L is 2
This gives a Landé g-factor of 0.8

Highest level:
J is given by 5/2
S is still 1/2
Therefore L is 3
This gives a Landé g-factor of 0.857…

At the ground state in lanthanum metal, I think it would lose its 5d electron. Therefore having a full 6s shell as its last shell. This would give S=0 (all electrons compensate each other). L=0 (same reasoning) and therefore J=0. This gives a Landé g-factor of 0

[jefdryck]

EDIT:
the L in highest level of 2D is still 2 (determined by the fact that it is a D multiplet), therefore the landé factor is 1.2

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