Slides

Deviations for Sm 3+ und Eu 3+
Reason: small L ⋅S - coupling for these ions.
Examplel: Sm 3+ (configuration 4f 5 )
5
Lz
4f
}
-3
5
S =
6 H
-2
2
5/2
-1
L = 2 + 3 = 5
0
1
5 5
J = L − S = 5 − =
2
2 2
3
The next higher term:
p J
p exp
Sm 3+ 0.845 1.74
Eu 3+ 0 3.4
5 7
J = L − S + 1 = + 1=
or
6 H
2 2
7/2
H. Zabel 3. Lecture: Local Moments
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