CHM 501

6. Find all the Russell-Saunders terms for the f2 configuration. Energy order them. For the ground state, find all the J values and energy order these states.

Answer

For f2:

L = 6 (ml = 3 for each electron) so ML = 6, 5, 4, … –5, –6

S = 1 (ms = ½ for each electron) do MS = 1, 0, –1

The total number of microstates = 14!/(2!)(12!) = 91

ML \ MS

1

0

6

 

(3+,3)

 

5

(3+,2+)

(3+,2) (3,2+)

 

4

(3+,1+)

(3+,1) (3,1+) (2+,2)

 

3

(3+,0+) (2+,1+)

(3+,0) (3,0+) (2+,1)(2,1+)

 

2

(3+,–1+) (2+,0+)

(3+,–1) (3,–1+) (2+,0)(2,0+) (1+,1)

 

1

(3+,–2+) (2+,–1+) (1+,0+)

(3+,–2) (2+,–1) (1+,0)

(3,–2+) (2,–1+) (1,0+)

 

0

(3+,–3+) (2+,–2+) (1+,–1+)

(3+,–3) (2+,–2) (1+,–1)

(3,–3+) (2,–2+) (1,–1+)

(0+,0)

 

     

ML = 5, MS = 1: 3H (33 microstates)

ML = 3, MS = 1: 3F (21 microstates)

ML = 1, MS = 1: 3P (9 microstates)

ML = 6, MS = 0: 1I (13 microstates)

ML = 4, MS = 0: 1G (9 microstates)

ML = 2, MS = 0: 1D (5 microstates)

ML = 0, MS = 0: 1S (1 microstate)

The above are listed in increasing energetic order (as least, that's the prediction).

For the 3H ground state, J = 6, 5, 4. Since the configuration is less than half-filled, the energy order is 3H4 (lowest energy) < 3H5 < 3H6.