6. Find all the Russell-Saunders terms for the f^{2} configuration. Energy order them. For the ground state, find all the J values and energy order these states.
Answer
For f^{2}:
L = 6 (m_{l} = 3 for each electron) so M_{L} = 6, 5, 4, … –5, –6
S = 1 (m_{s} = ½ for each electron) do M_{S} = 1, 0, –1
The total number of microstates = 14!/(2!)(12!) = 91
M_{L} \ M_{S} |
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6 |
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5 |
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4 |
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3 |
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2 |
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1 |
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0 |
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M_{L} = 5, M_{S} = 1: ^{3}H (33 microstates)
M_{L} = 3, M_{S} = 1: ^{3}F (21 microstates)
M_{L} = 1, M_{S} = 1: ^{3}P (9 microstates)
M_{L} = 6, M_{S} = 0: ^{1}I (13 microstates)
M_{L} = 4, M_{S} = 0: ^{1}G (9 microstates)
M_{L} = 2, M_{S} = 0: ^{1}D (5 microstates)
M_{L} = 0, M_{S} = 0: ^{1}S (1 microstate)
The above are listed in increasing energetic order (as least, that's the prediction).
For the ^{3}H ground state, J = 6, 5, 4. Since the configuration is less than half-filled, the energy order is ^{3}H_{4} (lowest energy) < ^{3}H_{5} < ^{3}H_{6}.