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Using the Hund rules, find the total angular monentum of the atom in the ground state whose partially filled subshell contains (a) three d electrons, (b) seven d electrons. |
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Answer» Solution :(a) The minimum spin angular momentum of three electrons can be `S=(3)/(2)`. This state is totally symmertic and hence the conjugate angular wavefunction must be antisymmetric By Pauli's exclusion priciple the totally antisymmetric state must have DIFFERENT MAGNETIC quantum numbers. It is easy to see that for `d` electrons the maximum value of the magnetic quantum number for orbital angular momentum`|M_(LZ)|=3` (from `2+1+0`). Higher value violate Pauli's priciple.THUS the state of highest orbital angular momentum consistent with Pauli's principle is `L=3`. The state of the ATOM is then `.^(4)F_(J)` where `J=L-S` by Hund's rule. Thus we get `.^(4)F_(3//2)` The magnitude of the angular momentum is ` ħ sqrt((3)/(2).(5)/(2))=( ħ)/(2)sqrt(15)` (b) Seven `d` electrons mean three holes. Then `S(3)/(2)` and `L=3` as before. But `J=L+S=(9)/(2)` by Hund's rule for more than half filled shell. Thus the state is `.^(4)F_(9//2)` Total angular momentum has the magnitude `ħsqrt((9)/(2).(11)/(2))=(3 ħ)/(2)sqrt(11)` |
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