p41_028 - 28. We use the results of problem 28 in Chapter...

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Unformatted text preview: 28. We use the results of problem 28 in Chapter 40. The Pauli principle requires that no more than two electrons be in the lowest energy level (at E 1 , 1 , 1 = 3( h 2 / 8 mL 2 )), but due to their degeneracies as many as six electrons can be in the next three levels ( E = E 1 , 1 , 2 = E 1 , 2 , 1 = E 2 , 1 , 1 = 6( h 2 / 8 mL 2 ), E = E 1 , 2 , 2 = E 2 , 2 , 1 = E 2 , 1 , 2 = 9( h 2 / 8 mL 2 ), and E = E 1 , 1 , 3 = E 1 , 3 , 1 = E 3 , 1 , 1 = 11( h 2 / 8 mL 2 )). Using Eq. 40-21, the level above those can only hold two electrons: E 2 , 2 , 2 = (2 2 + 2 2 + 2 2 )( h 2 / 8 mL 2 ) = 12( h 2 / 8 mL 2 ). And the next higher level can hold as much as twelve electrons (see part (e) of problem 28 in Chapter 40) and has energy E = 14( h 2 / 8 mL 2 ). (a) The configuration which provides the lowest system energy higher than that of the ground state has the first level filled, the second one with one vacancy, and the third one with one occupant: E first excited = 2 E 1 , 1 , 1 + 5 E + E...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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