p41_005 - X l =0 (2 l + 1) = 2 n 2 . (a) In this case n =...

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5. For a given quantum number n there are n possible values of l , ranging from 0 to n 1. For each l the number of possible electron states is N l =2(2 l + 1) (see problem 2). Thus, the total number of possible electron states for a given n is N n = n 1 X l =0 N l =2 n 1
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Unformatted text preview: X l =0 (2 l + 1) = 2 n 2 . (a) In this case n = 4, which implies N n = 2(4 2 ) = 32. (b) Now n = 1, so N n = 2(1 2 ) = 2. (c) Here n = 3, and we obtain N n = 2(3 2 ) = 18. (d) Finally, n = 2 N n = 2(2 2 ) = 8....
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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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