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Consider a modified quantum number scheme where: n = 1, 2, 3,...; l = 0 to (n+1); m = -l to +l (integers including 0); s = -1/4, -1/2, +1/2, +1/4 (four spin values). According to this scheme, what is the maximum number of electrons that can be associated with the second shell (n = 2)?

1. 64
2. 32
3. 16
4. 8

Correct answer: 64

Solution

For n=2, l = 0, 1, 2, 3 (since l goes from 0 to n+1 = 3). For each l, m has (2l+1) values and s has 4 values. l=0: 1*4 = 4. l=1: 3*4 = 12. l=2: 5*4 = 20. l=3: 7*4 = 28. Total = 4+12+20+28 = 64.

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