InterviewSolution
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InterviewSolution
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The maximum of electrons can have principal quantum number `n = 3` and spin quantum number `m_(s) = 1//2` is |
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Answer» `n = 3 rArr 1= 0,1,2` `{:(For l = 0 , m= 0 rArr m_(s) = - 1//2 and + 1//2),(For l = 1 , m= -1 rArr m_(s) = - 1//2 and + 1//2):}` `{:(0 rArr m_(s) = - 1//2 and + 1//2),(-1 rArr m_(s) = - 1//2 and + 1//2),("For l= 2,m= "-2 rArr m_(s) = - 1//2 and + 1//2),(-1 rArr m_(s) = - 1//2 and + 1//2),(0 rArr m_(s) = - 1//2 and + 1//2),(1 rArr m_(s) = - 1//2 and + 1//2),(2 rArr m_(s) = - 1//2 and + 1//2):}` Alermatively `n= 3` Number of electron `= 2n^(2) = 2 xx 3^(2) = 18 ` Number of electron with `m_(s) = - (1)/(2) = (18)/(2) = 9` |
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