A) \[3,2,1,+\frac{1}{2}\]
B) \[3,1,1,-\frac{1}{2}\]
C) \[~3,3,1-\frac{1}{2}\]
D) \[3,2,-2,+\frac{1}{2}\]
Correct Answer: C
Solution :
[c] Possible values of l and m depend upon the value of n \[l=0\,to\left( n-l \right)\] \[m=-l\] to \[+l\] through zero \[s=+\frac{1}{2}\] and \[-\frac{1}{2}\] Thus for \[n=3,\] \[l\] may be 0, 1 or 2; but not 3 \[m\] maybe \[-2,-1,0,+1\] or+2 s may be \[+\frac{1}{2}\] or \[-\frac{1}{2}\]You need to login to perform this action.
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