A) \[n=4,\,l=3,\,m=+1,\,s=+\frac{1}{2}\]
B) \[n=4,\,l=4,\,m=-4,\,s=-\frac{1}{2}\]
C) \[n=4,\,l=3,\,m=+4,\,s=+\frac{1}{2}\]
D) \[n=3,\,l=2,\,m=-2,\,s=+\frac{1}{2}\]
Correct Answer: A
Solution :
[a] For 4f orbital electron, \[n=4\] |
\[l=3\] (Because 0, 1, 2, 3) |
s, p, d, f |
m = + 3, + 2, +1, 0, - 1, - 2, - 3 |
s = +1/2 |
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