A) \[n=4,\,\,l=0,\,\,m=0,\,\,s=\frac{1}{2}\]
B) \[n=4,\,\,l=0,\,\,m=+2,\,\,s=\frac{1}{2}\]
C) \[n=4,\,\,l=1,\,\,m=+2,\,\,s=\frac{1}{2}\]
D) \[n=4,\,\,l=1,\,\,m=0,\,\,s=\frac{1}{2}\]
Correct Answer: A
Solution :
\[_{29}Cu=2,\,\,8,\,\,18,\,\,1\] \[=1{{s}^{2}},\,\,2{{s}^{2}}2{{p}^{6}},\,\,3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},\,\,4{{s}^{1}}\] \[\therefore \]Quantum numbers for\[4{{s}^{1}}\]electron are \[n=4,\,\,l=0,\,\,m=0,\,\,s=+\frac{1}{2}\]
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