Question

Arrange the electrons represented by following set of quantum numbers in decreasing order of energy.
(A) $n=4,l=0,m=0,s=+\frac{1}{2}$
(B) $n=3,l=1,m=1,s=-\frac{1}{2}$
(C) $n=3,l=2,m=0,s=+\frac{1}{2}$
(D) $n=3,l=0,m=0,s=-\frac{1}{2}$

Answer 1

(A) $n=4,l=0,m=0,s=$$\frac{+1}{2}\Rightarrow {4se}^{-}$

(B) $n=3,l=1,m=1,s=$$\frac{+1}{2}\Rightarrow {3pe}^{-}$

(C) $n=3,l=2,m=0,s=$$\frac{+1}{2}\Rightarrow {3de}^{-}$

(D) $n=3,l=0,m=0,s=$$\frac{-1}{2}\Rightarrow {3se}^{-}$

Therefore decreasing order of energy is:

$3d>4s>3p>3s$

(C) > (A) > (B) > (D)