Q. The electrons identified by quantum numbers $n = 4,\ l = 1$ $n = 4,\ l = 0$ $n = 3,\ l = 2$ $n = 2,\ l = 1$ Can be placed in order of increasing energy from the lowest to highest as
A) IV<II<III<I
B) II<IV<I<III
C) I<III<II<IV
D) III<I<IV<II
Explanation
(a)
For $n = 4,\ l = 1;\ 4p$
For$\ n = 4,\ l = 0;\ 4s$
For $n = 3,l\ = 2;3d$
For $n = 2,\ l = 1;2p$
The order of increasing energy is as
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2p < 4s < 3d < 4p$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i.e.,\ (IV) < (II) < (III) < (I)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.