Compare the energies of following sets of quantum numbers for multielectron system. (A) ( n =4,1=1 ) (B) ( n =4, l =2 ) (C) ( n =3,1=1 ) (D) ( n =3,1=2 ) (E) ( n =4,1=0 ) Choose the correct answer from the options given below :

Step-by-step solution

Energy level can be determined by comparing ( n + ) values (A) n =4, =1 ( n + )=5 (B) n =4, =2 ( n + )=6 (C) n =3, =1 ( n + )=4 (D) n =3, =2 ( n + )=5 (E) n =4, =0 ( n + )=4 For same value of (n+ ) , orbital having higher value of n , will have more energy. (B) >( A )>( D )>( E )>( C )