Question

1. The permissible set of all four quantum numbers for an electron are:

(a) n=2 1 =1 m=0,s=0

(b) n=31=2 m= +3 s=+1/2

(c)n = 3 1=0 m = 0 s = - 1/2

(d) n=2 l=3 m3+2 s=+1/2​

Answer 1

ANSWER:

• Option (c) n = 3, l = 0, m = 0, S = - 1/2 is correct.

GIVEN:

• (a) n = 2, l = 1, m = 0, S = 0

• (b) n = 3, l = 2, m = + 3, S = + 1/2

• (c) n = 3, l = 0, m = 0, S = - 1/2

• (d) n = 2, l = 3, m = +2, S = + 1/2

TO FIND:

• The permissible set of all four quantum numbers for an electron.

EXPLANATION:

(a) n = 2, l = 1, m = 0, S = 0

•°• n = 2

•°• when n = 2 possible values of l = 0, 1

•°• l = 1

•°• l < n it is correct

•°• when l = 1 possible values of m = - 1 , 0 , + 1

•°• m = 0

•°• It is also valid.

•°• But S = - 1/2 or + 1/2

•°• S ≠ 0

•°• option a) is incorrect.

$\rule{200}{1}$

(b) n = 3, l = 2, m = + 3, S = + 1/2

°•° n = 3

°•° when n = 3 possible values of l = 0, 1, 2

°•° l = 2

°•° l < n it is correct

°•° when l = 2 possible values of m = - 2, - 1 , 0 , + 1, + 2

°•° m = + 3

°•° It is not valid.

°•° Because m = +3 is not possible when l = 2

°•° option b) is incorrect.

$\rule{200}{1}$

(c) n = 3, l = 0, m = 0, S = - 1/2

•°• n = 3

•°• when n = 3 possible values of l = 0, 1, 2

•°• l = 0

•°• l < n it is correct

•°• when l = 0 possible value of m = 0

•°• m = 0

•°• It is also valid.

•°• S = - 1/2

•°• It is also correct because spin can be either + 1/2 or - 1/2.

•°• option c) is incorrect.

$\rule{200}{1}$

(d) n = 2, l = 3, m = +2, S = + 1/2

°•° n = 2

°•° when n = 2 possible values of l = 0, 1

°•° l = 3

°•° l > n it is incorrect

°•° option d) is incorrect.

$\rule{200}{1}$

NOTE : When single condition also dissatisfied we doesn't want to proceed further as it is already incorrect.