If there were 3 possible values   - 1 2 , 0 , + 1 2   for the spin magnetic quantum numbers, how many elements would there be in the   4 t h period of periodic table in 4s, 4p, 4d respectively.

Step-by-step solution

n = 4 l = 0 m l = 0 m s = - 1 2 , 0 , + 1 2 f o r e a c h m l i . e . , 3 i n 4 s l = 1 m l = - 1 , 0 , + 1 i . e . , 9 i n 4 p l = 2 m l = ± 2,1 , 0 i . e . , 15 i n 3 d Thus 4 s 3 , 4 p 3 and 3 d 15 i.e., in all 27 elements would have been in 4 t h period.