If there were three possible values `(-1//2,0,+1//2)` for the spin magnetic quantum number , ` m_s` how many elements would there be in the ` 4 th` period of periodic table in ` 4_s`, 3p, 4d` respectively ?

If there were three possible values `(-1//2,0,+1//2)` for the spin mag

1.	If there were three possible values `(-1//2,0,+1//2)` for the spin magnetic quantum number , ` m_s` how many elements would there be in the ` 4 th` period of periodic table in ` 4_s`, 3p, 4d` respectively ?
Answer» ` n= 4, l = 0, m_l =0, m_s =- 1/2 , 0, + 1/2` for each ` m_l i.e, 3` in `4s` ` l = 1, m_l = -1, 0, +1 i.e, 9` in `4 p` ` l= 2, m _l = +- 2 , +- 1, 0 i.e., 15` in `3 d` Thus ` 4 s^3 , 4 p^9` and ` 3d^(15) ` i.e., in all ` 27` elemnts would have been in ` 4th` period .

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If there were three possible values `(-1//2,0,+1//2)` for the spin magnetic quantum number , ` m_s` how many elements would there be in the ` 4 th` period of periodic table in ` 4_s`, 3p, 4d` respectively ?

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