Question

The maximum number of electrons that can have principal quantum number $n=3$ and spin quantum number $m_s=-\frac{1}{2}$ is:

• A. 8
• B. 9
• C. 7
• D. 6

Answer 1

The correct option is B 9

$Explanation:$

Given:

Principal quantum number $n=3$

Spin quantum number $m_s=\frac{-1}{2}$

The maximum number of electrons$=?$

No of orbital $=n^2=3^2=9$

The maximum number of electrons with $\frac{-1}{2}$ spin will be equal as no of orbital

$\therefore$ The maximum number of electrons$=9$

$Hencethecorrectanswerisoption\left(B\right).$