Question

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

Answer 1

When $n=3,l=0,1,2$ i.e., there are 3s, 3p and 3d orbitals. If all these orbitals are completely occupied as shown in the figure.

Total 18 electrons, 9 electrons with $s=+\frac{1}{2}$ and 9 with $s=-\frac{1}{2}$

Alternatively, in any $n^{th}$ orbit, there can be a maximum of $2n^2$ electrons.

Hence, when $n=3$, number of maximum electrons $=18$. Out of these 18 electrons, 9 can have spin $-\frac{1}{2}$ and remaining nine with spin $=+\frac{1}{2}$.