Maximum Number of Electrons with Given Principal and Spin Quantum Numbers

To find the maximum number of electrons that can have a principal quantum number $n = 3$ and a spin quantum number $m_s = -1/2$, we evaluate the number of available orbitals in the $n = 3$ main shell.

1. Azimuthal Quantum Number ($l$):
   For a given principal quantum number $n$, the azimuthal quantum number $l$ can take integral values ranging from $0$ to $n - 1$:
   $l = 0, 1, 2, \dots, (n - 1)$
   For $n = 3$:
   $l = 0 \; (3\text{s}), \quad l = 1 \; (3\text{p}), \quad l = 2 \; (3\text{d})$

2. Magnetic Quantum Number ($m_l$):
   For each value of $l$, the magnetic quantum number $m_l$ has $(2l + 1)$ possible values, representing the total number of orbitals in that subshell:

   • For $l = 0$ ($3\text{s}$ subshell): $2(0) + 1 = 1$ orbital ($m_l = 0$)
   • For $l = 1$ ($3\text{p}$ subshell): $2(1) + 1 = 3$ orbitals ($m_l = -1, 0, +1$)
   • For $l = 2$ ($3\text{d}$ subshell): $2(2) + 1 = 5$ orbitals ($m_l = -2, -1, 0, +1, +2$)

3. Total Number of Orbitals ($N_{\text{orbitals}}$):
   The total number of orbitals in the $n = 3$ shell is given by:
   $N_{\text{orbitals}} = n^2 = 3^2 = 1 + 3 + 5 = 9$

4. Spin Quantum Number ($m_s$):
   According to Pauli's Exclusion Principle, each orbital can accommodate a maximum of two electrons with opposite spins, i.e., one electron with $m_s = +1/2$ and one electron with $m_s = -1/2$.

   Since there are $9$ orbitals in total, each orbital can contribute exactly one electron having $m_s = -1/2$.

Therefore, the maximum number of electrons with $n = 3$ and $m_s = -1/2$ is $9$.