To answer this question, we need to consider the quantum numbers that describe electrons in atoms. Each electron in an atom is described by four quantum numbers:

1. The principal quantum number ($n$) dictates the energy level and size of the electron orbital.
2. The azimuthal (angular momentum) quantum number ($l$) defines the shape of the orbital. For a given $n$, $l$ can take any integer value from 0 to $n-1$.
3. The magnetic quantum number ($m_l$) describes the orientation of the orbital in space. For a given $l$, $m_l$ can take values from $-l$ to $+l$, including zero.
4. The spin quantum number ($m_s$) describes the direction of the electron's spin and can have a value of $+\frac{1}{2}$ or $-\frac{1}{2}$.

Given the principal quantum number, $n = 3$, and the spin quantum number, $m_s = -\frac{1}{2}$, we need to calculate the maximum number of electrons that can fit this criteria.

For $n = 3$, the possible values of $l$ are 0, 1, and 2 (s, p, and d orbitals respectively). Here's how the orbitals break down:

• When $l = 0$ (the 3s orbital), $m_l = 0$, so there's 1 orbital.
• When $l = 1$ (the 3p orbitals), $m_l$ can be -1, 0, or +1, providing 3 orbitals.
• When $l = 2$ (the 3d orbitals), $m_l$ can be -2, -1, 0, +1, or +2, giving 5 orbitals.

Each orbital can hold 2 electrons with opposite spins. Since we are specifically looking for electrons with $m_s = -\frac{1}{2}$, we can only count one electron per orbital. Therefore, we sum the total number of orbitals across the s, p, and d sublevels for $n = 3$:

3s: 1 orbital $ \times $ 1 electron (with $m_s = -\frac{1}{2}$) = 1 electron

3p: 3 orbitals $ \times $ 1 electron (with $m_s = -\frac{1}{2}$) = 3 electrons

3d: 5 orbitals $ \times $ 1 electron (with $m_s = -\frac{1}{2}$) = 5 electrons

If we sum these, we get:

1 + 3 + 5 = 9 electrons

Thus, the maximum number of electrons that can have the principal quantum number $n = 3$ and the spin quantum number $m_s = -\frac{1}{2}$ is 9.