The maximum number of electrons can have principal quantum number `n = 3` and spin quantum number `m_(s) = 1//2` is

Answer

To determine the maximum number of electrons that can have a principal quantum number \( n = 3 \) and a spin quantum number \( m_s = \frac{1}{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Principal Quantum Number**:
- The principal quantum number \( n \) indicates the energy level of an electron in an atom. Here, \( n = 3 \).

2. **Calculate the Number of Orbitals**:
- The total number of orbitals in a given energy level is given by the formula \( n^2 \).
- For \( n = 3 \):
\[ \text{Number of orbitals} = n^2 = 3^2 = 9 \]

3. **Determine the Maximum Number of Electrons per Orbital**:
- Each orbital can hold a maximum of 2 electrons, which can have opposite spins (\( m_s = +\frac{1}{2} \) and \( m_s = -\frac{1}{2} \)).
- However, since we are only considering the electrons with \( m_s = +\frac{1}{2} \), we will only count one electron per orbital.

4. **Calculate the Total Number of Electrons with Spin \( m_s = +\frac{1}{2} \)**:
- Since there are 9 orbitals and each can accommodate one electron with \( m_s = +\frac{1}{2} \):
\[ \text{Total electrons} = \text{Number of orbitals} = 9 \]

5. **Final Answer**:
- Therefore, the maximum number of electrons that can have \( n = 3 \) and \( m_s = +\frac{1}{2} \) is:
\[ \boxed{9} \]