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

Explanation

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 will follow these steps:

### Step 1: Identify the orbitals for \( n = 3 \)
The principal quantum number \( n = 3 \) corresponds to the following orbitals:
- 3s
- 3p
- 3d

### Step 2: Determine the maximum number of electrons in each orbital
- **3s orbital**: Can hold a maximum of 2 electrons.
- **3p orbital**: Can hold a maximum of 6 electrons.
- **3d orbital**: Can hold a maximum of 10 electrons.

### Step 3: Calculate the total number of electrons for each orbital
For each orbital, we need to consider how many can have the spin quantum number \( m_s = -\frac{1}{2} \):
- **3s**: Since it can hold 2 electrons, there will be 1 electron with \( m_s = -\frac{1}{2} \).
- **3p**: Since it can hold 6 electrons, there will be 3 electrons with \( m_s = -\frac{1}{2} \) (one for each of the three p orbitals).
- **3d**: Since it can hold 10 electrons, there will be 5 electrons with \( m_s = -\frac{1}{2} \) (one for each of the five d orbitals).

### Step 4: Sum the electrons with \( m_s = -\frac{1}{2} \)
Now, we add the number of electrons with \( m_s = -\frac{1}{2} \) from each orbital:
- From 3s: 1 electron
- From 3p: 3 electrons
- From 3d: 5 electrons

Total = \( 1 + 3 + 5 = 9 \)

### Final Answer
The maximum number of electrons that can have principal quantum number \( n = 3 \) and spin quantum number \( m_s = -\frac{1}{2} \) is **9**.

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