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

Explanation

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

### Step 1: Determine the number of orbitals for \( n = 3 \)
The number of orbitals in a given principal quantum number \( n \) can be calculated using the formula \( n^2 \).

\[ \text{Number of orbitals} = n^2 = 3^2 = 9 \]

### Step 2: Understand the electron capacity of orbitals
Each orbital can hold a maximum of 2 electrons, one with spin \( +\frac{1}{2} \) and the other with spin \( -\frac{1}{2} \).

### Step 3: Calculate the total number of electrons for \( n = 3 \)
Since we have 9 orbitals and each can hold 2 electrons, the total number of electrons that can occupy these orbitals is:

\[ \text{Total electrons} = \text{Number of orbitals} \times 2 = 9 \times 2 = 18 \]

### Step 4: Focus on the electrons with spin quantum number \( s = -\frac{1}{2} \)
Out of the total 18 electrons, half will have spin \( +\frac{1}{2} \) and half will have spin \( -\frac{1}{2} \). Therefore, the number of electrons with spin \( -\frac{1}{2} \) is:

\[ \text{Electrons with } s = -\frac{1}{2} = \frac{18}{2} = 9 \]

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

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