To solve the problem of calculating the maximum number of electrons in manganese (Mn, atomic number 25) that have the quantum numbers \( n = 3 \), \( m = 0 \), and \( s = +\frac{1}{2} \), we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Principal Quantum Number (n)**:
- Given \( n = 3 \), we know that we are looking at the third energy level, which includes the orbitals: \( 3s \), \( 3p \), and \( 3d \).
2. **Determine the Magnetic Quantum Number (m)**:
- Given \( m = 0 \), we need to identify which orbitals correspond to this magnetic quantum number.
- For the \( s \) orbital (\( l = 0 \)), \( m \) can only be \( 0 \).
- For the \( p \) orbital (\( l = 1 \)), \( m \) can be \( -1, 0, +1 \), so \( m = 0 \) corresponds to one of the \( p \) orbitals.
- For the \( d \) orbital (\( l = 2 \)), \( m \) can be \( -2, -1, 0, +1, +2 \), so \( m = 0 \) corresponds to one of the \( d \) orbitals.
3. **Identify the Spin Quantum Number (s)**:
- Given \( s = +\frac{1}{2} \), we are looking for electrons that have a spin of \( +\frac{1}{2} \).
4. **Count Electrons in Each Relevant Orbital**:
- **3s Orbital**: The \( 3s \) orbital can hold a maximum of 2 electrons. Since we are looking for \( s = +\frac{1}{2} \), we can have 1 electron with this spin.
- **3p Orbital**: The \( 3p \) orbital can hold a maximum of 6 electrons. The \( m = 0 \) state corresponds to one of the \( p \) orbitals, which can also have 1 electron with \( s = +\frac{1}{2} \).
- **3d Orbital**: The \( 3d \) orbital can hold a maximum of 10 electrons. The \( m = 0 \) state corresponds to one of the \( d \) orbitals, which can have 1 electron with \( s = +\frac{1}{2} \).
5. **Summing Up the Electrons**:
- From the \( 3s \) orbital: 1 electron (with \( s = +\frac{1}{2} \))
- From the \( 3p \) orbital: 1 electron (with \( s = +\frac{1}{2} \))
- From the \( 3d \) orbital: 1 electron (with \( s = +\frac{1}{2} \))
Therefore, the total number of electrons with the specified quantum numbers is:
\[
1 (from \, 3s) + 1 (from \, 3p) + 1 (from \, 3d) = 3
\]
### Final Answer:
The maximum number of electrons in \( \text{Mn} \) with \( n = 3 \), \( m = 0 \), and \( s = +\frac{1}{2} \) is **3**.
