To determine which set of quantum numbers describes the electron that is removed most easily from a potassium atom in its ground state, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Atomic Number and Electronic Configuration of Potassium:**
- Potassium (K) has an atomic number of 19.
- Its electronic configuration is:
\[
1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1
\]
- This shows that potassium has one electron in the 4s orbital, which is the outermost shell.
2. **Determine the Valence Electron:**
- The electron that is removed most easily from an atom is typically the valence electron, which is the outermost electron.
- In potassium, the valence electron is in the 4s orbital (4s^1).
3. **Assign Quantum Numbers to the Valence Electron:**
- The quantum numbers for the valence electron in the 4s orbital are:
- Principal quantum number (n): 4 (since it is in the 4th shell)
- Azimuthal quantum number (l): 0 (for s orbitals, l = 0)
- Magnetic quantum number (ml): 0 (for l = 0, ml can only be 0)
- Spin quantum number (ms): can be either +1/2 or -1/2 (we can choose +1/2 for this example)
Therefore, the set of quantum numbers for the valence electron is:
\[
n = 4, \, l = 0, \, m_l = 0, \, m_s = +\frac{1}{2}
\]
4. **Evaluate the Given Options:**
- We need to check the provided options against the quantum numbers we determined.
- The correct set of quantum numbers should match \( n = 4, l = 0, m_l = 0 \).
5. **Select the Correct Option:**
- After evaluating the options:
- Any option with \( n \neq 4 \) can be eliminated.
- The option with \( n = 4, l = 0, m_l = 0 \) and \( m_s = +\frac{1}{2} \) is the correct choice.
### Conclusion:
The correct set of quantum numbers that describes the electron which is removed most easily from a potassium atom in its ground state is:
\[
n = 4, \, l = 0, \, m_l = 0, \, m_s = +\frac{1}{2}
\]
