To determine which element has the minimum energy for the 2s orbital, we need to analyze the relationship between atomic number, electron energy levels, and orbital energies.
### Step-by-Step Solution:
1. **Understand the Energy Formula**: The energy of an electron in an orbital can be expressed using the formula:
\[
E = -\frac{13.6 \, Z^2}{n^2}
\]
where \(E\) is the energy, \(Z\) is the atomic number, and \(n\) is the principal quantum number (which is 2 for the 2s orbital).
2. **Identify the Elements and Their Atomic Numbers**:
- Hydrogen (H): Atomic number \(Z = 1\)
- Lithium (Li): Atomic number \(Z = 3\)
- Sodium (Na): Atomic number \(Z = 11\)
- Potassium (K): Atomic number \(Z = 19\)
3. **Calculate the Energy for Each Element**:
- For Hydrogen (H):
\[
E_H = -\frac{13.6 \times 1^2}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV}
\]
- For Lithium (Li):
\[
E_{Li} = -\frac{13.6 \times 3^2}{2^2} = -\frac{13.6 \times 9}{4} = -30.6 \, \text{eV}
\]
- For Sodium (Na):
\[
E_{Na} = -\frac{13.6 \times 11^2}{2^2} = -\frac{13.6 \times 121}{4} = -412.6 \, \text{eV}
\]
- For Potassium (K):
\[
E_{K} = -\frac{13.6 \times 19^2}{2^2} = -\frac{13.6 \times 361}{4} = -1225.3 \, \text{eV}
\]
4. **Analyze the Results**: The more negative the energy value, the lower the energy level. Therefore, we compare the calculated energies:
- \(E_H = -3.4 \, \text{eV}\)
- \(E_{Li} = -30.6 \, \text{eV}\)
- \(E_{Na} = -412.6 \, \text{eV}\)
- \(E_{K} = -1225.3 \, \text{eV}\)
5. **Determine the Minimum Energy**: Among these values, the least negative (highest energy) is for Hydrogen, and the most negative (lowest energy) is for Potassium. Thus, the energy of the 2s orbital is minimum for Potassium.
### Final Answer:
The element in which the energy of the 2s orbital is minimum is **Potassium (K)**.
