The energy of a hydrogen atom in the ground state is `-13.6 eV`. The eneergy of a `He^(+)` ion in the first excited state will be

Options

A. `-13.6` eV
B. `-27.2` eV
C. `-54.5` eV
D. `-6.8` eV

Answer

To find the energy of a \( \text{He}^+ \) ion in the first excited state, we can use the formula for the energy levels of hydrogen-like atoms:

\[ E_n = -\frac{13.6 \, \text{eV} \times Z^2}{n^2} \]

where:
- \( E_n \) is the energy of the electron in the nth energy level,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step-by-Step Solution:

1. **Identify the values of \( Z \) and \( n \)**:
- For \( \text{He}^+ \), the atomic number \( Z = 2 \).
- The first excited state corresponds to \( n = 2 \).

2. **Substitute the values into the formula**:
\[ E_2 = -\frac{13.6 \, \text{eV} \times (2)^2}{(2)^2} \]

3. **Calculate \( (2)^2 \)**:
\[ (2)^2 = 4 \]

4. **Substitute back into the equation**:
\[ E_2 = -\frac{13.6 \, \text{eV} \times 4}{4} \]

5. **Simplify the equation**:
\[ E_2 = -13.6 \, \text{eV} \]

6. **Conclusion**:
The energy of the \( \text{He}^+ \) ion in the first excited state is \( -13.6 \, \text{eV} \).

### Final Answer:
The energy of a \( \text{He}^+ \) ion in the first excited state is \( -13.6 \, \text{eV} \).