The ground state energy of hydrogen atom is `-13.6 eV`. The energy of second excited state of `He^(+)` ion in eV is

Options

A. `-27.2`
B. `-3.4`
C. `-54.4`
D. `-6.04`

Explanation

To find the energy of the second excited state of the He\(^+\) ion, we can use the formula for the energy levels of hydrogen-like atoms:

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

where:
- \(E\) is the energy of the state,
- \(Z\) is the atomic number,
- \(n\) is the principal quantum number (the shell number).

### Step-by-step Solution:

1. **Identify the Atomic Number (Z)**:
- For Helium (He), the atomic number \(Z = 2\).

2. **Determine the Principal Quantum Number (n)**:
- The ground state corresponds to \(n = 1\).
- The first excited state corresponds to \(n = 2\).
- The second excited state corresponds to \(n = 3\).

3. **Substitute the Values into the Formula**:
- Now, we substitute \(Z = 2\) and \(n = 3\) into the formula:
\[ E = -\frac{13.6 \, \text{eV} \cdot (2)^2}{(3)^2} \]

4. **Calculate \(Z^2\) and \(n^2\)**:
- \(Z^2 = 2^2 = 4\)
- \(n^2 = 3^2 = 9\)

5. **Plug in the Values**:
\[ E = -\frac{13.6 \, \text{eV} \cdot 4}{9} \]

6. **Perform the Multiplication**:
\[ E = -\frac{54.4 \, \text{eV}}{9} \]

7. **Final Calculation**:
\[ E = -6.04 \, \text{eV} \]

Thus, the energy of the second excited state of the He\(^+\) ion is \(-6.04 \, \text{eV}\).

### Final Answer:
\[ E = -6.04 \, \text{eV} \]