The ground state energy of hydrogen atom is -13.6eV . The energy of excited state `He^(+)` ion having principal quantum number n is , -6.04eV . The numerical value of n is ________.

Options

A. (a) 2
B. (b) 3
C. (c) 4
D. (d) 5

Explanation

To find the numerical value of the principal quantum number \( n \) for the excited state of the helium ion \( He^+ \), we can use the formula for the energy of an electron in a hydrogen-like atom:

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

Where:
- \( E \) is the energy of the electron,
- \( Z \) is the atomic number (for helium, \( Z = 2 \)),
- \( n \) is the principal quantum number.

### Step-by-step Solution:

1. **Identify the given values**:
- Ground state energy of hydrogen: \( -13.6 \text{ eV} \)
- Energy of excited state of \( He^+ \): \( -6.04 \text{ eV} \)
- Atomic number \( Z \) for helium: \( 2 \)

2. **Substitute the known values into the energy formula**:
\[ -6.04 = -\frac{13.6 \times 2^2}{n^2} \]

3. **Simplify the equation**:
\[ -6.04 = -\frac{13.6 \times 4}{n^2} \]
\[ -6.04 = -\frac{54.4}{n^2} \]

4. **Remove the negative signs**:
\[ 6.04 = \frac{54.4}{n^2} \]

5. **Cross-multiply to solve for \( n^2 \)**:
\[ 6.04 n^2 = 54.4 \]

6. **Isolate \( n^2 \)**:
\[ n^2 = \frac{54.4}{6.04} \]

7. **Calculate \( n^2 \)**:
\[ n^2 \approx 9 \]

8. **Take the square root to find \( n \)**:
\[ n = \sqrt{9} = 3 \]

Thus, the numerical value of \( n \) is \( 3 \).

### Final Answer:
The numerical value of \( n \) is \( 3 \).