The energy of a hydrogen atom in its ground state is `-13.6 eV`. The energy of the level corresponding to the quantum number `n = 2` (first excited state) in the hydrogen atom is

Options

A. `0.54 eV`
B. `-3.4 eV`
C. `-2.72 eV`
D. `-0.85 eV`

Answer

To find the energy of a hydrogen atom in its first excited state (quantum number \( n = 2 \)), we can use the formula derived from Bohr's theory of the hydrogen atom:

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

where:
- \( E_n \) is the energy of the electron in the nth level,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n \) is the principal quantum number.

### Step-by-Step Solution:

1. **Identify the parameters**:
- For hydrogen, the atomic number \( Z = 1 \).
- We need to find the energy for the first excited state, which corresponds to \( n = 2 \).

2. **Substitute the values into the formula**:
\[ E_2 = -\frac{13.6 \, (1)^2}{(2)^2} \, \text{eV} \]

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

4. **Substitute \( (2)^2 \) back into the equation**:
\[ E_2 = -\frac{13.6 \, (1)}{4} \, \text{eV} \]

5. **Perform the division**:
\[ E_2 = -\frac{13.6}{4} \, \text{eV} = -3.4 \, \text{eV} \]

6. **Conclusion**:
The energy of the level corresponding to the quantum number \( n = 2 \) in the hydrogen atom is \( -3.4 \, \text{eV} \).

### Final Answer:
The energy of the level corresponding to the quantum number \( n = 2 \) in the hydrogen atom is \( -3.4 \, \text{eV} \).