In a hydrogen atom, if energy of an electron in ground state is `- 13.6 eV`, then that in the `2^(nd)` excited state is :

Options

A. `1.51 eV`
B. `3.4 eV`
C. `6.04 eV`
D. `13.6 eV`

Explanation

To find the energy of an electron in the second excited state of a hydrogen atom, we can use the formula for the energy levels of hydrogen:

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

Where:
- \( E_n \) is the energy of the electron at the principal quantum number \( n \),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n \) is the principal quantum number.

### Step-by-Step Solution:

1. **Identify the Ground State Energy**:
The energy of the electron in the ground state (n=1) is given as \( -13.6 \, \text{eV} \).

2. **Determine the Principal Quantum Number for the Second Excited State**:
The second excited state corresponds to \( n = 3 \) (since the first excited state is \( n = 2 \) and the ground state is \( n = 1 \)).

3. **Substitute Values into the Energy Formula**:
We substitute \( Z = 1 \) and \( n = 3 \) into the energy formula:
\[ E_3 = -\frac{13.6 \, \text{eV} \cdot (1)^2}{(3)^2} \]

4. **Calculate the Denominator**:
Calculate \( (3)^2 = 9 \).

5. **Calculate the Energy**:
Now, substitute this value back into the equation:
\[ E_3 = -\frac{13.6 \, \text{eV}}{9} \]

6. **Perform the Division**:
\[ E_3 = -1.51 \, \text{eV} \]

7. **Final Answer**:
The energy of the electron in the second excited state is \( -1.51 \, \text{eV} \).

### Summary:
The energy of an electron in the second excited state of a hydrogen atom is \( -1.51 \, \text{eV} \).