If the ionization potential of hydrogen atom is 13.6 V, its energy in the n = 3 is approximately

Options

A. –1.14 eV
B. –1.51 eV
C. –3.4 eV
D. –4.53 eV

Answer

To find the energy of a hydrogen atom in the n = 3 state, we can use the formula derived from Bohr's theory of the hydrogen atom. The energy of an electron in the nth energy level is given by:

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

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 (in this case, \(n = 3\)),
- \(13.6 \, \text{eV}\) is the ionization potential of hydrogen.

### Step-by-Step Solution:

1. **Identify the values**:
- For hydrogen, \(Z = 1\).
- The principal quantum number, \(n = 3\).

2. **Plug the values into the formula**:
\[ E_3 = -\frac{1^2 \cdot 13.6 \, \text{eV}}{3^2} \]

3. **Calculate \(3^2\)**:
\[ 3^2 = 9 \]

4. **Substitute this value back into the equation**:
\[ E_3 = -\frac{1 \cdot 13.6 \, \text{eV}}{9} \]

5. **Calculate the energy**:
\[ E_3 = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

6. **Final result**:
The energy of the hydrogen atom in the n = 3 state is approximately \(-1.51 \, \text{eV}\).