The amount of energy emitted if electron falls from n = 3 to n = 2 , in hydrogen atom is

Options

A. 0.65 eV
B. 1.9 eV
C. 10.2 eV
D. 12.09 eV

Answer

To find the amount of energy emitted when an electron falls from n = 3 to n = 2 in a hydrogen atom, we can use the formula for the energy difference between two energy levels in a hydrogen atom:

\[ \Delta E = -13.6 \, \text{eV} \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \Delta E \) is the energy emitted,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (n = 2),
- \( n_2 \) is the higher energy level (n = 3).

### Step-by-Step Solution:

1. **Identify the values**:
- For hydrogen, \( Z = 1 \).
- \( n_1 = 2 \) (final state).
- \( n_2 = 3 \) (initial state).

2. **Substitute the values into the formula**:
\[ \Delta E = -13.6 \, \text{eV} \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

3. **Calculate \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \)**:
- \( \frac{1}{2^2} = \frac{1}{4} = 0.25 \)
- \( \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \)

4. **Find the difference**:
\[ \frac{1}{2^2} - \frac{1}{3^2} = 0.25 - 0.1111 = 0.1389 \]

5. **Substitute back into the equation**:
\[ \Delta E = -13.6 \, \text{eV} \cdot (0.1389) \]

6. **Calculate \( \Delta E \)**:
\[ \Delta E \approx -1.89 \, \text{eV} \]

7. **Final answer**:
The energy emitted when the electron falls from n = 3 to n = 2 in a hydrogen atom is approximately **1.89 eV** (the negative sign indicates energy is emitted).