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 the n = 3 level to the n = 2 level in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Levels**: The energy of an electron in a hydrogen atom at a given energy level n is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Calculate the Energy at n = 3**:
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.51 \, \text{eV} \]

3. **Calculate the Energy at n = 2**:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

4. **Calculate the Change in Energy (ΔE)**:
The change in energy when the electron transitions from n = 3 to n = 2 is given by:
\[ \Delta E = E_2 - E_3 \]
Substituting the values we calculated:
\[ \Delta E = (-3.4 \, \text{eV}) - (-1.51 \, \text{eV}) = -3.4 \, \text{eV} + 1.51 \, \text{eV} = -1.89 \, \text{eV} \]

5. **Interpret the Result**: The negative sign indicates that energy is emitted when the electron falls to a lower energy level. Therefore, the amount of energy emitted is:
\[ \Delta E = 1.89 \, \text{eV} \]

### Final Answer:
The amount of energy emitted when the electron falls from n = 3 to n = 2 in a hydrogen atom is approximately **1.89 eV**.