If an electron of a hydrogen atom jumps from `n = 2` to `n=1` then find the wavelength of released photon

Options

A. `121.5 nm`
B. `123.15 nm`
C. `125.15 nm`
D. `128.15 nm`

Answer

To find the wavelength of the photon released when an electron in a hydrogen atom jumps from the n=2 energy level to the n=1 energy level, we can use the Rydberg formula for hydrogen:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the emitted photon,
- \( R \) is the Rydberg constant for hydrogen, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \),
- \( n_1 \) is the lower energy level (1 in this case),
- \( n_2 \) is the higher energy level (2 in this case).

### Step-by-step Solution:

1. **Identify the values of \( n_1 \) and \( n_2 \)**:
- Here, \( n_1 = 1 \) and \( n_2 = 2 \).

2. **Substitute the values into the Rydberg formula**:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right) \]

3. **Calculate \( \frac{1}{4} \)**:
\[ \frac{1}{4} = 0.25 \]
Thus,
\[ 1 - \frac{1}{4} = 1 - 0.25 = 0.75 \]

4. **Substitute back into the equation**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.75 \]

5. **Calculate \( 1.097 \times 10^7 \times 0.75 \)**:
\[ \frac{1}{\lambda} = 0.82275 \times 10^7 \, \text{m}^{-1} \]

6. **Find \( \lambda \)** by taking the reciprocal:
\[ \lambda = \frac{1}{0.82275 \times 10^7} \approx 1.215 \times 10^{-7} \, \text{m} \]

7. **Convert to nanometers**:
\[ \lambda \approx 1.215 \times 10^{-7} \, \text{m} = 121.5 \, \text{nm} \]

### Final Answer:
The wavelength of the released photon is approximately **121.5 nm**.