Calculate the wavelength of radiation emited when an electron in a hydrogen atom makes a transition from an energy level with `n = 3` to a level with `n= 2`

Explanation

To calculate the wavelength of radiation emitted when an electron in a hydrogen atom transitions from an energy level with \( n = 3 \) to a level with \( n = 2 \), we can use the Rydberg formula for hydrogen:

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

where:
- \( \lambda \) is the wavelength of the emitted radiation,
- \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.1 \times 10^7 \, \text{m}^{-1} \),
- \( n_1 \) is the initial energy level (3 in this case),
- \( n_2 \) is the final energy level (2 in this case).

### Step-by-Step Solution:

1. **Identify the values of \( n_1 \) and \( n_2 \)**:
- Here, \( n_1 = 3 \) and \( n_2 = 2 \).

2. **Substitute the values into the Rydberg formula**:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

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. **Calculate \( \frac{1}{n_2^2} - \frac{1}{n_1^2} \)**:
\[ \frac{1}{n_2^2} - \frac{1}{n_1^2} = 0.25 - 0.1111 = 0.1389 \]

5. **Substitute back into the equation**:
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 0.1389 \]

6. **Calculate \( \frac{1}{\lambda} \)**:
\[ \frac{1}{\lambda} \approx 1.1 \times 10^7 \times 0.1389 \approx 1.538 \times 10^6 \, \text{m}^{-1} \]

7. **Calculate \( \lambda \)**:
\[ \lambda = \frac{1}{1.538 \times 10^6} \approx 6.5 \times 10^{-7} \, \text{m} \]

8. **Convert to nanometers**:
\[ \lambda \approx 6.5 \times 10^{-7} \, \text{m} = 650 \, \text{nm} \]

### Final Answer:
The wavelength of radiation emitted when an electron transitions from \( n = 3 \) to \( n = 2 \) in a hydrogen atom is approximately **650 nm**.