What is the wavelength of a photon emitted during a transition from `n = 5` state to the `n = 2` state in the hydrogen atom

Options

A. 434 nm
B. 234nm
C. 476nm
D. 244nm

Explanation

To find the wavelength of a photon emitted during a transition from the n = 5 state to the n = 2 state in a hydrogen atom, we can use the Rydberg formula for hydrogen. The formula for the wavelength (λ) of the emitted photon is given by:

\[ \frac{1}{\lambda} = R_H \left( \frac{Z^2}{n_f^2} - \frac{Z^2}{n_i^2} \right) \]

Where:
- \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.096 \times 10^7 \, \text{m}^{-1} \).
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)).
- \( n_f \) is the final energy level (in this case, \( n_f = 2 \)).
- \( n_i \) is the initial energy level (in this case, \( n_i = 5 \)).

### Step-by-Step Solution:

1. **Identify the values:**
- For hydrogen, \( Z = 1 \).
- \( n_f = 2 \)
- \( n_i = 5 \)

2. **Substitute the values into the Rydberg formula:**
\[ \frac{1}{\lambda} = R_H \left( \frac{1^2}{2^2} - \frac{1^2}{5^2} \right) \]

3. **Calculate \( \frac{1^2}{2^2} \) and \( \frac{1^2}{5^2} \):**
\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{5^2} = \frac{1}{25} = 0.04 \]

4. **Subtract the two fractions:**
\[ 0.25 - 0.04 = 0.21 \]

5. **Substitute back into the equation:**
\[ \frac{1}{\lambda} = R_H \times 0.21 \]
\[ \frac{1}{\lambda} = 1.096 \times 10^7 \times 0.21 \]

6. **Calculate \( \frac{1}{\lambda} \):**
\[ \frac{1}{\lambda} = 2.3036 \times 10^6 \, \text{m}^{-1} \]

7. **Take the reciprocal to find \( \lambda \):**
\[ \lambda = \frac{1}{2.3036 \times 10^6} \approx 4.344 \times 10^{-7} \, \text{m} \]

8. **Convert meters to nanometers (1 m = \( 10^9 \) nm):**
\[ \lambda \approx 434.4 \, \text{nm} \]

### Final Answer:
The wavelength of the photon emitted during the transition from \( n = 5 \) to \( n = 2 \) in a hydrogen atom is approximately **434.4 nm**.