What is the frequency of a photon emitted when an electron in a hydrogen atom transitions from n = 3 to n = 2? (R_H = 1.097 x 10^7 m^−1)

• A. 4.57 x 10^14 Hz
• B. 5.45 x 10^14 Hz
• C. 3.28 x 10^14 Hz
• D. 6.56 x 10^14 Hz

Answer: (A)

Verified step by step guidance

1. Identify the initial and final energy levels of the electron transition: n_initial = 3 and n_final = 2.
2. Use the Rydberg formula to calculate the wavelength of the emitted photon: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \), where \( R_H = 1.097 \times 10^7 \text{ m}^{-1} \).
3. Substitute the values of n_initial and n_final into the Rydberg formula to find \( \frac{1}{\lambda} \).
4. Calculate the wavelength \( \lambda \) by taking the reciprocal of the result from the previous step.
5. Use the speed of light equation \( c = \lambda \nu \) to find the frequency \( \nu \), where \( c = 3.00 \times 10^8 \text{ m/s} \). Rearrange to solve for \( \nu = \frac{c}{\lambda} \).