An electron in the n = 7 level of the hydrogen atom relaxes to a lower-energy level, emitting light of 397 nm. What is the value of n for the level to which the electron relaxed?

• A. n = 3
• B. n = 4
• C. n = 5
• D. n = 1

Answer: (B)

Verified step by step guidance

1. Identify the initial and final energy levels of the electron transition. The initial level is given as n = 7, and the final level is unknown, denoted as n_f.
2. Use the Rydberg formula to relate the wavelength of the emitted light to the energy levels: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \), where \( \lambda \) is the wavelength (397 nm), \( R_H \) is the Rydberg constant (1.097 x 10^7 m^-1), \( n_i \) is the initial energy level (7), and \( n_f \) is the final energy level.
3. Convert the wavelength from nanometers to meters for consistency in units: 397 nm = 397 x 10^-9 m.
4. Rearrange the Rydberg formula to solve for \( n_f \): \( \frac{1}{n_f^2} = \frac{1}{\lambda R_H} + \frac{1}{n_i^2} \). Substitute the known values into the equation.
5. Calculate \( n_f \) by evaluating the right side of the equation and finding the inverse square root. Compare the result to the given options (n = 3, n = 4, n = 5, n = 1) to determine the correct final energy level.