When an electron in a hydrogen atom moves from n = 8 to n = 2, light with what wavelength is emitted?

• A. 657 nm
• B. 93.8 nm
• C. 411 nm
• D. 487 nm
• E. 389 nm

Answer: (E)

Verified step by step guidance

1. Identify the initial and final energy levels of the electron transition: n_initial = 8 and n_final = 2.
2. Use the Rydberg formula to calculate the wavelength of the emitted light: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \), where \( R_H \) is the Rydberg constant for hydrogen (1.097 x 10^7 m^-1).
3. Substitute the values for n_initial and n_final into the Rydberg formula: \( \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{8^2} \right) \).
4. Calculate the difference in the fractions: \( \frac{1}{4} - \frac{1}{64} \), and simplify the expression.
5. Solve for \( \lambda \) by taking the reciprocal of the result from the Rydberg formula calculation to find the wavelength of the emitted light.