According to the quantum-mechanical model for the hydrogen atom, which electron transition produces light with the longer wavelength: 2p to 1s or 3p to 1s?

Verified step by step guidance

1. Identify the concept of electron transitions and how they relate to the emission of light. In the hydrogen atom, when an electron transitions from a higher energy level to a lower one, it emits a photon of light.
2. Recall that the energy of the emitted photon is related to the difference in energy levels between the initial and final states. The greater the energy difference, the shorter the wavelength of the emitted light.
3. Use the Rydberg formula to calculate the energy difference for each transition: \( \Delta E = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( R_H \) is the Rydberg constant, \( n_1 \) is the principal quantum number of the lower energy level, and \( n_2 \) is the principal quantum number of the higher energy level.
4. Calculate the energy difference for the 2p to 1s transition: \( n_1 = 1 \) and \( n_2 = 2 \).
5. Calculate the energy difference for the 3p to 1s transition: \( n_1 = 1 \) and \( n_2 = 3 \). Compare the energy differences to determine which transition emits light with a longer wavelength.