What is the energy of the photon emitted when an electron in a hydrogen atom falls from n = 4 to n = 2?
Q1. What is the energy of the photon emitted when an electron in a hydrogen atom falls from n = 4 to n = 2?
Answer: 2.55 eV
Explanation: Using energy level formula, the energy difference between n = 4 and n = 2 is 2.55 eV; option B is for n = 3 to n = 2.