The energy of the photon emitted when an electron jumps from n = 4 to n = 2 in a hydrogen atom is 2.55 eV. What is the wavelength of the emitted photon?
Q1. The energy of the photon emitted when an electron jumps from n = 4 to n = 2 in a hydrogen atom is 2.55 eV. What is the wavelength of the emitted photon?
Answer: 486 nm
Explanation: Use E = hc / λ, where E = 2.55 eV, to find λ = 486 nm.