The energy of a photon emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom is given by ΔE = 1.89 eV. What is the wavelength of the emitted photon?
Q1. The energy of a photon emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom is given by ΔE = 1.89 eV. What is the wavelength of the emitted photon?
Answer: 656 nm
Explanation: Use the formula E = hc / λ, where E = 1.89 eV = 1.89 * 1.6 * 10^-19 J, to find λ = hc / E = 656 nm.