Final answer:
When an electron in a hydrogen atom transitions from a higher energy level to a lower one (ni = 13 to nf = 3), the atom emits a photon, consistent with the processes of atomic excitations and de-excitations.
Step-by-step explanation:
When an electron in a hydrogen atom moves from a higher energy level (ni = 13) to a lower one (nf = 3), the atom emits a photon. This process is a part of atomic excitations and de-excitations, which can be explained using Bohr's model of the atom. According to the model, transitions to lower energy levels release energy in the form of electromagnetic radiation (photons), while transitions to higher energy levels require energy, typically absorbing a photon. This emission is observable in the hydrogen spectral lines, which are characteristic of the atom and correspond to specific wavelengths of light.
For the given transition, the energy of the emitted photon can be calculated using the equation ΔE = hf = Ei - Ef, where ΔE is the energy change, h is Planck's constant, f is the frequency of the emitted photon, Ei is the initial energy level, and Ef is the final energy level. In this case, the initial energy level is ni = 13 and the final energy level is nf = 3. According to the energy level equation En = (-13.6 eV/n²), the energy levels can be computed and the difference in energy will give us the energy of the photon that is emitted.