Final answer:
In the Bohr model of the hydrogen atom, the transition from the first excited state (n=2) to the ground state (n=1), part of the Lyman series, results in the emission of the lowest-energy photon. This is because the energy levels are proportional to 1/n² and transitions between closer energy levels emit photons with lower energy.
Step-by-step explanation:
The subject question is related to the Bohr hydrogen atom and its electronic transitions. Specifically, it explores which transition results in the emission of the lowest-energy photon. According to Bohr's theory, when an electron moves between energy levels, the atom emits or absorbs energy in the form of a photon. The energy of the photon is proportional to the difference between the initial and final energy states of the electron (En > Em). This relationship is given by the formula E = h(ν), where E is the energy, h is Planck's constant, and ν is the frequency of the emitted or absorbed photon.
The transition with the smallest energy difference will result in the emission of the lowest-energy photon. In the Bohr model, the energy levels are proportional to 1/n². So, the transition with the lowest energy emission will be the one between the closest energy levels. For example, the transition from the n=3 level to the n=2 level (part of the Balmer series) would produce a higher-energy photon compared to a transition from n=2 to n=1 (part of the Lyman series), because the energy levels are closer together in the latter.