Final answer:
The absorption of a 93.7-nm photon by a hydrogen atom increases its energy and significantly lowers the ionization energy required for n=6 compared to the ground state. The energy levels become less negative as 'n' increases, leading to a lower ionization energy as described by Bohr's theory.
Step-by-step explanation:
When a hydrogen atom absorbs a 93.7-nm photon, the energy of the atom increases. This absorption leads the atom to an excited state, specifically the fifth excited state or n=6. In this state, the ionization energy required to remove the electron decreases significantly compared to its requirement in the ground state (where n=1). While the ionization energy for a ground state hydrogen atom is approximately 13.6 eV, it is only 0.378 eV in the n=6 state, which is 36 times less energy.
The energy levels of an electron in a hydrogen atom are given by the formula En = -13.6 eV/n² where n is the principal quantum number. As the principal quantum number increases, the electron resides further from the nucleus and the energy levels become less negative, making it easier to ionize the atom. This behavior is well described by Bohr's theory of the hydrogen atom, which suggests that energy levels are proportional to 1/n², and the ionization energy corresponds to the energy needed to bring the electron from its current energy level to zero energy (free electron). Thus, when the hydrogen atom is in an excited state, it requires less additional energy to ionize it compared to the atom in its ground state.