Final answer:
An atom of hydrogen in an excited state has higher potential energy than in its ground state and requires less energy to become ionized. The principal quantum number, n, indicates the electron's energy level, and the atom can be ionized using energy that is specific to that level. To find n for a specific ionization energy, such as 0.850 eV, one would use the known energy levels of hydrogen.
Step-by-step explanation:
Compared to an atom of hydrogen in the ground state, an atom of hydrogen in an excited state has higher potential energy and can be ionized with less energy. In the Bohr model of the hydrogen atom, the ground state has the electron in the smallest orbit, with n=1 being the lowest energy level. When an atom absorbs energy, its electron moves to a higher energy level (or orbit), and the atom is then in an excited state.
If a hydrogen atom requires only 0.850 eV of energy to ionize, it is in an excited state. We can determine the principal quantum number, n, for this excited state atom using the energy levels known for the hydrogen atom. The energy required to ionize a hydrogen atom can be linked to the principal quantum number, which indicates the energy level the electron occupies before ionization.
The question asks for the n level if only 0.850 eV is required to ionize the atom, which means the atom is not in the ground state but an excited state. Since the ground state ionization energy is 13.6 eV, an energy less than this implies that the electron is at a higher energy level, and as n increases, the energy required to ionize decreases. The formula linking energy levels to ionization energy can be used to calculate the exact value of n.