Final answer:
The sentence is true; an excited state is less stable than a ground state. For a hydrogen atom that can be ionized with 0.850 eV, the principal quantum number n can be found using the Rydberg formula by comparing the given ionization energy to the electron's energy level in the excited state.
Step-by-step explanation:
The sentence stating that an excited state is less stable than a ground state is true. In the context of hydrogen atoms, the ground state is the most stable configuration where the electron is in the orbit closest to the nucleus (n=1). When hydrogen receives energy, the electron moves to an excited state by jumping to a higher energy level (n > 1), where it is less tightly bound to the nucleus and thus less stable. An atom in an excited state requires less energy to ionize because the electron is already at a higher energy level compared to when it is in the ground state.
To find n, the principal quantum number, for a hydrogen atom that can be ionized with 0.850 eV of energy when in its excited state, we will use the Rydberg formula for the energy levels of a hydrogen atom:
Energy = -13.6 eV (1/n2)
Given the ionization energy of 0.850 eV, we compare it to the energy of the electron in the nth state to find the corresponding n value. Since we're dealing with ionization, the final state is n = ∞ because the electron is removed from the atom. Normally, ionization energy from the ground state is 13.6 eV, but since it's already in an excited state, the value is less (0.850 eV). By using the energy difference (13.6 eV - 0.850 eV) and solving for n, we'll arrive at the principal quantum number of the excited state.