Final answer:
An atom with a ground state at 0 eV and an excited state at 2 eV would typically remain in the ground state unless it absorbs a photon with energy equal to the energy difference between these states. A photon with an energy of 1.5 eV would not be absorbed because energy levels are quantized; thus, an emission would correspond to the actual energy difference if the atom transitions from the excited state to the ground state.
Step-by-step explanation:
Regarding an atom with only two states: ground state at 0 eV and excited state at 2 eV, answering the hypothetical situation would depend on additional information that is not provided. Nonetheless, under normal circumstances, if no energy is added to the system, an atom would tend to remain in its ground state, the state of lowest possible energy. If a photon with an energy of 2 eV interacts with the atom, the atom could absorb this photon, and such energy would raise its electron to the excited state, in keeping with the concept that the atom can neither create nor destroy energy but can change energy from one form to another (conservation of energy).
However, if the energy of the photon is less than 2 eV, say 1.5 eV as in answer choice D, the atom cannot absorb it to reach the exact excited state since the energy levels are quantized, meaning the electron can only move between discrete energy levels. Thus, the atom would not absorb a 1.5 eV photon if the excited state is at 2 eV. If instead, the atom were already in the excited state, it could return to its ground state, and during this process, it would emit a photon of a specific energy equivalent to the energy difference between the excited state and the ground state. In the question posed, that emitted photon would have an energy of 2 eV—not 1.5 eV—due to the energy difference between the given states.