Atoms of some gas can be in three states with energies: -2.5 eV, -3.2 eV, -4.6 eV. If they are in a state

Atoms of some gas can be in three states with energies: -2.5 eV, -3.2 eV, -4.6 eV. If they are in a state with an energy of -3.2 eV, then what energy photons can the atoms of this gas emit?

For an electron in a state with an energy of -3.2 eV, only transitions with radiation to a state with an energy of -4.6 eV are possible (that is, from a state with a higher energy to a state with a lower energy). In this case, the energy of the emitted photon is equal to the difference between the energies of the electron before and after the transition:

ΔE = -3.2 eV – (-4.6 eV) = -3.2 eV + 4.6 eV = 1.4 eV.

Answer: 1.4 eV.