A certain material is kept at very low temperature. It is observed that when photons with energies between 0.2 and 0.9 eV strike the material, only photons of 0.4 eV and 0.7 eV are absorbed. Next, the material is warmed up so that it starts to emit photons. When it has been warmed up enough that 0.7 eV photons begin to be emitted, what other photon energies are also observed to be emitted by the material? Explain briefly.

The given data is listed below as,

• The energies of the photons are $0.2\mathrm{eV}\mathrm{and}0.9\mathrm{eV}.$
• The energies of the absorbed photons are $0.4\mathrm{eV}\mathrm{and}0.7\mathrm{eV}.$
• The energy of the photons when emitted from the warmed material is$0.7\mathrm{eV}.$

The law of Maxwell-Boltzmann distribution states that energy distribution occurs only between the distinguishable and the identical particles.

The Maxwell-Boltzmann law gives the photon energies that are to be emitted by the material.

The detected lines of absorption at the energy of$0.4\mathrm{eV}\mathrm{and}0.7\mathrm{eV}$ provides the position of the “first two states of energy” that is relative to the ground state. However, the absorbed photons take electrons from the ground state to different excited states.

While heating up the material, the electrons gain energy to move from the ground state to the other excited states and when the electrons come back to the ground states, the electrons emit photons.

As the law of Maxwell-Boltzmann distribution is mainly continuous, that shows some electrons will go to the first excited state that is$0.4eV$ and some electrons will also go from the first to the second excited state that is $0.3\mathrm{eV}.$

Thus, the photon energies observed to be emitted by the material are$0.3\mathrm{eV},04\mathrm{eV}\mathrm{and}0.7\mathrm{eV}.$