Radiative or non-radiative emission

Question

If a material absorbs photon, it's electron on outer valence band absorbs its energy and jumps to the higher energy level, when the band gap is similar to the photon's energy. If so, what determines the electron, to emit the energy either in radiative or non-radiative form (upon de-excitation)?

Answer 1

If a material absorbs photon, it's electron on outer valence band absorbs its energy and jumps to the higher energy level, when the band gap is similar to the photon's energy.

Firstly, how material absorbs photons depends on the type of materials. Atoms and molecules have discrete energy levels, and absorption results in electron moving to a higher energy level, satisfying condition: $$ E_f=E_i+\hbar\omega,$$ where $E_i, E_f$ are the energies of the initial and the final energy levels, whereas $\hbar\omega$ is the photon energy. Note that absorption is possible only if the initial level is already occupied by an electron.

In crystals, having regular periodic structure, energy levels are broadened into energy bands, so transitions occur between the bands. At normal temperatures the valence band is nearly fully occupied, and the conduction band is nearly empty, so the absorption is possible at any energy that is greater than the gap (i.e., the distance between the lowest energy in the conduction band and the highest energy in the valence band): $$\hbar\omega\geq E_g=E_c-E_v.$$ (Some subgap absorption is also possible due to impurities and exciton resonances.)

If so, what determines the electron, to emit the energy either in radiative or non-radiative form (upon de-excitation)?

In the first approximation probabilities of absorption and emission processes is equal: $w_{i\rightarrow f}=w_{f\rightarrow i}$. The rates of absorption and emission, i.e., how many electrons move to higher or to lower energy levels are however dependent on the initial occupation of the energy levels: $$ R_{i\rightarrow f}=w_{i\rightarrow f}N_i,\\ R_{f\rightarrow i}=w_{f\rightarrow i}N_f $$(for simplicity I ignore here constraints imposed by the Pauli exclusion principle, and the fact that the transition probabilities are proportional to the field intensity.)

In normal conditions the populations of the initial and the final states are related via a Boltzmann factor $$ \frac{N_f}{N_i}=e^{-\frac{E_f-E_i}{k_BT}}.$$ Thus, when we illuminate a material, the absorption rate is higher than the emission rate, and the overall absorption of radiation is observed. In typical photo-luminescence experiments, illumination is performed using a short pulse, and the emission of radiation observed after the illuminating pulse is over (alternatively, one can use constant illumination by a directed source, such as a laser, and observe luminescence emitted in a different direction.) The emission of radiation typically tries to restore the Boltzmann equilibrium, perturbed by the excitations of electrons via the absorption of photons.

Various non-radiative processes also become possible because now electrons are available to participate in them. E.g., Auger recombination involves relaxation of a conduction electron to a valence band, after transmitting energy to another conduction electron (rather than a photon.) Likewise, conduction electron may transition to valence band via giving its energy to lattice vibrations (phonons.)

If so, what determines the electron, to emit the energy either in radiative or non-radiative form (upon de-excitation)?

Whether radiative or non-ratiative relaxation occurs is a matter of the relative probabilities of various processes. In fact - unless we talk about state-of-the-art single atom experiments or the like - we always deal with a great number of electrons, part of which will relax via radiative processes, and part via non-radiative relaxation. If the rate of radiative transitions is $w_{f\rightarrow i}^{rad}$, while the rate of the non-radiative ones is $w_{f\rightarrow i}^{non-rad}$, we can write the following rate equations: $$ \frac{dN_f}{dt}=w_{i\rightarrow f}^{rad}N_i,\\ \frac{dN_i}{dt}=\left(w_{f\rightarrow i}^{rad}+ w_{f\rightarrow i}^{non-rad}\right)N_f. $$ In fact, the situation in crystals is more complicated, since we have to take account for a continuum of states in the valence and the conduction bands. Thus, while the initial and the final states might be located deep in the bands (if the excitation photon energy allows), both the holes and the electron would usually relax (mostly non-radiatively) towards the band edges, and only then recombine, emitting a photon close to the energy of the band gap (or non-radiatively). Moreover, with multiple electrons excited, nothing obliges an electron to recombine with the same hole that it originally came from (in fact, due to the particle indistinguishability, matching electrons and holes doesn't even make sense.)

Answer 2

The short answer is that radiative decay will happen whenever the energy excites from one band to the other, as intraband transitions have typically very low energies and can couple dispersively to phonons in the material. However, as pointed out by Jon Custer in the comments, the situation can get more complicated when impurities and defects are taken into account, as they can create nontrivial structures in the spectrum of the material.

As a rule of thumb, however, as long as the other dissipative channels of the material, such as phonons, magnons, or any other collective excitations possible, can be somehow suppressed and/or ignored, then the coupling to the electromagnetic field will dominate and the system will relax emitting a photon.