# Why is the exciton binding energy slightly smaller than the band gap energy?

In a semiconductor it requires energy equivalent to the band gap energy ($E_g$) to excite an electron to the conduction band. This gives rise to an exciton (conduction electron-valence hole pair). The energy released when these recombine (exciton binding energy, $E_B$) is said to be slightly lower than the band gap energy.

Here are my questions:

1. Why is $E_B$ lower than $E_g$?

2. Where does the energy difference $E_g - E_B$ go in the process of forming an exciton?

I have noticed this question as a possible duplicate, but the answer doesn't really say more than "It is very complicated". Is there no sensible way to get at least an intuitive understanding of these questions?

• Note that the exciton binding energy is the difference between quasiparticle band gap and the exciton excitation energy. In your text, you refer to exciton binding energy as it were exciton excitation energy. pubs.rsc.org/services/images/… – Mikael Kuisma Dec 7 '16 at 0:14

An exciton is an interaction between what were originally a free electron and a free hole. Through the Coulomb force, these pair to generate a pseudo-hydrogen like complex.

I might suggest looking at Jacques Pankove's excellent 'Optical Processes in Semiconductors', where excitons are introduced on page 12. Some relevant quotes to consider:

A free hole and a free electron as a pair of opposite charges experience a coulomb attraction. Hence the electron can orbit the hole as if this were a hydrogen-like atom...

The exciton can wander through the crystal (the electron and the hole are now only relatively free because they are associated as a mobile pair). Because of this mobility, the exciton is not a set of spatially localized states. Furthermore, the exciton states do not have a well-defined potential in the semiconductor's energy diagram. However, it is customary to use the conduction-band edge as a reference level and to make this edge the continuum state ($n=\infty$).

This 'custom' makes some sense, because the continuum state is a return of the electron and hole to their 'free' states, which are in the conduction and valence bands.

1. Why is $$E_{B}$$ lower than $$E_{g}$$?

There's no such thing that limits $$E_{B} < E_{g}$$. $$E_{B}$$ can be larger than $$E_{g}$$ in rare circumstances. When $$E_{B}$$ exceeds $$E_{g}$$, a macroscopic number of excitons will be spontaneously formed (without any excitations). This "ground" state is typically referred as an excitonic insulator. See Phys. Rev. 158, 462 (1967) for example. From Grosso & Pastori Parravicini, Solid State Physics, it turns out that $$E_{B} \approx 13.6\dfrac{m_{\text{ex}}}{m_{e}}\dfrac{1}{\varepsilon^{2}}\quad\text{(in eV)}$$ which is on the order of few meV in inorganic semiconductors, compared to the band gap of few eV. However, $$E_{g}$$ can be engineered in double quantum wells, for example, where indirect excitons are formed with an conduction-band electron in one well and a valence-band hole in another well. This way, $$E_{g}$$ can be made smaller than $$E_{B}$$. See Nat. Commun. 8, 1971 (2017).

1. Where does the energy difference $$E_{g}−E_{B}$$ go in the process of forming an exciton?

During the relaxation process, mostly phonons are taking energy away from the hot electrons so that excitons can be formed. A phonon with energy $$E_{\text{phonon}} = E_{g}−E_{B}$$ or many smaller-energy phonons can take the energy away. Defects or other radiative/non-radiative processes can, too.

• I think the key point is your statement "When $E_B$ exceeds $E_g$, a macroscopic number of excitons will be spontaneously formed". Essentially having $E_B < E_g$ means that your system is stable and will not create more excitons. – KF Gauss Apr 26 '19 at 3:50