Photoelectric absorption in semiconductors

In summary, the professor said that if you want to neglect the wave vector kph, you need to combine the conservation of momentum with the conservation of energy. However, the key is that the momentum has to be combined with the energy in the same way.
  • #1
MicheleC88
4
0
Hi everybody. I'm new here and, first of all, sorry for my bad english :-D

I'm studying photoelectric absorption in semiconductors.
The book (and professor too) says that, in the conservation law:

ki + kph = kf

(where ki and kf are wave vectors of initial and final electron state, and kph is the wave vector of incoming photon) we can neglect kph because it is ≈ 2π/λ, whereas ki and kf are ≈ 2π/a, and λ>>a. (a is the length of unitary cell in real space).
But I think that this assumption is good only if electron is at the edge of Brillouin Zone; if the initial and final electrons are near [itex]\Gamma[/itex]-point, they should have a very little wave vector, comparable with kph, making the approximation not valid.

what is wrong in my words?
 
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  • #2
Just do the math once and sketch the photon dispersion (energy versus k) and the electron dispersion (crystal electron or for simplicity even a free one) into the same graph with the correct dimensions. This is pretty instructive and will give you a good argument for why the approximation your professor gave you is a very good one pretty much everywhere.
 
  • #3
So, I think having understood from your words, the key is that the conservation of momentum has to be combined with the conservation of energy?
 
  • #4
Actually, that conservation law is valid only for the in-plane momentum, i.e. parallel to the surface of the material. The out-of-plane momentum is way more complicated than that.

Zz.
 
  • #5
I tried to make the following math.
from energy conservation: Eph≈Egap
(assuming that initial and final electrons are in proximity of, respectively, the maximum of VB and minimum of CB)

We know that kph = ω/c = Eph/([itex]\hbar[/itex]c)

So, from momentum conservation law:

|kf - ki | = |kph|≈Egap/([itex]\hbar[/itex]c)

Doing the calc (i assumed 1 eV for Egap):
|kf - ki | ≈ 10-4 angstrom-1, which is about 1 part of thousand of tipical size of Brillouin zone.

Are my reasoning correct to justify the assumpion kph=0 ?
 
Last edited:
  • #6
However, I think I made the math more complex than necessary:

ki - kf = kph = [itex]\frac{2π}{λ}\widehat{k}_{ph}[/itex]

(ki - kf) / (size of Brillouin zone) = [itex]\frac{2π}{λ}\frac{a}{2π}\widehat{k}_{ph}[/itex] << 1 [itex]\cdot[/itex] [itex]\widehat{k}_{ph}[/itex]

Thanks to all for the reply!
 

Related to Photoelectric absorption in semiconductors

1. What is photoelectric absorption in semiconductors?

Photoelectric absorption in semiconductors is a process in which a semiconductor material absorbs photons of light energy and converts it into electrical energy. This process is used in various electronic devices, such as solar cells, photodiodes, and LEDs.

2. How does photoelectric absorption occur in semiconductors?

Photoelectric absorption occurs when a photon of light with sufficient energy strikes a semiconductor material, causing an electron to be excited from its valence band to the conduction band. This creates a hole in the valence band and free electron in the conduction band, generating an electric current.

3. What factors affect photoelectric absorption in semiconductors?

The main factors that affect photoelectric absorption in semiconductors are the energy and intensity of the incident light, the band gap of the semiconductor material, and the properties of the material such as its thickness and impurities.

4. What are the applications of photoelectric absorption in semiconductors?

Photoelectric absorption in semiconductors has various applications in electronic devices. It is used in solar cells to convert light energy into electricity, in photodiodes to detect light, and in LEDs to emit light. It is also used in photoelectric sensors for detecting and measuring light intensity.

5. What is the difference between photoelectric absorption in semiconductors and metals?

In metals, photoelectric absorption occurs when photons strike the surface of the material, causing electrons to be emitted. In semiconductors, photoelectric absorption occurs within the material when photons are absorbed and create electron-hole pairs. Additionally, the absorption spectra of metals and semiconductors are different due to their different band structures.

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