# How electrons are excited in direct transitions?

by hokhani
Tags: electrons, transitions
 P: 269 When a photon is radiated to a direct gap semiconductor and an electron is excited from valence band minimum to conduction band maximum, the applied force on the electron is zero (because k isn't changed) but the electron acquires energy. What is the source of the energy obtained by the electron in this transition?
 P: 109 it's from valence band "max" to conduction band min. Think of it in terms of excitation in an atom. You're going from quantum number n to n+1. There's an increase in energy associated with that. It corresponds with being "further" away from the nucleus. This is the same story in bands. The valence band is "closer" to the nucleus and the conduction band is "farther" from the nucleus.
P: 3,593
 Quote by hokhani the applied force on the electron is zero (because k isn't changed) but the electron acquires energy. What is the source of the energy obtained by the electron in this transition?
The energy stems from the photon. I wouldn't say that the applied force is zero, as k isn't true momentum. The momentum of the photon is taken up by the lattice as a hole.

P: 269
How electrons are excited in direct transitions?

 Quote by DrDu I wouldn't say that the applied force is zero, as k isn't true momentum.
I can not give exactly any idea about photon force, But I know that in an electric field:
$Fexternal=d(\hbar k)/dt$
Emeritus
PF Gold
P: 29,238
 Quote by hokhani I can not give exactly any idea about photon force, But I know that in an electric field: $Fexternal=d(\hbar k)/dt$
Not sure what that has anything to do with this. We clearly know that a photon has momentum, so when it is absorbed, there has to be a momentum transfer. But as DrDu has stated, this is taken up by the lattice of the solids as a whole, and it is not manifested in the energy transition of the electron. This is not just an isolated electron encountering a photon.

The reverse is also true. An electron in the conduction band decaying back to the valence band can emit a photon. While the electron may not have change any of its crystal momentum, clearly a photon that is emitted has a momentum. The recoil momentum is once again taken up by the crystal lattice.

Zz.
PF Gold
P: 1,489
 Quote by hokhani $Fexternal=d(\hbar k)/dt$
You can't use BBcodes inside latex, you should use " _ " for the sub scripts.
P: 269
 Quote by adjacent You can't use BBcodes inside latex, you should use " _ " for the sub scripts.
Thank you. But using "_" I can only write one letter (for example: e) in the subscript and not more than one (for example: external).
PF Gold
P: 1,489
 Quote by hokhani Thank you. But using "_" I can only write one letter (for example: e) in the subscript and not more than one (for example: external).
Use _{Whatever you want} .
See,
$F_{external}=\frac{\text{d}(\hbar k)}{\text{d}t}$

What I wrote is:
$F_{external}=\frac{\text{d}(\hbar k)}{\text{d}t}$
You can also right-click on my latex> show math as> Tex commands
 Quote by hokhani I can not give exactly any idea about photon force, But I know that in an electric field: $F_\mathrm{external}=d(\hbar k)/dt$