How electrons are excited in direct transitions?

[hokhani]

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?

 

[rigetFrog]

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.

 

[DrDu]

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.

 

[hokhani]

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:
F_external=d(\hbar k)/dt

 

[ZapperZ]

I can not give exactly any idea about photon force, But I know that in an electric field:
F_external=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.

 

[adjacent]

F_external=d(\hbar k)/dt

You can't use BBcodes inside latex, you should use " _ " for the sub scripts.

 

[hokhani]

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).

 

[adjacent]

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:

[itex]F_{external}=\frac{\text{d}(\hbar k)}{\text{d}t}[/itex]

You can also right-click on my latex> show math as> Tex commands

 

[DrDu]

Thank you. But using "_" I can only write one letter (for example: e) in the subscript and not more than one (for example: external).

Enclose them in curly brackets or use _\mathrm{ext} if you want them roman style.

 

[ZapperZ]

I assume that your question has been satisfactorily answered (please acknowledge if it is, rather than let it hang) considering that we are now discussing how to do LaTex in this thread.

Zz.

 

[DrDu]

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

This holds for the motion of wavepackets in one band, not for interband transitions.

 

[ehild]

You can imagine the interband transition like an elastic collision, for example, with a photon and the electron. As the photon has momentum p=h/λ, the momentum of the electron also has to change. But the momentum of the photon is much less than that of the crystal momentum at the Brillouin zone boundary so the transition looks almost vertical. Exactly vertical transition can happen by the assistance of an other particle, with a phonon.


ehild