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Optical process in semiconductors

  • Thread starter Defennder
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Defennder
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1. Homework Statement
I'm trying to understand this part of my notes where the optical absorption in both indirect and direct bandgap semiconductors. This part specifically mathematically describes the change in energy and momentum of an electron in a semiconductor after it has absorbed energy from incident photon.

Qn: Why is [tex]\hbar\vec{q} << \hbar\vec{k}[/tex]?


2. Homework Equations

http://img222.imageshack.us/img222/9665/qnmj5.th.jpg [Broken]

This is the part I am referring to.

3. The Attempt at a Solution
 
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Answers and Replies

G01
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Hmm, I am currently taking Solid State myself and am currently learning this material, but I'll try to offer some insight.

I would think that [tex] \hbar\vec{q} << \hbar\vec{k}[/tex] would have to be true when you consider the wavelengths of typical electrons and photons.

Remember that the magnitude of the wave vector for either electron or photon is:

[tex] k=\frac{2\pi}{\lambda}[/tex]

Where, for electrons, lambda is the DeBroglie Wavelength.

An average optical photon has a wavelength of ~500nm, while the DeBroglie wavelength of an electron is in the picometer range, much, much smaller. So, how does this effect the relative values of k and q and thus, the values of [tex]\hbar k[/tex] and [tex]\hbar q[/tex]?
 
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Defennder
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Oh yeah, you're right. That should be the explanation. Thanks.
 

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