How does Band Gap and Refractive index relate to Wavelength?

AI Thread Summary
The discussion revolves around the relationship between band gap, refractive index, and wavelength in Gallium Nitride semiconductors. The band gap of 3.4 eV suggests a theoretical wavelength of approximately 365 nm, while the refractive index of 2.429 alters the light's velocity in the medium, affecting the wavelength. Calculations indicate that the wavelength in the medium could be around 150 nm, which does not align with the expected blue wavelength of approximately 445 nm. The confusion lies in reconciling these values to understand how they lead to the observed laser output. A clear explanation of how the refractive index influences the effective wavelength in practical applications is needed.
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Homework Statement


Nobel prize for physics for blue laser diode using Gallium Nitride with band gap of 3.4 eV and a refractive index of 2.429.
Explain how these parameters determine at what wavelength a Gallium Nitride semiconductor will laser at.

Homework Equations


E=\frac{hc}{\lambda}
Blue wavelength ## \approx 445 nm ##

The Attempt at a Solution


This equation will give the energy band gap wavelength. But how does the refractive index factor into this? I know it changes the velocity that the light moves through the medium.
I found this article: https://www.quora.com/Whats-the-relation-between-bandgap-the-extinction-coefficient-and-the-index-of-refraction
But I'm still not sure how to answer the question.
 
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When light goes from one medium to another, it's wavelength changes.
 
Using the equations I can get:
E=\frac{hc}{\lambda}
\lambda=\frac{hc}{E}=\frac{1240~eV~nm}{3.4~eV} \approx 365~nm
The velocity of the light changes in the medium by v=\frac{c}{n} replacing c in the above by the velocity in the medium gives
\lambda=\frac{hv}{E}=\frac{hc}{nE}=\frac{1240~eV~nm}{(2.429)(3.4)~eV} \approx 150~nm

I don't see how this gives blue light of 445 nm.
 
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