Quantum efficiency as a function of wavelength

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
gnurf
Messages
370
Reaction score
8
I'm trying to reproduce the plot in the attached figure. I know the band-gap energies, so I was hoping there was some simple way I could get the quantum efficiency as a function of wavelength. I read on wiki that Energy(eV) = 1240/wavelength(nm), so I mechanically plugged those in, and got

GaInP (Eg=1.85eV): 670 nm
GaAs (Eg=1.42eV): 873 nm
Ge (Eg=0.67eV): 1851 nm

Other than that the respective wavelengths came out in the right order, it didn't really help all that much. Is there some magical quantum mechanical formula I could drink in order to make that plot?

VthIs.png


EDIT: I should probably have posted this in the QM sub-forum. My apologies.
 
Last edited:
Physics news on Phys.org
E=h[itex]\upsilon[/itex]
E=[itex]\frac{hc}{\lambda}[/itex]
E=[itex]\frac{h\ =\ 4.135\ \times\ 10^{-15}\ eV\ s\\times\ 3\ \times\ 10^{17}\nm}{\lambda}[/itex]
E=[itex]\frac{1240}{\lambda}[/itex]