"From your data, is the bandgap of ZnSe direct or indirect?"

[chrononaut 114]

(urgent)

Hi,

This question was apart of an assignment sheet that I was given in 'Experimental Physics III' after having completed and obtained data for the practical called 'The Bandgap Energy of Semiconductor ZnSe'.
Cheers

Below is some screenshots of the (Matlab-processed) data we obtained, for some context.

We observed a cutt-off wavelength at approx λ = 475 nm, and then after calibration using known spectrum of Mercury, we calculated the bandgap of ZnSe with the equation E = hƒ = h*c/λ. Giving E = 2.63 eV. (Where the accepted value is 2.7 eV).

We obtained the data by focusing (with two lenses) the light from a given lamp onto the entrance slit of a monochromator, which came out of the exit slit onto a photodiode, with or without the ZnSe glass 'window'/filter/sample slotted infront of the photodiode. The photodiode was connected to a lock-in amplifier, and an oscilloscope to read the voltage off from.

I added in the orange dashed arrows on the plots to emphasize what the significance of Fig 2 is and what it means with respect to Fig 1. And am I correct in saying that Fig 2 is basically the 'transmission' or the transmittance, or something different?

Thanks

 

[Fred Wright]

A direct band gap material will absorb the light of wavelength equal to its band-gap whereas an in-direct band gap material will not. You found the band gap energy to be 2.63 eV so with $hc=1.239842 \frac{eV}{\mu m}$ $\lambda = \frac{1.2139842}{2.63}=.4714\mu m$. Do you see a strong absorption at that wavelength?

 

[Lord Jestocost]

Plot the absorption coefficient as function of light frequency. The functional dependence on frequency should differ for semiconductors with a direct or indirect band gap. I think the relevant equations can be found in textbooks on semiconductor physics or by searching the internet.

 

[chrononaut 114]

A direct band gap material will absorb the light of wavelength equal to its band-gap whereas an in-direct band gap material will not.

Thanks for the response, really appreciate it. Ok that makes sense, I think; so if it were an indirect bandgap it wouldn't 'line up' with the absorption peak since it has a different momentum value or something like that?

You found the band gap energy to be 2.63 eV so with $hc=1.239842 \frac{eV}{\mu m}$ $\lambda = \frac{1.2139842}{2.63}=.4714\mu m$. Do you see a strong absorption at that wavelength?

So, I think I see a strong absorption at that wavelength in the form of the minimum peak shown in Figure 2 of my original post? Is it correct to say that said peak in Fig 2 is an absorption peak?

Thanks

 

[chrononaut 114]

Plot the absorption coefficient as function of light frequency. The functional dependence on frequency should differ for semiconductors with a direct or indirect band gap. I think the relevant equations can be found in textbooks on semiconductor physics or by searching the internet.

Cheers. I found this expression for the absorption coefficient:

(from: http://www.pveducation.org/pvcdrom/absorption-coefficient)

It can easily be converted to a function of frequency by using λ = c/f .

But, here, 'k' is referred to as the 'extinction coefficient', which I couldn't find an expression for. Are you familiar with the term 'extinction coefficient'?
Thanks

 

[chrononaut 114]

I just found this version of an expression for the absorption coefficient:

(from: http://file.scirp.org/Html/2-7700668_17248.htm)

-But it doesn't specify what K is, rather it just says that it is a constant, so not really sure what to do with that
-Here E_g is the bandgap energy, as per usual
-Apparently n depends on the nature of the 'optical transition' (n = 1/2 for direct, n = 2 for indirect bandgap)

A larger screenshot from the website for a bit more context:

 

[Lord Jestocost]

Maybe, this will be of help: http://nptel.ac.in/courses/115102025/13 or [PDF]Optical properties of semiconductors