- #1
chrononaut 114
- 10
- 0
(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
Attachments
-
media%2Ff25%2Ff253eb59-4b7a-44ae-a1aa-c46e8fad9684%2FphpeFVIND.png8 KB · Views: 427
-
1_zpsfuegkzuv.png25 KB · Views: 551
-
2_zpselc7m5ay.png24.4 KB · Views: 602
-
media%2Ff25%2Ff253eb59-4b7a-44ae-a1aa-c46e8fad9684%2FphpeFVIND.png8 KB · Views: 1,236
-
2_zpselc7m5ay.png24.4 KB · Views: 1,232
-
1_zpsfuegkzuv.png25 KB · Views: 1,593
Last edited: