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Zinc selenide has a band gap of 2.58 eV. Over what range of wavelengths of visible light is it transparent?

2007-11-30 03:06:20 · 4 answers · asked by FER 2 in Science & Mathematics Physics

4 answers

2.58 eV = 2.58*1.6*10^-19J=4.128*10^-19J
wavelength = 6.62*10^-34*310^8/4.128*10^-19=481*10^-9m
this corresponds to blue color
So all light with wavelength lower than 481 nm can be absorbrd by this product. This product is trasparent for visble light with wavelength higher than 481 nm , green, yellow orange and red

2007-11-30 03:16:05 · answer #1 · answered by maussy 7 · 1 0

Can't remember exactly off the top of my head but about 420 (480?) - 720 nanometers

2007-11-30 03:14:09 · answer #2 · answered by tamarack58 5 · 1 0

If maussy is right (and with a name like that he has to be right), it would appear yellow. White - blue = yellow. Or red + green (+orange+blue/green) = yellow.

2007-11-30 06:06:47 · answer #3 · answered by za 7 · 0 0

Lost me

2007-11-30 03:08:31 · answer #4 · answered by Anonymous · 0 1

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