Question

for a semiconductor laser the band gap is 0.8 eV. what is the wavelength of light emitted from it.​

Answer 1

Answer:

bang gap(E)=0.8eV=0.8*1.6*10^(-19) J

we have

E=hc/λ

λ=hc/E

 =1.55*10^(-6)m

Answer 2

For a semiconductor laser the band gap is 0.8 eV.

We have to find the wavelength of light emitted from it.

band gap of a semiconductor is the minimum amount of energy required to free an electron from its bound state.

to find wavelength of light emitted from it is given by, E = hc/λ

where E is band gap, h is Plank's constant, c is the speed of light in vacuum.

here,

$E=8\times1.6\times10^{-19}J\\h=6.63\times10^{-34}Js\\c=3\times10^{8}ms^{-1}$

$8\times1.6\times10^{-19}=\frac{6.63\times10^{-34}\times3\times10^8}{\lambda}$

$\implies\lambda=\frac{6.63\times10^{-26}\times3}{8\times1.6\times10^{-19}}$

= 15.54 × 10^-7 m

= 1.554 μm

Therefore the wavelength emitted from it is 1.554μm.