Question

4. A photon of light can excite an electron from the valence band to the conduction band of a semiconductor. This process is called photoconduction. a. PbS has a band gap of 0.37 eV. What wavelength of light would be needed to start the photoconduction in this semiconductor? b. In the light meters of cameras one would need a semiconductor that operates efficiently in visible light, or at-550 nm. Would PbS be a suitable semiconductor for a light meter? Why or why not?

Answer 1

4.a)

Band gap energy = 0.37 eV = 0.37*1.6*10^-19 J = 0.592*10^-19 J

Let λ be the wavelength of light required

E = hc/λ

0.592*10^-19 = 6.626*10^-34*3*10⁸/λ

λ = 33.6*10^-7 m = 3360 nm

4.b)

The wavelength of the visible light is lesser than for PbS, so the energy generated will be more because λ is inversely proportional to energy. Hence, one can use PbS as a suitable semiconductor because it has lower band gap energy than visible light energy.