Answer:
a. 3.17*10¹⁹ b. 3.17*10¹⁴
Explanation:
a. Area A = 1cm²
Thickness h = 0.1cm
Energy of photon E = hf=hc/λ
Where f = frequency; c=speed of light; λ=wavelength = 6300Α=6300*10⁻¹⁰;
Planck's constant h = 6.626*10⁻³⁴ joule-seconds; speed of light c = 3*10⁸
Therefore E = (6.626*10⁻³⁴)*3*10⁸/6300*10⁻¹⁰=3.155*10¹⁹J
1Watt of light releases 3.17*10¹⁸ photons per second.
Volume of sample = Area * Thickness = 1*0.1=0.1cm³
Therefore, number of electron hole pairs that are generated per unit volume per unit time = 3.17*10¹⁸/0.1 = 3.17*10¹⁹photon/cm³-s
b. Steady state excess carrier concentration = 3.17*10¹⁹*10μs
=3.17*10¹⁹*10*10⁻⁶
=3.17*10¹⁴/cm³
