Question 4.1: A 0.46-μm- thick sample of GaAs is illuminated with monochro...
A 0.46-μm-thick sample of GaAs is illuminated with monochromatic light of hv = 2eV. The absorption coefficient α is 5 × 10^{4} cm^{-1} . The power incident on the sample is 10 mW.
(a) Find the total energy absorbed by the sample per second (J/s).
(b) Find the rate of excess thermal energy given up by the electrons
to the lattice before recombination (J/s).
(c) Find the number of photons per second given off from recombination
events, assuming perfect quantum efficiency.
(a) From Eq. (4–3),
\pmb{I}_{t}=\pmb{I}_{0}e^{-\alpha l} =10^{-2}exp(-5\times 10^{4}\times 0.46\times 10^{-4} )=10^{-2}e^{-2.3}=10^{-3}W
Thus the absorbed power is
10 – 1 = 9 mW = 9 × 10^{-3} J / s(b) The fraction of each photon energy unit which is converted to heat is
\frac{2-1.43}{2} =0.285Thus the amount of energy converted to heat per second is
0.285 × 9 × 10^{-3} = 2.57 ×10^{-3} J / s(c) Assuming one emitted photon for each photon absorbed (perfect
quantum efficiency), we have
Alternative solution: Recombination radiation accounts for 9 – 2.57 =
6.43 mW at 1.43 eV/photon.
\frac{6.43\times 10^{-3}}{1.6\times 10^{-19}\times 1.43} =2.81\times 10^{16}photons / s
