Question

A 0.5 m thick sample of um thick sample of In is illuminated with monochromatic light of hv=1.5 eV. The absorption coefficient 104 cm 1. The power incident on the sample is 15 mW. (a) Find the total energy absorbed by the sample per second (J/sec). (b) Find the rate of excess thermal energy given up by the electrons to the lattice before recombination (J/sec). (c) Find the number of photons per second given off from recombination events, assuming perfect quantum efficiency. (a)The transmitted intensity = le at = 15 x 10-3 exp (-104x102x0.5) = 9.1 mW Therefore, the absorbed power (15 9.1) mW = 5.9 mW = 5.9 10 3 J/sec. (b)Since the energy of the incident photon is greater than the band gap energy, hence, the excess energy of the excited electron will be dissipated as heat to the lattice. The fraction of energy converted to heat is given by (1.5 1.34)/1.5 = 0.107. Thus, the amount of energy converted to heat per second (c) For perfect quantum efficiency, one photon is emitted for each photon absorbed. Thus, the number of photons emitter per second or, alternately, recombination radiation accounts for 5.9 0.63 = 5.27 mW at 1.34 eV/photon.

Answer 1

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