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A semiconductor of thickness 0.5μm is illuminated with monochromatic light of hϑ = 2eV, and the absorption coefficient of semiconductor α = −5 × 10−4cm−1. If the incident power is 10mW (Assume semiconductor has perfect quantum efficiency) then number of photons given from recombination events per second is photons/sec
  • a)
    2.87 × 1016
  • b)
    3.56 × 1016
  • c)
    4.98 × 1016
  • d)
    5.57 × 1016
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A semiconductor of thickness 0.5μm is illuminated with monochromatic ...
I = I0e-αl
= 10-2 exp(-5 x 10-4 x 0.5 x 10-4)
= 0.82m W
Total energy absorbed = 10 - 0.82 = 9.18mW
Number of photons emitted
= 2.8687 x 1016 photons/sec
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Most Upvoted Answer
A semiconductor of thickness 0.5μm is illuminated with monochromatic ...
Number of Photons from Recombination Events per Second

Given data:
- Thickness of semiconductor (d) = 0.5 μm = 0.5 × 10^-4 cm
- Energy of monochromatic light (hϑ) = 2 eV
- Absorption coefficient of semiconductor (α) = -5 × 10^-4 cm^-1
- Incident power (P) = 10 mW

To find the number of photons given from recombination events per second, we can use the following formula:

Number of photons = (Power incident on the semiconductor / Energy per photon) × Quantum efficiency

Let's calculate each component step by step.

1. Power incident on the semiconductor (P_incident):
P_incident = P = 10 mW = 10 × 10^-3 W

2. Energy per photon (E_photon):
E_photon = hϑ = 2 eV × 1.6 × 10^-19 J/eV (converting eV to Joules)
E_photon = 3.2 × 10^-19 J

3. Quantum efficiency (η):
The semiconductor is assumed to have perfect quantum efficiency, which means all absorbed photons result in recombination events. Therefore, η = 1.

Now, we can substitute these values into the formula:

Number of photons = (P_incident / E_photon) × η

Number of photons = (10 × 10^-3 W) / (3.2 × 10^-19 J) × 1

Number of photons = (10 × 10^-3) / (3.2 × 10^-19) × 1 × 1 × 10^19

Number of photons = 3.125 × 10^16

Rounding off to two significant figures, we get:

Number of photons = 2.87 × 10^16 photons

Therefore, the correct answer is option 'A' - 2.87 × 10^16 photons.
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A semiconductor of thickness 0.5μm is illuminated with monochromatic light of hϑ = 2eV, and the absorption coefficient of semiconductor α = −5 × 10−4cm−1. If the incident power is 10mW (Assume semiconductor has perfect quantum efficiency) then number of photons given from recombination events per second is photons/seca)2.87 × 1016b)3.56 × 1016c)4.98 × 1016d)5.57 × 1016Correct answer is option 'A'. Can you explain this answer?
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