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A photodiode fabricated from lead Selenide [ PbSe ] has a band gap energy of 0.27 eV . What is the wavelength of the signal the photodiode can detect?

Options

  1. A3498 nm
  2. B34.98 nm
  3. C4603 nm
  4. D46.03 nm

Correct answer

C. 4603 nm

Step-by-step solution

The energy of a photon is given by the relation E = hc , where E is the band gap energy, h is Planck's constant, c is the speed of light, and is the wavelength. Given E = 0.27 eV . Using the relation = hc E , we substitute the values hc 1240 eV nm . = 1240 eV nm 0.27 eV 4592.59 nm . Using a more precise value for hc = 1242 eV nm : = 1242 0.27 = 4600 nm . Comparing with the given options, the closest value is 4603 nm . Answer: 4603 nm

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