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COMEDK · Physics · 30. Semiconductors

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

  1. A 3498 nm
  2. B 34.98 nm
  3. C 4603 nm
  4. D 46.03 nm
Verified Solution

Answer & Solution

Correct Answer

(C) 4603 nm

Step-by-step Solution

Detailed explanation

The energy of a photon is given by the relation \(E = \dfrac{hc}{\lambda}\), where \(E\) is the band gap energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. Given \(E = 0.27 \text{ eV}\). Using the relation…