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?
- A 3498 nm
- B 34.98 nm
- C 4603 nm
- D 46.03 nm
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…
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