Question:

If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is :

Planck's constant \(h = 6.63\times10^{-34}\) J.s. Speed of light \(c = 3\times10^{8}\) m/s

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The photodiode can detect a photon only if its energy is at least the band gap, so \(E_g = hc/\lambda_{max}\). Convert joules to eV.
Updated On: Nov 5, 2026
  • 1.5 eV
  • 2.0 eV
  • 3.1 eV
  • 1.1 eV
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The Correct Option is C

Solution and Explanation

The longest detectable wavelength corresponds to the minimum photon energy, equal to the band gap: \(E_g = \dfrac{hc}{\lambda_{max}}\).
\(E_g = \dfrac{6.63\times10^{-34}\times 3\times10^{8}}{400\times10^{-9}} = 4.97\times10^{-19}\) J.
In eV: \(E_g = \dfrac{4.97\times10^{-19}}{1.6\times10^{-19}} \approx 3.1\) eV.
Answer: (3) 3.1 eV
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