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JEE Main 2020 (3 Sep, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2020 (3 Sep, Shift 2)

Q73JEE Main 2020 · 3 SEP · SHIFT 2Easy

Special Purpose Diodes

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 ×\times 10–34 J.s. Speed of light c = 3 ×\times 108 m/s

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Answer: (c)

  1. Eg=hcλmaxE_g = \dfrac{hc}{\lambda_{max}}
  2. Eg=6.63×10−34×3×108400×10−9=4.97×10−19E_g = \dfrac{6.63\times10^{-34}\times3\times10^{8}}{400\times10^{-9}} = 4.97\times10^{-19} J =4.97×10−191.6×10−19= \dfrac{4.97\times10^{-19}}{1.6\times10^{-19}} eV ≈3.1\approx 3.1 eV
  3. So the answer is (c) 3.13.1 eV.
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