Semiconductors

JEE Main level, single correct, medium

127 of 209

Question 127 of 209

If a semiconductor photo diode can detect a photon with a maximum wavelength of 400nm400nm, then its band gap energy is: Planck's constant h=6.63×10−34J.sh=6.63\times {10}^{-34}J.s
Speed of light
c=3×108ms−1c=3\times {10}^{8}m{s}^{-1}

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Answer: (D) 3.1eV3.1eV

λ=400nm\lambda =400nm
Band gap
E0=hcλ=1237.5400=3.09eV{E}_{0}=\dfrac{hc}{\lambda }=\dfrac{1237.5}{400}=3.09eV

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