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
Solution
$E = {{hc} \over \lambda }$
<br><br>= $${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}$$
<br><br>= ${{1240} \over {400}}$ eV
<br><br>= 3.1 eV
About this question
Subject: Physics · Chapter: Electronic Devices · Topic: Semiconductors
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