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Physics
To detect light of wavelength 500 nm , the photodiode must be fabricated from a semiconductor of minimum bandwidth of
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Q. To detect light of wavelength $500 \, nm$ , the photodiode must be fabricated from a semiconductor of minimum bandwidth of
NTA Abhyas
NTA Abhyas 2022
A
$1.24 \, eV$
B
$0.62 \, eV$
C
$2.48 \, eV$
D
$3.2 \, eV$
Solution:
Minimum bandwidth $\Delta E$
$\Delta E=\frac{1240 \, eV nm}{500 \, nm}=2.48 \, eV$