Question -
Answer -
Energy band gap of the given photodiode, Eg =2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
The energyof a signal is given by the relation:
E =
Where,
h = Planck’sconstant
= 6.626 × 10−34 Js
c = Speedof light
= 3 × 108 m/s
E 
= 3.313 × 10−20 J
But 1.6 × 10−19 J = 1 eV
∴E = 3.313 × 10−20 J
The energy of a signal of wavelength 6000 nm is 0.207 eV,which is less than 2.8 eV − the energy band gap of a photodiode. Hence, thephotodiode cannot detect the signal.