(a) Calculation of energy of a photon of light
(b) Identification of photodiode
Why photodiode are operated in reverse bias 1 We have
‌E=hv=‌

hc

λ

‌=‌

6.63×10−34×3×108

600×10−9

J
‌=‌

19.89×10−26

6×10−7×1.6×10−19

‌eV
‌=‌

19.89

9.6

‌eV
‌=2.08‌eV
The band gap energy of diode D2(=2‌eV) is less than the energy of the photon.

Hence diode D2 will not be able to detect light of wavelength 600‌nm.

[Note: Some student may take the energy of the photon as 2‌eV and say that all the three diodes will be able is detect this right, Award them the mark for the last part of identification].
(b) A photodiode when operated in reverse bias, can measure the fractional change in carrier dominated reverse bias current with greater ease.