(a) Calculation of energy of a photon of light (b) Identification of photodiode Why photodiode are operated in reverse bias 1 We have $\;E=h v=\;\frac{h c}{\lambda}$ $\;=\;\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}\ \mathrm{J}$ $\;=\;\frac{19.89 \times 10^{-26}}{6 \times 10^{-7} \times 1.6 \times 10^{-19}}\ \mathrm{eV}$ $\;=\;\frac{19.89}{9.6}\ \mathrm{eV}$ $\;=2.08\ \mathrm{eV}$ The band gap energy of diode $D_2(=2\ \mathrm{eV})$ is less than the energy of the photon. Hence diode $D_2$ will not be able to detect light of wavelength $600\ \mathrm{nm}$. [Note: Some student may take the energy of the photon as $2\ \mathrm{eV}$ 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.