(a) Calculation of energy of a photon of light
(b) Identification of photodiode
Why photodiode are operated in reverse bias 1 We have
$\mathit{E}=\mathit{h}\mathit{v}=\frac{\mathit{h}\mathit{c}}{\mathit{\lambda }}$
$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{600\times {10}^{-9}}\mathrm{J}$
$=\frac{19.89\times {10}^{-26}}{6\times {10}^{-7}\times 1.6\times {10}^{-19}}\mathrm{eV}$
$=\frac{19.89}{9.6}\mathrm{eV}$
$=2.08\mathrm{eV}$
The band gap energy of diode ${\mathrm{D}}_2\left(=2\mathrm{eV}\right)$ is less than the energy of the photon.

Hence diode ${\mathrm{D}}_2$ 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.