(a) Energy corresponding to wavelength 600 nm

$\begin{array}{l}\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}\\ =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{600\times {10}^{-9}}\mathrm{J}=3.3\times {10}^{-19}\mathrm{J}\\ =\frac{3.3\times {10}^{-19}}{1.6\times {10}^{-19}}\mathrm{eV}=2.06\mathrm{eV}\end{array}$
The photon energy (E = 2.06 eV) is greater than the band gap for diode D₂ only. Hence, diode D₁ and D₃ will not be able to detect the given wavelength.
photo diode is operated reverse bias because in reverse bias it is easier to observe change in current
with change in light intensity.