E = \(\frac{hc}{\lambda}\)
E = \(\frac{6.64\times10^{-34}\times3\times10^8}{600\times10^{-9}}\)
E = \(\frac{19.92}{600}\times10^{-19}\)
E = \(\frac{0.0332\times10^{-19}}{1.6\times10^{-19}}\)
E = 0.020 ev
For the incident radiation to be detected by photo diode, energy of incident radiation photon should be greater then the band gap. In this case all band gap are greater then photodiode. So any diode will not able to detect light of wavelength.