Energy of the incident photon with a band gap of 6000 nm. E = $\frac{hc}{\lambda}$ J = $\frac{hc}{e\lambda}$ eV = $\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 6 \times 10^{-6}}$ = 0.207 eV The photodiode need an energy of 2.8 eV to give response to incident light. As E < $E_g$, the given photodiode cannot detect the radiation of wavelength 6000 nm.