In a photodiode, incident photon gives its energy to the electron which jumps the energy gap to conduct current.
Thus incident photon must have energy $\geq E_g$
For a photon of $\lambda=6000 \mathrm{~nm}$
$$
\begin{aligned}
\mathrm{E} &=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-9}}=3.3 \times 10^{-20} \mathrm{~J} \\
&=\frac{3.3 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV} \cong 2 \times 10^{-1} \mathrm{eV}=0.2 \mathrm{eV}
\end{aligned}
$$
Since energy of photon is less, hence it would not be detected.