No, the photodiode is unable to detect wavelengths of 6000 nm due to the following:

The energy bandgap of the given photodiode is given as $E_{g}=2.8 \mathrm{eV}$

The wavelength has the value $\lambda=6000 \mathrm{~nm}=6000 \times 10^{-9} \mathrm{~m}$

The energy of the signal is represented by the following relation:

$E=h c / \lambda$

In the equation, $h$ is the Planck’s constant $\left.=6.626 \times 10^{-34}\right\rfloor$ and $c$ is the speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$.

Putting the values in the equation, we get

$E=\left(6.626 \times 10^{-34} \times 3 \times 10^{8}\right) / 6000 \times 10^{-9}=3.313 \times 10^{-20} \mathrm{~J}$

But, $1.6 \times 10^{-19} \mathrm{~J}=1 \mathrm{eV}$

Therefore, $E=3.313 \times 10^{-20} \mathrm{~J}=3.313 \times 10^{-20} / 1.6 \times 10^{-19}=0.207 \mathrm{eV}$

A signal with a wavelength of 6000 nm has an energy of 0.207eV, which is less than the energy band gap of a photodiode, which is 2.8eV. As a result, the photodiode is unable to detect the signal.