$A$ $p-n$ photodiode is fabricated from a semiconductor with a band gap of $2.8 \; eV$. Can it detect a wavelength of $6000 \; nm$?

(NO) The energy band gap of the given photodiode is $E_g = 2.8 \; eV$.
The wavelength of the incident light is $\lambda = 6000 \; nm = 6000 \times 10^{-9} \; m$.
The energy of a photon is given by the relation $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \; Js$ and $c = 3 \times 10^8 \; m/s$:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-9}} \; J = 3.313 \times 10^{-20} \; J$.
Converting this energy into electron-volts $(eV)$:
$E = \frac{3.313 \times 10^{-20}}{1.6 \times 10^{-19}} \; eV \approx 0.207 \; eV$.
Since the energy of the incident photon $(0.207 \; eV)$ is less than the band gap energy of the photodiode $(2.8 \; eV)$,the photodiode cannot detect the signal of wavelength $6000 \; nm$.