(N/A) The solar radiation spectrum received by us is shown in the figure.
The maxima is near $1.5 \, eV$. For photo-excitation,$h\nu > E_g$. Hence,a semiconductor with a band gap $\sim 1.5 \, eV$ or lower is likely to give better solar conversion efficiency. Silicon has $E_g \sim 1.1 \, eV$ while for $GaAs$ it is $1.53 \, eV$.
In fact,$GaAs$ is better (in spite of its higher band gap) than $Si$ because of its relatively higher absorption coefficient. If we choose materials like $CdS$ or $CdSe$ $(E_g \sim 2.4 \, eV)$,we can use only the high-energy component of the solar energy for photo-conversion,and a significant part of the energy will be of no use.
The question arises: why do we not use a material like $PbS$ $(E_g \sim 0.4 \, eV)$ which satisfies the condition $h\nu > E_g$ for $\nu$ maxima corresponding to the solar radiation spectra? If we do so,most of the solar radiation will be absorbed on the top layer of the solar cell and will not reach in or near the depletion region. For effective electron-hole separation,due to the junction field,we want the photo-generation to occur in the junction region only.