A uniform monochromatic beam of light of wavelength $365\times {10}^{-9} \mathrm{m}$ and intensity ${10}^{-8} {\mathrm{Wm}}^{-2}$ falls on a surface having absorption coefficient 0.8 and work function 1.6 eV.

Let N be the number of photoelectrons per second.

$\therefore \text{ Intensity }=\mathrm{N}\times \text{ energy of one photon }\Rightarrow \mathrm{I}=\mathrm{N}\frac{\mathrm{hc}}{\mathrm{\lambda }}$

$\therefore$ Number of incident photons,

         $\begin{array}{l}{\mathrm{N}}_{\mathrm{i}}=\frac{\mathrm{I\lambda }}{\mathrm{hc}}\\ =\frac{{10}^{-8}\times 365\times {10}^{-9}}{6.63\times {10}^{-34}\times 3\times {10}^8}=18.35\times {10}^9\end{array}$

Number of photon ${\mathrm{N}}_{\mathrm{ab}}$ absorbed by the surface per unit area per unit time (i.e., absorbed flux),

         ${\mathrm{N}}_{\mathrm{ab}}=0.8\times 18.35\times {10}^9=1.47\times {10}^{10}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$

$\text{ Emitted flux }=\text{ No. of electrons emitted }{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}=1.47\times {10}^{10}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$

$\text{ Power absorbed per }{\mathrm{m}}^2=0.8\times \text{ Incident intensity }$

         $=0.8\times {10}^{-8}=8\times {10}^{-9} {\mathrm{Wm}}^{-2}$

From Einstein's photoelectric equation,

         $\begin{array}{l}{\mathrm{K}}_{max}=\mathrm{h\nu }-{\mathrm{W}}_0=\frac{\mathrm{hc}}{\mathrm{\lambda }}-{\mathrm{W}}_0\\ =\frac{1240}{365}-1.6\\ =3.4-1.6=1.80 \mathrm{eV}\end{array}$