A uniform monochromatic beam of light of wavelength $365\times {10}^{-9}$  m and intensity ${10}^{-8}{\text{ W/m}}^{\text{2}}$  falls on a surface having absorption coefficient 0.8 and work function 1.6 eV. If the rate of number of electrons emitted per m² is  $14.7\times {10}^y\text{,\hspace{0.17em}then\hspace{0.17em}}y=\_\_\_\_\_.$

The intensity of radiation I is defined as the energy passing per unit time per unit area normal to the direction of the beam. If N be the number of photons crossing unit area per unit time, then 
$I=N\times$ energy carried by one photon
 $I=N\frac{hc}{\lambda } \Rightarrow N=\frac{I\lambda }{hc}$
 If ni be the number of incident photons per unit area per unit time, then  
 $N_i=\frac{I_i\lambda }{hc}=\frac{{10}^{-8}\times 365\times {10}^{-9}}{6.62\times {10}^{-34}\times 3\times {10}^8}$    $\Rightarrow N_i=18.35\times {10}^9$
The number of photons absorbed Nab by the surface per unit area per unit time is given by 
Nab= absorption coefficient of surface  $\times N_i$

$$\begin{gathered} \Rightarrow N_{ab}=0.8\times 8.35\times {10}^9 \\ \Rightarrow N_{ab}=1.47\times {10}^{10}{m}^{-2}{s}^{-1} \end{gathered}$$
 Now assuming that each photon ejects only one electrons, the rate of electrons emitted per unit area is given by $N=N_{ab}=1.47\times {10}^{10}{m}^{-2}{s}^{-1}$