When a beam of $10.6 \mathrm{eV}$ photons of intensity $2.0 \mathrm{W}/{\mathrm{m}}^2$ falls on a platinum surface of area $1.0 \mathrm{x} {10}^{-4} {\mathrm{m}}^2$ and work function $5.6 \mathrm{eV}, 0.53\%$ of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is $\mathrm{y} \times {10}^{11}$ where the value of $\mathrm{y}$ is ________.

Incident energy

 $\begin{array}{rcl}\mathrm{E} & =& 10.6 \mathrm{eV} = 10.6 \mathrm{x} \left(1.6 \mathrm{x} {10}^{-19}\right) \mathrm{J} = 16.96 \mathrm{x} {10}^{-19} \mathrm{J}\\ \mathrm{Given} : \frac{\mathrm{Energy} \mathrm{incident}}{\mathrm{area} \mathrm{x} \mathrm{time}} & =& \raisebox{1ex}{$2\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^2$}\right.\\ \therefore \frac{\mathrm{Number} \mathrm{of} \mathrm{incident} \mathrm{photons}}{\mathrm{area} \mathrm{x} \mathrm{time}} & =& \frac{2}{ 16.96 \mathrm{x} {10}^{-19}} = 1.18 \mathrm{x} {10}^{18}\\ \therefore \frac{ \mathrm{incident} \mathrm{photons}}{\mathrm{time}} & =& \left(1.18 \mathrm{x} {10}^8\right) \mathrm{x} \mathrm{area}\\ & =& 1.18 \mathrm{x} {10}^{18} \mathrm{x} (1.0 \mathrm{x} {10}^{-4}) = 1.18 \mathrm{x} {10}^{14}\\ & & \\ \therefore \frac{\mathrm{Number} \mathrm{of} \mathrm{photoelectrons}}{\mathrm{time}} & =& \left(\frac{0.53}{100}\right) \mathrm{x} \left(1.18 \mathrm{x} {10}^{14}\right)\\ \mathrm{or} \mathrm{n} & =& 6.25 \mathrm{x} {10}^{11}\end{array}$