$\mathrm{A} \mathrm{beam} \mathrm{of} \mathrm{light} \mathrm{consists} \mathrm{of} \mathrm{four} \mathrm{wavelengths} 4000 \AA , 4800\AA , 6000 \AA \mathrm{and} 7000 \AA$, each of intensity $1.5 \mathrm{x} {10}^{-3} \mathrm{W} {\mathrm{m}}^{-2}$. The beam falls normally on an area ${10}^{-4} {\mathrm{m}}^2$, of a clean metallic surface of work function $1.9 \mathrm{eV}$. Assuming no loss of light energy (i.e., each capable photon emits one electron), the number of photoelectrons liberated per second is $\mathrm{y} \times {10}^{12}$, where the value of $\mathrm{y}$ is $\_\_\_\_\_\_\_$.

${\mathrm{E}}_1=\frac{12400}{4000} = 3.1 \mathrm{eV}, {\mathrm{E}}_2=\frac{12400}{4800} = 2.58 \mathrm{eV}, {\mathrm{E}}_3=\frac{12400}{6000} = 2.06 \mathrm{eV} \mathrm{and} {\mathrm{E}}_4=\frac{12400}{7000} = 1.77 \mathrm{eV}$

$$\begin{gathered} \mathrm{Energy} \mathrm{of} \mathrm{photon} \left({\mathrm{E}}_4\right) \mathrm{is} \mathrm{less} \mathrm{than} \mathrm{work} \mathrm{function}. \mathrm{Therefore}, \mathrm{light} \mathrm{of} \mathrm{wavelengths} 4000 \mathrm{\AA }, 4800 \mathrm{\AA }, \mathrm{and} 6000 \mathrm{\AA } \mathrm{can} \mathrm{only} \mathrm{emit} \mathrm{photoelectrons}. \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{photoelectrons} \mathrm{emitted} \mathrm{per} \mathrm{second} = \mathrm{Number} \mathrm{of} \mathrm{photons} \mathrm{incident} \mathrm{per} \mathrm{second} \end{gathered}$$ $$\begin{gathered} = \frac{{\mathrm{I}}_1{\mathrm{A}}_1}{{\mathrm{E}}_1} + \frac{{\mathrm{I}}_2{\mathrm{A}}_2}{{\mathrm{E}}_2} +\frac{{\mathrm{I}}_3{\mathrm{A}}_3}{{\mathrm{E}}_3} = \mathrm{IA}\left(\frac{1}{{\mathrm{E}}_1}+\frac{1}{{\mathrm{E}}_2}+\frac{1}{{\mathrm{E}}_3}\right) \\ =\frac{ \left(1.5 \mathrm{x} {10}^{-3}\right)\left({10}^{-4}\right)}{1.6 \mathrm{x} {10}^{-19}}\left(\frac{1}{3.1}+\frac{1}{2.58}+\frac{1}{2.06}\right) = 1.12 \mathrm{x} {10}^{12} \end{gathered}$$