A beam of light consists of four wavelength $4000\stackrel{\circ }{A}$,$4800\stackrel{\circ }{A}$,$6000\stackrel{\circ }{A}$ and $8000\stackrel{\circ }{A}$ each of intensity $1.5\times {10}^{-3}W{m}^{-2}$. The beam falls normally on an area ${10}^{-4}{m}^2$ of a clean metallic surface of work function $1.9eV$. Assuming no loss of light energy calculates the number of photoelectrons liberated per second.

$E_1=\frac{12375}{4000}=3.1eV$

$E_2=\frac{12375}{4800}=2.57eV$

$E_3=\frac{12375}{6000}=2.06eV\text{ and }$

$E_4=\frac{12375}{7000}=1.77eV$

Therefore, light of wavelengths $4000\stackrel{\circ }{A}$,$4800\stackrel{\circ }{A}$, and $6000\stackrel{\circ }{A}$ can emit photoelectrons.

So, number of photoelectrons emitted per second is

$n=\frac{I_1{A}_1}{E_1}+\frac{I_2{A}_2}{E_2}+\frac{I_3{A}_2}{E_3}$

$\Rightarrow n=IA\left(\frac{E_1{E}_2+E_2{E}_3+E_1{E}_3}{E_1{E}_2{E}_3}\right)$

$\Rightarrow n=\frac{\left(1.5\times {10}^{-3}\right)\left({10}^{-4}\right)}{1.6\times {10}^{-19}}\times \left[\frac{3.1\times 2.57+2.57\times 2.06+3.1\times 2.06}{3.1\times 2.57\times 2.06}\right]$

$\Rightarrow n=1.12\times {10}^{12}$