A point source is emitting 0.2 W of ultraviolet radiation at a wavelength of$\lambda =2537\stackrel{\circ }{\text{A}}$   .  This source is placed at a distance of 1.0 m from the cathode of a photoelectric cell. The  cathode is made of potassium (Work function$=2.22eV$  ) and  has a surface area of $4c{m}^2$   .According to classical theory, the time of exposure to the radiation required for  a potassium atom to accumulate sufficient energy to eject a photoelectron is$t_0$  .  Find  $t_0$ (Assume that radius of each potassium atom is$2\stackrel{\circ }{\text{A}}$  and the lone electron absorbs all energy incident on it).

 $=(1.3-0.5)=0.8eV$
(a) Energy required to eject a photoelectron is $\varphi =2.2eV$
    $=2.22\times 1.6\times {10}^{-19}J=3.552\times {10}^{-19}J$
According to the classical theory, energy flow is a continuous process and  a photo  electron will be ejected will be ejected if a potassium atom receives this amount of  energy over a length of time. The potassium atom is at a distance of 1.0 m from the  source.
Hence the intensity of ultraviolet radiation on the potassium surface is

 $I=\frac{W}{4\pi r^2}=\frac{0.2}{4\pi {\left(1\right)}^2}=\frac{0.2}{4\pi }J/m^2/s$      
Cross sectional area of one K atom is 

$A=\pi {(2\times {10}^{-10}m)}^2=4\pi \times {10}^{-20}{m}^2$  
     $t=\frac{3.552\times {10}^{-19}J}{\frac{0.2}{4\pi }J{m}^{-2}{s}^{-1}\times 4\pi \times {10}^{-20}{m}^2}=178s$