In a photo electric experiment the metallic surface has a threshold wave length of $5200\stackrel{\text{o}}{\text{A}}$. This surface is irradiated by a monochromatic light of wavelength $4500\stackrel{\text{o}}{\text{A}}$. If the photo electrons, are accelerated through a potential difference of 2 volt, final kinetic energy of the most energetic photo electrons will be 

$h\nu =\frac{hc}{\lambda }=\frac{12420}{4500}eV=2.76eV$

work function $\varphi =\frac{hc}{{\lambda }_0}=\frac{12420}{5200}eV=2.38eV$

$\therefore K_{\max }=$Maximum kinetic energy of photo electron

$=\left(2.76-2.38\right)eV=0.38eV$

When accelerated through a potential of 2 volt, additional energy gained by the photoelectrons = 2 eV

$\therefore$ Final kinetic energy of most energetic photo electrons = 0.38 eV + 2 eV = 2.38 eV