Explain photo electric effect.

Statement : “When light is allowed to fall on the metallic surface, electrons are ejected from the surface”. This effect is known as photo electric effect it is represented as follows
$\text{ i.e., }\mathrm{E}=\mathrm{h\vartheta }={\mathrm{W}}_0+\mathrm{K}.\mathrm{E}\dots$
Where $\text{ hข }$= Energy of incident photon
${\mathrm{W}}_0$ = work function = ${\mathrm{h}}_{ }{\mathrm{\vartheta }}_0$
KE = Kinetic Energy$=\frac{1}{2}{\mathrm{mv}}^2$
The equation (1) becomes
$\mathrm{h\vartheta }={\mathrm{h\vartheta }}_0+\mathrm{KE}\left(\text{ or }\right)\mathrm{h\vartheta }={\mathrm{h\vartheta }}_0+\frac{1}{2}{\mathrm{mv}}^2$
$\Rightarrow \mathrm{K}\cdot \mathrm{E}=\mathrm{h\vartheta }-{\mathrm{h\vartheta }}_0$
$\mathrm{K}.\mathrm{E}=\mathrm{h}\left[\mathrm{\vartheta }-{\mathrm{\vartheta }}_0\right]$
The photo electric effect is readily given by alkalimetals like K, Rb, CS. 
(or)
Statement : The ejection of electrons from the metal surface when light is fall on the metal surface. This is called photo electric effect.
This effect is represented as follows.
$\mathrm{h\vartheta }={\mathrm{W}}_0+\mathrm{K}.\mathrm{E}$
Where $\mathrm{h\vartheta }$ = Energy of incident light (photon).
${\mathrm{W}}_0 = \mathrm{work} \mathrm{function} = {\mathrm{h\vartheta }}_0$
K.E = kinetic Energy $=\frac{1}{2}{\mathrm{mv}}^2$
i,e $\mathrm{h\vartheta }={\mathrm{h\vartheta }}_0+\frac{1}{2}{\mathrm{mv}}^2$
$\mathrm{h\vartheta }={\mathrm{h\vartheta }}_0+\mathrm{KE}\Rightarrow \mathrm{KE}=\mathrm{h\vartheta }-{\mathrm{h\vartheta }}_0=\mathrm{h}\left(\mathrm{\vartheta }-{\mathrm{\vartheta }}_0\right)$