Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is : (Planck's constant $h=6.62\times {10}^{-34}J\cdot s$ )

 $hf=\varphi +e{V}_{stop}$
$\varphi$  is work function, which by definition is the minimum energy for the surface electron’s ejection.
Re-arranging,  $V_{stop}=\frac{h}{e}f-\frac{\varphi }{e}$
To find $\varphi$, point B is chosen because stopping potential is zero at this point.

 x intercept  $=\frac{\varphi }{h}=5.5\times {10}^{14}Hz$
$\therefore \varphi =\left(6.62\times {10}^{-34}\right)\left(5.5\times {10}^{14}\right)J$ 
 $=\frac{\left(6.62\times {10}^{-34}\right)\left(5.5\times {10}^{14}\right)}{1.6\times {10}^{-19}}eV$           $=2.276eV$