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

Identifying the threshold frequency from the x-intercept:

$$\text{At point } B(5.5, 0) \implies V_{\text{stop}} = 0 \implies f_0 = 5.5 \times 10^{14}\ \text{Hz}$$

Work function equation: $$\phi = h f_0$$

$$\phi = 6.62 \times 10^{-34}\ \text{J}\cdot\text{s} \times 5.5 \times 10^{14}\ \text{Hz} = 3.641 \times 10^{-19}\ \text{J}$$

$$\phi = \frac{3.641 \times 10^{-19}}{1.6 \times 10^{-19}}\ \text{eV} \approx 2.275\ \text{eV}$$