The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV , respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of

The incident photon energy must be greater than or equal to the metal’s work function for photo-emission to occur.

Photon energy relation:
$$E = \frac{hc}{\lambda}$$

Using $$hc = 1240\ \text{eV·nm}$$ (a convenient constant), we get for the given light of wavelength $$\lambda = 550\ \text{nm}$$:

$$E = \frac{1240}{550}\ \text{eV} \approx 2.25\ \text{eV}$$

Work functions:
Cesium: $$\phi_{Cs} = 1.9\ \text{eV}$$
Lithium: $$\phi_{Li} = 2.5\ \text{eV}$$

Comparison with photon energy:
$$E = 2.25\ \text{eV} \gt \phi_{Cs} = 1.9\ \text{eV}$$ ⇒ photo-electric emission is possible for Cs.
$$E = 2.25\ \text{eV} \lt \phi_{Li} = 2.5\ \text{eV}$$ ⇒ photo-electric emission is not possible for Li.

Therefore, the photo-electric effect occurs only with cesium.

Correct option: Cs only (Option C).