The incident photon's energy is calculated as: \[ E = \frac{1240}{\lambda} = \frac{1240}{550} \approx 2.25 \, \text{eV} \]. Photoelectric effect necessitates the photon energy exceeding the metal's work function.
For Cesium (work function = 1.9 eV):
As \( 2.25 \, \text{eV} > 1.9 \, \text{eV} \), photoelectric effect is feasible with Cs.
For Lithium (work function = 2.5 eV):
Since \( 2.25 \, \text{eV} < 2.5 \, \text{eV} \), photoelectric effect is not possible with Li.
Therefore, the solution is \( \boxed{\text{Cs only}} \).