Answer
Barium (2.5 eV) and lithium (2.3 eV)
Work Step by Step
In case of photo electric effect, when light of high enough frequency illuminates a metal surface, electrons can gain enough energy to escape the metal by absorbing photons in the illumination. The least energy required by an electron to escape from the metal surface is called work function $(\phi)$ of the metal.
Thus, $\phi=hf_0$, where h is the Planck's constant and $f_0$ is the cutoff frequency.
Here, the illuminating light is visible light whose energy is of around $2$ to $3$ $eV$.
Thus the range of $\phi$ should have to be of around $2$ to $3$ $eV$
Therefore, from the given options, barium (2.5 eV) and lithium (2.3 eV) are suitable for the photocell that operates via the photoelectric effect with visible light.
