To solve this problem, we need to determine which photosensitive surfaces can emit photoelectrons when exposed to electromagnetic radiation with energy 2.20 eV.$\\$The emission of photoelectrons is governed by the photoelectric effect, which states that photoelectrons are emitted if the energy of the incident radiation is greater than the work function of the material.$\\$Given:$\\$• Work function of Caesium (Cs): $\phi_{Cs} = 2.14$ eV$\\$• Work function of Potassium (K): $\phi_{K} = 2.30$ eV$\\$• Work function of Sodium (Na): $\phi_{Na} = 2.75$ eV$\\$• Energy of incident radiation: $E = 2.20$ eV$\\$Let's compare the energy of the incident radiation with the work functions:$\\$1. For Caesium (Cs):$\\ E = 2.20$ eV$> \phi_{Cs} = 2.14$ eV$\\$Since $E > \phi_{Cs},$ Caesium will emit photoelectrons.$\\$2. For Potassium (K):$\\ E = 2.20$ eV$< \phi_{K} = 2.30$ eV$\\$Since $E < \phi_{K},$ Potassium will not emit photoelectrons.$\\$3. For Sodium (Na):$\\ E = 2.20$ eV$< \phi_{Na} = 2.75$ eV$\\$Since $E < \phi_{Na},$ Sodium will not emit photoelectrons.$\\$Conclusion: Only Caesium (Cs) will emit photoelectrons when exposed to the given radiation.$\\$Therefore, the correct option is:$\\$Option 1: Cs only