Physics - Modern Physics Question with Solution | TestHub

For photoelectric emission to take place, the energy of the incident electromagnetic radiation must be greater than or equal to the work function of the photosensitive surface ($E\ge \Phi$).

Given incident energy $E=2.41\text{ eV}$.

Work function of Caesium (Cs) ${\Phi }_{\text{Cs}}=2.14\text{ eV}$. Since $2.41\text{ eV}>2.14\text{ eV}$, photoelectric emission will occur.

Work function of Potassium (K) ${\Phi }_{\text{K}}=2.30\text{ eV}$. Since $2.41\text{ eV}>2.30\text{ eV}$, photoelectric emission will occur.

Work function of Sodium (Na) ${\Phi }_{\text{Na}}=2.75\text{ eV}$. Since $2.41\text{ eV}<2.75\text{ eV}$, photoelectric emission will not occur.

Therefore, both K and Cs surfaces will emit photoelectrons.

Answer: Both K and Cs