COMEDK 2025 Morning Solved Paper

Concept:Photoelectric emission requires each photon to have energy greater than the metal’s work function, and the plate must be uncharged to avoid an opposing electric field.Explanation:For electrons to be emitted from a metal surface, the incident photon must provide enough energy to overcome the work function $\phi$.This condition is $h\nu \ge \phi$, where $h$ is Planck’s constant and $\nu$ is the photon frequency.Option A (positively charged plate) helps attract electrons, but if $h\nu < \phi$, no electrons are freed.Option B (negatively charged plate) creates a repulsive barrier, making emission harder.Option C (high intensity) increases the number of photons but does not change the threshold condition – if $h\nu < \phi$, no emission occurs regardless of intensity.For emission to always take place, two things are needed: the photon energy must exceed the work function, and the plate must have no net charge to avoid extra potential barriers.Option D states exactly this: the work function is less than the energy of a single photon and the plate is uncharged.Answer:Option D: The work function of the plate is less than the energy of a single photon and the plate is uncharged.