Photoelectric Effect + de-Broglie Wavelength Combined Question | JEE Main Concept 🚀
💡 Question A photoemissive substance is illuminated with radiation of wavelength λ i \lambda_i so that it releases electrons with de-Broglie wavelength λ e \lambda_e . The longest wavelength of radiation that can just emit photoelectrons (i.e., threshold wavelength) is λ 0 \lambda_0 . Find the expression for de-Broglie wavelength ( λ e \lambda_e ) of the emitted electrons in terms of λ i \lambda_i and λ 0 \lambda_0 . 🖼️ Question Image ✍️ Short Solution Step 1 — Photoelectric effect equation When light of wavelength λ i \lambda_i falls on a metal surface, the energy of the incident photon is: E = h c λ i The maximum kinetic energy of the emitted electron is: K max = h c λ i − h c λ 0 where h c λ 0 \frac{hc}{\lambda_0} is the work function ϕ \phi (minimum energy required to eject an electron). Step 2 — Relate kinetic energy and de-Broglie wavelength For an electron having de-Broglie wavelength λ e \lambda_e : K max = p 2 2...