In a photoemission experiment, Kₚ = 2KQ = 2Kᵣ. The same photon source is used for P and Q, but a different source for R. Work functions are 4.0, 4.5 and 5.5 eV, respectively. Find the photon energy used for R, in eV.
Physics · Photoelectric effect and dual nature of matter
JEE Advanced 2021

Question

In a photoemission experiment, Kₚ = 2KQ = 2Kᵣ. The same photon source is used for P and Q, but a different source for R. Work functions are 4.0, 4.5 and 5.5 eV, respectively. Find the photon energy used for R, in eV.

Physics · Photoelectric effect and dual nature of matter

JEE Advanced 2021

G1 Physics · Accepted Answer

Step 1: For P and Q with the same photon energy: Kₚ−KQ=φQ−φₚ=0.5 eV. Step 2: Given Kₚ=2KQ, KQ=0.5 eV and Kₚ=1.0 eV. Since Kᵣ=KQ=0.5 eV and φᵣ=5.5 eV, photon energy for R is 6.0 eV.