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← JEE Advanced 2021 · Paper 2
PhysicsInteger+4Question 19 of 57

JEE Advanced 2021 · Paper 2, Physics question 19

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals PP, QQ and RR are EPE_P, EQE_Q and ERE_R, respectively, and they are related by EP=2EQ=2ERE_P = 2E_Q = 2E_R. In this experiment, the same source of monochromatic light is used for metals PP and QQ while a different source of monochromatic light is used for the metal RR. The work functions for metals PP, QQ and RR are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal RR, in eV, is ____.

Answer and worked solution

Official answer: 6

P and Q (same photon energy hνh\nu):
EP=hν−4.0,EQ=hν−4.5.E_P=h\nu-4.0,\qquad E_Q=h\nu-4.5.
With EP=2EQE_P=2E_Q: hν−4=2hν−9h\nu-4=2h\nu-9, so hν=5h\nu=5 eV. Then EP=1E_P=1 eV and EQ=0.5E_Q=0.5 eV.

R. ER=EQ=0.5E_R=E_Q=0.5 eV, so the photon used for R has energy
5.5+0.5=6 eV.5.5+0.5=6\text{ eV}.