The light of two different frequencies whose photons have energies 3.8eV and 1.4eV respectively, illuminate a metallic surface whose work function is…

Question

The light of two different frequencies whose photons have energies 3.8eV and 1.4eV respectively, illuminate a metallic surface whose work function is 0.6eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

MARKS Content Team · Accepted Answer

Correct answer: (A) 2:1. Expression for maximum kinetic energy can be written as, KE_max=E-ϕ KE_(max, 1)=E_1-ϕ=3.8-0.6=3.2 eV, KE_(max, 2)=E_2-ϕ=1.4-0.6=0.8 eV KE_(max, 1)KE_(max, 2)=3.2/0.8=4⇒4=(12 mv)/1212 mv22 ⇒v12/v22=4 ⇒v_1v_2=2:1

The light of two different frequencies whose photons have energies $3.8 eV$ and $1.4 eV$ respectively, illuminate a metallic surface whose work function is $0.6 eV$ successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

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