Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitt…

Question

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be

Accepted Answer

Correct answer: D. (d) According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is K_ max= hv- _0 where hv is the energy of incident photon and _0 is the work function. But K_ max= 12mv_ max^2 ∴ 12mv_ max^2= hv- _0 As per question 12mv_ max1^2=1 eV-0.5 eV=0.5 eV … . . (i ) and 12mv_ max_2^2=2.5 eV-0.5 eV=2 eV … ( ii ) Dividing eqn. (i) by eqn. (ii), we get v_ max1 0.05em^2v_ max2 0.05em^2= 0.5 eV2 eV= 14 v_ max1 0.05emv_ max2 0.05em= 14= 12

$Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV . Ratio of maximum speeds of emitted electrons will be$

Solution

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