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Dual Nature Of Radiation And Matter

Question
CBSEENPH12047955

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

  • 1 : 4

  • 1 : 1

  • 1 : 5

  • 1 : 2

Solution

D.

1 : 2

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is

         Kmax = h ν  - ϕ0

where hu is the energy of incident photon and ϕ0 is the work function.

    Kmax = hv - ϕ0

∴   12 mvmax2  = hv - ϕ0

As per question

     12 mvmax2 = 1eV - 0.5 eV

                  = 0.5 eV               ....(i)

and 12 mvmax2 = 2.5 eV - 0.5 eV

                  = 2 eV                ......(ii)

Dividing eqn. (i) by eqn. (ii), we get

     vmax12vmax22 = 0.5 eV2 eV

              = 14

   vmax1vmax2 = 14

   vmax1vmax2 = 12

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