An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV . The proton captures the electrons and forms a hydrogen atom in second excited st

Question

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV . The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 A What is the maximum kinetic energy of the emitted photoelectron?

TestHub · Accepted Answer

Initially, energy of electron = + 3 eV finally, in 2 nd excited state, energy of electron = - ( 13 . 6 eV ) 3 2 = - 1 . 51 eV Loss in energy is emitted as photon, So, photon energy hc λ = 4 . 51 eV Now, photoelectric effect equation KE max = hc λ - ϕ = 4 . 51 - hc λ th = 4 . 51 eV - 12400 eVA 4000 A = 1 . 41 eV

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