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InterviewSolution
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When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy lamda_(A). The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is T_(B)=(T_(A) -1.50) eV. If the de Broglie wavelength of these photoelectrons is lamda_(B)=2lamda_(A), then select the correct statement statement (s) |
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