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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 (KE)A and de-Broglie wavelength is λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.7 eV is (KE)B, where (KE)B=(KE)A–1.5 eV. If the de-Broglie wavelength of these photoelectrons is λB(=2λA), then: |
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Answer» When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy (KE)A and de-Broglie wavelength is λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.7 eV is (KE)B, where (KE)B=(KE)A–1.5 eV. If the de-Broglie wavelength of these photoelectrons is λB(=2λA), then: |
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