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Sahay Sir > Question Answers > JEE Mains Physics 2002-2019 Solved Video Solutions > Chapter 18 - Atoms and Nuclei > When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA expressed in eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA−1.50eV). If the de-Broglie wavelength of these photoelectrons is λB=2λA, then

When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA expressed in eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA−1.50eV). If the de-Broglie wavelength of these photoelectrons is λB=2λA, then

Beta rays emitted by a radioactive material are. When photons of energy 4.25 eV strike the surface of a metal A