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Sahay Sir > Question Answers > JEE Mains Physics 2002-2019 Solved Video Solutions > Practice Set 1 > When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is

When photon of energy 4.0 eV strikes the surface of a metal A Which of the following gives a reversible operation?