BITSAT 2018 Slot 2 Solved Paper

When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is TB=(TA−1.50)eV. If the de-Broglie wavelength of these photoelectrons λB=2λA, then choose the correct statement(s).