When protons 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 – 1.50 eV). If the de-Broglie wavelength of these photoelectrons isλB=2λA, then