When photons of energy 4.25 eV strike the surface of a metal A, the ejected photo electrons have maximum kinetic energy Ta eV and de-Broglie wavelength λa . the maximum kinetic energy of photo electrons liberated from another metal B by photones of energy 4.7 eV is Tb = (Ta – 1.5) eV. If the De-Broglie wavelength of these photo electrons is λb = 2λa , then find
