When photons of energy 4.25 eV strike the surface of a metal A, the ejected photo electrons have maximum kinetic energy, TA (expressed in eV) and de Broglie wavelength lA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is TB=TA−1.50  eV. If the de Broglie wavelength of these photoelectrons is λB=2λA, then the work function of metal B is _______eV.