When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_A$ expressed in eV and de-Broglie wavelength $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from
another metal B by photons of energy 4.70 eV is $T_B=(T_A-1.50eV).$ If the de-Broglie wavelength of these photoelectrons is $\lambda_B=2\lambda_A,$ then