When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy `lamda_(A).` The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy `4.70 eV is T_(B)=(T_(A) -1.50) eV.` If the de Broglie wavelength of these photoelectrons is `lamda_(B)=2lamda_(A),` then select the correct statement statement (s)

Answer

`A=2.25eV,B=4.20 eV, T_(A)=2.00eV.`