When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectronshave maximum kinetic energy, T_(A) (expressed in eV) and de Broglie wavelength lambda_(A). The mximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is T_(B) = T_(A) - 1.50 eV. If the de Broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A), then which is not correct :

When photons of energy 4.25 eV strike the surface of a metal A, the ej

1.	When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectronshave maximum kinetic energy, T_(A) (expressed in eV) and de Broglie wavelength lambda_(A). The mximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is T_(B) = T_(A) - 1.50 eV. If the de Broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A), then which is not correct :
Answer» The WORK FUNCTION of A is 2.25 EV The work function of B is 3.70 eV `T_(A) = 2.00 eV` `T_(B) = 2.75 eV` ANSWER :D