When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, T1 (expressed in eV) and de-Broglie wavelength lambda2. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is T2 -T1 = 1.50eV. If the de Broglie wavelength of these photoelectrons is 2lamda= 2×lambda1 then which is not correct :1) The work function of A is 2.25 eV2) The work function of B is 3.70 eV3) T1 = 2.00eV4) T2= 2.75 eV

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 photoelectrons have maximum kinetic energy, T1 (expressed in eV) and de-Broglie wavelength lambda2. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is T2 -T1 = 1.50eV. If the de Broglie wavelength of these photoelectrons is 2lamda= 2×lambda1 then which is not correct :1) The work function of A is 2.25 eV2) The work function of B is 3.70 eV3) T1 = 2.00eV4) T2= 2.75 eV
Answer» The maximum KINETIC energy of photoelectrons LIBERATED from another metal B by photons of energy ... If the de Broglie wavelength of these photoelectrons is , then ... The work FUNCTION of B is 4.20 EV.