[" When photon of energy "4.25" eV strike the surface "],[" of a metal "A" ,the ejected photo electrons have "],[" maximum kinetic energy "T_(A)eV" and de-Broglie "],[" wave length "lambda_(A)" .The maximum kinetic energy of "],[" photo electrons liberated from another metal "B" by "],[" photon of energy "4.7" eV is "T_(B)=(T_(A)-1.5)eV" .If "],[" the de-Broglie wave length of these photo electrons "],[" is "lambda_(B)=2 lambda" then: "]

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 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 photon of energy 25eV strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy T_(A)eV and de Brogle wavelength lambda_(A) .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.76 eV is T_(B) = (T_(A) = 1.50) eV .If the de broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) then i. (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV III T_(A) = 2.0 eV IV. T_(B) = 3.5 eV

When photon of energy 25eV strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy T_(A)eV and de Brogle wavelength lambda_(A) .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.76 eV is T_(B) = (T_(A) = 1.50) eV .If the de broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) then i. (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV III T_(A) = 2.0 eV IV. T_(B) = 3.5 eV

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy T_(A) eV and De-broglie wavelength lambda_(A) . The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is T_(B) = (T_(A) - 1.50) eV if the de Brogle wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) , then