When photones of energy 4.0 eV fall on the surface of a metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` ( in eV) and a de-Broglie wavelength `lambda_(A)`. When the same photons fall on the surface of another metal B, the maximum kinetic energy of ejected photoelectrons is `T_(B) = T_(A) -1.5eV`. If the de-Broglie wavelength of these photoelectrons is `lambda_(B) =2 lambda _(A)`, then the work function of metal B is

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 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

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

When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, T_A (expressed in eV) and deBroglie wavelength lambda_A . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20V is T_B = T_A -1.50eV . If the deBroglie wavelength of those photoelectrons is lambda_B = 2lambda_A then

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