Photons of energy `2.0eV` fall on a metal plate and release photoelectrons with a maximum velocity `V`. By decreasing `lambda` and `25%` the maximum velocity of photoelectrons is doubled. The work function of the metal of the material plate in `eV` is nearly

Photon of energy 2.0 eV and wavelength lamda fall on a metal plate and release photoelectrons with a maximum velocity v. By decreasing lamda by25% the maximum velocity of photoelectrons is doubled .The work function of the material of the metal plate in eV is

A photon of energy 2.5 eV and wavelength ′ λ ′ falls on a metal surface and the ejected electrons have maximum velocity ′ v ′ . If the ′ λ ′ of the incident light is decreased by 20%, the maximum velocity of the emitted electrons is doubled. The work function of the metal is :

A photon of energy 2.5 eV and wavelenght lambda falls on a metal surface and the ejected electron have velocity 'v' . If the lambda of the incident light is decreased by 20% the maximum velocity of the emitted electrons is doubled. The work function of the metal is

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 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