When a radiation of energy 5eV falls on a surface,the emitted photoelectrons have a maximum kinetic energy of 3eV .The stopping potential is

Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV . The stopping voltage required for these electrons are

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 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 monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic are N and T respectively. If the intensity of radiation is 2 I, the number of emitted electrons and their maximum kinetic energy are respectively.

UV radiation of energy 4 eV falls on caesium surface whose photoelectric work function is 1.95 eV. The kinetic energy of the fastest photoelectrons is