When photon  of energy  `25eV` strike the  surface  of a metal A, the ejected photelectron
Channel: Doubtnut
Date: 2020-07-26

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`