A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. (b)Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.

Sol. a) When λ = 400 nm Energy of photon = hc/λ = 1240/400 = 3.1 eV This energy given to electron But for the first collision energy lost = 3.1 ev * 10% = 0.31 ev for second collision energy lost = 3.1 ev * 10% = 0.31 ev Total energy lost the two collision = 0.31 + 0.31 = 0.62 ev K.E. of photon electron when it comes out of metal = hc/λ – work function – Energy lost due to collision = 3.1 ev – 2.2 – 0.62 = 0.31 ev b) For the 3rd collision the energy lost = 0.31 ev Which just equative the KE lost in the 3rd collision electron. It just comes out of the metal Hence in the fourth collision electron becomes unable to come out of the metal Hence maximum number of collision = 4.