The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons (b) Stopping potential (c) maximum speed of the emitted photoelectrons?
(a). Work function, ϕo=2.14eV Frequency of light, ν=6×1014Hz Maximum K.E. is given by the photoelectric effect. K=hν−ϕo K=(1.6×10−196.626×10−34×6×1014)−2.14=0.345eV (b). For stopping potential, Vo, we can write the equation for kinetic energy as, K=eVo⇒Vo=eK =1.6×10−190.345×1.6×10−19=0.345V (c). The maximum speed of the emitted photoelectrons is v. Hence, the relation for kinetic energy can be written as, K=21mv2 m=9.1×10−31kg is the mass of electron. ⇒v2=2K/m ⇒v=3.323×105m/s=332.3km/s
