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Question

# The work function of caesium metal is 2.14 eV. When the light of frequency 6×1014Hz is incident on the metal surface, photoemission of electrons occurs. Calculate:(a) the maximum kinetic energy of the emitted electrons,(b) stopping potential,(c) the maximum speed of the emitted photoelectrons?

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Solution

## (a). Work function, ϕo=2.14eVFrequency of light, ν=6×1014HzMaximum K.E. is given by the photoelectric effect.K=hν−ϕoK=(6.626×10−34×6×10141.6×10−19)−2.14=0.345eV(b). For stopping potential, Vo, we can write the equation for kinetic energy as,K=eVo⇒Vo=Ke=0.345×1.6×10−191.6×10−19=0.345V(c). The maximum speed of the emitted photoelectrons is v.Hence, the relation for kinetic energy can be written as,K=12mv2m=9.1×10−31kg is the mass of electron.⇒v2=2K/m⇒v=3.323×105m/s=332.3km/s

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