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The work function of caesium metal is 2.14 eV. When light of frequency 6xx10^(14)Hz is incidennt on the metal surface, photoemission of electrons occus. What is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential, and (c) maximum speed of the emitted photoelectrons ? |
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Answer» Solution :Here work function `phi_(0)=2.14eV,` and FREQUENCY of incident radiation `v=6xx10^(14)Hz` `therefore`Energy of light photon `E=hv=6.63xx10^(-34)xx6xx10^(14)J=(6.63xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))eV=2.48eV` `therefore`Maximum kinetic energy of the emitted ELECTRONS `K_(max)=E-phi_(0)=2.48-2.14=0.34eV` (b) Stopping potential `V_(0)=(K_(max))/(e)=0.34V` (c) As `K_(max)=(1)/(2)mv_(max)^(2)=0.34eV=0.34xx1.6xx10^(-19)J=5.4xx10^(-20)`J `impliesv_(max)=SQRT((2K_(max))/(m))=sqrt((2xx5.4xx10^(-20))/(9.1xx10^(-31)))=3.44xx10^(5)ms^(-1) or 344km" "s^(-1)`. |
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