The work function of caesium is 2.14 eV. When light of frequency `6xx10^(14)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 and (c) maximum speed of the emitted photoelectrons. given , `h=6.63xx10^(-34)Js, 1eV=1.6xx10^(-19)J, c=3xx10^(8)m//s`.

Answer

1

Explanation

Given `phi_(0) = 2.14eV`
(1) According to photo electric equation, the maximum kinetic energy of electron.
`K_(max). = hv - phi_(0)`
Energy of incident photon
E = hv
`=(6.63xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))eV`
=2.49 eV
Thus,
`K_(max) = (2.49 - 2.14)eV`
`K_(max) = 0.35 eV = 0.56 xx 10^(-19) j`
(2) Stopping potential `V_(0)=(K_max)/(e)=(0.35eV)/(e)`
`V_(0)` = 0.35 volt
Maximum speed of electron `(V_(max))`.
`(1)/(2)xx9.1xx10^(-31)xxV_(m)^(2)=0.56 xx10^(-19)`
`rArrV_(m)=3.5xx10^(5)m//s`