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`.

Explanation

Given, work function, `m_0=2.14 eV`
`=2.14xx1.6xx10^(-19)J`
`=3.424 xx10^(-19)J`
Frequency of incident light , `v=6xx10^(14)Hz`
(a) Maximum kinetic energy of photoelectrons is
`K_(max)=hv-W_0`
`=6.62xx10^(-34)xx6xx10^(14)-3.424xx10^(-19)`
`=3.972xx10^(-19)-3.424xx10^^(-19)`
`=0.548xx10^(-19)J`
`=(0.548xx10^(-19)J)/(1.6xx10^(-19))eV`
= 0.34 eV
(b) Let `V_0` be the stopping potential . Then
`eV_0=K_(max)`
`:.V_0=(K_(max))/e=(0.548xx10^(-19))/(1.6xx10^(-19))`
or `V_0=0.342V`
(c) Also, if `v_(max)` is the maximum speed of emitted photoelectrons . Then,
`1/2mv_(max)^2=K_(max)`
`:.v_(max)^2=(2K_max)/m`
`=(2xx0.548xx10^(-19))/(9.1xx10^(-31))=0.12xx10^(12)`
`impliesv_(max)=0.345xx10^6ms^(-1)`
`=3.45 km s ^(-1)`