The work function of caesium metal 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?

Explanation

Work function `phi_(0)`=2.14 eV
Frequency of light `v_(0)=6xx10^(14)Hz`
`h=6.63xx10^(-34)` Js, l eV=`1.6xx10^(-19)J`
(a) Photoelectric equation ,
`K_(max)=hv-phi_(0)`
`=((6.63xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))-2.14)eV`
=(2.48625-2.14) eV `=0.34xx1.6xx10^(-19)J=0.544xx10^(-19)J`
(b) `K_(max)=eV_(0)`
`therefore V_(0)=(K_(max))/(e)=(0.34eV)/(e) therefore V_(0)=0.34 V`
(c )`K_(max)=0.34eV`
`therefore (1)/(2) mv_(max)^(2)=0.34xx1.6xx10^(-19)`
`therefore V_(max)^(2)=(2xx0.346xx1.6xx10^(-19))/(m)`
`=(2xx0.34xx1.6xx10^(-19))/(9.1xx10^(-31))`
`therefore v_(max)=sqrt((0.692xx1.6xx10^(-19))/(9.1xx10^(-31)))`
`therefore v_(max)=sqrt(0.12167xx10^(12))`
`=0.345775xx10^(6)~~345.8xx10^(3)m//s`
`therefore v_(max)=345.8 km//s`
Note:Textbook answer is different.