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

Work function of caesium metal, `phi_0=2.14 eV`
Frequency of light, `v=6.0xx10^(14) Hz`
(a)The maximum kinetic energy is given by the photoelectric effect as : `K=hv-phi_0`
Where, h=Planck's constant =`6.626xx10^(-34) Js`
`therefore K=(6.626xx10^(34)xx6xx10^(14))/(1.6xx10^(-19))- 2.14`
=2.485-2.140=0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV
(b)For Stopping potential `V_0` , we can write the equation for kinetic energy as :
`K=eV_0`
`therefore V_0=K/e` ltbr `=(0.345xx1.6xx10^(-19))/(1.6xx10^(-19))=0.345 V`
Hence, the stopping potential of the material is 0.345 V
(c )Maximum speed of the emitted photoelectrons=v
Hence, the realtion for kinetic energy can be written as :
`K=1/2 mv^2`
Where, m=Mass of electron =`9.1xx10^(-31)kg`
`v^2=(2K)/m`
`=(2xx0.345xx1.6xx10^(-19))/(9.1xx10^(-31))=0.1104 xx10^(12)`
`therefore v=3.323xx10^(5)` m/s = 332.3 km/s
Hence , the maximum speed of the emitted photoelectrons is 332.3 km/s