Question

A light whose frequency is equal to `6xx10^(14)Hz` is incident on a metal whose work function is `2eV(h=6.63xx10^(-34)Js,1eV=1.6xx10^(-19)J)`. The maximum energy of electrons emitted will be:
A. 2.49 eV
B. 4.49 eV
C. 0.49 eV
D. 5.49eV

Answer 1

Correct Answer - C
Absorbed energy=Threshold energy+Kinetic nergy of photoelectrons
Absorbed energy=hv
`=6.626xx10^(-34)xx6xx10^(14)`
`=3.9756xx10^(-19)J`
`=(3.9756xx10^(-19))/(1.6xx10^(-19))=2.49eV`
2.49=2eV+Kinetic energy of photoelectons
Kinetic energy of photoelectron=0.49eV