A light whose frequency is equal to 6xx1...

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:

2.49 eV

4.49 eV

0.49 eV

5.49eV

Text Solution

Verified by Experts

The correct Answer is:

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

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