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UV radiation of energy 4 eV falls on cae...

UV radiation of energy 4 eV falls on caesium surface whose photoelectric work function is 1.95 eV. The kinetic energy of the fastest photoelectrons is

A

`3.28xx10^(-19)J`

B

`4.88xx10^(-19)J`

C

`3.12xx10^(-19)J`

D

`6.4xx10^(-19)J`

Text Solution

Verified by Experts

The correct Answer is:
A
`KE_(max)=hv-phi`
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