The photoelectric work function of a metal surface is 2eV.When light of frequency `1.5xx10^(15)` Hz is incident on it, the maximum kinetic energy energy of the photo electrons ,approximately is

Options

A. 8eV
B. 6eV
C. 2eV
D. 4eV

Answer

D

Explanation

Work function ,`w=2eV=2xx1.6xx1010^(19)J`
Frequency ,`v=1.5xx10^(15)Hz.`
The expression for maximum kinetic energy is given as,
`(KE)_(max)=hv-w`
Where
h-Planck's constant `=6.6xx10^-(34)J`
`implies K.E_(max)=(6.6xx10^(-34))xx1.5xx10^(15))-(3.2xx10^(-19))`
`=9.9xx10^(-19)-3.2xx10^(-19)``=(6.7xx10^(19))/((1.6xx10^(19)))eV`
`K.E_(max)=4.2eV.`