The electric field of light wave is given as `vec(E)=10^(-3)cos((2 pi x)/(5xx10^(-7))-2pi xx 6xx10^(14)t) hat(x) (N)/(C)`. This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is:
Given, E (in eV) `= (12375)/(lambda ( "in" Å ))`

Options

A. 0.48 V
B. 2.48 V
C. 0.72 V
D. 2.0 V

Answer

A

Explanation

Angular frequency of electromagnetic wave can be taken from given wave function.
`omega=2pixx6xx10^(14) impliesf = 6 xx10^(14)Hz`
`lamda=c/f=(3xx10^8)/(6xx10^(14))=5000Å`
Energy of the photon :
`E=hv=(hc)/lamda=(12375)/5000eV=2.475eV`
From Einstein.s equation we may write the following:
`KE_(max)=E-phi`
`implieseV_S=E-phi`
`implieseV_S=2.475-2=0.475 eV`
`implies V_S=0.475V`