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

Answer

c

Explanation

Here `omega= 2pi xx 6 xx 10^(14)`
`implies f= 6 xx 10^(14) Hz`
Wavelength
`lambda= (c)/(f) = (3 xx 10^(8))/(6 xx 10^(14))= 0.5 xx 10^(-6)m= 5000Å`
Given `E= (12375)/(5000)= 2.48eV`
Using `E= W+ eV_(s)`
`= > 2.48 = 2+ eV_(s)`
or `V_(s)= 0.48V`