When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potential is `V`. If the same surface is illuminated with radiation of wavelength `2 lambda` , the stopping potential is `(V)/(4)`. The threshold wavelength surface is :

Options

A. `4 lambda`
B. `5 lambda`
C. `(5)/(2)lambda`
D. `3 lambda`

Answer

D

Explanation

`eV = (h c)/(lambda) - (h c)/(lambda_(0))` (1)
`(e V)/(4) = (h c)/(2lambda) -(h c)/(lambda_(0))` (2)
From equation (1) and (2)
on solving `lambda_(0) = 3 lambda`