When a certain metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential for photo electric current is `6 v_(0)`. When the same surface si illuminated with light of wavelength `2 lambda`, the stopping potential is `2v_(0)`. The threshold wavelength of this surface for photoelectric effect is -

Options

A. `6 lambda`
B. `4 lambda//3`
C. `4lambda`
D. `8lambda`

Answer

C

Explanation

`(e(6V_(0)))/(e(2V_(0))) = ((hc)/(lambda) - phi)/((hc)/(2lambda) - phi) rArr lambda_(0) = 4lambda`