When a metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential is `5 V_0`. When the same surface is illuminated with light of wavelength `3lambda`, the stopping potential is `V_0`. Then the work function of the metallic surface is:

Options

A. `(hc)/(6 lambda)`
B. `(hc)/(5lambda)`
C. `(hc)/(4lambda)`
D. `(2hc)/(4lambda)`

Answer

A

Explanation

`(hc)/(lambda) =5 eV_(0)+phi`
`(hc)/(3 lambda) =eV_(0) +phirArr (2hc)/(3lambda) = 4eV_(0) rArr phi=(hc)/(6 lambda) `