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`. If work function of the metallic surface is

Explanation

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