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 the light of wavelength `3lambda`, the stopping potential is `V_0`. Then, the work function of the metallic surface is

Options

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

Answer

A

Explanation

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