When a certain metallic surface is illuminated with monochromatic light of wavelength `lamda`, the stopping potential for photoelectric current is `3V_0` and when the same surface is illuminated with light of wavelength `2lamda`, the stopping potential is `V_0`. The threshold wavelength of this surface for photoelectrice effect is

Options

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

Answer

C

Explanation

`(hc)/(lambda) =phi +e.3 V_(0)` ….(i)
`(hc)/(2lambda) =phi + e.V_(0)` ….(ii)
3(ii) - (i) `implies (hc)/(2 lambda) =3 phi - phi =2 phi =2. (hc)/(lambda_(0))`
`lambda_(0) =4 lambda`