When light of wavelength `lambda` is incident on photosensitive surface, the stopping potential is V. When light of wavelength `3lambda` is incident on same surface, the stopping potential is `V/6`Thereshould wave length for the surface is

Options

A. `2lambda`
B. `3lambda`
C. `4lambda`
D. `5lambda`

Answer

D

Explanation

According to question, using Einstein's photoelectric equation `(hc)/lambda-phi=eV`
In first case, for lambda stopping potential is V i.e. equation becomes
`(hc)/lambda-phi=eV`
Similarly in second case, for 31 stopping potential is `V/6`
i.e. equation becomes `(hc)/(3lambda)-phi=(eV)/6`
Rightarrow `(2hc)/(lambda)-phi=eV`
From Eqs. (i) and (ii) by subtracting
`(2hc)/(lambda)=eV+6phi``(hc)/lambda=eV+phi`
`(---)/((hc)/lambda=5phi)`
and `phi=(hc)/lambda_0` lt brgt Thus `lambda_0= 5lambda`