When a photosensitive surface is illuminated with light of wavelength `lambda`, the stopping potential is `V_(0)`. But when light of wavelength `2lambda` is incident on the same surface, the stopping potential is `(V_(0))/(4)`. What is the threshold wavelength for the surface ?

Options

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

Answer

B

Explanation

`eV_(0)=h(v-v_(0))=hC(1/lambda-1/lambda_(0))` ...(i)
and `(eV_(0))/(4)=hC((1)/(2lambda)-(1)/(lambda_(0)))` ...(ii)
Dividing (i) by (ii) we get,
`4/1=((1)/(lambda)-(1)/(lambda_(0)))/((1)/(2lambda)-(1)/(lambda_(0))) :.(4)/(2lambda)-(4)/(lambda_(0))=(1)/(lambda)-(1)/(lambda_(0))`
`(1)/(lambda)=(3)/(lambda_(0)) :. lambda_(0)=3lambda`