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

First case, `eV=h(v-v_(0))=h[(1)/(lambda)-(1)/(lambda_(0))]` ...(1)
Second case, `(eV)/(6)=h[(1)/(3lambda)-(1)/(lambda_(0))]` ...(2)
Dividing (1) by (2), we get
`6/1=((1)/(lambda)-(1)/(lambda_(0)))/((1)/(3lambda)-(1)/(lambda_(0)))`
`:.(6)/(3lambda)-(6)/(lambda_(0))=(1)/(lambda)-(1)/(lambda_(0))`
`:.(1)/(lambda)=(5)/(lambda_(0)):.lambda_(0)=5lambda`