When a certain metallic surface is illuminated with mono chromatic light of wavelength `lambda`, the stopping potential for photoelectric current is `3V_(0)`. When the same surface is illuminated with light of wavelength `2lambda` the stopping potential is `V_(0)`. The threshold wavelength for this surface for photoelectric effect is.

Options

A. `4lambda`
B. `lambda/4`
C. `lambda/6`
D. `6lambda`

Answer

A

Explanation

For the photo emission `eV_(s)=h(v-v_(0))=hc(1/lambda-1/lambda_(0))`
`:.V_(s)=(hc)/(e)[1/lambda -1/lambda_(0)]`
In the first case, `3V_(0)=(hc)/(e)[1/lambda -1/lambda_(0)]` ...(1)
and in the second case, `V_(0)=(hc)/(e)[(1)/(2lambda)-(1)/(lambda_(0))]` ...(2)
`:.3V_(0)=(3hc)/(e)[(1)/(2lambda)-(1)/(lambda_(0))]` ...(2a)
Equating (1) and (2a) we get
`(hc)/(e) [(1)/(lambda)-(1)/(lambda_(0))]=(3hc)/(e)[(1)/(2lambda)-(1)/(lambda_(0))]=(3hc)/(e)[(1)/(2lambda)-(1)/(lambda_(0))]`
`(1)/(lambda)-(1)/(lambda_(0))=(3)/(2lambda)-(3)/(lambda_(0))`
`:.(2)/(lambda_(0))=(3)/(2lambda)-(2)/(2lambda)=(1)/(2lambda)`
`:.lambda_(0)=4lambda`