When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potential is `V`. If the same surface is illuminated with radiation of wavelength `2 lambda` , the stopping potential is `(V)/(4)`. The threshold wavelength surface is :

Options

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

Answer

d

Explanation

Case(i), `eV=(hc)/(lambda)-(hc)/(lambda_(0))..........(i)`
Case (ii), `(eV)/4=(hc)/(2lambda)-(hc)/(lambda_(0))`
or `eV=(4hc)/(2lambda)-(4hc)/(lambda_(0)).......(ii)`
From (i) and (ii),
`(hc)/(lambda)-(hc)/(lambda_(0))=(4hc)/(2lambda)-(4hc)/(lambda_(0))`
`1/lambda-1/(lambda_(0))=2/lambda-4/(lambda_(0)) or 3/(lambda_(0)) =1/lambda or lambda_(0)=3lambda`