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

Answer

A

Explanation

`(hc)/(lambda)=(hc)/(lambda_(p))+eV.(i)`
`(hc)/(2lambda)=(hc)/(lambda_(O))+(eV)/4.......(ii)`
`(hc)/(2lambda)=(hc)/(lambda_(O))+1/4((hc)/(lambda)-(hc)/(lambda_(O)))`
`1/(2lambda)-1/(4lambda)=1/(lambda_(O))-1/(4lambda_(O))`
`1/(4lambda)=3/(4lambda_(O)) implies lambda_(O)=3 lambda`