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

Options

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

Answer

D

Explanation

`K_("max") = (hc)/lamda - (hc)/(lamda_0) `
`eV = (hc)/lamda - (hc)/(lamda_0) " "...(i)`
`(eV)/(4) = (hc)/(2lamda) - (hc)/lamda_0 " "...(ii)`
From (i) and (ii)
`implies 4 = (1/lamda -1/lamda_0)/(1/(2lamda)-1/lamda_0)implies lamda_0 =3lamda`