When a centimeter thick surface is illuminated with light of wavelength `lamda`, the stopping potential is V. When the same surface is illuminated by light of wavelength `2lamda`, the stopping potential is `(V)/(3)`. Threshold wavelength for the metallic surface is

Options

A. `(4 lambda)/3`
B. `4 lambda`
C. `6 lambda`
D. `(8 lambda)/3`

Answer

B

Explanation

`(hC)/(lamda) =phi +eV`....(i)
`(hC)/(2 lamda) =phi+ (eV)/(3)`....(ii)
3. II-I
`rArr ((3)/(2)-1)(hc)/(lamda)=2 phi rArr phi =(hc)/(4 lamda)`
`:. lamda_(th) = 4 lamda`