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

Answer

B

Explanation

`eV=hc((1)/(lamda)-(1)/(lamda_0))`
`(eV)/(3)=hc((1)/(2lamda)-(1)/(lamda_0))`
Dividing Eq. (i) and (ii), we get `lamda_0=4lamda`