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

Answer

A

Explanation

Einstein `P.E.` equation
Case I : `eV = (hc)/(lambda) - (hc)/(lambda_(0))` ….(i)
Case II : `eV//4 = (hc)/( 2lambda) - (hc)/(lambda_(0))` …(ii)
From equations (i) and (ii)
`rArr 4 = ((1)/(lambda))/((1)/(2 lambda)) - ((1)/(lambda_(0)))/((1)/(lambda))` On solving `lambda_(0) = 31`