When radiation of wavelength `lambda ` is incident on a metallic surface , the stopping potential is `4.8 "volts"` . If the same surface is illuminated with radiation of double the wavelength , then the stopping potential becomes `1.6 "volts"`. Then the threshold wavelength for the surface is

Options

A. `2lambda`
B. `4lambda`
C. `6lambda`
D. `8lambda`

Answer

B

Explanation

By using `(hc)/(e ) ((1)/(lambda)-(1)/(lambda_(0)))=V_(0)`
`rArr (hc)/(e )((1)/(lambda)-(1)/(lambda_(0)))=4.8 " " `….(i)
`and (hc)/(e ) ((1)/(2lambda) - (1)/(lambda_(0))) = 1.6 " " ` …(ii)
From equation (i) by (ii),
`(((1)/(lambda)-(1)/(lambda_(0))))/(((1)/(2lambda)-(1)/(lambda_(0))))=(4.8)/(1.6) rArr lambda_(0) = 4lambda`