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

Answer

B

Explanation

`(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`