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

Answer

B

Explanation

`eV_0 =(hc)/lamda -(hc)/lamda_0`
`implies (hc)/(e) (1/lamda-1/lamda_0)=4.8 " "...(i)`
`(hc)/(e) (1/(2lamda)-1/lamda_0)=1.6 " "...(ii)`
From (i) and (ii)
`(1/lamda-1/lamda_0)/(1/(2lamda)-1/lamda_0)=(4.8)/(1.6)implieslamda_0=4lamda`