When a metallic surface is illuminated with radiation of wavelength `lamda` , the stopping potential is V . If the same surface is illuminated with radiation of wavelength `2lamda` , the stopping potential is `V/4` . The threshold wavelength for the metallic surface is

Options

A. `5/2 lamda`
B. `3lamda`
C. `4lamda`
D. `5lamda`

Answer

B

Explanation

According to Einstein.s photoelectric equation,
`eV_s =(hc)/lamda -(hc)/lamda_0`
`:.` As per equation , `eV =(hc)/lamda - (hc)/lamda_0 " "....(i)`
`(eV)/4 =(hc)/(2lamda)-(hc)/lamda_0 " "...(ii)`
From equations (i) and (ii) , we get
`(hc)/(2lamda)-(hc)/((4lamda))=(hc)/lamda_0 -(hc)/(4lamda_0 )implies (hc)/(4lamda)=(3hc)/(4lamda_0 )" or " lamda_0 = 3lamda`