A certain metallic surface is illuminated with monochromatic light of wavelength `lamda` . The stopping potential for photoelectric current for this light is `3V_0` . If the same surface is illuminated with light of wavelength `2 lamda` , the stopping potential is `V_0` . The threshold wavelength for this surface for photoelectric effect is

Options

A. `lamda/4`
B. `lamda/6`
C. `6lamda`
D. `4lamda`

Answer

D

Explanation

According to Einstein.s photoelectric equaiton ,
`eV_s =(hc)/lamda -(hc)/lamda_0`
where `V_s` = Stopping potential , `lamda` = Incident wavelength
`lamda_0` = Threshold wavelength
or `V_s =(hc)/e(1/lamda-1/lamda_0)`

For the second case
`3V_0 =(hc)/e(1/lamda-1/lamda_0)" "....(i)`
For the second case
`V_0 =(hc)/e(1/(2lamda)-1/lamda_0)" "....(ii)`
Divide eqn. (i) by (ii) ,we get
`3=((1/lamda-1/lamda_0))/((1/(2lamda)-1/lamda_0))`
`3/(2lamda)-3/lamda_0=1/lamda-1/lamda_0" or " 1/(2lamda)=2/(lamda_0)" or " lamda_0 = 4lamda`