When a metallic surface is illuminated by a light of wavelength `lamda`, the stopping potential for the photoelectric current is 3V. When the same surface is illuminated by light of wavelength `2 lamda`, the stopping potential is 1V, the threshold wavelength for this surface is

Options

A. `4 lamda`
B. `3.5 lamda`
C. `3 lamda`
D. `2.75 lamda`

Answer

A

Explanation

`(hc)/lamda- (lamda c)/lamda_0= 3`
`(hc)/(lamda.)-(hc)/lamda_0=1`
Dividing Eq. (i) by Eq. (ii)
`(1/lamda-1/lamda_0)/(1/(2 lamda)-1/lamda_0)=3`
`(2(lamda_0-lamda))/(lamda_0-2lamda)=3`
`2 lamda_0-2 lamda=3 lamda_0-6lamda`
`lamda_0=4 lamda`