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

Options

A. `6 lambda`
B. `4 lambda`
C. `(lambda)/(4)`
D. `(lambda)/(6)`

Answer

B

Explanation

Einstein's photoelectric equation can be written as
`(1)/(2) mv^(2) = hv - varphi rArr eV_( s) = ( hc)/(lambda) - varphi`
`3 eV_(0) = (hc)/(lambda) - varphi` …..(i)
`eV_(o) = (hc)/(2 lambda) - varphi` ….(ii)
`3 e V_(o) = (3 hc)/( 2 lambda) - 3 varphi`
Multiplying eqn. (ii) by (iii) and subtracting it from eqn. (i)
`varphi = (hc)/( 4 lambda)`
So , threshold wavelength , `lambda_(th) = (hc)/(varphi) = (hc)/(hc//4 lambda) = 4 lambda`