When a certain metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential for photo electric current is `6 v_(0)`. When the same surface si illuminated with light of wavelength `2 lambda`, the stopping potential is `2v_(0)`. The threshold wavelength of this surface for photoelectric effect is -

Options

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

Answer

B

Explanation

From photoelectric equation,
`hv=W+eV_(0)` (where, W=work function)
so, `(hc)/(lamda)=W+3eV_(0)` . . . (i)
also, `(hc)/(2lamda)=W+eV_(0)`
`implies (hc)/(lamda)=2W+2eV_(0)` . . . (ii)
Subtracting eq. (i) from (ii) , we get
`0=W-eV_(0) implies W=eV_(0)`
from eq. (i), `(hc)/(lamda)=eV_(0)+3eV_(0)=4eV_(0)`
the threshold wavelength is given by
`lamda_(th)=(hc)/(W)=(4eV_(0)lamda)/(eV_(0))=4lamda`.