When a piece of metal is illuminated by monochromatic light of wavelength `lambda` then the stopping potential for photoelectric current is 2.5 `V_0`. When the same surface is illuminated by light of wavelength 1.5 `lambda`, then the stopping potential becomes `V_0`. Find the value of threshold wavelength for photoelctric emission.

Explanation

As , `eV_0=(hc)/lambda-phi_0`,
so `2.5V_0e=(hc)/lambda-phi_0.....(i)`
and `V_0e=(hc)/(1.5lambda)-phi_0.....(ii)`
From (i),
`eV_0=(hc)/(2.5lambda)-(phi_0)/2.5......(iii)`
From (ii) and (iii),
`(hc)/(1.5lambda)-phi_0=(hc)/(2.5lambda)-(phi_0)/2.5`
or `(hc)/lambda[1/1.5-1/2.5]=phi_0(1-1/2.5)=1.5/2.5 phi_0=3/5phi_0`
of `(hc)/lambda[10/15-10/25]=3/2 (hc)/(lambda_0)`
or `(hc)/lambda[4/15]=3/5 (hc)/(lambda_0)`
or `lambda_0=9/4 lambda=2.25lambda`