A metallic surface is irradiated by a monochromatic light of frequency `v_(1)` and stopping potential is found to be `V_(1)`. If the light of frequency `v_(2)` irradiates the surface, the stopping potential will be

Options

A. `V_(1) + h/e (upsilon_(1) + upsilon_(2)) `
B. `V_(1) + h/e (upsilon_(2) + upsilon_(1)) `
C. `V_(1) + h/e (upsilon_(2) - upsilon_(1)) `
D. `V_(1) - h/e (upsilon_(1) + upsilon_(2)) `

Answer

B

Explanation

Maximum kinetic energy , `K_(max) = 1/2 mv^(2) = eV_s`
where `V_s` is the stopping potential
According to Einstein.s photoelectric effect
`hupsilon_1 = phi_0+eV_1" "...(i)`
`hupsilon_2 = phi_0+eV_2" "...(ii)`
` :. h (upsilon_(1) -upsilon_(2) )=e(V_1-V_2) " "` (From (i) and (ii) )
`h/e (upsilon_(1) - upsilon_(2))=V_1 -V_(2) " or " V_(2) = V_(1) +h/e (upsilon_(2) - upsilon_(1))`