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

Answer

B

Explanation

(b) : Maximum kinetic energy
`K_(max)=(1)/(2)mv^(2)=eV_(o)`
where `V_(o)` is the stopping potential.
According to Einstein's photoelectric equation
`hv_(1)=phi_(0)+eV_(1)` . . .(i)
`hv_(2)=phi_(0)+eV_(2)` . . .(ii)
`:.h(v_(1)-v_(2))=e(V_(1)-V_(2))`
`(h)/(e)(v_(1)-v_(2))=V_(1)-V_(2) or V_(2)=V_(1)+(h)/(e)(v_(2)-v_(1))`