# Photoelectric effect stopping potential

$q_eV_s = hf - \phi$

My question is... suppose we are testing the photoelectric effect. One plate is illuminated. We have applied the stopping potential. Suppose an electron leaves one plate with $KE = hf - \phi$ Now by the time it reaches the other plate, it should have 0 kinetic energy right?

My question is about current... current should be 0 in this case, but charge has traveled from one plate to the other... so I'm confused as to how current can be 0.

Also, I'm having trouble thinking about the electron barely reaching the other side... vs being pulled back to the first plate before it can... are these situations somehow equivalent current wise?

Appreciate any clarification.

• For a voltage higher than the stopping plate, do you understand what the electron does? Now lower the voltage a bit - what happens now? What happens when the voltage is such that the electron comes to a halt (KE=0) 1 angstom above the surface. Is there current flow now? – Jon Custer Oct 15 '14 at 23:12
• Stopping potential is negative. So a voltage lower than stopping potential (more negative), the electron will come to a stop before reaching the other side, then come back to the initial plate. So current flow is zero. – Ameet Sharma Oct 15 '14 at 23:17
• So is there a discontinuity in current right at the stopping potential? More negative than the stopping potential gives 0 current. But does current suddenly jump if it's right at the stopping potential... – Ameet Sharma Oct 15 '14 at 23:31

• the Kinetic energy actually gets to Zero at the next plate. it means the $V=0$ therefor the electron can not enter the circuit. so even the Highest energy photo electron couldn't reach the other plate to enter the circuit and create a current therefor $I=0$ and that voltage is called stoppage Voltage.