Why is there a stopping potential

Stop potential of the photoelectric effect

qeV.s = hf-ϕ

My question is ... suppose we are testing the photoelectric effect. A plate is illuminated. We applied the stop potential. Suppose an electron leaves one plate with K.E. = hf-ϕ Now when 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 the charge has migrated from one plate to another ... so I'm confused how current can be 0.

Also, I have trouble thinking of the electron barely reaching the other side ... instead of being pulled back to the first plate before it can ... are these situations any equivalent in terms of current?

Appreciate any clarification.

Jon Custer

At a voltage higher than the stop plate, do you understand what the electron is doing? Now lower the tension a little - what happens now? What happens if the voltage is such that the electron 1 angstome comes to a standstill above the surface (KE = 0)? Is there current flow now?

Ameet Sharma

The stop potential is negative. At a voltage lower than the stop potential (more negative), the electron will stop before reaching the other side and then return to the starting plate. The current flow is therefore zero.

Ameet Sharma

So is there a discontinuity in the current right at the stop potential? More negative than the stop potential gives 0 current. But does the current suddenly jump when it is exactly at the stop potential ...


You're right - the electron could make it to the other plate. But then it has zero velocity and it still feels the electric field. So it will "fall back" to the disk it came from - and the net current is zero.

In reality, the speed distribution of the electrons is continuous (they don't all travel straight to the other plate at the maximum possible speed), so you will see a drop in current as you increase the deceleration potential.


Some notes:

  • These kinetic energy that you are using is the maximum Amount of kinetic energy among the electrons. it means that maybe few of them have that amount of energy, other photoelectrons with lower kinetic energies (hence lower speed) simply accelerate back to the plate from which they were shot.

  • The kinetic energy actually reaches the next plate zero . it means that V. = 0 therefore the electron cannot enter the circuit. Even the photoelectron with the highest energy could not reach the other plate to enter the circuit and generate a current for it i = 0 and this voltage is called the stop voltage.