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.
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.
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.
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