# How is photo-current related to stopping voltage in the Photoelectric Effect?

Lately, I have been revising the Photoelectric Effect as part of my school curriculum, but there is a disagreement or at least I am confused by the following two statements I have encountered. I saw a question similar to mine by sausage, but it treated the problem based on a simulation, and the answer given was more related to the simulation rather than the concept itself.

"Greater intensity at a particular frequency means a greater number of photons per second absorbed, and thus a greater number of electrons emitted per second and a greater photo-current"

" The intensity doesn’t appear in equation eV0 = hf - Ф, so V0 is independent of intensity. "

I can understand both statements separately, but I cannot see how they hold at the same time. Suppose we have two electrodes placed in an evacuated glass tube connected by a battery. Then the cathode is illuminated, and electrons will be emitted and photo-current will be generated. According to the first statement for a greater intensity of light the magnitude of the photo-current will increase. So, wouldn't we need to generate a larger electric field with opposite direction(cathode to anode) to stop the flow of electrons and reduce the photo-current to zero? Does this mean that also the stopping voltage would increase, by disagreeing with the second statement?

• Won't a given voltage stop a given photoelectron from a given color of light, independent of the number of accompanying similar photoelectrons? – DJohnM Nov 25 '17 at 22:55

The magnitude of the required electric field is set by the energy of a given electron released from the metal. Specifically, it's the electric field such that the potential energy difference between the two plates is greater than the kinetic energy of any individual electron.

Turning up the intensity increases the number of electrons, but does not increase any individual electron's energy. Therefore, it does not affect the required electric field.

Stopping the electron flow only has to do with their velocity not their number. You can see it in the relation "eU=E" where e is the elementary charge, U the tension of the current and E the kinetic energy of the electron (don't forget to quantize your variables to make your formulas understandable).

And we now that the kinetic energy of an electron is proportional to the frequency of the light hitting the cathode. Thus the voltage you need to apply to stop the electrons is proportional to the frequency of the light and not it's intensity.

I hope my answer helped, don't hesitate if you have any question :)

• "Thus the current you need to apply to stop the electrons is proportional to the frequency of the light and not it's intensity"...This seems to contradict the rest of your answer , and reality... – DJohnM Nov 25 '17 at 23:35
• Could you develop please? I'd like to know where I am wrong thank you. – Gornemant Nov 26 '17 at 8:43
• It's the voltage, not the current... – DJohnM Nov 26 '17 at 20:05