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?