# Schematic design of the apparatus (photoelectric effect)

I think I understand the basic idea of photoelectric effect but there are two things in the schematic diagram of the apparatus for the investigation of the photoelectric effect which I do not understand.

Firstly, the emitting electrode is connected to a positive terminal of a battery. This implies there is an electron deficiency there, so where are the 'loose' electrons are coming from? I understand there might be some to start with but with the intensive light shone on the emitter shouldn't the emitter run out of those loosely bound electrons quite soon?

Second problem is what happens at the collecting electrode. It is connected to the negative terminal of the battery to create the retarding potential difference and I understand as the photoelectrons travel across the gap they experience a repulsive force. Now the book says that those with NOT enough kinetic energy will be stopped and no current will be registered. I am however interested in details as to what happens to those who do make it through. I suppose they would not make it all the way up to a galvanometer, so how exactly is the current produced? Do the photoelectrons 'push' the free electrons in the metal by repulsive force, just like it happens in a regular wire which is connected between the terminals of the battery? What eventually happens to those free electrons which are located near the negative terminal - do they get pushed 'inside' the battery - but this looks as if the battery is connected with the wrong polarity? Also I do not understand why the photoelectrons could push the free electrons in the metal - there are many more free electrons and their combined repulsive force should be much greater, shouldn't it?

This I do not understand.

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The simplest answer is that the diagram is wired backwards, electrons will flow to the positive terminal. One could argue that the wired as it is, that the intent is to provide a potential barrier to the electron flow so only those with an excess of energy will get to the other terminal and pump energy into the battery. However, the excess energy of the photon typically goes into heating the substrate and does not go into the kinetic energy of the electron.

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This is simply incorrect. The bias in the diagram is exactly the one you want and need to measure the maximum kinetic energy of the liberated electrons. Further, it is a good thing that the number of conduction electrons on the anode is reduced because you want to interact only with bound (as opposed to free) electrons; otherwise you don't have a fixed potential barrier to overcome which allows you to detect the quantum nature of the energy transfer. –  dmckee Mar 9 '13 at 22:32
$$Q = C U = n q,$$
where $C$ is the capacitance of plates' arrangement (a constant), $U$ is the voltage of the battery (a constant), $n$ is an excess or deficiency number of electrons and $q$ is the charge of one electron (a constant). So in order the situation is in equilibrium you need the excess of $n$ electrons on collector plate and the deficiency of $n$ electrons on emitter plate.