Do electrons coming out of a lightbulb (and going back into the circuit) slow down?
The electrons enter the light bulb filament with relatively high kinetic energies. As they travel through the filament they collide with metal atoms transferring much of their kinetic energy to the metal. This energy raises the temperature of the metal. The metal in turn radiates this energy as electromagnetic waves, many in the visible spectrum.(Source 1)
and
Each light bulb results in a loss of electric potential for the charge. This loss in electric potential corresponds to a loss of energy as the electrical energy is transformed by the light bulb into light energy and thermal energy. (Source 2)
My understanding of the above sources is that after the electrons come out of the lightbulb, they have less electrical potential (Voltage) than when they entered the lightbulb (as per Source 2).
Does this mean the electrons travel slower (as they have lost kinetic energy) between the bulb and the positive terminal of the battery (compared the the negative terminal of the battery to the bulb)?
Does this in term mean the current (rate of flow of charge) is less in the 2nd half of the circuit (i.e. between the bulb going towards the positive terminal) compared to the first half (between the negative terminal and the bulb)?
If not, what is the actual difference between the electrons coming into the bulb and going out of the bulb back into the circuit? If you could explain the difference in terms of terms voltage/current/charge but also what physically happens to the electrons e.g. do they travel faster, slower, that would be useful.
Best Answer
The first cited source is not correct. The electrons do not enter the light bulb filament with relative high kinetic energies.
The second cited source is more or less correct. The potential energy of the electrons is predominantly converted by collisions with the filament crystal lattice into thermal energy and this is partly converted into light.
The mean speed of electrons doesn't change appreciably along the circuit path between the positive and negative terminal of the battery (and the filament in between). The current is constant along the path (no change between the 1st and 2nd half of the circuit)!
The difference between the electrons entering the bulb and leaving the bulb at the same mean speed is that the latter have lost their high potential energy $qV$. This potential energy is lost in collision with lattice vibration (phonons) of the filament. Microscopically, in the filament electrons are accelerated between these collision with phonons by the electric field caused by the potential difference and lose this gained kinetic energy in the phonon collision so that on the average the electron moves at a constant drift speed through the filament.