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)


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.

  • 2
    $\begingroup$ They have, over the course of moving through the filament, converted electric potential energy (power supply) to kinetic energy (acceleration in the field for a short bit) to thermal energy (scattering from lattice atoms). After scattering, they start accelerating in the field again, only to blunder in to another atom and lose energy. They do not enter with a kinetic energy higher than they leave with, but are at a higher potential. $\endgroup$
    – Jon Custer
    Oct 26, 2016 at 20:48

3 Answers 3

  1. The first cited source is not correct. The electrons do not enter the light bulb filament with relative high kinetic energies.

  2. 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.

  3. 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)!

  4. 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.


The first statement "The electrons enter the light bulb filament . . . " is not really correct.

At one end of the bulb the free electrons have a higher electric potential energy than at the other end.
As the free electrons travel though the bulb with a constant average speed they lose electric potential energy and that energy is converted into heat and light.

The mechanism for this conversion of energy is that free electrons find themselves in an electric field provided by the power supply.

The electric field accelerates a free electron, speeds up thus gaining kinetic energy and then the free electron collides with a lattice ion and transfers some of its kinetic energy to the lattice ion.
As a result the lattice ion vibrates more - its temperature has gone up.

The free electron is again accelerated by the electric field provided by the power supply, the free electron speeds up thus gaining kinetic energy and then the free electron collides with a lattice ion and transfers some of its kinetic energy to the lattice ion.

And so the process repeats itself with the free electron moving through the filament of the bulb with an average speed which is called the drift speed.

Imagine that you are sliding down a slope which has alternate stripes which have a friction free surface and then a surface which produces friction.
As you slide down you accelerated on the friction free surface losing gravitational potential energy and gaining kinetic energy. This is equivalent to the free electron accelerating between collisions. When on the surface which provides friction you lose some of your kinetic energy which then becomes heat. This is equivalent to the free electron colliding with a lattice ion.

So as they go round the circuit the free electrons have a drift speed which depends on a number of factors given by this equation.

$I = neAv_{\rm d}$ where $I$ is the current. $n$ is the number of charge carriers per unit volume, $e$ is the charge on a free electron, $A$ is the area of the conductor and $v_{\rm d}$ is the drift speed of the free electrons.

So in their passage around a circuit the free electrons are continually losing electric potential energy and at the same time gaining kinetic energy and then losing that gained kinetic energy when colliding with a lattice ion.
It is not the case that at one end of the bulb the free elections have a lot of kinetic energy which they lose whilst going through the bulb.

The only difference between the free electrons entering a bulb and leaving the bulb is that on leaving they have less electric potential energy, their drift speed is the same throughout the bulb.

  • 1
    $\begingroup$ Thank you for the explanations and analogies of how the potential energy is converted to heat and light and why the electron speeds up and - They add so much to my understanding of both the answer to the question and the overall concept. $\endgroup$
    – K-Feldspar
    Oct 26, 2016 at 23:19
  • $\begingroup$ So, can I conclude the following: 1. The current is the same on the three parts of the circuits (before the bulb, inside the bulb and after the bulb). 2. Individual electrons are indeed slower inside the bulb than otherwise (because they are constantly colliding with the lattice) and thus for (1) to hold we need more electrons per area inside the bulb. Is this correct? $\endgroup$
    – gota
    Feb 14, 2018 at 10:39
  • $\begingroup$ $I = nevA$ tells you all you need to know about about the electrons where $I$ is the current, $ n$ is the number of electrons per unit volume, $e$ is the charge on an electron, $v$ is the drift speed of the electrons and $A$ is the cross sectional area of the conductor. $\endgroup$
    – Farcher
    Feb 14, 2018 at 11:39

Several models have been proposed in previous answers for the transfer of electric potential energy to heat and light within the filament.

But I don't think that they addressed the most perceptive part of your question.   You asked, what is the actual difference between electrons coming into the bulb and going out of the bulb?   Any model for the energy transfer within the bulb must be consistent with the answer to that question.   And the answer is that they are spaced further apart when they leave the bulb, and are drifting a little faster.

With the greater spacing between them, they must drift faster in order to maintain the same flow rate around the circuit.  And they must be spaced further apart, because potential energy of any kind is always the energy associated with the position within a field.   Electrons forced to be closer together have had energy transferred to the field between them by force exerted through a distance (as is always the case for energy transfer).   When the electrons move apart, the energy is transferred out of the field, again by force exerted through a distance.   Voltage is often referred to as electrical pressure, and that is a very apt term.   Pressure is energy per unit volume.

In voltage sources, electrons are compressed at one end and rarefied at the other by various means.   When a circuit is completed, the electric field is not some kind of disembodied phantom that appears instantaneously, it is established as a wave of pressure which propagates through the circuit, compressing all of the electrons.

We know that the voltage drops along the length of a bulb filament or other uniform physical resistor in proportion to the length.   this implies that spacing and drift speed increase continuously through the length of the filament.   That's why I say that any model of the particle by particle mechanism for transfer of electric potential energy to thermal and light energy in a resistor must explain the spacing and drift speed increases

  • $\begingroup$ If the wire is electrically neutral, how can the electrons in one part be "spaced further apart" than in another? $\endgroup$
    – BowlOfRed
    Oct 27, 2016 at 17:37
  • $\begingroup$ We're not talking about a just a wire. We're talking about a length of wire which doesn't resist the flow of current significantly, then an bulb filament which does resist the flow of current, and then another wire which doesn't resist the flow of current significantly. $\endgroup$
    – D. Ennis
    Oct 27, 2016 at 17:44
  • $\begingroup$ I read your answer as the electrons in the wire at a higher voltage (leaving the bulb) are spaced further apart than the electrons in the lower voltage wire. Do you mean this is true for all the wire at that potential, or only some region near the filament/resistor? I didn't understand that part. $\endgroup$
    – BowlOfRed
    Oct 27, 2016 at 17:48
  • $\begingroup$ No the electrons enter the bulb at a higher voltage than they leave it with. I mean that the whole wire before the bulb has closer spacing of electrons than the whole wire after it. $\endgroup$
    – D. Ennis
    Oct 27, 2016 at 17:56
  • $\begingroup$ But if the whole wire has a different spacing of electrons, doesn't that affect the total charge? The bulk spacing can't be different than the bulk spacing of protons (which are fixed) and still be electrically neutral. $\endgroup$
    – BowlOfRed
    Oct 27, 2016 at 18:00

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