2
$\begingroup$

In an electric circuit, how does the excitation of the free electrons to higher energy levels translate into net forward motion of the electrons to the positive terminal? My concept of electrons being excited by the electric field is that they have a higher orbital but they are still localized around the nucleus of a few atoms (especially in a more resistive element). But I know there must be a small net forward motion for there to be current.

I need to marry the classical model of the colliding electrons and atoms of the medium with the Quantum principles involving the excitation of charges to higher energy bands and emission of photons.

Also, a second question is what causes the electric field in the conductor to be practically zero (compared to that in the resistor) in a non - electrostatic model (i.e., when there is a current).

Thank you so much for your help. I have scoured the internet for insight but I have not found a site that marries the classical to modern theory clearly for me.

$\endgroup$
2
  • 1
    $\begingroup$ The question Are free electrons in a metal really free is related and might make interesting reading. $\endgroup$ Feb 2 '15 at 17:04
  • $\begingroup$ I agree with Steven that the E field is not zero unless we are in an ideal conductor.. My disagreement is with his statement that the field around the circuit is the same (resistors and conductors). This cannot be for components in series if we understand that the potential difference is the integral of the E field with respect to distance as seen by the charges. Also, i do not understand the principle of the conservation of density. Pls elaborate. Thank you . $\endgroup$
    – CuriousJo
    Feb 3 '15 at 16:04
1
$\begingroup$

There are bound electrons and there are free electrons - the former interact with radiation and electric fields and the like by bouncing around energy levels and emitting photons (dissipating energy, as in a resistive element), the latter do actually move around and are strictly not localized around the nucleus of a few atoms, and largely preserve energy through a wire, for example.

There is no inconsistency between the quantum model of bound electrons and the idea of a free electron - the latter just has its own quantum model (:

The electric field is definitely 0 in a conductor. Not sure why Steeven seems to think that current would stop flowing without the presence of a net field. It doesn't take any energy to move electrons through a voltage drop of 0 as is the case in a wire. They just keep moving along to prevent charge building up. You could think of it really loosely as the maintenance of flow/average velocity through some horizontal segment of plumbing. It's not due to any net gravitational force (electric field) the water is (electrons are) subject to in that area, it's just maintaining the conservation of density (charge) at the ends, where other things are going on.

$\endgroup$
0
$\begingroup$

I need to marry the classical model of the colliding electrons and atoms of the medium with the Quantum principles involving the excitation of charges to higher energy bands and emission of photons.

My answer to this question explains something about the concept of the conduction band.

Think of it this way. In all the filled shells (including shells that will be filled by to atoms joining into a lattice or a molecule e.g.), all slots are taken. There are no more spots for electrons to move to - no vacancies. Therefor, no electrons can move (because of the exclusion principle, there can never be to electrons at the same exact state). These are called the valence bands.

The unfilled shells make up the energy states in the conduction band. It is called the conduction band because it is not full, so electrons can move from one state to an empty one. This gives freedom of movement within the band and throughout the same energy state stretching over many atoms in the lattice. The Wave function of the free electrons spreads over many atoms - they are not "bound" to a specific atom anymore. This is a conductor.

For a very little force (that is, a very little electric field) the electrons can be pushed from one energy state to another within the band. When this electric field is applied the charges will move (a current will flow).

Especially in semiconductors, more electrons can be moved up to the conduction band be being excited from addition of energy (light, heat etc.). Here conductivity can rise from higher temperature (to some extend). In usual metallic conductors, their conductivity will usually decrease with temperature, since more collisions will take place and less freedom (essentially less energy states to move to because of many already taken slots) is the result.

Also, a second question is what causes the electric field in the conductor to be practically zero (compared to that in the resistor) in a non - electrostatic model (i.e., when there is a current).

The electric field is not zero in a conductor. If it was, current would stop flowing, unless it was a superconductor. The electric field in a circuit is caused by e.g. a battery. This provides a potential difference from one end of circuit to another (over all the circuit components). A potential difference is essentially a willingness of electrons to move from higher to less potential energy. This is the case since charges set up an electric field throughout that will drag them along.

In a resistor this electric field (seen as a potential difference over it) is still present. It is just harder for the charges to move through the material.

$\endgroup$
1
  • $\begingroup$ I have doubts regarding your statement that the electric field is the same throughout the circuit through all components. The integral of the electric field over length of the medium gives the potential difference between two points and we assume practically that the potential difference between two points of a conductor is zero. Again this is compared to the difference that would be seen across a resistor for the same current. $\endgroup$
    – CuriousJo
    Feb 2 '15 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.