When an electron is said to be in conduction band is it completely detached from the atom to act as free electron and conduct electricity as per free electron sea model?Do a jump from valence to conduction actually imply ionisation of atoms?Please clarify me about conduction band.

  • 2
    $\begingroup$ From a solid state physics perspective (including semiconductor physics), both the valence band(s) and the conduction band(s) are bands of the crystal structure - the wave functions extend through the crystal and are not associated with any single atom, only the crystal. This applies to semiconductors, insulators, and metals equally. $\endgroup$
    – Jon Custer
    Jan 31, 2020 at 21:17
  • $\begingroup$ Why don’t you add answers? I always see you writing comments. $\endgroup$
    – boyfarrell
    Jan 31, 2020 at 22:30
  • $\begingroup$ It is not a free electron because it is still confined to the material. It’s a free electron in the sense that it can move around within the material (Crystal). So it is not ionised. If it were ionised then the material would have net positive charge. $\endgroup$ Feb 1, 2020 at 4:57
  • $\begingroup$ @user3518839 What do you mean by positive charge on material?How can there be a positive charge on the material as whole when there is no loss of electrons involved. $\endgroup$
    – Sharad1
    Feb 1, 2020 at 15:57

3 Answers 3


In crystals the bands are formed from the atomic orbitals mixing with each other so as to make (by level splitting) new energy states that, instead of being localized around atoms, are spread over the whole crystal. In individual atoms there's an infinite number of energy levels between which an electron can transition when its atom is excited.

Ionization of atoms involves getting at least one electron of an atom to an energy corresponding to unbounded motion. The smallest energy level which an electron must get into to make the atom ionized is known is the vacuum level. If an electron is excited from e.g. $2s$ to $4p$ level, it still remains bound, so it's not ionization.

When you bring many atoms closer together, so that their discrete levels start splitting, one of the former energy levels splits into the valence band, and another into the conduction band. And there are still lots of (formerly discrete) levels above these bands, each of which is also split into some band. The energy state corresponding to the vacuum level also splits into a band, top of which is now the vacuum level of the crystal.

Now, excitation from the valence band into conduction band is analogous to excitation from one bound state to another, like the above mentioned example of $2s\to 4p$. And ionization is getting an electron to the vacuum level of the crystal. This means that such an electron can leave the bulk of the crystal and never return back, unlike the electrons in conduction or valence bands.


The electrons at the bands are not free. What happens is that their wave functions can't be adequately described considering only the Coulomb potential of one nucleus. But they can't also be adequately described as a free particle. In a way their wave function combine both situations. According to Bloch's theorem:

\begin{equation} \psi(\boldsymbol r) = u(\boldsymbol r)e^{-i\boldsymbol{k.r}} \end{equation}

The first part is a periodic function with the periodicity of the lattice. The electrons are more probably found around each nucleus due to it. The exponential represents a free particle.

So the atoms are not ionized because the electrons are there around.

Instead of thinking of atomic orbitals as 5s for example for each atom, think of all electrons from such orbitals belonging to a crystal orbital, called band.


First of all you have to take into account that the quantum description of electrons in a solid is at a great extent quite far from everyday life physical intuition. Moreover, the one-particle description of electronic states is just an approximation of the many-body electronic wavefunction which puts all the electrons on the same foot (accordingly to their indistinguishability).

With these warnings, the Bloch's description of the one-electron states corresponds to delocalized wavefunctions for all the states (even for core states). However, one has to take into account that Bloch's description of the energy bands is not unique. An alternative description in term of the so called Wannier's orbitals is available and allows a more meaningful description of the localized character of electronic states.

Even without goingg to Wannier's picture, more physical insight can be gained by observing that only partly filled bands contribute to electric conduction. Therefore in a metal only the conduction band allows a non zero electronic current. In this sense electrons in conduction band can be considered free to move as soon an external field is applied. Notice that this does not imply that their wavefunction is the free particle wavefunction. Presence of nuclei and of other electrons in lower lying filled bands does contribute to modify the wavefunctions from the simple plane wave form. So, success of free electron model to describe electronic properties of a few simple metals (usually, s- or p- bonded metals) does not imply that in these cases electrons really behave like plane waves. For a complete agreement with experiments it is necessary to take into account the exact form of the electronic orbitals.

About considering a jump from valence to conduction band as an ionization process, it is attempting to use the analogy and sometimes it is possible to find descriptions along this line, although, strictly speaking, it is plainly wrong: it is as if in molecular physics one would speak about ionization to describe any electronic transition to excited states.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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