Is there any simple way to understand why alloy have more resistance than metals?

My teacher ask this, I answer that, there might be more free electrons in metals than an alloy, but she said you are not accurate, So what is better answer to this question?


3 Answers 3


I tend to agree with Sanya in that I am not sure about the universality of this. There might of course be instances where this is the case.

A pure metal has a periodic lattice of ions. There is then a conduction band of electrons that fills the space between the ions. These electrons have wave vectors in the reciprocal space. In space the occurrence of ions at locations according to elementary translation vectors $\vec t_i$ in the $x,~y,z$ with $$ \vec T~=~\sum_{i=1}^3n_i\vec t_i $$ the reciprocal basis is given by $$ \vec t_i~=~2\pi\vec n_i\frac{\epsilon_{ijk}t_jt_k}{\epsilon_{ijk}t_it_jt_k}. $$ so the reciprocal lattice vector is $$ \vec G~=~\sum h_i\vec t_i, $$ where $\vec G_i~=~h_i\vec t_i$ The volume associated with this vector is a Brillouin zone. The volume of the Brillouin zone is $$ Vol~=~\epsilon_{ijk}G_iG_jG_k. $$

An electron in the lattice with wave vector $\vec k$ passes between Brillouin zones with $\vec k'~=~\vec k~+~\vec G$. We then have for a near perfect lattice lots of electrons passing through Brillouin zones accordingly and they then constructively interfere as waves. We can now see that if we have an imperfect lattice there may now be more destructive interference of waves.

This may not be universally the case. For a semiconductor crystalline lattice the introduction of impurities increases its conductivity. This is because this doping permits more movement of electrons from the valence bands associated with the ions to the conduction band. So for semiconductors exactly the opposite can happen; the introduction of impurities with various properties can increase conductivity.

It may be the case that very pure metals have in most cases more conductivity. Of course pure copper and iron are very ductile and weak. As a result we introduce impurities, carbon in iron or zinc in copper (brass). There is also the question of what happens with lattice dislocations. If you take a piece of copper tubing you find it is very flexible. However, if you bend it back and forth it becomes stiff. This is because lots of lattice imperfections result which block further motion. Similarly with iron dislocations, usually due to impurities make it harder. How these play a role I am not certain of.

  • $\begingroup$ I predict that this answer will get the most upvotes and picked as answer $\endgroup$
    – user121963
    Jun 27, 2016 at 14:27

First of all, I want to see an (experimental) proof that any metal has a higher resistance than any alloy (at any pressure, temperature and volume).
What I presume your teacher might have wanted to hear is something along the following lines: a perfect, perfectly static crystal would be, if I remember correctly, perfectly transparent to an electron, so there would be no electric resistance. The reasons for resistance can be found in lattice vibrations (e.g. thermal) and impurities, defects etc. These latter could be expected to be more numerous in alloys.
I do however, as stated above, not believe in the initial claim.

To save what has been added in the comment section by Jon Custer: "Of course, for the myriad of ordered alloys things might be different." This supports the assumption that the question in its generality is probably not even asking about a true "fact".

  • $\begingroup$ I would like to add to this answer. In a more precise way, one can say that the energy difference between the valence and conduction band of electrons increases due to the impurities added which makes it n alloy. The reason for this increase in energy difference might be that the energy of a valence electron in the atom of the metal is different than that of the impurity metal atom. $\endgroup$
    – lattitude
    Jun 27, 2016 at 12:42
  • 1
    $\begingroup$ @lattitude - That makes no particular sense in a metal - the conduction occurs in a partially filled band, so there are no valence electrons to be concerned about in the process. But, Sanya is generally right for random alloys. Of course, for the myriad of ordered alloys things might be different. $\endgroup$
    – Jon Custer
    Jun 27, 2016 at 14:13
  • $\begingroup$ @JonCuster as stated above, this is only a hypothesis of what kind of generalised answer might be expected. Personally, I think a question like this should not be asked because it is ill defined. $\endgroup$
    – Sanya
    Jun 27, 2016 at 14:15
  • $\begingroup$ @Sanya - Oh, I fully agree with you. The simple answer is additional scattering, and that is the most general case. Things like ordered alloys may be different. $\endgroup$
    – Jon Custer
    Jun 27, 2016 at 14:16

The short answer would be the empirical Matthiessen's rule: the total resistivity of a crystalline metallic specimen is the sum of the resistivity due to thermal agitation of the metal ions of the lattice and the resistivity due to the presence of imperfections in the crystal (scattering).

There are of course deviations from that rule:

  • it assumes that scattering by imperfections and by phonons are independent of each other, which is only approximately true.
  • it ignores changes to the band structure of the metal and to the phonon spectrum and distribution. These are expected in concentrated alloys.
  • parallel conduction by two groups of electrons is not taken into account. In alloys of monovalent elements this two-band effect can already be significant at low concentrations.

See Electrical resistivity of ten selected binary alloy systems (Ho et al.) for a more detailed overview.


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