3
$\begingroup$

I am currently studying Principles of Optics: Electromagnetic Theory of Propagation, Interference and Diffraction of Light, 7th edition, by Max Born and Emil Wolf. Chapter 1.1.2 Material equations says the following:

Metals are very good conductors, but there are other classes of good conducting materials such as ionic solutions in liquids and also in solids. In metals the conductivity decreases with increasing temperature. However, in other classes of materials, known as semiconductors (e.g. germanium), conductivity increases with temperature over a wide range.

An increasing temperature means that, on average, there is greater mobility of the atoms that constitute the metal. And since conductivity is due to the movement of electrons in the material, shouldn't this mean that conductivity increases as temperature increases?

$\endgroup$
1

4 Answers 4

6
$\begingroup$

In a metal, the number of charge carriers is almost unaffected by temperature. There is an enormous number of free electrons per unit volume (compared to a semiconductor like germanium), so Fermi Dirac statistics applies.

As the temperature increases, the scattering events of the electrons with the phonons increase because the number of phonon increases. These scattering events increase the resistivity of the material, hence the conductivity lowers.

Edit to answer the comment "Can you please clarify what you mean by "scattering events"?" : A scattering event in this context is an interaction between two quasiparticles, the electron (in a solid) and a phonon. They both carry energy and momentum, can interact and thus the electrons can have their momentum drastically changed and their energy somewhat changed, after an interaction with a phonon.

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. Can you please clarify what you mean by "scattering events"? $\endgroup$ Commented Jun 28, 2020 at 15:01
  • $\begingroup$ I've edited my answer @ThePointer. $\endgroup$ Commented Jun 28, 2020 at 18:51
2
$\begingroup$

The characteristic feature of metals is that the valence electrons of the atoms delocalise across the crystal lattice- this is an intrinsically quantum mechanical phenomenon. In essence, the electrons propagate as plane waves and this delocalisation lowers the energy of the electrons. As you increase the temperature, very few electrons are promoted to higher energy plane-wave states, the effect of the lattice vibrations also become quite important. As you increase the temperature of the crystals, you can excite quantised vibrational modes of the crystal lattice- these are called phonons. Electrons can scatter off phonons, resulting in decreased mobility. Since the phonons carry momentum (actually crystal momentum), collisions with electrons will alter the momentum of the electron. This effect reduces the average number of electrons with momentum directed in the direction of an applied electric field, thus decreasing the conductivity.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. I did recently learn about phonons and the idea of "effective mass" from my semiconductor physics textbook, and I think your answer did a good job in connecting this with what I read. $\endgroup$ Commented Jun 28, 2020 at 19:11
0
$\begingroup$

In metals, increase in temperature decreases average time between collision of charge carriers which increases the resistivity and therefore conductivity decrease.(also increase in temperature don't affect the no of charge carriers in metals).

$\endgroup$
0
$\begingroup$

The key point is that thermal motion disrupts the periodicity of the potential. As stated in ch. 26 of Ashcroft and Mermin: "Bloch electrons in a perfect periodic potential can sustain an electric current even in the absence of any driving electric field; i.e., their conductivity is infinite. The finite conductivity of metals is entirely due to deviations in the lattice of ions from perfect periodicity. The most important such deviation is that associated with the thermal vibrations of the ions about their equilibrium positions, for it is an intrinsic source of resistivity, present even in a perfect sample free from such crystal imperfections as impurities, defects, and boundaries."

$\endgroup$

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