Thermal properties of materials
The lattice thermal conductivity of an insulating single crystal decreases roughly in proportion to 1/T as temperature rises, at temperatures comparable to or above the Debye temperature. What is the main reason?
- AAbove the Debye temperature the heat capacity falls inversely with temperature, so phonons carry less heat
- BThe number of phonons grows, and with it the phonon–phonon scattering that does not conserve total momentum (umklapp processes)
- CThermal expansion of the lattice increases phonon scattering at crystal surfaces and grain boundaries, shortening the mean free path
Show answer and explanation
Answer: B. The number of phonons grows, and with it the phonon–phonon scattering that does not conserve total momentum (umklapp processes)
Among phonon collisions, normal processes, which conserve the total momentum, hardly impede the flow of heat, whereas umklapp processes, in which the total momentum changes by a reciprocal lattice vector, act against the heat flow and are the main source of thermal resistance. At high temperatures the number of phonons grows in proportion to temperature, so the mean free path shortens roughly as 1/T, and since the heat capacity is nearly constant, κ ∝ 1/T. Above the Debye temperature the heat capacity stays near 3R per mole of atoms; a sharp drop happens on the low-temperature side instead. Scattering at surfaces and grain boundaries hardly depends on temperature and dominates at very low temperatures, where the phonon mean free path is long; there κ falls with decreasing temperature, as the heat capacity does.