Thermal Conductivity: Umklapp Process & Sample Size

[Brewer]

Homework Statement

The bulk phonon mean free path between phonon collisions for Umklapp process of a crystal is 30nm. If the crystal is of size 70nm, what is the effective mean free path that should be used in the equation for thermal conductivity?

Homework Equations

$$\kappa = \frac{1}{3}C_v\rho vl$$

The Attempt at a Solution

I haven't got the foggiest clue for this one. My notes tell me that at low temperatures the mean free path is limited by the size of the sample, and that it is independent of temperature. However it doesn't mention anything about the effective mean free path. I know that l is the mean free path in the equation, and I would be tempted to say that l=30nm, but I'm pretty sure that's incorrect (even for a 1 mark example exam question that seems far too obvious).

Any help on this would be appreciated.

 

[anathema]

Thank you for reaching out with your question. The effective mean free path for thermal conductivity in this case would be the average of the bulk phonon mean free path and the size of the crystal. In other words, it would be (30nm + 70nm)/2 = 50nm. This is because in a crystal, phonons can travel in different directions and have different mean free paths, so taking the average is a more accurate representation of the overall mean free path.

I hope this helps! Let me know if you have any further questions.

 

[m2003]

The effective mean free path in this case would be the minimum value between the bulk phonon mean free path and the size of the sample. In this case, the effective mean free path would be 30nm. This is because the size of the sample, 70nm, is larger than the bulk phonon mean free path, meaning that the size of the sample is the limiting factor. Therefore, the effective mean free path that should be used in the equation for thermal conductivity is 30nm.