How does the Einstein model explain the low heat capacity of diamond?

[latentcorpse]

http://www.ph.ed.ac.uk/~pmonthou/Statistical-Mechanics/documents/SM9.pdf

In the third last paragraph of p36, we are told that the Einstein model discussed in this lecture succesfully explains the low heat capacity of diamond - i can't however see how it does, or at least find an explanation anywhere in that lecture.

I follw the argument that at low T, x is large and so the thermal energy is small in comparison to the energy difference between the ground state and the first excited state but how does that help explain the low heat capacity of diamond - surely all we can gather from this is that it's going to be statistically more likely to find particles in the diamond lattice in the ground state?

 

[Mapes]

For diamond, \kappa is relatively large (p34), so \omega is large, so T^* is large, so T/T^* is small, so c_v is small.

 

[latentcorpse]

what's the relationship between \omega and T*?

and how did u conclude that C_v was going to be small as a result of that ratio being small?

 

[Mapes]

From the bottom of p35, T^*\propto \omega; from the graph on the bottom right of p36, c_v decreases with increasing T^*.

 

[miketem]

You have to look at the properties of diamond. It's SP3 hybrid bond, thus it make a non-polar molecule. We can see that it has no free electron moving inside the lattice and by calculate the total energy inside the 3 dimensional crystal with it's own Madelug constant, we found it's a good heat conductance and good isolator (bandgap 5.5 at 300K). No free electron no electron collision, tightly bond so less translation, the contribute more to vibrational.