# The Debye temperature for diamond

To simplify the calculation, let's assume that the average speed of sound in the diamond is simply $v_s=\sqrt{E/\rho}\simeq1.414\times10^4 \ \text{m/s}$, and the Debye frequency

$$\omega_D=v_s\left( \frac{6 \pi^2 N}{V} \right)^{1/3}\simeq3.087\times10^{14} \ \text{Hz}$$

Here I used the density of diamond to calculate $N/V$:

$$\frac{N}{V} = \frac{N_A \ \rho(\text{Diamond})}{M_{atom}(\text{C})} \simeq\frac{(6.02\times10^{23}\text{/mol})\times(3.5\ \text{g/cm^3})}{12\ \text{g/mol}} =1.756\times10^{29}\ \text{m^{-3}}$$

So the calculated approximate Debye temperature for diamond is

$$\Theta_D=\frac{\hbar \omega_D} {k_B}\simeq2357.7\ \text{K}$$

Which is almost exactly $2^{1/3}$ times the experimental Debye temperature for diamond, which is $1860 \ \text{K}$. So if we just count half of the atoms in the diamond crytsal, the calculation is almost precise. Is this a coincidence, or due to some characteristics of the diamond structure?

• Yeah I'm definitely over-thinking this. My estimate of $v_s$ has a error of ~20%, so anything "precise" would be "coincidental". – arax Oct 29 '14 at 15:52