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?

up vote 1 down vote accepted

Several sources give a higher (experimental) value of the Debye temperature for diamond - about 2220K: http://www.sbfisica.org.br/bjp/download/v03/v03a03.pdf , http://www.cvd-diamond.com/properties_en.htm , http://www.chm.bris.ac.uk/motm/diamond/diamprop.htm .

  • 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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