# Debye Temperature for Copper

I am trying to calculate the Debye temperature, $$\theta_D$$, of copper using the following:

$$\theta_D = \frac{\hbar v_s}{k_B} \left( \frac{6\pi^2N}{V} \right)^{1/3}$$

I have the following values: $$\rho = 8900$$ kgm$$^{-3}$$, $$v_s = 3800$$ms$$^{-1}$$, atomic mass $$M_a=63.5$$gmol$$^{-1}$$. Now, the speed of sound, $$v_s$$ is not correct for copper according to online tables, and it seems to be closer to $$4600$$ms$$^{-1}$$.

However, I also know that the Debye temp for copper is about $$343$$K.

Using the fact that,

$${N\over V} = \frac{N_A\rho}{M_a} = 8.44\times10^{27}$$

where $$N_A$$ is Avogadro's number, I get,

\begin{align} \theta_D &= \frac{(1.055\times10^{-34})(4600)}{(1.381\times10^{-23})}\cdot \left( 6\pi^2\cdot8.44\times10^{27} \right)^{1/3} \\\\ &=279K \end{align}

Which just isn't right. And using the value for $$v_s$$ provided in the question gives an even lower answer of $$230$$K...which isn't right either.

Am I missing something here?

• The number density should of the order 10^28.
– nasu
Commented Oct 6, 2018 at 11:43
• @nasu doing that gives an answer of around 600K, which is even worse! Commented Oct 6, 2018 at 20:38

There are few things going on here. The first is that you seem to be mixing units for density and the molar mass, using kg in one case, and g in the other. If you fix that, you will correctly get a number density on the order of $$10^{28}$$. However, you still won't find good agreement with the $$~345K$$ value you expect. Why is this?

Well, there's a second and subtler thing going on, which is that you are using a single speed of sound. In reality, the speed of sound is different in the (one) longitudinal and (two) transverse directions. If you instead use a mean speed calculated through $$\bar{v}_s = 3^\frac{1}{3}\left( \frac{1}{v^3_{\mathrm{transverse}}}+\frac{2}{v^3_{\mathrm{longitudinal}}}\right)^{-\frac{1}{3}}$$ you'll get a lot closer to the experimental value. You'll note that this isn't an ordinary average velocity, it's simply a constant defined in the derivation of the Debye temperature.

The third thing worth mentioning is that the speed of sound is going to depend on how your sample of copper was made. The value $$v_s=3800$$m/s is the longitudinal sound speed in thin copper rods, whereas $$v_s=4600$$m/s is a (rather low) value for the bulk material.

• Nice, clear, thorough answer! In need of some solid state "oversight": Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?
– uhoh
Commented Oct 7, 2018 at 2:37
• Much appreciated! Commented Oct 7, 2018 at 5:29
• @uhoh Thank you! I had a look at your question, but don't know the answer offhand. Commented Oct 7, 2018 at 17:58

Great post. And thanks Anyon for the clarifications. I will post the complete clear solution for that problem with links and numbers and while doing that I will correct a small mistake that Anyon did.

First, concerning the density and molar mass of copper, I have $$\rho = 8960 \text{ kg/m}^{3}$$ and $$M_a = 63.546 \text{ u}$$ which I took from wikipedia https://en.m.wikipedia.org/wiki/Copper

This gives me a density of \begin{align} {N \over V} & = \frac{1000 \cdot N_A \cdot \rho}{M_a} \\ & = \frac{1000\cdot6.023\times10^{28}\cdot8960}{63.546} \\ & = 8.49\times10^{28} \text{ atom per m}^3 \end{align}

Now, as Anyon said, we have 2 transverse and 1 longitudinal sound waves, which have the speeds $$v_{\mathrm{transverse}} = 2325 \text{ m/s}$$ and $$v_{\mathrm{longitudinal}} = 4760 \text{ m/s}$$ which I took from https://www.engineeringtoolbox.com/amp/sound-speed-solids-d_713.html

Now, when applying the formula for $$\theta_D$$, one should take an "average" speed of sound. Here is where Anyon made a small mistake. Since we have 2 transverse and 1 longitudinal the average speed is obtained by the formula

$$\bar{v}_s = 3^\frac{1}{3}\left( \frac{2}{v^3_{\mathrm{transverse}}}+\frac{1}{v^3_{\mathrm{longitudinal}}}\right)^{-\frac{1}{3}}$$

Plugging in the numbers, we get

\begin{align} \bar{v}_s & = 3^\frac{1}{3}\left( \frac{2}{2325^3}+\frac{1}{4760^3}\right)^{-\frac{1}{3}} \\ & = 2611.69 \text{ m/s} \end{align} Now using the formula $$\theta_D = \frac{\hbar v_s}{k_B} \left( \frac{6\pi^2N}{V} \right)^{1/3}$$ And plugging in the numbers we have, we get \begin{align} \theta_D & = \frac{1.055\times10^{-34}\cdot2611.69}{1.38\times10^{-23}}\left(6\pi^2\cdot8.49\times10^{28}\right)^{1/3} \\ & = 342 \text{ K} \end{align}

Which is within the expected value of 343 K.