What is the atom density of aluminium at 600K?

[alfredbester]

Hi, I appear to have lost some mass/density.

Q: Calculate the atom density, n, of aluminium at 600k.

Data given (some of it not relevant to this part of the ? but i'll post it all).

The debye temperature $$T_D = 428 K$$ . The velocity of sound at room temp, $$v = 5100m s^{-1}$$. The interatomic spacing a = 405pm, and its relative atomic mass is 27.

I found the atom density n to be:

$$n = (KT_D / \hbar v )^3.(1 / 6\pi^2)$$

I used $${\omega_m}^3 = 6\pi^2 v^3 n$$, and the fact $${\omega_m} = K T_D / \hbar$$ to get my equations for n.
I assumed the atom density was the same at 600k as at room temperature (the way the question was worded I couldn't see any other method). Plugging the numbers in I found $$n = 2.24x10^{28} m^{-3}$$.

Then I'm asked to compare the density found with aluminiums true density of $$2700 kg m^{-3}$$ and explain any difference.
My density is just the atom density multiplied by the atomic mass (assuming the mass is just contained within the aluminium).
Therefore
$$/rho = n m(amu) = n = 2.24x10^{28} * (27 / 6.022x10^{26})) = 1000 kg m^{-3}$$.

There in lies my problem. I'd expect my approximations to overestimate the density if anything.

 

[alfredbester]

Is there anyway my density could be right? Maybe some assumption of the debye model I'm not aware of. I've got the formulas straight out the textbook I don't see how it could be wrong and it's the right order of magnitude.

 

[alfredbester]

Just had a thought is the atom density n, the number of molecules vibrating at $${\omega_m}$$ / volume. Guess that would explain why the density is significantly lower, since many atoms won't be vibrating at $${\omega_m}$$.