Number of phonons - Which length should I use?

[unscientific]

Homework Statement

(a) Find debye frequency.
(b) Find number of atoms.

Homework Equations

The Attempt at a Solution

Part(a)
[/B]
Density of states is given by
$$g(\omega) = \frac{3V\omega^2}{2 \pi^2 c^3} = N \left[ \frac{12 \pi \omega^2}{(2\pi)^2 n c^3} \right] = 9N \frac{\omega}{\omega_D^3}$$
Debye frequency is given by
$$\omega_D^3 = 6 \pi^2 n c^3$$

Part(b)
The number of atoms $N$ is related to occupation number $n(\vec k)$ by
$$N = \sum\limits_{k} n(\vec k) = \int \frac{g(\omega)}{e^{\beta \hbar \omega} - 1} d\omega$$
$$N = \frac{3V}{2 \pi^2 c^3} \int_0^{\infty} \frac{\omega^2}{e^{\beta \hbar \omega} - 1} d\omega$$
$$N =\frac{3V}{2 \pi^2 c^3} \left( \frac{1}{\beta \hbar}\right)^3 \int_0^{\infty} \frac{x^2}{e^x -1} dx$$
$$N = \frac{3V}{2\pi^2 c^3} \left(\frac{k}{\hbar} \right)^3 \cdot 2.404 \cdot T^3$$

Which volume should I use at this point? Should I use $(0.409nm)^3$ or should I use $(180nm)^3$?
Using the former gives $2.07$ which is exactly the number of lattice points of an FCC lattice. Using the latter gives $1.8 \times 10^8$.

 

[mfb]

"present within the nanocrystal" -> (180nm)³

 

[unscientific]

"present within the nanocrystal" -> (180nm)³

Using the lattice constant gives exactly $N\approx 2$. Could I take that my answers are right?

 

[mfb]

Exactly 2? That is odd, it depends on temperature and there is no reason why this specific temperature should give exactly 2.

 

[unscientific]

Exactly 2? That is odd, it depends on temperature and there is no reason why this specific temperature should give exactly 2.

Considering this is an FCC, does 2 phonons = 2 lattice points?

 

[mfb]

What happens if you take a different temperature, like 100 K or 300 K?

 

[unscientific]

What happens if you take a different temperature, like 100 K or 300 K?

True. So Number of phonons $\neq$ number of lattice points?

 

[mfb]

Sure.
There can be a temperature where the numbers are similar, but that is a meaningless coincidence.