# Homework Help: Number of phonons - Which length should I use?

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1. Apr 27, 2015

### unscientific

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

Part(a)

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 $1.8 \times 10^8$.

2. Apr 29, 2015

### Staff: Mentor

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

3. Apr 30, 2015

### unscientific

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

4. May 1, 2015

### Staff: Mentor

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

5. May 2, 2015

### unscientific

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

6. May 2, 2015

### Staff: Mentor

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

7. May 3, 2015

### unscientific

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

8. May 3, 2015

### Staff: Mentor

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