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Number of phonons - Which length should I use?

  1. Apr 27, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Find debye frequency.
    (b) Find number of atoms.
    2012_B6_Q2.png
    2. Relevant equations


    3. The attempt at a solution

    Part(a)

    Density of states is given by
    [tex]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}[/tex]
    Debye frequency is given by
    [tex]\omega_D^3 = 6 \pi^2 n c^3 [/tex]

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

    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. jcsd
  3. Apr 29, 2015 #2

    mfb

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    "present within the nanocrystal" -> (180nm)3
     
  4. Apr 30, 2015 #3
    Using the lattice constant gives exactly ##N\approx 2##. Could I take that my answers are right?
     
  5. May 1, 2015 #4

    mfb

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    Exactly 2? That is odd, it depends on temperature and there is no reason why this specific temperature should give exactly 2.
     
  6. May 2, 2015 #5
    Considering this is an FCC, does 2 phonons = 2 lattice points?
     
  7. May 2, 2015 #6

    mfb

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    What happens if you take a different temperature, like 100 K or 300 K?
     
  8. May 3, 2015 #7
    True. So Number of phonons ##\neq## number of lattice points?
     
  9. May 3, 2015 #8

    mfb

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