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Threshold temperature { Failure of delong petit law}

  1. Dec 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi all, I have been trying to go over this problem for sometime now.. yes I have easily browsed over +200 articles, lectures , etc on google.

    The question is related to delong petit law and wants me to find out the temperature at which it fails i.e when quantum effects prevail ( In this it's aluminium).




    2. Relevant equations

    Aluminum = 27g/mol

    3. The attempt at a solution

    At higher temperature specific heat of solids is roughly ~3R but all this changes of course with drop in temperature , Cv tends to -> 0.

    Now I have made few assumptions and have ended up with two values of temperature one is 161kelvin and other being about 50k which I are probably too high. Anyways...

    From what I know is that as temperature drops so does the no. of degrees of freedom.
    According to equipartition at room temperature we get 3RT but it's my intuitive feeling that at lower temperatures D of freedom is 2 ( due to vibration ? )
    So over all energy is = KbT , another variation of it can be :
    Δ E = hω.

    Now here is one of my attempt.
    Δ E = hω
    KbT = hω

    T = hω / kb (i)

    w = 1/2pi * {square root of k/m}
    w = 3.36*10^12 Hz plug this int eq. (i)
    T = hw/kb
    T = ~161kelvin...
    On the previous part spring constant 'k' was given as 20 N/m however I am not sure if I am meant to use it for this question since in the previous bit temperature was '300k'.


    My other approach was to use: 24.9 > dE/dt
    24.9 > KbT/ (t-300)

    P.S: I have not slept through the night it's 9 am now, feedback of any kind will be appreciated. Thanks
     
    Last edited: Dec 25, 2011
  2. jcsd
  3. Dec 25, 2011 #2

    ehild

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    Dulong-Petit's law is valid for kBT>>[itex]\hbar[/itex]ω where ω is the angular frequency defined by ω=sqrt(k/m).

    As the atoms of a crystal can only move about their equilibrium positions, all possible 3N motions of the N atoms in the crystal are coupled vibrations - waves with frequency from 0 to the 2ω0 and energy [itex]\hbar[/itex]ω. See:http://en.wikipedia.org/wiki/Phonon
    The higher-energy lattice modes are not excited at law temperatures. The equipartition principle -that all vibrations have kTB energy is not valid for T<[itex]\hbar[/itex]ω/ kB, so Dulong-Petit's law fails.

    What was given in the problem? Was it the fundamental frequency of the lattice mode or the spring constant?

    ehild
     
  4. Dec 26, 2011 #3
    Let me post the whole question.
    An atom i lies in a chain of atoms, between two neighbouring atoms j
    and k. Assume, for simplicity, that j and k can be regarded as fixed, while atom i
    vibrates.
     
    Last edited: Dec 26, 2011
  5. Dec 26, 2011 #4

    ehild

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    You can consider the spring constant unchanged with temperature.

    ehild
     
  6. Dec 26, 2011 #5
    .....
     
    Last edited: Dec 26, 2011
  7. Dec 26, 2011 #6

    ehild

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    Your result is about correct. The spring constant depends slightly with temperature, as the mean interatomic distance changes. But you do not need to take it into account and you can not as no details were given.

    ehild
     
  8. Dec 26, 2011 #7
    I see. Thanks for all your help, I appreciate it ( wish I could do anything in return, feel guilty)but there's another part which I don't seem to understand. For part ii) where the question asks me to find the % of the root mean square amplitude relative to inter atomic distance. Now I assume what the question wants me to find is the inter atomic distance 'r' which I could get by letting : 1/2kx^2 = P.E
    where x = (r-r0).
     
  9. Dec 26, 2011 #8
    Sorry a repost
     
    Last edited: Dec 26, 2011
  10. Dec 26, 2011 #9

    ehild

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    Read the problem text:
    Find the mean-square amplitude of the vibration at 300 K.

    ehild
     
  11. Dec 26, 2011 #10
    I did that but I wasn't sure. Now that you have assured me.. it does make sense.. they did say 'mean' which would obviously be the amplitude on a graph. Thanks for all your help!
     
  12. Dec 26, 2011 #11

    ehild

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    The atoms are vibrating so the interatomic distance changes. Till the vibration is SHM the mean separation is r0.
    Determine the rms amplitude of the vibration, equal to A/sqrt(2). You get it from the energy of the oscillator.


    ehild
     
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