Threshold temperature { Failure of delong petit law}

[ibysaiyan]

Homework Statement

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).

Homework Equations

Aluminum = 27g/mol

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

 

[ehild]

Dulong-Petit's law is valid for k_BT>>[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 kT_B energy is not valid for T<[itex]\hbar[/itex]ω/ k_B, 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

 

[ibysaiyan]

Dulong-Petit's law is valid for k_BT>>[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 kT_B energy is not valid for T<[itex]\hbar[/itex]ω/ k_B, 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

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.

(i) Assuming that equipartition holds, show that the root mean square amplitude of
vibration xrms at temperature T of the atom is B k T
D
, where D is the spring constant
of an isolated pair of atoms.
(ii) In aluminium (atomic weight 27 gm/mole), what percentage of the mean
interatomic separation (0.29 nm) is xrms at 300K. Assume D= 20 Nm-1.
(iii) A three dimensional solid can be modeled of as a network of atoms, which can
now move in three dimensions. What is the resulting molar heat capacity at high
temperatures? [This is the Dulong-Petit law].
(iv) Estimate, in the case of aluminium, the temperature at which you expect to see
deviations from this result, due to quantum effects (characteristic energy D E = homework ),
on cooling the sample. Explain why this occurs.

 

[ehild]

You can consider the spring constant unchanged with temperature.

ehild

 

[ibysaiyan]

You can consider the spring constant unchanged with temperature.

ehild

...

 

[ehild]

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

 

[ibysaiyan]

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

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).

 

[ibysaiyan]

Sorry a repost

 

[ehild]

For part ii) where the question asks me to find the % of the root mean square amplitude relative to inter atomic distance.

Read the problem text:

In aluminium (atomic weight 27 gm/mole), what percentage of the mean
interatomic separation (0.29 nm) is xrms at 300K.

Find the mean-square amplitude of the vibration at 300 K.

ehild

 

[ibysaiyan]

Read the problem text:

Find the mean-square amplitude of the vibration at 300 K.

ehild

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!

 

[ehild]

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