Does Quantum Molar Internal Energy Converge to Classical at High Temperatures?

  • Thread starter Thread starter plonker
  • Start date Start date
  • Tags Tags
    Classical Quantum
plonker
Messages
2
Reaction score
0

Homework Statement



I'm so confused please help :\

Show that the contribution to the total energy from molar internal energy Um reverts to the classical expression at high T.

Homework Equations



Classical: Um = 3NakT Quantum Um = 3NAhv/e^(hv/kT)-1

The Attempt at a Solution



Manipulating variables- E=hv
Quantum rearranging: hv/kT= ln(3Nahv)-ln(Um)
Very confused on what is meant by total energy though. Is that supposed to be E+Um?
 
Last edited:
Physics news on Phys.org
Pardon my ignorance (I have never heard of this in all my years in physics), but what is Um?
 
My teacher said it was "internal energy, U" but while in the context of the failures of classical physics in terms of heat capacities. Apparently Einstein calculated the contribution of the oscillation of the atoms to the total molar energy of metal and obtained the quantum equation above in place of the classical one? I'm so nervous for this course now :\
 
I am assuming that for high T you can manipulate e^{\frac{h \nu}{k T}}. Try this.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top