Temperature as a function of vibration energy quanta

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SUMMARY

The discussion centers on calculating temperature as a function of vibration energy quanta for a cluster of four copper atoms modeled as independent harmonic oscillators. The energy of each oscillator is defined by the equation Evib = ¯hω(n + 1/2), where ω is derived from the spring constant (k ≈ 120 N/m) and the mass of the copper atom. Key calculations include the number of microstates Ω(n) = ((n+11)!)/(n!11!), entropy S(n) = k*ln(((n+11)!)/(n!11!)), and the relationship between temperature T and entropy through T = dEint/dS, assuming n is significantly larger than the number of oscillators.

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  • Understanding of harmonic oscillators in quantum mechanics
  • Familiarity with statistical mechanics concepts such as microstates and entropy
  • Knowledge of thermodynamic relationships involving temperature and entropy
  • Basic calculus for differentiation and factorial functions
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  • Explore the derivation of the canonical ensemble in statistical mechanics
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Students and researchers in physics, particularly those focusing on quantum mechanics and statistical mechanics, as well as anyone interested in the thermodynamic properties of small atomic clusters.

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Homework Statement


Consider a small cluster of four copper atoms. Assume, that each atom can oscillate around its equilibrium position independently of the other atoms. Let us model each direction of vibration as a harmonic oscillator, whose energy is quantized
Evib = ¯hω(n +1/2), where ω = sqrt(k/m), k is the ’spring constant’ of the bond between the atoms (k ≈ 120 N/m) and m is the mass of the copper atom. Compute as the function of vibration energy quanta n = 0,1,2,3,... the (a) number of microstates Ω
(b) entropy S
(c) temperature T
(d) heat capacity C / atom

Homework Equations

The Attempt at a Solution


I have already functions for the number of microstates and the entropy S:
Ω(n) = ((n+11)!) : (n!11!)
S(n) = k*ln (((n+11)!) : (n!11!)
Now I try to find the function T(n) and started with:

1/T = dS/dEint (dS = change in entropy, dEint = change in internal energy)

Can I put this equations as T = dEint/dS ?
 
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Yes you can do that, but you'll have to put E(int) in terms of S. It may help if you're allowed to make the assumption that n >> # of oscillators.
 
Last edited:

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