Finding 2D Fermion Gas U/N with Temperature & Area

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SUMMARY

The discussion focuses on calculating the energy per particle, U/N, for a two-dimensional gas of N fermions with mass M, in thermal equilibrium at temperature T and area A. The user correctly applies the formula U = ∑(n_i ε_i) and derives the relationship between the chemical potential and the energy levels using the expression exp(μβ) = n/λ, where n = N/V and λ = mkT/2πħ². The challenge lies in expressing U/N in terms of the area per particle, a = A/N, which the user initially finds complex but ultimately confirms their calculations are accurate.

PREREQUISITES
  • Understanding of statistical mechanics, particularly Fermi-Dirac statistics
  • Familiarity with thermodynamic concepts such as temperature, energy, and chemical potential
  • Knowledge of quantum mechanics, specifically the properties of fermions
  • Basic proficiency in mathematical summation and exponential functions
NEXT STEPS
  • Explore the derivation of Fermi-Dirac distribution and its applications in statistical mechanics
  • Study the implications of temperature on the behavior of fermionic systems in two dimensions
  • Investigate the relationship between area per particle and thermodynamic properties in 2D systems
  • Learn about the role of the chemical potential in quantum gases and its calculation methods
USEFUL FOR

This discussion is beneficial for physics students, researchers in statistical mechanics, and anyone studying quantum gases and their thermodynamic properties.

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Homework Statement

For a gas of N fermions with mass M in 2D in a region of area A in thermal equilibrium at temperature T, we are asked to find ##U/N## in fuction of ##T## and ##a=A/N##.

The attempt at a solution

I used ##U=\sum(<n_i>\epsilon_i) = \sum(\exp(\beta(\mu-\epsilon_i))\epsilon_i = \exp(\mu\beta)\sum(\exp(-\epsilon_i\beta)\epsilon_i##
from which ##\exp(\mu\beta)=n/\lambda## where ##n=N/V## and ##\lambda=mkT/2\pi\hbar^2##

Now as you can tell expressing this in terms of ##a## is awkward to say the least, did I go wrong in the calculations somewhere?
 
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Never mind, it was right all along I just got confused for some reason.
 

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