Question

Derive an expression for ground state energy of a completely degenerate FD gas. For
copper, take N/V =8.5 x10^28 m^-3 , h=6.62x 10^-34 Js and Me = 9.1x 10^-31 kg to calculate Fermi
energy

Answer 1

Answer on Question#50678, Physics, Molecular Physics | Thermodynamics

In order to find expression of ground state energy of completely degenerate Fermi-Dirac gas one has to know:

1. The density of states in terms of energy for electrons $dN(E) = \sqrt{2m^3}\frac{V}{\pi^2\hbar^3}\sqrt{E} dE$ , which is the number of energy levels which are between energy interval $E,E + dE$

2. The Fermi-Dirac distribution $f(E) = \frac{1}{e^{\frac{E - \mu}{\hbar T}} - 1}$ , which gives the average number of fermions in state with energy $E$ .

3. The Fermi energy, which is the maximum energy of occupied particles under zero temperature. Actually, under zero temperature the Fermi-Dirac distribution changes to the following step function:

The Fermi energy is easily obtained by equaling the number of particles to its corresponding expression in terms of density of states and Fermi-Dirac distribution under zero temperature (the integral expression is also understandable from the picture above):

$N = \int_{0}^{E_{F}} dN(E) = \frac{(2m)^{3/2} V E_{F}^{3/2}}{3\pi^{2}\hbar^{3}}$ , from where $E_{F} = \frac{\hbar^{2}}{2m} (3\pi^{2})^{2/3} \left( \frac{N}{V} \right)^{2/3}$ . It is expressed in terms of $\frac{N}{V}$ , $\hbar$ and $m$ .

In order to calculate the ground state energy of one particle, one has to evaluate the mean value $\bar{E} = \int_{0}^{\infty} E f(E) dN(E)$ , which is much more simple to evaluate in case of zero temperature:

$\bar{E} = \int_{0}^{E_{F}} E dN(E) = \sqrt{2m^{3}} \frac{V}{\pi^{2}\hbar^{3}} \int_{0}^{E_{F}} E^{3/2} dE = (2m)^{3/2} \frac{V}{5\pi^{2}\hbar^{3}} E_{F}^{5/2} = \frac{3}{5} N E_{F}$ using expression for Fermi energy.

Hence, for one particle in completely degenerate ideal Fermi gas, the ground state energy is $E_0 = \bar{E} = \frac{3}{5} E_F$ .

Using given numeric values, first calculate Fermi energy: $E_{F} = 4.45 \cdot 10^{-17} J$ . Hence the ground state energy is $E_{0} = \frac{3}{5} E_{F} = 2.67 \cdot 10^{-17} J$ .

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