What physical conditions lead electrons in a metal to form a degenerated Fermi gas? Show that the density of states for an ideal two-dimensional Fermi gas of particle with spin 1/2 and mass m that occupies area A is:
Obtain a relationship between the number of electrons per unit area and the Fermi energy E_f at 0 K.
Show that the average electron energy for the two dimensional gas at 0 K is E_f/2.
Obtain the temperature dependence, at low temperatures, of the heat capacity of unit area of the Fermi gas. Explain why the magnitude of the heat capacity is much smaller than the value predicted for a classical two-dimensional gas. Obtain the temperature dependence, at low temperature of the contribution of the lattice vibrations to the heat capacity of a very thing film of a metal.
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Electrons in metals form a degenerate Fermi gas, because the density of electrons is far larger than the number of quantum states per unit volume with energy below k T. Then, since each electron must occupy a different quantum state, most electrons will have energies far larger than k T. The thermal energy of k T which is available for thermal fluctuations is then only available for the electrons in the top energy levels around k T of the maximum energy level occupied. Only a small fraction of all the electrons are in these quantum states, so most electrons do not participate in thermal processes.
The density of states of a 2d Fermi gas can be computed as follows. The total number of quantum states in an area A and area Ap in momentum space is given by:
2 A Ap/h^2
Here the factor 2 comes from the two possible spin states the spin 1/2 particle can have. The area in momentum space between the magnitude of the momentum of P and P + dP is given by 2 pi P dP (i.e. the circumference times the width). The number of quantum states with momenta between P and P + dP is thus given by:
4 pi A/h^2 P dP
Since 2 P dP can be written as d(P^2), we can write this as:
2 pi A/h^2 d(P^2 )
Since P^2/(2m) is the energy E, we can rewrite this in terms of the range in energy dE:
4 pi m A/h^2 dE
The density of states is therefore:
4 pi m A/h^2 = m A/(pi hbar^2)
So, we see that the density of states does not depend in the energy in the 2d case. The total number of states below some arbitrary energy E is thus obtained by simply multiplying the density of states by E:
N(E) = m A/(pi hbar^2) E (1)
If we have M electrons at 0 K then these M electrons will occupy the M lowest energy levels. So, this means that the Fermi energy Ef will be such that N(Ef) = M:
M = m A/(pi hbar^2) Ef --------->
M/A = m/(pi hbar^2) Ef
Since the density of states does not depend on the energy, the electrons are uniformly distributed as a function of energy between zero and Ef, therefore the average energy of the electrons is Ef/2.
To compute the heat capacity of the 2d electron gas, we must consider the Fermi-Dirac distribution at finite temperature. We can write the total energy E and the number of particles N in terms of the temperature and the chemical potential as:
E = integral over e of e g(e) de/[exp(beta (e - mu)) + 1]
N = integral over e of g(e) de/[exp(beta (e - mu)) + 1]
g(e) = m A/(pi hbar^2)
is the density of states obtained earlier in the problem. By "e" we denote the single particle electron energy (we use the lower case to avoid confusion with the total energy E). The integrals over e are from zero to infinity.
To compute the heat capacity near T = 0, we need to know how E behaves near zero. The above integral expression for E, gives E in terms of T and mu, but mu will depend on T also. So, we need to consider the expression for N in terms of mu and T also and solve for mu in terms of N and T.
So, let's first evaluate ...
We compute the heat capacity of electrons confined to a two dimensional area at zero temperature. The electrons are modeled as a non-interacting Fermi gas. We also derive the contribution to the heat capacity from lattice vibrations (phonons) in the two dimensional system. All the computations, including the evaluations of the integrals over the Fermi distribution functions using contour integrals methods, are worked out explicitly.