1 Fermi-Dirac distribution and the Fermi-level Density of states tells us how many states exist at a...
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- Slide 1
- 1 Fermi-Dirac distribution and the Fermi-level Density of
states tells us how many states exist at a given energy E. The
Fermi function f(E) specifies how many of the existing states at
the energy E will be filled with electrons. The function f(E)
specifies, under equilibrium conditions, the probability that an
available state at an energy E will be occupied by an electron. It
is a probability distribution function. E F = Fermi energy or Fermi
level k= Boltzmann constant = 1.38 10 23 J/K = 8.6 10 5 eV/K T=
absolute temperature in K
- Slide 2
- 2 Fermi-Dirac distribution: Consider T 0 K For E > E F : For
E < E F : EEFEEF 0 1 f(E)
- Slide 3
- 3 If E = E F then f(E F ) = If then Thus the following
approximation is valid: i.e., most states at energies 3kT above E F
are empty. If then Thus the following approximation is valid: So, 1
f(E) = Probability that a state is empty, decays to zero. So, most
states will be filled. kT (at 300 K) = 0.025eV, E g (Si) = 1.1eV,
so 3kT is very small in comparison. Fermi-Dirac distribution:
Consider T > 0 K
- Slide 4
- 4 Temperature dependence of Fermi-Dirac distribution
- Slide 5
- 5
- Slide 6
- 6 Equilibrium distribution of carriers Distribution of carriers
= DOS probability of occupancy = g(E) f(E) (where DOS = Density of
states) Total number of electrons in CB (conduction band) = Total
number of holes in VB (valence band) =
- Slide 7
- Properties of a Fermion gas
- Slide 8
- The internal energy of a gas of N fermions
- Slide 9
- Integration by parts (I) In calculus, integration by parts is a
rule that transforms the integral of products of functions into
other, hopefully simpler, integrals. The rule arises from the
product rule of differentiation. If u = f(x), v = g(x), and the
differentials du = f '(x) dx and dv = g'(x) dx; then in its
simplest form the product rule is
- Slide 10
- Integration by parts (II) In the traditional calculus
curriculum, this rule is often stated using indefinite integrals in
the form As a simple example, consider Since ln x simplifies to 1/x
when differentiated, we make this part of ; since 1/x 2 simplifies
to 1/x when integrated, we make this part of g. The formula now
yields
- Slide 11
- At T = 0, U = (3/5)N F, this energy is large because all the
electrons must occupy the lowest energy states up to the Fermi
level. The average energy of a free electron in silver at T = 0 is
The mean kinetic energy of an electron, even at absolute zero, is
two orders of magnitude greater than the mean kinetic energy of an
ordinary gas molecule at room temperature.
- Slide 12
- Heat capacity The electronic heat capacity C e can be found by
taking the derivative of Equation (19.18): For temperatures that
are small compared with the Fermi temperature, we can neglect the
second term in the expansion compared with the first and
obtain
- Slide 13
- Thus the electronic specific heat capacity is 2.2 x 10 -2 R.
This small value explains why metals have a specific heat capacity
of about 3R, the same as for other solids. It was originally
believed that their free electrons should contribute an additional
(3/2) R associated with their three translational degrees of
freedom. Our last calculation shows that the contribution is
negligible. The energy of the electrons changes only slightly with
temperature (dU/dT is small) because only those electrons near the
Fermi level can increase their energies as the temperature is
raised, and there are precious few of them.
- Slide 14
- At very low temperatures the picture is different. From the
Debye theory, C v is proportional to T 3 and so the heat capacity
of a metal takes the form C v = AT + BT 3, where the first term is
the electronic contribution and the second is associated with the
crystal lattice. At sufficiently low temperatures, the AT term can
dominate, as the sketch of Figure 19.9 indicates. Figure 19.9
Sketch of the heat capacity of a metal as a function of temperature
showing the electronic and lattice contributions.
- Slide 15
- S = 0 at T = 0, as it must be. The Helmholtz function F = U -TS
is The fermion gas pressure is found from
- Slide 16
- For silver we find that N/V = 5.9 x10 28 m -3 and T F =
65,000K. Thus P = 2/5 *5.9*10 28 *(1.38*10 -23 ) (6.5*10 4 ) =
2.1*10 10 Pa = 2.1*10 5 atm. Given this tremendous pressure, we can
appreciate the role of the surface potential barrier in keeping the
electrons from evaporating from the metal.
- Slide 17
- 19.5 Applications to White Dwarf Stars The temperature inside
the core of a typical star is at the order of 10 7 K. The atoms are
completely ionized at such a high T, which creates a hugh electron
gas The loss of gravitational energy balances with an increase in
the kinetic energy of the electrons and ions, which prevent the
collapse of star!
- Slide 18
- Example: The pressure of the electron gas in Sirius B can be
calculated with the formula Using the following numbers MassM =
2.09 10 30 kg Radius R= 5.57 10 6 m VolumeV= 7.23 10 20 m 3
- Slide 19
- Assuming that nuclear fusion has ceased after all the core
hydrogen has been converted to helium! The number nucleons = Since
the ratio of nucleons and electrons is 2:1 there are electrons
- Slide 20
- Therefore, T(=10 7 K) is much smaller then T F. i.e. is a valid
assumption ! Thus: P can be calculated as
- Slide 21
- A white dwarf is stable when its total energy is minimum For
Since can be expressed as Where
- Slide 22
- For gravitational energy of a solid With In summary To find the
minimum U with respect to R
- Slide 23
- 19.7 a) Calculate Fermi energy for Aluminum assuming three
electrons per Aluminum atom.
- Slide 24
- 19.7b) Show that the aluminum at T=100 K, differs from F by
less than 0.01%. (The density of aluminum is 2.69 x 10 3 kg m - 3
and its atomic weight is 27.)
- Slide 25
- 19.7c) Calculate the electronic contribution to the specific
heat capacity of aluminum at room temperature and compare it to 3R.
Using the following equation
- Slide 26
- 19.13. Consider the collapse of the sun into a white dwarf. For
the sun, M= 2 x 10 30 kg, R = 7 x 10 8 m, V= 1.4 x 10 27 m 3.
Calculate the Fermi energy of the Suns electrons.
- Slide 27
- (b) What is the Fermi temperature? (c) What is the average
speed of the electrons in the fermion gas (see problem 19-4).
Compare your answer with the speed of light.