Show that the average kinetic energy of a conduction...

Key Concepts

Fermi Energy

Fermi energy is the highest occupied energy level of a system of fermions at absolute zero temperature. In the context of conduction electrons in metals, it represents the upper limit of electron energies in the ground state, determining many electronic and thermal properties of the material. Calculations involving the Fermi energy allow one to relate microscopic energy distributions to macroscopic measurable quantities.

Density of States

The density of states describes the number of available quantum states per unit energy interval that electrons can occupy. It is a crucial concept for calculating various thermodynamic and transport properties in solid state physics, including the total energy and average kinetic energy of electrons. In the free electron model, the density of states function shapes the distribution of electrons at different energy levels.

Free Electron Model

The free electron model treats conduction electrons in metals as free particles that do not interact significantly with one another or with the lattice ions, except for obeying the Pauli exclusion principle. This model provides a simplified framework for analyzing electron behavior at 0 K, where electrons fill up energy states up to the Fermi energy, enabling straightforward application of quantum statistics to derive properties like the average kinetic energy.

Fermi-Dirac Statistics

Fermi-Dirac statistics govern the distribution of fermions, such as electrons, over energy states in thermal equilibrium, taking into account the Pauli exclusion principle. At absolute zero, this distribution becomes a step function where all states with energy below the Fermi energy are occupied and those above are empty. This statistical framework is essential for calculating integrals that yield the average kinetic energy and other properties of an electron gas.

Transcript

00:01 So this question asks us to show that the average kinetic energy of a conduction electron in a metal at zero kelvin is equal to 3 over 5 multiplied by the fermi energy shown by the expression in this red box here.

00:13 We are given an equation for the average kinetic energy with this big integral to solve.

00:18 We are given the number density of states given by this expression here.

00:23 And we also know that the number density of the electrons is given by this expression here.

00:28 So the first point to note is that we are only interested in states with values between zero and the fermi energy.

00:38 Because for the purposes of this, it's impossible to have any particles in states above that value, because the fermi energy represents sort of the highest energy level that we can go to at a temperature of zero.

00:49 Another thing to note is that the denominator in the expression for capital n is actually the fermi direct distribution, which we will call lowercase f .e.

01:05 And the key thing to know about the fermi direct distribution is that it always takes a value of 1 at 0 kelvin between the range of zero and the fermi energy here.

01:18 We can use that to our advantage because it simplifies the integral quite a lot.

01:23 So we'll start by saying that also apologies in advance about my handwriting.

01:29 I'm trying to write this out with a mouse, that our average kinetic energy will be equal to 1 over our number density n subscript e, multiplied by the integral, and remember we're only interested in the values with the maximum of the fermi energy and 0.

01:47 Taking into consideration that our fermi direct distribution is 1, this just comes out as e multiplied by, well, we have our constant c times by e to the half.

02:03 And as you can see, that's the only thing that's left in this expression for the number density of states, obviously with respect to e.

02:15 So bringing our constant c out to the outside of the integral and combining our e factors, we end up with an expression that we have to integrate for e to the three over the three.

02:27 2 and that comes out as well our constants will stay there where they are so we have c over n e and now we need to add one to this power which will make it 5 over 2 and divide by that power so that ends up with a 2 over 5 at the front when we evaluate this expression at zero it will just be 0 so we only need to consider the first term with e f in so remember we've added 1 to the power so all that comes out will be ef to the 5 over 2...

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