Free Electron Model in Solids: Lecture Notes, Slides of Solid State Physics

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Solid State Physics

FREE ELECTRON MODEL

Lecture 14

A.H. Harker

Physics and Astronomy UCL

6 The Free Electron Model

6.1 Basic Assumptions

In the free electron model, we assume that the valence electrons can be treated as free, or at least moving in a region constant potential, and non-interacting. We’ll examine the assumption of a constant po- tential first, and try to justify the neglect of interactions later.

From atomic theory, we are used to the idea that different electronic functions must be orthogonal to each other (remember we used this idea in discussing the short-range repulsive part of interatomic po- tentials) i.e. if ψc(r) is a core function and ψv(r) is a valence function ∫ ψc(r)ψv(r)dr = 0.

Let’s see how orthogonality might be achieved for a slowly-varying wave.

To achieve orthogonality:

we need high spatial frequency (large k ) components in the wave. Large k → large energy.

So finally we assume that the attractive potential of the ion cores can be represented by a flat-bottomed potential.

We go further, and assume that the potential is deep enough that we can use a simple ‘particle-in-a-box’ model – the free electron model.

So we can use the result from section 4.4 (Lecture 10) that the number of states with modulus of wavevector between k and k + dk is

g(k)dk =

V

8 π^3

4 πk^2 dk =

V

2 π^2

k^2 dk.

For electrons dE dk

ℏ^2 k m

ℏ^2

m

2 mE ℏ^2

m

2 E.

We also need to include a factor of 2 for spin up and spin down.

g(E) = 2g(k)

dk dE = 2

V

2 π^2

k^2

dk dE = 2

V

2 π^2

2 mE ℏ^2

m ℏ

2 E

V m π^2 ℏ^3

2 mE.

Note that as V increases, so does the density of states.

6.1.3 The Fermi Energy

Remember that the Fermi distribution function n(E)

n(E) =

exp((E − EF )/kBT ) + 1 at absolute zero is 1 up to the Fermi energy EF. Suppose the volume V contains Ne electrons. Then we know

Ne =

0

g(E)n(E)dE

∫ EF

0

g(E)dE

V

2 m^3 π^2 ℏ^3

∫ EF

0

E dE

V

2 m^3 π^2 ℏ^3

2 EF^3 /^2

so

EF =

ℏ^2

2 m

3 π^2 Ne V

6.1.4 Orders of magnitude

For a typical solid, the interatomic spacing is about 2. 5 × 10 −^10 m. Assume each atom is in a cube with that dimension, and releases one valence electron, giving an electron density Ne/V ≈ 6 × 1028 m−^3. Putting in the numbers, we find

• EF ≈ 9 × 10 −^19 J = 6 eV ;
• TF ≈ 70 , 000 K ;
• kF ≈ 1. 2 × 1010 m−^1 , comparable with the reciprocal lattice spacing
  2. 5 × 1010 m−^1 ;

We can also estimate the electron velocity at the Fermi energy:

vF =

ℏkF m

≈ 1. 4 × 106 m s−^1 ,

which is fast, but not relativistic.

total energy of electrons =

∫ EF

0

E g(E) dE =

NeEF,

so that the average energy per electron is^35 EF.

6.1.5 The Fermi surface

In later sections we shall talk a good deal about the Fermi surface. This is an constant-energy surface in reciprocal space (k-space) with energy corresponding to the Fermi energy. For the free electron gas, this is a sphere of radius kF.

The current

J = BT 2 exp

φ kBT

with a theoretical value

B =

emkB^2 πℏ^2

= 1. 2 × 106 A m−^2 K−^2.

Experimentally the exponential dependence is confirmed, with simi- lar values for B.

6.2.2 Field emission

A large applied field alters the potential outside the metal enough to allow electrons to tunnel out.

6.2.3 Photoemission

A photon with energy greater than the work function can eject an electron from the metal.

6.2.4 X-ray emission (Auger spectroscopy)

A high-energy electron incident on a metal may knock out an elec- tron from a core state (almost unchanged from the atomic state). An alectron from the band can fall into the empty state, emitting an x- ray.