VẬT LÝ CHẤT RẮN 065 the hall effect and thermal conductivity

Chapter 6 Free Electron Fermi Gas The Hall Effect Chapter 6 Free Electron Fermi Gas The Hall Effect and Thermal Conductivity Motion in Magnetic Field • Last time, we looked at what happens when you ap[.]

Chapter 6: Free Electron Fermi Gas: The Hall Effect and Thermal

Conductivity

Motion in Magnetic Field

• Last time, we looked at what happens when you apply an electric

field to a Fermi gas

• Now, we are going to look at what happens when you apply a

magnetic field

• For the application of an electric field, we found that the Fermi

sphere is displaced by ∆k

F

ky

E e

k dt

d a

m

h r

r

−

=

∆

=

=

kx (for a field in the x-direction)

Effect of Collisions

• We also showed that the electrons undergo collisions in a material after a collision time τ This means that the equation we wrote

above has to be modified:

• This second term, which takes collisions into account, acts like a frictional force Since there is no derivative in this form, it says

that the force is proportional to ∆k/τ, which is like saying it is

proportional to the velocity of the particles (k ~ momentum)

• Other frictional forces act this way, like friction due to air

resistance (this is proportional to the velocity, and explains why a terminal velocity exists Similarily, it explains why the electrons’ velocity doesn’t grow without limit when you apply an electric field)

( ) k dt

d

h

r

∆

⎟

⎠

⎞

⎜

⎝

⎛ +

=

τ

1

Electric and Magnetic Fields

• Using this expression for the force, let’s go back to our original

equation (which included both electric and magnetic fields):

• Now, assuming that mv = ħ∆k (ie the electrons velocities are shifted

by this amount due to the electric/magnetic field) we have:

( )

( E v B )

e

B

v c

E e

k dt

d F

r r

r

r r

r

r h

r

× +

−

=

⎟

⎠

⎞

⎜

⎝

⎛ + ×

−

=

∆

⎟

⎠

⎞

⎜

⎝

⎛ +

(SI units)

⎟

⎠

⎞

⎜

⎝

−

=

⎟

⎠

⎞

⎜

⎝

c

E e

v dt

d m

τ

An example: The Hall Effect

• To illustrate how this equation works, lets take a metal and apply a static magnetic field B in the z-direction, and

a static electric field Ex in the x-direction.

• How do the electrons move? (the result of this

experiment is called the Hall Effect)

y

Ex

x

The Hall Effect

• Consider the motion of the electrons in the x, y, and z direction

• Initially: v = (vx,vY,vz), E = (Ex, 0, 0), B = (0, 0, B)

• Apply the fields, and observe what happens

⎟

⎠

⎞

⎜

⎝

⎛ +

−

=

⎟

⎠

⎞

⎜

⎝

⎛ +

→

⎟

⎠

⎞

⎜

⎝

⎛ + ×

−

=

⎟

⎠

⎞

⎜

⎝

⎛ +

c

E e

v dt

d m B

v c

E e

v dt

d m

τ τ

r r r

r

x

y

Ex

B

Initial state: electrons start to flow

in the –y direction This creates a field EY

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

B

v c

e

v dt

d

m 1 Y 1 x

τ

(net force in –x direction)

(net force in the –y direction)

We need to consider what happens after this initial electron drift to the bottom part of the metal slab (electrons build up down here)

The Hall Effect

• The electrons will build up in the lower part of the metal, generating an

electric field EY

• This will occur until the motion reaches a steady state – that is, when the

forces are balanced in the y-direction (the Lorentz force from the magnetic

field, and the Electric force from the build-up of the electrons will be equal

and cancel eachother out)

y

x

Ex

(electrons build up down here)

EY

⎟

⎠

⎞

⎜

⎝

⎛ −

−

=

→

⎟

⎠

⎞

⎜

⎝

⎛ +

−

=

→

⎟

⎠

⎞

⎜

⎝

⎛ −

−

=

⎟

⎠

⎞

⎜

⎝

⎛ +

⎟

⎠

⎞

⎜

⎝

⎛ +

−

=

⎟

⎠

⎞

⎜

⎝

⎛ +

B

v c

E e mv

B

v c

E e mv

B

v c

E e

v dt

d m

B

v c

E e

v dt

d m

x Y

Y

Y x

x

x Y

Y

Y x

x

1 1

1 1

1 1

τ τ τ

τ

Steady-state conditions: d/dt part = 0 →

(forces are balanced)

Steady state equations:

EY (from

Lorentz

Force)

The Hall Effect

• These equations of motion are often expressed as:

• For electrons moving in the z-direction, we have: vz = 0 (they don’t feel the magnetic field), unless there is an electric field and then we have: vz = -(eτ/m)Ez

• The constant ωC = (eB/mc) is called the cyclotron frequency (see the next assignment to find out why)

• The Hall Effect occurs when we apply an electric field in the

x-direction and a magnetic field in the z-x-direction (which generates,

as well, an electric field in the y-direction EY) What happens?

x c Y

Y Y

c x

m

e v

v

E m e

v = − τ − ω τ ; = − τ + ω τ

The Hall Effect

• If the current can’t flow out of the metal in the y-direction, then in the steady state, the velocity of the electrons in the y-direction must be zero (vY = 0)

• Using the equations we just derived, we then have:

• And: so putting it together:

e

v m

e

v m

E

v

E m

e v

x c x

c Y

x c Y

Y

ω τ

τ ω

τ ω τ

=

=

→

= +

−

τ

τ

e

v m E

E m

e v

x x

x x

−

=

→

−

=

x

mc eB

The Hall Effect

• So, what happens is that we get a net flow of electrons in the x-direction, but

we also get an electric field set up in the y-direction.

• Hall Effect measurements compare the ratio of the field created in the

y-direction, to the current in the x-y-direction, and the magnetic field in the

z-direction

• The Hall coefficient is defined by:

(from steady state solution)

nec m

B E ne

mc E

eB B

j

E R

x

x x

Y H

1 /

/

−

=

=

τ

τ

(CGS units)

(SI units)

x

y

Ex

(in steady state, no of e - is

constant at bottom)

EY

RH = − 1

(from last lecture)

EY

(electric charge) (density of conduction electrons)

The Hall Effect

Metal RH

(exp)

No of carriers /atom

RH (theory)

Li -1.89 1 el -1.48

Na -2.619 1 el -2.603

K -4.946 1 el -4.944

Rb -5.6 1 el -6.04

Cu -0.6 1 el -0.82

Ag -1.0 1 el -1.19

Au -0.8 1 el -1.18

Al 1.136 1 hole 1.135

In 1.774 1 hole 1.780

• Why is this an important

measurement?

• It relates 3 simple, easily

measured quantities (the

current density, the electric

field in y-direction, and the

magnetic field in the

z-direction) to the density of

conduction electrons (which

we can calculate)

• Also, it gives some strange

results!

• In particular, RH changes

• Why?

Electrons and holes

• The Hall Effect was an important experiment historically because it suggested that a carrier could have a positive charge

• These carriers are “holes” in the electron sea (and thus, being the absence of an electron, they have a net

positive charge) These were first explained by

Heisenburg.

• We can’t explain why this would happen with our free

electron theory (but it arises naturally in band theory)

• Note: the conditions we derived for the steady state can

be invalid for several conditions (ie when there is a

distribution of collision times) But in general, it is a very powerful tool for looking at properties of materials

Thermal Conductivity

• Last chapter, we found that for phonons, the thermal conductivity is

κ = 1/3 Cvl (heat capacity/unit volume = C, v is the velocity of

phonons, l is the mean free path)

• The exact same theory can be applied for free electrons (just like phonons, they move with a certain velocity, and have a mean free path)

• Most of the mobile electrons are at the Fermi energy, so εF = ½ mvF2

• We also have, using Cel = 1/2 π2Nk2T/εf and l = vf τ

m

T nk

l

v mv

T k

V

N Cvl

el

f f

el

3

2

1 2

1 3

1 3

1

2 2

2

2 2

τ

π κ

π κ

=

→

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

=

=

(so, κel ~ T at low temperatures) (note: n = electron density)

Thermal Conductivity

• What carries the heat in a metal? The electrons or

phonons?

• In pure metals, the electronic part is much greater than the phonons (density of free electrons is high)

• If the metal is impure, the mean free path of the

electrons will be lower, so the phonon part may dominate

• Note that the thermal conductivity of a metal should look like the specific heat at low temperatures:

C ~ γT + βT3 κ ~ AT + BT3

(different constants, but same functional form

at low temperatures)

Wiedemann-Franz Law

• In the early days, scientists did not know if what carried heat and what

carried electrical current in metals was the same particle (ie the electron)

• One test of this was to take the ratio of the thermal conductivity to the

electrical conductivity:

• Note that the Lorentz number is just a constant – the terms of m and n

cancel It should be the same for all metals and equal to 2.45 x 10 -8 WΩ/K 2

(using the Free Electron Fermi Gas model in SI units)

• How well does this hold up?

LT

T e

k m

ne

m Tn

k

=

⎟

⎠

⎞

⎜

⎝

⎛

=

=

2 2

2

2 2

3 /

3

τ

τ

π σ

L = Lorentz number)

Wiedemann-Franz Law

L L Metal (O deg C) (100 deg C)

• This model works incredibly

well for most metals

(L ~ 2.45 W Ω/K2)

• Therefore, it appears that

electrons carry both heat and

current (which should be

expected)

• At low temperatures, the

Lorentz number decreases

slightly (our model starts to

break down)

• Over all though, it is very

successful

(L in units of 10 8 W Ω/K 2 )