• One Dimensional Materials

• Electronic structure in 1D

• Electrical transport in 1D

• Most of this stuff is elementary quantum mechanics:

• Variations on infinite potential well

• Quantum mechanics, SM McMurry

• Quantum transport, Supriyo Datta

• Intro to Solid State Physics (8th Edition) C Kittel, (with PMcEuen)

Ch 18

PY4P04 Nanoscience: One Dimensional

Nanostructures

• What is a 1D Material?

• Strictly speaking a 1D material is one where electrons have

freedom to move in 1D ONLY.

• Ie a perfectly linear array of atoms

• However such structures are virtually non-existent

• In reality we have quasi 1D materials

• These are materials that are small but finite in 2 dimensions

• Take a 3D structure and shrink its size in the x, and then the y

direction

x z

y

Lx

Ly

Lx, Ly< 100 nm

• This can be done in 2 ways

• Top down by etching bulk semiconductors

• OR Bottom up

• This means building up one dimensional structures atom

by atom

We can do this directly by moving

atoms with a scanning tunnelling

microscope, but only at small scales

because it is just not practical!

We need to convince atoms to

arrange themselves into useful

structures!!!!!!Self assembly

Types of quasi 1D material

• There are a

wide range of

these materials

available today

ZnO nanowires

grown through

evaporation of zinc

powders

Gold nanowires

grown by self

assembly of Au

clusters templated

by surface terraces

ZnO nano-swords

Ni nanowires

templated by Alumina

membranes

CdSe nanowires grown in solution

Also 1D versions or “normal semiconductors”

Si nanowires grown by chemical

vapour deposition

Si nanowires grown by molecular

beam epitaxy

In 1991 carbon tubes were found in a machine built to make C60: Carbon Nanotubes

Most common 1D material has to be the

carbon nanotube

Roll up graphene sheets

Graphite consists of Graphene sheets that

are held together by van der Waals forces.

Each graphene sheet is formed from sp2

bonded carbon atoms in a hexagonal array

A graphene sheet can be rolled up to form a

single walled nanotube

How are they formed?

http://www.nanoonline.net/articles/ swntproperties.htm

A graphene sheet can be rolled up into a cylinder, diameter D in an

infinite number of ways.

To make the nanotube, take the strip defined by the dotted lines and

roll around the direction defined by T, such that DB and AO

The ends of the nanotube are then capped by hemispherical carbon

cages

A nanotube is rolled around an axis ie [OB]

D

A given nanotube can be

identified uniquely by its

circumference vector (red), C.

This vector is at an angle to the

a1 unit vector. is the chiral angle

C is specified by how many unit

vectors in the a1 and a2 directions

it equals:

Here

This is represented as (4,2)

The vector T represents the

length of the nanotubes unit cell.

21ˆ2ˆ4ˆ aaC

21ˆˆˆ anamC

Roll-up vectors

Nanotube type

Rolling up a nanotube like this gives a specific nanotube specified by

(n,m)

Nanotubes are characterised according to 3 basic types:

Armchair tubes: n=m, =30o

Zigzag tubes: m or n =0, =0o

Chiral tubes: All others, 0o < < 30o

Armchair: (m,n)=(8,8)

Zigzag: (m,n)=(10,0)

a1

a2

(5,0)

(5,5)

(4,3)

(2,2)

120o

223mnmnad cc

length C-C

nm14.0 cca

mn

m

2

3tan 1

NB does not uniquely

specify the nanotube

Diameters: 0.7 < D < 1.5 nm Lengths: 0.2 < L < 10,000 m

Tube types are often specified by a

diagram like this one.

Each tube type is represented by a

(non-green) dot

The circumference vector goes from

the green dot to the dot representing

the tube type.

m and n can be counted off

We can easily calculate the nanotube diameter and chiral angle

for a given (n,m) [cosine rule]

1 2 3 CCa a a

• Can also get Multi-walled

nanotubes (MWNT)

• Concentric cylinders of graphene

separated by 0.35 nm

• Each separate shell tends to have a

different (n,m)

Nanotube Synthesis

carbon + catalyst + energy => nanotubes)

I. Arc vapourisation of a catalyst-containing electrode

II. Laser ablation of a catalyst-containing target

III. Decomposition of vapour phase catalyst/carbon

mixtures at high temperature (a.k.a. CVD - chemical

vapour deposition)

Most commom type of CVD process:

Recently developed commercial process - ‘HiPco’ (High

Pressure Carbon monOxide (CO)).

Proceeds via decomposition of acetylene and Fe(CO)5 under

conditions of high pressure and temperature in the presence

of excess CO.

I. Apparatus for arc vapourisation

synthesis of SWNTs showing

enlarged hollow catalyst-containing

anode.

II. Schematic depiction of laser ablation

apparatus.

III. CVD apparatus for forming

nanotubes via decomposition of

acetylene and Fe(CO)5 using argon as

a carrier gas.

Synthesis apparatus

(A)SWNT imaged by transmission electron microscopy (TEM).

(B)TEM image of a MWNT made up of 10 concentric SWNT cylinders

(C) and (D) TEM and scanning electron microscope (SEM) images of SWNT raw

material.

(E) and (F) TEM images of SWNT bundles. Note the ordered 2D lattice in (F).

(G) and (I) STM images of discrete and bundled SWNTs.

Microscopy

of carbon

nanotubes

SWNT

MWNT

Hata et al, Science 306, 1362 (2004)

How long can they be?

Electronic structure

Nanotubes can be metallic

or semiconducting

depending on (n,m)!!!

If

Where p is an integer, then

the nanotube is metallic: 1/3

tubes metallic

Otherwise the NT is

semiconducting with

bandgap Eg.

Eg depends on diameter, D.

pnm

3

D

aE cc

g

2 2.7 eV

Ouyang M et al. Science 292 (2001) 702.

E-Ef (eV)

• When you apply a force to a material it stretches.

• We usually write the stress, and the extension, as

Mechanical Properties 101

L0

A

F

L0+L

A

A

F

0

L

L

The stiffness, or resistance to stretching, is

given by the Youngs Modulus, Y or E.

The strength, F, is the stress at fracture.

The ductility is the strain at fracture, F

ddY 0.00 0.02 0.04 0.060

2x1010

4x1010

6x1010

F

(P

a)

Y

F

F

FY

For a brittle material:

Y

• How strong are nanotubes?

• We need to work out the force per area required to break

all the bonds around the circumference of a nanotube.

acc

2accSin60

There are N bonds around a circumference

602 Sina

DN

cc

The total force to break the NT is FT:

FCC is the force to break 1 C-C bond CCT NFF

The total stress to break the NT

(strength) is A

NFCCF

A is the area of

the bonds to be

broken

Where d is the

thickness of the

NT wall

DdA

3 3

CC CCF

cc cc

F FD

Dda a d

• What is the force required to break the C-C bond?

When a Force, FCC, stretches a bond by lF just before breaking,

The work done is the bond energy, ECC: Fcccc lFE

/

3

CC FF

cc

E l

a d

F F ccl a

• For a C-C bond, Ecc ~3105J/mol or

510-19J per bond, F~20%

• acc=0.14 nm and d=0.35 nm

This gives our estimate as

F~200 GPa

Cf High Strength Steel Wire: F=2.4 GPa, Y=210 GPa

23

CCF

F cc

E

a d

1000F

F

Y GPa

,

0,

NT FFF

cc NT

Ll

a L

We assume the strain in

one bond is the same as

the overall strain at failure

Yu et al Science, 287, 637 (2000)

Y F

Nanotubes ~ 1000 GPa ~ 100 GPa Carbon fibre 200-700 GPa 1.5-5 GPa

High Strength Steel Wire 210 GPa 2.4 GPa

The mechanical properties can be measured in a modified electron microscope

They are the strongest materials known to man!!!!!!!

The stress strain curves are linear up to fracture indicating a brittle

material. We can measure Y and F. They are close to our estimate.

Current carrying ability

712 10 /NT

lS m

R A

76 10 /Ag S m

However Jmax=21010 A/cm2!!

This is about 100 times better than

copper, silver or gold

This is because electrons in

nanotubes don’t scatter Joule

heating is very low

MWNT

Wei BJ, APL, 2001, 350, 6-14

Optical properties

of nanotubes

The optical properties of SWNTs are

dominated by the spikes in the DOS.

Photons with energy E>0.8 eV (< 1500nm)

can excite electrons from the valence band

to the conduction band

For nanotubes, this means VB and CB for

the same subband

Fermi’s golden rule tells us that the

absorption co-efficient is proportional to the

DOS at the energy the photon brings the

electron to: (E) g(E)

This means the absorption is large for

transitions between spikes

Called S11, S22 etc transitions

S11 S22

Thus the absorption spectrum for one tube would be a

set of spikes.

The lowest energy spike would be at an energy equal

to the energy difference between the first subbands of

the VB and CB

The second lowest energy spike would be at an

energy equal to the energy difference between the

second subbands of the VB and CB etc

S11 S22

HOWEVER, no sample ever has one

tube type.

They always have a range of (n,m)

Thus you get an array of sets of spikes

0.5 1.0 1.5 2.0 2.5 3.0 3.50.00

0.05

0.10

0.15

0.20

Absorb

ance (

au)

Photon energy (eV)

Nanotubes

S11 S22

M11

Red:S11 Green, S22 Blue, M11

What happens to the excited electronic state after an electron is

promoted to the conduction band?

S11 S22

Semiconducting nanotubes can photoluminesce!

For about 1% of the photons absorbed, the electron

and hole recombine by emitting a photon

Before emission of a photon, the electron (and

hole) decay to the bottom (top) of the lowest

subband by emission of phonons.

Thus for any semiconducting SWNT, the emission

energy is the same as the S11 absorption energy

For a set of nanotubes, the emission peaks occur

at the same energy as the S11 absorption peaks,

no matter what energy the sample is excited at.

Photoluminescence Excitation

The Pl intensity is proportional to the absorption co-efficient, , of the emitting

tube type. (more photons absorbed: more photons emitted)

We can measure the Pl intensity at one energy, say 1 eV while scanning the

excitation wavelength from 1-3 eV, for example.

The Pl will scale in proportion to the absorbance.

This will give a PLE, or excitation spectrum. For a simple material:

However, if a material has 2 or more emitting components, we can detect at

the emission energy for one component only. Scanning the excitation

wavelength gives us the absorbance spectrum of that component alone.

We can measure the absorption spectrum of an individual tube type in a

mixture!

PLEI

500 1000 15000.0

0.2

0.4

Ab

so

rba

nce

Wavelength [nm]

PLE spectrum of a nanotube sample gives the absorption coefficient of 1

nanotube only

In fact there is an even better way to pick out both the absorption and

emission of individual nanotubes.

Make photoluminescence measurements on a mixture of nanotubes, usually

dispersed using surfactants.

The difference is that instead of measuring one PL spectrum with a fixed

excitation energy, say 2 eV, many spectra are measured with many

excitation energies ie 1.05 eV, 1.1 eV, 1.15 eV……..2.9 eV, 2.95 eV, 3.0 eV

Absorption

PLE S22

S33

The beauty of this is that we have data on:

1 The emission intensity at all emission energies for a given excitation

energy (PL)

2 The emission intensity for one emission energy for all excitation energy

(PLE)

We have this data for all tubes in the sample. This data is generally plotted

as a Pl contour map. Each peak represents PL from a different nanotube

type - Tells you what semiconducting tubes you have in a sample

Can also plot as a 3D intensity profile

Excitation

Wavelength

(nm) Emission energy (eV)

Electronic Properties of 1D structures

Particle in a 1-D Box (infinite square well)

Nanoscience is about systems where one or more dimensions are confined to the nanoscale

The electrons in a material are confined by the strong attraction to the atomic nuclei.

In some materials electron shells are unfilled you can have (nearly) free electrons metals

However the electrons are still trapped within the material

Model this by a potential energy of zero in the material and infinity outside Infinite square well

This confinement results in the formation of quantised energy levels.

Solve Schroedinger equation for 1-D

infinite square well, length L.

Inside the well the potential energy: V(x)=0

x L 0

V

2 2

22V E

m x

2 2

22E

m x

Trial Solution:

What is k? Use Boundary Conditions

ASin kx

2 2 2 2 2 2 2

2 22 2 8n

k n n hE

m m L mL

2n

n xSin

L L

kL n

• What about real systems?

• How about Hexene (C6H8)

• Each C has 3 valance electrons ( electrons) involved in bonding leaving 1 free electron ( electron)

• Treat the electrons (1 from each C) as completely free ie a free electron gas.

• These then fill up the

energy levels (2 per level, )

Alkenes:

0 0.2 0.4 0.6 0.8 1.0

9

4

1

x /L

E (

h2/8

mL

2)

, Ef

C

H

C

H

C

H

C

H

C

H

C

H

H H

a

C

H

C

H

C

H

C

H

C

H

C

H

H H

a

Alkenes in general

• What about alkenes in general? H[CH]NH

There are N free electrons

The length is L=(N-1)a

The Fermi Energy is the energy of the top most filled level

That is: En when n=N/2 (1 e- per C, 2 e- per level)

If N big then: independent of N

C

H

C

H

C

H

C

H

C

H

C

H

H H

a

N=6

2 2 2 2

228 8 1n

n h n hE

mL m N a

2 2 2 2

2 22

( / 2)

32 18 1F

N h N hE

ma Nm N a

2

232F

hE

ma

We can work out the Fermi energy for a long alkene (polyacetylene)

a0.15 nm Ef 4.18 eV above the bottom

of the well

If all the electrons are below Ef 4.18 eV for any chain length, then the

energy levels must get closer together for larger and larger N.

Lets work out the spacing between 2 adjacent levels

2

232F

hE

ma

2 2 2 2

228 8 1n

n h n hE

mL m N a

2 2 22 2

1 2 2 2

1 2 1

8 1 8 1 8 1n n

n h n hn hE E E

m N a m N a m N a

• The spacing increases with n

• Lets work out a representative spacing

ie the energy difference between Ef and the level below.

The Fermi energy is when n=N/2 or N=2n

This gets smaller as N gets bigger!!!!!

The levels get closer together as N !!!!!!!!

2

2

2 1

8 1

n hE

m N a

2 2

2 2

1

8 18 1

N h hE

ma Nm N a

16.7

1

eVE

N

0

5

10

15

0 10 20 30 40 50

N

E

(eV

)

• At some point the levels are so close together it is hard to tell them apart

• They have become a band.

• When does this happen?

• We could imagine that this is when the spacing between the top 2 levels is kT (cf Kubo criterion)

• Then we have or

• At room temp this is N670 or L 670a 100 nm

This suggests that we see energy bands (like in bulk materials) for 1D objects bigger than 100nm

BUT

Quantised energy levels for objects smaller than 100nm

2

28 1

hE kT

ma N

2

21

8

hN N

ma kT

Density of States

• Thus: when we have 1D objects longer than 100 nm we need to

consider energy bands and so think in terms of Density of States.

Total number of states per

unit energy per unit

length including spin

degeneracy

• Where is the Fermi Energy?

We calculate it in exactly the

same way as we would do in 3D

0 1 2 3 4 5

0

5

10

15

20

25

giD

(e

V-1n

m-1

)

E (eV)

E

mEg D

12)(1

• The available states are filled until all electrons are

allocated.

• The energy of the highest state is then the Fermi

Energy.

• The available states that get filled are those with k

values between -kf and +kf.

• We know that

• Where each state is indexed by n

• This means, the amount of k space per state is

2xk L n

2xk

L

2x

nk

L

• So the total number of electrons, N, is the total amount of k space

divided by the amount of k space per electron (times 2 for spin)

Or

• kf is the maximum value of k for an (ground state) electron in our

material (at 0K)

• But remember, for a wave

• Thus is the smallest occupied (at 0K)

electron wavelength

• From above

• If each atom contributes 1 electron then

• And we have

2 2

2 /

fkN

L

2f

Nk

L

2 /k 2 /f fk

4 /f L N

/ 1/N L a

4f a

4a

Back to the Fermi Energy

• As we saw, when the potential energy, V=0

Thus

• As before

2 2

2

kE

m

2 2

2

f

f

kE

m

22

2 2f

NE

m L

22

32f

h NE

m L

2

232f

hE

ma

0 1 2 3 4 5

0

5

10

15

20

25

giD

(e

V-1n

m-1

)

E (eV)

Ef

N/L is the linear free electron density

If each atom contributes 1 free

electron, then N/L=1/a, giving

• Another way to find Ef is to remember that

• This follows from the definition of g1D(E) as the number

of states per unit energy per unit length

• Integrating gives

dEEgL

NfE

D )(0

1

dEEm

E

mdE

L

Nff EE

0

2/1

0

212

fEm

L

N2

2

Rearranging and remembering that

N/L=1/a gives 2

232f

hE

ma

• What about finite temperatures?

• At T>0 it is no longer the case

that all electrons are in states

below Ef and all states above Ef

are empty.

• Thermal occupation

• Then the number of electrons per

unit length in states between E

and E+dE is

• Where

Fermi-Dirac function

1( ) ( )e DN f E g E dE

0 1 2 3 4 50.01

0.1

1

10

f(E

)giD

(E)

(eV

-1n

m-1

)

E (eV)

Ef

1/exp

1)(

kTEEEf

f

4.1 4.2 4.3 0.00

0.05

0.10

0.15

0.20

0.25

f(E

)g iD

(E)

( eV

-1 nm

-1 )

E (eV)

E f

g iD (E)

Peierls Distortion!!!! • Beware!!!!!!

• The example of a long alkene

(polyacetylene) being a metal is WRONG!!

• This is because 1D metals tend to be

unstable

• They can lower their energy by a process

known as bond alternation.

• The array of single bonds distorts into and

array of single and double bonds, doubling

its period.

• There is no longer a free electron per carbon

as they have gone into double bonds

resulting in the formation of a bandgap at Ef

• The extra strain energy is more than

compensated by the energy reduction due to

the formation of the gap

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

C

H

k

E

kf

Ef

k

E

kf

Ef Eg

Everything has some size in the lateral dimensions !!!!!!!!

Even a carbon nanotube has a width (diameter) of at least 0.5nm.

These are known as Quasi 1D materials

How do we model the electronic structure?

We can model a quasi 1D material as a very long rod with a rectangular

cross section

There are virtually no ID materials!!!!!

x

z y

Ly Lx

Lz

V(x)=0 for 0xLx, 0yLy, 0zLz

V = everywhere else set up Schroedinger Eqn

Solving the Schroedinger equation gives the energies of any free

electrons (ie not tightly bound to atoms)

2 2 2 2

2 2 22E

m x y z

( , , ) ( ) ( ) ( )x y zx y z x y z

8

x y z x y z

x y z

Sin k x Sin k y Sin k zL L L

2 2 2 2 2 2

2 2 28 8 8

x y z

x y z

n h m h p hE E E E

mL mL mL

We generally solve this by separation of variables where we assume

the wavefunction is the product of three wavefunctions each

dependent only on x,y or z.

Then the Schroedinger eqn becomes

In 3D the Schroedinger

Eqn, when V is zero, is

Electron

energies:

Remember, in our

quasi 1D material. Lx

is VERY big so the

spacings between the

“x” energy levels are

very small.

2 2 2 2 2 2

2 2 28 8 8

x y z

x y z

n h m h p hE E E E

mL mL mL

As we will see, this

allows us to explain

the presence of

quantum subbands

Model a nanotube as a hollow cylinder with infinitesimally thin

walls

In cylindrical polar co-ords (z=z, x=rCos and y=rSin)

2 2 2 2

2 2 22E

m x y z

2 2 2 2

2 2 2 2

1 1

2E

m r r r r z

2 2 2

2 2 2

1

2E

m R z

r

z R

Lz

• But here r=R = constant so

derivatives w.r.t. r disappear

• As before ( , ) ( ) ( )zz z

2 2

2 2 2 2

1 2( ) ( ) ( ) ( )z z

mEz z

R z

2 2

2 2 2 2

1 1 1 2z

z

mE

R z

As before we separate the variables

2 2

2

2 2

22

2 2

2

2

l

z zz z z

mR Em

mEk

z

NB

We have seen this equation before:

The solution is

zSinkL

z

z

z

2

EEE z

Boundary

conditions give

2

22

8

/

z

z

zz

mL

hnE

Lnk

• What about

• This is slightly different

• Unlike z there is no special value of where is zero

• One solution to the equation above is

(1 e- moving clockwise +

1 e- moving anti-clockwise)

• To find ml we use “periodic bound conditions”. Ie the wavefunction must

be the same if you around the circumference by 1 revolution: ie

2

2

2

2

2 2lm

EmR

ll imim

BeAe

)2()(

22 llll imimimim

BeAeBeAe

22 llllll imimimimimimeBeeAeBeAe

This is only true if ml=0, 1, 2, 3….. 12

lime

• What about the energy?

• We can understand this another way

• The circumference of the tube must be equal to an integral number

of wavelengths of the wavefunction or there will be destructive

interference effects.

2

2

22

EmR

ml 22

22

2

22

82 Rm

mh

mR

mE ll

lmR 2 But for a wave

2k

lmRk

2

2222

22 mR

m

m

kE l

Same as ABOVE

2

22

2

222

,22 mR

m

mL

nEEE l

z

zmn l

• The total energy is

• Remember for a nanotube, Lz is very long

• This means that the energy levels are very close together

2

222

2

22

2

222

2

22

,2222 z

l

z

lmn

L

n

mmR

m

mL

n

mR

mE

l

kz

2

m

kkE z

mzm ll 2)(

22

This means that for every ml we

have a band of energy levels

associated with different kzs

The minimum energy of each band

is then when kz = 0 ie ml

•As Lz gets very big, n/Lz starts to look like a

continuous function

•Ie kz becomes continuous Band

•However R is small so ml2/R2 is big

quantized levels

• This means that there is an energy band associated with every one

of the quantised energy levels for the direction

• The lowest level of each band is ml

• These are known as Quantum subbands.

• Before we said that

• Now, for quasi 1D materials we need to say

• Ie the DOS of the subband scales with the energy above the

associated quantised level.

1( ) ( , )

l

T D l

m

g E g m E•The total density is states is just the sum

over all the subbands (each with different

ml.

For E>ml

For E< ml, g1D=0, for that subband

E

mEg D

12)(1

lm

DE

mEg

12)(1

m

kkE z

mzm ll 2)(

22

2 2

22l

lm

m

mR

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0

100

200

300

400

500

ml=4ml=3

ml=2

gT(e

V-1nm

-1)

E (eV)

ml=1Calculated low

energy DOS for

a R=1 nm

nanotube

2

22

2mR

mlml

2

22

42

4

mR

2

22

32

3

mR

2

22

22

2

mR

2

22

12

1

mR

1( ) ( , )l

T D l

m

g E g m E

The Fermi energy can be found from dEEgN

fE

T )(0

Spikes are when E=ml

and are called van

Hove Singularities

lm

DE

mEg

12)(1

• How accurate is this?

•Our model suggests all nanotubes

are metallic.

•We have used a very simple particle

in a box model.

•We have not even considered atomic

structure

It is possible to do more sophisticated

calculations, taking into account the

way the sheet of graphene is rolled

up.

At low energy the DOS looks very

similar to ours.

However there are big differences

close to Ef.

NT can be metals or

semiconductors!!!!!

Ouyang M et al. Science 292 (2001) 702.

• Does this match experiment?

• We can do Scanning Tunnelling Microscopy measurements to image the nanotube.

• Here an atomically sharp tip is brought close to the sample, A voltage is applied. Electrons tunnel between sample and tip.

• As the tip is moved across the sample more electrons tunnel from regions of high electron density (orbitals)

• Generate image

• From the image we can tell that this is an (8,8) nanotube

• We can measure tunnelling current for different volatages.

• dI/dVg(E)

• We see spikes!!!!

• Experiment matches Theory!!!!!

Theory

Exp

Ouyang M et al. Science

292 (2001) 702.

Optical absorption spectra also show

sharp peaks

• Optical transitions (light absorption) can

occur between conduction and valence

band states of semiconducting

nanotubes.

• Fermi’s golden rule says optical

absorbance is proportional to the DOS

In an experimental

spectrum we see many

many sharp peaks,

each due to transitions

between pairs of DOS

peaks in a wide range

of nanotube types.

• Shine in a high energy (low

wavelength) photon and an

electron is promoted to the

conduction band.

• This then gives off phonons to

decay to the bottom of the CB

before decaying back to the

valence band by emitting a

photon.

• We can measure the

photoluminescence (PL)

spectrum which consists of

sharp spikes because of the

spike in the DOS at the

bandedge.

• Measuring the PL spectrum for

many incident wavelengths

and you build up a contour

map.

• Spikes in this map appear for

individual (n,m) types.

Excitation

Emission

800 900 1000 1100 1200 1300 1400

0

50000

100000

150000

200000

250000

300000

Ph

oto

lum

inesce

nce

in

ten

sity

Emission wavelength (nm)

Excite=655 nm

Photoluminescence also shows

spikes in the DOS

Nanoscale Electrical Transport

• Imagine a 1D object, of length L (L<100 nm).

• It has a set of energy levels given by

• Imagine two large metal contacts are brought close to each end. Each

contains a band of states filled up to Ef

• If they are brought into intimate (atomic) contact, the barriers between

them disappear and electrons can flow from material to material

• This happens until all the Fermi energies coincide

2

222

2mL

nEn

Ef1 Ef2

n=1

n=2

n=3

The case for

L=5a

(N=6)

What happens if a voltage, V, is applied between the metal

contacts?

• The Fermi energies of the contacts separate by an amount

• What happens then?

• This means that the 1D channel is in contact with 2 materials with 2

different Fermi energies.

• The Fermi energies want to equalise to lower the total energy.

• They generally do this by electrons flowing towards lower energy till

the Fermi energies are equalised

eVEE ff 21

Ef1

Ef2

n=1

n=2

n=3 eV

•This means electrons move from the top energy level in the channel to

material 2 to reduce their energy

•But electrons can also reduce their energy by moving from material 1 to

the channel

•However, this does not equalise the energies because the Fermi energies

are held apart by the applied voltage

This results in a net electron flow A CURRENT

This will happen so long as there is a channel energy level between Ef1and

Ef2

Ef1

Ef2

n=1

n=2

n=3 eV

•How do we calculate current flow?

•Contact 1 wants to give the channel electrons until the probability of

occupancy in the channel is equal to the probability of occupancy for an

energy level in the metal at the same energy as the channel ().

•Contact 2 wants to take electrons from the channel until the probability of

occupancy in the channel is equal to the probability of occupancy for an

energy level in contact 2 at the same energy as the channel ().

•Remember the probability of occupancy is given by the Fermi Function

•The thing is, the two metal contacts have different Efs and so different Fermi

functions and so different occupancies at a given energy!!!!!!!

Each metal contact is trying to set the channel occupancy at a different level!!

1/exp

1)(

1

1

kTE

ff

1/exp

1)(

2

2

kTE

ff

1/exp

1)(

kTEEEf

f

Channel

energy =

Ef1

Ef2 e.g.

f()=0.87

if eV=4kT

eV

f(E)

f(E)

e.g.

f()=0.13

if eV=4kT

• This means that contact 1 wants to pump electrons into the channel

• Channel 2 wants to take them out

• Contact 1 wants the probability of occupancy of the upper level of

the channel to be f1() and so the population to be 2f1() (spin up

and spin down)

• Contact 2 wants the probability of occupancy of the upper level of

the channel to be f2() and so the population to be 2f2() (spin up

and spin down)

• Say the average population is 2fCh

• We might imagine the electron flow into the channel across the left

junction to be

• And the electron flow into the channel across the right junction to be

1 12 ( ) 2 ChI f f

2 22 ( ) 2 ChI f f

• Putting in proportionality constants

• At steady state, the net inflow into the channel will be zero, and so

• Solving for fCh, gives

• The total current is given by

11 1

22 2

2

2

Ch

Ch

eI f f

eI f f

1 2 0I I 1 1 2 2

1 2

Ch

f ff

1 2I I I

1 1 1 2 2 1 2

1 1 2

1 2 1 2

2 2( ) ( )

e f f eI f f f

It will become clear later why we write the

proportionality constants as and 1 /e 2 /e

Rem: depends on the voltage 1 2( ) ( )f f

• This equation is interesting

• It means that current will only flow IF f1() f2()

• This is important. If the energy level in the channel closest to Ef1 and Ef2

is very high in energy (>> Ef1, Ef2), then

1 2

1 2

1 2

2( ) ( )

eI f f

1 2( ) ( ) 0f f

However, if it is very low in energy (<<

Ef1, Ef2), then

1 2( ) ( ) 1f f

In both these cases, I0

This means that I is only appreciable

for energy levels between Ef1 and Ef2

NB the state can be either filled or empty and

still pass current

• What are 1 and 2?

• The currents flowing into the channel from each

contact are

• This means the number of electrons per second

entering from the left is

• If the probability of occupancy of the level in metal 1

at energy is high (f1()1) and the channel level at

is initially empty (fCh=0)

Then the maximum initial rate of electron flow into the

channel will be

• The minimum time per electron to enter the channel

will be the inverse of the rate

11 1

22 2

2

2

Ch

Ch

eI f f

eI f f

1 11

2Ch

If f

e

1 12I

e

1

12

• Similarly we can show that the rate of escape of an electron to

the right hand contact is

• The min time taken to escape is then approximately

• Then 1 or 2 can be considered the lifetime of the electron in

the channel.

• If 1 and 2 are big, then electrons can get in and out easily:

They don’t stay long in the channel

• But the Uncertainty principle says that if the lifetime is finite

then there must be a spread in the energy of the level

• But this means that

• That means connection of the channel to contact 2 broadens

the energy level to a width of (at least) 2

• Connection to both electrodes broadens the level to a width of

2 22I

e

2

22

22

E

2

2 2

1 1

2 2 / 2E

1 2

• Remember

• If eV=Ef1-Ef2 is greater than a few kT then

f1()1 and f2()0 In this case

1 2

1 2

1 2

2( ) ( )

eI f f

1 2( ) ( ) 1f f

If the contacts are identical then

Under these circumstances, the current is:

This suggests that the better the

connection to the electrodes (higher 1),

the higher the current, independent of

voltage

1 2

1eI

• This couldn’t be true because we know IV !!

• What have we missed?

• Remember we said earlier that I is only appreciable for energy levels

between Ef1 and Ef2.

• This is the key

Ef1

Ef2

If we increase 1 then we also increase the

width of the energy level.

This means more of the level will lie outside

the bounds set by Ef1 and Ef2 and so will not

contribute to the current.

This means we need to calculate the current,

realising it only flows through the part of the

broadened energy level that lies between Ef1

and Ef2.

• To do this we need to express the broadening of the line

in terms of Density of states

• The exact expression for the current is

• However it is easier to use an approximation

• Previously we expressed the current as

• The correct current is approximately this current times the fraction of

DOS between Ef1 and Ef2

• We can assume the line broadens such that we can express the

density of states as a Lorentzian function

Where is the FWHM and is the mean energy

2 2

2( )

( ) ( / 2)D E

E

1 21 2

1 2

2( ) ( ) ( )

eI f E f E D E dE

1 2

1 2

1 2

2( ) ( )

eI f f

• Then

• Remember if eV>>kT then

• Taking in which case

• Then

• If the channel is well connected to the electrodes then 1 and 2 are big and the level is broad

1

21 21 2

1 2

( )2

( ) ( )

( )

f

f

E

E

D E dEe

I f f

D E dE

1 2( ) ( ) 1f f

2 2

2( )

( ) ( / 2)D E

E

( ) 1D E dE

1

2

1 2

2 2

1 2

2 2

( ) ( / 2)

f

f

E

E

eI dE

E

• If the channel is well connected to the electrodes then 1

and 2 are big and the level is broad

1 and 2 small 1 and 2 big

• When 1 and 2 are small we can approximate the DOS

as constant between Ef1 and Ef2.

Ef1

Ef2

Ef1

Ef2

• We will use the value of D(E) at E= while realising that this is not

the best approximation in the world

• Then

• And so

Where we note that

If the contacts are identical such that and we remember

that

2( )D

1 2

1 2

1 2 1 2

2 2

( )f f

eI E E

1 2

1 2

1 2f fE E eV

22eI V

h

Ohms law says

Where G is the conductance

1I V GV

R

Thus nanosized objects have

quantised conductance

2

0

2eG

h

NB: This is exact as all

the approximations

cancel!!!

• This means that a nanosized object has a limit to the

amount of current it can carry even in the absence of

scattering!!!

• Also the conductance is independent of sample length!!

(unlike Ohms law)

• Is this true? Experiment says so

• This is known as ballistic transport

• What about a long 1D object where the energy levels are

so close together that there is effectively a band?

• We can think as the contacts as reservoirs (1) and acceptors (2) of

charge

• Of those electrons in contact 1 travelling right-wards as many as can

be accommodated will enter the nanotube

• By accommodated we mean they can only go where there are

states to take them

• In addition electrode 2 takes as many electrons as possible from the

nanotube

1 2 - +

• In contact 1 (at T=0) there are only electrons below Ef1.

• In the nanotube there are only available states above the

nanotube Fermi energy, Ef,NT

• Only half of these states are associated with electrons moving

to the right, (ie states with k>0)

• Thus half the states in the nanotube between Ef1 and Ef,NT get

filled from contact 1 with right moving electrons

• Contact 2 has only available states above Ef2.

• Thus contact 2 can take electrons from all states in the

nanotube above Ef2 filled by electrons moving to the right.

• That is half the states between Ef1 and Ef2.

• Only these right moving electrons can enter contact 2

E

x

Electrode 1

Filled states

Electrode 2

Filled states

Electrode 1

Empty states

Electrode 2

Empty states

Nanotube

Filled states

Nanotube

Empty states

Electrons with k>0

Electrons with k>0

• In 3D the current density is given by

Where n is the carrier density (m-3) and v is the velocity

• In 1D the equivalent equation is for the current

Where n is the number of carriers per unit length.

Here n is the number of states per unit length, from which

contact 2 can take electrons.

That is half (right travelling) those states between Ef1 and

Ef2.

Remember

That means n is just half the number states in the energy

interval eV above the energy Ef2.

J nev

I nev

1 2f fE E eV

• Imagine a very small voltage, dV is applied

• Remember the density of states is the number of states per unit energy per

unit length.

• Then, the number of carriers, dn is

• Which means the current is

• Remember that

• An electron with kinetic energy E has a velocity given by

• Therefore the DOS can be written as a function of the electron velocity

1 ( )

2

D fg Edn edV

E

mEg D

12)(1

1 ( )

2

Df f

g EdI ev dn ev edV

2

2

f

f

mvE

1

4( )D f

f

g vhv

The vf’s cancel giving as before 22dI eG

dV h

• This means that a perfectly conducting one-dimensional channel has a finite conductance that depends only on fundamental constants, e and h.

• If more than 1 energy level (or more than 1 subband for a very long object) lay between Ef1 and Ef2 than we would get a contribution of 2e2/h from each channel (ie each level or subband).

• These act like conductors in parallel and so the conductance's add.

• Thus if n channels have energy between Ef1 and Ef2 then the total conductance is

• The conductance is quantised

22ne /hTG

What if the 1D object is not perfectly conducting??

• If the object is not perfectly conducting the conductance can be lower than 2e2/h

• What happens if there is a defect in the channel?

• Lets go back to our model 1D system, our chain of C atoms

• What if we replace one of them by a Si atom?

• An electron in the vicinity if this Si atom will experience a different local potential energy

• This will scatter electrons (see McMurry p68 or any other QM text)

• These scattered electrons are scattered backwards and so are not transmitted through the channel

• The transmission probability, T <1

• This allows us to write the conductance as

Where T is the probability of transmission

(transmission co-efficient) of the electron

through the channel

If there are n channels between Ef1 and Ef2 then

This is the famous Landauer formula.

Because the resistance is the inverse of the

conductance we can write the channel

resistance as (considering only 1 level, n=1)

This can be rewritten as

22eG T

h

22n

n

eG T

h

2

1

2

hR

e T

2

(1 )

2

h T TR

e T

• The Resistance is

• Remember, (1-T)=, the reflection co-efficient.

• This means

• So R has two components, one constant and one

controlled by properties of the channel (T, )

• These components are interpreted as a contact resistance,

RC and the intrinsic resistance of the channel itself, RI.

• The latter can be considered as a resistance, intrinsic

to the material

2

(1 )

2

h T TR

e T

2 2 22 2 2C I

h T h hR R R

e T e e T

• The contact resistance is the resistance associated with

electrons getting in and out of the material and is fixed

• As we have just seen the channel resistance is controlled by

the properties of the channel, ie the probability of

transmission

• Transmission can be impeded if electrons can be scattered.

• Electrons can be scattered elastically by the presence of

potential wells or barriers in the channel

• For a single potential barrier (or well) the transmission can

easily be calculated (seeMcMurray p68, Scattering by a 1D

square well)

• Things are more interesting when there are 2 potential

barriers.

• Imagine if our chain of C atoms has 2 silicon atoms and hence 2 potential barriers

• The presence of these barriers can cause scattering

• We can model these as square barriers

• Now we need to work out the transmission probability for this system

Resonant Tunnelling

V(x)

x

0

1 2

0 L

• Imagine an electron incident from the left.

• It is described by a wavefunction

• When it encounters barrier 1 part of its amplitude is

transmitted and part is reflected

• The fraction transmitted is given by

• The fraction reflected is given by

• These quantities are complex as there may be a phase

change on reflection/transmission. In addition they

depend on the physical properties of the barrier.

ikxAe

1

1 1

it t e

1

1 1

ir r e

V(x)

x 0

1 2 ikxAe

ikx

t Be1ikx

r Ce1

0 L

• The transmitted wave encounters the second barrier

• When it encounters barrier 2 part of its amplitude is

transmitted and part is reflected

• The fraction transmitted is given by

• The fraction reflected is given by

• However, the reflected bit can be re-reflected back off

barrier 1 (see red arrow)!!! This adds a further

contribution to t1.

2

2 2

it t e

2

2 2

ir r e

V(x)

x 0

1 2

ikx

t Be1 ikx

t De2

ikx

r Ce2

0 L

3

ikx

r Ee

• Remember, what we want is the transmission probability

• This is the probability that an electron starting on the left gets to the

right

• Now we use the expressions for reflected and transmitted fractions of

the wavefunction to work out D

• D is the fraction of B transmitted

• C is the fraction of B reflected

• B is the fraction of A transmitted + the fraction of C re-reflected (ie E).

• We also need to take into account the extra phase accumulated by

r2 compared to t1 as (at a given position x) it has travelled an extra

distance of 2(L-x) and hence has phase greater by 2k(L-x)

2

2

A

DT

kLiikxkLiikxxikLikikxxLxik eeeee 2222)(2

This means

CrAtB

BerC

BtD

kLi

11

2

2

2

Aerr

ttD

kLi2

21

21

1

Putting these together gives

V(x)

x 0

1 2

ikx

t Be1 ikx

t De2

ikx

r Ce2

0 L

3

ikx

r Ee

Aerr

ttD

kLi2

21

21

1 But remember t1, t2, r1, r2

are complex

kLikLi errerrrr

tt

A

DT

2

21

2*

2

*

1

2

2

2

1

2

2

2

1

2

2

1

1

1 1

it t e

1

1 1

ir r e

2

2 2

it t e

2

2 2

ir r e

Cosrrrr

ttT

21

2

2

2

1

2

2

2

1

21212 kL

• This means the transmission is a maximum when Cos=1

• That means when

• Imagine for a moment there is no phase change on reflection ie 1 and 2 are zero

• Then the transmission is a max when

• This is when

• This is when we have constructive interference of all the reflected waves

• When the transmission is maximised the electrons are said to tunnel resonantly through the barrier

Cosrrrr

ttT

21

2

2

2

1

2

2

2

1

21

2122 kLn

kLn

Ln 2

/2k

• When the barriers are identical then t1= t2= t and r1= r2= r

• In this case, at resonance

• But remember the probability is the amplitude squared and

the electron must be reflected or transmitted

Therefore and hence, T=1

When does resonance occur? When k=n/L

But remember Therefore resonance occurs for

electron energies given by:

2241 rtT

122 rt

m

kE

2

22

2

222

Re2mL

nE s

These are the same as the energy levels of

the potential well defined by the barriers!!

This is a general situation: Tunnelling through a confined electron

system is always enhanced for energies matching the bound electron

energies

Incoherent addition of waves: Ohms Law

• The previous analysis relied upon us keeping track of the phase of the scattered electron. This added an ei2kL term

• However, in real materials the electrons can be scattered inelastically by phonons.

• This results in a change in their energy and hence in their momentum

• Thus, every time an electron is scattered by a phonon, k changes, which means the phase changes.

• If the phase changes unpredictably we need to treat the electrons as incoherent.

• This happens even in the absence of defects as long as T>0

• As electrons travel through a 1D material there is more and

more chance they will be backscattered by interaction with a

phonon, defect etc

• Thus the longer the channel is, the lower the transmission!

• To see this, imagine a long channel, which we consider as

consisting of 2 sections, 1 and 2

• We give each section a probability of transmission, T and a

probability of reflection, R.

• The probability that the electron is transmitted though the

first and then the second section at the first attempt is T1T2

T2=1-2 T1=1-1

T1T2

T12 1T2

T1 2 1 2 1T2

• However an electron that is transmitted by section 1 but

reflected by section 2 can the be reflected back towards the

right by section 1 and then transmitted by section 2. (see red

arrows)

• The probability of this is T1R2R1T2

• The same thing can happen with 4 reflections (blue arrows)

with probability T1R2R1R2R1T2

• In fact the electron can be transmitted with any even number

of reflections

• The total transmission is

• This is just a well known series

• NB we don’t have to account for phase as we are considering

incoherent electrons

.....12

212121 TTT

21

21

1

TTT

• Remember that

• Rearranging gives

Remember that the transmission must be a function of length

ie T=T(L)

If the length of section 1 is L1, of section 2 is L2 and the total

length is L1+L2.

Then

The following function fits perfectly

T 1

1111

21

TTT

2121

21

TTTT

TTT

1)(

1

)(

1

)(

1

221121

LTLTLLT

( )T LL

Is a constant of order to the mean free path

and represents the length where the

probability of transmission is 0.5

• Remember the conductance per channel is

• We can now write the conductance as a function of length

• The resistance is then

• Again this is the sum of contact and intrinsic resistances

• But can be rearranged to give

• This gives the resistance as

22eG T

h

Lh

eLG

22)(

L

e

h

e

hL

e

hLR

222 222)(

LT

TT

TL

1

Te

h

e

hLR

22 22)(

Which is what we had before.

C IR R R

What about Ohms Law?

• This says, for a macroscale 3d conductor

• So far we have an expression for the

intrinsic resistance of a 1D material (per

channel)

A

LR

L

e

hLRI 22

)(

L

e

h

nLR TI 2,

2

1)(

However, remember we said that if you has a quasi 1D material, each

subband near the Fermi energy would contribute to the conductance.

The bigger the material, the more subbands contribute..

•If there are n subbands and they all contribute equally, then we

can think of them as resistors in parallel, giving a total intrinsic

resistance, RI,T

,

1 1 1 1......

I T I I I I

n

R R R R R

• How many subbands are there?

• Only the subbands or modes in the energy range E where

contribute to the current.

• For a rectangular cross section material of width, a, and height, a, the

subbands are at energy

2n a cA

A

L

ce

hLR TI

2,2

)(

Where c is a

constant, A is

area. Therefore

nece

h 1

2 2

This is Ohms law,

where is the

resistivity

2 22 2

2( )

8x z

hE n n

ma

2

2 2

1

1 8

2 ( 1)f ff f x z

N ma

E E E h n n

2

2 2

8

2 ( 1)f fx z

N man E pkT

E h n n

Close to Ef, the number of bands per unit energy is N/E

The number of subbands in the range E is then

nxf and nxf represent the bands near Ef

1 2( ) ( ) 0f f

P~10

Applications

1 Nano electronics: Beating Moores Law

It will be difficult to keep making Si transistors smaller and

smaller.

Need new technology

Transistor feature size

Carbon Nanotube Transistors

•Individual NTs can be laid directly

onto contacts via nano-

manipulation and their transport

characteristics studied

(A) Output and (B) transfer characteristics of a

CNFET with cobalt electrodes See Ouyang, Huang and Lieber,

Accts. Chem. Res. 35, 1018-1024 (2002)

Drain

VDS

VGS

• Between 2 metals, the

energy profile looks

approximately like this

Nanotube applications:

Field Emission

Vacuum level

Ef Ef

• However apply a

voltage, V, and the

picture changes

• In the field between the

metals the electron

potential energy, VPE(x)

is

(blue line)

• This allows electrons

to tunnel out of the

left hand metal

• Field emission

( )PEV x eEx

: Workfunction

Ef

Ef

V

- +

Vacuum

level

Vacuum level

eV

e-

x

VPE

Remember, the potential energy of an e- between nanotube and metal

is

Therefore

What if you replace the left metal with a nanotube?

2

0

1( )

4

qE r

rWhere q is the charge on the sphere and r is

the distance from the centre of the sphere

( )PEV x eEx

Gauss’ Law says

that the Electric field

in the vicinity of the

surface of a charged

sphere (radius R) is

given by

( )PEdV xeE

dx

Far from the nanotube, E=V/L

Ef

Ef

V

- +

Vacuum

level

Vacuum level

eV

e-

x

VPE

L

However, close to the

nanotube,

is very BIG, because r is

very small

2

0

( ) 1

4

PEdV x qe

dx r

( )PEdV xeE

dx

( )PEdV x Ve

dx L

[Here q=CV]

As for a

normal metal

Makes it very easy

for electrons to

tunnel out

Transparent Electrode

So how do the CNT FEDs work ?

Samsung CNT display

Electrons are

emitted by

nanotubes grown on

a metal base

Current controlled

by a gate

Emitted electrons hit

phosphor-coated

transparent

electrode (ITO)

Can make very

bright displays

Probably most

realistic short term

application of NTs

Composites

• Nanotubes are very strong and stiff (see page 18)

• Add them to a plastic to form a composite. It should be stronger

and stiffer than the base plastic

• Nothing new, composites have been around for 4000 years

• Never before with such strong fillers however

F-111 Fighter Bomber Mud Brick Wall in N’gev Desert

Straw Carbon Fibres

• How strong and how stiff?

• The strength and stiffness of the composite are found from

the weighted average of the plastic and nanotube properties

• Say nanotubes are added to a plastic such that the nanotube

volume fraction is Vf.

• The plastic volume fraction is then 1-Vf.

• The weighted averages for strength and stiffness are:

1

1

Comp NT f Poly f

Comp NT f Poly f

V V

Y Y V Y V

Typically:

YNT=1000GPa, YPoly=1GPa

NT=50GPa, Poly=0.05GPa

This gives scope for large reinforcement at low nanotube

volume fraction

0 1 2 3 4 5 6 70

100

200

300

400

Str

ess, ,

(MP

a)

Strain, , (%)

0.6%

0.3%

0.15%

0.08%

Plastic

Vf

Mechanical properties of composites for different Vf

0.000 0.002 0.004 0.0060.0

0.1

0.2

0.3

0.4

0.5

C

omp

(GPa

)

Fractional volume Vf

0.000 0.002 0.004 0.0060

2

4

6

8

10

YC

om

p (

GP

a)

Volume Fraction, Vf

4 increase in both strength

and stiffness with less than

1% nanotubes

Coleman et al, Adv. Func. Mat. 14, 8, 791-798 (2004)

Nanotube

reinforced

tennis

racquet….

http://www.tennis-

warehouse.com/descpageRCB

AB-BPDP.html