Lecture 5

Tunneling

An electron of such an energy will never appear here!

classically

Ekin= 1 eV

0 V -2 V x

Potential barriers and tunneling

According to Newtonian mechanics, if the total energy is E, a particle that is on the left side of the barrier can go no farther than x=0. If the total energy is greater than U0, the particle can pass the barrier.

Tunneling – quantum approach

Schroedinger eq. for region x>L

EUdx

dm 02

22

2

)(2

022

2

EUm

dx

d

Solution: xAex )(

Potential barriers and tunneling

)(2

)(2

022

022 EU

mAeEU

meA xx

Two solutions: )(2

021 EUm

or )(2

022 EUm

Normalization condition: 1)(0

dxx

Solution: xAex 2)(

The probability to find a particle in the region II within

xxEUm

Axpr

002

20 )(

22exp)(

x

Potential barriers and tunneling

xxEUm

Axpr

002

20 )(

22exp)(

xAex 2)(

Potential barriers and tunneling

example

Let electrons of kinetic energy E=2 eV hit the barrier height of energy U0= 5 eV and the width of L=1.0 nm. Find the percent of electrons passing through the barrier?

LEU

mUE

UE

II

Tpad

trans )(2

2exp116 000 T=7.1·10-8

insulator

semiconductor

metalA

If L=0.5 nm.then T=5.2 ·10-4!

Scanning tunneling electron miscroscope

LeI 2

)(2

0 EUm

gdzie

Scanning tunneling electron miscroscope

Scanning tunneling electron miscroscope

Scanning tunneling electron miscroscope

Image downloaded from IBM, Almaden, Calif.It shows 48 Fe atoms arranged on a Cu (111) surface

Scanning tunneling electron miscroscope

a particle decay

Approximate potential - energy function for an a particle in a nucleus.

Tunneling

Nuclear fusion ( synteza ) is another example of tunneling effect

E.g. The proton – proton cycle

Young’s double slit experiment

a) constructive interference

For constructive interference along a chosen direction, the phase difference must be an even multiple of

msind m = 0, 1, 2, …d

b) destructive interference For destructive interference along a chosen direction, the phase difference must be an odd multiple of

21msind m = 0, 1, 2, …

a, b, c – computer simulation

d - experiment

Electron interference

Franhofer Diffraction

a dy

dysin2

d

sin2 a

Re

Im

ER

R

maxER

20E

2

1I )cos1(R2

2

1 20

2/sin2E 2

2max

0

22max0 2/

2/sinE

22max /sin

/sinsinI

a

a-1.0 -0.5 0.0 0.5 1.0

0.0

0.2

0.4

0.6

0.8

1.0

rela

tive

inte

nsity

[I/I

max

]

diffraction angle []

Electron Waves

• Electrons with 20eV energy, have a wavelength of about 0.27 nm

• This is around the same size as the average spacing of atoms in a crystal lattice

• These atoms will therefore form a diffraction grating for electron “waves”

dNi=0.215nmdiffraction

nm165.0sin d

de Broglie

m

peVba 2

2

nm167.02

bameV

h

p

h

C.J.Davisson and L.G.Germer

Resolution

Rayleigh’s criterion:

When the location of the central maximum of one image coincides with the the location of the first minimum of the second image, the images are resolved.

For a circular aperture:

D22.1min

Electron Microscope