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Lecture 5
Tunneling
classically
An electron of such an
energy will never appear
here!
Ekin= 1 eV
0V
-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
2 d 2

 U 0  E
2
2m dx
Solution:

d 2  2m
 2 (U 0  E )
2
dx

( x)  Ae x
Potential barriers and tunneling
2 x
A e
2m
2m
x
2
 2 (U 0  E ) Ae
   2 (U 0  E )


Two solutions:    2m (U  E ) or
1
0
2
2m
 2   2 (U 0  E )



 ( x ) dx  1
Normalization condition:
Solution:
The probability to find
a particle in the region
II within x
2 x
0
 ( x )  Ae


2m
pr ( x0 )  A exp  2 
(U 0  E ) x0  x
2



2
Potential barriers and tunneling

2 x
 ( x )  Ae


2m
pr ( x0 )  A exp  2  2 (U 0  E ) x0  x



2
Potential barriers and tunneling
example
A
metal
insulator
semiconductor
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?
T


I trans
E 
E 
2m
1 
 exp  2
 16
(U 0  E ) L 
I pad
U0  U0 



If L=0.5 nm.then T=5.2 ·10-4!
T=7.1·10-8
Scanning tunneling electron
miscroscope
I  e 2L
gdzie  
2m
(U0  E )

Scanning tunneling electron miscroscope
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
 particle decay
Approximate potential - energy
function for an  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 

d

d sin   m
m = 0, 1, 2, …
b) destructive interference
For destructive interference along a
chosen direction, the phase difference
must be an odd multiple of 
d sin   m  12   
m = 0, 1, 2, …
Electron interference
a, b, c – computer simulation
d - experiment
Franhofer Diffraction

2 sin 
dy

2a sin 


1
1
I   0 E 2  0  2R 2 (1  cos ) 
22
2
E 
 0   max   2 sin 2  / 2  
  
 sin  / 2  
 0 E 2max  
 
 / 2 
2
 I 2max
 sin a sin  /   


 a sin  /  
2
R  E max
Im
R

R
E

Re
1.0
0.8
relative intensity [I/Imax]
a dy
d 
0.6
0.4
0.2
0.0
-1.0
-0.5
0.0
diffraction angle []
0.5
1.0
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”
C.J.Davisson and L.G.Germer
diffraction
dNi=0.215nm
  d sin   0.165nm
de Broglie
eVba
p2

2m
h
h
 
 0.167 nm
p
2meVba
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:
 min

 1.22 
D

Electron Microscope