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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 2L 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 2a 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