Low Energy Electron Diffraction(LEED)Chem 6a Winter 2005California Institute of TechnologyPrepared by: Bruce M. [email protected] mrobsr’nn’PElectron Scattering Off of a 2-D Lattice IRjm= ja1+ ma2j,m = integers()()()()000obs jmjmik r Rik Rjmsobs jmobs jmerefk,krRψψ′⋅−⋅⎛⎞⎜⎟′=⎜⎟−⎝⎠rrrrrrrrrrElastic Scattered electron wavefunction()()()000obsjmik rik k Rjmobsefk,kerψ′⋅′−⋅⎛⎞′≈⎜⎟⎜⎟⎝⎠rrrrrrrr2ˆknπλ′′=rwithn is a unit vector from the origin, O to the observation point, P. n’is a unit vector from the j,mlattice point to the observation point, P and n0is the unit vector along the propagation direction of the incident electronsa1a2ORj mrobsr’nn’P()00ik rinc.reψψ⋅=rrrIncident electron wavefunction002ˆknπλ=rwithand2hmEλ=Electron Scattering Off of a 2-D Lattice II• If the incident spot size << |robs| then we may make the approximations :() ()2111MMtot jmsobs s obsmjrrψψ===∑∑rr()() ()2102 010011obsik rMMimk k a i jk k amjobsefk,k e erψ⋅⎡⎤⎡ ⎤−⋅ −⋅⎣⎦⎣ ⎦==⎛⎞=⎜⎟⎜⎟⎝⎠∑∑rrrr rrrrrrr• Sum over j,m• These sums may be evaluated directly. The scattered intensity is α |ψstot |2thus()()()()()()22210 1 20 22002201 0 211221122totsobsobssin M k k a sin M k k afk,krrsin k ka sin k kaψψ⎛⎞⎛ ⎞−⋅−⋅⎜⎟⎜ ⎟⎝⎠⎝ ⎠=⎛⎞⎛⎞−⋅ −⋅⎜⎟⎜⎟⎝⎠⎝⎠rrrrrrrrrrrr rrrrNote that the scattered intensity depends on the energy and direction of the incident wave through the function f. This gives information on the atomic arrangement of the surface unit cell()()00jmfk,k f k,k′≈rr rr2ˆkk nπλ′≈=rrandElectron Scattering Off of a 2-D Lattice III• Maxima occur when denominators approach zero giving the Laue Conditions()012mnkk a mπ−⋅=rrr()022mnkk a nπ−⋅=rrror()01mnˆˆnn amλ−⋅=r()02mnˆˆnn a nλ−⋅=r( m, n = integers )• The Laue Conditions constrain the “allowed” directions, nmn, for interference maxima• Usually choose normal incidence i.e. n0·ai= 01mnNamλ⋅=rr2mnNanλ⋅=rrwhere Nmnis the projection of nmnin the plane of the sample and 0 ≤ |Nmn| ≤ 1Solving for the Direct Lattice from the Laue Conditions• We define the reciprocal lattice throughij ijbaδ⋅=rr• Nmnis expressed in terms of the reciprocal lattice vectors so as to reproduce the Laueconditions• Therefore the Nmn, are proportional to vectors in the reciprocal lattice. For a given electron energy, the maximum values for m and n are limited since we must have()12mnNmbnbλ=+rrra1b1b2a2()1201mb nbλ≤+≤rrMore on Solving for the Direct Lattice()()mnmnmn zdNtanDNθ==rr()221mnmn zNN+=r()mn mnmn zˆˆnNNz=+rReferring to the figure at the right we may obtain the following relationshipsso that()()21mnmnmndDNdD=⎛⎞+⎜⎟⎝⎠rrrDdmnrmnsampleθˆzN10N01diffraction patternUsing the definitions on the previous page one may first solve for the directions of a1and a2and then solve for absolute magnitudes. For all dimensions needed in the calculations, see next pageNote: To obtain an absolute length scale, one needs to measure some feature on the vacuum chamber (e.g. the circular glass window, the radius of the e-gun) in order to calibrate the CCD camera images. See detailed LEED procedure for more on this.Top view of UHV chamber and sample mountf = b + c/2 + (x-x0) –LD = f-ax0 = x-micrometer reading when the sample rod is at the center of the chamber = 8.11 mm = 0.3193 inches on the black numberingx = x-micrometer reading of the sample rod when LEED data is takenc = 12.0 inchesL = 2.538 inches ± 0.020 in.a = 5.30 inchesb = 4.5625 inches ± 0.0156 in.R = 2.70 inchesAll of the relevant dimensions needed to calculate the sample-to-screen distance, D, are shown in the schematic to the left. The center of the chamber is O and the center of the spherical screen is O’abcfRxx0LOO’DFor the most accurate calculation of bond distances, the sample should be positioned at O’ so that D = R. This occurs when x = 7.49 mm on the black