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Published byTiara Babbitt Modified about 1 year ago

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1. Example for KL1L2,: The binding energy of K(1s) L1(2s) and L2(2p1/2) can be found as h -E1s-W, hv-E2s-W, hv-E2p1/2-W. Then Ekin for KL1L2 is (h -E1s-W)-(hv-E2s-W)-(hv-E2p1/2-W)-W=E2s+Esp1/2- E1s-h

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From the two figures in binding energy scale, what can be determine is the Auger and photoemission (with the same values at both figures). 4, 5, 6 are Auger and rest are photoemission. Then you need to think about the Fe electronic structure, 26 electrons, it should be 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6. As 4s3d electrons are very close to each other and become valence band, you need to find the rest of energy levels, small binding energy is for shallower energy level (bigger n and l values), then you can give notation for them. One thing need to know is the 2p will be split into 2p3/2 and 2p1/2, with about 1:2 intensity difference p 3s 2p3/2 2s 2p1/2 2)

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While 1s is too deep, therefore we can not see it. One thing keep in mind is that the distance between energy levels are bigger and bigger from right to left when the n and L increase, there is only exception is 2p3/2 and 2p1/2, while they have the same n and L number only j are different. The intensity difference between 2p3/2 and 2p1/2 are due to the number of the electronic states avaiable, for 2p3/2 it is 2j+1=2x3/2+1=4 and for 2p1/2 is 2, therefore the intensity ratio between 2p1/2 and 2p3/2 peaks are: 1:2. Now we need to think about how to solve the Auger electrons. From the Figure, you can get the binding energy of all the wanted energy levels and also the Auger electron kinetic energy, then you can calculate the what can be possible energy level involved in the Auger process and obtain the notation for them. For example, peak 4 is about 660 eV binding energy in first Figure, with 1254 eV photon energy, you can obtain the kinetic energy should be about 594 eV. The binding energy for 2s is about 850 eV and two 2p levels are about 710 eV, 3p are about 60 eV and 3s are about 90 eV, the only possible is =590 eV, others are too far, so the notation is L23M23M23. You can find the other is L23M23M45, and L23M45M45. (M45 is Valence band, if you write L23M23V or L23VV should be ok). 2p Fermi 3p

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3. If you know the different probe depths of electron and x-ray, then the angular dependence is easy to be found. First case: cos ( Second case: cos ( cos (45 O- for c(2X2) the coverage is ½ Let intensity of photoemission from substrate be I s0. The substrate photoemission intensity, I s = [(1 - coverage) Is 0 ] + [coverage I attenuated ] = 1/2Is 0 + 1/2Is 0 e -a/

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Let intensity of photoemission from very thick absorbate be I a0. The absorbate photoemission intensity, I a = coverage I a0 (1-e -a/ )= ½ I a0 (1-e -a/ ) As Iso and Iao is determined by S factor, if they have the same S factor Iso/Iao = 1, Therefore, it can be simplified as: (1/2 + 1/2e -a/ )/( 1/2 (1-e -a/ )) of course you can simplify more the formula.

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5, I give another superstructure as example. a) You need to find a primitive lattice cell for the superstructure with the two lattice parameter b1 and b2 have the same angles as the structure a1 and a2, which is 90 degree in this case. b1 b2 Also need to remember that b1 and b2 have the same order as a1 and a2, b1 is right side of b2. Then you can get them as shown. Then you can get the relationship of b1 (b2) with respect to a1 and a2. b1=a1+a2 b2=-2a1+2a2 Therefore the Matrix notation is

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The Woods notation will be The coverage is easy, while the b1Xb2 will give the area of the superstructure, which is 4, and there is only 1 atom of superstructure in the primitive cell, therefore, the coverage is ¼. b) You need to find the relationship between superstructure lattice parameter and substrate lattice parameter in Reciprocal space. The relationship between the real space lattice parameter and reciprocal space one, you must remember., if you do not want to calculate the how to calculate the M* from M. It is very easy to remember, while in the reciprocal space, you need have the same order of two lattice parameter: a1* must at the right side of a2*, from right hand rule, to get a a1* (vector) from a2xn, and a2* from nxa1. This is keep true for b1* and b2*.

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First we try to get b1* b1*=2πb2 x n/׀(b1 x b2)׀ While, b1=a1+a2, b2= -2a1+2a2, ׀(b1 x b2)׀=4a1 2, therefore, b1*= 2π(-2a1+2a2) x n/4a1 2, While a1*=2πa2 x n/׀(a1 x a2)׀= 2πa2 x n/a1 2 and a2*=2πn x a1/׀(a1 x a2)׀= 2πn x a1/a1 2 you can get b1*=1/2 a2*+1/2 a1* For b2* b2*= 2πn x b1/׀(b1 x b2)׀ = 2πn x (a1+a2)/4a1 2 =1/4 a2* - 1/4 a1* Notice the sign! Then you know the b1* and b2*, then it is easy to draw reciprocal lattices, first draw the a big square for the substrate then try to draw the superstructure accordingly.

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b1* b2* (0 0) (0 1) (1 0) (1 1) superstructure substrate One thing you can check, that the short lattice parameter direction in real space always become longer and longer become shorter. If you see not this way, then you must be wrong!

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As said that you can also determine the b1* and b2* from matrix notation, while, M* is (M -1 ) T transpose inverse matrix, there is mii= mjj*/detM* mij=- mji*/detM* mii*= mjj/detM Mij= -mji/detM And detM=4 m11*= m22/4= 2/4= 1/2 m22*= m11/4= 1/4 M12*= -m21/4= 2/4 = 1/2 M21*= -m12/4 = -1/4 Or M is M* is This in fact give you the same results as before b1*=1/2 a2*+1/2 a1* b2*=-1/4 a1* + 1/4 a2* Then it is the same to draw the Figures.

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