10. Appendix

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660 <strong>Appendix</strong> B coefficient of oxygen atoms to Si atoms is one. Typically an exposure of ∼1 L is needed to cover a surface with a monolayer of adatoms. Students are urged to repeat the calculation for a (111) and a (110) Si surface. Solution to Problem 8.2 We refer for notation to Fig. 8.9. The average radius of the analyzer is Ra. The applied voltage is Va, the kinetic energy of the electrons under consideration is Ee. The electron band pass energy, being non-relativistic, is obtained by equating the electron centripetal force mv 2 /Ra to the force eVa/2¢Ra exerted by the electric field on the electron (of charge e and mass m): mv 2 Ra eVa 2¢Ra i.e., Ee eVa 4 Ra ¢Ra For some typical values: Va 1V,Ra 10 cm and ¢Ra 0.5 cm we find that Ee 5eV. The resolution of the hemispherical analyzer is defined by the change in the band pass energy ¢Ee induced by a small change ‰Ra in Ra: From the above result ¢Ee/‰Ra (eVa/4¢Ra). In case of the typical hemispherical analyzer values given above the resolution is 0.05 eV/mm. Solution to Problem 8.5 A model of the zincblende crystal structure would be helpful in visualizing this problem Let us first consider a (100) surface. It involves atoms of only one of the two sublattices discussed in Chap. 2 as constituting diamond or zincblende crystals. Hence no operations transforming an atom in one of the sublattices into one in the other can exist and a single flat surface layer must have a symmorphic space group. One may at first sight think that C4, a fourfold rotation about the [100] axis, is a symmetry operation. This is not the case: the atoms have tetrahedral, not fourfold symmetry because of the sp 3 bonding. The only symmetry operations of the diamond (Table 2.5) or the zincblende (Table 2.3) point groups that survive when we consider the (100) surface are: a C2 rotation around the [100] axis and two reflections across the (011) and (01-1) planes. The point group is C2v whose characters are shown in Table 2.14. Similar considerations concerning a (111) surface lead to the point group C3v. Note, however, that there are two possible non-equivalent flat (111) surfaces. One of them has “dangling bonds” sticking out of the crystal perpendicular to (111). In the other, each atom has three “dangling bonds” at an angle
Solution to Problem 8.9 661 of 19.5 ◦ with the surface. Show that the former is energetically favorable when cleaving a crystal parallel to one of the equivalent (111) surfaces. The primitive translation vectors of a (100) surface are (a/2)(011) and (a/2)(01-1). The two corresponding reciprocal lattice vectors can be obtained from (2.9) by defining a3 a(001). Using these two vectors it is easy to draw the corresponding Brillouin zone. The groups of the k-vector at the center of the Brillouin.zone. are also C2v and C3v. Solution to Problem 8.9 Let us start with the simplest case, that of a zincblende crystal (point group at °: Td). We will discuss the hexagonal faces first and then the tetragonal faces. The point in a tetragonal face with the highest symmetry is the X-point: (2°/a0)(100). The corresponding group of the k-vector is D2d (Table 2.15). The slope of a given band perpendicular to this face is proportional to the expectation value of px according to (2.35). It is easy to check with Table 2.15 that px has X3 symmetry. All bands are non-degenerate (we are neglecting spin at this point) at the X point except those of X5 symmetry which are twofold degenerate (see Fig. 2.14. In this figure the X5 bands are split into X6 and X7 by spin-orbit interaction. In the absence of spin-orbit interaction these bands remain doubly degenerate along the ¢ direction because of time reversal invariance). Any nondegenerate bands will approach the X-point along ¢ with zero slope because the product of two nondegenerate representations is the identity representation, X1 in our case. Since px belongs to X3, its expectation value for a nondegenerate representation must vanish. Since the X5 degeneracy does not split along ¢, both bands belonging to it act as a single band which also end at the X point with zero slope. The only other high symmetry point on the (100) face is the W-point, (1, 0, 1 2 ). Its point group is S4 [2.4]. It is easy to see that the matrix elements of px do not vanish by symmetry at this point and thus no one-dimensional van Hove singularities occur there. Let us consider now a face perpendicular to the [111] axis. The highest symmetry point of k-space on this surface is L. The corresponding group is C3v (see Table 2.12). The component of p along the [111] direction is invariant under all operations of C3v and, therefore, we cannot use an argument similar to that used for the (100) face to prove that the bands have zero slope perpendicular to the (111) face. An examination of Figs. 2.14 and 2.15, however, suggests that this is still the case. In order to prove it, we use the following arguments. Let us move along the § axis, perpendicular to the (111) face, by an infinitesimally small amount ‰. The resulting point we label as (apple/a0)(1 ‰)(111). Using time reversal symmetry we can show that the energy of a given band at this point is equal to that at (apple/a0)(1 ‰)(111). Adding to
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Appendix A: Pioneers of Semiconduct
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Ultra-Pure Germanium 555 Ultra-Pure
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References Ultra-Pure Germanium 557
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Two Pseudopotential Methods: Empiri
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The Early Stages of Band-Structures
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Cyclotron Resonance and Structure o
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References Cyclotron Resonance and
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Optical Properties of Amorphous Sem
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Optical Spectroscopy of Shallow Imp
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Page 21 and 22:
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On the Prehistory of Angular Resolv
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The Discovery and Very Basics of th
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The Birth of the Semiconductor Supe
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Number of papers 500 200 100 50 20
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Appendix B: Solutions to Some of th
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Solution to Problem 2.6 585 transfo
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⎧ ⎫ 2apple ⎪⎨ cos (y z)
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Partial solution to Problem 2.8 589
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Solution to Problem 2.16 Solution t
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Solution to Problem 2.20 597 to the
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Therefore the result of I ′ on
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Solution to Problem 3.2(c) Solution
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Solution to Problem 3.5 603 Similar
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Solution to Problem 3.7 (b and c) 6
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u3. The above equation can be reduc
Page 57 and 58: Solution to Problem 3.8 (a, c and d
Page 59 and 60: Solution to Problem 3.11(a,b and c)
Page 61 and 62: Solution to Problem 3.16 613 (1) Al
Page 63 and 64: Solution to Problem 3.17 Solution t
Page 65 and 66: Solution to Problem 3.17 617 forces
Page 67 and 68: Solution to Problem 4.1 619 the abo
Page 69 and 70: C dz ′ z ′ i° Solution to Pro
Page 71 and 72: R -R A y E’-E y E’-E R x Soluti
Page 73 and 74: Solution to Problem 5.3(a) 625 tion
Page 75 and 76: Solution to Problem 5.5 627 To obta
Page 77 and 78: Ùm ∞ NLOq 0 2 apple dq 0 ∞
Page 79 and 80: plus Jz ·Fz Solution to Problem 5
Page 81 and 82: Solution to Problem 6.7 633 axes. F
Page 83 and 84: which can be found in standard text
Page 85 and 86: Solution to Problem 6.11 637 The co
Page 87 and 88: Solution to Problem 6.19(a) 639 if
Page 89 and 90: Solution to Problem 6.19(a) 641 k.
Page 91 and 92: Solution to Problem 6.19(a) 643 The
Page 93 and 94: Solution to Problem 6.21 645 by a s
Page 95 and 96: Solution to Problem 6.21 647 °25
Page 97 and 98: Equating the two expressions for Nn
Page 99 and 100: Solution to Problem 7.5 651 erties
Page 101 and 102: A Short Cut to the Calculation of t
Page 103 and 104: The diagram for the process labeled
Page 105 and 106: Solution to Problem 7.12 657 or sca
Page 107: Solution to Problem 8.1 Solution to
Page 111 and 112: Solution to Problem 8.10 663 corres
Page 113 and 114: Solution to Problem 9.2 Solution to
Page 115 and 116: Solution to Problem 9.4 667 Table 9
Page 117 and 118: Solution to Problem 9.14 669 Again
Page 119: Solution to Problem 9.14 671 Note t
Page 122 and 123: 674 Appendix C A4.1.2 Historical Ba
Page 124 and 125: 676 Appendix C from the slopes of t
Page 126 and 127: 678 Appendix C Fig. A4.3 The electr
Page 128 and 129: 680 Appendix C Fig. A4.5 (a)-(d)The
Page 130 and 131: 682 Appendix C tion of alloying (se
Page 132 and 133: 684 Appendix C 17 -3 Hall Concentra
Page 135 and 136: Appendix D: Recent Developments and
Page 137 and 138: Chapter 1 689 ing up the glass to a
Page 139 and 140: Chapter 2 691 Popov, V. N., Lambin,
Page 141 and 142: Chapter 2 693 and zinc-blende group
Page 143 and 144: Chapter 3 695 Inelastic x-ray Scatt
Page 145 and 146: Chapter 4 697 results from a better
Page 147 and 148: Chapter 5 699 K. Alberi, K. M. Yu,
Page 149 and 150: Chapter 5 701 P. Ramvall, N. Carlss
Page 151 and 152: Chapter 6 Ab initio calculations of
Page 153 and 154: Chapter 6 705 Strain-induced birefr
Page 155 and 156: Chapter 7 707 S. Baroni, S. de Giro
Page 157 and 158: Chapter 7 709 W. Windl, K. Karch, P
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Chapter 8 711 Strain Shift Coeffici
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Chapter 9 713 Z. J. Zhao, F. Liu, L
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Chapter 9 715 S. F. Ren, W. Cheng,
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Chapter 9 717 V. P. Gusynin, S. G.
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720 References tions II. Misfitting
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722 References 2.30 R. M. Wentzcovi
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724 References 3.23 R. M. Martin: E
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726 References Nye J. F.: Physical
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728 References Schubert E. F.: Dopi
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730 References Ferry D. K.: Semicon
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732 References 6.28 S. Logothetidis
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734 References 6.73 H. D. Fuchs, C.
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736 References 6.116 D. E. Aspnes:
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738 References Chapter 7 7.1 W. Hei
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740 References 7.43 G. Finkelstein,
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742 References 7.82 J. R. Sandercoc
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744 References 7.122 D. C. Reynolds
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746 References 8.27 A. Goldmann, J.
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748 References Electronic and Surfa
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750 References 9.31 A. Pinczuk, G.
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752 References 9.68 J. P. Palmier:
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Subject Index A ·-Sn (gray tin) 95
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C c(2 × 8)(111) surface of Ge - di
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Dielectric function 117, 246, 253,
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Emission rate vs frequency in Ge 34
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Galena (PbS) 1 Gamma-ray detectors
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Insulators 1 Integral quantum Hall
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- - vs T 222 - temperature dependen
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- frequency 253, 306, 335 - - of va
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Resonant Raman profile 403 Resonant
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- - LA phonons 512 - - LO phonons 5
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- structure 142 - symmetry operatio
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10. Appendix

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