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Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1By D. A. Neamen Problem Solutions_Chapter 15Problem Solutions1.1(a) fcc: 8 corner atomsatom 6 face atoms atoms Total of 4 atoms per unit cell(b) bcc: 8 corner atomsatom 1 enclosed atom =1 atom Total of 2 atoms per unit cell(c) Diamond: 8 corner atomsatom 6 face atomsatoms 4 enclosed atoms = 4 atoms Total of 8 atoms per unit cell_1.2(a) Simple cubic lattice: Unit cell vol 1 atom per cell, so atom vol Then Ratio (b) Face-centered cubic lattice Unit cell vol 4 atoms per cell, so atom vol Then Ratio (c) Body-centered cubic lattice Unit cell vol 2 atoms per cell, so atom vol Then Ratio (d) Diamond lattice Body diagonal Unit cell vol 8 atoms per cell, so atom vol Then Ratio _1.3(a) ; From Problem 1.2d, Then Center of one silicon atom to center of nearest neighbor (b) Number density cm(c) Mass density grams/cm_1.4(a) 4 Ga atoms per unit cell Number density Density of Ga atoms cm 4 As atoms per unit cell Density of As atoms cm(b) 8 Ge atoms per unit cell Number density Density of Ge atoms cm_1.5 From Figure 1.15(a) (b) _1.6 _1.7 (a) Simple cubic: (b) fcc: (c) bcc: (d) diamond: _1.8(a) (b)(c) A-atoms: # of atoms Density cm B-atoms: # of atoms Density cm_1.9(a)# of atoms Number density cmMass density gm/cm(b)# of atoms Number density cmMass density gm/cm_1.10 From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground, Volume = 0.74 cm_1.11(b)(c) Na: Density cm Cl: Density cm(d) Na: At. Wt. = 22.99Cl: At. Wt. = 35.45 So, mass per unit cell Then mass density grams/cm_1.12(a) Then Density of A: cm Density of B: cm(b) Same as (a)(c) Same material_1.13 (a) For 1.12(a), A-atoms Surface density cm For 1.12(b), B-atoms: Surface density cm For 1.12(a) and (b), Same material(b) For 1.12(a), A-atoms; Surface density cm B-atoms; Surface density cm For 1.12(b), A-atoms; Surface density cm B-atoms; Surface density cm For 1.12(a) and (b), Same material_1.14(a) Vol. Density Surface Density (b) Same as (a)_1.15(i) (110) plane (see Figure 1.10(b) (ii) (111) plane (see Figure 1.10(c) (iii) (220) plane Same as (110) plane and 110 direction (iv) (321) plane Intercepts of plane at 321 direction is perpendicular to (321) plane_1.16 (a) (b) _1.17 Intercepts: 2, 4, 3 (634) plane_1.18(a)(b)(c)_1.19(a) Simple cubic (i) (100) plane:Surface density cm(ii) (110) plane:Surface density cm(iii) (111) plane:Area of plane where Now So Area of plane cm Surface density cm(b) bcc (i) (100) plane:Surface density cm(ii) (110) plane:Surface density cm(iii) (111) plane:Surface density cm(c) fcc (i) (100) plane:Surface density cm(ii) (110) plane:Surface density cm(iii) (111) plane:Surface density cm_1.20(a) (100) plane: - similar to a fcc: Surface density cm(b) (110) plane: Surface density cm(c) (111) plane: Surface density cm_1.21 (a) #/cm cm(b) #/cm cm(c)(d) # of atoms Area of plane: (see Problem 1.19) Area cm #/cm = cm _1.22 Density of silicon atoms cm and 4 valence electrons per atom, so Density of valence electrons cm_1.23 Density of GaAs atoms cm An average of 4 valence electrons per atom, So Density of valen

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