Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 1 Exercise Solutions TYU 1.2 (a) Number of atoms per (100) lattice plane Ex 1.1 (a) Number of atoms per unit cell Surface Density (b) Volume Density = cm (b) Number of atoms per (110) lattice plane cm _______________________________________ Surface Density Ex 1.2 Intercepts of plane; p=1, q=2, s=2 Inverse; cm (c) Number of atoms per (111) lattice plane Multiply by lowest common denominator, plane _______________________________________ Lattice plane area Ex 1.3 (a) Number of atoms per (100) plane where Surface Density cm (b) Number of atoms per (110) plane Then lattice plane area Surface Density Surface Density cm cm _______________________________________ Test Your Understanding Solutions _______________________________________ TYU 1.1 TYU 1.3 Number of atoms per unit cell (a) For (100) planes, distance (b) For (110) planes, distance Volume Density cm _______________________________________ Radius _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU 1.4 (a) 8 corner atoms (b) 6 face-centered atoms (c) 4 atoms totally enclosed _______________________________________ TYU 1.5 Number of atoms in the unit cell Volume Density cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 1 Problem Solutions Then 1.1 (a) fcc: 8 corner atoms atom 6 face atoms atoms Ratio Total of 4 atoms per unit cell (b) bcc: 8 corner atoms atom 1 enclosed atom =1 atom (d) Diamond lattice Total of 2 atoms per unit cell (c) Diamond: 8 corner atoms atom Body diagonal 6 face atoms atoms 4 enclosed atoms = 4 atoms Unit cell vol Total of 8 atoms per unit cell _______________________________________ 8 atoms per cell, so atom vol 1.2 (a) Simple cubic lattice: Then Unit cell vol 1 atom per cell, so atom vol Ratio Then _______________________________________ Ratio 1.3 (b) Face-centered cubic lattice (a) ; From Problem 1.2d, Unit cell vol Then 4 atoms per cell, so atom vol Center of one silicon atom to center of Then nearest neighbor (b) Number density Ratio cm (c) Body-centered cubic lattice (c) Mass density Unit cell vol grams/cm _______________________________________ 2 atoms per cell, so atom vol Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) 1.4 (a) 4 Ga atoms per unit cell (b) Number density (c) A-atoms: # of atoms Density of Ga atoms cm Density 4 As atoms per unit cell Density of As atoms cm cm (b) 8 Ge atoms per unit cell B-atoms: # of atoms Number density Density Density of Ge atoms cm _______________________________________ cm 1.5 _______________________________________ From Figure 1.15 1.9 (a) (a) # of atoms (b) Number density cm _______________________________________ Mass density 1.6 gm/cm (b) _______________________________________ # of atoms 1.7 (a) Simple cubic: Number density (b) fcc: cm (c) bcc: Mass density (d) diamond: gm/cm _______________________________________ _______________________________________ 1.8 1.10 From Problem 1.2, percent volume of fcc Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ atoms is 74%; Therefore after coffee is Surface density cm ground, Volume = 0.74 cm For 1.12(a) and (b), Same material _______________________________________ 1.11 (b) (b) For 1.12(a), A-atoms; (c) Na: Density Surface density cm cm Cl: Density cm B-atoms; (d) Na: At. Wt. = 22.99 Surface density Cl: At. Wt. = 35.45 So, mass per unit cell cm For 1.12(b), A-atoms; Surface density Then mass density cm grams/cm B-atoms; _______________________________________ Surface density cm 1.12 (a) For 1.12(a) and (b), Same material _______________________________________ Then Density of A: 1.14 (a) Vol. Density cm Density of B: Surface Density cm (b) Same as (a) _______________________________________ (b) Same as (a) 1.15 (c) Same material (i) (110) plane _______________________________________ (see Figure 1.10(b)) (ii) (111) plane 1.13 (see Figure 1.10(c)) (iii) (220) plane (a) For 1.12(a), A-atoms Same as (110) plane and [110] direction Surface density (iv) (321) plane cm Intercepts of plane at For 1.12(b), B-atoms: Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ [321] direction is perpendicular to (321) plane _______________________________________ So 1.16 (a) Area of plane (b) cm Surface density _______________________________________ cm 1.17 (b) bcc Intercepts: 2, 4, 3 (i) (100) plane: Surface density cm (634) plane _______________________________________ (ii) (110) plane: 1.18 Surface density (a) cm (b) (iii) (111) plane: (c) Surface density _______________________________________ cm (c) fcc 1.19 (i) (100) plane: (a) Simple cubic Surface density cm (i) (100) plane: (ii) (110) plane: Surface density Surface density cm (ii) (110) plane: cm (iii) (111) plane: Surface density cm Surface density (iii) (111) plane: cm Area of plane _______________________________________ where 1.20 (a) (100) plane: - similar to a fcc: Now Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ = cm Surface density cm (b) (110) plane: _______________________________________ Surface density 1.22 Density of silicon atoms cm and cm 4 valence electrons per atom, so Density of valence electrons cm _______________________________________ (c) (111) plane: 1.23 Density of GaAs atoms Surface density cm cm _______________________________________ An average of 4 valence electrons per atom, So 1.21 Density of valence electrons cm _______________________________________ 1.24 (a) #/cm (a) cm (b) (b) #/cm _______________________________________ 1.25 cm (a) Fraction by weight (c) (b) Fraction by weight (d) # of atoms Area of plane: (see Problem 1.19) _______________________________________ 1.26 Volume density cm Area So cm cm We have Then #/cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.27 Volume density cm So cm We have Then _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 2 Exercise Solutions or eV Ex. 2.1 Then (a) eV, eV, eV J (b) or eV (b) J J or or eV eV _______________________________________ Then eV Ex 2.2 eV (a) eV _______________________________________ kg-m/s m Ex 2.4 or (c) J kg-m/s Now Set Then = J or eV _______________________________________ Ex 2.3 or m (a) (a) m J or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ % kg-m/s (b) m (b) or % _______________________________________ J or eV _______________________________________ Ex 2.5 (a) m Then TYU 2.2 (a) eV s (b) (b) Same as part (a), s _______________________________________ _______________________________________ TYU 2.3 (a) Ex 2.6 From Example 2.6, we have eV = m meV, meV, meV _______________________________________ Test Your Understanding TYU 2.1 (b) (a)