part01

Chapter 1: Semiconductors: A General Introduction

Source note. This chapter follows Pierret, Semiconductor Device Fundamentals, Chapter 1 (book pp. 3-22). Where the OCR scan lost a figure or an overbar, I reconstruct it from the physics and say so explicitly.

The big question of this chapter. Almost every phone, laptop, car ignition and LED on Earth is built from a handful of strange solids that are neither good conductors nor good insulators. What are these solids made of, how clean must they be, how are their atoms arranged, how do we give names to the planes and directions inside them, and how on Earth do we grow a perfect crystal the size of a baseball bat? That is the whole chapter. Think of it as the “meet the cast and see the stage” chapter: later chapters will let the actors (electrons and holes) perform.

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1.1 General Material Properties

The overwhelming majority of solid-state devices on the market today are made from semiconductors. So before any device physics, let us ask: what do semiconducting materials look like in terms of (1) composition, (2) purity and (3) structure?

1.1.1 Composition

Table 1.1 lists the atomic compositions of semiconductors you are likely to meet in the device literature. There are three big families.

Classification	Symbol	Name
(1) Elemental	Si	Silicon
	Ge	Germanium
(2) Compounds
(a) IV-IV	SiC	Silicon carbide
(b) III-V	AlP	Aluminum phosphide
	AlAs	Aluminum arsenide
	AlSb	Aluminum antimonide
	GaN	Gallium nitride
	GaP	Gallium phosphide
	GaAs	Gallium arsenide
	GaSb	Gallium antimonide
	InP	Indium phosphide
	InAs	Indium arsenide
	InSb	Indium antimonide
(c) II-VI	ZnO	Zinc oxide
	ZnS	Zinc sulfide
	ZnSe	Zinc selenide
	ZnTe	Zinc telluride
	CdS	Cadmium sulfide
	CdSe	Cadmium selenide
	CdTe	Cadmium telluride
	HgS	Mercury sulfide
(d) IV-VI	PbS	Lead sulfide
	PbSe	Lead selenide
	PbTe	Lead telluride
(3) Alloys
(a) Binary	${\mathrm{Si}}_{1-x}{\mathrm{Ge}}_x$
(b) Ternary	${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}\mathrm{As}$ (or ${\mathrm{Ga}}_{1-x}{\mathrm{Al}}_x\mathrm{As}$)
	${\mathrm{Al}}_x{\mathrm{In}}_{1-x}\mathrm{As}$
	${\mathrm{Cd}}_{1-x}{\mathrm{Mn}}_x\mathrm{Te}$
	${\mathrm{GaAs}}_{1-x}{\mathrm{P}}_x$
	${\mathrm{Ga}}_x{\mathrm{In}}_{1-x}\mathrm{As}$
	${\mathrm{Ga}}_x{\mathrm{In}}_{1-x}\mathrm{P}$
	${\mathrm{Hg}}_{1-x}{\mathrm{Cd}}_x\mathrm{Te}$
(c) Quaternary	${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}{\mathrm{As}}_y{\mathrm{Sb}}_{1-y}$
	${\mathrm{Ga}}_x{\mathrm{In}}_{1-x}{\mathrm{As}}_y{\mathrm{P}}_{1-y}$

(The scan of the last rows is garbled; the entries above are the standard ones the book lists.)

The alloy subscript. In an alloy formula the $x$ (or $y$) is a fraction between 0 and 1. For instance ${\mathrm{Al}}_{0.3}{\mathrm{Ga}}_{0.7}\mathrm{As}$ means 3 Al atoms and 7 Ga atoms for every 10 As atoms. The Al and Ga together share one group of lattice sites; As occupies the other.

Who are the stars?

• Silicon (Si) is, by a wide margin, king. Thanks largely to the maturity of its fabrication technology, it totally dominates the market: most discrete devices and integrated circuits (ICs), including the CPU in your computer and the ignition module in a car, are silicon.
• Gallium arsenide (GaAs) has superior electron transport (electrons move faster) and special optical properties, so it appears in laser diodes and high-speed ICs.
• The rest live in “niche” roles that are almost always high-speed, high-temperature, or optoelectronic (light-related).

Because of Si’s dominance, the text focuses on Si, gives GaAs comparable attention where feasible, and brings in others as needed.

Why is the list so short? Think of the periodic table as a huge menu. You could combine elements in countless ways, yet only a tiny group give semiconductors. Look at the abbreviated periodic chart (Table 1.2):

II	III	IV	V	VI
4 Be	5 B	6 C	7 N	8 O
12 Mg	13 Al	14 Si	15 P	16 S
30 Zn	31 Ga	32 Ge	33 As	34 Se
48 Cd	49 In	50 Sn	51 Sb	52 Te
80 Hg	81 Tl	82 Pb	83 Bi	84 Po

(Table 1.2: abbreviated periodic chart; the column numbers are the Roman numerals, the numbers are atomic numbers.)

Except for the IV-VI compounds, every semiconductor in Table 1.1 is either built from Column IV elements, or from a pair of elements sitting equal distances on either side of Column IV:

• Column III (Ga) + Column V (As) $\to$ III-V compound GaAs.
• Column II (Zn) + Column VI (Se) $\to$ II-VI compound ZnSe.
• A fractional mix of Column III elements (Al and Ga) + Column V (As) $\to$ alloy ${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}\mathrm{As}$.

The hidden reason (an intuition you should keep). Average the valence electrons: Ga has 3, As has 5; $(3+5)/2=4$. Zn has 2, Se has 6; $(2+6)/2=4$. Whatever the recipe, the average is four valence electrons per atom. That magic number controls the chemical bonding, as we will see when we find that each atom has four neighbors, and it is why the semiconductor club has such exclusive membership. Think of it like a dance where every dancer must hold exactly four hands: you can swap partners of different “hand-counts” only as long as the average stays at four.

1.1.2 Purity

As Chapter 2 will explain, extremely minute traces of impurity atoms called dopants can drastically change a semiconductor’s electrical properties. So the composition must be controlled with extraordinary care, and modern semiconductors are among the purest solids ever made.

In silicon the unintentional dopant content is routinely below

$$\frac{1\text{ impurity atom}}{10^9\text{Si atoms}}$$

A way to feel this number. Plant a forest of maple trees from coast to coast and border to border across the entire United States (Alaska included), one tree every 50 ft. An impurity level of one part in $10^9$ would correspond to finding only about 25 crabapple trees in that whole continental maple forest. That is the purity of the raw material.

Do not be confused: this refers to unintentional, undesired impurities. Dopant atoms are purposely added at levels ranging from about one part in $10^8$ up to one impurity atom per $10^3$ semiconductor atoms to control the electrical behavior. So the recipe is: make it almost perfectly pure, then sprinkle in exactly the amount of “spice” you want.

1.1.3 Structure

The spatial arrangement of atoms strongly influences material properties. Figure 1.1 (three sketches of atoms) classifies solids into three broad types.

Figure 1.1 (described). (a) Amorphous: dots scattered with no repeating pattern. (b) Polycrystalline: several patches, each of which is an ordered array, but each patch is tilted relative to its neighbors. (c) Crystalline: one perfectly regular array filling the whole picture.

Type	Description	Analogy
Amorphous	No recognizable long-range order; the arrangement in one section looks different from any other section.	A pile of marbles dumped into a box.
Polycrystalline	Composed of crystalline subsections (grains) that are disjointed or misoriented relative to each other.	A patchwork quilt of tiled floors laid by workers who never agreed on orientation.
Crystalline	Atoms form an orderly 3-D array; you can reproduce the arrangement in any section from any other section.	A perfectly stacked orchard, trees in flawless rows in every direction.

Examples from real devices: an amorphous-Si thin-film transistor is the switch inside liquid crystal displays (LCDs); polycrystalline-Si gates are used in MOSFETs. But in the vast majority of devices, the active region sits in a crystalline semiconductor. Hence the next section.

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1.2 Crystal Structure

Since working semiconductors are typically crystalline, we want a more detailed picture of the atomic arrangement in the principal semiconductors. The plan of this section: (1) how to describe atom positions in crystals (unit cells); (2) practice with simple 3-D lattices; (3) the semiconductor lattices themselves; (4) Miller indices, a shorthand for planes and directions.

1.2.1 The Unit Cell Concept

Definition. A unit cell is a small portion of a crystal that could be used to reproduce the crystal.

Figure 1.2 (described). (a) A sample 2-D lattice of atoms. (b) A unit cell: a square of side $a$ with a quarter-atom at each corner. (c) The original lattice rebuilt by copying the cell and stacking copies edge to edge. (d) An alternative, shifted choice of unit cell that works equally well.

Think of wallpaper: you never need the whole wall’s pattern, only one repeating tile. Copy the tile and tessellate it, and the wall appears. That tile is the unit cell.

Two common confusions:

1. Unit cells are not unique. The cell of Fig. 1.2(d) specifies the lattice just as well as that of Fig. 1.2(b). Any tile that repeats to fill space works.
2. A unit cell need not be primitive (the smallest possible cell). It is usually better to use a slightly larger cell with orthogonal (right-angle) sides rather than a primitive cell with slanted sides, especially in 3-D, where non-cubic cells are hard to describe and visualize.

1.2.2 Simple 3-D Unit Cells

Semiconductor crystals are three-dimensional, so we use 3-D cells.

Figure 1.3 (described). Four cubes with side length (the lattice constant) $a$: (a) simple cubic: atoms at the 8 corners; (b) the “pedantically correct” simple cubic showing only the $1/8$ of each corner atom inside the cell; (c) body-centered cubic (bcc): corners plus one atom at the cube center; (d) face-centered cubic (fcc): corners plus one atom at the center of each of the 6 faces.

Simple cubic. An equal-sided box with an atom at each corner. Only $1/8$ of each corner atom lies inside the cell (the other seven octants belong to neighboring cells). Stack copies like nursery blocks to get the simple cubic lattice: planes of atoms separated by the lattice constant $a$, and viewed from above, atoms in each plane sit directly over atoms in the plane below.

bcc adds one atom at the cube center. fcc adds one atom at the center of each face, of which only half lies inside the cell.

Counting atoms per cell (verify these yourself!):

$$\text{simple cubic: }8\times \frac{1}{8}=1$$ $$\text{bcc: }8\times \frac{1}{8}+1=2$$ $$\text{fcc: }8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$$

Intuition for the counting: a corner atom is shared by 8 cubes (so contributes $1/8$), a face atom is shared by 2 cubes (so $1/2$), and a body-center atom belongs to one cube (so $1$).

Exercise 1.1 (computer-based, visualization)

Task. The book’s software contains a directory MacMolecule with ASCII input files for a Macintosh program (also named MacMolecule, from the University of Arizona) that draws “ball-and-stick” 3-D colour renderings of molecules, unit cells and lattices. The files visualize the simple cubic, bcc and fcc cells and the diamond and zincblende cells.

What to do and why. The first display is a $z$-direction view, so it looks nearly two-dimensional. Rotate the model by grabbing and dragging its edges to see the depth. Advanced users may edit or write input files. There is no equation or MATLAB code here; the exercise trains 3-D intuition, which is exactly the skill you need for the diamond lattice in the next subsection. (If you do not have a Mac, any free crystal viewer or a short script that plots atom coordinates can serve the same purpose; see the MATLAB code in Exercise 1.3 for a starting point on coordinates.)

1.2.3 Semiconductor Lattices

Now we position atoms in the real semiconductors.

Diamond lattice (Si and Ge). Figure 1.4(a) shows the diamond lattice unit cell (named after diamond, carbon, which is also a Column IV element).

Figure 1.4 (described). (a) A cube with atoms at all corners and face centers (like fcc) plus four additional atoms inside the cube. (b) The zincblende cell drawn with GaAs: same geometry, but two kinds of atoms (Ga and As). (c) An enlarged top corner of (a) showing one atom at the cube corner bonded to four nearest neighbors, which sit at the vertices of a tetrahedron; the corner-to-neighbor distance is one quarter of the body diagonal; marks $a/2$ indicate half the cube side. Cube side lengths: $a=5.43\text{A˚}$ for Si and $a=5.65\text{A˚}$ for GaAs at $T=300\text{K}$.

Where are the four interior atoms? In coordinates $(x,y,z)$ in units of $a$, with the origin at the top-front-left corner and measuring along the body diagonal:

• One interior atom lies along the body diagonal, exactly $1/4$ of the way down from the top-front-left corner.
• The other three are displaced by the same quarter-body-diagonal vector from the front, top and left face-center atoms, respectively.

Cleaner way to see it: the diamond lattice is just two interpenetrating fcc lattices. The corner and face atoms belong to one fcc lattice; the four interior atoms belong to a second fcc lattice, displaced by one-quarter of a body diagonal along the body-diagonal direction relative to the first.

Atoms per diamond cell: fcc contributes 4, the interior gives 4 more:

$$N_{\text{diamond}}=4+4=8$$

Zincblende lattice (most III-V, including GaAs). Essentially identical to diamond, except the lattice sites are shared between two different atoms. Ga occupies one of the interpenetrating fcc sublattices; As occupies the other.

Why bother? A useful application: atomic density. With the unit cell we can do geometry. At room temperature the Si lattice constant is $a=5.43\text{A˚}$ (where $1\text{A˚}=10^{-8}\text{cm}$). There are 8 atoms per cell and the cell volume is $a^3$, so

$$n_{\text{Si}}=\frac{8}{a^3}$$ $$=\frac{8}{(5.43\times 10^{-8}\text{cm}{)}^3}$$ $$\approx 5\times 10^{22}{\text{atoms/cm}}^3$$

(almost exactly; the precise value is $4.9968\times 10^{22}$ as the MATLAB run below shows). Similar calculations give atomic radii, distances between atomic planes, and so on.

The key physical fact. For our purposes the most important message is (Fig. 1.4(c)) that atoms in the diamond and zincblende lattices have four nearest neighbors. The chemical bonding (“crystalline glue”) in the major semiconductors is therefore dominated by the attraction between each atom and its four closest neighbors, matching the four valence electrons per atom from Section 1.1.1. Each atom shares one electron with each of four neighbors: four bonds, four electrons, a stable arrangement. Store this away; Chapter 2 builds on it.

Exercise 1.2

Problem. If the lattice constant of Si is $a=5.43\times 10^{-8}$ cm, what is the distance $d$ from the center of one Si atom to the center of its nearest neighbor?

Solution.

The atom at the top-front corner and the interior atom lying one quarter of the way down the body diagonal are nearest neighbors (Fig. 1.4 caption).

The body diagonal of a cube of side $a$ is

$$L_{\text{diag}}=\sqrt{a^2+a^2+a^2}=\sqrt{3}a\text{(Pythagoras in 3-D)}$$

The nearest neighbor sits a quarter of the way along it:

$$d=\frac{1}{4}{L}_{\text{diag}}=\frac{\sqrt{3}}{4}a$$ $$=\frac{\sqrt{3}}{4}(5.43\times 10^{-8}\text{cm})$$ $$=0.4330\times 5.43\times 10^{-8}\text{cm}$$ $$\boxed{d\approx 2.35\times 10^{-8}\text{cm}=2.35\text{A˚}}$$

Exercise 1.3 (computer-based, MATLAB)

Task. Write a MATLAB program that computes the number of atoms per cm${}^3$ in a cubic crystal. Use the input function to enter the number of atoms per unit cell and the lattice constant $a$ for a specific crystal. List the program and record the output for silicon.

Logic. The atom density is (atoms per cell) divided by (cell volume):

$$n=\frac{N}{a^3}$$

If $a$ is entered in angstroms, convert to cm: $a_{\text{cm}}=a_{\text{A˚}}\times 10^{-8}$, so $a_{\text{cm}}^3=a_{\text{A˚}}^3\times 10^{-24}$, and therefore

$$n=\frac{N(10^{24})}{a_{\text{A˚}}^3}{\text{cm}}^{-3}$$

That is exactly the line in the program (the factor $10^{24}$ does the unit conversion so the user can type the convenient number 5.43).

% Exercise 1.3
% Computation of the number of atoms/cm^3 in a cubic lattice
N = input('input number of atoms/unit cell, N = ');
a = input('lattice constant in angstrom, a = ');
atmden = N*(1.0e24)/(a^3)   % number of atoms/cm^3 (no semicolon, so MATLAB prints it)

Program output for Si:

input number of atoms/unit cell, N = 8
lattice constant in angstrom, a = 5.43
atmden =
   4.9968e+22

Check by hand: $5.43^3=160.10$, so $8\times 10^{24}/160.10=4.9968\times 10^{22}{\text{cm}}^{-3}$. Matches.

1.2.4 Miller Indices

Device-grade silicon comes as thin, round wafers (Fig. 1.5): 150 mm (6”) and 200 mm (8”) diameter wafers about 0.625 mm and 0.725 mm thick. The face is polished and etched to a mirror finish. Crucially:

• The wafer surface is carefully pre-oriented along a specific crystallographic plane, e.g. a $(100)$ plane.
• A flat or notch ground along the periphery marks a reference direction in the surface plane, e.g. a $[011]$ direction (Fig. 1.5 labels the $(100)$ surface and the $[011]$ directions for the notch and the flat).

Why care? Surface orientation affects processing steps and the characteristics of devices. The flat is used, for example, to orient device arrays on the wafer so that, when the wafer is sawed up (separated into individual dice), you get high yield without shattering chips along the “wrong” crystal direction. Miller indices are the standard language for planes and directions.

Indexing procedure for planes (four steps). I carry out a sample with the plane of Fig. 1.6(a).

Figure 1.6 (described). (a) A plane cutting the $x$-axis at $1a$, the $y$-axis at $2a$ and the $z$-axis at $3a$. (b) The $(001)$ plane, parallel to the $xy$ plane, cutting only the $z$-axis at $a$. (c) The $(2\bar{2}1)$-type plane of the text; see the caution below.

Step	Procedure	Sample (Fig. 1.6(a))
1	Set axes along the unit-cell edges. Note where the plane intercepts each axis; divide each intercept by the cell length along that axis; record the normalized intercepts in order $x,y,z$.	$1,2,3$
2	Invert: form $[1/\text{intercept}]$s.	$1,\frac{1}{2},\frac{1}{3}$
3	Multiply by the smallest number that turns the set into the smallest possible whole numbers.	$6,3,2$
4	Enclose in parentheses.	$(632)$

Step 3 detail: the least common multiple of the denominators $1,2,3$ is $6$, so $6\times (1,\frac{1}{2},\frac{1}{3})=(6,3,2)$.

Special facts.

1. A plane parallel to an axis intercepts it at infinity, so the reciprocal is $0$. Example: Fig. 1.6(b) intercepts the axes at $\infty ,\infty ,1$, inverse $0,0,1$, so it is a $(001)$ plane.
2. If the plane crosses the negative side of an axis, a minus sign (written as an overbar) goes over that index. The Fig. 1.6(c) plane, with one negative intercept, is the type of plane written $(2\bar{2}1)$ in the book (the overbar was lost in the scan; the idea is that the plane has a negative intercept on one axis).
3. Equivalent planes. In the diamond lattice the six planes that coincide with the cube faces contain identical atomic arrangements, so crystal symmetry makes $(100),(010),(001),(\bar{1}00),(0\bar{1}0),(00\bar{1})$ indistinguishable. A group of equivalent planes is written with braces: $\{100\}$.
4. Planes through the origin cannot be indexed (their intercepts would be $0$ and the reciprocals infinite). Move the origin to a lattice point not on the plane; this is legitimate because parallel planes are equivalent.

Figure 1.7 (described). (a) Unit cubes showing the $(100)$ plane (the face perpendicular to $x$), the $(110)$ plane (a diagonal slice parallel to $z$, hitting $x$ and $y$ at $a$), and the $(111)$ plane (the triangle through $(a,0,0),(0,a,0),(0,0,a)$). (b) Direction vectors $[100]$ (along $x$), $[110]$ (face diagonal in the $xy$ plane), $[111]$ (body diagonal).

Directions. Treat a direction as a vector:

1. Draw a vector of arbitrary length along the direction.
2. Find its projections (components) on the axes.
3. Multiply to get the smallest whole numbers (length changes, direction does not).
4. Enclose in square brackets $[]$.

Table 1.3 Miller convention summary.

Convention	Interpretation
$(hkl)$	Crystal plane
$\{hkl\}$	Equivalent planes
$[hkl]$	Crystal direction
$⟨hkl⟩$	Equivalent directions

Practical tips.

• The inverse job (given indices, picture the plane) is easy for the low-index set, e.g. $(111)$, $(110)$, $[001]$. Memorize those.
• For cubic crystals, the direction $[hkl]$ is normal to the plane $(hkl)$. E.g. $[110]\perp (110)$.

Exercise 1.4

Problem. For a cubic crystal lattice: (a) determine the Miller indices of the plane and direction vector shown in Fig. E1.4(a); (b) sketch the plane and direction vector characterized by $(011)$ and $[011]$.

Solution (a). From the figure, the plane intercepts the $x$, $y$, $z$ axes at $-1,1,2$ (in units of $a$). The direction vector has projections $2a,a,0$.

Plane:

$$\text{intercepts: }(-1,1,2)$$ $$[1/\text{intercept}]\text{s: }\left(-1,1,\frac{1}{2}\right)$$ $$\times 2\text{ (lowest whole numbers): }(-2,2,1)$$ $$\boxed{(\bar{2}21)}$$

Direction:

$$\text{projections: }(2,1,0)\text{(already whole, no common factor)}$$ $$\boxed{[210]}$$

Solution (b). For the $(011)$ plane:

$$[1/\text{intercept}]\text{s}=(0,1,1)\Rightarrow \text{intercepts}=(\infty ,1,1)$$

The plane meets the $y$-axis at $a$, the $z$-axis at $a$, and is parallel to the $x$-axis. In a cubic crystal, $[011]$ is normal to the $(011)$ plane, so the direction vector is drawn from the origin to the point $(0,a,a)$ (perpendicular to the plane, which itself is the slanted rectangle containing the $x$ direction and the line from $(0,a,0)$ to $(0,0,a)$). These are the sketches of Fig. E1.4(b).

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1.3 Crystal Growth

1.3.1 Obtaining Ultrapure Si

Where does device silicon come from? Not sandstone deposits? No. A by-product of South African diamond mines? No. Scooped from the ocean floor by special submarines? No. Although Si is the second most abundant element in the Earth’s crust and a component of many compounds (chiefly silica, impure $\mathrm{Si}{\mathrm{O}}_2$, and the silicates, Si + O + another element), silicon never occurs alone in nature. Single-crystal Si is a man-made material.

So step one is to separate Si from its compounds and purify it (Fig. 1.8).

Figure 1.8 (flow chart described).

$$\text{Silica}\underset{\text{in presence of C}}{\overset{\text{reduced}}{\to }}\text{very impure Si}\stackrel{\text{chlorinated}}{\to }{\text{SiCl}}_4\text{(liquid)}\stackrel{\text{distilled, etc.}}{\to }{\text{ultrapure SiCl}}_4\underset{\text{atmosphere}}{\overset{{\text{reduced in H}}_2}{\to }}\text{ultrapure polycrystalline Si}$$

(In the scan the intermediate labels read “SiC”; the text clarifies that the chlorinated products are $\mathrm{SiC}{\mathrm{l}}_4$ or $\mathrm{SiHC}{\mathrm{l}}_3$.)

1. Reduction. Silica is heated with carbon in an electric furnace. Carbon pulls the oxygen away from $\mathrm{Si}{\mathrm{O}}_2$ (reduces it), leaving impure (low-grade, “ferrosilicon”) Si.
2. Chlorination. The impure Si is chlorinated to give $\mathrm{SiC}{\mathrm{l}}_4$ or $\mathrm{SiHC}{\mathrm{l}}_3$, which are liquids at room temperature.
3. Distillation. Liquids are much easier to purify than solids (standard methods like repeated distillation exist). The result is ultrapure $\mathrm{SiC}{\mathrm{l}}_4$ or $\mathrm{SiHC}{\mathrm{l}}_3$. Turning the solid into a liquid is the clever maneuver: it is like washing a dusty rug by unraveling it into thread instead of beating it.
4. Chemical reduction. The ultrapure halide is reduced to elemental Si, e.g. by heating in hydrogen:

$$\mathrm{SiC}{\mathrm{l}}_4+2{\mathrm{H}}_2⟶4HCl+\mathrm{Si}$$

1.3.2 Single-Crystal Formation

The silicon so obtained is ultrapure but polycrystalline. To make the large single crystals used in devices, we use the Czochralski method.

Figure 1.9 (described). (a) Photo of a computer-controlled Czochralski crystal puller. (b) Schematic: a closed chamber filled with argon (Ar) with vacuum pumps and a gas valve; inside, a graphite crucible holding a silica (quartz) crucible with the Si melt, surrounded by a graphite heater and a thermal shield; a crucible shaft below rotates the crucible; above, a seed shaft carries a seed holder with a seed crystal that dips into the melt; a viewing port with an optical system lets operators monitor; exhaust gas is Ar + SiO + CO. (c) Photo of the resulting silicon ingot.

Steps.

1. Place ultrapure polycrystalline Si in a quartz crucible and heat it in an inert atmosphere (argon) to form a melt.
2. Take a small seed crystal of single-crystal Si whose bottom face normal is carefully aligned along a chosen direction (typically a $⟨111⟩$ or $⟨100⟩$ direction), clamp it to a metal rod, and dip it into the melt.
3. Once thermal equilibrium is reached, lower the temperature of the melt near the seed. Silicon freezes onto the seed, and the added material is a structurally perfect extension of the seed’s lattice.
4. Slowly rotate and withdraw the seed. More and more Si freezes onto the bottom of the growing crystal.

The result is a large, cylindrical single crystal, an ingot, routinely up to 200 mm (8 in) in diameter and 1 to 2 m long. Wafers (Fig. 1.5) are made by slicing the ingot into thin discs with a diamond-edged saw.

Intuition: it is like growing rock candy on a string. The string (seed) tells the new sugar (silicon) exactly how to line up, so the grown crystal continues the same orientation forever. The slow rotation averages out temperature irregularities so the ingot grows round and uniform.

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1.4 Summary

• Semiconductor compositions: elemental (Si, Ge), compounds (III-V, II-VI, etc.) and alloys. On average there are four valence electrons per atom.
• Device-quality semiconductors are typically single crystals of very high compositional purity.
• Si crystallizes in the diamond lattice; GaAs in the zincblende lattice; in both cases each atom has four nearest neighbors.
• Miller indices are the accepted means of identifying planes and directions within a crystal.
• Large single crystals of device-grade Si are commonly produced by the Czochralski method.

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Problems (Chapter 1)

Problem information table (as in the book).

Problem	Complete after	Difficulty	Points (weighting)	Short description
1.1	1.4	-	10 (a-h: 1 each, i: 2)	Short answer review
1.2	1.2.3	1	8	Unit cell for alloy
1.3	-	-	10 (5 each part)	Ge atomic density
1.4	-	-	8 (4 each part)	Nearest-neighbor distances
1.5	1.2.4	2-3	15 (a-2, b-2, c-3, d-3, e-5)	Atoms/cm${}^2$ on planes
1.6	-	1	8 (2 each subpart)	Determine Miller indices
1.7	-	1	16 (2 each part)	Given (hkl), sketch plane
1.8	-	1	16 (2 each part)	Given [hkl], sketch direction
1.9	-	1	8 (4 each part)	Perpendicular directions
1.10	-	2	10	Direction on wafer surface
1.11	-	2	10 (a-3, b-2, c-3, d-2)	Combination question
1.12	-	2	12 (2 each part)	Equivalent planes/directions
1.13	1.2.3	3	20 (5 each part)	% of volume occupied

Problem 1.1 Quick Quiz

Problem. Answer concisely. (a) Name (i) one elemental semiconductor and (ii) one III-V compound semiconductor. (b) Difference between crystalline and polycrystalline? (c) Word definition of unit cell. (d) How many atoms in a simple cubic, bcc, fcc and diamond unit cell? (e) $1\text{A˚}=$ ? cm. (f) In terms of $a$, distance between nearest neighbors in a simple cubic lattice? (g) How many nearest neighbors in the diamond and zincblende lattices? (h) What is being indicated by the bracket sets $()$, $[]$, $\{\}$, $⟨⟩$ in Miller indexing? (i) Describe the Czochralski method.

Solution.

(a) (i) Silicon (Si) or germanium (Ge). (ii) GaAs (also InP, GaP, etc.).

(b) A crystalline solid has atoms in one orderly 3-D array throughout; a polycrystalline solid is made of many crystalline grains that are misoriented/disjointed relative to each other, so order exists only within each grain.

(c) A unit cell is a small portion of a crystal that can be used to reproduce the entire crystal by stacking copies of it.

(d) Counting corner ($1/8$), face ($1/2$), interior (1):

$$\text{simple cubic: }8\left(\frac{1}{8}\right)=1,\text{bcc: }8\left(\frac{1}{8}\right)+1=2$$ $$\text{fcc: }8\left(\frac{1}{8}\right)+6\left(\frac{1}{2}\right)=4,\text{diamond: }4(\text{fcc})+4(\text{interior})=8$$

(e) $1\text{A˚}=10^{-8}\text{cm}$.

(f) In a simple cubic lattice the nearest neighbors are one cell edge apart: $d=a$.

(g) Four nearest neighbors in both diamond and zincblende lattices.

(h) $(hkl)$: a specific crystal plane; $[hkl]$: a specific crystal direction; $\{hkl\}$: a set of equivalent planes; $⟨hkl⟩$: a set of equivalent directions.

(i) Ultrapure polycrystalline Si is melted in a quartz crucible in an inert atmosphere. A small oriented single-crystal seed on a rod is dipped in the melt. The melt near the seed is cooled slightly so Si freezes onto the seed as a perfect extension of its lattice. The seed is slowly rotated and withdrawn, pulling out a large cylindrical single-crystal ingot (up to about 200 mm diameter, 1-2 m long) that is later sawed into wafers.

Problem 1.2 Unit cell of ${\mathrm{Al}}_{0.5}{\mathrm{Ga}}_{0.5}\mathrm{As}$

Problem. The GaAs unit cell is pictured in Fig. 1.4(b). Describe (or sketch) the unit cell for ${\mathrm{Al}}_{0.5}{\mathrm{Ga}}_{0.5}\mathrm{As}$.

Solution.

The zincblende cell contains 8 atoms: 4 on one fcc sublattice and 4 on the other. In GaAs, the 4 Ga atoms sit on the sublattice (cube corners and face centers, if we choose that one) and the 4 As atoms sit on the other (the four interior positions).

In ${\mathrm{Al}}_x{\mathrm{Ga}}_{1-x}\mathrm{As}$ the Group III sublattice is shared between Al and Ga, and the As sublattice stays purely As. With $x=0.5$:

$$N_{\text{group III sites}}=4,N_{\mathrm{Al}}=x\cdot 4=0.5\times 4=2,N_{\mathrm{Ga}}=(1-x)\cdot 4=2,N_{\mathrm{As}}=4$$

So the cell is the same zincblende cube (side about 5.65 Å, nearly the same as GaAs), with 4 As atoms on one fcc sublattice and, on the other fcc sublattice, 2 Al and 2 Ga atoms that are randomly distributed over those 4 sites. (For a sketch: replace two of the four Ga atoms in Fig. 1.4(b), for instance two of the face-center or corner Ga atoms, with Al.) On average, of course, every Group III site is “half Al, half Ga”.

Problem 1.3 Germanium atomic density

Problem. (a) The lattice constant of Ge at room temperature is $a=5.65\times 10^{-8}$ cm. Determine the number of Ge atoms per cm${}^3$. (b) Run the Exercise 1.3 program to verify your result from (a).

Solution (a). Ge has the diamond structure: 8 atoms per cell.

$$n=\frac{8}{a^3}$$ $$a^3=(5.65\times 10^{-8}{)}^3=180.36\times 10^{-24}{\text{cm}}^3=1.8036\times 10^{-22}{\text{cm}}^3$$ $$n=\frac{8}{1.8036\times 10^{-22}{\text{cm}}^3}$$ $$\boxed{n\approx 4.44\times 10^{22}{\text{atoms/cm}}^3}$$

Solution (b). Run the program with N = 8, a = 5.65:

input number of atoms/unit cell, N = 8
lattice constant in angstrom, a = 5.65
atmden =
   4.4356e+22

since $8\times 10^{24}/5.65^3=8\times 10^{24}/180.36=4.4356\times 10^{22}$. This agrees with part (a).

Problem 1.4 Nearest-neighbor distances

Problem. In terms of the lattice constant $a$, what is the distance between nearest-neighbor atoms in (a) a bcc lattice? (b) an fcc lattice?

Solution (a), bcc. The center atom touches the 8 corner atoms. The distance from the cube center to a corner is half a body diagonal:

$$d_{\text{bcc}}=\frac{1}{2}\sqrt{a^2+a^2+a^2}=\frac{\sqrt{3}}{2}a\approx 0.866a$$

(Check that this beats the corner-to-corner distance $a$: $0.866a<a$, so yes it is the nearest.)

Solution (b), fcc. A corner atom and the atom at the center of an adjacent face are separated by half a face diagonal:

$$d_{\text{fcc}}=\frac{1}{2}\sqrt{a^2+a^2}=\frac{\sqrt{2}}{2}a=\frac{a}{\sqrt{2}}\approx 0.707a$$

(Again less than $a$.)

Problem 1.5 Atoms per cm${}^2$ on wafer surfaces

Problem. The surface of a Si wafer is a $(100)$ plane. (a) Sketch the placement of Si atoms on the surface. (b) Determine the number of atoms per cm${}^2$ at the surface. (c)/(d) Repeat (a) and (b) for a $(110)$ surface. (e) Write a MATLAB program that computes the surface density of atoms on $(100)$ planes of cubic crystals, with the number of atoms centered on the $(100)$ face of the unit cell and the lattice constant in angstrom as inputs; confirm the result of (b).

Solution (a). The $(100)$ plane is one face of the cubic cell, a square of side $a$. In the diamond lattice, the atoms with centers on that face are the four corner atoms and the one atom at the face center (a square pattern with an additional atom in the middle; no interior atoms lie in the face). The interior atoms are at $a/4$ and $3a/4$ depths, not on the face.

Solution (b). Atoms per face (corner atoms are shared among 4 squares in the plane, so each counts $1/4$):

$$N_{\text{face}}=4\left(\frac{1}{4}\right)+1=2$$

Area of the face:

$$A=a^2=(5.43\times 10^{-8}\text{cm}{)}^2=2.949\times 10^{-15}{\text{cm}}^2$$ $${\sigma }_{(100)}=\frac{2}{a^2}=\frac{2}{2.949\times 10^{-15}{\text{cm}}^2}$$ $$\boxed{{\sigma }_{(100)}\approx 6.78\times 10^{14}{\text{atoms/cm}}^2}$$

Solution (c). The $(110)$ plane is a rectangle cutting the cube diagonally. With $x+y=a$ it contains the four points $(a,0,0),(0,a,0),(a,0,a),(0,a,a)$, so its sides are $a$ (along $z$) and $\sqrt{2}a$ (along the face diagonal). Atoms with centers on this plane:

• the 4 corner points of the rectangle, which are fcc lattice sites: each shared by 4 rectangles in the plane, so $4\times \frac{1}{4}=1$;
• two face-center atoms located at the midpoints of the two long edges, e.g. $(a/2,a/2,0)$ and $(a/2,a/2,a)$, each shared by 2 rectangles: $2\times \frac{1}{2}=1$;
• two of the diamond interior atoms, $(3a/4,a/4,3a/4)$ and $(a/4,3a/4,3a/4)$, which both satisfy $x+y=a$ and lie fully inside the rectangle: $2$.

(For the sketch, draw a rectangle $a$ tall by $\sqrt{2}a$ wide with atoms at the four corners, the midpoints of the top and bottom edges, and two atoms inside, positioned at one-quarter and three-quarters of the width, at three-quarters of the height.)

Solution (d).

$$N_{(110)}=1+1+2=4,A=\sqrt{2}{a}^2$$ $${\sigma }_{(110)}=\frac{4}{\sqrt{2}{a}^2}=\frac{2\sqrt{2}}{a^2}=\frac{2.828}{2.949\times 10^{-15}{\text{cm}}^2}$$ $$\boxed{{\sigma }_{(110)}\approx 9.59\times 10^{14}{\text{atoms/cm}}^2}$$

Solution (e), MATLAB. The surface density is (atoms on the face)/(face area). With $a$ in angstrom, $a^2$ in cm${}^2$ equals $a_{\text{A˚}}^2\times 10^{-16}$, so $\sigma =N_{\text{face}}\times 10^{16}/a_{\text{A˚}}^2$.

% Problem 1.5(e)
% Surface density of atoms on (100) planes of cubic crystals
N = input('number of atoms with centers on the (100) face, N = ');
a = input('lattice constant in angstrom, a = ');
sdens = N*(1.0e16)/(a^2)   % atoms/cm^2

Run for Si:

number of atoms with centers on the (100) face, N = 2
lattice constant in angstrom, a = 5.43
sdens =
   6.7831e+14

This confirms the part (b) result ($6.78\times 10^{14}{\text{cm}}^{-2}$).

Problem 1.6 Determining Miller indices

Problem. Record all steps. (a) As shown in Fig. P1.6(a), a crystalline plane has intercepts of $1a$, $3a$ and $1a$ on the $x$, $y$ and $z$ axes ($a$ = cubic cell side). (i) Miller index notation for the plane? (ii) Miller indices of the direction normal to the plane? (b) Assuming a cubic structure, determine the Miller indices of (i) the plane and (ii) the vector pictured in Fig. P1.6(b).

Assumption note: the figure labels in the scan are garbled. I read Fig. P1.6(a) as intercepts $(1,3,1)\times a$, and Fig. P1.6(b) as a plane with intercepts $x=a$, $y=a$, $z=-2a$ together with a vector whose projections are $(a,a,-a/2)$. If your figure differs, apply exactly the same steps to the values you read.

Solution (a)(i).

$$\text{normalized intercepts: }(1,3,1)$$ $$[1/\text{intercept}]\text{s: }\left(1,\frac{1}{3},1\right)$$ $$\times 3:(3,1,3)$$ $$\boxed{(313)}$$

Solution (a)(ii). In a cubic crystal the normal to $(hkl)$ is $[hkl]$:

$$\boxed{[313]}$$

Solution (b)(i).

$$\text{intercepts: }(1,1,-2)$$ $$[1/\text{intercept}]\text{s: }\left(1,1,-\frac{1}{2}\right)$$ $$\times 2:(2,2,-1)$$ $$\boxed{(22\bar{1})}$$

Solution (b)(ii).

$$\text{projections: }\left(1,1,-\frac{1}{2}\right)\times 2=(2,2,-1)$$ $$\boxed{[22\bar{1}]}$$

(consistent: for a cubic crystal this direction is normal to the plane of part (i)).

Problem 1.7 Sketching planes

Problem. Assuming a cubic crystal system, sketch the following planes: (a) $(001)$, (b) $(111)$, (c) $(123)$, (d) $(\bar{1}10)$, (e) $(010)$, (f) $(\bar{1}\bar{1}\bar{1})$-type plane as printed, (g) $(2\bar{2}1)$-type plane as printed, (h) $(0\bar{1}0)$-type plane as printed.

Assumption note: overbars were lost in the OCR for parts (f)-(h). I take (f) $=(\bar{1}11)$, (g) $=(\bar{2}21)$, (h) $=(0\bar{1}0)$, and give the general recipe so any other sign choice follows immediately.

Method for every plane. Invert the indices to get intercepts, $x=a/h,y=a/k,z=a/l$ (an index $0$ means parallel to that axis), then mark the three intercepts on the axes and join them. Negative indices put the intercept on the negative axis.

Part	$(hkl)$	Intercepts $(x,y,z)$ in units of $a$	Description of the sketch
(a)	$(001)$	$(\infty ,\infty ,1)$	The top face of the unit cube: horizontal square at $z=a$, parallel to the $xy$ plane.
(b)	$(111)$	$(1,1,1)$	The equilateral triangle joining $(a,0,0)$, $(0,a,0)$ and $(0,0,a)$.
(c)	$(123)$	$(1,\frac{1}{2},\frac{1}{3})$	Triangle joining $(a,0,0)$, $(0,a/2,0)$, $(0,0,a/3)$; a steep plane.
(d)	$(\bar{1}10)$	$(-1,1,\infty )$	A vertical plane parallel to $z$, cutting the $-x$ axis at $-a$ and the $+y$ axis at $+a$.
(e)	$(010)$	$(\infty ,1,\infty )$	The face at $y=a$, parallel to the $xz$ plane.
(f)	$(\bar{1}11)$	$(-1,1,1)$	Triangle joining $(-a,0,0)$, $(0,a,0)$, $(0,0,a)$.
(g)	$(\bar{2}21)$	$(-\frac{1}{2},\frac{1}{2},1)$	Triangle joining $(-a/2,0,0)$, $(0,a/2,0)$, $(0,0,a)$.
(h)	$(0\bar{1}0)$	$(\infty ,-1,\infty )$	The face at $y=-a$, parallel to $xz$ (the opposite face from (e)).

If a plane would pass through the origin (all indices not zero cannot, but a sketch from shifted indices might), translate the origin to a neighboring lattice point first, as noted in special fact 4.

Problem 1.8 Sketching directions

Problem. Assuming a cubic system, use an appropriately directed arrow to identify each direction.

Assumption note: the printed list is scrambled by the OCR and its overbars are lost, so I treat the eight directions as (a) $[\bar{1}\bar{1}\bar{1}]$, (b) $[\bar{1}00]$, (c) $[0\bar{1}1]$, (d) $[00\bar{1}]$, (e) $[\bar{1}10]$, (f) $[\bar{2}\bar{1}0]$, (g) $[\bar{1}21]$, (h) $[1\bar{1}0]$. The method works for any triple.

Method. The direction $[hkl]$ is the arrow from the origin to the point $(h,k,l)$ (in any common unit). Draw the arrow from the origin to that point.

Part	Direction	Arrow from origin to	Description
(a)	$[\bar{1}\bar{1}\bar{1}]$	$(-1,-1,-1)$	Body diagonal, pointing toward the rear-bottom-left corner.
(b)	$[\bar{1}00]$	$(-1,0,0)$	Along the negative $x$-axis.
(c)	$[0\bar{1}1]$	$(0,-1,1)$	Face diagonal of the $yz$ face, toward $-y$ and $+z$.
(d)	$[00\bar{1}]$	$(0,0,-1)$	Straight down the $-z$ axis.
(e)	$[\bar{1}10]$	$(-1,1,0)$	Face diagonal in the $xy$ plane toward $-x,+y$.
(f)	$[\bar{2}\bar{1}0]$	$(-2,-1,0)$	In the $xy$ plane, twice as far along $-x$ as along $-y$.
(g)	$[\bar{1}21]$	$(-1,2,1)$	Component $-1$ in $x$, $+2$ in $y$, $+1$ in $z$.
(h)	$[1\bar{1}0]$	$(1,-1,0)$	Face diagonal in the $xy$ plane toward $+x,-y$.

Problem 1.9 Perpendicular directions

Problem. Identify two crystalline directions in a cubic crystal that are perpendicular to (a) the $[100]$ direction, (b) the $[111]$ direction.

The note in the book: for directions $[h_1{k}_1{l}_1]$ and $[h_2{k}_2{l}_2]$ in a cubic crystal,

$$\cos \theta =\frac{h_1{h}_2+k_1{k}_2+l_1{l}_2}{\sqrt{(h_1^2+k_1^2+l_1^2)(h_2^2+k_2^2+l_2^2)}}$$

so perpendicularity ($\cos \theta =0$) requires

$$h_1{h}_2+k_1{k}_2+l_1{l}_2=0$$

Solution (a). With $[h_1{k}_1{l}_1]=[100]$:

$$1\cdot h_2+0\cdot k_2+0\cdot l_2=0\Rightarrow h_2=0$$

Any $[0kl]$ works. Two choices:

$$\boxed{[010]\text{and}[001]}$$

(also $[011]$, $[01\bar{1}]$, and so on).

Solution (b). With $[111]$:

$$h_2+k_2+l_2=0$$

Two simple solutions:

$$[1\bar{1}0]:1-1+0=0✓[01\bar{1}]:0+1-1=0✓$$ $$\boxed{[1\bar{1}0]\text{and}[01\bar{1}]}$$

(also $[11\bar{2}]$: $1+1-2=0$.)

Problem 1.10 Grooves on a wafer surface

Problem. As in Fig. 1.5, $[011]$ is the direction in the surface plane normal to the major flat when the wafer surface is $(100)$. To construct a device structure, a parallel set of grooves must be etched in the $(100)$ surface plane along the $[01\bar{1}]$ direction (OCR lost the overbar; I assume $[01\bar{1}]$). Sketch how the grooves are oriented relative to the flat and explain.

Solution.

The surface is $(100)$, so every in-plane direction $[0kl]$ has $h=0$. Check that $[01\bar{1}]$ is in the surface:

$$h=0✓$$

Angle with the flat’s normal direction $[011]$:

$$\cos \theta =\frac{(0)(0)+(1)(1)+(-1)(1)}{\sqrt{(0+1+1)(0+1+1)}}=\frac{0}{2}=0\Rightarrow \theta =90^{\circ }$$

Since $[011]$ is perpendicular to the flat (normal to it), a direction perpendicular to $[011]$ within the surface is parallel to the flat. Sketch: draw the circular wafer with the flat at the bottom edge; draw parallel grooves as straight lines running parallel to the flat (left-right). (If the problem actually asked for the grooves along $[011]$ itself, they would be perpendicular to the flat: top-to-bottom lines.)

Problem 1.11 Combination question (single atom at the cube center)

Problem. A crystalline lattice is characterized by the cubic unit cell of Fig. P1.11, which has a single atom positioned at the center of the cube. (a) What is the name of the lattice generated by the cell? (b) Number of atoms per unit volume (in terms of $a$). (c) If the crystal has a $(110)$ surface plane, number of atoms per unit area whose centers lie on it. (d) A direction vector is drawn through the center of the atom in the unit cell; specify its Miller indices.

Assumption for (d): the vector drawn in the figure is lost; I take it to pass from the cube corner (origin) through the atom at the center, i.e. along the body diagonal. Any other vector is handled by the projection method.

Solution (a). An atom at the center of the cube and nothing else. Translating the cell by $a/2$ along each axis puts the atom at a corner and gives atoms at the 8 corners: this is the simple cubic lattice (the atoms are separated by $a$ in each direction).

Solution (b). One atom per cell of volume $a^3$:

$$\boxed{n=\frac{1}{a^3}}$$

Solution (c). The $(110)$ plane is a rectangle of sides $a$ and $\sqrt{2}a$ whose four corners are lattice points (each atom shared by four rectangles; no other lattice points are on it):

$$N=4\left(\frac{1}{4}\right)=1,A=\sqrt{2}{a}^2$$ $$\boxed{{\sigma }_{(110)}=\frac{1}{\sqrt{2}{a}^2}\approx \frac{0.707}{a^2}}$$

Solution (d). From the corner at the origin to the center at $(a/2,a/2,a/2)$:

$$\text{projections: }\left(\frac{a}{2},\frac{a}{2},\frac{a}{2}\right)\times \frac{2}{a}=(1,1,1)$$ $$\boxed{[111]}$$

Problem 1.12 Equivalent planes and directions

Problem. For a cubic lattice, how many equivalent planes (a)-(c) or equivalent directions (d)-(f) belong to each designation: (a) $\{100\}$, (b) $\{110\}$, (c) $\{111\}$, (d) $⟨100⟩$, (e) $⟨110⟩$, (f) $⟨111⟩$?

Solution. Count all index permutations with all sign choices (the book counts $(100)$ and $(\bar{1}00)$ as different planes, as in its remark about the six cube faces).

(a) $\{100\}$: one index is $\pm 1$ and two are $0$. Choose which index is nonzero (3 ways) times sign (2):

$$3\times 2=\boxed{6}$$

(b) $\{110\}$: two indices $\pm 1$, one $0$: choose which is zero (3) times signs ($2^2=4$):

$$3\times 4=\boxed{12}$$

(c) $\{111\}$: all indices $\pm 1$:

$$2^3=\boxed{8}$$

(d) $⟨100⟩$: same counting as (a): $\boxed{6}$.

(e) $⟨110⟩$: same as (b): $\boxed{12}$.

(f) $⟨111⟩$: same as (c): $\boxed{8}$.

(Equivalent directions: the 6 along the cube edges, 12 along face diagonals, 8 along body diagonals. If one counts only distinct orientations up to reversal, divide each by 2: 3, 6, 4.)

Problem 1.13 Packing fractions

Problem. Treating atoms as rigid spheres with radii equal to one half the distance between nearest neighbors, show that the ratio of volume occupied by atoms to the total available volume is (a) $\pi /6$ ($52\%$) for simple cubic, (b) $\sqrt{3}\pi /8$ ($68\%$) for bcc, (c) $\sqrt{2}\pi /6$ ($74\%$) for fcc, (d) $\sqrt{3}\pi /16$ ($34\%$) for diamond.

General set-up. If the nearest-neighbor distance is $d$, the sphere radius is $r=d/2$, the volume of one sphere is $\frac{4}{3}\pi r^3$, and the fraction is

$$f=\frac{N\cdot \frac{4}{3}\pi r^3}{a^3}$$

with $N$ atoms per cell.

(a) Simple cubic. $N=1$, $d=a$, $r=a/2$.

$$f=\frac{1\cdot \frac{4}{3}\pi {\left(\frac{a}{2}\right)}^3}{a^3}=\frac{\frac{4}{3}\pi \cdot \frac{1}{8}}{1}=\frac{\pi }{6}=0.5236\Rightarrow \boxed{52\%}$$

(b) bcc. $N=2$, $d=\frac{\sqrt{3}}{2}a$ (Problem 1.4), so $r=\frac{\sqrt{3}}{4}a$.

$$r^3=\frac{3\sqrt{3}}{64}{a}^3$$ $$f=\frac{2\cdot \frac{4}{3}\pi \cdot \frac{3\sqrt{3}}{64}{a}^3}{a^3}=2\cdot \frac{4\cdot 3\sqrt{3}\pi }{3\cdot 64}=2\cdot \frac{\sqrt{3}\pi }{16}=\frac{\sqrt{3}\pi }{8}=0.6802\Rightarrow \boxed{68\%}$$

(c) fcc. $N=4$, $d=\frac{\sqrt{2}}{2}a$, $r=\frac{\sqrt{2}}{4}a$.

$$r^3=\frac{2\sqrt{2}}{64}{a}^3=\frac{\sqrt{2}}{32}{a}^3$$ $$f=\frac{4\cdot \frac{4}{3}\pi \cdot \frac{\sqrt{2}}{32}{a}^3}{a^3}=4\cdot \frac{\sqrt{2}\pi }{24}=\frac{\sqrt{2}\pi }{6}=0.7405\Rightarrow \boxed{74\%}$$

(d) Diamond. $N=8$, $d=\frac{\sqrt{3}}{4}a$ (Exercise 1.2), $r=\frac{\sqrt{3}}{8}a$.

$$r^3=\frac{3\sqrt{3}}{512}{a}^3$$ $$f=\frac{8\cdot \frac{4}{3}\pi \cdot \frac{3\sqrt{3}}{512}{a}^3}{a^3}=8\cdot \frac{\sqrt{3}\pi }{128}=\frac{\sqrt{3}\pi }{16}=0.3401\Rightarrow \boxed{34\%}$$

Take-away. Diamond is a very “empty” lattice (only 34 percent full) compared to fcc (74 percent, the densest possible packing of equal spheres). The emptiness is the price of strong, directional four-fold bonding: atoms hold onto exactly four neighbors rather than packing as tightly as possible. This sets the stage for Chapter 2, where those four bonds determine how electrons behave.