Worked examples — Crystal lattice structure of silicon

This page is the drill hall for the silicon lattice topic. The parent note built the ideas; here we push every idea to its edges — every quadrant of the geometry, every degenerate input, every limiting value, and the two kinds of word problems (real-world + exam twist).

Nothing new is assumed. Every symbol used below is re-earned the moment it appears.

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The scenario matrix

Before working anything, let's list every case class this topic can throw at you. Each worked example below is tagged with the cell it covers.

#	Case class	What makes it tricky	Example
A	Bond angle from vectors (standard)	dot product of two "up" bonds	Ex 1
B	Bond angle — a different pair (sign check)	two bonds both pointing "down", is the angle the same?	Ex 2
C	Degenerate input: angle of a bond with itself	$\theta =0$, cosine $=+1$	Ex 3
D	Opposite-direction limit: what would $180^{\circ }$ need?	why silicon can't have collinear bonds	Ex 4
E	Counting atoms in the unit cell	shared corners/faces vs interior	Ex 5
F	Length scales: bond length from lattice constant	body-diagonal geometry, $\sqrt{3}$	Ex 6
G	Atomic density (real-world number)	8 atoms per $a^3$, unit conversion Å→cm	Ex 7
H	Word problem: doping ratio	compare dopant count to Ex 7 density	Ex 8
I	Exam twist: packing fraction of diamond cubic	combine radius, count, volume — the "gotcha"	Ex 9

Two tools power almost everything:

Recall What is

$|\mathbf{u}|$ and why the square root? $|\mathbf{u}|$ is the arrow's length. By Pythagoras in 3D, length $=\sqrt{u_1^2+u_2^2+u_3^2}$. We divide by the two lengths so that $\cos \theta$ depends only on direction, never on how long we drew the arrows.

We reuse the parent note's four bond vectors from the central atom (at the origin) to alternate cube corners: $\mathbf{a}=(1,1,1),\mathbf{b}=(1,-1,-1),\mathbf{c}=(-1,1,-1),\mathbf{d}=(-1,-1,1).$

Figure s01 — A cube (white edges) with the yellow central silicon atom at the origin and four blue rods reaching to alternate corners $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$. Those four rods are the tetrahedral bonds; every angle example below simply picks two of them.

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Group 1 — Bond angle, every sign case

Steps:

1. Compute the dot product: $\mathbf{a}\cdot \mathbf{b}=(1)(1)+(1)(-1)+(1)(-1)=1-1-1=-1.$

   • Why this step? The dot product is the only number we need to unlock the angle — it's the numerator of $\cos \theta$.

2. Compute the lengths: $|\mathbf{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3},|\mathbf{b}|=\sqrt{1^2+(-1{)}^2+(-1{)}^2}=\sqrt{3}.$

   • Why this step? We must strip out arrow-length so only direction survives.

3. Assemble the cosine: $\cos \theta =\frac{-1}{\sqrt{3}\cdot \sqrt{3}}=\frac{-1}{3}.$

   • Why this step? This is the definition — now $\theta$ is trapped.

4. Undo the cosine. ==$\arccos$== ("arc-cosine") answers "which angle has this cosine?" — it is the inverse machine that turns a cosine value back into an angle: $\theta =\arccos \left(-\frac{1}{3}\right)\approx 109.47^{\circ }.$

   • Why this step? We have $\cos \theta$; we want $\theta$ itself. $\arccos$ is the exact undo-button.

Verify: $\cos (109.47^{\circ })\approx -0.3333=-\frac{1}{3}$. ✓ And it lands between $90^{\circ }$ and $180^{\circ }$, matching "obtuse." ✓

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Steps:

1. Dot product: $\mathbf{c}\cdot \mathbf{d}=(-1)(-1)+(1)(-1)+(-1)(1)=1-1-1=-1.$

   • Why this step? Even though both vectors have negative entries, the products of matching signs behave predictably — signs are not a special case, they just multiply.

2. Compute the lengths: $|\mathbf{c}|=\sqrt{(-1{)}^2+1^2+(-1{)}^2}=\sqrt{3},|\mathbf{d}|=\sqrt{(-1{)}^2+(-1{)}^2+1^2}=\sqrt{3}.$

   • Why this step? Same as Ex 1 — squaring kills the minus signs, so both "negative-corner" bonds still have length $\sqrt{3}$. We divide by these to isolate direction.

3. Assemble the cosine and undo it: $\cos \theta =\frac{-1}{\sqrt{3}\cdot \sqrt{3}}=\frac{-1}{3}\Rightarrow \theta =\arccos (-\frac{1}{3})\approx 109.47^{\circ }.$

   • Why this step? This is the definition of $\cos \theta$ again; then $\arccos$ turns the cosine value back into the actual angle — the exact undo-button used in Ex 1.

Verify: identical arithmetic to Ex 1 → $-\frac{1}{3}$ → $109.47^{\circ }$. ✓

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Steps:

1. Dot product of $\mathbf{a}$ with itself: $\mathbf{a}\cdot \mathbf{a}=1^2+1^2+1^2=3.$

   • Why this step? A vector dotted with itself gives its length squared — this is where $|\mathbf{a}|=\sqrt{3}$ comes from.

2. Divide: $\cos \theta =\frac{3}{(\sqrt{3}{)}^2}=\frac{3}{3}=1.$

   • Why this step? The cosine formula needs the dot product divided by both lengths; here both lengths are $|\mathbf{a}|$, so the denominator is $|\mathbf{a}{|}^2=(\sqrt{3}{)}^2=3$. Dividing gives the cosine of the self-angle.

3. Undo: $\theta =\arccos (1)=0^{\circ }.$

   • Why this step? A cosine of exactly $+1$ is the maximum a cosine can reach — it corresponds to no separation at all.

Verify: $\cos (0^{\circ })=1$. ✓

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Steps:

1. Set $\theta =180^{\circ }$. Then $\cos (180^{\circ })=-1$ (the minimum possible cosine).

   • Why this step? $180^{\circ }$ is the extreme "point away" case; cosine bottoms out at $-1$.

2. For our unit-ish bonds ($|\mathbf{u}|=|\mathbf{v}|=\sqrt{3}$), this needs $\frac{\mathbf{u}\cdot \mathbf{v}}{3}=-1\Rightarrow \mathbf{u}\cdot \mathbf{v}=-3.$

   • Why this step? Solve the cosine equation backwards for the required dot product.

3. But the most negative any of our pairs reaches is $-1$ (Ex 1), giving $\cos \theta =-\frac{1}{3}$, not $-1$. So no silicon bond pair is collinear.

   • Why this step? It shows the tetrahedral angle sits strictly between $90^{\circ }$ and $180^{\circ }$ — never the degenerate extreme.

Verify: $-\frac{1}{3}>-1$, so $109.47^{\circ }<180^{\circ }$ strictly. ✓

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Group 2 — Counting and length scales

Figure s02 — A flattened schematic of one unit cell. Red dots are shared corner atoms (each split among 8 cubes, weight $\frac{1}{8}$), green dots are shared face atoms (split between 2 cubes, weight $\frac{1}{2}$), and yellow dots are the 4 interior atoms that belong entirely to this cell (weight $1$). Read off who owns which atom as we count.

Steps:

1. Corners. Each of the 8 corner atoms is shared by 8 cubes meeting at that corner, so only $\frac{1}{8}$ belongs here: $8\times \frac{1}{8}=1.$

   • Why this step? A corner atom is "communal property" — split it fairly (the red dots in the figure).

2. Faces. Each of the 6 face-centre atoms is split between the 2 cubes sharing that face: $6\times \frac{1}{2}=3.$

   • Why this step? Same fairness rule, but a face touches only 2 cubes (the green dots).

3. Interior. The 4 tetrahedral atoms sit fully inside — shared with nobody: $4\times 1=4.$

   • Why this step? An interior atom (the yellow dots) has no neighbour cube to share with, so its whole self — weight $1$ — counts for this cell alone. These are the atoms that make diamond cubic different from plain FCC.

4. Total: $N=1+3+4=8.$

   • Why this step? Number density and every later count need the whole-cell atom total, so we add the three fairly-weighted contributions into one number.

Verify: $8(\frac{1}{8})+6(\frac{1}{2})+4=1+3+4=8.$ ✓

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Figure s03 — A single cube of edge $a$ with axes $x,y,z$. The red line is the full body diagonal (corner to opposite corner, length $\sqrt{3}a$); the thick green segment is its first quarter — from the yellow central atom to the blue nearest-neighbour atom. That green segment is exactly the bond we measure.

Steps:

1. Body diagonal length. A cube of edge $a$ has a body diagonal (corner to opposite corner) of $d_{\text{diag}}=\sqrt{a^2+a^2+a^2}=\sqrt{3}a.$

   • Why this step? Nearest neighbours in diamond cubic sit along the body diagonal (the red line) — the two interpenetrating FCC lattices are offset by $\frac{1}{4}$ of it. Pythagoras in 3D gives the diagonal.

2. Bond = quarter diagonal. Let $\ell$ denote the bond length — the straight-line distance between two bonded (nearest-neighbour) atoms. The offset is $\frac{1}{4}$ of the body diagonal, so $\ell =\frac{\sqrt{3}}{4}a.$

   • Why this step? That $\frac{1}{4}$ offset is the definition of the diamond-cubic structure (parent note); the green segment in the figure is exactly this quarter.

3. Plug in $a=5.43\text{A˚}$: $\ell =\frac{\sqrt{3}}{4}\times 5.43\approx \frac{1.732}{4}\times 5.43\approx 2.35\text{A˚}.$

   • Why we now plug in numbers? We have the exact symbolic relation $\ell =\frac{\sqrt{3}}{4}a$; substituting the measured $a=5.43\text{A˚}$ turns it into the concrete physical bond length quoted in tables — the number we can compare to experiment.

Verify: $\frac{\sqrt{3}}{4}\times 5.43=2.351\dots \text{A˚}$. ✓

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Steps:

1. Cell volume: $a^3=(5.43\times 10^{-8}\text{cm}{)}^3=1.601\times 10^{-22}{\text{cm}}^3.$

   • Why this step? Each unit cell occupies volume $a^3$; density is atoms ÷ volume. We convert Å→cm now ($1\text{A˚}=10^{-8}\text{cm}$) so the answer comes out in the standard ${\text{cm}}^{-3}$ used for doping.

2. Divide atoms by volume: $n=\frac{8}{1.601\times 10^{-22}}\approx 5.0\times 10^{22}{\text{atoms/cm}}^3.$

   • Why this step? Number density means "atoms per unit volume," so by its very definition $n=N/a^3$: take the whole-cell atom count from Ex 5 and divide by the cell's volume from Step 1.

Verify: $8/(5.43\times 10^{-8}{)}^3=4.996\times 10^{22}$. ✓

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Group 3 — Word problem and exam twist

Steps:

1. Ratio of dopants to silicon (using Ex 7's $n=5.0\times 10^{22}$): $\frac{n_D}{n}=\frac{1\times 10^{16}}{5.0\times 10^{22}}=2\times 10^{-7}.$

   • Why this step? "What fraction is replaced" is literally dopants ÷ hosts.

2. Interpret: $2\times 10^{-7}=$ one dopant per $5\times 10^6$ silicon atoms. $\frac{1}{2\times 10^{-7}}=5\times 10^6.$

   • Why this step? Reciprocal of a fraction = "one per how many."

Verify: $10^{16}/(5\times 10^{22})=2\times 10^{-7}$ and its reciprocal $=5\times 10^6$. ✓

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Steps:

1. Atoms touch along the bond. Let $\ell$ be the bond length — the distance between two nearest-neighbour atoms (re-earned here: same quantity as Ex 6, $\ell =\frac{\sqrt{3}}{4}a$). If we model atoms as hard spheres of radius $r$ that just touch their nearest neighbour, then two radii span exactly one bond: $2r=\ell =\frac{\sqrt{3}}{4}a\Rightarrow r=\frac{\sqrt{3}}{8}a.$

   • Why this step? Packing fraction models atoms as hard spheres touching their nearest neighbours; the bond length sets the radius.

2. Volume of one sphere: $V_{\text{atom}}=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(\frac{\sqrt{3}}{8}a\right)}^3.$

   • Why this step? Sphere volume formula; we'll multiply by the atoms-per-cell count next.

3. Total atom volume uses the $N=8$ atoms per cell from Ex 5: $V_{\text{atoms}}=8\cdot \frac{4}{3}\pi {\left(\frac{\sqrt{3}}{8}a\right)}^3.$

   • Why this step? The packing fraction needs the volume of all the cell's atoms, not just one — so multiply the single-sphere volume by the 8 atoms Ex 5 counted.

4. Packing fraction = total atom volume ÷ cube volume $a^3$: $\text{PF}=\frac{8\cdot \frac{4}{3}\pi {\left(\frac{\sqrt{3}}{8}\right)}^3{a}^3}{a^3}=8\cdot \frac{4}{3}\pi \cdot \frac{3\sqrt{3}}{512}=\frac{\pi \sqrt{3}}{16}\approx \mathrm{0.340.}$

   • Why this step? Dividing filled volume by the whole cube gives the fraction that is solid. The $a^3$ cancels — packing fraction is a pure number, independent of scale. That is the elegant "gotcha": you never needed $a$ at all.

Verify: $\frac{\pi \sqrt{3}}{16}=0.3401\dots \approx 0.34$. ✓ (Well below the $0.74$ of close-packed metals, as forecast — silicon is a loose lattice.)

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Recall Which cell did each example cover?

Ex 1 ::: A — standard bond angle Ex 2 ::: B — different pair, same angle (sign check) Ex 3 ::: C — degenerate, $\theta =0$ Ex 4 ::: D — $180^{\circ }$ limit is impossible Ex 5 ::: E — atom counting = 8 Ex 6 ::: F — bond length $=\frac{\sqrt{3}}{4}a$ Ex 7 ::: G — density $5\times 10^{22}$ Ex 8 ::: H — doping ratio word problem Ex 9 ::: I — packing fraction $\approx 0.34$

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Connections

• Parent topic (Hinglish)
• Diamond cubic structure
• Face-centred cubic (FCC) lattice
• Valence electrons and the octet rule
• Doping of silicon
• Energy band gap in semiconductors
• Holes and electrons as charge carriers
• Single-crystal silicon wafer manufacturing
• Miller indices and crystal planes