Exercises — Crystal lattice structure of silicon

"Group IV" is literally the label that counts outer-shell electrons: a group-IV atom has 4 valence electrons. Each electron pairs up with one electron donated by a neighbour, so silicon reaches out to 4 neighbours — one shared pair per bond. See Valence electrons and the octet rule. Answer: 4 valence electrons, 4 nearest neighbours.

Silicon forms the diamond cubic structure: two interpenetrating face-centred cubic (FCC) lattices, offset from each other by one quarter of the cube's body diagonal. See Diamond cubic structure. Answer: diamond cubic = two FCC lattices offset by ¼ body diagonal.

A tetrahedron (a 4-cornered "tent"), with bond angle $\approx 109.5^{\circ }$. Answer: tetrahedron, $109.5^{\circ }$.

WHAT: density = atoms per unit volume. Each cube holds 8 atoms and occupies volume $a^3$. WHY $a^3$: the unit cell is a cube of side $a$, so its volume is edge cubed. $n=\frac{N}{a^3}=\frac{8}{(5.43\times 10^{-8}{)}^3{\text{cm}}^3}$ Cube the edge: $(5.43\times 10^{-8}{)}^3=1.601\times 10^{-22}{\text{cm}}^3$. $n=\frac{8}{1.601\times 10^{-22}}\approx 5.00\times 10^{22}{\text{atoms/cm}}^3$ Answer: $n\approx 5.0\times 10^{22}{\text{atoms/cm}}^3$.

WHY $\frac{\sqrt{3}}{4}$: the two atoms of a bond sit at a cube corner and at a point $\frac{1}{4}$ of the way along the body diagonal (that ¼ offset is the diamond-cubic shift). The full body diagonal of a cube of edge $a$ has length $\sqrt{3}a$ (Pythagoras in 3D: $\sqrt{a^2+a^2+a^2}$). A quarter of it is $\frac{\sqrt{3}}{4}a$. $\ell =\frac{\sqrt{3}}{4}(5.43)=\frac{1.7321}{4}\times 5.43=0.4330\times 5.43\approx 2.35\text{A˚}$ Answer: $\ell \approx 2.35\text{A˚}$.

• 8 corners, each shared by 8 cubes → $8\times \frac{1}{8}=1$.
• 6 faces, each shared by 2 cubes → $6\times \frac{1}{2}=3$.
• 4 interior atoms (the tetrahedral-site atoms), owned wholly by this cell → $4\times 1=4$. $N=1+3+4=8.$ Answer: $N=8$.

WHICH TOOL & WHY: to find the angle between two directions we use the dot product, because the dot product is defined so that $\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos \theta$ — it is the one tool that converts two vectors directly into the cosine of their angle. No other single operation does that. WHAT the dot product is: multiply matching components and add: $\mathbf{a}\cdot \mathbf{b}=a_x{b}_x+a_y{b}_y+a_z{b}_z$. $\mathbf{a}\cdot \mathbf{b}=(1)(1)+(1)(-1)+(1)(-1)=1-1-1=-1$ WHAT the lengths are: $|\mathbf{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$, and by symmetry $|\mathbf{b}|=\sqrt{3}$. $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{-1}{\sqrt{3}\cdot \sqrt{3}}=\frac{-1}{3}$ WHY arccos: $\cos \theta$ tells us "how aligned"; to recover the angle itself we invert cosine — $\arccos$ answers "which angle has this cosine?" $\theta =\arccos \left(-\frac{1}{3}\right)\approx 109.47^{\circ }$ WHAT IT LOOKS LIKE (figure s01): the cosine is negative, so the angle is obtuse — the two bonds lean away from each other, which is exactly what four mutually-repelling bonds do to spread out maximally. Answer: $\theta =\arccos (-1/3)\approx 109.47^{\circ }$.

$\mathbf{a}\cdot \mathbf{c}=(1)(-1)+(1)(1)+(1)(-1)=-1$ Same dot product $-1$, same lengths $\sqrt{3}$, so $\cos \theta =-\frac{1}{3}$ again → $109.47^{\circ }$. WHY it must repeat: every one of the four vectors has exactly two sign-flips relative to any other, so each pairwise dot product is $(+1)+(-1)+(-1)=-1$. That uniformity is what "regular tetrahedron" means — every bond angle identical. Answer: yes, all six pairwise angles equal $109.47^{\circ }$.

WHAT & WHY: density (g/cm³) ÷ molar mass (g/mol) = moles per cm³; multiply by $N_A$ (atoms per mole) → atoms per cm³. This is a completely independent route to $n$ (bulk measurement, not geometry), so agreement cross-validates the lattice model. $n=\frac{\rho N_A}{M}=\frac{(2.33)(6.022\times 10^{23})}{28.09}$ $n=\frac{1.403\times 10^{24}}{28.09}\approx 5.00\times 10^{22}{\text{atoms/cm}}^3$ This matches the geometric $n\approx 5.0\times 10^{22}$ from L2.1 — two independent methods, same number. That agreement is why we trust the 8-atoms-per-cell picture. Answer: $n\approx 5.0\times 10^{22}{\text{atoms/cm}}^3$ — consistent.

WHY compare to $n$: the parent note stressed $n\approx 5\times 10^{22}$ is the yardstick for doping. Fraction doped = dopant density ÷ silicon density. $f=\frac{N_D}{n}=\frac{1\times 10^{17}}{5.0\times 10^{22}}=2\times 10^{-6}$ That is 2 parts per million. One dopant per $\frac{1}{f}=\frac{5.0\times 10^{22}}{1\times 10^{17}}=5\times 10^5=500,000\text{silicon atoms.}$ WHAT IT MEANS: an astonishingly tiny sprinkle — half a million host atoms per impurity — is enough to change conductivity dramatically, precisely because pure silicon has so few free carriers to begin with. Answer: $f=2\times 10^{-6}$ (2 ppm); ~1 dopant per 500,000 Si atoms.

WHY $\frac{a}{\sqrt{2}}$: within one FCC sub-lattice, the closest atoms are a corner and an adjacent face-centre, separated by half a face diagonal. A face diagonal of a square of side $a$ is $\sqrt{2}a$; half of it is $\frac{\sqrt{2}}{2}a=\frac{a}{\sqrt{2}}$. $d_2=\frac{5.43}{\sqrt{2}}=\frac{5.43}{1.4142}\approx 3.84\text{A˚}$ Compare: $d_2\approx 3.84\text{A˚}>\ell \approx 2.35\text{A˚}$. ✓ So the 4 tetrahedral bonds are genuinely the closest atoms; the 12 next-nearest are ~1.6× farther and are not directly bonded. Answer: $d_2\approx 3.84\text{A˚}$, larger than the $2.35\text{A˚}$ bond length.

At $0\text{K}$: every valence electron is locked in one of the 4 covalent bonds. The regular lattice creates a band gap $E_g\approx 1.12\text{eV}$ — a forbidden energy zone. With no thermal energy, no electron can jump the gap, so zero free carriers → perfect insulator. At $300\text{K}$: thermal vibration breaks a few bonds, promoting electrons across the gap and leaving holes behind. The free-electron density $n_i\approx 10^{10}{\text{cm}}^{-3}$. Connect to atom count: $\frac{n_i}{n}=\frac{10^{10}}{5.0\times 10^{22}}=2\times 10^{-13}$ Only about 1 in $5\times 10^{12}$ atoms has donated a free electron — a trillion-fold smaller than the atom count. That "almost none, but not exactly none" is precisely the controllable in-between state that makes silicon a semiconductor. Answer: ratio $\approx 2\times 10^{-13}$; free electrons are ~1 per $5\times 10^{12}$ atoms.

(a) Simple cubic has atoms only at the 8 corners, each shared by 8 cubes: $N=8\times \frac{1}{8}=1$ atom per cell. (b) Simple cubic gives each atom 6 nearest neighbours (along $\pm x,\pm y,\pm z$). But group IV supplies only 4 valence electrons → only 4 shared pairs possible. Six neighbours cannot each get a full shared bond; the octet rule (Valence electrons and the octet rule) is violated. The bonding count forces 4 neighbours, not 6. (c) Neighbour angle in simple cubic is $90^{\circ }$ (adjacent axes are perpendicular). That is less than the tetrahedral $109.5^{\circ }$ — the bonds would be more crowded, higher-energy, and not the arrangement that maximally separates 4 electron pairs. Silicon avoids it. Answer: (a) 1 atom/cell; (b) only 4 valence electrons ⇒ 4 bonds, not 6; (c) $90^{\circ }<109.5^{\circ }$.

Group IV → 4 valence electrons → shares 1 with each of 4 neighbours to complete an octet → 4 covalent bonds → bonds repel (VSEPR) → tetrahedron at $109.5^{\circ }$ → tiled through space = diamond cubic lattice (two offset FCC) → periodicity of that lattice creates the energy band gap ($E_g\approx 1.12\text{eV}$) → gap small enough to excite carriers, large enough to control → semiconductor → but band structure needs unbroken periodicity, so defects/grain boundaries ruin it → therefore chips use single-crystal wafers. Answer: group IV → 4 e⁻ → octet sharing → 4 bonds → VSEPR tetrahedron ($109.5^{\circ }$) → diamond cubic → band gap → semiconductor → single-crystal requirement.

Nearest-neighbour count and geometry? ::: 4 neighbours, tetrahedron, $109.47^{\circ }$. Atomic density of Si? ::: $\approx 5.0\times 10^{22}{\text{cm}}^{-3}$. Bond length in terms of $a$? ::: $\frac{\sqrt{3}}{4}a\approx 2.35\text{A˚}$. Doping at $10^{17}{\text{cm}}^{-3}$ = how many ppm? ::: $2$ ppm (one per $500,000$ Si atoms). Why single-crystal wafers? ::: Band gap needs unbroken lattice periodicity.