Covalent bonding in silicon crystals: Exercises — Hardware deep dive

Before we start, here is the one toolbox every numeric problem draws from. Read it once; we point back to it constantly.

Formula The intrinsic-carrier tool

$n_{i} \propto e^{- E_{g} / (2 k_{B} T)}$

$n_{i}$ = intrinsic carrier concentration = how many free electrons (and equally many holes) exist per cubic centimetre in pure silicon, purely because heat broke some bonds.
$E_{g}$ = band gap = the energy in electron-volts (eV) needed to snap one covalent bond and free its electron.
$k_{B}$ = Boltzmann constant = $8.617 \times 1 0^{- 5} eV/K$ . It is the "exchange rate" between temperature and energy: multiply it by a temperature and you get the typical thermal energy jiggling each particle.
$T$ = absolute temperature in kelvin (K).
$e^{x}$ = the exponential function. Here it answers "what fraction of bonds happen to have enough thermal energy to break?" — a fraction that shrinks fast as the required energy $E_{g}$ grows or as the available thermal budget $k_{B} T$ shrinks.

Why the exponential and not, say, a straight line? Thermal energy is shared randomly among trillions of jiggling atoms. The chance that one particular bond randomly collects an amount $\geq E_{g}$ of that energy is governed by Boltzmann statistics, whose answer is always an exponential $e^{- energy / k_{B} T}$ . No other simple curve matches nature here.

Why the factor of 2 in the denominator? Breaking one bond makes two carriers at once — a free electron and a hole. The cost $E_{g}$ is effectively split between the two populations, and the algebra of the mass-action law $n p = n_{i}^{2}$ puts a $2$ under $E_{g}$ for $n_{i}$ itself.

To turn a proportionality ( $\propto$ ) into a number, we always take a ratio of the same material at two temperatures, so the unknown constant in front cancels: $\frac{n _{i} ( T _{2} )}{n _{i} ( T _{1} )} = \frac{e ^{- E_{g} / (2 k_{B} T_{2})}}{e ^{- E_{g} / (2 k_{B} T_{1})}} = exp [\frac{E _{g}}{2 k _{B}} (\frac{1}{T _{1}} - \frac{1}{T _{2}})] .$ We derive this identity once here so no later solution has to re-explain it.

Figure — Covalent bonding in silicon crystals

Level 1 — Recognition

Can you recall and identify the core facts?

L1.1

State silicon's number of valence electrons and its coordination number in the crystal, and say what "coordination number" means in one line.

Recall Solution

Valence electrons = 4. Silicon is group IV; its outer shell ( $n = 3$ ) holds $3 s^{2} 3 p^{2} = 4$ electrons.
Coordination number = 4. This is the number of nearest neighbours an atom is directly bonded to.
What it looks like: each atom sits at the centre of a tetrahedron, with one bond reaching out to each of the 4 corner atoms.

L1.2

How many electrons does a silicon atom "see" in its outer shell once it is inside the crystal, and why is that number the target?

Recall Solution

Own contribution: $4$ valence electrons.
Shared in: one electron from each of $4$ neighbours $= + 4$ .
Total $= 4 + 4 = 8$ — a full octet, the most stable arrangement. Reaching 8 is the whole reason the atom bonds at all.

L1.3

Give silicon's band gap at 300 K, and name the two extreme categories it sits between.

Recall Solution

$E_{g} \approx 1.12 eV$ at 300 K.
Below it are conductors/metals (effectively no gap — always free electrons); far above it are insulators (e.g. diamond, $E_{g} \approx 5.5 eV$ ). Silicon's in-between gap makes it a semiconductor.

Level 2 — Application

Plug the facts into a single formula or count.

L2.1

In a small silicon fragment of $N$ atoms (interior atoms only, ignore surface), how many covalent bonds exist? Each bond is shared by two atoms.

Recall Solution

Each atom owns 4 bonds, but each bond is counted by both atoms it connects, so we double-count.
Bonds $= \frac{4 N}{2} = 2N$ .
Why divide by 2: if I list "4 bonds per atom" for all $N$ atoms I get $4 N$ , but bond A–B appears once in A's list and once in B's list — halving removes the duplicate.

L2.2

Compute the dimensionless exponent $E_{g} / (2 k_{B} T)$ for silicon at $T = 300 K$ . Use $E_{g} = 1.12$ eV, $k_{B} = 8.617 \times 1 0^{- 5}$ eV/K.

Recall Solution

$\frac{E _{g}}{2 k _{B} T} = \frac{1.12}{2 \times ( 8.617 \times 1 0 ^{- 5} ) \times 300} = \frac{1.12}{0.051702} \approx 21.66 .$ What this number means: the required bond-breaking energy is about 21.7 times the average thermal jiggle. That is why so few bonds break at room temperature — the exponent lands in $e^{- 21.7}$ , a tiny fraction.

L2.3

At 300 K, the fraction of bonds broken scales as $e^{- E_{g} / (2 k_{B} T)}$ . Using the exponent from L2.2, express this fraction as a power of ten.

Recall Solution

$e^{- 21.66} = 1 0^{- 21.66/ l n 10} = 1 0^{- 21.66/2.3026} \approx 1 0^{- 9.4} .$ So only about one bond in $1 0^{9.4}$ (roughly four billion) is broken — vanishingly few, which is exactly why pure silicon is a poor conductor at room temperature.

Level 3 — Analysis

Compare, take ratios, reason about trends.

L3.1

Silicon's $n_{i}$ at 300 K is $\sim 1 0^{10} cm^{- 3}$ . Estimate $n_{i}$ at 400 K.

Recall Solution

Use the ratio identity from the toolbox so the unknown constant cancels: $\frac{n _{i} ( 400 )}{n _{i} ( 300 )} = exp [\frac{E _{g}}{2 k _{B}} (\frac{1}{300} - \frac{1}{400})] .$

$\frac{E _{g}}{2 k _{B}} = \frac{1.12}{2 \times 8.617 \times 1 0 ^{- 5}} = 6499 K$ .
$\frac{1}{300} - \frac{1}{400} = 0.0033 \overline{3} - 0.0025 = 0.0008 \overline{3} K^{- 1}$ .
Exponent $= 6499 \times 0.0008333 \approx 5.42$ .
Ratio $= e^{5.42} \approx 225$ .
$n_{i} (400) \approx 225 \times 1 0^{10} \approx 2.3 \times 1 0^{12} cm^{- 3}$ .

What it looks like / means: a mere 100 K rise multiplies free carriers by ~200×. Conduction in silicon is fiercely temperature-sensitive — the physical root of thermal runaway in devices.

L3.2

Carbon (diamond, group IV, same bonding) has $E_{g} \approx 5.5$ eV. Compare its broken-bond fraction to silicon's at 300 K as an order of magnitude.

Recall Solution

Same $2 k_{B} T = 0.051702$ eV as L2.2. Diamond's exponent: $\frac{5.5}{0.051702} \approx 106.4.$ Broken fraction $\sim e^{- 106.4} \approx 1 0^{- 46.2}$ . Ratio to silicon ( $1 0^{- 9.4}$ ): $\frac{1 0 ^{- 46.2}}{1 0 ^{- 9.4}} = 1 0^{- 36.8} .$ Interpretation: diamond breaks fewer bonds than silicon by a factor of about $1 0^{37}$ . Identical bonding, but the bigger gap makes diamond an insulator while silicon semiconducts. The size of $E_{g}$ , not the bond type, decides the category.

Level 4 — Synthesis

Combine several ideas into a new result.

L4.1

A student measures silicon's $n_{i}$ at two temperatures and finds it grows by a factor of exactly 10 between $T_{1} = 300$ K and some higher $T_{2}$ . Estimate $T_{2}$ .

Recall Solution

Set the ratio to 10 and solve for $T_{2}$ : $10 = exp [\frac{E _{g}}{2 k _{B}} (\frac{1}{T _{1}} - \frac{1}{T _{2}})] .$ Take natural log of both sides: $ln 10 = \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{1}} - \frac{1}{T _{2}}), 2.3026 = 6499 (\frac{1}{300} - \frac{1}{T _{2}}) .$

$\frac{1}{300} - \frac{1}{T _{2}} = \frac{2.3026}{6499} = 3.543 \times 1 0^{- 4}$ .
$\frac{1}{T _{2}} = 0.0033 \overline{3} - 0.0003543 = 0.0029790$ .
$T_{2} = \frac{1}{0.0029790} \approx 335.7 K$ .

Takeaway: raising $T$ by only ~36 K (to about 63 °C) tenfolds the intrinsic carriers. This is why device engineers obsess over cooling.

L4.2

Bonds-and-carriers bookkeeping. When one bond breaks in intrinsic silicon, it produces one free electron and one hole. If a sample momentarily has $1 0^{12} cm^{- 3}$ free electrons from thermal generation, how many holes does it have, and why?

Recall Solution

$1 0^{12} cm^{- 3}$ holes — exactly equal.
Why: every broken bond makes the pair together; there is no way to make a free electron without simultaneously leaving a hole. In pure (intrinsic) silicon, $n = p = n_{i}$ . This is charge conservation made visible.

Level 5 — Mastery

Full multi-step reasoning, edge cases, and "why" chains.

L5.1 — The limiting cases

Describe $n_{i}$ in three limits and connect each to bond-breaking: (a) $T \to 0$ K, (b) $E_{g} \to 0$ , (c) $E_{g}$ very large. Use the toolbox.

Recall Solution

(a) $T \to 0$ K: exponent $E_{g} / (2 k_{B} T) \to + \infty$ , so $e^{- \infty} \to 0$ , giving $n_{i} \to 0$ . Physically: zero thermal energy → no bond ever breaks → every electron stays locked in a covalent bond → silicon is a perfect insulator at absolute zero. This is the exact statement the parent note makes.
(b) $E_{g} \to 0$ : exponent $\to 0$ , so $e^{0} = 1$ ; the broken-bond fraction saturates and $n_{i}$ is huge for any $T > 0$ . Physically: a "gap of zero" means bonds cost nothing to break → behaves like a metal (free electrons always available).
(c) $E_{g}$ very large (e.g. diamond, 5.5 eV): exponent is enormous, $n_{i} \to$ astronomically small. Physically: an insulator at ordinary temperatures — bonds essentially never break.

The unifying picture: the single knob $E_{g} / (2 k_{B} T)$ slides silicon continuously between metal-like and insulator-like behaviour. Semiconductors are exactly the materials whose knob sits in the tunable middle.

Argue from energy cost why silicon shares 4 electrons rather than donating 4 (like a metal) or grabbing 4 (like a halogen would need). No formula required — reason it.

Recall Solution

Silicon has 4 valence electrons and needs 8 for a full shell.
Donate all 4 (become $Si^{4 +}$ ): stripping four electrons costs a rapidly rising ionisation energy (each successive electron is harder to remove) — far too expensive.
Grab 4 (become $Si^{4 -}$ ): cramming four extra electrons onto one atom piles up mutual repulsion — also far too expensive.
Share: contribute one electron to each of 4 bonds and receive one back from each neighbour → the atom reaches 8 while never fully losing or gaining charge. The energy cost of sharing is the lowest of the three routes.
Contrast: sodium (1 valence electron) donates cheaply → metallic; chlorine (7) grabs one cheaply → ionic. Silicon's middle count makes sharing the winner. This is why the crystal is a covalent network.

L5.3 — Full synthesis: temperature, carriers, and conduction

A pure silicon sample is cooled from 400 K to 300 K. Using results above, state (i) roughly what happens to $n_{i}$ , (ii) what happens to its conductivity qualitatively, and (iii) tie it back to bonds.

Recall Solution

(i) From L3.1, going 300 → 400 K multiplied $n_{i}$ by ~225×. Reversing (400 → 300 K) divides $n_{i}$ by ~225, i.e. drops it from $\sim 2.3 \times 1 0^{12}$ to $\sim 1 0^{10} cm^{- 3}$ .
(ii) Conductivity $σ$ scales with carrier count, so cooling makes intrinsic silicon much less conductive — opposite to a metal, whose conductivity rises on cooling. This inverse temperature behaviour is a fingerprint of a semiconductor.
(iii) Cooling removes thermal energy, so fewer bonds have enough to break; more electrons stay locked in their covalent bonds → fewer free electrons and fewer holes → less current. Every macroscopic effect traces straight back to how many covalent bonds are intact.

Recall One-line self-check

Why is the factor of 2 in $2 k_{B} T$ non-negotiable for $n_{i}$ ? ::: Because one broken bond makes an electron–hole pair, and $n_{i}$ inherits half the gap in the exponent (mass-action law $n p = n_{i}^{2}$ ).

Connections

Intrinsic vs extrinsic semiconductors — where $n_{i}$ and the $n = p$ result come from
Electron-hole pair generation and recombination — the pair-making these problems count
Band theory of solids — turns "broken bond" into "electron in the conduction band"
Doping with donors and acceptors — how to raise carriers without heat
Diamond cubic crystal structure — the 4-neighbour geometry behind the counts
Ionic vs covalent vs metallic bonding — the energy argument in L5.2