Band gap and its meaning for conductivity: Exercises

Throughout, we reuse three constants. Let us pin them down once so no symbol is ever unexplained.

The single formula that powers most of this page: $n \propto e^{- E_{g} / (2 k_{B} T)}, σ \propto e^{- E_{g} / (2 k_{B} T)}$ Here $n$ is the intrinsic carrier concentration (how many mobile electrons per cubic metre) and $σ$ is the conductivity. Both share the same exponential, so any ratio of $n$ equals the same ratio of $σ$ .

Figure — Band gap and its meaning for conductivity

L1 — Recognition

Q1.1 Read the gap off the band diagram

A crystal has its valence-band top at $E_{v} = 4.0 eV$ and its conduction-band bottom at $E_{c} = 5.1 eV$ (energies measured from some fixed zero). State $E_{g}$ and classify the material.

Recall Solution

WHAT: The band gap is by definition the height of the forbidden zone. $E_{g} = E_{c} - E_{v} = 5.1 - 4.0 = 1.1 eV .$ Classify: $1.1 eV$ lies in the $0.1$ – $3 eV$ window → this is a semiconductor (essentially silicon, $E_{g}^{Si} = 1.12$ eV). Look at figure s01: the gap here is the middle-sized orange bar.

Q1.2 Which cannot conduct, and why?

You are told: material A has bands that overlap; material B has a completely full valence band and a $9 eV$ gap; material C has a half-filled band. Which one carries essentially no current at room temperature? Give the one-line reason.

Recall Solution

Answer: material B. Why: conduction needs an empty state next to an occupied one so an electron can shift. A completely full band is a packed parking lot — no empty adjacent spot, no motion — and the next empty band is $9 eV$ away, far beyond the $\sim 0.026 eV$ of thermal energy. So B is an insulator. A (overlap → partly filled → metal) and C (half-filled → metal) both conduct freely.

L2 — Application

Q2.1 Thermal energy at a temperature

Compute $k_{B} T$ in eV at $T = 300 K$ and at $T = 77 K$ (liquid-nitrogen temperature).

Recall Solution

$k_{B} T = (8.617 \times 1 0^{- 5} eV/K) \times T$ .

At $300 K$ : $8.617 \times 1 0^{- 5} \times 300 = 0.02585 eV \approx 0.026 eV$ .
At $77 K$ : $8.617 \times 1 0^{- 5} \times 77 = 0.006635 eV \approx 0.0066 eV$ . Meaning: cooling to $77 K$ shrinks the available thermal jiggle by a factor $300/77 \approx 3.9$ . Fewer electrons will manage the jump.

Q2.2 Boltzmann suppression for silicon

For silicon ( $E_{g} = 1.12 eV$ ) at $300 K$ , evaluate the exponent $\frac{E _{g}}{2 k _{B} T}$ and the factor $e^{- E_{g} / (2 k_{B} T)}$ .

Recall Solution

WHY the $2$ : the Fermi level sits mid-gap, so an electron only climbs $E_{c} - E_{F} = E_{g} /2$ to reach the conduction band. That is why the exponent carries $E_{g} /2$ , not $E_{g}$ . $\frac{E _{g}}{2 k _{B} T} = \frac{1.12}{2 \times 0.02585} = \frac{1.12}{0.05170} = 21.66.$ $e^{- 21.66} \approx 3.9 \times 1 0^{- 10} .$ So only about 4 in ten billion electrons (relative to the prefactor scale) make the crossing — tiny, but non-zero, which is exactly what makes silicon a semiconductor and not an insulator.

Q2.3 Same calculation for diamond

Repeat Q2.2 for diamond, $E_{g} = 5.5 eV$ , at $300 K$ . Comment.

Recall Solution

$\frac{E _{g}}{2 k _{B} T} = \frac{5.5}{0.05170} = 106.4, e^{- 106.4} \approx 6 \times 1 0^{- 47} .$ Comment: this is unimaginably small — effectively zero carriers. Diamond is an insulator not because its physics is different, but because the same exponential formula, fed a big gap, collapses to nothing.

L3 — Analysis

Q3.1 Heating a semiconductor

By what factor does the intrinsic carrier concentration $n$ of silicon change when $T$ goes from $300 K$ to $400 K$ ? (Ignore the weak prefactor; use only the exponential.)

Recall Solution

Step 1 — collapse the constants. Define the temperature scale $\frac{E _{g}}{2 k _{B}} = \frac{1.12}{2 \times 8.617 \times 1 0 ^{- 5}} = 6499 K \approx 6500 K .$ Why: it turns the messy exponent into one number with units of kelvin, so the ratio is clean. Step 2 — take the ratio. $\frac{n ( 400 )}{n ( 300 )} = \frac{e ^{- 6500/400}}{e ^{- 6500/300}} = e^{6500 (\frac{1}{300} - \frac{1}{400})} .$ $\frac{1}{300} - \frac{1}{400} = \frac{400 - 300}{300 \times 400} = \frac{100}{120000} = 8.333 \times 1 0^{- 4} K^{- 1} .$ $= e^{6500 \times 8.333 \times 1 0^{- 4}} = e^{5.417} \approx 225.$ Meaning: a $100 K$ warm-up multiplies the carrier population by ~225×. Conductivity of a semiconductor rises steeply with temperature.

Q3.2 The metal behaves oppositely

Your friend says: "By the same logic, heating copper (a metal) should also raise its conductivity a lot." Explain in words why this is wrong, referencing figure s02.

Recall Solution

In a metal there is no gap to cross — the conduction band is already partly full, so $n$ barely changes with $T$ . What does change is scattering: higher $T$ means the lattice ions vibrate more violently (bigger displacement), electrons collide more often, mobility drops, so $σ$ decreases. See Temperature dependence of resistivity in metals and Conductivity, mobility and drift current. Figure s02 contrasts the two: the semiconductor curve (teal) shoots up with $T$ ; the metal curve (orange) gently sags.

L4 — Synthesis

Q4.1 Find the gap from two measurements

A mystery semiconductor's conductivity rises by a factor of 50 when heated from $300 K$ to $350 K$ . Estimate its band gap $E_{g}$ (in eV).

Recall Solution

WHAT we know: $\frac{σ ( 350 )}{σ ( 300 )} = 50 = e^{\frac{E _{g}}{2 k _{B}} (\frac{1}{300} - \frac{1}{350})}$ . Step 1 — the temperature bracket. $\frac{1}{300} - \frac{1}{350} = \frac{350 - 300}{300 \times 350} = \frac{50}{105000} = 4.762 \times 1 0^{- 4} K^{- 1} .$ Step 2 — take logs (the tool that undoes an exponential — we want the exponent, so we invert $e^{x}$ ): $ln 50 = \frac{E _{g}}{2 k _{B}} \times 4.762 \times 1 0^{- 4} .$ $ln 50 = 3.912.$ Step 3 — solve for $E_{g}$ . $\frac{E _{g}}{2 k _{B}} = \frac{3.912}{4.762 \times 1 0 ^{- 4}} = 8214 K .$ $E_{g} = 2 \times (8.617 \times 1 0^{- 5}) \times 8214 = 1.416 eV .$ Sanity check: $\approx 1.42 eV$ — this is essentially GaAs! We recovered a real material's gap from two resistance readings.

Q4.2 Two knobs at once

Silicon is cooled to $200 K$ . By what factor does $n$ drop from its $300 K$ value, and does this push silicon toward "metal-like" or "insulator-like" behaviour?

Recall Solution

Using the scale $E_{g} /2 k_{B} = 6499 K$ from Q3.1: $\frac{n ( 200 )}{n ( 300 )} = e^{6499 (\frac{1}{300} - \frac{1}{200})} .$ $\frac{1}{300} - \frac{1}{200} = \frac{200 - 300}{60000} = - 1.6667 \times 1 0^{- 3} K^{- 1} .$ $= e^{6499 \times (- 1.6667 \times 1 0^{- 3})} = e^{- 10.83} \approx 1.98 \times 1 0^{- 5} .$ So $n$ falls by a factor $\approx 5.1 \times 1 0^{4}$ — carriers plummet by about $50, 000 \times$ . Cooling makes silicon behave more like an insulator (very few carriers). This is why cooled intrinsic semiconductors are nearly non-conducting, and why real devices rely on doping rather than intrinsic carriers.

L5 — Mastery

Q5.1 The "activation energy" plot (Arrhenius reasoning)

An engineer measures $σ$ at several temperatures and plots $ln σ$ against $1/ T$ . She gets a straight line of slope $- 7500 K$ . From ONLY that slope, what is the band gap? Sketch (via figure s03) why the plot is a straight line and what its slope encodes.

Recall Solution

WHY this tool — the logarithm and the reciprocal axis: starting from $σ \propto e^{- E_{g} / (2 k_{B} T)}$ , taking $ln$ gives $ln σ = const - \frac{E _{g}}{2 k _{B}} \cdot \frac{1}{T} .$ This is exactly the form $y = c + m x$ with $y = ln σ$ , $x = 1/ T$ , and slope $m = - \frac{E _{g}}{2 k _{B}}$ . A curved exponential becomes a straight line — that is why we plot $ln σ$ vs $1/ T$ : the slope directly hands us the gap. See figure s03: the red dashed line's steepness is the gap. Extract $E_{g}$ : $- \frac{E _{g}}{2 k _{B}} = - 7500 K \Rightarrow E_{g} = 2 \times (8.617 \times 1 0^{- 5}) \times 7500 = 1.293 eV .$ So $E_{g} \approx 1.29 eV$ .

Q5.2 Why the observed slope can double

A subtle point: sometimes the measured slope gives an " $E_{g}$ " that is exactly half the true gap. Explain the physics, using the definition of the exponent.

Recall Solution

The intrinsic law is $n \propto e^{- E_{g} / (2 k_{B} T)}$ , so the carrier slope is $- E_{g} / (2 k_{B})$ and one recovers the full $E_{g}$ (as in Q5.1). But some textbooks/experiments plot with the exponent written as $e^{- E_{a} / (k_{B} T)}$ , calling $E_{a}$ the activation energy. Matching the two forms: $\frac{E _{a}}{k _{B} T} = \frac{E _{g}}{2 k _{B} T} \Rightarrow E_{a} = \frac{E _{g}}{2} .$ So an "activation energy" of $0.56 eV$ measured for silicon is not a contradiction — it is $E_{g} /2 = 1.12/2$ . The lesson: always check whether the exponent in your model carries the $2$ or hides it inside a redefined $E_{a}$ . The factor of 2 (from the mid-gap Fermi level) is the single most-confused number in this entire topic.

Recall Verify Q5.2 numerically

If $E_{g} = 1.12 eV$ for Si, then $E_{a} = 0.56 eV$ . A slope-based plot using $e^{- E_{a} / (k_{B} T)}$ has slope $- E_{a} / k_{B} = - 0.56/ (8.617 \times 1 0^{- 5}) = - 6499 K$ — the same $6499 K$ scale we met in Q3.1. Everything is consistent.

Recall Self-test summary (cover the right side)

Formula linking $n$ , $E_{g}$ , $T$ ::: $n \propto e^{- E_{g} / (2 k_{B} T)}$ Value of $k_{B} T$ at 300 K ::: $\approx 0.026 eV$ Silicon exponent $E_{g} / (2 k_{B} T)$ at 300 K ::: $\approx 21.7$ Temperature scale $E_{g} /2 k_{B}$ for Si ::: $\approx 6500 K$ $n (400) / n (300)$ for Si ::: $\approx 225$ Why metals lose conductivity when heated ::: more lattice vibration → more scattering → lower mobility How to get $E_{g}$ from a $ln σ$ vs $1/ T$ slope ::: $E_{g} = - 2 k_{B} \times (slope)$

Connections

Parent: Band gap and conductivity
Fermi level and Fermi-Dirac distribution
Intrinsic vs extrinsic semiconductors
Doping and carrier concentration
Conductivity, mobility and drift current
Temperature dependence of resistivity in metals
Direct vs indirect band gap
Formation of energy bands from atomic orbitals