Direct vs indirect band gap materials: Exercises — Hardware deep dive

This page is your self-test. Each problem states cleanly what to find, then hides a complete worked solution inside a collapsible callout. Problems climb from L1 Recognition (just name the thing) to L5 Mastery (chain several ideas together). Every number you are asked to compute is machine-checked.

Before starting, recall the toolbox from the parent note the topic note:

Crystal momentum $ℏ k$ — an electron's "position along the horizontal axis" of the $E$ – $k$ picture. See Band Theory Basics.
Photon momentum $q_{p h} = 2 π / λ$ — tiny.
Phonon — a lattice vibration that carries lots of momentum but little energy. See Phonons and Lattice Vibrations.
Vertical (direct) transition: $k_{c} \approx k_{v}$ .

Useful constants used throughout: $h = 6.626 \times 1 0^{- 34} J\cdots, ℏ = 1.055 \times 1 0^{- 34} J\cdots, c = 3.00 \times 1 0^{8} m/s, 1 eV = 1.602 \times 1 0^{- 19} J .$

A handy shortcut you will reuse: the photon energy in electron-volts for a wavelength in nanometres is $E [eV] \approx \frac{1240}{λ [ nm ]} .$

Figure — Direct vs indirect band gap materials

Level 1 — Recognition

Recall Solution L1.1

Read straight off the extrema positions:

GaAs — CBM and VBM both at $Γ$ ( $k = 0$ ) → direct.
Si — CBM near the $X$ point ( $k \neq = 0$ ) → indirect.
GaN — direct.
Ge — CBM at the $L$ point ( $k \neq = 0$ ) → indirect.
GaP — indirect.
InP — direct.

Mnemonic: the "Ga_As/Ga_N/InP" light-emitters are direct; the group-IV workhorses Si and Ge (plus GaP) are indirect.

Recall Solution L1.2

Panel A = direct: the electron drops straight down (vertical blue arrow), a photon alone conserves momentum.
Panel B = indirect: the electron must move sideways in $k$ (pink dashed arrow) as well as down. A phonon supplies that sideways momentum; the photon supplies (most of) the energy.

Level 2 — Application

Recall Solution L2.1

Use $λ [nm] = 1240/ E [eV]$ : $λ = \frac{1240}{1.42} \approx 873 nm .$ What this is: near-infrared, just past the red edge of vision (~700 nm) — invisible to the eye, which is why GaAs is used in IR remotes and lasers rather than for visible LEDs.

Recall Solution L2.2

Step 1 — photon momentum (what/why): the wavevector measures how much crystal momentum a photon can donate. $q_{p h} = \frac{2 π}{873 \times 1 0 ^{- 9} m} \approx 7.20 \times 1 0^{6} m^{- 1} .$ Step 2 — zone half-width: $k_{B Z} = \frac{π}{0.565 \times 1 0 ^{- 9} m} \approx 5.56 \times 1 0^{9} m^{- 1} .$ Step 3 — ratio: $\frac{q _{p h}}{k _{B Z}} \approx \frac{7.20 \times 1 0 ^{6}}{5.56 \times 1 0 ^{9}} \approx 1.30 \times 1 0^{- 3} .$ What it looks like: on the horizontal axis of figure s01, the photon's push is about 1/1000 of the way across the panel — invisibly small. That is exactly why only a vertical drop is allowed optically.

Level 3 — Analysis

Recall Solution L3.1

Recall the Tauc forms:

Direct: $α \propto (h ν - E_{g})^{1/2}$ , so $α^{2} \propto (h ν - E_{g})$ → $α^{2}$ linear.
Indirect: $α \propto (h ν - E_{g})^{2}$ , so $α^{1/2} \propto (h ν - E_{g})$ → $α^{1/2}$ linear.

Dataset P: $α^{2}$ linear ⇒ direct gap, $E_{g} = 1.42$ eV ⇒ GaAs. Dataset Q: $α^{1/2}$ linear ⇒ indirect gap, $E_{g} \approx 1.12$ eV ⇒ Silicon.

Why the trick works: the power law becomes a straight line only when you raise $α$ to the reciprocal of its exponent, and the line's x-intercept reads off $E_{g}$ (see the two straight lines in figure s02).

Recall Solution L3.2

What/why: $d = 1/ α$ is the distance over which light intensity falls to $1/ e$ — a "how thick must my absorber be" number. $d_{GaAs} = \frac{1}{1 0 ^{4} cm ^{- 1}} = 1 0^{- 4} cm = 1 μ m,$ $d_{Si} = \frac{1}{1 0 ^{2} cm ^{- 1}} = 1 0^{- 2} cm = 100 μ m .$ Ratio: $d_{Si} / d_{GaAs} = 100$ . Interpretation: Silicon (indirect, weak phonon-assisted absorption) needs ~100× more thickness than direct GaAs — the reason Silicon Solar Cells are relatively thick wafers while GaAs cells can be thin films.

Level 4 — Synthesis

Recall Solution L4.1

(a) Zone half-width: $k_{B Z} = \frac{π}{0.543 \times 1 0 ^{- 9}} \approx 5.79 \times 1 0^{9} m^{- 1},$ $Δ k = 0.85 \times 5.79 \times 1 0^{9} \approx 4.92 \times 1 0^{9} m^{- 1} .$ (b) $\frac{Δ k}{q _{p h}} = \frac{4.92 \times 1 0 ^{9}}{6 \times 1 0 ^{6}} \approx 8.2 \times 1 0^{2} \approx 820.$ (c) The photon can supply only ~1/820 of the needed momentum, so a phonon — which lives near the zone edge and carries momentum of order $k_{B Z}$ — must be created or absorbed to make up the difference. Requiring electron and photon and phonon to meet simultaneously is the rare three-body event that makes silicon a poor emitter. See Recombination Mechanisms.

Recall Solution L4.2

Why two branches: to jump indirectly, the electron can either absorb an existing phonon (borrowing energy $E_{p h}$ , so a smaller photon suffices) or emit one (paying $E_{p h}$ , needing a bigger photon). $E_{abs} = E_{g} - E_{p h} = 1.12 - 0.058 = 1.062 eV,$ $E_{emit} = E_{g} + E_{p h} = 1.12 + 0.058 = 1.178 eV .$ Separation: $E_{emit} - E_{abs} = 2 E_{p h} = 0.116 eV$ . What it looks like: a small "step-in-step" shape in $α$ vs $h ν$ — two thresholds $2 E_{p h}$ apart — a fingerprint of indirect, phonon-assisted absorption. Relevant to Optical Absorption in Semiconductors.

Level 5 — Mastery

Recall Solution L5.1

(a) $E_{g} = 1240/650 \approx 1.91 eV$ . So a red emitter needs $E_{g} \approx 1.9$ eV (materials like AlGaInP hit this). (b) It must be direct. LED emission is the reverse of absorption — an electron drops across the gap and hands its energy to a photon. A photon carries negligible momentum, so this only happens efficiently when the CBM and VBM share $k$ (direct). In an indirect material the drop needs a phonon too → rare → weak light. (c) Silicon's emitted wavelength: $λ_{Si} = \frac{1240}{1.12} \approx 1107 nm .$ Two independent failures:

Wrong colour: 1107 nm is infrared, not red — you can't see it at all.
Wrong gap type: silicon is indirect, so recombination is phonon-assisted, slow, and mostly non-radiative (heat) — even the IR it emits is extremely faint.

Fixing only one problem is not enough; a good visible LED needs both a direct gap and the right $E_{g}$ . See LEDs and Laser Diodes.

Recall Solution L5.2

Why Beer's law: light dies exponentially with depth; 90% absorbed means 10% transmitted, so $e^{- α t} = 0.10$ . Solve for $t$ : $t = \frac{- l n ( 0.10 )}{α} = \frac{l n 10}{α} = \frac{2.303}{α} .$ Silicon ( $α = 1 0^{3} cm^{- 1}$ ): $t_{Si} = \frac{2.303}{1 0 ^{3} cm ^{- 1}} = 2.303 \times 1 0^{- 3} cm \approx 23.0 μ m .$ GaAs ( $α = 1 0^{4} cm^{- 1}$ ): $t_{GaAs} = \frac{2.303}{1 0 ^{4} cm ^{- 1}} \approx 2.30 μ m .$ Comment: GaAs needs ~10× less material at this $α$ ; in reality near the band edge silicon's $α$ drops far lower still, so practical wafers are ~100–200 µm. The exponential Beer's law is the quantitative face of "indirect = weak absorption = thick cell."

Recall Self-check: did you climb every level?

L1 name types ::: GaAs/GaN/InP direct; Si/Ge/GaP indirect — set by extrema position, not gap size. L2 photon momentum ::: $q_{p h} / k_{B Z} \sim 1 0^{- 3}$ , so only vertical transitions are optically allowed. L3 Tauc plots ::: $α^{2}$ linear ⇒ direct; $α^{1/2}$ linear ⇒ indirect; intercept = $E_{g}$ . L4 phonon role ::: phonon supplies the ~800× larger momentum jump; only splits the edge by $\pm E_{p h}$ in energy. L5 design + Beer ::: need direct gap AND correct $E_{g}$ ; 90% absorption ⇒ $t = ln 10/ α$ .