Visual walkthrough — Compound semiconductors (GaN, GaAs, SiC) overview

Step 1 — What is a "band" and what is the "gap"?

WHAT. Inside a solid, electrons are not allowed to have just any energy. They can only sit in certain allowed energy ranges, called bands. Between two bands there is a forbidden zone — no electron is allowed to have an energy inside it. That forbidden zone is the bandgap.

WHY. We start here because every symbol later ($E_g$, "falling," "photon") only makes sense once you can see the two shelves and the empty gap between them. Nothing on this page is meaningful without this picture.

PICTURE. Think of two shelves on a wall. The lower shelf is the valence band — the "at-rest" home of electrons, where they are stuck to atoms. The upper shelf is the conduction band — where an electron is free to roam and carry current. The wall gap between the shelves is $E_g$, measured in electron-volts (eV).

Step 2 — The electron falls, and something must carry the energy away

WHAT. Suppose an electron is sitting up on the conduction shelf (we put it there with a battery or with heat). It "wants" to drop back down to the empty spot on the valence shelf. When it drops, it loses exactly $E_g$ of energy. That lost energy cannot vanish — energy is conserved. It leaves as a tiny packet of light called a photon.

WHY. This is the pivot of the whole derivation: it links a material property ($E_g$) to a light property (the photon's energy). We invoke conservation of energy — not any fancier tool — because that alone fixes the photon's energy.

PICTURE. The electron (blue dot) drops from the top shelf to the bottom shelf. The height it falls is $E_g$. Out flies an orange wiggly arrow — the photon — carrying away that exact energy.

Step 3 — What is a photon's energy? Introducing frequency

WHAT. A photon is a ripple in the electromagnetic field. Like any wave, it wiggles up and down; the number of full wiggles per second is its frequency, written $\nu$ (Greek "nu"), measured in hertz (Hz = wiggles/second). Nature's rule (found by Max Planck) is: a photon's energy is proportional to its frequency.

WHY. We need a bridge from "energy" (Step 2) to "color." Color is not energy directly — color is wavelength. Frequency is the stepping stone between them. So we introduce $\nu$ because it is the quantity that energy is directly tied to.

PICTURE. A slow, lazy wave (few wiggles per second, low $\nu$) carries little energy — that's red/infrared. A fast, tight wave (many wiggles, high $\nu$) carries lots — that's blue/violet.

Step 4 — From frequency to wavelength (the thing our eyes read as color)

WHAT. Instead of counting wiggles per second, we can measure the length of one wiggle in space — that is the wavelength $\lambda$ (Greek "lambda"), measured in nanometres (nm). Frequency and wavelength are two views of the same wave, tied together because all light travels at the same speed $c$.

WHY. Our eyes and the LED-color chart speak wavelength, not frequency. So we must convert $\nu \to \lambda$. The tool that does it is the universal wave relation $c=\lambda \nu$ — chosen because it is the only thing linking how long a wiggle is to how many pass per second.

PICTURE. Imagine wave crests sailing past a fixed post at fixed speed $c$. If the crests are far apart (big $\lambda$), few pass per second (small $\nu$). If crests are packed tight (small $\lambda$), many pass per second (big $\nu$). Speed = spacing × rate.

Step 5 — Substitute and assemble the master formula

WHAT. We now have three true statements and we chain them: $E_g=E_{\text{photon}}=h\nu =h\cdot \frac{c}{\lambda }.$ Solve for $\lambda$.

WHY. Each substitution replaces one symbol by something we already earned: Step 2 gave $E_{\text{photon}}=E_g$; Step 3 gave $E_{\text{photon}}=h\nu$; Step 4 gave $\nu =c/\lambda$. No new tool is needed — just plugging in and rearranging. This is why the earlier steps had to come first: we may only use symbols we've defined.

PICTURE. A "chain of equals," each link labelled with the step that justifies it, ending at the clean result.

Step 6 — Where the number 1240 comes from (unit-crunching)

WHAT. Physicists give $E_g$ in eV and want $\lambda$ in nm. If we bundle $h$ and $c$ and convert their units, the messy constants collapse into one friendly number: $hc\approx 1240\text{eV⋅nm}.$

WHY. We do this so we never have to touch $10^{-34}$ and $10^8$ again. It's not new physics — it's the same $hc$, just pre-converted into the units engineers actually use.

PICTURE. Two ugly tiles ($6.626\times 10^{-34}\text{J⋅s}$ and $3\times 10^8\text{m/s}$), fed through a unit-converter box, out pops the tidy tile $1240\text{eV⋅nm}$.

Let's compute $hc$ once to trust the number: $hc=(6.626\times 10^{-34}\text{J⋅s})(3.0\times 10^8\text{m/s})=1.988\times 10^{-25}\text{J⋅m}.$ Convert joules→eV (divide by $1.602\times 10^{-19}$) and metres→nm ($\times 10^9$): $hc=\frac{1.988\times 10^{-25}}{1.602\times 10^{-19}}\times 10^9\text{eV⋅nm}\approx 1240\text{eV⋅nm}.✓$

Step 7 — Plug in the three headline materials

WHAT. Drop each material's $E_g$ (from the parent table) into $\lambda =1240/E_g$.

WHY. This is the payoff: it turns three atomic numbers into three colors and proves the parent's central claim about why GaN was needed.

PICTURE. A spectrum bar. Each material's arrow lands at its computed wavelength — Si and GaAs fall in the invisible infrared, GaN lands right in visible blue.

Step 8 — The edge and degenerate cases (don't get ambushed)

Every formula has boundaries. Walk them so you're never surprised.

Case A — Indirect-gap materials (Si, SiC). The formula still gives a wavelength, but almost no photon actually comes out. Why? In an indirect material the electron and the empty valence spot sit at different momenta, so the electron cannot just drop straight down — it must simultaneously borrow a phonon (a lattice vibration) to shed momentum. That three-body coincidence is rare, so light emission is feeble. Lesson: $\lambda =1240/E_g$ predicts the color if it emits; directness decides whether it emits.

Case B — $E_g\to$ very large (deep UV). As $E_g$ grows, $\lambda =1240/E_g\to$ smaller and smaller (UV, then X-ray). $\lambda$ never goes to zero for real materials because $E_g$ is finite, but the trend tells you ultra-wide-gap materials (e.g. AlN, ~6 eV) emit deep UV.

Case C — $E_g\to 0$ (a "metal"). The formula blows up: $\lambda =1240/0\to \infty$. Physically, a zero gap means the shelves touch — there is no clean fall, no fixed photon energy. Such a material is a metal/semimetal, not a light emitter. Lesson: the formula only applies when there is a gap to fall across.

Case D — Room-temperature broadening. Real LEDs don't emit one razor-sharp $\lambda$; heat smears the electron energies, so you get a band of colors around $1240/E_g$. The formula gives the center of that band, not a single line.

The one-picture summary

Everything above, compressed: an electron falls a height $E_g$ → emits a photon of energy $E_g$ → that energy fixes frequency ($E=h\nu$) → speed of light fixes wavelength ($c=\lambda \nu$) → so $\lambda =hc/E_g=1240/E_g$. Bigger gap ⇒ bluer light. Silicon lands in the infrared and, being indirect, stays dark; GaN's wide direct gap reaches blue.

Active recall