Lesson 5 · 30 min

Metals and bands: conductors, semiconductors and colour

Add atoms one by one to a chain of molecular orbitals until the levels merge into a band; use the band gap to explain metals, semiconductors and insulators — and why cadmium sulfide is yellow.

In this lesson

Explain how N atomic orbitals combine into N molecular orbitals that merge into a band when N is large
Tell metals, semiconductors and insulators apart by how their bands are filled
Explain what n-type and p-type doping do
Calculate the absorption-edge wavelength from a band gap and explain the colour of semiconductors

Copper wire carries current across continents, while diamond is a superb insulator; silicon sits in between and conducts better when warmed. Colour tells the same story: cadmium sulfide is the bright pigment “cadmium yellow”, cadmium selenide is nearly black-red, zinc oxide is snow white, and a lump of silicon has a grey-black metallic sheen. All of these differences come from one idea — energy bands.

From two atoms to a lump of metal

The 3s orbitals of two sodium atoms combine into two molecular orbitals: a lower bonding orbital and a higher antibonding one, $2|\beta|$ apart. Four atoms in a chain give four molecular orbitals, six give six … N atomic orbitals always combine into N molecular orbitals. The lowest has the same sign on every atom (fully bonding), the highest changes sign between every pair of neighbours (fully antibonding), and the rest lie in between.

Predict first

In a chain of 1000 sodium atoms, how are the energies of the molecular orbitals built from the 3s orbitals distributed?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Still just two levels: bonding and antibonding 1000 levels spreading ever wider without limit 1000 levels packed into a band of finite width, with almost no spacing All levels equal the original 3s energy

E_k = \alpha + 2\beta\cos\frac{k\pi}{N+1}, \qquad k = 1, 2, \dots, N

Molecular-orbital energies of a Hückel chain of N atoms. As N → ∞ the levels fill a band of width

4|\beta|

$\alpha$: energy of the isolated atomic orbital (Coulomb integral)
$\beta$: resonance integral between neighbouring orbitals, negative
$N$: number of atoms in the chain
$k$: orbital label; orbital k has k − 1 nodes

Hückel chain · tight-binding model

From molecular orbitals to bands

Atoms in the chain, N 8

Valence electrons per atom

Orbitals per atom

Show orbital k k = 1

Levels: 8
Spread of levels: 3.759|β| (limit 4|β|)
Level spacing at mid-band: 0.695|β|
Smallest excitation energy: 0.695|β| The least energy that moves one electron into a vacancy; in a partly filled band it goes to zero as N grows.

The top band is only partly filled: empty levels sit right above the filled ones, so electrons move with a nudge of energy — a metal.

Smallest excitation energy against N

Keep one valence electron per atom and drag N from 1 to 60: the gap between the highest filled and lowest empty level shrinks to zero. Then switch to two electrons: with s orbitals alone the band is full; add the p orbitals and shrink the s–p separation until the bands overlap — the reason magnesium is still a metal.

Metals, semiconductors and insulators

Each sodium atom contributes one electron. The 3s band of N atoms has N levels and room for 2N electrons, so it is only half full. Right above the highest filled level lie empty ones, and the slightest electric field lets electrons move into them and drift along the field — that is a metal. The familiar “sea of electrons” pictures exactly these electrons, delocalised through the whole lump of metal.

If a band is exactly full and the next band above lies across a band gap $E_\mathrm g$ , electrons need at least $E_\mathrm g$ of energy before they can conduct:

a large gap (diamond, 5.47 eV): thermal motion at room temperature can hardly excite any electrons — an insulator;
a modest gap (silicon 1.12 eV, germanium 0.66 eV): a few electrons are thermally excited into the conduction band, each leaving a hole in the valence band — a semiconductor, which conducts better the hotter it gets.

At room temperature $k_\mathrm BT = 0.026$ eV, and the number of thermally excited carriers scales roughly as $e^{-E_\mathrm g/2k_\mathrm BT}$ : silicon has about $10^{36}$ times as many as diamond.

Pure silicon has too few carriers to be very useful. Add a trace of phosphorus (one part per million, say): each P atom has one more valence electron than Si, bound by only about 0.045 eV, and at room temperature nearly all of these electrons enter the conduction band — an n-type semiconductor, conducting mainly by electrons. Add boron, with one electron fewer, and it leaves an empty level about 0.045 eV above the top of the valence band that takes up electrons, filling the valence band with holes — a p-type semiconductor. Put n-type and p-type side by side and you have the heart of diodes, solar cells and transistors: the p–n junction.

Why cadmium sulfide is yellow

Semiconductors absorb light through the same process that lets them conduct: a photon with more energy than $E_\mathrm g$ lifts an electron from the valence band into the conduction band and is absorbed; a photon with less energy cannot, and passes through or is scattered back. The gap is an energy threshold, and its absorption edge lies at

\lambda_\text{edge} = \frac{hc}{E_\mathrm g} = \frac{1240\ \mathrm{eV\,nm}}{E_\mathrm g}

Light shorter than

\lambda_\text{edge}

is absorbed; longer wavelengths survive and set the colour.

$h$: Planck constant
$c$: speed of light
$E_\mathrm g$: band gap

Idealised step edge · colour computed from the spectrum

The band gap sets the colour

Band gap E_g 2.42 eV · Cadmium sulfide

Absorption edge λ = hc/E_g: 512 nm
Class: semiconductor
Intrinsic carriers at 300 K (relative to Si): 1.2 × 10⁻¹¹

Computed body colour rgb(255 238 0)

Pick ZnO, CdS, GaP, CdSe and Si in turn: the absorption edge moves from the ultraviolet into the visible and on into the infrared, and the colour goes from white to yellow, orange, dark red and finally black. Drag the slider to find the gap at which colour first appears.

Worked example

The colour of cadmium sulfide

CdS has a band gap of 2.42 eV at 300 K. Which visible light does it absorb, and what colour does it look?

Concept check

Which is colourless (white) in white light: ZnO (3.37 eV) or CdS (2.42 eV)?

ZnO: its edge is in the ultraviolet, so it absorbs no visible light CdS: its smaller gap absorbs less light Both are yellow Both are black

Try the calculation

Silicon has a band gap of 1.12 eV. What is its absorption-edge wavelength?

Key takeaways

N atomic orbitals make N molecular orbitals; for large N they merge into a band about 4|β| wide.
A partly filled band → metal; a full band separated from an empty one by a gap → semiconductor (about 0.5–3 eV) or insulator.
Thermally excited carriers scale as e^(−E_g/2k_BT); phosphorus doping gives n-type, boron doping p-type silicon.
Photons above E_g are absorbed: the edge at λ = 1240/E_g nm sets the colour — ZnO white, CdS yellow, CdSe dark red, Si black.

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Data and references

S. M. Sze, K. K. Ng, Physics of Semiconductor Devices, 3rd ed. (Wiley, 2007), Appendix G — band gaps; donor and acceptor levels in Si.
C. Kittel, Introduction to Solid State Physics, 8th ed. (Wiley, 2005), Ch. 7–8, Table 8.1 — CdS, CdSe.
Ü. Özgür et al., “A comprehensive review of ZnO materials and devices”, J. Appl. Phys. 98, 041301 (2005).
P. A. Cox, The Electronic Structure and Chemistry of Solids (Oxford University Press, 1987), Ch. 1–4.
R. Hoffmann, Solids and Surfaces: A Chemist’s View of Bonding in Extended Structures (VCH, 1988).
C. Wyman, P.-P. Sloan, P. Shirley, JCGT 2(2), 1 (2013) — analytic CIE 1931 colour-matching functions used for the computed colours.