---
title: "Electrons in Solids"
book: "University Physics — Year 3"
subject: physics
language: en
chapter: 24
exercises: 12
source: https://one-course.com/books/physics/5/en/chapter/24-electrons-in-solids
---

# Chapter 24 — Electrons in Solids

Copper conducts electricity a million billion billion times better than quartz — the widest range of any physical property, commanded by materials that look like grey lumps either way. The scaffolding of the last chapter explains none of it; the *electrons* poured into it explain all of it. This chapter runs the century’s three passes at the problem, each inheriting the last’s wreckage: Drude’s classical pinball (1900), which gets Ohm’s law right and the heat capacity scandalously wrong; Sommerfeld’s Fermi gas, which is [Chapter 19](https://one-course.com/books/physics/5/en/chapter/19-quantum-statistics#ch-b3-quantum-statistics) cashing its cheque; and Bloch’s insight that an electron wave in a periodic [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) Bragg-reflects like any other wave — opening *[band gaps](#thm-b3-electrons-in-solids-bloch)* that sort all solids into [metals](#def-b3-electrons-in-solids-classes), [insulators](#def-b3-electrons-in-solids-classes), and the narrow-gap middle children, [semiconductors](#def-b3-electrons-in-solids-classes), on which the whole electronic age is built. The chapter ends inside a solar panel.

## 24.1 Drude’s pinball metal

**Proposition 24.1 (The Drude conductivity).**

Treat a [metal](#def-b3-electrons-in-solids-classes)’s valence electrons as a classical gas, density $n$, bouncing off obstacles every $\tau$ seconds. A field $\vect E$ accelerates each between collisions; the average *[drift velocity](#prop-b3-electrons-in-solids-drude)* is $\vect v_{\text{d}} = -e\tau\vect E/m_{\text{e}}$, tiny beside the thermal motion. The current density $\vect j =
-ne\vect v_{\text{d}}$ then obeys Ohm’s law locally, $\vect j = \sigma\vect E$, with

$$
\sigma = \frac{ne^2\tau}{m_{\text{e}}} .
$$

Copper’s measured $\sigma$ demands $\tau \approx 2.5 \times 10^{-14}\,\mathrm{s}$ — and the model’s two glories follow: it explains why resistors heat (the field’s work is thermalised each collision), and it predicts the Wiedemann–Franz law, that $\kappa/\sigma T$ is the same constant for all [metals](#def-b3-electrons-in-solids-classes) — electrons carry both charge and heat ([Exercise 24.5](#exo-b3-electrons-in-solids-5)).

**Proof.** Between collisions $\dot{\vect v} = -e\vect E/m_{\text{e}}$; starting afresh at each collision, the mean velocity acquired in a mean time $\tau$ is $-e\tau\vect E/m_{\text{e}}$. Then $\vect j =
-ne\vect v_{\text{d}} = (ne^2\tau/m_{\text{e}})\vect E$. ∎

![Drude’s picture: an electron ricochets at 106\, m/ s, while the field superimposes a drift of fractions of a millimetre per second (). Ohm’s law drops out; the model’s failures taught more still.](https://one-course.com/images/onecourse/chapters/physics-5/b3-electrons-in-solids/fig-d284274830a3.svg)

*Drude’s picture: an electron ricochets at $\sim10^{6}\,\mathrm{m}/\mathrm{s}$, while the field superimposes a drift of fractions of a millimetre per second ([Exercise 24.2](#exo-b3-electrons-in-solids-2)). Ohm’s law drops out; the model’s failures taught more still.*

**Remark 24.2 (Sommerfeld’s rescue).**

Drude’s gas should add $\tfrac32k_{\text{B}}$ per electron to a [metal](#def-b3-electrons-in-solids-classes)’s heat capacity; experiment finds a hundred times less. [Chapter 19](https://one-course.com/books/physics/5/en/chapter/19-quantum-statistics#ch-b3-quantum-statistics) already acquitted the electrons: they are a *degenerate Fermi gas*, $T_{\text{F}} \sim
8 \times 10^{4}\,\mathrm{K}$, and only the thermal shell within $k_{\text{B}}T$ of the Fermi surface can respond — to heat, and to scattering. Sommerfeld’s re-run of Drude with Fermi–Dirac statistics keeps the conductivity formula (with $\tau$ now the lifetime of Fermi-surface electrons, whose $10^{6}\,\mathrm{m}/\mathrm{s}$ is $v_{\text{F}}$, not thermal speed), fixes the heat capacity to its measured $T/T_{\text{F}}$ sliver, and even repairs Wiedemann–Franz’s numerical constant. One mystery survives every free-electron model: why quartz will not conduct at all.

## 24.2 Bloch: the lattice speaks

**Theorem 24.3 (Bloch states and band gaps).**

In a perfect periodic potential, the stationary states are *Bloch waves* $\psi_k(x) = u_k(x)\eu^{\iu kx}$ — plane waves dressed with the [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice)’s own periodicity $u_k$ — which propagate *without scattering*: a perfect [crystal](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) has zero resistance, and real resistance comes from imperfections (vibrations, impurities). But not all energies survive. At $k = \pi/a$ the electron’s wavelength is $2a$: Bragg’s condition on the [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) itself ([Theorem 23.6](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#thm-b3-crystalline-solids-dispersion) made the same discovery for sound). The reflected and incident waves form two standing patterns — charge piled on the ions (lower energy) or between them (higher) — so the free-electron parabola $E = \hbar^2k^2/2m_{\text{e}}$ tears open: an interval of forbidden energies, the *[band gap](#thm-b3-electrons-in-solids-bloch)* $E_{\text{g}}$, separates a filled-out *band* of allowed states from the next. A [crystal](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice)’s electronic identity is its ladder of bands and gaps.

**Proof.** *Admitted at this level.* ∎

![The nearly-free electron. Away from the zone edge the electron barely notices the lattice; at k = π/a its own Bragg reflection splits the parabola, forbidding the energies in the gap. Sound waves did exactly this in the last chapter — same lattice, same interference, new tenant.](https://one-course.com/images/onecourse/chapters/physics-5/b3-electrons-in-solids/fig-16b22cb1629d.svg)

*The nearly-free electron. Away from the zone edge the electron barely notices the [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice); at $k = \pi/a$ its own Bragg reflection splits the parabola, forbidding the energies in the gap. Sound waves did exactly this in the last chapter — same [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice), same interference, new tenant.*

**Definition 24.4 (Metals, insulators, semiconductors).**

Fill the bands with the [crystal](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice)’s electrons, two per state ([Chapter 14](https://one-course.com/books/physics/5/en/chapter/14-identical-particles#ch-b3-identical-particles)), coldest first. If the topmost occupied band is *partly* filled, infinitesimal energy can shift electrons into net motion: a *metal* (sodium: one valence electron, half a band; divalent metals conduct through overlapping bands). If it is *exactly* full, with a wide gap above, no small push can rearrange anything: an *insulator* (diamond, $E_{\text{g}} = 5.5\,\mathrm{eV}$). If the gap is small — $1.12\,\mathrm{eV}$ in silicon — thermal agitation hoists a few electrons across, each leaving a mobile *[hole](#def-b3-electrons-in-solids-doping)* below: a *semiconductor*, whose carrier count $\propto\eu^{-E_{\text{g}}/2k_{\text{B}}T}$ doubles every few degrees. Hence the great divide: metals conduct worse when heated (more vibrations to scatter off), semiconductors better (exponentially more carriers) — one sign flip that identifies a material’s class in a single measurement.

![Band filling decides everything. A partly filled band conducts; a full band under a wide gap cannot; a narrow gap lets temperature (or light, or doping) negotiate — the semiconductor’s entire usefulness is that its conductivity is adjustable.](https://one-course.com/images/onecourse/chapters/physics-5/b3-electrons-in-solids/fig-45ec09f57aa1.svg)

*Band filling decides everything. A partly filled band conducts; a full band under a wide gap cannot; a narrow gap lets temperature (or light, or [doping](#def-b3-electrons-in-solids-doping)) negotiate — the [semiconductor](#def-b3-electrons-in-solids-classes)’s entire usefulness is that its conductivity is *adjustable*.*

## 24.3 Engineering the gap: doping and the junction

**Definition 24.5 (Doping).**

Replace one silicon atom in a million by phosphorus (five valence electrons): four bonds are satisfied and the fifth electron, bound by a mere $45\,\mathrm{meV}$ ([Exercise 24.10](#exo-b3-electrons-in-solids-10)), is free at room temperature. Such *donors* make an *n-type* [semiconductor](#def-b3-electrons-in-solids-classes), conduction by electrons; boron (three electrons) is an *acceptor*, grabbing a bond electron and releasing a mobile *hole* — a missing electron that moves, responds to fields, and carries positive charge as genuinely as a bubble carries buoyancy: *p-type*. Doping swings the carrier density — and hence conductivity — across six orders of magnitude at will: the knob that turns sand into circuitry.

**Proposition 24.6 (The p–n junction).**

Join p-type to n-type. Electrons spill toward the [holes](#def-b3-electrons-in-solids-doping) and annihilate them near the interface, exposing a *[depletion zone](#prop-b3-electrons-in-solids-junction)* of fixed ionised dopants whose double layer builds an internal field — equilibrium at the built-in voltage $V_{\text{bi}} = (k_{\text{B}}T/e)
\ln(N_{\text{a}}N_{\text{d}}/n_{\text{i}}^2) \approx
0.7\,\mathrm{V}$ in silicon. The junction then rectifies: forward bias lowers the barrier and current grows as $\eu^{eV/k_{\text{B}}T}$; reverse bias raises it and only a leakage $I_0$ flows:

$$
I = I_0\big(\eu^{eV/k_{\text{B}}T} - 1\big) .
$$

This asymmetry is the *diode* — and, run in its variants, the LED (recombining pairs emit gap-energy photons), the solar cell (gap-energy photons create pairs that the built-in field sweeps out: [Problem 24.1](#pb-b3-electrons-in-solids-1)), and, doubled into sandwiches, the transistor: the switch of which a modern chip prints hundreds of billions.

**Proof.** *Admitted at this level.* ∎

![The diode law. Forward voltage is repaid exponentially; reverse voltage buys only the leakage I_0. One junction rectifies; two make a transistor; a square metre of them, sunlit, makes a power plant.](https://one-course.com/images/onecourse/chapters/physics-5/b3-electrons-in-solids/fig-fcd4c3115e60.svg)

*The diode law. Forward voltage is repaid exponentially; reverse voltage buys only the leakage $I_0$. One junction rectifies; two make a transistor; a square metre of them, sunlit, makes a power plant.*

![A finished silicon wafer: hundreds of chips, billions of junctions each, printed into one doped crystal — band theory as the twentieth century’s most consequential manufacturing recipe.](https://one-course.com/images/onecourse/chapters/physics-5/b3-electrons-in-solids/img-bb4e74add7d4.jpg)

*A finished silicon wafer: hundreds of chips, billions of junctions each, printed into one doped [crystal](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) — band theory as the twentieth century’s most consequential manufacturing recipe.*

## 24.4 Exercises

**Exercise 24.1 ★.**

Drude’s copper. $n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$, resistivity $\rho = 1.7 \times 10^{-8}\,\Omega\,\mathrm{m}$. Compute (a) $\tau =
m_{\text{e}}/ne^2\rho$; (b) the mean free path using the Fermi velocity $v_{\text{F}} = 1.6 \times 10^{6}\,\mathrm{m}/\mathrm{s}$; (c) that path in [lattice constants](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) ($a = 3.61\,\text{Å}$); (d) explain why so long a flight already hints that ions themselves do not do the scattering.

**Solution of Exercise 24.1.**

(a) $\tau = m_{\text{e}}/ne^2\rho = 2.5 \times 10^{-14}\,\mathrm{s}$. (b) $\ell =
v_{\text{F}}\tau \approx 39\,\mathrm{nm}$. (c) About 110 [lattice constants](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice). (d) An electron sailing past a hundred ions without scattering cannot be bouncing off the ions themselves — the [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) must be transparent to it, exactly what Bloch’s theorem ([Theorem 24.3](#thm-b3-electrons-in-solids-bloch)) later proves.

**Exercise 24.2 ★.**

The snail in the wire. A $1\,\mathrm{mm}^{2}$ copper wire carries $10\,\mathrm{A}$. (a) Compute the [drift velocity](#prop-b3-electrons-in-solids-drude). (b) How long would an electron take to cross a $1\,\mathrm{m}$ lamp cord? (c) Why does the lamp nevertheless light instantly? (d) Compare $v_{\text{d}}$ with $v_{\text{F}}$ and comment on Drude’s pinball picture.

**Solution of Exercise 24.2.**

(a) $v_{\text{d}} = I/neA = 7.3 \times 10^{-4}\,\mathrm{m}/\mathrm{s}$ — under a millimetre per second. (b) About 23 minutes. (c) The *field* establishes itself along the wire at nearly light speed and sets the whole electron sea drifting at once: the marchers are slow, the order to march is not. (d) $v_{\text{d}}/v_{\text{F}} \sim
10^{-9}$: conduction is an imperceptible bias on a violent quantum motion, not a calm classical flow.

**Exercise 24.3 ★.**

Sorting by bands. Classify, with the band-filling rule: (a) sodium (one valence electron); (b) magnesium (two — yet a [metal](#def-b3-electrons-in-solids-classes): what must its bands do?); (c) diamond ($E_{\text{g}} = 5.5\,\mathrm{eV}$); (d) silicon ($1.12\,\mathrm{eV}$) — and state for each the sign of $\dd\rho/\dd T$.

**Solution of Exercise 24.3.**

(a) Half-filled band: [metal](#def-b3-electrons-in-solids-classes), $\dd\rho/\dd T > 0$. (b) Two electrons would fill its band exactly — magnesium conducts because its full band *overlaps* the next empty one: [metal](#def-b3-electrons-in-solids-classes), $\dd\rho/\dd T > 0$. (c) Full band, $5.5\,\mathrm{eV}$ gap: [insulator](#def-b3-electrons-in-solids-classes) (formally $\dd\rho/\dd T < 0$, but with essentially no carriers to count). (d) [Semiconductor](#def-b3-electrons-in-solids-classes): $\dd\rho/\dd T < 0$, the exponential sign flip that betrays the class.

**Exercise 24.4 ★.**

Gaps and photons. (a) What photon wavelength matches silicon’s gap, and in which spectral region does silicon become transparent? (b) Same for diamond: why is it clear in the visible? (c) A gallium nitride LED has $E_{\text{g}} =
3.4\,\mathrm{eV}$: what colour edge does that set, and why did blue LEDs need this material? (d) Why does a red LED die dark rather than glow faintly white when overdriven?

**Solution of Exercise 24.4.**

(a) $\lambda = hc/E_{\text{g}} = 1.1\,\text{µ}\mathrm{m}$: silicon is transparent in the infrared beyond that — which is why infrared cameras can be lensed in silicon. (b) $225\,\mathrm{nm}$, in the ultraviolet: no visible photon can be absorbed, so diamond is water-clear. (c) $365\,\mathrm{nm}$: a gap wide enough for blue ($2.8\,\mathrm{eV}$) exists in nitrides and almost nowhere else practical — the blue LED waited decades on materials growth. (d) Its gap fixes its photon: overdriving adds heat, not gap width, and the diode cooks dark — an LED’s colour is a material constant, not a brightness setting.

**Exercise 24.5 ★★.**

Wiedemann–Franz. (a) From copper’s $\kappa =
400\,\mathrm{W}/(\mathrm{m}\,\mathrm{K})$ and $\sigma = 5.9 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ at $300\,\mathrm{K}$, compute $\kappa/\sigma T$. (b) Compare with the Sommerfeld value $L = \pi^2k_{\text{B}}^2/3e^2 =
2.44 \times 10^{-8}\,\mathrm{W}\,\Omega/\mathrm{K}^{2}$. (c) Explain in one sentence why one kind of carrier ties the two conductivities. (d) Cooking pans and their handles: use the law to explain a kitchen’s material choices.

**Solution of Exercise 24.5.**

(a) $\kappa/\sigma T = 400/(5.9\times10^7\times300) =
2.3 \times 10^{-8}\,\mathrm{W}\,\Omega/\mathrm{K}^{2}$. (b) Within ten percent of $L$. (c) The same Fermi-surface electrons carry both the charge and the heat, so their scattering time cancels in the ratio. (d) The steel pan conducts heat to the food because its electrons move; the wooden handle, an [insulator](#def-b3-electrons-in-solids-classes) in both senses, keeps them — and the heat — out of your hand.

**Exercise 24.6 ★★.**

The heat-capacity acquittal. (a) Drude predicts $\tfrac32nk_{\text{B}}$ from the electrons: with copper’s $n$, what fraction would that add to the [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice)’s $3nk_{\text{B}}$ (one conduction electron per atom)? (b) [Chapter 19](https://one-course.com/books/physics/5/en/chapter/19-quantum-statistics#ch-b3-quantum-statistics) instead gives $C_{\text{el}}
\sim \tfrac{\pi^2}{2}nk_{\text{B}}(T/T_{\text{F}})$: evaluate the suppression $T/T_{\text{F}}$ at $300\,\mathrm{K}$ ($T_{\text{F}} = 8.1 \times 10^{4}\,\mathrm{K}$). (c) Why does the electronic term nevertheless *win* at liquid-helium temperatures (recall the [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice)’s $T^3$)? (d) What measurement, plotted as $C/T$ against $T^2$, untangles the two?

**Solution of Exercise 24.6.**

(a) Fifty percent extra — flagrantly absent from experiment. (b) $T/T_{\text{F}} \approx 0.004$: the electronic term is throttled to under a percent. (c) The [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice)’s $T^3$ collapses faster than the electrons’ $T$: below a few kelvin the “negligible” electrons are all that is left. (d) $C/T =
\gamma + \beta T^2$: the intercept $\gamma$ weighs the electrons, the slope $\beta$ the [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) — one graph, both tenants.

**Exercise 24.7 ★★.**

The Fermi surface meets the zone. (a) For copper, compute the Fermi wavelength $\lambda_{\text{F}} = 2\pi/k_{\text{F}}$ with $k_{\text{F}} = (3\pi^2n)^{1/3}$. (b) Compare with $2a$: how close is the Fermi surface to the Bragg condition? (c) Explain physically why the two standing waves at $k = \pi/a$ (charge on ions versus between them) must differ in energy. (d) Which experimental fact of [Exercise 24.1](#exo-b3-electrons-in-solids-1) does Bloch’s no-scattering theorem finally explain?

**Solution of Exercise 24.7.**

(a) $k_{\text{F}} = (3\pi^2n)^{1/3} = 1.36 \times 10^{10}\,\mathrm{m}^{-1}$: $\lambda_{\text{F}} = 4.6\,\text{Å}$. (b) $2a =
7.2\,\text{Å}$: the same order — copper’s Fermi surface reaches toward the zone boundary, and the gap-opening physics happens *at* the energies that matter. (c) One standing wave piles its charge on the positive ions (lower electrostatic energy), the other between them (higher): same wavelength, two energies — the gap. (d) The 110-lattice-constant free path: Bloch waves do not scatter off a perfect [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice), so only vibrations and impurities remain to resist.

**Exercise 24.8 ★★.**

The exponential thermometer. Intrinsic silicon has $n_{\text{i}}
\propto \eu^{-E_{\text{g}}/2k_{\text{B}}T}$. (a) Compute the carrier-density ratio between $400\,\mathrm{K}$ and $300\,\mathrm{K}$. (b) Hence sketch a thermistor’s resistance against temperature and contrast it with a platinum wire’s. (c) Why the factor 2 in the exponent (what is created *in pairs*)? (d) Estimate the temperature at which silicon electronics fails because intrinsic carriers swamp a $5 \times 10^{21}\,\mathrm{m}^{-3}$ [doping](#def-b3-electrons-in-solids-doping) ($n_{\text{i}}(300\,\text{K}) \approx 1 \times 10^{16}\,\mathrm{m}^{-3}$).

**Solution of Exercise 24.8.**

(a) $E_{\text{g}}/2k_{\text{B}} = 6500\,\mathrm{K}$: $\exp[6500(1/300 - 1/400)] \approx 230$. (b) The thermistor’s resistance plunges exponentially — a steep, sensitive curve; platinum’s climbs gently and linearly (more phonons). One is a thermometer by carrier count, the other by scattering. (c) Carriers are born in electron–hole *pairs*, and the equilibrium $np = n_{\text{i}}^2$ splits the gap’s cost between the two — hence $E_{\text{g}}/2$. (d) $n_{\text{i}}$ reaches $5 \times 10^{21}\,\mathrm{m}^{-3}$ near $760\,\mathrm{K}$: the [doping](#def-b3-electrons-in-solids-doping) drowns and the circuit forgets its design. Real devices give up earlier — their reverse leaks, doubling every ten kelvin, misbehave first.

**Exercise 24.9 ★★.**

[Doping](#def-b3-electrons-in-solids-doping) arithmetic. Silicon has $5 \times 10^{28}\,\mathrm{atoms}/\mathrm{m}^{3}$. (a) One phosphorus per million silicon atoms: what carrier density, and what ratio to $n_{\text{i}} \approx 1 \times 10^{16}\,\mathrm{m}^{-3}$? (b) By what factor has one ppm of dirt changed the conductivity? (c) Explain why [semiconductor](#def-b3-electrons-in-solids-classes) fabrication first purifies to parts per *billion* before [doping](#def-b3-electrons-in-solids-doping) deliberately. (d) Estimate the average distance between [donors](#def-b3-electrons-in-solids-doping) and compare with the $2.4\,\mathrm{nm}$ orbit of [Exercise 24.10](#exo-b3-electrons-in-solids-10).

**Solution of Exercise 24.9.**

(a) $5 \times 10^{22}\,\mathrm{m}^{-3}$ — five million times $n_{\text{i}}$. (b) The same factor $\sim5\times10^6$: one speck per million atoms owns the conductivity outright. (c) Because accidental ppm would do the same uninvited: only a [crystal](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) pure to ppb has a conductivity that belongs to the designer. (d) $d \sim n^{-1/3}
\approx 27\,\mathrm{nm}$, ten times the $2.4\,\mathrm{nm}$ orbit: [donors](#def-b3-electrons-in-solids-doping) are isolated hydrogens; push the [doping](#def-b3-electrons-in-solids-doping) a hundredfold and the orbits touch — an impurity band, and eventually a [metal](#def-b3-electrons-in-solids-classes).

**Exercise 24.10 ★★★.**

The [donor](#def-b3-electrons-in-solids-doping) as a [hydrogen atom](https://one-course.com/books/physics/5/en/chapter/11-the-hydrogen-atom#thm-b3-hydrogen-atom-levels). The fifth phosphorus electron orbits its $+e$ ion *inside* silicon: screen Coulomb by $\varepsilon_{\text{r}} = 11.7$ ([Chapter 22](https://one-course.com/books/physics/5/en/chapter/22-electromagnetism-in-matter#ch-b3-electromagnetism-in-matter)) and lighten the electron to its effective band mass $m^* = 0.26\,m_{\text{e}}$. (a) Scale the hydrogen results of [Chapter 11](https://one-course.com/books/physics/5/en/chapter/11-the-hydrogen-atom#ch-b3-hydrogen-atom): show $E = 13.6\,\mathrm{eV}\times(m^*/m_{\text{e}})/
\varepsilon_{\text{r}}^2$ and evaluate. (b) Compare with $k_{\text{B}}T$ at $300\,\mathrm{K}$: are [donors](#def-b3-electrons-in-solids-doping) ionised? (c) Scale the [Bohr radius](https://one-course.com/books/physics/5/en/chapter/11-the-hydrogen-atom#thm-b3-hydrogen-atom-levels) the same way. (d) The orbit spans dozens of [lattice](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) cells: explain why that self-consistently justifies using $\varepsilon_{\text{r}}$ and $m^*$ at all.

**Solution of Exercise 24.10.**

(a) $E = 13.6\times0.26/11.7^2 \approx 26\,\mathrm{meV}$ (the measured phosphorus value, $45\,\mathrm{meV}$, keeps the scale honest). (b) Comparable to $k_{\text{B}}T = 26\,\mathrm{meV}$: essentially all [donors](#def-b3-electrons-in-solids-doping) are ionised at room temperature — the premise of [Definition 24.5](#def-b3-electrons-in-solids-doping). (c) $a =
0.53\,\text{Å}\times\varepsilon_{\text{r}}/(m^*/
m_{\text{e}}) \approx 2.4\,\mathrm{nm}$. (d) An orbit spanning dozens of cells sees the [crystal](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) as a smooth medium — which is precisely the condition under which a bulk $\varepsilon_{\text{r}}$ and a band-averaged $m^*$ are legitimate: the approximation certifies itself.

**Exercise 24.11 ★★★.**

Junction numbers. A silicon diode has $N_{\text{a}} =
N_{\text{d}} = 1 \times 10^{22}\,\mathrm{m}^{-3}$, $n_{\text{i}} =
1 \times 10^{16}\,\mathrm{m}^{-3}$, $T = 300\,\mathrm{K}$. (a) Compute $V_{\text{bi}} = (k_{\text{B}}T/e)\ln(N_{\text{a}}N_{\text{d}}/
n_{\text{i}}^2)$. (b) From the diode law, compute the ratio of currents at $+0.5\,\mathrm{V}$ and $-0.5\,\mathrm{V}$. (c) Why does $I_0$ — hence the reverse leak — roughly double every $10\,\mathrm{K}$? (d) A bridge of four such diodes turns AC into DC: sketch the circuit’s idea in words.

**Solution of Exercise 24.11.**

(a) $V_{\text{bi}} = 0.0259\ln(10^{12}) \approx 0.71\,\mathrm{V}$. (b) $eV/k_{\text{B}}T = 19.3$: ratio $\eu^{19.3} \approx
2\times10^{8}$ — the diode is a one-way street to eight digits. (c) $I_0 \propto n_{\text{i}}^2 \propto
\eu^{-E_{\text{g}}/k_{\text{B}}T}$: the same exponential as [Exercise 24.8](#exo-b3-electrons-in-solids-8), hence the rule-of-thumb doubling. (d) The four diodes form a diamond: each half-cycle, whichever pair is forward-biased steers the current through the load in the *same* direction — AC in, bumpy DC out.

**Exercise 24.12 ★★★.**

Designing the solar gap. (a) Photons below $E_{\text{g}}$ pass through; photon energy above $E_{\text{g}}$ is lost as heat within picoseconds. Explain the resulting trade-off in choosing $E_{\text{g}}$. (b) The optimum near $1.3\,\mathrm{eV}$ caps single-junction efficiency at about a third (Shockley–Queisser): where does silicon ($1.12\,\mathrm{eV}$) stand? (c) Why do tandem stacks (a wide-gap cell atop a narrow-gap one) beat the cap? (d) Why is a *hot* solar panel a worse one? (Two chapter-honest reasons: the diode law’s $I_0$, and the gap’s slight shrinkage with $T$.)

**Solution of Exercise 24.12.**

(a) Narrow the gap and more photons clear it but each delivers only the small gap energy; widen it and each photon pays more but fewer qualify: harvest $\times$ voltage peaks in between. (b) Just below the $1.3\,\mathrm{eV}$ optimum: silicon’s ideal ceiling is $\sim30\,\%$ — close enough that its cheapness wins. (c) The wide-gap top cell takes the blue at high voltage, passing the red to the narrow-gap cell below: each photon is harvested near its own energy, thermalisation shrinks, and the stack’s ceiling climbs toward the forties. (d) Heat raises $I_0$ exponentially, dragging down the open-circuit voltage $\sim(k_{\text{B}}T/e)\ln(I_{\text{L}}/I_0)$; and the gap itself narrows slightly, trading voltage for current it cannot fully recover.

![Replica of the first transistor (Bell Laboratories, 1947, public domain): two gold contacts pressed onto a sliver of germanium. Band theory’s first machine — and, by descent, all the others.](https://one-course.com/images/onecourse/chapters/physics-5/b3-electrons-in-solids/img-0a49a6d4e4f8.jpg)

*Replica of the first transistor (Bell Laboratories, 1947, public domain): two gold contacts pressed onto a sliver of germanium. Band theory’s first machine — and, by descent, all the others.*

## 24.5 Problem: The rooftop referendum

**Problem 24.1.**

*The rooftop referendum.* Your neighbour is deciding whether to roof their house with photovoltaic panels and has appointed you, the physicist, as arbiter. The candidate panel: silicon, $2\,\mathrm{m}^{2}$, sixty cells in series, rated $400\,\mathrm{W}$ under the standard $1000\,\mathrm{W}/\mathrm{m}^{2}$ of noon sunlight.

**Part I — Sunlight meets the gap.**

1. Sunlight is roughly a $5800\,\mathrm{K}$ thermal spectrum ( [Chapter 20](https://one-course.com/books/physics/5/en/chapter/20-photons-and-phonons#ch-b3-photons-phonons) ): compute the energy of its peak photons (Wien) in eV.
2. Silicon’s gap is $1.12\,\mathrm{eV}$ : what is the longest wavelength a silicon cell can harvest?
3. Roughly a fifth of the sun’s power arrives in photons below the gap. What happens to it in the panel?
4. A $2.5\,\mathrm{eV}$ green photon is absorbed: how much of its energy survives as electron–hole pair energy, and where does the rest go, and how fast?
5. Combine items 3 and 4 into the spectrum’s verdict: about half the incident power is gone before any electronics begins. State the two loss channels in one sentence each.
6. Why does a photon need $\ge E_{\text{g}}$ at all — what forbids absorbing two half-gap photons in quick succession in ordinary silicon?

**Part II — The junction as engine.**

7. Each cell is a [p–n junction](#prop-b3-electrons-in-solids-junction) . Explain in three sentences how the [depletion zone](#prop-b3-electrons-in-solids-junction) ’s built-in field turns a created pair into external current — which carrier goes which way, and why they do not simply recombine.
8. With $N_{\text{a}} = N_{\text{d}} = 1 \times 10^{22}\,\mathrm{m}^{-3}$ and $n_{\text{i}} = 1 \times 10^{16}\,\mathrm{m}^{-3}$ , compute $V_{\text{bi}}$ at $300\,\mathrm{K}$ .
9. A working cell delivers about $0.6\,\mathrm{V}$ . Sixty in series: the panel’s operating voltage?
10. From the $400\,\mathrm{W}$ rating, deduce the operating current.
11. Estimate the photon flux (photons per second) the panel absorbs usefully, taking $\sim2\,\mathrm{eV}$ per absorbed photon, and compare with the electron flux your current implies: what fraction of absorbed photons yields a collected electron?
12. The rated efficiency: $400\,\mathrm{W}$ from $2000\,\mathrm{W}$ incident. Reconcile 20 % with Part I’s “half lost before electronics”: where do the remaining thirty points go?
13. Why does the cell deliver $0.6\,\mathrm{V}$ and not the full $1.12\,\mathrm{V}$ of the gap? (Name the culprit in the diode law.)

**Part III — Real roofs.**

14. A cloudless winter noon at latitude $50^\circ$ delivers light at $30^\circ$ elevation. Compute the geometric factor relative to normal incidence on a horizontal panel.
15. The panel datasheet lists $-0.4\,\%/\text{K}$ of output: a black roof panel reaches $65\,{}^{\circ}\mathrm{C}$ in summer. How much of the rating survives?
16. Explain the temperature loss with [Exercise 24.12](#exo-b3-electrons-in-solids-12) (d).
17. A chimney shades one cell of the sixty. Why can that strangle the whole series string — and what does the cell’s own diode nature do to the poor shaded cell?
18. Manufacturers wire a *bypass diode* across each sub-string: explain its job in one sentence.
19. Averaged over days and weather, the roof yields $15\,\%$ of rated power. Estimate the yearly energy from the $400\,\mathrm{W}$ panel in kilowatt-hours.

**Part IV — The verdict.**

20. At $0.25\,\mathrm{euros}$ per kWh, what does the panel earn per year? With an installed cost of $300\,\mathrm{euros}$ , what is the payback time?
21. The panel took roughly $500\,\mathrm{kWh}$ to manufacture (silicon is purified by melting): how long until it has repaid its own energy?
22. Your neighbour asks why the panel cannot be “just made black” to catch the sub-gap fifth. Answer with band theory in two sentences.
23. They ask next why not stack a second, smaller-gap panel beneath: answer with [Exercise 24.12](#exo-b3-electrons-in-solids-12) (c) — and name the practical catch.
24. Thirty years on, the panel has faded to 85 % of its rating. Which microscopic villains of this chapter (recall what limits $\tau$ , and what junctions fear) plausibly age a cell?
25. Deliver the arbiter’s summary in four lines: gap physics sets the harvest, the junction converts it, series wiring and temperature tax it, and the ledger — energy and euros — closes in the panel’s favour.

**Solution of Problem 24.1.**

**1.** $\lambda_{\text{peak}} \approx 500\,\mathrm{nm}$: about $2.5\,\mathrm{eV}$. **2.** $hc/E_{\text{g}} \approx 1.1\,\text{µ}\mathrm{m}$. **3.** It sails through the cells (no final state to absorb into) and ends as heat in the backing — warming the roof, not the wires. **4.** $1.12\,\mathrm{eV}$ survives as the pair; the excess $1.4\,\mathrm{eV}$ drains into phonons within picoseconds — faster than any circuit could intercept. **5.** Sub-gap transparency ($\sim20\,\%$ of the power) and above-gap thermalisation ($\sim30\,\%$): the spectrum is taxed half before the junction sees a single electron. **6.** Absorption is a single-photon quantum jump needing a real final state; at half the gap there is none to pause in, and two-photon events are negligible at sunlight’s photon densities. **7.** Pairs created in or near the [depletion zone](#prop-b3-electrons-in-solids-junction) feel the built-in field: electrons are swept to the n side, [holes](#def-b3-electrons-in-solids-doping) to the p side, and the field separates them faster than they can find each other to recombine — charge accumulates until an external circuit relieves it as current. **8.** $V_{\text{bi}} = 0.0259\ln(10^{12}) \approx
0.71\,\mathrm{V}$. **9.** $60\times0.6\,\mathrm{V} = 36\,\mathrm{V}$. **10.** $I = 400/36 \approx 11\,\mathrm{A}$. **11.** Panel-wide, above-gap light is $\sim1400\,\mathrm{W}$ at $\sim2\,\mathrm{eV}$ each: $4\times10^{21}$ photons per second — but the sixty cells are in *series*, so the $11\,\mathrm{A}$ (an electron flux $I/e \approx
7\times10^{19}\,\mathrm{s}^{-1}$) passes through each cell, whose own share of photons is $4\times10^{21}/60 \approx
7\times10^{19}$ per second: nearly every absorbed photon yields a collected electron. The quantum efficiency is excellent; the losses are energetic, not numeric. **12.** After the spectrum’s 50 %, the cell pays the voltage deficit ($0.6/1.12 \approx 54\,\%$) and a few points of reflection and recombination: $0.5\times0.54 \approx 27\,\%$, trimmed to the rated 20 %. **13.** The leakage $I_0$: the open-circuit voltage $(k_{\text{B}}T/e)\ln(I_{\text{L}}/I_0)$ stops where photocurrent and diode leak balance — about $0.6\,\mathrm{V}$, well short of the gap. **14.** $\sin30^\circ = 0.5$: half the noon rating from geometry alone — before clouds. **15.** $\Delta T = 40\,\mathrm{K}$: $-16\,\%$, leaving $\approx335\,\mathrm{W}$ — the sunniest days are not the best days per watt. **16.** Heat inflates $I_0$ exponentially, and the open-circuit voltage — a logarithm’s rebuke — falls a fraction of a percent per kelvin; the slightly shrunken gap finishes the job. **17.** Series wiring forces one common current: the shaded cell, generating none, is driven into reverse bias by its fifty-nine colleagues and dissipates their power as a hot spot — the string throttles to the weakest cell. **18.** The bypass diode gives the current a forward path around the shaded sub-string, sacrificing its voltage instead of the whole panel’s output. **19.** $400\,\mathrm{W}\times0.15\times8760\,\mathrm{h} \approx
530\,\mathrm{kWh}$ per year. **20.** $\approx130\,\mathrm{euros}$ per year: payback in roughly $300/130 \approx 2.5$ years, then two decades of profit. **21.** $500/530 \approx$ one year: the panel repays its manufacturing energy about as fast as its price. **22.** Blackness is not a choice but a [band structure](#thm-b3-electrons-in-solids-bloch): absorption needs an empty state one photon-energy above a full one, and below the gap silicon simply has none to offer — paint cannot add states. **23.** Stack a wide-gap cell on top to take the blue at high voltage and pass the red down: the tandem beats the single-junction ceiling. The catch: the series stack must match currents (and prices) between layers. **24.** Whatever shortens $\tau$ and poisons junctions: UV-created defects and in-diffused impurities scatter and trap carriers, moisture corrodes contacts, and hot spots age the diodes — the slow entropy of a [crystal](https://one-course.com/books/physics/5/en/chapter/23-crystalline-solids#def-b3-crystalline-solids-lattice) asked to sit in the sun for thirty years. **25.** The gap harvests half the sun; the junction turns pairs into $36\,\mathrm{V}$ of ordered current; series strings, shade and summer heat take their cut; and at $\sim530\,\mathrm{kWh}$ a year against a one-year energy and three-year money payback, the arbiter rules: roof it.