---
title: "Particle Diffusion"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 24
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/24-particle-diffusion
---

# Chapter 24 — Particle Diffusion

Drop a crystal of dye into still water and watch: a coloured cloud forms around it, grows, softens at its edges, and spreads — in a minute over a millimetre, in an hour over a centimetre, in a week over the glass. Nothing pushes the dye; it is carried by the ceaseless jostling of the molecules, which sends each particle on a random walk and, on average, from where there are many to where there are few. This is *diffusion*, the slowest and most universal of transports: it feeds every cell with oxygen, dopes every transistor with boron, hardens steel with carbon, and lets a perfume cross a room — though, as we shall see, not in the time one might think. This chapter gives diffusion its law (Fick), its equation (from a particle balance), its characteristic solutions and time scales ($L \sim \sqrt{Dt}$), and its microscopic origin in the random walk — which also explains why the [diffusion equation](#thm-b2-particle-diffusion-equation), unlike every equation of mechanics, knows the direction of time. The next chapter will reuse all of it for heat.

![Ink released in still water: the sharp cloud blurs and spreads as its molecules diffuse — over millimetres in a minute, centimetres in an hour.](https://one-course.com/images/onecourse/chapters/physics-4/b2-particle-diffusion/img-5f0218bb2714.jpg)

*Ink released in still water: the sharp cloud blurs and spreads as its molecules diffuse — over millimetres in a minute, centimetres in an hour.*

## 24.1 Particle density, flux and Fick’s law

**Definition 24.1 (Density and particle current).**

For a species of particles (molecules, ions, atoms in a solid) the *number density* $n(M,t)$ is the number of particles per unit volume around $M$ (in $\mathrm{m}^{-3}$; the molar concentration is $c = n/N_A$). The *particle current density* $\vect{j}_N$ is the vector such that the number of particles crossing an oriented surface element $\dd\vect S$ in $\dd t$ is $\vect{j}_N\cdot\dd\vect S\,\dd t$ (in $\mathrm{m}^{-2}\,\mathrm{s}^{-1}$): the *particle flux* through a surface $S$ is $\Phi_N = \iint_S \vect{j}_N\cdot\dd\vect S$, the number of particles per second through $S$. For particles carried by a fluid moving at $\vect v$, $\vect{j}_N = n\vect v$ (convection); diffusion is the transport that remains in a fluid at rest.

**Theorem 24.2 (Fick’s law).**

In a medium at rest, where the density is not uniform, a particle current appears, proportional and opposite to the [gradient](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#def-b2-maxwell-equations-operators) of the density:

$$
\vect{j}_N = -D\,\vect{\operatorname{grad}}\,n ,
$$

where the *diffusion coefficient* $D > 0$ (in $\mathrm{m}^{2}/\mathrm{s}$) depends on the diffusing species, on the medium and on the temperature. Particles go *down* the density [gradient](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#def-b2-maxwell-equations-operators), from the denser to the rarer regions, at a rate set by $D$.

**Proof.** Phenomenological (a law of experience, like Ohm’s): linear in the [gradient](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#def-b2-maxwell-equations-operators) for small [gradients](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#def-b2-maxwell-equations-operators), isotropic in an isotropic medium, and with the sign that experience imposes. The random-walk model of [Section 24.4](#sec-24-4) derives it, and $D$ with it, from the molecular motion. ∎

**Example 24.3 (Orders of magnitude of DDD).**

Gases: $D \sim 1 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$ (water vapour in air $2.5 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$, a perfume molecule $5 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$). Liquids: $D \sim 1 \times 10^{-9}\,\mathrm{m}^{2}/\mathrm{s}$ (oxygen in water $2 \times 10^{-9}\,\mathrm{m}^{2}/\mathrm{s}$, sugar $5 \times 10^{-10}\,\mathrm{m}^{2}/\mathrm{s}$, a protein $1 \times 10^{-10}\,\mathrm{m}^{2}/\mathrm{s}$). Solids: tiny and steeply increasing with temperature, $D = D_0\,\eu^{-E_{\text{a}}/k_BT}$ (boron in silicon: $1.5 \times 10^{-17}\,\mathrm{m}^{2}/\mathrm{s}$ at $1100\,{}^{\circ}\mathrm{C}$, unmeasurably small at room temperature — which is why a transistor, once made, keeps its doping profile for decades). Four orders of magnitude from gas to liquid, eight or more from liquid to solid.

## 24.2 The particle balance and the diffusion equation

**Theorem 24.4 (Local particle balance).**

If the particles are neither created nor destroyed,

$$
\frac{\partial n}{\partial t} + \operatorname{div}\vect{j}_N = 0 ;
$$

with a source creating $\sigma$ particles per unit volume and time (a chemical reaction, an absorption), $\partial_t n +
\operatorname{div}\vect{j}_N = \sigma$. In one dimension (density and current depending on $x$ only): $\partial_t n + \partial_x j_N = 0$.

**Proof.** Take the slab between $x$ and $x + \dd x$, of section $S$: it holds $n\,S\,\dd x$ particles; in $\dd t$, $j_N(x)S\,\dd t$ enter through the left face and $j_N(x + \dd x)S\,\dd t$ leave through the right one, so $\partial_t n\,S\,\dd x\,\dd t = -(j_N(x+\dd x) - j_N(x))S\,\dd t =
-\partial_x j_N\,\dd x\,S\,\dd t$. In three dimensions the same count on a small box gives the [divergence](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#def-b2-maxwell-equations-operators) (the flux out of a closed surface per unit volume, [Chapter 11](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#ch-b2-maxwell-equations)), or directly: for any fixed volume $V$, $\dd/\dd t \iiint_V n\,\dd\tau = -\iint_{\partial V}
\vect{j}_N\cdot\dd\vect S = -\iiint_V \operatorname{div}\vect{j}_N\,\dd\tau$ by the Ostrogradski theorem. ∎

![The one-dimensional balance: the slab between x and x + x gains what enters on the left and loses what leaves on the right; the difference is the change of n inside.](https://one-course.com/images/onecourse/chapters/physics-4/b2-particle-diffusion/fig-041f27e38723.svg)

*The one-dimensional balance: the slab between $x$ and $x + \dd
x$ gains what enters on the left and loses what leaves on the right; the difference is the change of $n$ inside.*

**Theorem 24.5 (Diffusion equation).**

Combining Fick’s law and the balance, for uniform $D$,

$$
\frac{\partial n}{\partial t} = D\,\Delta n + \sigma ,
\qquad\text{in one dimension}\quad
\frac{\partial n}{\partial t} = D\,\frac{\partial^2 n}{\partial x^2} + \sigma .
$$

This *diffusion equation* is linear (solutions superpose), first order in time and second order in space — not a [wave equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation): it has no propagation speed, but a characteristic relation between length and time,

$$
L \sim \sqrt{D t}, \qquad t \sim \frac{L^2}{D} :
$$

diffusing over twice the distance takes four times as long.

**Proof.** $\partial_t n = -\operatorname{div}(-D\,\vect{\operatorname{grad}}\,n) +
\sigma = D\operatorname{div}\vect{\operatorname{grad}}\,n + \sigma = D\Delta
n + \sigma$. For the scales, compare $n/t$ with $Dn/L^2$. ∎

**Remark 24.6 (Diffusion is irreversible).**

Change $t$ into $-t$ in the [wave equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation), $\partial_t^2 u = c^2
\partial_x^2 u$, and it is unchanged: a film of a wave run backwards is a possible wave. Do the same in the [diffusion equation](#thm-b2-particle-diffusion-equation) and the sign of the left side flips: $n(x,-t)$ is *not* a solution. A cloud that spreads is natural; a cloud that spontaneously gathers into a crystal is not — diffusion has an arrow of time, the arrow of the second law ([Chapter 25](https://one-course.com/books/physics/4/en/chapter/25-heat-conduction#ch-b2-heat-conduction) computes the entropy it creates). The equation also shows that diffusion smooths: where $n$ is locally a maximum ($\partial_x^2 n < 0$) it decreases, where it is a minimum it increases.

**Example 24.7 (How long does it take?).**

$t \sim L^2/D$. Oxygen across a cell, $L = 10\,\text{µ}\mathrm{m}$ in water: $10^{-10}/2 \times 10^{-9} = 0.05\,\mathrm{s}$ — diffusion feeds a cell with ease; across $1\,\mathrm{mm}$ of tissue: $500\,\mathrm{s}$ — too slow, which is why nothing living is thicker than a fraction of a millimetre without blood vessels. Sugar across a cup of unstirred tea, $L = 5\,\mathrm{cm}$: $2.5 \times 10^{-3}/5 \times 10^{-10} = 5 \times 10^6\,\mathrm{s}$, two months — stir. A perfume across a room by diffusion alone, $5\,\mathrm{m}$ in air: $25/10^{-5} = 3 \times 10^{6}\,\mathrm{s}$, a month; you smell it within a minute because the air moves: *convection* carries, diffusion only completes the last millimetres.

## 24.3 Characteristic solutions

**Proposition 24.8 (Stationary regime: the membrane).**

Between two reservoirs held at $n_1$ and $n_2$, across a membrane of thickness $e$ and area $S$, the stationary density is linear,

$$
n(x) = n_1 + (n_2 - n_1)\frac{x}{e}, \qquad
j_N = D\,\frac{n_1 - n_2}{e}, \qquad
\Phi_N = \frac{n_1 - n_2}{R_{\text{d}}}, \quad R_{\text{d}} = \frac{e}{DS}:
$$

the membrane has a *diffusive resistance* $e/DS$, exactly as a wire has $\ell/\gamma S$ — resistances in series add, and a thin membrane of small $D$ can still dominate.

**Proof.** Stationary, one-dimensional, no source: $\partial_x^2 n = 0$, so $n$ is affine; $j_N = -D\,\dd n/\dd x$ is uniform. ∎

**Proposition 24.9 (Stationary regime: spherical geometry).**

Around a sphere of radius $a$ whose surface is held at $n_0$, in a medium where $n \to 0$ far away, the stationary density is

$$
n(r) = n_0\,\frac{a}{r}, \qquad
\Phi_N = 4\pi D a\,n_0 :
$$

the total current released (or absorbed, with the signs reversed) grows with the *radius*, not the surface, of the sphere — the geometry of a diffusive sink is that of an electrostatic capacitance.

**Proof.** $\Delta n = \frac{1}{r^2}\frac{\dd}{\dd r}(r^2\,\dd n/\dd r) = 0$ gives $n = A + B/r$; the boundary conditions fix $A = 0$, $B = n_0 a$; then $\Phi_N = -4\pi r^2 D\,\dd n/\dd r = 4\pi D n_0 a$ at every $r$ (no accumulation in the stationary regime). ∎

![Stationary profiles. Left: across a membrane the density is linear and the current uniform. Right: around a sphere the density falls as 1/r; the total current 4π D a n_0 is proportional to the radius.](https://one-course.com/images/onecourse/chapters/physics-4/b2-particle-diffusion/fig-648ab3fd0fc2.svg)

*Stationary profiles. Left: across a membrane the density is linear and the current uniform. Right: around a sphere the density falls as $1/r$; the total current $4\pi D a n_0$ is proportional to the radius.*

**Proposition 24.10 (The spreading Gaussian).**

$N$ particles per unit area released at $t = 0$ on the plane $x = 0$ of an infinite medium spread as

$$
n(x,t) = \frac{N}{\sqrt{4\pi D t}}\,\exp\Big(-\frac{x^2}{4Dt}\Big) :
$$

a Gaussian of standard deviation $\sigma(t) = \sqrt{2Dt}$, of constant area $N$ (the particles are conserved) and decreasing peak $\propto
1/\sqrt t$. In three dimensions, $N$ particles released at a point spread as $n = N\,(4\pi Dt)^{-3/2}\exp(-r^2/4Dt)$, with $\langle r^2
\rangle = 6Dt$. A pulse of particles released at the *surface* of a half-space (which they cannot leave) gives twice the expression above for $x > 0$.

**Proof.** Substitute: with $u = x^2/4Dt$, $\partial_t n = n\,(-1/2t + u/t)$ and $D\,\partial_x^2 n = D\,n\,(-1/2Dt + x^2/4D^2t^2) = n\,(-1/2t + u/t)$; equal. The normalisation $\int n\,\dd x = N$ follows from $\int
\eu^{-x^2/4Dt}\dd x = \sqrt{4\pi Dt}$, and $\int x^2 n\,\dd x/N = 2Dt$. Uniqueness (given the initial pulse) is admitted. ∎

![The Gaussian solution at three times: the width grows as √ t, the peak falls as 1/√ t, the area (the number of particles) stays the same.](https://one-course.com/images/onecourse/chapters/physics-4/b2-particle-diffusion/fig-a02209e1d70a.svg)

*The Gaussian solution at three times: the width grows as $\sqrt t$, the peak falls as $1/\sqrt t$, the area (the number of particles) stays the same.*

**Proposition 24.11 (Constant surface concentration).**

A half-space $x > 0$ initially empty, whose surface is held at the density $n_0$ from $t = 0$ (a gas in contact with a solid that dissolves it), fills as

$$
n(x,t) = n_0\,\operatorname{erfc}\Big(\frac{x}{2\sqrt{Dt}}\Big),
\qquad
\operatorname{erfc}(u) = \frac{2}{\sqrt\pi}\int_u^{\infty}\eu^{-s^2}\dd s,
$$

the *complementary error function*, which falls from $1$ at $u =
0$ to $0.16$ at $u = 1$ and $0.005$ at $u = 2$: the penetration depth is again $\sim 2\sqrt{Dt}$, and the total number of particles absorbed per unit area is $2n_0\sqrt{Dt/\pi}$.

**Proof.** Look for $n = f(u)$ with $u = x/2\sqrt{Dt}$: the equation becomes $f'' + 2uf' = 0$, so $f' \propto \eu^{-u^2}$ and $f$ is an error function; the conditions $f(0) = n_0$, $f(\infty) = 0$ select $\operatorname{erfc}$. The absorbed number is $\int_0^\infty n\,\dd x =
n_0\,2\sqrt{Dt}\int_0^\infty\operatorname{erfc}(u)\dd u = 2n_0\sqrt{Dt/\pi}$. ∎

**Example 24.12 (Hardening steel).**

A steel part held at $900\,{}^{\circ}\mathrm{C}$ in a carbon-rich gas absorbs carbon at its surface; with $D = 5 \times 10^{-12}\,\mathrm{m}^{2}/\mathrm{s}$ for carbon in hot iron, four hours give $2\sqrt{Dt} = 0.5\,\mathrm{mm}$: a hard skin of half a millimetre on a tough core — the *case hardening* of gears and [bearings](https://one-course.com/books/physics/4/en/chapter/1-rigid-body-mechanics#def-b2-rigid-body-mechanics-pivot), timed with the erfc profile.

## 24.4 The microscopic picture: random walk

**Proposition 24.13 (Random walk and the diffusion coefficient).**

A particle that makes a step of length $\ell$ in a random direction every $\tau$ (a molecule between two collisions) has, after $N =
t/\tau$ steps, a mean square displacement

$$
\langle x^2\rangle = N\ell^2 = \frac{\ell^2}{\tau}\,t
\quad\text{(one dimension)}, \qquad
\langle r^2\rangle = 3\langle x^2\rangle \quad\text{(three dimensions)}.
$$

Comparing with the Gaussian solution, $\langle x^2\rangle = 2Dt$:

$$
D = \frac{\ell^2}{2\tau} \quad\text{(1D)}, \qquad
D = \frac{\ell^2}{6\tau} = \frac{\ell\,v^*}{6}\ \sim\ \frac{\ell\,v^*}{3}
\quad\text{(3D, with } v^* = \ell/\tau\text{; the kinetic theory gives } \tfrac13) .
$$

The walker’s distance grows as $\sqrt t$, not $t$: to go twice as far it needs four times as many steps.

**Proof.** $x = \sum_i \epsilon_i\ell$ with independent $\epsilon_i = \pm1$ (1D): $\langle x^2\rangle = \sum_{i,j}\langle\epsilon_i\epsilon_j\rangle\ell^2 =
N\ell^2$, the cross terms averaging to zero. The distribution of $x$ for large $N$ tends to a Gaussian (the central limit theorem, whose proof belongs to the probability course) — which is why the macroscopic law is the [diffusion equation](#thm-b2-particle-diffusion-equation); Fick’s law itself follows from counting the walkers crossing a plane from both sides: $j_N \approx
-\tfrac12\ell v^*\,\partial_x n$ (a more careful average gives $1/3$ in three dimensions). ∎

![A random walk of 400 steps of length : the walker has wandered only some 20 from its start — the √ N law that makes diffusion so slow over long distances and so fast over short ones.](https://one-course.com/images/onecourse/chapters/physics-4/b2-particle-diffusion/fig-d913d7707ee0.svg)

*A random walk of $400$ steps of length $\ell$: the walker has wandered only some $20\ell$ from its start — the $\sqrt N$ law that makes diffusion so slow over long distances and so fast over short ones.*

**Example 24.14 (From molecules to DDD).**

Air at room temperature: mean free path $\ell \approx 70\,\mathrm{nm}$, mean speed $v^* \approx 500\,\mathrm{m}/\mathrm{s}$: $D \approx \ell v^*/3 =
1.2 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$, as measured. To diffuse $1\,\mathrm{m}$ a molecule needs $N = (L/\ell)^2 = 2 \times 10^{14}$ collisions, i.e. $N\ell/v^*
\approx 3 \times 10^{4}\,\mathrm{s}$ — eight hours for a trip it would make in $2\,\mathrm{ms}$ if it flew straight. In a liquid, a sphere of radius $a$ buffeted by the molecules obeys the Stokes–Einstein relation $D =
k_BT/6\pi\eta a$ (Einstein 1905): for sugar, $a \approx 0.4\,\mathrm{nm}$ in water, $D = 5 \times 10^{-10}\,\mathrm{m}^{2}/\mathrm{s}$; for a $1\,\text{µ}\mathrm{m}$ grain, $4 \times 10^{-13}\,\mathrm{m}^{2}/\mathrm{s}$, a displacement of a micrometre per second — the Brownian motion that Perrin measured to count Avogadro’s number. In a solid an atom jumps to a neighbouring site only when a thermal fluctuation supplies the activation energy: $D = D_0\,\eu^{-E_{\text{a}}/k_BT}$, doubling every few tens of kelvins near $1000\,{}^{\circ}\mathrm{C}$.

**Method 24.15 (Diffusion estimates).**

(1) Identify $D$ (gas $10^{-5}$, liquid $10^{-9}$, solid Arrhenius). (2) Time or distance: $t \sim L^2/D$, $L \sim \sqrt{Dt}$ (with the exact $\sqrt{2Dt}$ for the Gaussian width, $2\sqrt{Dt}$ for the erfc depth). (3) Stationary: linear in a slab ($R_{\text{d}} = e/DS$), $1/r$ around a sphere ($\Phi = 4\pi D a\,\Delta n$), $\ln r$ around a cylinder. (4) Transient: Gaussian for a pulse, erfc for a held surface; superpose. (5) Ask whether convection does not dominate. (6) Microscopic check: $D \sim \ell v^*/3$.

## 24.5 Exercises

**Exercise 24.1 ★.**

Diffusion times $L^2/D$: a sugar cube at the bottom of a $5\,\mathrm{cm}$ cup ($D = 5 \times 10^{-10}\,\mathrm{m}^{2}/\mathrm{s}$); perfume across a $5\,\mathrm{m}$ room ($D =
5 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$); oxygen across $1\,\mathrm{mm}$ of tissue and across a $10\,\text{µ}\mathrm{m}$ cell ($D = 2 \times 10^{-9}\,\mathrm{m}^{2}/\mathrm{s}$). Which of these does diffusion actually do?

**Solution of Exercise 24.1.**

$L^2/D$: sugar $5 \times 10^6\,\mathrm{s}$ (two months); perfume the same; tissue $500\,\mathrm{s}$; cell $0.05\,\mathrm{s}$. Diffusion really does the last one (and, barely, the second to last); convection does the rest.

**Exercise 24.2 ★.**

A membrane $1\,\text{µ}\mathrm{m}$ thick and $1\,\mathrm{cm}^{2}$ in area, $D =
1 \times 10^{-11}\,\mathrm{m}^{2}/\mathrm{s}$ inside it, separates a solution at $1\,\mathrm{mol}/\mathrm{m}^{3}$ from pure water. [Current density](https://one-course.com/books/physics/4/en/chapter/10-charges-currents-and-conduction#def-b2-charges-currents-conduction-densities), molar flux and number of molecules per second; diffusive resistance; with two such membranes in series.

**Solution of Exercise 24.2.**

$j = D\Delta n/e = 1 \times 10^{-5}\,\mathrm{mol}/\mathrm{m}^{2}/\mathrm{s}$; $\times 1 \times 10^{-4}\,\mathrm{m}^{2}$: $1 \times 10^{-9}\,\mathrm{mol}/\mathrm{s}$, $6 \times 10^{14}$ molecules per second; $R_{\text{d}} = e/DS = 1 \times 10^{9}\,\mathrm{s}/\mathrm{m}^{3}$; two in series: half.

**Exercise 24.3 ★.**

A thin layer of dye ($N = 1 \times 10^{20}\,\mathrm{m}^{-2}$) is released in water, $D = 1 \times 10^{-9}\,\mathrm{m}^{2}/\mathrm{s}$. Width $\sqrt{2Dt}$ and peak density after $1\,\mathrm{s}$, $1\,\mathrm{h}$, $1\,\mathrm{day}$. When has the peak fallen to $1\%$ of its value at $1\,\mathrm{s}$?

**Solution of Exercise 24.3.**

$\sqrt{2Dt}$: $45\,\text{µ}\mathrm{m}$, $2.7\,\mathrm{mm}$, $13\,\mathrm{mm}$; peak $N/\sqrt{4\pi
Dt}$: $9 \times 10^{23}\,\mathrm{m}^{-3}$, $1.5 \times 10^{22}\,\mathrm{m}^{-3}$, $3 \times 10^{21}\,\mathrm{m}^{-3}$; $1\%$ after $10^4$ times longer: $1 \times 10^{4}\,\mathrm{s}$.

**Exercise 24.4 ★.**

Air molecules: $\ell = 70\,\mathrm{nm}$, $v^* = 500\,\mathrm{m}/\mathrm{s}$. Estimate $D$; number of collisions and time to diffuse $1\,\mathrm{m}$; compare with the straight flight. Same for a molecule in water ($\ell \approx
0.1\,\mathrm{nm}$, $v^* \approx 500\,\mathrm{m}/\mathrm{s}$): is the estimate right, and why not exactly?

**Solution of Exercise 24.4.**

$D \approx 1.2 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$; $N = (L/\ell)^2 = 2 \times 10^{14}$; $N\ell/v^*
= 3 \times 10^{4}\,\mathrm{s}$ against $2\,\mathrm{ms}$. Water: the estimate gives $2 \times 10^{-8}\,\mathrm{m}^{2}/\mathrm{s}$, ten times the measured values — in a liquid a molecule rattles in the cage of its neighbours and its successive steps are anti-correlated; the effective step is shorter.

**Exercise 24.5 ★★.**

*Oxygen in a tissue.* A slab of tissue of thickness $2a$ consumes oxygen at the uniform rate $q$ (per unit volume); its two faces are held at $n_0$. (a) Write the stationary equation with the sink and solve it. (b) Condition for oxygen to reach the centre. (c) $D = 2 \times 10^{-9}\,\mathrm{m}^{2}/\mathrm{s}$, $n_0 = 0.2\,\mathrm{mol}/\mathrm{m}^{3}$, $q = 0.01\,\mathrm{mol}/\mathrm{m}^{3}/\mathrm{s}$: maximum $a$. (d) Conclude on the spacing of capillaries.

**Solution of Exercise 24.5.**

(a) $Dn'' = q$: $n = n_0 - q(a^2 - x^2)/2D$. (b) $n(0) \ge 0$: $a \le
\sqrt{2Dn_0/q}$. (c) $0.28\,\mathrm{mm}$. (d) No cell can be farther than a few hundred micrometres from a capillary (in practice $50$–$100\,\text{µ}\mathrm{m}$ in active tissue).

**Exercise 24.6 ★★.**

*A dissolving grain.* A sugar sphere of radius $a = 1\,\mathrm{mm}$ in still water keeps its surface at the saturation density $n_{\text{s}}
= 5800\,\mathrm{mol}/\mathrm{m}^{3}$; far away the water is pure; $D =
5 \times 10^{-10}\,\mathrm{m}^{2}/\mathrm{s}$. (a) Stationary profile and total current. (b) Number of moles in the grain (density $1590\,\mathrm{kg}/\mathrm{m}^{3}$, molar mass $342\,\mathrm{g}/\mathrm{mol}$) and dissolution time (take the current constant). (c) Why is the true time longer, and why does stirring help so much? (d) Show that for a very long cylinder the stationary profile is logarithmic and the problem has no solution in an infinite medium.

**Solution of Exercise 24.6.**

(a) $n = n_{\text{s}}a/r$, $\Phi = 4\pi D a n_{\text{s}} = 3.6 \times 10^{-8}\,\mathrm{mol}/\mathrm{s}$. (b) $1.95 \times 10^{-5}\,\mathrm{mol}$; $540\,\mathrm{s}$. (c) The $1/r$ shell takes $\sim a^2/D \approx 2000\,\mathrm{s}$ to build and the grain shrinks; stirring replaces the shell of thickness $\sim a$ by a [boundary layer](https://one-course.com/books/physics/4/en/chapter/4-viscous-flows#prop-b2-viscous-flows-bl) $\delta \ll a$ and multiplies the flux by $a/\delta$. (d) $(rn')' = 0$: $n = A + B\ln r$, which cannot vanish at infinity: no stationary state — the cloud of a cylinder keeps growing (logarithmically).

**Exercise 24.7 ★★.**

*Conservation and irreversibility.* (a) From the equation, show that $\int n\,\dd x$ is constant and that $\dd\langle x^2\rangle/\dd t =
2D$ (integrate by parts, $n \to 0$ at infinity). (b) Show that $n(x,-t)$ does not satisfy the equation. (c) What becomes of the Gaussian solution for $t < 0$? (d) Show that $\int n^2\,\dd x$ can only decrease: diffusion flattens.

**Solution of Exercise 24.7.**

(a) $\dd/\dd t\int n = D[n']_{-\infty}^{\infty} = 0$; $\dd/\dd t\int x^2 n = D\int
x^2 n'' = 2D\int n = 2DN$. (b) $\partial_t[n(x,-t)] = -D\,\partial_x^2 n$: the sign flips. (c) For $t < 0$ the “width” $2Dt$ is negative: no such state — the pulse cannot be unspread. (d) $\dd/\dd t\int n^2 = 2D\int nn''
= -2D\int n'^2 \le 0$.

**Exercise 24.8 ★★.**

*Carburising.* Carbon in hot iron, $D = 5 \times 10^{-12}\,\mathrm{m}^{2}/\mathrm{s}$ at $900\,{}^{\circ}\mathrm{C}$; surface held at $n_0$. (a) Check that $n_0
\operatorname{erfc}(x/2\sqrt{Dt})$ solves the equation. (b) Depth at which $n = 0.1\,n_0$ after $4\,\mathrm{h}$ ($\operatorname{erfc}(1.16) = 0.1$). (c) Time for twice the depth. (d) At $950\,{}^{\circ}\mathrm{C}$, $D$ is $2.5$ times larger: time saved.

**Solution of Exercise 24.8.**

(a) [Proposition 24.11](#prop-b2-particle-diffusion-erfc). (b) $x = 2 \times 1.16\sqrt{Dt}
= 0.62\,\mathrm{mm}$. (c) $4\times$: $16\,\mathrm{h}$. (d) $4/2.5 = 1.6\,\mathrm{h}$.

**Exercise 24.9 ★★.**

*Stokes–Einstein.* $D = k_BT/6\pi\eta a$, water $\eta =
1 \times 10^{-3}\,\mathrm{Pa}\,\mathrm{s}$, $300\,\mathrm{K}$. (a) $D$ for $a = 1\,\text{µ}\mathrm{m}$ and the root-mean-square displacement in $1\,\mathrm{s}$, $1\,\mathrm{min}$. (b) $D$ for a protein, $a = 3\,\mathrm{nm}$. (c) Sugar has $D = 5 \times 10^{-10}\,\mathrm{m}^{2}/\mathrm{s}$: effective radius. (d) How does measuring $\langle x^2\rangle$ of a grain under a microscope give Avogadro’s number?

**Solution of Exercise 24.9.**

(a) $D = 2.2 \times 10^{-13}\,\mathrm{m}^{2}/\mathrm{s}$; $\sqrt{2Dt}$: $0.66\,\text{µ}\mathrm{m}$, $5\,\text{µ}\mathrm{m}$. (b) $7 \times 10^{-11}\,\mathrm{m}^{2}/\mathrm{s}$. (c) $a = k_BT/6\pi\eta D =
0.44\,\mathrm{nm}$. (d) $\langle x^2\rangle = 2Dt = RTt/3\pi\eta a N_A$: every quantity but $N_A$ is measured.

**Exercise 24.10 ★★★.**

*Time lag of a membrane.* A membrane of thickness $e$, initially empty, is exposed on one face to $n_1$ at $t = 0$, the other face kept at $0$. (a) What is the final stationary flux? (b) Argue that the flux on the far face rises over a time $\sim e^2/D$ (the exact lag is $e^2/6D$). (c) A drug patch with $e = 20\,\text{µ}\mathrm{m}$, $D =
1 \times 10^{-13}\,\mathrm{m}^{2}/\mathrm{s}$ through the skin’s outer layer: lag time. (d) Why is the lag useful to measure $D$ and the stationary flux to measure $D \times$ solubility?

**Solution of Exercise 24.10.**

(a) $Dn_1S/e$. (b) The particles need $\sim e^2/D$ to cross. (c) $e^2/6D = 670\,\mathrm{s}$, eleven minutes. (d) The lag gives $D$ alone; the stationary flux gives $D \times$ (solubility): two measurements, two unknowns.

**Exercise 24.11 ★★★.**

*The perfect absorber.* A sphere of radius $a$ absorbs every particle that touches it, in a medium at $n_\infty$ far away. (a) Stationary profile and total current captured. (b) A bacterium, $a =
1\,\text{µ}\mathrm{m}$, in a sugar solution at $n_\infty = 6 \times 10^{20}\,\mathrm{m}^{-3}$, $D = 5 \times 10^{-10}\,\mathrm{m}^{2}/\mathrm{s}$: molecules captured per second. (c) Its surface is covered by $N_{\text{r}}$ small absorbing receptors of radius $s$, the rest reflecting: each receptor alone captures $\approx
4Ds n_\infty$ (a disc); show that the whole cell captures nearly the maximum once $N_{\text{r}} s \gg a$, i.e. with a tiny fraction of its surface covered. (d) Comment: why cells can afford thousands of different receptors.

**Solution of Exercise 24.11.**

(a) $n = n_\infty(1 - a/r)$, $\Phi = 4\pi D a n_\infty$. (b) $3.8 \times
10^6$ per second. (c) Receptors act like conductances in parallel and then in series with the spherical shell: $\Phi \approx 4\pi Dan_\infty
\cdot N_{\text{r}}s/(N_{\text{r}}s + \pi a)$; half the maximum for $N_{\text{r}}s = \pi a$: with $s = 1\,\mathrm{nm}$, $N_{\text{r}} \approx
3000$, covering $N_{\text{r}}s^2/4a^2 \approx 10^{-3}$ of the surface. (d) Each kind of receptor needs a negligible area for near-maximal capture: a cell can watch thousands of substances at once.

**Exercise 24.12 ★★★.**

*Solving it on a grid.* Divide space into cells of size $\Delta x$ and time into steps $\Delta t$; write $n_i^{k+1} = n_i^k + \alpha\,(n_{i+1}^k
- 2n_i^k + n_{i-1}^k)$ with $\alpha = D\Delta t/\Delta x^2$. (a) Justify it from the equation. (b) Interpret it as a random walk when $\alpha =
1/2$. (c) Show that for $\alpha > 1/2$ a density alternating $+,-,+,-$ from cell to cell grows: instability. (d) For the boron profile of the problem below ($D = 1.5 \times 10^{-17}\,\mathrm{m}^{2}/\mathrm{s}$, $\Delta x = 10\,\mathrm{nm}$), the largest stable $\Delta t$ and the number of steps for one hour.

**Solution of Exercise 24.12.**

(a) Forward difference in $t$, centred second difference in $x$. (b) $\alpha = 1/2$: $n_i^{k+1} = (n_{i-1}^k + n_{i+1}^k)/2$ — every particle hops left or right with probability $1/2$. (c) For $n_i = (-1)^i$, $n^{k+1} =
(1 - 4\alpha)n^k$: $|1 - 4\alpha| > 1$ when $\alpha > 1/2$. (d) $\Delta t \le
\Delta x^2/2D = 3.3\,\mathrm{s}$; about $1100$ steps.

## 24.6 Problem: A doped wafer and a scented room

**Problem 24.1.**

Weekend problem — diffusion in a solid and in a gas

**Part I — Drive-in of boron in silicon.** Boron diffuses in silicon with $D = D_0\,\eu^{-E_{\text{a}}/k_BT}$, $D_0 =
7.6 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$, $E_{\text{a}} = 3.46\,\mathrm{eV}$; $k_B =
8.62 \times 10^{-5}\,\mathrm{eV}/\mathrm{K}$. A dose $Q = 1 \times 10^{18}\,\mathrm{m}^{-2}$ of boron has been deposited in a very thin layer at the surface of a wafer whose background doping is $n_B = 1 \times 10^{21}\,\mathrm{m}^{-3}$; it is then heated (“driven in”) at $1100\,{}^{\circ}\mathrm{C}$ for one hour. The surface reflects the boron (no escape).

1. Compute $D$ at $1100\,{}^{\circ}\mathrm{C}$ and at $1000\,{}^{\circ}\mathrm{C}$ . By what factor does it change over these $100\,\mathrm{K}$ ?
2. Why is the profile after the drive-in $n(x,t) = (Q/\sqrt{\pi  Dt})\exp(-x^2/4Dt)$ and not the Gaussian of [Proposition 24.10](#prop-b2-particle-diffusion-gaussian) ? Check that its integral over $x > 0$ is $Q$ .
3. Surface concentration after one hour.
4. The *junction depth* $x_j$ is where $n = n_B$ : compute it.
5. How does $x_j$ change if the drive-in lasts four hours? (Careful: the surface concentration changes too.)
6. Relative change of $D$ for a $10\,\mathrm{K}$ error in temperature; temperature control needed for $x_j$ to $1\,\%$ .
7. At room temperature, $D$ : estimate the time for the profile to move by one atomic spacing, and conclude.

**Part II — Predeposition.** The dose itself was put in at $950\,{}^{\circ}\mathrm{C}$ from a gas that holds the surface at the solubility limit $n_0 = 2 \times 10^{26}\,\mathrm{m}^{-3}$ for $30\,\mathrm{min}$.

8. $D$ at $950\,{}^{\circ}\mathrm{C}$ and $\sqrt{Dt}$ for $30\,\mathrm{min}$ .
9. Profile at the end of the predeposition; depth at which $n =  n_B$ ( $\operatorname{erfc}(3.2) \approx 6 \times 10^{-6}$ ).
10. Dose introduced; compare with the $Q$ of Part I.
11. Why does one predeposit at a lower temperature and drive in at a higher one?
12. Justify that during the drive-in the predeposited layer can be treated as infinitely thin.

**Part III — A perfume in a room.** A drop of perfume ($1\,\mathrm{mg}$, molar mass $150\,\mathrm{g}/\mathrm{mol}$) evaporates at once in the corner of a still room at $300\,\mathrm{K}$, $1\,\mathrm{bar}$. Take the perfume molecule’s collision diameter with air as $d =
0.5\,\mathrm{nm}$ and the [number density](#def-b2-particle-diffusion-flux) of air $n_{\text{a}} = P/k_BT$.

13. Number of molecules released; [number density](#def-b2-particle-diffusion-flux) of air.
14. Mean free path $\ell = 1/(\sqrt2\,\pi d^2 n_{\text{a}})$ of the perfume molecule, and its mean speed $v^* = \sqrt{8RT/\pi M}$ .
15. Estimate $D = \ell v^*/3$ .
16. Time to diffuse $5\,\mathrm{m}$ ; and if the air drifts at $0.1\,\mathrm{m}/\mathrm{s}$ ?
17. Write the three-dimensional Gaussian for the cloud (a corner: the walls reflect, multiply by $8$ ). At a point $1\,\mathrm{m}$ away, at what time is the concentration maximal?
18. Maximum [number density](#def-b2-particle-diffusion-flux) there; compare with a perception threshold of $1 \times 10^{13}\,\mathrm{m}^{-3}$ .
19. The cloud’s edge: at what distance is the density $10^{-6}$ of its central value after one day?
20. Number of collisions a perfume molecule suffers in a day.

**Part IV — The walker and the arrow of time.**

21. A walker makes $N$ steps $\pm\ell$ on a line: mean and mean square displacement; show that $\langle x^2\rangle = 2Dt$ identifies $D = \ell^2/2\tau$ .
22. Probability that the walker returns exactly to its start after $N = 2$ , $4$ , $6$ steps; comment on the trend.
23. Film a diffusing cloud and run the film backwards: what do you see, and which equation is violated?
24. Explain in one paragraph how the reversible motion of molecules produces the irreversible [diffusion equation](#thm-b2-particle-diffusion-equation) .
25. Summarise: the length scale, the time scale, and the one quantity that distinguishes a gas, a liquid and a hot solid.

**Solution of Problem 24.1.**

**1.** $k_BT = 0.118\,\mathrm{eV}$, $E_{\text{a}}/k_BT = 29.2$: $D =
1.6 \times 10^{-17}\,\mathrm{m}^{2}/\mathrm{s}$; at $1000\,{}^{\circ}\mathrm{C}$: $1.6 \times 10^{-18}\,\mathrm{m}^{2}/\mathrm{s}$ — a factor $10$.

**2.** The reflecting surface is handled by the mirror image of the pulse, doubling the amplitude on $x > 0$; $\int_0^\infty = (Q/\sqrt{\pi Dt})
\cdot \tfrac12\sqrt{4\pi Dt} = Q$.

**3.** $Dt = 5.8 \times 10^{-14}\,\mathrm{m}^{2}$; $\sqrt{\pi Dt} = 0.43\,\text{µ}\mathrm{m}$; $n_{\text{s}}
= 2.3 \times 10^{24}\,\mathrm{m}^{-3}$.

**4.** $\eu^{-x^2/4Dt} = 4.3 \times 10^{-4}$: $x_j^2 = 4Dt \times 7.75$, $x_j =
1.3\,\text{µ}\mathrm{m}$.

**5.** $n_{\text{s}}$ halves, the logarithm drops to $7.05$, $4Dt$ quadruples: $x_j = 2.6\,\text{µ}\mathrm{m}$ — less than doubled.

**6.** $\delta D/D = (E_{\text{a}}/k_BT)\,\delta T/T = 21\%$; $x_j \propto
\sqrt{Dt}$ roughly, so $1\%$ on $x_j$ needs $2\%$ on $D$: $1\,\mathrm{K}$.

**7.** $E_{\text{a}}/k_BT = 134$: $D \approx 10^{-62}\,\mathrm{m}^{2}/\mathrm{s}$; one spacing ($0.25\,\mathrm{nm}$) in $a^2/D \sim 10^{43}\,\mathrm{s}$: frozen.

**8.** $E_{\text{a}}/k_BT = 32.8$: $D = 4.3 \times 10^{-19}\,\mathrm{m}^{2}/\mathrm{s}$; $\sqrt{Dt} =
28\,\mathrm{nm}$.

**9.** $n_0\operatorname{erfc}(x/2\sqrt{Dt})$; $n_B/n_0 = 5 \times 10^{-6}$: $x \approx 6.4\sqrt{Dt} = 0.18\,\text{µ}\mathrm{m}$.

**10.** $2n_0\sqrt{Dt/\pi} = 6.3 \times 10^{18}\,\mathrm{m}^{-2}$: six times $Q$ — the dose grows as $\sqrt t$ and is set by the time.

**11.** Low temperature: small $D$, dose controlled by time, shallow; high temperature: the fixed dose is driven deep quickly.

**12.** $0.18\,\text{µ}\mathrm{m}$ against $1.3\,\text{µ}\mathrm{m}$: thin.

**13.** $4 \times 10^{18}$ molecules; $n_{\text{a}} = 2.4 \times 10^{25}\,\mathrm{m}^{-3}$.

**14.** $\ell = 37\,\mathrm{nm}$; $v^* = 206\,\mathrm{m}/\mathrm{s}$.

**15.** $D \approx 2.5 \times 10^{-6}\,\mathrm{m}^{2}/\mathrm{s}$.

**16.** $L^2/D = 10^7\,\mathrm{s}$, months; drifting: $50\,\mathrm{s}$.

**17.** $n = 8N(4\pi Dt)^{-3/2}\eu^{-r^2/4Dt}$; maximal at $t^* = r^2/6D =
6.7 \times 10^{4}\,\mathrm{s}$, eighteen hours.

**18.** $n_{\max} = 8N(4\pi Dt^*)^{-3/2}\eu^{-3/2} \approx 2 \times 10^{18}\,\mathrm{m}^{-3}$, far above threshold: smelt, eventually.

**19.** $r^2 = 4Dt\ln 10^6 = 12\,\mathrm{m}^{2}$: $3.5\,\mathrm{m}$.

**20.** $v^*t/\ell \approx 5 \times 10^{14}$.

**21.** $\langle x\rangle = 0$, $\langle x^2\rangle = N\ell^2 = \ell^2 t/\tau$; $D = \ell^2/2\tau$.

**22.** $\binom{N}{N/2}/2^N$: $1/2$, $3/8$, $5/16$ — decreasing (as $1/\sqrt{\pi N/2}$): the walker wanders off as $\sqrt N$.

**23.** The cloud gathers itself into a point — never seen; it violates the [diffusion equation](#thm-b2-particle-diffusion-equation), not the laws of mechanics.

**24.** Each collision is reversible, but the initial state (all particles together) is exceptional: almost every microscopic history from it spreads, and the reverse needs a conspiracy of all the velocities. The random-walk average keeps only what is typical and discards that information; irreversibility is statistical.

**25.** $L \sim \sqrt{Dt}$, $t \sim L^2/D$; $D \sim \ell v^*/3$, i.e. the step between collisions: $10^{-5}$, $10^{-9}$, $10^{-17}\,\mathrm{m}^{2}/\mathrm{s}$.