University Physics — Year 2 · Bachelor Year 2

24Particle 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, 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.

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 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, 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 for small gradients, isotropic in an isotropic medium, and with the sign that experience imposes. The random-walk model of Section 24.4 derives it, and $D$ with it, from the molecular motion. ∎

Example 24.3 (Orders of magnitude of $D$ )

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 (the flux out of a closed surface per unit volume, Chapter 11), 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. — 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: 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, $\partial_t^2 u = c^2 \partial_x^2 u$ , and it is unchanged: a film of a wave run backwards is a possible wave. Do the same in the diffusion equation 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 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. — 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. — 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, 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; 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. — 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 $D$ )

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

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, molar flux and number of molecules per second; diffusive resistance; with two such membranes in series.

Solution

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

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

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

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

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 $\delta \ll a$ 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

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

Solution of Exercise 24.8.

(a) Proposition 24.11. (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

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

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

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

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).

Compute $D$ at $1100\,{}^{\circ}\mathrm{C}$ and at $1000\,{}^{\circ}\mathrm{C}$ . By what factor does it change over these $100\,\mathrm{K}$ ?
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? Check that its integral over $x > 0$ is $Q$ .
Surface concentration after one hour.
The junction depth $x_j$ is where $n = n_B$ : compute it.
How does $x_j$ change if the drive-in lasts four hours? (Careful: the surface concentration changes too.)
Relative change of $D$ for a $10\,\mathrm{K}$ error in temperature; temperature control needed for $x_j$ to $1\,\%$ .
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 $30\,\mathrm{min}$ .

$D$ at $950\,{}^{\circ}\mathrm{C}$ and $\sqrt{Dt}$ for $30\,\mathrm{min}$ .
Profile at the end of the predeposition; depth at which $n = n_B$ ( $\operatorname{erfc}(3.2) \approx 6 \times 10^{-6}$ ).
Dose introduced; compare with the $Q$ of Part I.
Why does one predeposit at a lower temperature and drive in at a higher one?
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 of air $n_{\text{a}} = P/k_BT$ .

Number of molecules released; number density of air.
Mean free path $\ell = 1/(\sqrt2\,\pi d^2 n_{\text{a}})$ of the perfume molecule, and its mean speed $v^* = \sqrt{8RT/\pi M}$ .
Estimate $D = \ell v^*/3$ .
Time to diffuse $5\,\mathrm{m}$ ; and if the air drifts at $0.1\,\mathrm{m}/\mathrm{s}$ ?
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?
Maximum number density there; compare with a perception threshold of $1 \times 10^{13}\,\mathrm{m}^{-3}$ .
The cloud’s edge: at what distance is the density $10^{-6}$ of its central value after one day?
Number of collisions a perfume molecule suffers in a day.

Part IV — The walker and the arrow of time.

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

Solution

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, 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: $10^{-5}$ , $10^{-9}$ , $10^{-17}\,\mathrm{m}^{2}/\mathrm{s}$ .

Terms defined in this chapter

Density and particle current

See all 393 terms in the glossary