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Diffusion and Brownian motion

Learning objectives

  • Outcome

    State definitions of diffusion and Brownian motion and give one observable example of each.

  • Outcome

    Explain Brownian motion as the result of unbalanced impacts of surrounding molecules on a suspended particle.

  • Outcome

    Predict how temperature, particle size, and particle mass affect the rate of diffusion and the amplitude of Brownian jitter.

  • Outcome

    Estimate the diffusion time over a distance LL using t∼L2/(2D)t \sim L^2/(2D).

Understand the idea

Diffusion and Brownian Motion: How Particle Chaos Creates Order

Text

Particles of a gas, liquid, or solution move continuously and chaotically; their mean kinetic energy increases with temperature, for example ⟨Ek⟩=32kBT\langle E_k \rangle = \frac{3}{2}k_B T for an ideal gas.

Diffusion is the spontaneous transport of a substance from a region of higher concentration to a region of lower concentration: random walks carry particles out of a dense region more often than back into it. Brownian motion is the jitter and displacement of a large suspended particle (for example, a dust grain in water) seen under a microscope, caused by unbalanced impacts of many molecules of the medium.

Both phenomena are direct consequences of the molecular-kinetic model: visible order (equalized concentration, steady chaotic jitter) is born from disordered collisions. The diffusion coefficient DD increases with temperature and decreases as particle size and the viscosity of the medium increase. For a random walk the mean square displacement grows linearly with time, ⟨x2⟩=2Dt\langle x^2 \rangle = 2Dt, so diffusion over large distances is very slow, t∼L2/Dt \sim L^2/D.

Worked examples

Direct

Problem

A drop of ink is placed in still water at 20 ∘C20\ ^\circ\text{C}. After some time the water is uniformly tinted. Explain the spreading of the ink in terms of particle motion, and state what kind of motion a single ink particle undergoes.

Solution

Water molecules and ink particles move and collide constantly. In the region of the drop the concentration of ink particles is higher, so random walks carry particles out of this region more often than back into it: a net flow from high to low concentration appears — diffusion. No external stirring is required. A single ink particle undergoes Brownian motion: it jitters and moves along a jagged path because the impacts of water molecules on it do not balance.

Application

Problem

The diffusion coefficient of oxygen in water at 20 ∘C20\ ^\circ\text{C} is about D=2.1×10−9 m2/sD = 2.1\times10^{-9}\ \text{m}^2/\text{s}. Estimate the time for oxygen molecules to diffuse a mean distance L=1 mmL = 1\ \text{mm}, using t∼L2/(2D)t \sim L^2/(2D).

Solution

Convert the distance to metres: L=1 mm=1×10−3 mL = 1\ \text{mm} = 1\times10^{-3}\ \text{m}. Then

L2=(1×10−3)2=1×10−6 m2,L^2 = (1\times10^{-3})^2 = 1\times10^{-6}\ \text{m}^2,

t∼L22D=1×10−62×2.1×10−9≈2.4×102 s≈4 min.t \sim \frac{L^2}{2D} = \frac{1\times10^{-6}}{2\times2.1\times10^{-9}} \approx 2.4\times10^{2}\ \text{s} \approx 4\ \text{min}.

This is an order-of-magnitude estimate: the mean square displacement grows as ⟨x2⟩=2Dt\langle x^2 \rangle = 2Dt, hence t∼L2/Dt \sim L^2/D. In living tissue oxygen transport is speeded up by convection and blood flow, so the real time can be shorter.

Transfer

Problem

A pollen grain of radius rr is suspended in water. Two samples are prepared: one at 20 ∘C20\ ^\circ\text{C} and one at 40 ∘C40\ ^\circ\text{C}. (a) Predict how the observed Brownian jitter differs between the samples. (b) A second grain of twice the mass is placed in the warm sample; explain whether its jitter is larger or smaller than that of the first grain at the same temperature.

Solution

(a) On heating, the mean speed of water molecules and the collision rate increase, so the unbalanced impulses received by the particle become more frequent and stronger. The particle diffusion coefficient D=kBT/(6πηr)D = k_B T/(6\pi \eta r) is proportional to TT, so at 40 ∘C40\ ^\circ\text{C} the jitter and displacement over the same time are noticeably larger than at 20 ∘C20\ ^\circ\text{C} (all else being equal).

(b) From the Maxwell–Boltzmann distribution for a suspended particle, 12m⟨v2⟩=32kBT\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}k_B T, hence vrms=3kBT/mv_{\text{rms}} = \sqrt{3k_B T/m}. A grain of twice the mass at the same temperature has a smaller root-mean-square speed, and its diffusion coefficient D=kBT/(6πηr)D = k_B T/(6\pi \eta r) is also smaller if the mass grows with size (radius). Therefore the jitter of the more massive grain will be smaller.

Practice

Text

Observation (household, safe). Pour water into two clean glasses: one at room temperature, one warm from the tap (not boiling). Add one drop of food colouring to each and, without stirring, record every 30 s the time until the colour is uniform; sketch the colour front. Compare the times and explain the difference by temperature.

Simulation (for the molecular scale). Open an interactive simulation (for example, PhET “Diffusion” or “Gas Properties”) and track a single particle’s path: check that it is jagged and random, and that the concentration equalizes by itself. Do not drink the coloured water; do not heat the water above warm tap water.

Visual reference

Purpose

Show how disordered microscopic walks produce a directed macroscopic flow in diffusion, and how unbalanced molecular impacts make a large particle jitter.

Representation

Two-panel sketch.

(a) On the left, a region of high concentration of small circles (molecules), on the right a region of low concentration; arrows of random walks point in all directions, but the net flow arrow points to the right, and below the sketch a graph of concentration c(x)c(x) decreases from left to right and flattens with time.

(b) A large circle (the Brownian particle) is surrounded by small molecules; impact arrows from different sides have unequal lengths, and their resultant F⃗\vec F is shown separately; below, a jagged particle trajectory (random walk) is drawn in a plane.

Real-world examples

Sugar dissolving in tea

Scenario

A 4 g sugar cube is dropped into 200 mL of tea at 80 °C. The sweetness is noticeable after about a minute even without stirring, while in the same volume of iced tea at 5 °C it takes minutes. Under a magnifier, the edge of a sugar grain blurs into the water long before the cube disappears.

Connection

Sugar molecules leave the crystal and cross the thin unstirred water layer about 0.1 mm0.1\ \text{mm} thick next to its surface by diffusion alone. With D≈5⋅10−10 m2/sD \approx 5\cdot10^{-10}\ \text{m}^2/\text{s} for sucrose in water, the random-walk relation ⟨x2⟩=2Dt\langle x^2\rangle = 2Dt gives t≈x2/(2D)=(10−4)2/(2⋅5⋅10−10)≈10 st \approx x^2/(2D) = (10^{-4})^2/(2\cdot5\cdot10^{-10}) \approx 10\ \text{s} — exactly the timescale on which the taste appears. Heating the tea raises the mean kinetic energy of the water molecules and lowers viscosity, so DD grows and the same layer is crossed faster. Mixing the remaining 200 mL in the cup is done by convection when you stir: diffusion alone over 5 cm would need hours.

Why a phone battery charges faster when warm

Scenario

A lithium-ion phone battery stores lithium in graphite particles about 2 μm2\ \mu\text{m} in radius. Charging pushes lithium ions into those particles; a full charge takes about 1 hour at 20 °C and about 40 minutes at 35 °C, while below 0 °C the phone may refuse to fast-charge at all.

Connection

Inside a graphite particle there is no stirring: each lithium ion moves by random hops between sites, so the filling time of a particle is set by the random-walk relation t≈r2/(2D)t \approx r^2/(2D). With r=2 μmr = 2\ \mu\text{m} and D≈10−14 m2/sD \approx 10^{-14}\ \text{m}^2/\text{s} at room temperature, t≈(2⋅10−6)2/(2⋅10−14)≈200 st \approx (2\cdot10^{-6})^2/(2\cdot10^{-14}) \approx 200\ \text{s} — one of the steps that limits the charging current. Because DD grows with temperature, warming the cell speeds up the hops; the same relation explains why cold slows charging and why manufacturers limit the current.

Road salt attacking a bridge

Scenario

Steel reinforcement in a road bridge lies about 50 mm below the concrete surface. In winter, salt from the road dissolves and forms chloride ions at the surface. With a chloride diffusion coefficient of about 10−12 m2/s10^{-12}\ \text{m}^2/\text{s} in sound concrete, the ions need roughly 40 years to reach the steel in noticeable concentration — which is why bridges are designed for a 50–100 year service life and are inspected for corrosion.

Connection

This is diffusion in a solid: chloride ions do not flow as a stream but perform a random walk through the pore water between cement grains. The penetration depth follows from ⟨x2⟩=2Dt\langle x^2\rangle = 2Dt, so t=x2/(2D)=(0.05)2/(2⋅10−12)≈1.25⋅109 s≈40 yearst = x^2/(2D) = (0.05)^2/(2\cdot10^{-12}) \approx 1.25\cdot10^9\ \text{s} \approx 40\ \text{years}. Because depth grows only as t\sqrt{t}, doubling the service life needs four times the time — thick concrete cover or waterproofing membranes, rather than a slightly thicker surface layer, are what actually delay corrosion.