Describe and explain Brownian motion in terms of random molecular bombardment

Cambridge IGCSE Physics 0625 – Particle Model: Brownian Motion

2.1.2 Brownian Motion – What the syllabus expects

• Brownian motion is observable evidence for the kinetic particle model of matter.
• Fluids (liquids and gases) consist of molecules moving randomly, colliding with each other and with any suspended particle.
• The incessant, random molecular impacts cause the erratic motion of tiny suspended particles.

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What you will learn

• Definition of Brownian motion (core).
• Explanation in terms of random molecular bombardment (core).
• Key factors that affect the motion – temperature, particle size, viscosity (core).
• Simple classroom experiment that demonstrates the phenomenon (core).
• Experimental evidence that supports the particle model (core).
• Common misconceptions (core).
• Core‑only summary – what you must be able to write in the exam.
• Extension material for higher‑ability work – diffusion, Stokes‑Einstein relation and a worked example.

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Definition (Core)

Brownian motion is the erratic, random movement of very small particles (typically a few µm in size) when they are suspended in a fluid (liquid or gas). The motion can be seen with a microscope and provides direct, macroscopic evidence for the kinetic particle model of matter.

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Explanation in terms of random molecular bombardment (Core)

According to the particle model:

• A fluid is made up of a huge number of molecules that are in constant, random motion.
• These molecules collide with each other and with any solid particle that is present.

When a suspended particle is in the fluid it is continually struck by surrounding molecules. The impacts are:

• Random in direction – no preferred side of the particle is hit more often.
• Unequal in magnitude – molecular speeds vary, so each collision delivers a different impulse.
• Very frequent – many billions of collisions occur each second.

The resultant force on the particle therefore fluctuates rapidly, giving the particle a series of tiny accelerations. The net effect is a jittery, apparently “random’’ path – the observed Brownian motion.

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Key factors that influence the motion (Core)

1. Temperature: Higher temperature → higher average molecular speed (\(\langle v\rangle\propto\sqrt{T}\)) → more energetic collisions → more vigorous Brownian motion.
2. Particle size (or mass): For a given collision force, a smaller particle experiences a larger acceleration, so the motion is more noticeable for smaller particles.
3. Fluid viscosity: In a more viscous fluid the collisions are damped more quickly, reducing the apparent displacement.

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Simple classroom experiment (Core)

Objective: Observe Brownian motion and see the effect of temperature.

1. Fill a clean glass slide with a thin layer of water.
2. Add a few drops of milk (or a pinch of finely ground pepper). The tiny fat droplets / pepper particles act as the suspended particles.
3. Cover with a cover slip to avoid convection currents.
4. Place the slide on a low‑power microscope (×40–×100). Focus on a region where individual particles can be seen.
5. Record the motion for 30 s using a smartphone camera or by sketching the path.
6. Warm the slide gently (e.g., by holding a warm finger on the edge) and repeat the observation. The particles should move more vigorously.

This activity directly satisfies the syllabus requirement to “describe an experiment that demonstrates Brownian motion” and reinforces the temperature‑dependence discussed above.

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Experimental evidence supporting the particle model (Core)

Observation	Interpretation in terms of the particle model
Microscopic particles suspended in a still liquid move erratically.	Continuous, random impacts from moving fluid molecules.
Motion becomes more vigorous when the temperature is raised.	Higher kinetic energy → faster, more forceful molecular collisions.
Smaller particles show larger, more visible displacements.	Each collision imparts a greater acceleration to a particle of lower mass.
Brownian motion is observed even when the fluid appears perfectly still.	Driven by microscopic molecular motion, not by macroscopic convection currents.

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Common misconceptions (Core)

• It is not caused by convection currents; it occurs even in a still fluid.
• The suspended particle does not have its own source of energy – the motion is entirely due to external molecular impacts.
• Brownian motion is not limited to particles visible with the naked eye; the same principle operates at the atomic and molecular scale.

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Core summary – what you must be able to write in the exam

• Brownian motion = observable evidence for the kinetic particle model.
• It arises from random, incessant molecular bombardment of suspended particles.
• Increasing temperature → more vigorous motion; decreasing particle size → more noticeable motion; higher viscosity → less motion.
• The motion is completely random – there is no preferred direction.

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Extension (Optional – for higher‑ability work)

Relation to diffusion – mean‑square displacement

In one dimension the average of the square of the displacement after a time \(t\) is

\[ \langle x^{2}\rangle = 2Dt \]

where \(D\) is the diffusion coefficient (units m² s⁻¹).

Stokes‑Einstein relation

\[ D = \frac{k_{\mathrm B}T}{6\pi\eta r} \]

• \(k_{\mathrm B}=1.38\times10^{-23}\,\text{J K}^{-1}\) – Boltzmann’s constant
• \(T\) – absolute temperature (K)
• \(\eta\) – dynamic viscosity of the fluid (Pa s)
• \(r\) – radius of the spherical particle (m)

Worked example (extension)

Problem: A spherical particle of radius \(r = 0.5\,\mu\text{m}\) is suspended in water at \(20^{\circ}\text{C}\) (\(T = 293\text{ K}\)). Water has a viscosity \(\eta = 1.0\times10^{-3}\,\text{Pa·s}\). Calculate the mean‑square displacement \(\langle x^{2}\rangle\) after \(t = 10\) s.

Solution:

1. Calculate the diffusion coefficient using the Stokes‑Einstein equation: \[ D = \frac{(1.38\times10^{-23})(293)}{6\pi(1.0\times10^{-3})(0.5\times10^{-6})} \approx 4.3\times10^{-13}\,\text{m}^{2}\text{s}^{-1}. \]
2. Insert \(D\) into the mean‑square displacement formula: \[ \langle x^{2}\rangle = 2Dt = 2(4.3\times10^{-13})(10) \approx 8.6\times10^{-12}\,\text{m}^{2}. \]
3. Take the square root to obtain a typical linear displacement: \[ \sqrt{\langle x^{2}\rangle}\approx 9.3\times10^{-6}\,\text{m}=9.3\,\mu\text{m}. \]

The particle is expected to wander roughly 9 µm from its starting point after 10 s – a distance easily observable under a microscope.

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Suggested diagram (for classroom use)