The Brownian Motion Formula describes the random, erratic movement of microscopic particles suspended in a fluid medium, and the key expression for the mean square displacement is given by \ ( \langle x^2 \rangle = 2Dt \), where D is the diffusion constant and t is time. This concept is covered in NCERT Physics and Chemistry for Class 11 and Class 12, and it appears in JEE Main, JEE Advanced, and NEET in the context of kinetic theory, diffusion, and statistical mechanics. This article covers the complete Brownian Motion Formula, its derivation, a full formula sheet, three solved examples at progressive difficulty, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Brownian Motion Formulas at a Glance
Quick reference for the most important Brownian motion and diffusion formulas.
- Mean Square Displacement (1D): \( \langle x^2 \rangle = 2Dt \)
- Mean Square Displacement (3D): \( \langle r^2 \rangle = 6Dt \)
- Einstein–Stokes Diffusion Constant: \( D = \frac{k_B T}{6 \pi \eta r} \)
- Einstein Relation: \( D = \mu k_B T \)
- Root Mean Square Displacement: \( x_{rms} = \sqrt{2Dt} \)
- Diffusion Flux (Fick’s First Law): \( J = -D \frac{dC}{dx} \)
- Boltzmann Factor: \( k_B = 1.38 \times 10^{-23} \) J/K
What is Brownian Motion Formula?
The Brownian Motion Formula quantifies the random thermal motion of small particles immersed in a liquid or gas. In 1827, botanist Robert Brown observed pollen grains suspended in water moving in an irregular, unpredictable path. This phenomenon remained unexplained for nearly 80 years until Albert Einstein provided a mathematical framework in 1905.
Einstein showed that the random collisions of fluid molecules with a suspended particle cause it to move erratically. The displacement of the particle over time follows a statistical pattern. The mean square displacement (MSD) is the primary measurable quantity. It grows linearly with time, which is the hallmark of diffusive motion.
In NCERT Physics Class 11 (Chapter 13: Kinetic Theory) and Class 12 (Chapter 9: Ray Optics and related thermodynamics contexts), Brownian motion is discussed as direct evidence for the kinetic theory of matter. The Brownian Motion Formula is also central to Class 11 Chemistry (Chapter 5: States of Matter). Understanding this formula builds a strong foundation for thermodynamics and statistical mechanics at the JEE and NEET level.
Brownian Motion Formula — Expression and Variables
The primary Brownian Motion Formula for mean square displacement in one dimension is:
\[ \langle x^2 \rangle = 2Dt \]
For three-dimensional motion, the formula extends to:
\[ \langle r^2 \rangle = 6Dt \]
The diffusion constant D is calculated using the Einstein–Stokes equation:
\[ D = \frac{k_B T}{6 \pi \eta r} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \langle x^2 \rangle \) | Mean square displacement (1D) | m² |
| \( \langle r^2 \rangle \) | Mean square displacement (3D) | m² |
| D | Diffusion constant | m²/s |
| t | Time elapsed | s (seconds) |
| \( k_B \) | Boltzmann constant | J/K (= 1.38 × 10²³ J/K) |
| T | Absolute temperature | K (Kelvin) |
| \( \eta \) | Dynamic viscosity of the fluid | Pa·s (N·s/m²) |
| r | Radius of the suspended particle | m (metres) |
| \( x_{rms} \) | Root mean square displacement | m |
Derivation of the Brownian Motion Formula
Einstein derived the mean square displacement formula using the theory of random walks combined with thermodynamic principles. Here is a step-by-step outline:
Step 1: Consider a particle undergoing random collisions in a fluid. Each collision imparts a small, random impulse to the particle.
Step 2: Over a large number of collisions in time t, the net displacement x is the sum of many small random steps. By the central limit theorem, the distribution of x is Gaussian.
Step 3: Einstein used the diffusion equation \( \frac{\partial P}{\partial t} = D \frac{\partial^2 P}{\partial x^2} \), where P(x,t) is the probability density of finding the particle at position x at time t.
Step 4: Solving this equation with an initial delta-function condition gives a Gaussian solution. Computing \( \langle x^2 \rangle = \int x^2 P(x,t)\,dx \) yields \( \langle x^2 \rangle = 2Dt \).
Step 5: Einstein then connected D to measurable fluid properties using Stokes’ drag law, giving \( D = \frac{k_B T}{6\pi\eta r} \). This is the Einstein–Stokes relation.
Complete Physics Formula Sheet: Brownian Motion and Diffusion
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Mean Square Displacement (1D) | \( \langle x^2 \rangle = 2Dt \) | D = diffusion constant, t = time | m² | Class 11, Ch 13 |
| Mean Square Displacement (3D) | \( \langle r^2 \rangle = 6Dt \) | D = diffusion constant, t = time | m² | Class 11, Ch 13 |
| Root Mean Square Displacement | \( x_{rms} = \sqrt{2Dt} \) | D = diffusion constant, t = time | m | Class 11, Ch 13 |
| Einstein–Stokes Diffusion Constant | \( D = \frac{k_B T}{6\pi\eta r} \) | k⊂B; = Boltzmann const, T = temp, η = viscosity, r = radius | m²/s | Class 11, Ch 13 |
| Einstein Relation (Mobility) | \( D = \mu k_B T \) | μ = mobility of particle, k⊂B; = Boltzmann const, T = temperature | m²/s | Class 11, Ch 13 |
| Fick’s First Law of Diffusion | \( J = -D \frac{dC}{dx} \) | J = diffusion flux, C = concentration, x = position | mol/(m²·s) | Class 11, Ch 5 (Chem) |
| Fick’s Second Law of Diffusion | \( \frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2} \) | C = concentration, t = time, x = position | mol/m³ | Class 11, Ch 5 (Chem) |
| Average Kinetic Energy per Molecule | \( \langle KE \rangle = \frac{3}{2} k_B T \) | k⊂B; = Boltzmann constant, T = absolute temperature | J | Class 11, Ch 13 |
| Stokes’ Drag Force | \( F_{drag} = 6\pi\eta r v \) | η = viscosity, r = radius, v = velocity | N | Class 11, Ch 10 |
| Boltzmann Constant | \( k_B = \frac{R}{N_A} \) | R = gas constant (8.314 J/mol·K), N⊂A; = Avogadro number | J/K | Class 11, Ch 13 |
Brownian Motion Formula — Solved Examples
Example 1 (Class 11-12 Level)
Problem: A pollen grain undergoes Brownian motion in water. The diffusion constant of the grain is \( D = 5 \times 10^{-13} \) m²/s. Calculate the root mean square displacement of the grain after 10 seconds.
Given:
- Diffusion constant: \( D = 5 \times 10^{-13} \) m²/s
- Time: \( t = 10 \) s
Step 1: Write the formula for root mean square displacement:
\[ x_{rms} = \sqrt{2Dt} \]
Step 2: Substitute the given values:
\[ x_{rms} = \sqrt{2 \times 5 \times 10^{-13} \times 10} \]
Step 3: Simplify inside the square root:
\[ x_{rms} = \sqrt{10^{-11}} = \sqrt{10^{-11}} \approx 3.16 \times 10^{-6} \text{ m} \]
Answer
The root mean square displacement of the pollen grain after 10 seconds is approximately \( 3.16 \times 10^{-6} \) m, or about 3.16 micrometres (μm).
Example 2 (Class 12 / Advanced Level)
Problem: A spherical nanoparticle of radius \( r = 50 \) nm is suspended in water at temperature \( T = 300 \) K. The dynamic viscosity of water is \( \eta = 1 \times 10^{-3} \) Pa·s. Calculate (a) the diffusion constant D and (b) the mean square displacement after \( t = 1 \) s.
Given:
- \( r = 50 \) nm \( = 50 \times 10^{-9} \) m \( = 5 \times 10^{-8} \) m
- \( T = 300 \) K
- \( \eta = 1 \times 10^{-3} \) Pa·s
- \( k_B = 1.38 \times 10^{-23} \) J/K
- \( t = 1 \) s
Step 1: Apply the Einstein–Stokes formula for the diffusion constant:
\[ D = \frac{k_B T}{6\pi\eta r} \]
Step 2: Substitute all values:
\[ D = \frac{1.38 \times 10^{-23} \times 300}{6 \times 3.1416 \times 1 \times 10^{-3} \times 5 \times 10^{-8}} \]
Step 3: Calculate the numerator: \( 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21} \) J
Step 4: Calculate the denominator: \( 6 \times 3.1416 \times 10^{-3} \times 5 \times 10^{-8} = 9.42 \times 10^{-10} \) N·s/m
Step 5: Divide to get D:
\[ D = \frac{4.14 \times 10^{-21}}{9.42 \times 10^{-10}} \approx 4.39 \times 10^{-12} \text{ m}^2/\text{s} \]
Step 6: Calculate mean square displacement for t = 1 s:
\[ \langle x^2 \rangle = 2Dt = 2 \times 4.39 \times 10^{-12} \times 1 = 8.78 \times 10^{-12} \text{ m}^2 \]
Answer
(a) The diffusion constant is \( D \approx 4.39 \times 10^{-12} \) m²/s. (b) The mean square displacement after 1 second is \( \langle x^2 \rangle \approx 8.78 \times 10^{-12} \) m².
Example 3 (JEE/NEET Level)
Problem: Two particles A and B undergo Brownian motion in the same fluid at the same temperature. Particle A has radius r and particle B has radius 2r. After time t, the mean square displacement of particle A is \( \langle x_A^2 \rangle \). Find the ratio \( \langle x_A^2 \rangle / \langle x_B^2 \rangle \). Also determine which particle diffuses faster and explain why.
Given:
- Radius of A: r
- Radius of B: 2r
- Same fluid, same temperature T, same time t
Step 1: Write the diffusion constant for each particle using the Einstein–Stokes formula:
\[ D_A = \frac{k_B T}{6\pi\eta r}, \quad D_B = \frac{k_B T}{6\pi\eta (2r)} = \frac{k_B T}{12\pi\eta r} \]
Step 2: Observe that \( D_A = 2D_B \). A smaller particle has a larger diffusion constant.
Step 3: Write the mean square displacement for each:
\[ \langle x_A^2 \rangle = 2D_A t, \quad \langle x_B^2 \rangle = 2D_B t \]
Step 4: Form the ratio:
\[ \frac{\langle x_A^2 \rangle}{\langle x_B^2 \rangle} = \frac{2D_A t}{2D_B t} = \frac{D_A}{D_B} = \frac{k_B T / (6\pi\eta r)}{k_B T / (12\pi\eta r)} = 2 \]
Step 5: Interpret the result. Since \( \langle x_A^2 \rangle = 2 \langle x_B^2 \rangle \), particle A (smaller radius) diffuses farther in the same time. A smaller particle experiences less viscous drag. Therefore it moves more freely through the fluid.
Answer
The ratio \( \langle x_A^2 \rangle / \langle x_B^2 \rangle = 2 \). Particle A (smaller radius r) diffuses faster because it has a larger diffusion constant. The diffusion constant is inversely proportional to the particle radius, as given by the Einstein–Stokes equation.
CBSE Exam Tips 2025-26
- Memorise both forms of MSD: Use \( \langle x^2 \rangle = 2Dt \) for 1D problems and \( \langle r^2 \rangle = 6Dt \) for 3D problems. CBSE questions often specify the dimension.
- Know the Einstein–Stokes formula: The expression \( D = k_B T / (6\pi\eta r) \) is frequently tested in Class 11 kinetic theory questions. Write it correctly with \( 6\pi \) in the denominator.
- Connect to kinetic theory: CBSE 2025-26 board papers often link Brownian motion to the kinetic theory of gases. Be ready to explain why higher temperature increases D.
- Units are critical: Always convert nm to m and °C to K before substituting. A wrong unit conversion costs full marks in numerical problems.
- We recommend practising at least five numerical problems on the Brownian Motion Formula before your board exam. Focus on problems that ask you to find D first and then use it to find MSD.
- Short-answer questions: CBSE frequently asks “What is Brownian motion?” as a 2-mark question. Our experts suggest mentioning Robert Brown, the role of molecular collisions, and the linear dependence of MSD on time.
Common Mistakes to Avoid
- Using the wrong MSD formula for 3D: Many students use \( \langle r^2 \rangle = 2Dt \) for three-dimensional motion. The correct formula is \( \langle r^2 \rangle = 6Dt \). The factor of 6 comes from three independent spatial dimensions (2D per dimension).
- Confusing D with velocity: The diffusion constant D has units of m²/s, not m/s. It is not a speed. Students sometimes treat it as the speed of the particle, which is incorrect.
- Forgetting to convert temperature to Kelvin: The Einstein–Stokes formula requires absolute temperature T in Kelvin. Always add 273.15 when the temperature is given in Celsius.
- Incorrect power of 10 in calculations: When substituting \( k_B = 1.38 \times 10^{-23} \) J/K, students often mishandle the exponent. Write out each power of 10 separately before combining.
- Ignoring the linear relationship with time: Some students assume MSD grows as \( t^2 \) (like ballistic motion). Brownian motion gives \( \langle x^2 \rangle \propto t \) (linear), not \( t^2 \). This distinction is important in JEE conceptual questions.
JEE/NEET Application of Brownian Motion Formula
In our experience, JEE aspirants encounter the Brownian Motion Formula in two main contexts: kinetic theory of gases and statistical mechanics. NEET aspirants see it in the context of colloidal systems and biophysics. Here are the key application patterns:
Application Pattern 1: Ratio Problems Involving Particle Size
JEE frequently tests the inverse relationship between particle radius and diffusion constant. If particle radius doubles, D halves, and MSD halves for the same time. This is a direct consequence of \( D = k_B T / (6\pi\eta r) \). Expect questions that give two particles of different sizes and ask for the ratio of their displacements or diffusion constants.
Application Pattern 2: Temperature Dependence of Diffusion
Both JEE and NEET test the effect of temperature on Brownian motion. Since \( D \propto T \), doubling the absolute temperature doubles the diffusion constant and therefore doubles the mean square displacement for a fixed time. Questions may ask you to compare MSD at 300 K and 600 K. The answer is that MSD doubles. This also connects to the kinetic energy formula \( \langle KE \rangle = \frac{3}{2} k_B T \).
Application Pattern 3: Distinguishing Brownian Motion from Ballistic Motion
A classic JEE conceptual question asks students to identify the type of motion from an MSD-vs-time graph. For Brownian (diffusive) motion, \( \langle x^2 \rangle \propto t \) gives a straight line through the origin. For ballistic motion, \( \langle x^2 \rangle \propto t^2 \) gives a parabola. Students must recognise this distinction. In our experience, this graph-based question appears in JEE Main approximately once every two years.
For further reading on related kinetic theory concepts, visit the official NCERT website at ncert.nic.in, where you can access the Class 11 Physics textbook Chapter 13 on Kinetic Theory.
FAQs on Brownian Motion Formula
Explore more related formula articles on ncertbooks.net to strengthen your physics preparation. Study the Pressure Formula to understand how fluid properties connect to Brownian motion. Review the Average Force Formula to see how random impulses drive particle displacement. Visit our complete Physics Formulas hub for a full list of Class 11 and Class 12 formulas organised by chapter. You can also explore the Power Formula and the Banking of Road Formula for more mechanics and dynamics practice.