Brownian Motion Formula: Cause, Formula and Examples

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Brownian motion or Brownian movement is the random or irregular motion of small particles in the fluids. The irregular motion of smoke or dust particles in air is the example of Brownian motion.

  • This motion was discovered by botanist Robert Brown in 1827.
  • While looking through a microscope, he observed that pollen grains in water move in an irregular manner.
  • That was surprising because a force is required to start a motion and cause changes in direction.
  • In 1905, Albert Einstein published a paper demonstrating that the random forces generated by the thermally excited water molecules cause the motion of the grains.
  • This explanation was experimentally verified by Jean Perrin in 1908, for which he was awarded the Nobel prize in 1926.

Key Terms: Brownian motion, viscosity, force, density, Diffusion constant, Boltzmann constant, Avagadro’s number.


Cause of Brownian Motion

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Brownian motion can be defined as the irregular or uncontrolled movement of particles in fluid because of their constant collision with other fast moving molecules.

  • The suspended particles are extremely large in size as compared to the molecules of the fluids and these particles are continuously bombarded with the molecules of the fluids from all sides.
  • If the size of the suspended particles are sufficiently large then, an equal number of molecules strike the particles from each side at each instant.
  • For the smaller particles, the number of molecules striking the various sides of the particle at any instant may not be equal.
  • Hence, the particles are subjected to an unbalanced force at each instant, causing them to move randomly. This motion is called Brownian motion.
  • A low frequency sound is produced during the brownian motion of the particles, which is known as Brownian noise. It is also known as Brown noise or Red noise.

Brownian Motion

Brownian Motion


Brownian Motion Formula

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Brownian motion of the particles suspended in a fluid can be calculated by a parameter known as the Diffusion constant (D) and it is given by

D=μkBT

Where, 

  • µ = mobility of the particles in the fluid,
  • kB = 1.38 x 10-23 JK-1, is Boltzmann constant and,
  • T = Temperature of the fluid.

But, mobility of the particle of radius r in a fluid of coefficient of viscosity ɳ, is given by

\(μ= \frac{1}{6πrη}\) 

D= \(\frac{k_BT}{6πrη}\)

Also, Boltzmann constant, kB=\(\frac{R}{N_A}\)

D= \(\frac{RT}{6πrηN_A}\)

Where,

  • R = 8.314 J mol -1-1, is Universal gas constant and 
  • NA = 6.02 x 1023 mol -1, is Avagadro’s number.

Also, Root mean square speed of the suspended particles in Brownian motion is given by

vrms= \(\sqrt{\frac{3k_BT}{m}}\)

Where, m is the mass of the suspended particle.


Solved Examples

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Ques. What is the root mean square speed of smoke particles each of mass 5 × 10-17 kg in their Brownian motion in air at N.T.P ?

Ans. The root mean square speed of the particle in Brownian motion is given by

vrms\(\sqrt{\frac{3k_BT}{m}}\)

Given, mass of the smoke particle, m = 5 x 10-17 kg

We have, kB = 1.38 x 10-23 JK-1

T = 273 K

Therefore, vrms= √3 x 1.38 x 10-23  x 273 / 5 x 10-17

∴ vrms= √2.26 x 10-4 

vrms=1.5 x 10-2 m/s

Ques. Calculate the temperature of the surroundings if the diffusion constant of a Brownian particle is 0.5 × 10-6 m2/s, the radius is 1.2 x 10-10 m, and fluid viscosity is 18 x 10-6 Pascal-second. (3 marks)

Ans. We know,

Diffusion constant, D= \(\frac{k_BT}{6πrη}\)

Given, Fluid viscosity, ɳ = 18 x 10-6 Ps

Diffusion constant, D = 0.5 x 10-6 m2/s

Radius of the Brownian particle, r = 1.2 x 10-10 m

Therefore, temperature of the surrounding is given by

T= 6πrηD / kB

T=6 x 3.14 x 1.2 x 10-10 x18 x10-6  x 0.5 x 10-6 / 1.38 x 10-23=1474 K


Factors Affecting Brownian Motion

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These are the factors which affects Brownian motion of the particles in a fluid:

  • Temperature of the fluid: Brownian motion of a particle increases with the increase in temperature and decreases with the decrease in temperature of the fluid.
  • Viscosity of the fluid: Brownian motion of the particles decreases with the increase in viscosity of the fluid in which particles are suspended and vice versa.
  • Density of the fluid: Brownian motion of the particles increases with the decrease in density of the fluid and vice versa.
  • Size of the particles: Brownian motion of the particles of bigger size is less to that of the particles of smaller size.

Things to Remember

  • The random or irregular motion of particles suspended in a fluid is called Brownian motion.
  • It was discovered by Robert Brown in 1827.
  • The unbalanced random force due to molecules of the fluid on the suspended particles causes it to move in an irregular manner.
  • Brownian motion can be calculated by parameter diffusion constant, which is given by 

D=RT6rNA

  • Brownian motion of the particles in a fluid depends on the size of the particle, density, viscosity and temperature of the fluid.
  • Motion of the pollen grain in water, movement of dust particles in a room, diffusion of pollutants in air are the Brownian motion examples.

Sample Questions

Ques. Estimate the root mean square speed of the suspended particles in Brownian motion, if the particle mass is 10-6 kg and the temperature of the liquid is 27°C. (3 marks)

Ans. The root mean square speed of the particle in Brownian motion is given by

vrms\(\sqrt{\frac{3k_BT}{m}}\)

Given, mass of the particle, m = 10-6 kg

We have, kB = 1.38 x 10-23 JK-1

T = 27°C = 27+ 273 = 300 K

Therefore, vrms= √3 x 1.38 x 10-23 x 300 / 10-6

∴ vrms= √1.242 x 10-14

vrms= 1.11 x 10-7 m/s

Ques. Define Brownian movement with examples. (2 marks)

Ans. Brownian movement is the random or zig-zag movement of the suspended particles in a fluid (liquid or gas).

Motion of the pollen grain in still water and motion of dust particles in a room are the examples of Brownian motion.

Ques. The Brownian motion is due to
(A) Temperature fluctuations within the liquid phase
(B) Attraction and repulsion between charges on the colloidal particles
(C) Impact of the molecules of the dispersion medium on the colloidal particles
(D) Convective current (2 marks)

Ans. Correct option is (C)

Generally, a dispersion medium is a solvent over which the dispersed phase is distributed. Brownian motion is due to the impact of the molecules of the dispersion medium on the colloidal particles.

Ques. Calculate the root mean square speed of the suspended particles in Brownian motion, if the particle mass is 2 x 10-5 kg and the temperature of the liquid is 127°C. (3 marks)

Ans. The root mean square speed of the particle in Brownian motion is given by

vrms\(\sqrt{\frac{3k_BT}{m}}\)

Given, mass of the particle, m = 2 x 10-6 kg

We have, kB = 1.38 x 10-23 JK-1

T = 127°C = 127+ 273 = 400 K

Therefore, vrms= √3 x 1.38 x 10-23 x 400 / 2 x 10-5

∴ vrms= √8.28 x 10-16 

vrms= 2.9 x 10-8 m/s

Ques. How does the density of the fluid affect the Brownian motion of a particle? (1 marks)

Ans. Brownian motion of the particles increases with the decrease in density of the fluid and vice versa.

Ques. What causes Brownian motion? (2 marks)

Ans. The suspended particles collide with the large number of fluid molecules. If the number of molecules of the fluid striking the various sides of the particle at any instant may not be equal, then the particles are subjected to an unbalanced force causing them to move randomly. This causes Brownian motion of the particles.

Ques. Calculate the temperature of the surroundings if the diffusion constant of a Brownian particle is 1.5 × 10-7 m2/s, the radius is 1.7 x 10-10 m, and fluid viscosity is 18 x 10-6 Pascal-second. (3 marks)

Ans. We know,

Diffusion constant, D= \(\frac{k_BT}{6πrη}\)

Given, Fluid viscosity, ɳ = 18 x 10-6 Ps

Diffusion constant, D = 1.5 x 10-7 m2/s

Radius of the Brownian particle, r = 1.7 x 10-10 m

Therefore, temperature of the surrounding is given by

T= 6πrηD / kB

T=6 x 3.14 x 1.7 x 10-10 x 18x 10-6 x 1.5 x 10-7 / 1.38 x 10-23 = 626 K

Ques. What is the root mean square speed of smoke particles each of mass 4 × 10-14 kg in their Brownian motion in air at N.T.P ? (3 marks)

Ans. The root mean square speed of the particle in Brownian motion is given by

vrms\(\sqrt{\frac{3k_BT}{m}}\)

Given, mass of the smoke particle, m = 4 x 10-14 kg

We have, kB = 1.38 x 10-23 JK-1

T = 273 K

Therefore, vrms =√ 3 x 1.38 x 10-23 x 273 / 4 x 10-14

∴ vrms = √ 28.25 x 10-10

vrms =5.31 x 10-5 m/s

Ques. How does the temperature affect the Brownian motion? (1 marks)

Ans. Brownian motion of a particle increases with the increase in temperature and decreases with the decrease in temperature of the fluid.

Ques. Estimate the root mean square speed of the suspended particles in Brownian motion, if the particle mass is 3 x 10-5 kg and the temperature of the liquid is 100°C. (3 marks)

Ans. The root mean square speed of the particle in Brownian motion is given by

vrms\(\sqrt{\frac{3k_BT}{m}}\)

Given, mass of the particle, m = 3 x 10-5 kg

We have, kB = 1.38 x 10-23 JK-1

T = 100°C = 100+ 273 = 373 K

Therefore, vrms= √3 x 1.38x 10-23 x 373 / 3 x 10-5

∴ vrms= √ 514.7 x 10-18 

vrms = 2.26 x 10-8 m/s

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