Brownian Motion Formula

Brownian Movement is defined as the random movement of particles inside a fluid. It is the random zig-zag motion of a particle, which is typically observed under a high-power ultra-microscope. It can be interpreted as the uncontrolled or irregular movement of particles in a fluid caused by constant collision with other fast-moving molecules. The random movement of a particle is typically observed to be stronger in smaller particles, less viscous liquids, and higher temperatures. When these particles move, they collide with one another. The Brownian motion explains the random movement of tiny particles floating in fluids.

Brownian Motion Formula

The Brownian motion is calculated using a parameter known as the diffusion constant. Its formula is given by the ratio of the product of gas constant and temperature to the product of six pi times Avogadro’s number, the viscosity of the fluid, and the radius of the particle. It is denoted by the symbol D. It is a unitless quantity as it is the ratio of the same quantities and hence has no dimensional formula.

D = RT/6πrηN_a

or

D = k_BT/6πrη

Where,

• D is the diffusion constant,
• k_B is the Boltzmann constant with the value of 1.381 × 10^-23 J/K
• R is the gas constant with the value of 8.314 J K^-1 mol^-1,
• T is the temperature of the surroundings,
• π is a constant with the value of 3.14,
• r is the radius of Brownian particle,
• η is the fluid viscosity,
• N_a is the Avogadro’s number, that is, 6.06 x 10²³ mol^-1.

Sample Problems

Problem 1: Calculate the diffusion constant of a Brownian particle if its radius is 2 m, fluid viscosity is 0.056 Pa, and temperature is 300 K.

Solution:

We have,

T = 300

r = 2

η = 0.056

Using the formula we have,

D = k_BT/6πrη

= (1.381 × 10^-23 × 300) / (6 × 3.14 × 2 × 0.056)

= 1.96 × 10^-21

Problem 2: Calculate the diffusion constant of a Brownian particle if its radius is 3 m, fluid viscosity is 0.068 Pa, and temperature is 250 K.

Solution:

We have,

T = 250

r = 3

η = 0.068

Using the formula we have,

D = k_BT/6πrη

= (1.381 × 10^-23 × 250) / (6 × 3.14 × 3 × 0.068)

= 8.98 × 10^-22

Problem 3: Calculate the temperature of the surroundings if the diffusion constant of a Brownian particle is 7.5 × 10^-22, the radius is 2.5 m, and fluid viscosity is 0.087 Pa s.

Solution:

We have,

D = 7.5 × 10^-22

r = 2.5

η = 0.087

Using the formula we have,

D = k_BT/6πrη

=> T = 6πDrη/k_B

= (6 × 3.14 × 7.5 × 10^-22 × 2.5 × 0.087) / (1.381 × 10^-23)

= 22.25 × 10

= 222.5 K

Problem 4: Calculate the temperature of the surroundings if the diffusion constant of a Brownian particle is 6.8 × 10^-22, the radius is 4 m, and fluid viscosity is 0.062 Pa s.

Solution:

We have,

D = 6.8 × 10^-22

r = 4

η = 0.062

Using the formula we have,

D = k_BT/6πrη

=> T = 6πDrη/k_B

= (6 × 3.14 × 6.8 × 10^-22 × 4 × 0.062) / (1.381 × 10^-23)

= 23 × 10

= 230 K

Problem 5: Calculate the radius of the Brownian particle if the diffusion constant of a Brownian particle is 5.28 × 10^-22, the temperature is 400 K, and fluid viscosity is 0.051 Pa.

Solution:

We have,

D = 5.28 × 10^-22

T = 400

η = 0.051

Using the formula we have,

D = k_BT/6πrη

=> r = k_BT/6πDη 

= (1.381 × 10^-23 × 400) / (6 × 3.14 × 5.28 × 10^-22 × 0.051)

= 108 × 10^-1

= 10.8 m

Problem 6: Calculate the radius of the Brownian particle if the diffusion constant of a Brownian particle is 4.87 × 10^-22, the temperature is 350 K, and fluid viscosity is 0.091 Pa.

Solution:

We have,

D = 4.87 × 10^-22

T = 350

η = 0.091

Using the formula we have,

D = k_BT/6πrη

=> r = k_BT/6πDη

= (1.381 × 10^-23 × 350) / (6 × 3.14 × 4.87 × 10^-22 × 0.091)

= 57.8 × 10^-1

= 5.78 m

Problem 7: Calculate the fluid viscosity if the radius of the Brownian particle is 6 m, the diffusion constant is 3.5 × 10^-22, and the temperature is 500 K.

Solution:

We have,

D = 3.5 × 10^-22

T = 500

r = 6

Using the formula we have,

D = k_BT/6πrη

=> η = k_BT/6πDr

= (1.381 × 10^-23 × 500) / (6 × 3.14 × 3.5 × 10^-22 × 6)

= 1.74 × 10^-1

= 0.174 Pa s