## Brownian Motion Formula :

Diffusion Constant Formula D is equal to the product of the gas constant R _{(JK-1 mol-1)} in Joules per Kelvin per mole and the temperature T in Kelvin is divided by Avogadro’s number N_{a} _{(mol-1) }in mole and viscosity of the liquid 6πɳ _{(}_{ N m−2)} in newton meter square and the radius of the Brownian particle a. Hence the Diffusion Constant Formula has been written as,

**D = R _{(JK-1 mol-1)}T_{(k)} / N_{a} _{(mol-1)}6πɳ_{(}_{ N m−2)} a .**

**(or)**

Diffusion Constant Formula D is equal to the product of the Boltzmann’s constant k_{B} _{(J/K) }in joules per Kelvin and the temperature in T_{(k)} is divided by viscosity of the liquid _{ }6πɳ _{(}_{ N m−2)} newton meter square in Brownian particle a. .hence the Diffusion Constant Formula has been written as,

**D= k _{B} _{(J/K)}T_{(k)} / _{ }6πɳ _{(}_{ N m−2)}a .**

Where,

R → gas constant in 8.314 JK^{-1 }mol^{-1}

Na →referred to as Avogadro’s number in 6.06 x 10 ^{23} mol^{-1}

T → temperature in (k)

η → viscosity of the liquid ( N m−2)

a→ radius of the Brownian particle

kB = R/NA is Boltzmans constant in 1.381*10^{-23} J/K

π is a constant with the value in 3.14

Considering a large Brownian particle immersed in a fluid of much smaller particles (atoms). The radius of the Brownian particle is typically 10−9m < a < 5 × 10−7m.

## Sample Sums :

### Sum1 :

Calculate the diffusion constant of a Brownian particle if its radius is 3 m, fluid viscosity is 0.046 Pas, and temperature is 200 K.

### Solution :

We have,

T = 200

r = 3

η = 0.046

Using the formula we have,

D = k_{B}T/6πrη

= (1.381 × 10^{-23} × 200) / (6 × 3.14 × 3 × 0.046)

= 1.06 × 10^{-21}.

### Sum2 :

Calculate the diffusion constant of a Brownian particle if its radius is 2 m, fluid viscosity is 0.058 Pas, and temperature is 150 K.

### Solution :

We have,

T = 150

r = 2

η = 0.058

Using the formula we have,

D = k_{B}T/6πrη

= (1.381 × 10^{-23} × 150) / (6 × 3.14 × 2 × 0.058)

= 9.47 × 10^{-22}.