Brownian Motion Formula and Brownian Motion Sums

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 Na (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) / Na (mol-1)6πɳ( N m−2) a .

(or)

Diffusion Constant Formula  D is equal to the product of the Boltzmann’s constant kB (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= kB (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 = kBT/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 = kBT/6πrη

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

= 9.47 × 10-22.

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