Brownian Motion

The suspended particle in colloidal solution are moving in to and fro motion, this random motion of particle in solution is called Brownian motion. Brownian motion holds the kinetic theory of gaseous system.

Einstein’s theory of the translational Brownian motion

Einstein’s theory of translational Brownian motion explains the physical nature of the phenomenon. The Brownian particles due to their random motion tend to diffuse into the medium in the course of time. Einstein related the diffusion coefficient to Brownian motion. There are two ways to calculated the diffusion coefficient. They are:

1. From the irregular motion of the suspended particles
2. From the difference in osmotic pressure between different parts due to difference in the concentration of the suspended particles caused due to difference in concentration of the particles.

Calculation of coefficient of diffusion from random molecular motion
let us suppose that the molecular concentration is not uniform; but has gradient causing diffusion of the particles. Consider a cylinder with its axis parallel to x-axis and its end faces S₁ and S₂ separated by a distance \(\Delta \). Let A be the cross section area of the cylinder and n₁ and n₂ denote the molecular concentration at ends S₁ and S₂
then the number of particles crossing the end face S₂ \( \tau = \frac 12 n_1\Delta A\).

Similarly, the number of particles crossing the end fae S₁ to the left in time \(\tau\),

\begin{align*} &= \frac 12 n_2 . \Delta A\\ \text {hence the excess of particles crossing a middle layer to the right} \\ &= \frac 12 (n_1 –n_2)\Delta A. \end{align*}
If \(\frac {dn}{dx}\) is the gradient of molecular concentration, then according to definition of diffusion coefficient we must have

\begin{align*} \frac 12(n_1 – n_2)\Delta A &= -D \frac {dn}{dx}\tau A \\ \text {But }\: n_1 – n_2 &= - \Delta \frac {dn}{dx}\\ \frac 12 \left (-\Delta \frac {dn}{dx}\right ) \Delta A = -D\frac {dn}{dx}\ \tau A. \\ \therefore D &= \frac {\Delta ^2}{2\tau} \dots (i) \\\end{align*}

Calculation of coefficient of diffusion from the difference in osmotic pressure between different parts due to difference in concentration of the suspended particles.
Let the osmotic pressure acting on the end faces S₁ and S₂ of the cylinder be P₁ and P₂. From the gas laws.

\begin{align*} p_1 &= n_1kT \\ p_2 &= n_2kT\\ \end{align*}
Due to this difference of pressure, the cylinder will experience a force is equal to

\begin{align*} (p_1 – p_2)A &= (n_1kT – n_2kT)A \\ &= (n_1 – n_2)kTA \end{align*}

If n is the mean concentration, then this force acts on \(n.\: A\Delta\) particles ontained in the cylinder. Thus the force acting on the single particles is given by

\begin{align*} F &= \frac {(n_1 –n_2)kTA}{nA\Delta}\\ &= \frac {kT}{n}.\frac {n_1 –n_2}{\Delta} \\ &= \frac {kT}{n} \left (-\frac {dn}{dx}\right) \\ &= -\frac {kT}{n} . \frac {dn}{dx} \\ \end{align*}according to stroke’s law the particle being suspended in viscous medium and assumed spherical will have a steady velocity v given by

\begin{align*} f &= 6\pi \eta r v\\ &= -\frac {kT}{n}.\frac {dn}{dx}\end{align*}

Where r is the radius of the particle and \(\eta\) is the coefficient of viscosity

$$ \therefore nv = - \frac {kT}{6\pi \eta r}.\frac {dn}{dx} $$

This expression represents the number of particles moving to the right per unit area.

According of diffusion coefficient

\begin{align*} nv &= -D. \frac {dn}{dx} \\ \therefore -D . \frac {dn}{dx} &= - \frac {kT}{6\pi \eta r}.\frac {dn}{dx}\\ D &= \frac {KT}{6\pi\eta r} \\ D &= \frac {RT}{R}. \frac {1}{6\pi \eta r} \dots (2)\left (\text {since}\: k = \frac RN\right ) \\ \end{align*}

where N is Avogadro number.

Equating (1) and (2), we get

\begin{align*} \Delta ^2 &= \frac {RT}{N}. \frac {1}{3\pi \eta r}\tau. \dots (3)\end{align*}

By getting all the values N is calculated.

Bibliography

S.S. Singhal, J.P. Agarwal, Satya Prakash. heat and thermodynamics and statistical physics. pragati prakashan, 2010.

—. Heat and Thermodynamics and Statistical Physics. Pragati Prakashan, 2010.

Vatsyayan, Dr. Rakesh Ranjan. Refresher Course in Physics. kathmandu: Surya Book Traders, 2015.