Unit 4 – Part B: ELECTRON THEORY OF MATERIALS

1. Using the classical free electron theory, derive the mathematical expression for electrical conductivity and thermal conductivity of metals and hence deduce the Wiedemann-Franz law.

1. Electrical Conductivity ($ \sigma $)

Definition: It is defined as the quantity of electric charge ($ Q $) conducted per unit time ($ t $) across unit area ($ A $) of the solid for unit applied electric field ($ E $).

$$ \sigma = \frac{Q}{tAE} $$

Step 1: Acceleration When an electric field $ E $ is applied, the force on an electron is $ F = eE $. According to Newton's law, $ F = ma $. $$ a = \frac{eE}{m} $$

Step 2: Drift Velocity ($ v_d $) Drift velocity is the product of acceleration and relaxation time ($ \tau $): $$ v_d = a\tau = \frac{eE\tau}{m} $$

Step 3: Current Density ($ J $) If $ n $ is the free electron density, current density is: $$ J = ne v_d $$ Substituting $ v_d $: $$ J = \frac{ne^2E\tau}{m} $$

Step 4: Final Expression From Ohm's law, $ J = \sigma E $. Comparing the terms:

$$ \sigma = \frac{ne^2\tau}{m} $$

2. Thermal Conductivity ($ K $)

Definition: The amount of heat conducted per unit area per unit time maintained at unit temperature gradient.

[Image of thermal conductivity mechanism in metals]

Step 1: Heat Transfer Analysis Consider a rod with hot end A (Temp $ T $) and cold end B (Temp $ T-dT $). Average Kinetic Energy at A = $ \frac{3}{2}kT $. Average Kinetic Energy at B = $ \frac{3}{2}k(T-dT) $.

Step 2: Net Heat Flow ($ Q $) The number of electrons crossing per unit area is $ \frac{1}{6}nv $. The net energy transferred per unit area per second is: $$ Q = \frac{1}{6}nv \left[ \frac{3}{2}kT - \frac{3}{2}k(T-dT) \right] \times 2 $$ (Factor of 2 accounts for transfer from A to B and deficit from B to A) $$ Q = \frac{1}{2}nv k dT $$

Step 3: Final Expression By definition, $ Q = K \frac{dT}{\lambda} $. Equating the two expressions for $ Q $: $$ K \frac{dT}{\lambda} = \frac{1}{2}nv k dT $$ $$ K = \frac{1}{2}nv k \lambda $$ Since $ \lambda = v\tau $:

$$ K = \frac{1}{2}nv^2 k \tau $$

3. Wiedemann-Franz Law

Statement: The ratio of thermal conductivity to electrical conductivity is directly proportional to the absolute temperature.

Derivation Ratio of $ K $ to $ \sigma $: $$ \frac{K}{\sigma} = \frac{\frac{1}{2}nv^2 k \tau}{\frac{ne^2\tau}{m}} = \frac{1}{2} \frac{mv^2 k}{e^2} $$ We know Kinetic Energy $ \frac{1}{2}mv^2 = \frac{3}{2}kT $. Substituting this: $$ \frac{K}{\sigma} = \frac{3}{2}kT \frac{k}{e^2} $$ $$ \frac{K}{\sigma} = \frac{3}{2} \left( \frac{k}{e} \right)^2 T $$ $$ \frac{K}{\sigma} = L T $$

$ L = \frac{K}{\sigma T} $ (Lorentz Number)

2. What is Density of States? Derive an expression for the density of states and carrier concentration in a metal.

1. Definition: Density of States, Z(E)

The Density of States, denoted by $ Z(E) $ or $ N(E) $, is defined as the number of available energy states per unit energy interval per unit volume of the metal.

Mathematically:

$$Z(E) dE = \frac{dN}{V}$$

Where $ dN $ is the number of energy states available between energy $ E $ and $ E + dE $.

2. Derivation of Density of States

Step 1: Energy Levels in 3D Consider a metal piece as a cube of side $ L $ and volume $ V = L^3 $. According to quantum mechanics (particle in a box), the energy levels of a free electron are given by: $$E = \frac{h^2}{8mL^2} (n_x^2 + n_y^2 + n_z^2)$$ Let $ n^2 = n_x^2 + n_y^2 + n_z^2 $, where $ n $ is the radius vector in "n-space". $$E = \frac{h^2 n^2}{8mL^2} \quad \implies \quad n^2 = \frac{8mL^2E}{h^2}$$ $$n = \sqrt{\frac{8mL^2E}{h^2}} \quad \text{--- (1)}$$

Step 2: Number of States in Sphere Consider a sphere of radius $ n $ in quantum number space.

The total volume of the sphere is $ \frac{4}{3}\pi n^3 $.
Since $ n_x, n_y, n_z $ must be positive integers, we only consider the positive octant (1/8th) of the sphere.
According to Pauli's Exclusion Principle, each state can hold 2 electrons (spin up and spin down).

The total number of energy states $ N $ within a sphere of radius $ n $ is: $$N = 2 \times \frac{1}{8} \times \frac{4}{3}\pi n^3$$ $$N = \frac{\pi n^3}{3} \quad \text{--- (2)}$$

Step 3: Substitute Energy Substitute the value of $ n $ from eq (1) into eq (2): $$N = \frac{\pi}{3} \left( \sqrt{\frac{8mL^2E}{h^2}} \right)^3$$ $$N = \frac{\pi}{3} \left( \frac{8mL^2E}{h^2} \right)^{3/2}$$ Since $ V = L^3 $: $$N = \frac{\pi}{3} V \frac{(8m)^{3/2}}{h^3} E^{3/2}$$

Step 4: Differentiate to find Z(E) The number of states within the energy interval $ E $ and $ E+dE $ is found by differentiating $ N $ with respect to $ E $: $$dN = \frac{\pi V}{3 h^3} (8m)^{3/2} \left( \frac{3}{2} E^{1/2} \right) dE$$ $$dN = \frac{\pi V}{2 h^3} (2\sqrt{2}m)^{3} E^{1/2} dE$$ Simplifying, we get the Density of States expression:

$$Z(E) = \frac{dN}{dE} = \frac{4\pi}{h^3} (2m)^{3/2} E^{1/2} V$$

(Note: Often expressed per unit volume by setting $ V=1 $).

3. Carrier Concentration in Metals

Definition The carrier concentration ($ n_c $) is the total number of electrons per unit volume. It is calculated by integrating the product of the Density of States $ Z(E) $ and the Fermi-Dirac probability function $ F(E) $. $$n_c = \int_0^{\infty} Z(E) F(E) dE$$

At Absolute Zero (T = 0K) At $ T = 0K $:

For $ E < E_F $, the probability $ F(E) = 1 $ (All states are filled).
For $ E > E_F $, the probability $ F(E) = 0 $ (All states are empty).

Therefore, the integration limits become 0 to $ E_F $: $$n_c = \int_0^{E_F} Z(E) \cdot 1 \cdot dE$$

Calculation Substitute $ Z(E) $ (per unit volume, i.e., $ V=1 $): $$n_c = \int_0^{E_F} \frac{4\pi}{h^3} (2m)^{3/2} E^{1/2} dE$$ $$n_c = \frac{4\pi}{h^3} (2m)^{3/2} \left[ \frac{E^{3/2}}{3/2} \right]_0^{E_F}$$ $$n_c = \frac{4\pi}{h^3} (2m)^{3/2} \frac{2}{3} E_F^{3/2}$$

$$n_c = \frac{8\pi}{3h^3} (2mE_F)^{3/2}$$

3. (i) Write a short note on Tight Binding Approximation for energy band in solids. (ii) Derive the expression for effective mass of an electron.

(i) Tight Binding Approximation

Introduction

The Tight Binding Approximation (TBA) is a method used to calculate the electronic band structure of solids. Unlike the free electron model, which assumes electrons move freely, the TBA assumes that electrons are tightly bound to their respective atoms, similar to electrons in isolated atoms.

Key Assumptions

Proximity: The electrons spend most of their time in the vicinity of the atomic nuclei.
Weak Overlap: The wavefunctions of electrons on neighboring atoms overlap only slightly. This overlap acts as a small perturbation.
Atomic Similarity: The crystal potential near an atom is approximated by the potential of a single isolated atom.

Wavefunction Formation (LCAO)

The crystal wavefunction $ \psi_k(r) $ is constructed as a linear combination of atomic orbitals (LCAO). If $ \phi(r) $ represents the atomic orbital for a single atom at the origin, the wavefunction for the entire crystal satisfying Bloch's theorem is:

$$\psi_k(r) = \sum_{n} e^{ik \cdot R_n} \phi(r - R_n)$$

Where $ R_n $ is the position vector of the $ n^{th} $ atom and $ k $ is the wave vector.

Result: Energy Bands

Due to the interaction between neighboring atoms (perturbation), the discrete atomic energy levels split into continuous energy bands. The width of these bands depends on the extent of the overlap between the orbitals. Stronger overlap leads to wider bands, while weaker overlap results in narrower bands.

(ii) Derivation of Effective Mass

Concept

An electron in a crystal lattice does not move as a free particle because it interacts with the periodic potential of the lattice. To account for these internal forces without complex calculations, we assign the electron a hypothetical mass called the Effective Mass ($ m^* $).

Step 1: Group Velocity The velocity of an electron wave packet (group velocity $ v_g $) in a band is given by: $$v_g = \frac{d\omega}{dk}$$ Since Energy $ E = \hbar \omega $, we have $ \omega = E/\hbar $. Therefore: $$v_g = \frac{1}{\hbar} \frac{dE}{dk} \quad \text{--- (1)}$$

Step 2: Acceleration The rate of change of velocity (acceleration $ a $) is: $$a = \frac{dv_g}{dt} = \frac{d}{dt} \left( \frac{1}{\hbar} \frac{dE}{dk} \right)$$ Using the chain rule $ \frac{d}{dt} = \frac{dk}{dt} \frac{d}{dk} $: $$a = \frac{1}{\hbar} \frac{d^2E}{dk^2} \frac{dk}{dt} \quad \text{--- (2)}$$

Step 3: Applied Force When an external force $ F $ is applied to the electron, the work done is equal to the increase in energy $ dE $ in time $ dt $: $$dE = F \cdot v_g \cdot dt$$ $$\frac{dE}{dt} = F \cdot v_g$$ Also, from quantum mechanics: $ \frac{dE}{dt} = \frac{dE}{dk} \frac{dk}{dt} $. Equating these implies: $$F = \hbar \frac{dk}{dt} \implies \frac{dk}{dt} = \frac{F}{\hbar} \quad \text{--- (3)}$$

Step 4: Comparison with Newton's Law Substitute eq (3) into eq (2): $$a = \frac{1}{\hbar} \frac{d^2E}{dk^2} \left( \frac{F}{\hbar} \right)$$ $$a = \frac{1}{\hbar^2} \frac{d^2E}{dk^2} F$$ $$F = \left[ \frac{\hbar^2}{\frac{d^2E}{dk^2}} \right] a$$ Comparing this with Newton's Second Law, $ F = m^* a $, we get:

$$m^* = \frac{\hbar^2}{\frac{d^2E}{dk^2}}$$

Physical Significance

The effective mass is inversely proportional to the curvature of the energy band $ \frac{d^2E}{dk^2} $.
Flat bands (low curvature) $\rightarrow$ High effective mass (heavy electrons).
Curved bands (high curvature) $\rightarrow$ Low effective mass (light electrons).

4. Write a short note on: (i) Fermi-Dirac Distribution, (ii) Fermi energy at T=0K and T>0K, and (iii) Significance of Fermi energy.

(i) Fermi-Dirac Distribution Function, F(E)

Definition

The Fermi-Dirac distribution function represents the probability that an electron will occupy a particular energy state $ E $ at a given absolute temperature $ T $. It is based on the Pauli Exclusion Principle, which states that no two fermions (electrons) can occupy the same quantum state.

Formula The probability function $ F(E) $ is given by:

$$F(E) = \frac{1}{1 + e^{(E - E_F) / k_B T}}$$

Where:

$ E $ = Energy of the state
$ E_F $ = Fermi Energy level
$ k_B $ = Boltzmann constant
$ T $ = Absolute Temperature

(ii) Fermi Energy at Different Temperatures

Case 1: At Absolute Zero (T = 0K) At $ T = 0K $, the behavior depends on whether the energy $ E $ is greater or less than the Fermi energy $ E_F $:

If $ E < E_F $: The term $ (E - E_F) $ is negative. $$F(E) = \frac{1}{1 + e^{-\infty}} = \frac{1}{1+0} = 1$$ (100% probability of occupation)
If $ E > E_F $: The term $ (E - E_F) $ is positive. $$F(E) = \frac{1}{1 + e^{+\infty}} = \frac{1}{1+\infty} = 0$$ (0% probability of occupation)

Conclusion: At 0K, all energy levels below $ E_F $ are completely filled, and all levels above $ E_F $ are completely empty. Thus, Fermi energy is the maximum energy possessed by an electron at absolute zero.

Case 2: At Temperature T > 0K As temperature increases, some electrons gain thermal energy and move to higher energy states.
Consider the specific energy state where $ E = E_F $: $$F(E) = \frac{1}{1 + e^{(E_F - E_F) / k_B T}} = \frac{1}{1 + e^0}$$ $$F(E) = \frac{1}{1 + 1} = \frac{1}{2} = 0.5$$ Conclusion: At any temperature above absolute zero, the Fermi energy level is the energy state which has a 50% probability of being occupied by an electron.

(iii) Significance of Fermi Energy

Reference Level: It acts as a reference energy level which separates occupied states from unoccupied states at 0K.
Probability Marker: It represents the energy state with a 50% probability of occupancy at temperatures greater than 0K.
Classification of Materials: The position of the Fermi level determines the electrical conductivity of the material (whether it is a conductor, semiconductor, or insulator).
Velocity Calculation: It helps in calculating the Fermi velocity of electrons, which is crucial for understanding transport properties like electrical and thermal conductivity.

5. (i) Contrast the term degenerate and non-degenerate states. (ii) The thermal conductivity of a metal is 123.93 Wm⁻¹K⁻¹. Find the electrical conductivity and Lorentz number when the metal possesses relaxation time 10⁻¹⁴ sec at 300K. (Density of electron = 6 × 10²⁸ per m³)

(i) Contrast: Degenerate vs. Non-Degenerate States

In quantum mechanics, a state is defined by a specific set of quantum numbers and a corresponding wave function.

Feature	Degenerate States	Non-Degenerate States
Definition	States are called degenerate when multiple different wave functions (eigenfunctions) correspond to the same energy level (eigenvalue).	States are called non-degenerate when one specific wave function corresponds to one unique energy level.
Energy Mapping	One Energy Value $\rightarrow$ Many States (Wave functions).	One Energy Value $\rightarrow$ One State (Wave function).
Quantum Numbers	Particles have different sets of quantum numbers but possess the same total energy.	Particles having different sets of quantum numbers possess different total energies.
Probability Density	The probability of finding a particle may differ for the different states, even though energy is the same.	The probability distribution is unique to that specific energy level.
Example	In a Hydrogen atom, the 2s and 2p orbitals have the same energy (ignoring fine structure) but different shapes. The energy level n=2 is degenerate.	The ground state (1s) of a Hydrogen atom is non-degenerate (ignoring spin).
Symmetry	Usually arises due to high symmetry in the system.	Usually arises in systems with lower symmetry or when perturbations lift degeneracy.

(ii) Numerical Problem

Given Data

Thermal Conductivity, $ K = 123.93 \, W m^{-1} K^{-1} $
Relaxation Time, $ \tau = 10^{-14} \, s $
Temperature, $ T = 300 \, K $
Electron Density, $ n = 6 \times 10^{28} \, m^{-3} $
Mass of electron, $ m = 9.11 \times 10^{-31} \, kg $
Charge of electron, $ e = 1.6 \times 10^{-19} \, C $

1. Calculation of Electrical Conductivity ($\sigma$)

The formula for electrical conductivity is:

$$\sigma = \frac{n e^2 \tau}{m}$$

Substituting the values:

$$\sigma = \frac{(6 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times (10^{-14})}{9.11 \times 10^{-31}}$$ $$\sigma = \frac{6 \times 10^{28} \times 2.56 \times 10^{-38} \times 10^{-14}}{9.11 \times 10^{-31}}$$

Grouping powers of 10:

$$\text{Numerator} = (6 \times 2.56) \times 10^{(28 - 38 - 14)} = 15.36 \times 10^{-24}$$ $$\sigma = \frac{15.36 \times 10^{-24}}{9.11 \times 10^{-31}}$$ $$\sigma = 1.686 \times 10^{7}$$

$$\sigma = 1.686 \times 10^7 \, \Omega^{-1}m^{-1} \, (\text{or } S/m)$$

2. Calculation of Lorentz Number (L)

According to the Wiedemann-Franz Law:

$$\frac{K}{\sigma T} = L$$ $$L = \frac{K}{\sigma T}$$