Dulong-Petit Law

-heat capacity of a solid: Cv = 3R

Magnetic Moments of Atoms

-individual atoms have magnetic moments -if the electron shell is not full then there are unpaired spins which contribute to the magnetic moment -the orbital motion of electrons also generates a magnetic moment -most atoms in the periodic table are magnetic as individual atoms, but not as whole solids

Molar Magnetic Susceptibility Definition

χ = dM/dH M - magnetisation of the sample, i.e. dipole moment per unit volume H - magnetic field

Magnetic Susceptibility and Types of Magnets

-a positive χ indicates a paramagnet -a negative χ indicates a diamagnet

How many elements are metals, insulators or semiconductors?

-by plotting resistivity against atomic number, it is immediately clear that most elements have a low resistivity and are therefore metals -there are only four elemental semiconductors

Schrodinger Equation for Free Electrons

HΨ = EΨ Ψ - wave function E - energy H - Hamiltonian -in this equation, H is an operator which can be applied to the wave function, the resulting function will contain information about the energy

Hamiltonian for Free Electrons

H = -ħ²/2m ∇² + V(r) -where V(r) is the ionic potential -for free electrons V(r)=0 so H = -ħ²/2m ∇²

Wave Function for Free Electrons

-choosing a plane wave function and working in one dimension: Ψ (r) = A e^(i|k.|r) -this can be justified by noting that operating with the momentum operator (-iħ∇) gives eigenvalues of ħk which is obviously the momentum of a free electron -also A = V^(-1/2) where V is the volume of the box containing the electron

Normalising

∫ Ψ* Ψ d|r = 1 -where Ψ* is the complex conjugate -integrating from negative infinity to positive infinity -also Ψ* Ψ gives the probability of locating a particle -from negative to positive infinity the probability must be 1 since the particles does exist somewhere

How to show that electrons can be represented as plane waves of the form Ψ = A e^(i|k.|r)

-apply the momentum operator: ^p = ħ/i *∇ -to the wave function in a volume of 1 which means A=1: ħ/i *∇ e^(i|k.x) = ħk * e^(ik.x) -this gives momentum as p=ħk -using the Schrodinger equation to find energy: - ħ²/2m * ∇² Ψ= EΨ = ħ²k²/2m Ψ = E Ψ => p²/2m = E -as these results for momentum and energy are correct, the wave function is correct for an electron

Periodic Boundary Conditions

-consider an empty box of side L -periodic conditions allow functions such that: Ψ(x) = Ψ(x+L) -these boundary conditions mean that every distance L the wave function has the same value -these boundary conditions restrict the plane waves that can exist in the box

Possible K Values Derivation

-choose: e^(ikx.L) = e^(iky.L) = e^(ikz.L) = 1 -in order for e^z=1 where z is a complex number, z=2πin, where n is an integer -this gives: kx = 2π*nx/L -and similarly for ky and kz -then finding the modulus of k: k² = (2π/L)² (nx² + ny² + nz²)

Picturing Possible k Values

-imagine k space, as three dimensional space with kx on the x axis ky on y and k on z -the possible k values are whole number coordinates where the distance between the numbers in 2π/L

How many electrons can occupy the same k value?

-as electrons are fermions, only one electron (or zero electrons) can occupy a particular quantum state -however wave vector, k, alone does not determine a quantum state -a spin up electron and a spin down electron can occupy each k value

How to count the number of states?

-draw a sphere with radius kmax over the allowed states centred at (0,0,0) in k space -calculate the volume of the sphere: V = 4/3 π kmax³ -calculate the volume per k value: (2π/L)³ -divide the total volume by the volume per k value to get the number of possible k values: 1/2 * L³kmax³/3π² -to account for the spin up and spin down states, double this number to get the total number of states: N = L³kmax³/3π² -replace L³ with V, and divide by V to obtain number density, n=N/V : n = kmax³/3π² -identifying kmax with kf, the Fermi wave vector we obtain: kf³ = 3π²n

Derive the free electron density of states

-calculate the volume of a sphere in k space: 4/3 * π * kf³ -divide by the volume per point in k space (2/L)³ : 1/2*L³kf³/3π² = 1/2 * Vkf³/3π² -multiply by two for spin up and spin down states to get the total number of states: N = Vkf³/3π² -number of states per unit volume in k space can be found by dividing by V -use the equation E = ħ²kf²/2m to write N in terms of E : N = V*(2mE)^(3/2) / 3π²ħ³ -differentiate with respect to E to find the density of states; dN/dE = g(E) = V(2m)^(3/2)*√E / 2π²ħ³

Define g(E)

-total number of states with energy E (not number if occupied states) -by definition g(E) = dN/dE

Define g(E)dE

-total number of states with energy between E and E+dE

Describe the graph of g(E) against E

-since g(E) = V(2m)^(3/2)*√E / 2π²ħ³ , the graph follow a √E shape -this graph does not change significantly with temperature -the graph shows that there are more possible states per energy at higher energy levels eventhough we know that electrons obey the Pauli exclusion principle -this is because there are many sets of k values that give the same magnitude of momentum and therefore the same energy, but are moving in a different direction, but only two states (spin up/down) can occupy each energy

How are the electron states filled?

-at every energy there can be two electrons, one spin up and one spin down -the states are filled from the lowest energy state up with two electrons per energy

Why does temperature have almost no effect on the electron density of states?

-electron states are filled from the lowest energy state upwards -reducing the temperature to T=0K does not effect the electrons since there is no available lower energy state for an electron to move to if you were to take energy away from it -therefore nothing happens

How to find the number of free electrons per unit volume in a given material Density Method

-find number of valence electrons Z from the periodic table e.g. first column elements donate 1 electron, 2nd column electrons donate 2 electrons etc. -look up the density ρ -find the atomic mass (mass of 1 mole of the material) M -recall Advogadro's Number Na -then sub into: n = Na*Z*ρ / M

How to find the number of free electrons per unit volume in a given material Unit Cell Method

-find number of valence electrons Z from the periodic table e.g. first column elements donate 1 electron, 2nd column electrons donate 2 electrons etc. -find N, the number of atoms per unit cell, 4 for a face centred cubic -look up ao, the length of the unit cell -sub into: n = Z * N / ao³

How to calculate the number of possible states with the Fermi energy

N = V*(2mE)^(3/2) / 3π²ħ³ => dN/N = 3/2 * dE/E dN/dE = 3/2 * N/E => g(Ef) = 3/2 * N/Ef

What stops electron energy from changing with temperature?

-when there are no available states the electron energy cannot change -at most temperatures kT is so small in comparison to Ef that changing the temperature has no measurable impact on the energy of the electrons

How to find the total number of states given the density of states?

N = ∫ g(E) dE -where the integral is taken form 0 to infinity

Electrochemical Potential

look up

Fermi-Dirac Distribution

f(E) = 1 / (e^[(E-Ef)/kb*T] + 1) -f tells you the probability that a state with energy E will be occupied at a given temperature T

Formula for the Number of Occupied States

Nocc = g(E)*f(E)dE

Heat Capacity Equation

C = dE/dT = π²/3*g(Ef)*kb²*T ≈ 2*g(Ef)*kb²*T

Magnetic Moment of a Free Electron

μ = -g*μb*s -where g is the Lande g-factor which is 2 for an electron -and μb is the Bohr magneton, eћ/2m -and s is the spin quantum number 1/2 for an electron -so for an electron: μ = μb

Magnetisation

-the dipole moment per unit volume; M = μ/V = g(Ef)*μb²*B

Magnetic Suscetibility

-how easy it is to magnetise a material, a higher number means it is easier; Χ = dM/dH = μo*μb²*g(Ef)

Spin Resolved Density of States in a Magnetic Field Description

-g(E) has been split into spin up and spin down states -in a magnetic field the energy of a dipole is given by: E = -μ*B -for a free electron μ=μb -so electrons with their spin aligned with the field will have their energy lowered by μb*B -and electrons with their spin opposite to the field will have their energy increased by μb*B -to normalise the Fermi energy, make it the same for spin up and spin down electrons, some electrons will flip spins giving a net difference of -μb to +μb which =2μb -since there are now different numbers of spin up and spin down electrons the sample has a magnetic moment: m = μb [Nu - Nd]