Beyond Free-Electron Theory

Question 1

What is an orbital?

Answer 1

• in quantum mechanics, an atomic orbital is a wave function that describes the of either one or a pair of electrons in an atom
• atomic orbitals are designated by a combination of numerals e.g. 1s, 2p, 3d, 4f
• the numbers indicate the energy levels as well as relative distance from the nucleus
• the letters s, p, d and f designate the shape of the orbital, a consequence of the electron’s angular momentum

Question 2

Describe the s and p orbitals

Answer 2

• s orbitals are spherical and centred on the nucleus, a 1s electron is almost entirely confined to a spherical region close to the nucleus, a 2s electron is restricted to a somewhat larger sphere
• a p orbital has the approximate shape of a pair of lobes on opposite sides of the nucleus
• an electron in a p orbital has equal probability of being found in either half
• no p orbitals exist in the first energy level but there is a set of three in each of the higher levels

Question 3

Describe the d and f orbitals

Answer 3

• in all but the first two principal levels, there is a set of five d orbitals
• in all but the first three principal levels there is a set of 7 f orbitals
• all have complicated orientations

Question 4

Radial Wave Functions for Outer Orbitals

d vs s & p

Answer 4

• s and p levels extend much further away from the nucleus than d
• this means that the overlap with the next atom’s orbitals will be much greater for s and p levels than for the d levels

Question 5

Pseudopotential

Answer 5

• a potential wave function
• in this context it is used to remove oscillations of radial wave functions near the nucleus that are a computational inconvenience and have no real impact on calculations involving interaction between atoms

Question 6

Energy Level Scheme for Two Atoms at Equilibrium Separation

Answer 6

• two isolated atoms are brought closer together
• they will reach an equilibrium separation distance where the force between the two atoms is zero
• the outer levels have overlapped, and when two wave functions overlap the two independent levels split into two new energy levels, one higher and one lower
• the magnitude of the split, the difference between the upper and lower energies is proportional to the overlap
• note that the lowest levels will not split as they do not overlap with the neighbouring atom

Question 7

Energy Level Scheme - Adding More Atoms to the Model

Answer 7

• if the separation is the same for each pair, then the splitting or bandwidth will be constant
• but the number of levels in the bandwidth will be equal to the number of atoms

Question 8

Energy Level Scheme

Bragg Reflection

Answer 8

-at critical k values Bragg reflection of electrons arises and these reflections interfere constructively with incident waves to produce standing waves

Question 9

Energy Level Scheme

Gap

Answer 9

• there are two forms of standing wave with squared modulus that will correspond to electrons accumulating wither between the atoms or sited on the atoms
• those sited on the atoms have lower energy so at the critical values of k, there will emerge a gap
• since the standing waves aren’t propagating, group velocity is zero, they describe states that do not exist which is why the range of forbidden energies is called a gap

Question 10

Energy Level Scheme

Critical k values

Answer 10

k = nπ/a

-where n is any positive integer

Question 11

How to derive the critical k values?

Answer 11

-consider a one dimensional chain of atoms bonded by the interaction between their potentials
-what happens to an electron trying to propagate through the chain of atoms
-treat each atom as a point scatterer and the electrons as plane waves
-using knowledge of x-ray diffraction in crystals realise there is an analogue to Bragg’s Law, nλ = 2dsinθ byt for electrons in the crystal
-when the wavelength of the electrons matches the interatomic separation, Bragg reflection will occur:
nλ = 2a
-and since λ=2π/k :
k = nπ/a

Question 12

Brillouin Zone

Definition

Answer 12

• the range of energies between the critical k values where Bragg reflection occurs
• i.e. the first Brillouin zone is the range of energies between the first two critical values of k

Question 13

Where are the gaps in relation to the Brillouin zones?

Answer 13

• the gaps fall at zone boundaries, i.e. in between the Brillouin zones
• note that in general the gaps are not all the same size

Question 14

Shape of the Curve at the Zone Boundary

Answer 14

-the definition of the group velocity:
vg = dω/dk
= 1/ℏ dE/dk
-so the group velocity is given by the tangent to the E(k) curve
-since the group velocity has to go to zero at the zone boundary (the states are standing waves at that point), the E(k) curve has to meet the zone at right angles

Question 15

Effective Mass

Equation

Answer 15

m* = ℏ² [d²E/dk²]^(-1)

Question 16

The Kronnig-Penny Model

Description

Answer 16

• replaces the real ionic potential with a series of wells and barriers
• the height of each barrier is Vo, the width of the well is a and the width of the barrier is b

Question 17

The Kronnig–Penny Model

Bloch’s Theorem

Answer 17

-the appropriate wave functions are clearly:

ψ(x) = Ae^(ikx) + Be^(-ikx) , 0

Question 18

The Kronnig–Penny Model

Boundary Conditions

Answer 18

ψB(a):
Ae^(ika) + Be^(-ika) = 
(Ce^(-κb) + De^(κb))*e^(iq(a+b))
ψ'B(a):
ikAe^(ika) -ikBe^(-ika) = 
(-κCe^(-κb) + κDe^(κb))*e^(iq(a+b))
-we can solve this as a 4x4 determinant to find A, B, C, D
-we can then write the function that describes the energy levels as a function of q, the dispersion relation of the Kronnin-Penny Model

Question 19

The Kronnig-Penny Model

Dispersion Relation

Answer 19

cos(q(a+b)) =
cos(ka)cosh(κb) + 1/2(κ/k - k/κ)sin(ka)sinh(κb)
-the LHS can only take values from -1 to +1
-and if we choose ka=nπ then the right hand side -> ±cosh(κb) but the magnitude of cosh(κb)>1 for all states in the well so there are no solutions for k=nπ/a
-which is what we found using the Bragg reflection argument

Question 20

How many states in the first Brillouin zone?

Answer 20

2N states exactly fill the first Brillouin zone

Question 21

Brillouin Zones

Metals

Answer 21

• since 2n states exactly fill the first Brillouin zone, we can infer that elements with odd valences will half fill Brillouin zones
• if the Fermi energy is halfway up the available states in a Brillouin zone, then there are many states immediately adjacent to Ef so the element is a metal

Question 22

Brillouin Zones

Semiconductors and Insulators

Answer 22

• even valence elements should fill the Brillouin zones
• these elements should be either insulators or semicondutors depending on the size of the gap
• e.g. Si and Ge are both semiconductors since they both have a valence of 4 and the gap is about 1ev or less
• they would be insulators if the gap was large enough

Question 23

Chemical Potential vs Fermi Energys

Answer 23

• for metals, Ef lies at the top of a Brillouin zone
• for insulators or semiconductors ‘Ef’ lies halfway up the gap but by definition Ef separates the full and empty states and there are no no states in the gap
• this ‘Ef’ is actually the chemical potential, semiconductors and insulators don’t really have an Ef although the term is sometimes used interchangeably with chemical potential
• for metals Ef = chemical potential since Ef lies at the zone boundary anyway

Question 24

Semimetal

Answer 24

• there are exceptions to the odd valence = metal, even valence = insulator or semiconductor rule
• e.g. Ba has a valence of 2 and yet is a metal but not a good one
• the density of states at Ef is very low in Ba but there isn’t a clear gap, and conductivity is lo hence the name semi metal

Question 25

Density of States and Band Structure

Answer 25

-if you integrate the band structure or all values of k in 3D, you get the density of states

Question 26

How to split up the periodic table

Answer 26

Group 1 - alkali metals, one valence electron, partially filled bands
Group 2 - even number of valence electrons but overlapping bands so metals
Group 4 - semiconductors, insulators, filled bands
Transition Metals
Noble Metals - one valence electron, partially filled bands so metals

Question 27

Overlapping Bands

Answer 27

• bands overlapping means that an electron can be in the second band be in an energy state lower than an energy state in the lower band
• this means that some of the states (the lower energy ones) in the second band will be filled before the uppermost states in the first band
• this leads to partially filled bands