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APPUNTI DEL CORSO:
MATERIA CONDENSATA
ANNO 2018/2019
DOCENTE PROF. SERGIO CAPRARA
LINGUA INGLESE
F=U-TS
Free energy of the system at low temperatures entropy gives low contribution to the free energy and the system tends to be ordered: this is what we know as a crystal, an ordered state where atoms and molecules occupy fixed positions.
An ideal crystal is infinite and the atoms that form it are fixed with a constant distance among them. The Bravais lattice is a mathematical model for ideal crystals.
Suppose we have 3 different vectors not in the same plane, called ɛ1, ɛ2, ɛ3, we can define R = n1ɛ1 + n2ɛ2 + n3ɛ3 where n ε N. The collection of all R is the Bravais Lattice.
In 1D there is just one BL
-----------------------------
In 3D there are 4 crystal families that form 5 different BL. All the BL have translational symmetry under the period of the lattice. There are also other symmetries mirror symmetry, rotational symmetry etc. Crystal families are listed from the less symmetric to the most symmetric.
2) If |a| ≠ |b| and θ ≠ 90°, it has no other punctual symmetries (they leave one point invariant) than the inversion (fR ε BL). So -R ε BL). This is called Monoclinic Lattice
3rd. ORTHOROMBIC
|a| ≠ |b| ≠ |c| θ1 = θ2 = θ3 = 90°
Orthorombic is primitive. From this we can obtain a base centered. If we put a point in the center of all faces we have a FACE CENTERED. If we put a point in the middle of diagonals we obtain a BODY-CENTERED ⟶ 4 BL
4th. TETRAGONAL
|a| = |b| ≠ |c| θ1 = θ2 = θ3 = 90°
We can only decorate it with a body-centered ⟶ 3 BL
6th. RHOMBOHEDRAL
|a| = |b| = |c| θ1 = θ2 = θ3 ≠ 90° (it's a cube compressed or stressed)
It's only primitive. ⟶ 1 BL
5th. HEXAGONAL
We have an hexagonal in the plane and whatever we want in the height.
|a| = |b| θ1 = 120 θ2 = θ3 = 90°
It's only primitive ⟶ 1 BL
7th. CUBE
Starting from the primitive cell, we can obtain a body-centered or a face centered ⟶ 3 BL
We can prove the previous formula starting from
V = |b1.(b2xb3)|
Let's see some examples:
1 - Cubic Lattice
|a1| = |a2| = |a3|
a1.(a2xa3) = a33(x ^ x z ^) = a33x(x ^) = a3
b1 = 2a3z2(y x z) = 2a x
In the same way b2 = 2ay; b3 = 2az
The reciprocal cell has a different order, the new volume is 2a3 according to the general formula
2 - FCC
fcc is not a primitive cell
If we choose as primitive vectors
a1 = a2(y ^ z ^)
a2 = a2(x ^ z ^)
a3 = a(x ^ y ^)
then we have a primitive cell.
Let's calculate the volume of the primitive unit cell
a1(a2xa3) = (a2)3((x ^ + z ^)(x ^)(x ^))
= (0 + z ^ y ^)x ^) = (32)(z ^ y ^) = a34
The volume of the primitive cell is 14 the volume of the conventional cell, we could know because the conventional cell
Von Laue Formulation
To have positive constructive interference
d cosθ + d cosθ' = nλ
d · (^n - ^n') = nλ
^k = 2π/λ ^n wave vector of incoming ray
^k' = 2π/λ ^n' wave vector of outcoming ray
2π/λ dr (^n - ^n') = 2π/λ nλ ⟹ d · (^k - ^k') = 2πn
If we consider all the lattice instead of just two points:
^R · (^k - ^k') = 2πn
This last formula tells us that we have constructive interference when ^k - ^k' is a reciprocal lattice vector.
Bragg and Von Laue formulation are equivalent, but the second one makes evident the importance of the reciprocal lattice. So, through the x-rays, we are investigating the reciprocal lattice.
Since the scattering is elastic |^k| = |^k'|
^K - ^k' - ^k ∈RL ^k · ^k' = (^k - ^K) · ^k' (^k' + ^k = ^k' - ^k)
^k' + ^k - 2π/k + ^k =
use k = |K| = |K'| e poi q.v.d. qua crota ombra. imombra.
Some more information about scattering, let's consider X-ray with k wave vector.
EWALD CONSTRUCTION
All the reciprocal lattice points that stand on ewald sphere give constructive interference.
This construction suggests several methods to investigate the crystal structure.
DEBYE-SCHERRER
X-ray beam
(diagram)
sample of polycristal powder or grains D >> λ
This technique consists in rotating the sample in order to have a spherical average. It is possible to prove that the condition for constructive interference will be
K = 2ksinφθ
How can I study the structure of a crystal with x-ray, if I can only have information about the magnitude of K and nothing about his orientation?
EX p 108 n: 1
Powder specimen of 3 diff. monoatomic cubic crystal are analyzed with D.S. means. It is known that one sample is FCC, body centh and one diamond (we will see it next lesson)
We will see 4 different spots
Try with R̅: 2π/a (-1, ±2, 1)
ℓ = aℎ/k12 + ℓ2 + c2
= 2/5
ℓ = 2/5 a = 4/5 Å ≈ 0.8 Å
k̅' = 2π/a (-1, ±2, 0)
= 3π/a (5/3, ±2, 0)
= 27π/33 (±3, 0, ±2)
So we will see
This spots will be weaker than the λ ±13 Å ones because ĥ is bigger
DIAMOND
Diamond is pure carbon, like graphene, but it has Sp3 hybridization. The structure of the lattice is fcc + basis so it is not a pure BL. We can interpret it as two compensating fcc shifted by 1/4 along the main diagonal.
The basis vectors are
d1 = 0
d2 = a/4 (⅕ + ⅕ + ⅕)
Each carbon is surrounded by 4 carbons forming a tetraog
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