Transcript Lecture 3a - web page for staff
ENE 311
Lecture 3
Bohr’s model
• Niels Bohr came out with a model for hydrogen atom from emission spectra experiments.
• The simplest Bohr’s model is that the atom has a positively charged nucleus and negatively charged electrons. • The total charge of all electrons is equal to that of the nucleus. Therefore, the whole atom is neutral.
Bohr’s model
Bohr made some statements about electrons as 1. Electrons exist in certain stable and orbit circularly around the nucleus without radiating.
Bohr’s model
2. The electron could shift to a higher or lower level of energy by gaining or losing energy equal to the difference in the energy levels.
Bohr’s model
3. The angular momentum p orbit is of electron in an
p
n ; n = 1,2,3,...
Bohr’s model
• Hydrogen atom is the best model to study Schrödinger’s equation in 3 dimensions. • It consists of an electron and a proton. The potential energy of electron at distance r from the proton is
e
2 4 0
r
where ε 0 = permittivity of free space = 8.85 x 10 -12 F/m
Hydrogen atom
We consider only time-independent Schrödinger’s equation since we are now interested in the probability of electron to be found at a distance r from the nucleus and its allowed energy levels.
2 2
m
2
V
E
Hydrogen atom
Use spherical coordinate (r , , ), ψ only depends on r.
2 2
x
2 2
y
2 2
z
2 2
r
2 2
r
r
2 2
m
2
r
2 2
r
r
E
e
2 4 0
r
0
Hydrogen atom
A solution of this equation is Ψ = exp(-c 0
r) c
0 4 2
e m
0 The energy of hydrogen can be calculated as
E
me
4 8 0 2
h
2
Hydrogen atom
• Bohr further said that a stationary orbit is determined by the quantization of energy represented by 4 quantum numbers (n, l, m, s).
• Any combination of quantum number specifies one allow state or energy level which is “
Pauli exclusion principle
”.
Hydrogen atom
For hydrogen atom, quantized energy is
E n
me
4 8 0 2
h
2 1 .
n
2 13.6 eV
n
2 *For n = 1, this is called a “ground state”.
Basic Crystal Structure
Solids may be classified into 3 groups: - Amorphous - Polycrystalline - Crystalline
Basic Crystal Structure
1.
Amorphous:
Atoms are randomly arranged (formlessness).
Basic Crystal Structure
2.
Polycrystalline:
regions.
Atoms have many small regions. Each region has a well organized structure but differs from its neighboring
Basic Crystal Structure
3.
Crystalline:
Atoms are arranged in an orderly array.
Basic Crystal Structure
• A collection of point periodically arranged is called a
lattice
. • Lattice contains a volume called a
unit cell
which is representative of the entire lattice. • By repeating the unit cell throughout the crystal, one can generate the entire lattice.
Unit Cell
• Unit cell has an atom at each corner shared with adjacent cells. • This unit cell is characterized by integral multiples of vectors such as
a
,
b
, and
c
called basis vectors. • Basis vectors are not normal to each other and not necessarily equal in length.
Unit Cell
Every equivalent lattice point in the crystal can be expressed by • Two-dimensional case:
R
=
m
a
+ n
b
• Three-dimensional case:
R
=
m
a
+ n
b
+p
c
where m , n, and p are integers
Unit Cell
The smallest unit cell that can be repeated to form the lattice is called a
primitive cell
shown below.
Unit Cell
Ex. 2-D lattice r = 3a+2b
Basic Crystal Structure
• There are three basic cubic-crystal structures: (a) simple cubic (b) body-centered cubic, and (c) face-centered cubic.
Simple Cubic (SC)
• Each corner of cubic lattice is occupied by an atom which is shared equally by eight adjacent unit cells.
• The total number of atoms is equal to 8 8 1 atom/unit cell
Simple Cubic (SC)
Atomic packing factor(PF) = volume of atom in unit cell volume of unit cell
PF
1 4 3
r
3
a
3 4 3
r
3 3 6 0.52
For the case of maximum packing (atoms in the corners touching each other), about 52%of the SC unit cell volume is filled with hard spheres, and about 48% of the volume is empty.
Body-centered cubic (BCC)
• An additional atom is located at the center of the cube. • The total number of atoms is equal to 8 1 8 2 atoms/unit cell
Body-centered cubic (BCC)
PF
2 4 3
r
3
a
3 8 3 4
r
3
r
3 3 3 0.68
8 Ex. Ba, Ce, Cr, Mo, Nb, K
Face-centered cubic (FCC)
Ex. Al, Ag, Au, Cu, Ne • One atom is added at each of six cubic faces in addition to the eight corner atoms.
• The total number of atoms is equal to 8 1 8 6 1 2 4 atoms/unit cell
PF
4 4 3
r
3
a
3 16 3 2 2
r r
3 3 2 0.742
6
The Diamond Structure
• This structure is like the FCC crystal family. • It results from the interpenetration of two FCC lattices with one displaced from the other by one-quarter of the distance along a body diagonal of the cube or a displacement of .
The Diamond Structure
• If a corner atom has one nearest neighbor in the body diagonal, then it will have no nearest neighbor in the reverse direction. • Therefore, there are eight atoms in one unit cell.
• Examples for this structure are Si and Ge.
Zinc blende structure
• For the Zinc blende structure, this results from the diamond structure with mixed atoms such that one FCC sublattice has column III (or V) atoms and the other has Column V (or III) atoms. • This is a typical structure of III-V compounds.
Ex. GaAs, GaP, ZnS, CdS
Fourteen Bravais lattices
Lattice systems
Lattice System
Triclinic Monoclinic Orthorhombic Tetragonal Cubic Hexagonal Trigonal
Unit cell properties
a a b c b c
β
= β = 90
°
a b c a = b c = β =
= 90 °
= β =
= 90 °
a = b = c a = b c a = b = c = β =
= 90 °
= β = 90
° ,
= 120 °
= β =
90 °
Crystal Planes and Miller Indices
• The crystal properties along different planes are different. • The device characteristics (not only mechanical properties but also electrical properties) are dependent on the crystal orientation. • The way to define crystal planes is to use
Miller indices
.
• The Miller indices are useful to specify the orientation of crystal planes and directions.
Crystal Planes and Miller Indices
Miller indices can be obtained using this following procedure: 1. Determine the intercepts of the plane with crystal axes (three Cartesian coordinates).
2. Take the reciprocals of the numbers from 1 .
3. Multiply the fractions by the least common multiple of the intercepts.
4. Label the plane in parentheses (hkl) as the Miller indices for a single plane.
Crystal Planes and Miller Indices
Ex. Find the Miller indices of the below plane.
Crystal Planes and Miller Indices
• • • • Intercepts are a, 3a, 2a: (132) Take the reciprocals of these intercepts: 1 , 1/3,1/2 Multiply the fractions by the least common multiple, which in this case is 6, we get 6,2,3 . Label the plane as (623)
Crystal Planes and Miller Indices
The Miller indices of important planes in a cubic crystal.
Crystal Planes and Miller Indices
There are some other conventions as: 1.
For a plane that intercepts the negative x-axis, for example, 2. {
hkl
}: For planes of equivalent symmetry depending only on the orientation of the axes,
Crystal Planes and Miller Indices
3. [hkl]: For a crystal direction, such as [010] for the y-axis. By definition, the [010] direction is normal to (010) plane, and the [111] direction is perpendicular to the (111) plane.
Crystal Planes and Miller Indices
4. <
hk
l>: For a full set of equivalent directions, e.g. <100> for [100], [010], [001], 100 00 1
Density of crystal
• Crystal with a lattice constant “a” and “n” atoms within unit cell, then the weight of unit cell is equal to the weight of atoms per unit cell.
• The density of the crystal will be expressed as
nM N A
.
1
a
3
Density of crystal
nM N A
.
1
a
3 where = density of crystal [g/cm3] M = atomic weight [g/mole] NA = Avogadro’s number = 6.022 x 10 23 atoms/mole
n a
3 = number of atoms per unit volume
Density of crystal
Ex. At 300 K the lattice constant for Si is 5.43 Å. Calculate the number of Si atoms per cubic centimeter and the density of Si at room temperature. Note: Atomic weight of Si = 28.09 g/mole
Density of crystal
Ex. At 300 K the lattice constant for Si is 5.43 Å. Calculate the number of Si atoms per cubic centimeter and the density of Si at room temperature. Note: Atomic weight of Si = 28.09 g/mole Sol n There are eight atoms per unit cell. Therefore, there are 8/a 3 = 8/(5.43 x 10 -8 ) 3 = 5 x 10 22 atoms/cm 3
nM N A
.
1
a
3 22 5 10 atoms/cm 3 28.09 g/mole 23 2.33g/cm 3
Density of crystal
Ex. Find the density of Cu in [g/cm 63.5 g/mole.
3 ]. Cu has an atomic radius of 1.278 Å and atomic weight of
Density of crystal
Ex. Find the density of Cu in [g/cm 63.5 g/mole.
3 ]. Cu has an atomic radius of 1.278 Å and atomic weight of • Sol n Cu has an FCC structure.
a
4
r
2 3.615
Å 2 Number of atoms/unit cell is 4.
nM N A
.
1
a
3 23 -8 3 8.928g/cm 3