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 of electron in an
orbit is
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
V (r )  
e2
4 0 r
where ε0 = 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
2m
2  V  E
Hydrogen atom
Use spherical coordinate (r,,), ψ only depends
on r.
2
2
2
2







 2 
2
  2  2  2  2 
x
y
z
r
r r
  2 2 

 2 
2m  r
r r
2
 
e2 
  0
E 
4 0 r 
 
Hydrogen atom
A solution of this equation is
Ψ = exp(-c0r)
c0 
e2 m
4 2 0
The energy of hydrogen can be calculated as
me4
E 2 2
8 0 h
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
me4 1
13.6 eV
En   2 2 . 2  
8 0 h n
n2
*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: Atoms have many small
regions. Each region has a well organized
structure but differs from its neighboring
regions.
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 = ma + nb
• Three-dimensional case:
R = ma + nb +pc
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
1
8   1 atom/unit cell
8
Simple Cubic (SC)
Atomic packing factor(PF) =
volume of atom in unit cell
volume of unit cell
4
 4

1   r 3    r 3 
3
3
    0.52
PF   3   
a
(2r )3
6
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
 1
 8    1  2 atoms/unit cell
 8
Body-centered cubic (BCC)
4
 8

2    r3    r3 
3
  3
  3  0.68
PF 
3
a3
8
 4r 


 3
Ex. Ba, Ce, Cr, Mo, Nb, K
Face-centered cubic (FCC)
• 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
 1  1
 8     6    4 atoms/unit cell
 8  2
Ex. Al, Ag, Au, Cu, Ne
4
  16

4    r3    r3 
3
 3
  2  0.742
PF 
3
a3
6
2 2r


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 a 3 / 4 .
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
Unit cell properties
abc
β
Monoclinic
abc
 = β = 90°  
Orthorhombic
abc
 = β =  = 90°
Tetragonal
a=bc
 = β =  = 90°
Cubic
a=b=c
 = β =  = 90°
Hexagonal
a=bc
 = β = 90°,  = 120°
Trigonal
a=b=c
 = 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.
 hkl  For a plane that intercepts the
negative x-axis, for example,  100 
2. {hkl}: For planes of equivalent symmetry
depending only on the orientation of the axes,
e.g. {100} for (100), (010), (001),  0 10  ,  00 1 ,  100 
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. <hkl>: For a full set of equivalent directions, e.g.
<100> for [100], [010], [001],  100 , 0 10 , 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
 
 NA
 1
. 3
 a
Density of crystal
 nM
 
 NA
 1
. 3
 a
where  = density of crystal [g/cm3]
M = atomic weight [g/mole]
NA = Avogadro’s number = 6.022 x 1023 atoms/mole
n = number of atoms per unit volume
a3
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
Soln There are eight atoms per unit cell.
Therefore, there are 8/a3 = 8/(5.43 x 10-8)3
= 5 x 1022 atoms/cm3
 nM
 
 NA
 1 5 1022 atoms/cm3  28.09 g/mole
3

2.33g/cm
. 3 
23
a
6.02

10
atoms/mole

Density of crystal
Ex. Find the density of Cu in [g/cm3]. Cu has an
atomic radius of 1.278 Å and atomic weight of
63.5 g/mole.
Density of crystal
Ex. Find the density of Cu in [g/cm3]. Cu has an
atomic radius of 1.278 Å and atomic weight of
63.5 g/mole.
• Soln Cu has an FCC structure.
a
4r 4 1.278

 3.615
2
2
Å
Number of atoms/unit cell is 4.
 nM
 
 NA

 1
. 3
 a
4  63.5
6.02 1023   3.615 10-8 
3
 8.928g/cm3