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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.

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ene 311

ENE 311

Lecture 3

bohr s model
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 model1
Bohr’s model

Bohr made some statements about electrons as

  • Electrons exist in certain stable and orbit circularly around the nucleus without radiating.
bohr s model2
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 model3
Bohr’s model

3. The angular momentum p of electron in an orbit is

bohr s model4
Bohr’s model
  • Hydrogen atom is the best model to study Schrödinger’s equationin 3 dimensions.
  • It consists of an electron and a proton. The potential energy of electron at distance r from the proton is

where ε0 = permittivity of free space = 8.85 x 10-12 F/m

hydrogen atom
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.

hydrogen atom1
Hydrogen atom

Use spherical coordinate (r,,), ψ only depends

on r.

hydrogen atom2
Hydrogen atom

A solution of this equation is

Ψ = exp(-c0r)

The energy of hydrogen can be calculated as

hydrogen atom3
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 atom4
Hydrogen atom

For hydrogen atom, quantized energy is

*For n = 1, this is called a “ground state”.

basic crystal structure
Basic Crystal Structure

Solids may be classified into 3 groups:

  • Amorphous
  • Polycrystalline
  • Crystalline
basic crystal structure1
Basic Crystal Structure
  • Amorphous: Atoms are randomly arranged (formlessness).
basic crystal structure2
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 structure3
Basic Crystal Structure

3. Crystalline: Atoms are arranged in an orderly array.

basic crystal structure4
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
  • 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 cell1
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 cell2
Unit Cell

The smallest unit cell that can be repeated to form the lattice is called a primitive cell shown below.

unit cell3
Unit Cell

Ex. 2-D lattice

r = 3a+2b

basic crystal structure5
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
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
simple cubic sc1
Simple Cubic (SC)

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
Body-centered cubic (BCC)
  • An additional atom is located at the center of the cube.
  • The total number of atoms is equal to
body centered cubic bcc1
Body-centered cubic (BCC)

Ex. Ba, Ce, Cr, Mo, Nb, K

face centered cubic fcc
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

Ex. Al, Ag, Au, Cu, Ne

the diamond structure
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 structure1
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
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

crystal planes and miller indices
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 indices1
Crystal Planes and Miller Indices

Miller indices can be obtained using this following procedure:

  • Determine the intercepts of the plane with crystal axes (three Cartesian coordinates).
  • Take the reciprocals of the numbers from 1.
  • Multiply the fractions by the least common multiple of the intercepts.
  • Label the plane in parentheses (hkl) as the Miller indices for a single plane.
crystal planes and miller indices2
Crystal Planes and Miller Indices

Ex.Find the Miller indices of the below plane.

crystal planes and miller indices3
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 indices4
Crystal Planes and Miller Indices

The Miller indices of important planes

in a cubic crystal.

crystal planes and miller indices5
Crystal Planes and Miller Indices

There are some other conventions as:

  • For a plane that intercepts the negative x-axis, for example,
  • {hkl}: For planes of equivalent symmetry depending only on the orientation of the axes, e.g. {100} for (100), (010), (001), , ,
crystal planes and miller indices6
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 indices7
Crystal Planes and Miller Indices

4. <hkl>: For a full set of equivalent directions, e.g. <100> for [100], [010], [001], , ,

density of crystal
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
density of crystal1
Density of crystal

where = density of crystal [g/cm3]

M = atomic weight [g/mole]

NA = Avogadro’s number = 6.022 x 1023 atoms/mole

= number of atoms per unit volume

density of crystal2
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 crystal3
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

density of crystal4
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 crystal5
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.

Å

Number of atoms/unit cell is 4.