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# ENE 311 - PowerPoint PPT Presentation

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

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

• 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

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

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 atom

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

on r.

Hydrogen atom

A solution of this equation is

Ψ = exp(-c0r)

The energy of hydrogen can be calculated as

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

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

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

Ex. Al, Ag, Au, Cu, Ne

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

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:

• 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 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:

• 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 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], , ,

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 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 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

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.

Å

Number of atoms/unit cell is 4.