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

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 electronsas

1 . Electrons exist in certain stable and orbit circularly around the nucleus without radi

ating.

Bohr’s model

2. The electron could shift to a higher or low er level of energy by gaining or losing en

ergy equal to the difference in the energylevels.

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 S chrödinger’s equation in3 dimensions.

• It consists of an electron and a proton. The p otential energy of electron at distance r fro

m the proton is

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

2

0

( )4

eV r

r

Hydrogen atom

- We consider only time independent Schrödi nger’s equation since we are now interested

in the probability of electron to be found at a distance r from the nucleus and its allowe d energy levels.

22

2V E

m

Hydrogen atom

Use spherical coordinate (r,, ), ψ only depend s

on r. 2 2 2 22

2 2 2 2

2

x y z r r r

2 2 2

20

20

2 4

eE

m r r r r

Hydrogen atom

A solution of this equation is

Ψ -= exp( c0

r)

The energy of hydrogen can be calculated as

2

0 204

e mc

4

2 208

meE

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 specif ies 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”.

4

2 2 2 20

1 13.6 eV.

8n

meE

h n n

Basic Crystal Structure

Solids may be classified into3 groups:

- Amorphous

- Polycrystalline

- Crystalline


1 .Amorphous: Atoms are randomly arra nged (formlessness).


2. Polycrystalline: Atoms have many sm all regions. Each region has a well organ

ized structure but differs from its neighb oring regions.


3.Crystalline: Atoms are arranged in a

n orderly array.


• 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 cry stal, 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


• - There are three basic cubic crystal structures:

(a) simple cubic

- (b) body centered cubic, and

- (c) face centered cubic.

Simple Cubic (SC)

18 1 atom/unit cell

8

• 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)volume of atom in unit cell

Atomic packing factor(PF) = volume of unit cell

3 3

3 3

4 41

3 30.52

(2 ) 6

r rPF

a r

For the case of maximum packing (at oms in the corners touching each oth

er), about 52% of the SC unit cell volu me is filled with hard spheres, and ab out 48% of the volume is empty.

Body-centered cubic (BCC)

• An additional atom is located at the center of the cube.


18 1 2 atoms/unit cell

8

Body-centered cubic (BCC)

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

3 3

33

4 82

33 30.68

84

3

r rPF

a r

Face-centered cubic (FCC)

• One atom is added at each of six cubic faces in addition to the eight corner atoms.


1 18 6 4 atoms/unit cell

8 2

3 3

33

4 164

23 30.742

62 2

r rPF

a r

Ex. Al, Ag, Au, Cu, Ne

The Diamond Structure

• This structure is like the FCC crystal family.

• It results from the inter penetration of two FCC

lattices with one displa ced from the other by o

- ne quarter of the distan ce along a body diagon

al of the cube or a displ acement of . 3 / 4a

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 eig ht 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 Unit cell properties

Triclinic 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.


Miller indices can be obtained using this followi ng procedure:

1 . Determine the intercepts of the plane with crys tal axes (three Cartesian coordinates).

2. Take the reciprocals of the numbers from1.

3. Multiply the fractions by the least common mul tiple of the intercepts.

4. Label the plane in parentheses (hkl ) as the Mill er indices for a single plane.


Ex. Find the Miller indices of the below plane.


• Intercepts are a,3 a,2a:

(132)

• Take the reciprocals of these intercepts:

1,1/3,1/2

• Multiply the fractions b y the least common mu

ltiple, which in this cas e is6 , we get 6,2,3 .

• Label the plane as(623)


The Miller indices of important planes in a cubic crystal.


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, e.g. {100} for (100), (010), (001), , ,

hkl 100

100 010 001


3. [hkl ]: For a crystal direction, such as[010] - for the y axis. By definition, the[010] direction is normal to (010) plane, a

nd the [111] direction is perpendicular to the (111) plane.


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

3

1.

A

nM

N a

Density of crystal

where = density of crystal [g/cm3]

M = atomic weight [g/mole]

NA = ========== ====== = 6.022 x 1023 atoms/mole

= number of atoms per unit volume

3

1.

A

nM

N a

3

n

a

Density of crystal

Ex. At 300 K the lattice constant for Si is 5.43 Å. Calculate the number of Si atoms per cub ic centimeter and the density of Si at room t

emperature. 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 cub ic centimeter and the density of Si at room t

emperature. 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

22 33

3 23

1 5 10 atoms/cm 28.09 g/mole. 2.33g/cm

6.02 10 atoms/moleA

nM

N a

Density of crystal

Ex. Find the density of Cu in [g/cm3 ]. Cu has a n atomic radius of 1.278 Å and atomic weig ht of 63.5 g/mole.

Density of crystal

Ex. Find the density of Cu in [g/cm3 ]. Cu has a n atomic radius of 1.278 Å and atomic weig ht of 63.5 g/mole.

• Soln Cu has an FCC structure.

Å

Number of atoms/unit cell is 4.

4 4 1.2783.615

2 2

ra

3

3323 -8

1.

4 63.58.928g/cm

6.02 10 3.615 10

A

nM

N a

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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