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The Science and Engineering of

Materials,

Chapter 2 – Atomic Structure

Objectives of Chapter 2

• The goal of this chapter is to describe the underlying

physical concepts related to the structure of matter.

• To examine the relationships between structure of

atoms-bonds-properties of engineering materials.

• Learn about different levels of structure i.e. atomic

structure, nanostructure, microstructure, and

macrostructure.

• Nanotechnology
• Micro-electro-

mechanical (MEMS)

systems-Airbag sensors

• Nanostructures

Figure 2.

Section 2.

The Structure of Materials:

Technological Relevance

Level of Structure Example of Technologies

Atomic Structure Diamond – edge of cutting tools

Atomic Arrangements: Lead-zirconium-titanate Long-Range Order [ Pb ( Zr x Ti 1-x )] or PZT – (LRO) gas igniters

Atomic Arrangements: Amorphous silica - fiber Short-Range Order optical communications (SRO) industry

Figures 2.2 – 2.

Table 2.1 Levels of Structure

Section 2.

The Structure of the Atom

 The atomic number of an element is equal to the number of electrons or protons in each atom.

 The atomic mass of an element is equal to the average number of protons and neutrons in the atom.

 The Avogadro number of an element is the number of atoms or molecules in a mole.

 The atomic mass unit of an element is the mass of an atom expressed as 1/12 the mass of a carbon atom.

Calculate the number of atoms in 100 g of silver.

Example 2.1 SOLUTION

The number of silver atoms is =

mol g

g  atomsmol

=5.58  1023

Example 2.

Calculate the Number of Atoms in Silver

Example 2.2 SOLUTION

The radius of a particle is 1.5 nm.

Volume of each iron magnetic nano-particle

= (4/3)(1.5  10 -7^ cm)^3

= 1.4137  10 -20^ cm^3

Density of iron = 7.8 g/cm^3. Atomic mass of iron is 56 g/mol.

Mass of each iron nano-particle

= 7.8 g/cm^3  1.4137  10 -20^ cm^3

= 1.102  10 -19^ g.

One mole or 56 g of Fe contains 6.023  1023 atoms, therefore, the number of atoms in one Fe nano-particle will be 1186.

Example 2. Dopant Concentration In Silicon Crystals

Silicon single crystals are used extensively to make computer chips. Calculate the concentration of silicon atoms in silicon, or the number of silicon atoms per unit volume of silicon. During the growth of silicon single crystals it is often desirable to deliberately introduce atoms of other elements (known as dopants) to control and change the electrical conductivity and other electrical properties of silicon. Phosphorus (P) is one such dopant that is added to make silicon crystals n -type semiconductors. Assume that the concentration of P atoms required in a silicon crystal is 10^17 atoms/cm^3. Compare the concentrations of atoms in silicon and the concentration of P atoms. What is the significance of these numbers from a technological viewpoint? Assume that density of silicon is 2.33 g/cm^3.

Example 2.3 SOLUTION (Continued)

Significance of comparing dopant and Si atom concentrations: If we were to add phosphorus (P) into this crystal, such that the concentration of P is 1017 atoms/cm^3 , the ratio of concentration of atoms in silicon to that of P will be

(5  1022 )/(10^17 )= 5  105. This says that only 1 out of 500,000 atoms of the doped crystal will be that of phosphorus (P)! This is equivalent to one apple in 500,000 oranges! This explains why the single crystals of silicon must have exceptional purity and at the same time very small and uniform levels of dopants.

• Quantum numbers are the numbers that assign electrons in an atom to discrete energy levels.
• A quantum shell is a set of fixed energy levels to which electrons belong.
• Pauli exclusion principle specifies that no more than two electrons in a material can have the same energy. The two electrons have opposite magnetic spins.
• The valence of an atom is the number of electrons in an atom that participate in bonding or chemical reactions.
• Electronegativity describes the tendency of an atom to gain an electron.

Section 2.3 The Electronic Structure

of the Atom

© 2003 Brooks/Cole Publishing / Thomson Learning™

Figure 2.9 The complete set of quantum numbers for each of the 11 electrons in sodium

Using the electronic structures, compare the electronegativities of calcium and bromine.

Example 2.4 SOLUTION

The electronic structures, obtained from Appendix C, are:

Ca: 1 s^22 s^22 p^63 s^23 p^6 4 s^2

Br: 1 s^22 s^22 p^63 s^23 p^63 d^10 4 s^24 p^5

Calcium has two electrons in its outer 4 s orbital and bromine has seven electrons in its outer 4 s 4 p orbital. Calcium, with an electronegativity of 1.0, tends to give up electrons and has low electronegativity, but bromine, with an electronegativity of 2.8, tends to accept electrons and is strongly electronegative. This difference in electronegativity values suggests that these elements may react readily to form a compound.

Example 16. Comparing Electronegativities

• III-V semiconductor is a semiconductor that is based on group 3A and 5B elements (e.g. GaAs).
• II-VI semiconductor is a semiconductor that is based on group 2B and 6B elements (e.g. CdSe).
• Transition elements are the elements whose electronic configurations are such that their inner “d” and “f” levels begin to fill up.
• Electropositive element is an element whose atoms want to participate in chemical interactions by donating electrons and are therefore highly reactive.

Section 2.4 The Periodic Table