FANDAMENTALS OF MATERIALS SCIENCE AND
ENGINEERING
(For the beginners)
By Gizachew Berhanu
Chapter one: Basic Concepts of Materials
1.1. Introduction
Dear reader, welcome to the introductory chapter “basic concepts of materials”. In this chapter, we try to discuss
the nature and development of materials science and engineering. It introduces to you what a Materials Science and
Engineering is, its nature and its historical development. After some definitions about what materials science and
engineering is, we go to classify materials engineering according to their nature and their various categories of
applications. Moreover, we emphasize on the importance of the structure-property relationship of materials.
Learning objectives
At the end of this chapter, reader should be able to . . .
1. Evaluate how much he/she knows, and how much he/she does not know about materials
2. Describe the subject of materials science and engineering as a scientific discipline.
3. Reason out why need to study materials Science and Engineering.
4. Cite the primary classification of solid materials.
5. Give distinctive features of each group of materials.
6. Give some applications of different types of materials.
7. Cite six basic properties of materials
8. Discuss the relationship among structure, properties, processing and performance of materials
9. Briefly discuss biomaterials, smart materials and nanomaterials.
1.2. Definition: - What are materials?
Everything found on the earth that surrounds us is a matter. The origin of the word “matter” is Latin word matri
for mother. Matter is anything that has mass and takes up space. Mass is a measurement of the amount of matter
in an object. Everything, however, is not made of matter. For example, heat, light, radio waves, and magnetic
fields are some things that are not made of matter. Every scientific discipline concerns itself with matter. Of all
matter surrounding us, a portion comprises materials. What do we mean by materials?
According to oxford dictionary, materials may be defined as substances of which something is composed or made.
This definition is broad, from an engineering application point of view. The type of matter that human being
produced by themselves for their daily activities and use with to manufacture another interested tools is called
materials. According to this definition a naturally found rock is not a material, intrinsically; however, if it
is used in aggregate (concrete) by humans, it becomes a material [8].
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Natural rock
Aggregation of rocks for concrete materials
Fig 1.1 concrete work
In the same way log of tree lying on the ground is not a material. However, if it is manufactured as bed, chairs,
tables, doors etc and used by humans, it becomes a material.
b) Seat table
a) Log of a tree
Fig 1.2 wood work
The skin that taken from an animal becomes material after it processed in either industry or traditional skills and
changed to coat, bag, belt and many other things. Further, we can restrict the definition of materials to matters
useful to mankind. Even here, the range is too broad for the purposes of engineering. For instance, a large number
of things to mankind, such as food, medicines, explosives, chemicals, water, steel, plastics and concrete only some
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of which qualify as engineering materials. Therefore, the word ‘materials’ here does not refer to all matter in the
Universe. If this were so, it would include all the physical sciences and the life sciences— from astronomy to
zoology! By including the word ‘inanimate’ in the definition, we can exclude the life sciences from our purview.
We then have to be more specific and define materials as that part of inanimate matter which is useful to the
engineer in the practice of his profession. In the currently understood sense of the term, materials refer only to
solid materials, even though it is possible to quote a number of examples of liquid and gaseous materials such as
sulphuric- acid and steam, which are useful to the engineer.
1.3 Historical Development of materials.
All of us live in a world of dynamic change, and materials are no exception. The advancement of civilization has
historically depended on the improvement of materials to work with. Transportation, housing, clothing,
communication, recreation, and food production virtually every segment of our everyday lives is influenced to one
degree or another by materials [1].
The development and advancement of societies have been intimately tied to the members’ ability to produce and
manipulate materials to fill their needs. Prehistoric humans were restricted to naturally accessible materials such as
stone, wood, bones, skin and fur [1].
Over time, they discovered techniques for producing materials that had properties superior to those of the natural
ones; these new materials included pottery and various metals. Furthermore, it was discovered that the properties of
a material could be altered by heat treatments (annealing) and by the addition of other (carbon to steel) substances.
Early civilizations have been designated by the level of their materials development moved from the materials
Stone Age into Bronze Age then Iron Age to present information age.
Note that this advance did not take place uniformly everywhere. In our country Ethiopia, agriculture comprises a
large part of our present economy. The production and processing of materials into finished goods constitutes a
large part of present economy in developed countries.
Engineers design most manufactured products and the processing systems required for their production. Since
products require materials, engineers should be knowledgeable about the internal structure and properties of
materials so that they can choose the most suitable ones for each application and develop the best processing
methods.
Research and development engineers create new materials or modify the properties of existing ones. Design
engineers use existing, modified, or new materials to design and create new products and systems. Sometimes
design engineers have a problem in their design that requires a new material to be created by research scientists and
engineers. For example, engineers designing a high-speed civil transport will have to develop new high-temperature
materials that will withstand temperatures as high as 1800℃[3].
Research is currently underway to develop new ceramic-matrix composites, refractory intermetallic compounds,
and single-crystal superalloys for this and other similar applications [2].
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We must remember that materials usage and engineering designs are constantly changing. This change continues to
accelerate. No one can predict the long-term future advances in material design and usage. Many years ago, many
people would not have believed that someday computers would become a common household item similar to a
telephone, television or a refrigerator.
And today, we still find it hard to believe that someday space travel will be commercialized and we may even
colonize Mars. Nevertheless, science and engineering push and transform our most unachievable dreams to reality.
The search for new materials goes on continuously. More recently, the field of nonomaterials has attracted a great
deal of attention from scientists and engineers all over the world. Novel structural, chemical, and mechanical
properties of nonomaterial have opened new and exciting possibilities in the application of these materials to a
variety of engineering and medical problems [3]. In many cases what was impossible yesterday is a reality today!
1.4 What is Materials Science and Engineering?
The word ‘science’ in the phrase refers to the physical sciences, in particular to chemistry and physics. As we
confine ourselves mainly to solids in materials science, the subject is related to solid state chemistry and solid state
physics. Materials science is primarily concerned with the search for basic knowledge about the internal structure,
properties, and processing of materials. The word ‘engineering’ indicates that the engineering usefulness of the
matter under study is always kept in mind, irrespective of whether the basic laws of science can be applied
rigorously or not. Materials engineering is mainly concerned with the use of fundamental and applied knowledge
of materials so that the materials can be converted into products needed or desired by society.
The term materials science and engineering combines both materials science and materials engineering and is the
subject matter of this text. In general, Materials science and Engineering is an interdisciplinary field of science and
engineering that studies and manipulates the composition and structure of materials across length scales to control
materials properties through synthesis and processing [2].
The term composition means the chemical make-up of a material. The term structure means a description of the
arrangement of atoms, as seen in chapter 4 in detail. The term property refers to a material trait in terms of the kind
and magnitude of response to a specific imposed stimulus. Definitions of properties are made independent of
material shape and size unless the size of the material is extremely small. Important properties of solid materials
may be grouped into six different categories:
1) Mechanical properties: - refers to deformation to an applied load or force; examples include elastic modulus
(stiffness), strength, and toughness.
2) Electrical properties:- For electrical properties, such as electrical conductivity and dielectric constant, the
stimulus is an electric field.
3) Thermal properties: - The thermal behavior of solids can be represented in terms of heat capacity and thermal
conductivity.
4) Magnetic properties: - demonstrate the response of a material to the application of a magnetic field.
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5) Optical properties:-The stimulus is electromagnetic or light radiation; index of refraction and reflectivity are
representative optical properties.
6) Deteriorative properties: - Relate to the chemical reactivity of materials.
Materials scientists and engineers not only deal with the development of materials, but also with the synthesis
and processing of materials and manufacturing processes related to the production of components. The term
“synthesis” refers to how materials are made from naturally occurring or man-made chemicals.
The term “processing” means how materials are shaped into useful components to cause changes in the properties
of different materials.
One of the most important functions of materials scientists and engineers is to establish the relationships between a
material or a device’s properties and performance and the microstructure of that material, its composition, and the
way the material or the device was synthesized and processed. In materials science, the emphasis is on the
underlying relationships between the synthesis and processing, structure, and properties of materials. In materials
engineering, the focus is on how to translate or transform materials into useful devices or structures.
1.5 Why we study Materials Science and Engineering?
One of the most fascinating aspects of materials science involves the investigation of a material’s structure. The
structure of materials has a profound influence on many properties of materials, even if the overall composition
does not change. The changes in the material’s properties are due to a change in its internal structure. The structure
at the microscopic scale is known as the microstructure. If we can understand what has changed microscopically,
we can begin to discover ways to control the material’s properties.
A thorough knowledge of materials science and engineering will make you a better engineer and designer [1].
Materials science underlies all technological advances and an understanding of the basics of materials and their
applications will not only make you a better engineer, but also will help you during the design process. In order to
be a good designer, you must learn what materials will be appropriate to use in different applications. You need to
be capable of choosing the right material for your application based on its properties, and you must recognize how
and why these properties might change over time and due to processing. Any engineer can look up materials
properties in a book or search databases for a material that meets design specifications, but the ability to innovate
and to incorporate materials safely in a design is rooted in an understanding of how to manipulate materials
properties and functionality through the control of the material’s structure and processing techniques [2]. The most
important aspect of materials is that they are enabling; materials make things happen. For example, in the history of
civilization, materials such as stone, iron, and bronze played a key role in mankind’s development. In today’s fastpaced world, the discovery of silicon single crystals and an understanding of their properties have enabled the
information age.
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1.6 Types of Materials
Solid materials have been conveniently grouped into the following basic categories based primarily on chemical
makeup and atomic structure.
1. Metals and Alloys
2. Ceramics, glasses and glass-ceramics
3. Polymers
4. Semiconductors
5. Composites
Materials in each of these groups possess different structures and properties. The differences in strength, which are
compared in Figure 1-3, illustrate the wide range of properties from which engineers can select. Since metallic
materials are extensively used for load-bearing applications, their mechanical properties are of great practical
interest [2].
Figure 1.3 Representative strengths of various categories of materials.
An alternative way of classifying materials is according to the three major areas in which they are used:
(i)Structures
(ii)Machines
(iii) Devices.
Structures (not to be confused with the internal structure of a material) refer to the objects without moving parts
erected by engineers, such as a concrete dam, a steel melting furnace, a suspension bridge and an oil refinery tower.
Machines include lathes, steam and gas turbines, engines, electric motors and generators.
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Devices are the most recent addition to engineering materials and refer to such innovations as a transistor, a
photoelectric cell, piezoelectric pressure gauges, ceramic magnets and lasers.
Invariably, in each category of applications, we find materials from all the three groups described above. To give
some examples, an aircraft structure is built of aluminium alloys and plastics; a steel melting furnace is built of
refractory oxides and structural steel; safety helmets are made of glass-reinforced plastics. Similarly, we have
metal-oxide semiconductors. ASTU Research Park in Fig. 1.4 shown below depicts this interplay between material
groups and categories of applications.
Window glass
Reinforcement metals
Ceramic floor
Polymer tire
Fig. 1.4 Adama sciences and Technology University Research Park building
1. Metals and Alloys: Metals are solid materials that are normally combinations of metallic elements. 91 of the 118
elements in the periodic table are metals. Typically hard, opaque, shiny, and features good electrical and thermal
conductivity. Metals are generally malleable that is, they can be hammered or pressed permanently out of shape
without breaking or cracking as well as fusible (able to be fused or melted) and ductile (able to be drawn out into a
thin wire)[4]. Metals and alloys have relatively high strength, high stiffness, and shock resistance. They are
particularly useful for structural or load-bearing applications. Although pure metals are occasionally used, alloys
provide improvement in a particular desirable property or permit better combinations of properties [3]. An alloy is a
metal that contains additions of one or more metals or non-metals. Steels and bronze are examples of alloys [2].
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a) If no metals, it is difficult to have such kind of renaissance dam
b) If no metal, no SINOTRUCK
Fig 1.5 roles of metals and alloys in real world
2. Ceramics: Ceramics can be defined as inorganic crystalline materials. These types of materials are generally
compounds between metallic and nonmetallic elements and include compounds such as oxides, nitrides, and
carbides [1]. Typically they are insulating and resistant to high temperatures and harsh environments. Beach sand
and rocks are examples of naturally occurring ceramics. Traditional ceramics are used to make bricks, tableware,
toilets, bathroom sinks, refractories (heat-resistant material), and abrasives. Advanced ceramics are materials made
by refining naturally occurring ceramics and other special processes [5].
Advanced ceramics are used in substrates that house computer chips, sensors, capacitors, wireless communications,
inductors, and electrical insulation. Some ceramics are used as barrier coatings to protect metallic substrates in
turbine engines. In general, due to the presence of porosity (small holes), ceramics do not conduct heat well; they
must be heated to very high temperatures before melting. Ceramics are strong and hard, but also very brittle.
Glasses and Glass-Ceramics: A glass is defined by ASTM as “an inorganic product of fusion which has been
cooled to rigid condition without crystallization” [6]. Glass is an amorphous material, often, but not always, derived
from a molten liquid. The term “amorphous” refers to materials that do not have a regular, periodic arrangement of
atoms. Amorphous materials will be discussed in Chapter 4. The fiber optics industry is founded on optical fibers
based on high purity silica glass. Glasses are also used in houses, cars, computer and television screens, and
hundreds of other applications.
Glasses can be thermally treated (tempered) to make them stronger. Forming glasses and nucleating (forming)
small crystals within them by a special thermal process creates materials that are known as glass-ceramics [6].
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Glass-ceramic
ceramic
Glass bottles
Fig 1.6 applications of glasses, glass-ceramic and ceramics materials
3. Polymers: Polymers are typically organic materials based upon carbon and hydrogen. They are very large
molecular structures. Usually they are low density and are not stable at high temperatures. They are produced using
a process known as polymerization. Polymeric materials include rubber and many types of adhesives. Polymers
typically are good electrical and thermal insulators although there are exceptions such as the semiconducting
polymers. They are typically not suitable for use at high temperatures [1]. Many polymers have very good
resistance to corrosive chemicals. Polymers are used in many applications, including electronic devices.
4. Semiconductors: Semiconductors have electrical properties intermediate between metallic conductors and
ceramic insulators. Electrical properties are strongly dependent upon small amounts of impurities. Silicon,
germanium, and gallium arsenide-based semiconductors such as those used in computers and electronics are part of
a broader class of materials known as electronic materials which have enabled the information age [2]. In some
semiconductors, the level of conductivity can be controlled to enable electronic devices such as transistors, diodes,
etc., that are used to build integrated circuits. In many applications, we need large single crystals of
semiconductors. These are grown from molten materials. Often, thin films of semiconducting materials are also
made using specialized processes [2].
5. Composite Materials: Composites consist of more than one material type. The constituents keep their properties
and the overall composite will have properties different from each of them [2]. Most composite materials consist of
a selected filler or reinforcing material and a compatible resin binder to obtain the specific characteristics and
properties desired. Usually, the components do not dissolve in each other, and they can be physically identified by
an interface between them. Many different combinations of reinforcements and matrices are used to produce
composite materials. Fiberglass, a combination of glass and a polymer, is an example. Concrete and plywood are
other familiar composites [1]. Many new combinations include ceramic fibers in metal or polymer matrix. The
main idea in developing composites is to blend the properties of different materials. These are formed from two or
more materials, producing properties not found in any single material. The glass fibers make the polymer stiffer,
without significantly increasing its density. With composites, we can produce lightweight, strong, ductile,
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temperature-resistant materials or we can produce hard, yet shock-resistant, cutting tools that would otherwise
shatter. Advanced aircraft and aerospace vehicles rely heavily on composites such as carbon fiber-reinforced
polymers (Fig.1.7). Sports equipment such as bicycles, golf clubs, tennis rackets, and the like also make use of
different kinds of composite materials that are light and stiff. Two outstanding types of modern composite materials
used for engineering applications are fiberglass-reinforcing material in a polyester or epoxy matrix and carbon
fibers in an epoxy matrix [2].
Fig.1.7 Overview of the wide variety of composite parts used in the Airplane transport. This airplane has advanced
composites (Source: airlineworld.wordpress.com)
1.7 Recent Advanced engineering materials
Biomaterials: As a science, biomaterials are about fifty years old. It can be defined as any systemically,
pharmacologically inert substance or combination of substances utilized for implantation within or incorporation
with a living system to supplement or replace functions of living tissues or organs [7]. It must be compatible, not
toxic, biodegradable and strong enough for handling a pressure during performance. Our bones and teeth are made,
in part, from a naturally formed ceramic known as hydroxyapatite [2]. A number of artificial organs, bone
replacement parts, cardiovascular stents, orthodontic braces, and other components are made using different
plastics, titanium alloys, and nonmagnetic stainless steels.
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Figure 1.8 All the preceding materials metals, ceramics, polymers, composites, and semiconductors may
be used as biomaterials.
Smart Materials: A smart material can sense and respond to an external stimulus such as a change in
temperature, the application of a stress, or a change in humidity or chemical environment [2]. Usually a smart
material-based system consists of sensors and actuators that read changes and initiate an action. The sensory
component detects a change in the environment, and the actuator component performs a specific function or a
response. For instance, some smart materials change or produce color when exposed to changes in temperature,
light intensity, or an electric current.
Some of the more technologically important smart materials that can function as actuators are shape-memory
alloys and piezoelectric ceramics [1].
Shape-memory alloys are metal alloys that, once strained, revert
back to their original shape upon an increase in temperature above
a critical transformation temperature. The change in shape back to
the original is due to a change in the crystal structure (dislocation)
above the transformation temperature.
Figure1.9 Illustration of Shape-memory alloys are metals that,
after having been deformed, revert to their original shape when
temperature is changed. (Source:www.memo.com)
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Actuators may also be made of piezoelectric materials that produce an electric field when exposed to a mechanical
force. Conversely, a change in an external electric field will produce a mechanical response in the same material.
Such materials may be used to sense and reduce undesirable vibrations of a component through their actuator
response. Once a vibration is detected, a current is applied to produce a mechanical response that counters the effect
of the vibration.
Nanomaterials: In previous physics courses one might familiar with the term nano representing for prefix 10-9
unit and symbolized by “n”. The same is true for this subtopic “nanomaterials” that we will discuss at much small
size desired materials. Nanomaterials are generally defined as those materials that have a characteristic length scale
such as particle diameter, grain size, layer thickness, and etc. less than 100 nm [3]. Nanomaterials can be metallic,
polymeric, ceramic, electronic, or composite.
In this respect:
 Ceramic powder aggregates of less than 100 nm in size,
 Bulk metals with grain size less than 100 nm,
 Thin polymeric films with thickness less than 100 nm, and
 Electronic wires with diameter less than 100 nm are all considered as nanomaterials or
nanostructured materials.
Figure 1.10 A material class that has fascinating properties
and tremendous technological promise is the nanomaterials,
which may be any one of the four basic types metals,
ceramics, polymers, or composites. (Source: www.nano.com)
Xnm
At the nanoscale, the properties of the material are neither that of the molecular or atomic level nor that of the bulk
material [1]. The development of scanning probe microscopes enables scientists to observe individual atoms and
molecules at nano-scale to design and build new structures from their atomic-level constituents. Metallurgists have
always been aware that by refining the grain structure of a metal to ultrafine (submicron) levels, its strength and
hardness increases significantly in comparison to the coarse-grained (micron-size) bulk metal. For example,
nanostructured pure copper has yield strength six times that of coarse-grained copper [3].
Materials exhibited physical and chemical dramatic change properties as particle size approaches atomic
dimensions [3]. For example, materials that are opaque in the macroscopic domain may become transparent on the
nanoscale; some solids become liquids, chemically stable materials become combustible, and electrical insulators
become conductors. Generally, properties may depend on size in this nanoscale domain. Some of these effects are
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quantum mechanical in origin, whereas others are related to surface phenomena the proportion of atoms located on
surface sites of a particle increases dramatically as its size decreases [1].
The early applications of nanomaterials were as chemical catalysts and pigments. The future applications of
nanomaterials are only limited to the imagination and one of the major obstacles in fulfilling this potential is the
ability to efficiently and inexpensively produce these materials. For instance the manufacturing of orthopaedic and
dental implants from nanomaterials with better biocompatibility characteristics, better strength, and better wear
characteristics than metals. One such material is nanocrystalline zirconia (zirconium oxide), a hard and wear
resistant ceramic that is chemically stable and biocompatible [2]. This material can be processed in a porous form,
and when it is used as implant material, it allows for bone to grow into its pores, resulting in a more stable fixation.
Nanomaterials may also be used in producing paint or coating materials that are significantly more resistant to
scratching and environmental damage. Also, electronic devices such as transistors diodes and even lasers may be
developed on a nanowire. Such materials science advancements will have both technological and economical
impact on all areas of engineering and industries.
1.8 Structure–Property Relationships in Materials
Until recently, it has been the practice in a course on engineering materials to list the composition,
treatment, properties and uses of as many materials as possible. The number and variety of engineering
materials and applications have increased tremendously in recent years. Now we have more than a
thousand types of steel alone, each with a specific composition, thermal and mechanical history.
Therefore, it is impossible to describe an adequate number of engineering materials in one course.
Moreover, our knowledge of the internal structure of materials and how this structure correlates with the
properties has rapidly advanced in recent decades. So, it is more interesting and appropriate to study some
of the key factors that determine the structure–property relationships, rather than go for a fully descriptive
account of a large number of materials. This approach is adopted in this book. The discussion of a
structure-dependent property is usually followed by typical applications.
The levels of structure which are of the greatest interest in materials science and engineering are the
microstructure, the substructure and the crystal structure. The chemical, mechanical, electrical and
magnetic properties are among the most important engineering properties. We first develop the basic
concepts pertaining to the levels of structure. These include concepts in equilibrium and kinetics, the
geometry of crystals, the arrangement of atoms in the unit cell of crystalline materials, the sub-structural
imperfections in crystals, and the microstructure of single phase and multi-phase materials. We then
discuss how changes in the structure are brought about and how they can be controlled to the best possible
advantage. Solid state diffusion and control of phase transformations by heat treatment are the main topics
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here. In the latter half of the book, corrosion among chemical properties, elastic and plastic deformation
among mechanical properties and several electrical and magnetic properties are discussed with numerous
examples of typical engineering materials.
The gross composition of a material is important in determining its structure. Yet, for a given gross
composition, radical changes in the structure and properties can be brought about by subtle changes in the
concentration and distribution of minute quantities of impurities. The same may also be possible by a thermal or
a mechanical treatment that involves no change in the overall composition of the material. Materials Science and
Engineering deals more with this kind of changes rather than with the effect of gross composition on the
properties.
1.9 Design and Selection
Material engineers should be knowledgeable of various classes of materials, their properties, structure,
manufacturing processes involved, environmental issues, economic issues, and more. As the complexity of the
component under consideration increases, the complexity of the analysis and the factors involved in the
materials selection process also increase. Consider the materials selection issues for the frame and forks of a
bicycle. The selected material must be strong enough to support the load without yielding (permanent
deformation) or fracture. The chosen material must be stiff to resist excessive elastic deformation and fatigue
failure (due to repeated loading). The corrosion resistance of the material may be a consideration over the life of
the bicycle. Also, the weight of frame is important if the bicycle is used for racing: It must be lightweight. What
materials will satisfy all of the above requirements? A proper materials selection process must consider the
issues of strength, stiffness, weight, and shape of the component (shape factor) and utilize materials selection
charts in order to determine the most suitable material for the application. The detailed selection process is
outside the scope of this textbook, but we use this example as an exercise in identifying various material
candidates for this application. It turns out that a number of materials may satisfy the strength, stiffness, and
weight considerations including some aluminum alloys, titanium alloys, magnesium alloys, steel, carbon fiber
reinforced plastic (CFRP), and even wood. Wood has excellent properties for our application but it cannot be
easily shaped to from a frame and the forks. Further analysis shows CFRP is the best choice; it offers a strong,
stiff, and lightweight frame that is both fatigue and corrosion resistant. However, the fabrication process is
costly. Therefore, if cost is an issue, this material may not be the most suitable choice. The remaining materials,
all metal alloys, are all suitable and comparatively easy to manufacture into the desired shape. If cost is a major
issue, steel emerges as the most suitable choice. On the other hand, if lower bicycle weight is important, the
aluminum alloy emerges as the most suitable material. Titanium and magnesium alloys are more expensive than
both aluminum and steel alloys and are lighter than steel; they, however, do not offer significant advantages
over aluminum.
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Summary
Terminologies
Materials: a substance from which something is composed.
Engineering Materials: Materials used to produce technical products. Part of inanimate matter which is
useful to the engineer in the practice of his profession.
Materials Science: is primarily concerned with the search for basic knowledge about the internal
structure, properties, and processing of materials.
Materials engineering: is mainly concerned with the use of fundamental and applied knowledge of
materials so that the materials can be converted into products needed or desired by society.
Materials science and Engineering: is an interdisciplinary field of science and engineering that studies
and manipulates the composition and structure of materials across length scales to control materials
properties through synthesis and processing.
Structure: description of the arrangement of atoms in the materials.
Properties: refers to a material trait in terms of the kind and magnitude of response to a specific imposed
stimulus. Mechanical, electrical, magnetic, thermal, optical and deteriorative are the basic important
properties of materials.
Synthesis: refers to how materials are made from naturally occurring or man-made chemicals.
Processing: refers to how materials are shaped into useful components to cause changes in the properties
of different materials.
Classification of materials
Solid materials are classified into three basic categories. Metals, Ceramics and Polymers.
Metals are solid materials that are normally combinations of metallic elements. Typically hard, opaque,
shiny, and features good electrical and thermal conductivity.
Ceramics: compounds between metallic and nonmetallic elements and include compounds such as
oxides, nitrides, and carbides. Typically they are insulating and resistant to high temperatures and harsh
environments; Ceramics are strong and hard, but also very brittle.
Glasses and glass-ceramics: ASTM defined glass as “an inorganic product of fusion which has been
cooled to rigid condition without crystallization”.
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Glass-ceramics: formed by thermal processes with controlled crystallization of small crystals within
glass materials.
Polymeric materials:materials consisting of long molecular chains or networks of low weight
elements such as carbon, hydrogen, oxygen, and nitrogen. Most polymeric materials have low electrical
conductivities.
Semiconductors: type of materials in which electrical conductivity is intermediate between conductors
and insulators.Electrical properties are strongly dependent upon small amounts of impurities. Silicon,
germanium, and gallium arsenide-based semiconductors such as those used in computers and
electronics are part of a broader class of materials known as electronic materials.
Composites:are materials composed of the combination of at least two different materials for the seek
of distinctive desired property. The constituents keep unique properties and the overall composite will
have properties different from each of them.
Biomaterials:Nondrug materials that can be used to treat enhance or replace any tissue, organ, or
function in an organism.
Smart-materials: a recent advanced environmental sensitive material which can be composed any of
the above material. It consists sensor to stimulate the environmental change and actuators for the
respond.
Nanomaterials: are materials that have characteristic scale length less than 100nm.
References
1) Materials science and Engineering an introduction; William. D. Callister, Jr. David.G. Rethwisch 8th ed. (2012).
2) The science and Engineering of Materials; Donald R. Askeland, Pradeep P.Fulay, Wendelin J.Wright.
6th ed (2010).
3) An introduction to materials Engineering and science; Brain S. Mitchell,3rd,(2004)
4) Pro. Hae-Geon Lee, extractive metallurgy lecture note
5) Dr. Kalid Ahmed, fundamentals of ceramics lecture note
6) Prof. Heo Jong, glass and glass-ceramics lecture note
7) Prof. GM.Kim , biomaterials lecture note
8) Mechanical behavior of materials, MareMeyers and Krishan Chawla 2nd ed. (2009).
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Problems
1) Identify whether it is engineering materials or not from the following alternatives.
a) Beach sand
d) Gel
b) Laser
e) Forklift
c) Cotton
f) Pin
2. Define the following terminologies with less than 25 words.
a) Engineering materials
f) Metal
b) Properties
g) Sensor
c) Synthesis
h) Piezoelectric material
d) Composition
i) Actuator
e) Structure
j) Memory shape
3. Compare and contrast the twins given below.
a) Metals, ceramics
b) glass, glass-ceramics
d) Materials science, materials engineering
c) ceramics, polymers
e) materials scientist, materials Engineer
4. What do we mean by the phrase “information age”?
5. Compare and contrast differences and similarities of “Structures”, “Devices” and “Machines “as option
categories of engineering materials.
6. Among elements of this set which one is the latest? {Bronze Age, Stone Age, Iron Age}
7. Cite the type of material which is playing the great role in information Age.
8. Concrete is the aggregation of course sand with cement and steel bar is used as reinforcement as the same time.
What do you think about why people do like that?
9. (a) In what class of materials does GaAs (gallium arsenide) belong? (b) What are its desirable properties? (c)
What are its applications in electronic industries?
10. Nickel-base superalloys are used in the structure of aircraft turbine engines. What are the major properties of
this metal that make it suitable for this application?
11. A new Research Park which is the first in its kind in Ethiopia had been constructing at Adama Science and
Technology University since this material was being written. The construction company charged to construct this
Research Park was Tekleberhan p.l.c. Project Company. It was observed that the majority workers of the company
occasionally wore protective helmet hat at their work place. (a) Name the important criteria for selecting materials
to use in a protective helmet. (b) Identify materials that would satisfy these criteria. (c) Why would a solid metal
helmet not be a good choice?
17
By Gizachew Berhanu
12. Assume you are a materials engineer working for a certain fiber optics cable company in Addis Ababa. The
owner of the company assigns you to design a fiber optics cable as long as the distance between earth and moon. a)
What type of raw materials would you expect the company should provide? b) Estimate the radius of the fiber if the
company provides 5g amount of raw material with density of water? c) What kind of solid materials would you
categorize the new fiber cable you designed and what type of recent advanced technology the company you are
working for is developing?
13. Suppose you would like to produce a composite material based on aluminum having a density of 1.5 g/cm3.
Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.56
g/cm3, into the aluminum be a likely possibility? Explain. (Aluminum has a density of 2.7 g/cm3).
14. Write one paragraph about why single crystal silicon is currently the material of choice for microelecronics
applications. Write a second paragraph about potential alternatives to single-crystal silicon for solar cell
applications. Provide a list of the references or websites that you used. You must use at least three references.
15. Ethio Telecom displaced that currently it could provide mobile telephone service to approximately 50 million
subscribers. In contrary, more than 80% of the people in the country are rural area dwellers where the utility of
electricity is a series problem. Many people must go the most probable nearest town to recharge their mobile phone
and this again delay their work time or the mobile will be switched off until they will have a time to go and
recharge it. To solve this problem at least partially, the eastern Zone industry found at Dukem town wishes to
manufacture shoe sole that could generate electricity during just walking to at least enable the peoples to recharge
their mobile phone from walking energy. Assume the manager of the industry raise this idea to you a material
engineer to suggest him what type of material he could select and related working principles. Write the appropriate
material type that could success the industry and aided this people by recharging their cell phone from their shoes
sole during walking.
16. An insurance company in Addis Ababa daily faced to repay for its members due to the destructive property and
lost human life they exposed by car accident. The company gets into a series anxiety that it will run with empty
capital if the situation continues with the current rate. Many times the accident could occur due to car collision and
the plastic part of the car remains safety and the metal part would be deformed. Due to this, the company determine
to decrease at least the expense for destructed car by order future car’s metal part must made from advanced metal
alloy that could remember its original shape and easily fixed if some aid like temperature is introduced from outside
after deformation and sign an agreement with TOYOTA car manufacturing company. Write the appropriate
material that satisfy the desired condition to safe the insurance company from critical crises it would worried about.
18
By Gizachew Berhanu
Chapter Two: Structure of Atoms
2.1 Introduction
In chapter one, we discussed that primitive peoples often limited to the naturally occurring materials in their
environment and historical eras have been closely associated with materials from which important objects have
been made. As civilization and technology have developed, the range of engineering materials expanded. The Stone
Age gave way to the Bronze Age, which in turn was followed by the Iron Age. These labels were chosen much
later, through the lens of history, and it may be dangerous to try to characterize our own time period. But it isn’t
hard to imagine that future archaeologists or historians might label the late 20th century and early 21 st century as
the Polymer and semiconductor Age. We also added that the success or failure of many engineering activities
depends on the selection of engineering materials whose properties match the specific requirements of application.
If the match was not a good one, compromises were required.
Materials could now be processed and their properties could be altered and possibly enhanced. The alloying or heat
treatment of metals or firing of ceramics is examples of techniques that can be substantially alter the properties of a
material. Fewer compromises were required and enhanced design possibilities emerged. Products became more
sophisticated. While the early successes in altering materials were largely the results of trial and error, we now
recognize that the engineering properties of a material are a direct result of the structure of that material. Changes in
properties, therefore, are the direct change in the material structure.
Since all materials are made up of the same basic components- protons, neutrons and electrons, it is amazing that so
many different materials exist with such widely varying properties. This variation is explained by many possible
combinations of these units in a macroscopic assembly. The subatomic particles, listed above, combine in different
arrangements to form the various elemental atoms, each having a nucleus of protons and neutrons surrounded by a
proper number of electrons to maintain charge neutrality.
The arrangement of electrons surrounding the nucleus affects the electrical, magnetic and optical properties as well
as how the atoms bond to one another. Atomic bonding then produces a higher level of structure, which may be in
the form of molecule, crystal or amorphous aggregate. This structure and the imperfections that may be present
have a profound effect on mechanical properties. As a result of the ability to control structures through processing
and the ability to develop new structures through techniques such as composite materials, engineers now have at
their disposal a wide variety of materials with an almost unlimited range of properties. The properties of these
materials depend on all levels of their structure from subatomic to macroscopic. Therefore, it is important for us,
the engineers to understand the entire structure spectrum and the way the basic structure and changes in that
structure will affect properties. For the sake of this thorough understanding of structure-property relationship, this
chapter introduces atomic structure, electrons in atoms, periodic table and atomic bonding that you might be
familiar with them from your previous general chemistry. Good reading, go ahead!
19
By Gizachew Berhanu
Objectives of this chapter
At the end of this chapter, the student should be able to

Understand about the discovery of electron, proton and neutron and their characteristics

Describe Thomson, Rutherford, and Bohr atomic models

Understand dual nature of electrons

Understand the important features of the quantum model of atoms

Understand nature of electromagnetic radiation and Planck’s quantum theory

Explain the photoelectric effect and describe features of atomic spectra

State the de Broglie relation and Heisenberg uncertainty principle

Define an atomic orbital in terms of quantum numbers

State Aufbau principle, Pauli exclusion principle, and Hund’s rule of maximum multiplicity

Write electronic configurations of atoms

Understand why atoms are react

Explain the formation of different types of bonds

Classify atomic bonding in terms of their nature

Calculate the bonding force that hold the whole crystals of the system

Explain the different types of carbon hybridization
2.2 Fundamental concepts of atoms
What is an atom? Why study atom?
History of atom
 A Greek philosopher in 500B.C. Democritus coined a term atom meaning “uncuttable.” According to this
hypothesis if you have mango and you cut your mango and keeping cut it repeatedly and to the last point you
will find extremely small mango which you cannot cut. Your last mango that you are unable to cut further is
regarded as an atom.

In 1805, almost 2304 years later John Dalton proposed atomic theory. Unlike Democritus, John Dolton had
some experiment to prove the existence of atom. His introduction of atomic theory marks the inception of a
modern era in chemical thinking. According to this theory, all matter is composed of very small particles called
atoms. The atoms were regarded to be structureless, hard, impenetrable particles which cannot be subdivided.
Dalton’s ideas of the structure of matter were born out by a considerable amount of subsequent experimental
evidences towards the end of the nineteenth century.
 In 1850 Faraday found cathode ray tube.
20
By Gizachew Berhanu
 In 1904 Sir J. J. Thomson proposed the first definite theory as to the internal structure of the atom. According
to this theory the atom was assumed to consist of a sphere of uniform distribution of about 10−10 𝑚 positive
charge with electrons embedded in it such that the number of electrons equal to the number of positive charges
and the atom as a whole is electrically neutral. This model of atom could account the electrical neutrality of
atom, but it could not explain the results of gold foil scattering experiment carried out by Rutherford.
 In 1909 Rutherford conducted gold foil experiment and proposed new atomic model. He conducted a scattering
experiment in 1911 to find out the arrangement of electrons and protons. He bombarded a thin gold foil with
particles emanating from radium.
 In 1912, Bohr proposed new atomic model. Though offering a satisfactory model for explaining the spectra of
hydrogen atom, could not explain the spectra of multi-electron atom. In Bohr model an electron is regarded as
charged particle.
 In 1924, De Broglie hypothesis talks about dual nature of electron.
 In 1926, Schrödinger proposed an equation called Schrödinger equation to describe the electron distributions in
space and the allowed energy levels in atoms.
 In 1927, Heisenberg proposed Heisenberg uncertainty principle
 In 1932, Chadwick found neutron and quantum model of atom which is mostly correct and recent widely
accepted model. The entire above model had issues incorrect model and the only correct model is quantum
model.
What is an atom?
The atom is a basic unit of matter that consists of a dense central nucleus surrounded by a cloud of negatively
charged electrons. Most of the atom, about 99.9% is empty space. Early in the twentieth century, it has been proved
that an atom consists of smaller particles such as electrons, protons and neutrons. The proton, a positively charged
particle, is present in the central part of the atom called nucleus. The electron, a negatively charged particle, is
present around the nucleus. The neutron, a neutral particle, is also present in the nucleus of the atom. Since the
atom is electrically neutral, the number of positive charges on the nucleus is exactly balanced by an equal number
of orbital electrons. Every electron carries a charge of −1.602 × 10 -19C, whereas every proton carries a charge of
+1.602 × 10−19 C. So for an atom to remain neutral, the numbers of electrons and protons must be equal. Because
neutrons have no charge, the number of neutrons present is not restricted by the requirement for electrical
neutrality. For most elements, the number of neutrons can vary from one atom to another, as we will see.
21
By Gizachew Berhanu
2.2.1 Atomic Number (Z) and Atomic Mass (A)
The presence of positive charge on the nucleus is due to the proton in the nucleus. The number of protons present in
the nucleus is equal to atomic number (Z). Please make sure that atomic number is always equal to number of
protons but not always number of electron though it is the case if the atom is neutral.
𝐀𝐭𝐨𝐦𝐢𝐜 𝐧𝐮𝐦𝐛𝐞𝐫 (𝐙) = 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐩𝐫𝐨𝐭𝐨𝐧𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐧𝐮𝐜𝐥𝐞𝐮𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐚𝐭𝐨𝐦
Most of the mass of the atom is contained within the nucleus. Mass of the nucleus is due to protons and neutrons.
Protons and neutrons present in the nucleus are collectively known as nucleons. The total number of nucleons is
termed as mass number (A) of the atom. The mass of each proton and neutron is 1.67 x10-24 g, but the mass of
each electron is only 9.11x10-28 g or mass of proton is equal to about 1833 times mass of electron. Atomic mass is
also the mass in grams of the Avogadro constant NA of atoms. The quantity NA = 6.022 X 1023 atoms per mol is
the number of atoms or molecules in a mole. Therefore, the atomic mass has units of g/mol. An alternative unit for
atomic mass is the atomic mass unit, or amu, which is 1/ 12 the mass of carbon 12 (i.e., the carbon atom with
twelve nucleons—six protons and six neutrons). As an example, one mole of iron contains 6.022 X 1023 atoms and
has a mass of 55.847 g, or 55.847 amu. Calculations including a material’s atomic mass and the Avogadro’s
constant are helpful to understanding more about the structure of a material.
𝐌𝐚𝐬𝐬 𝐧𝐮𝐦𝐛𝐞𝐫 (𝐀) = 𝐌𝐚𝐬𝐬 𝐨𝐟 𝐩𝐫𝐨𝐭𝐨𝐧 + 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐧𝐞𝐮𝐭𝐫𝐨𝐧
Atom representation =
𝐀
𝐙𝐗
where X represent an element
Isobars and isotopes
Isobars are the atoms with same mass number but different atomic number. For instance
14
6C
and 147N. Here note
that the number of proton is different and the property is also different since number of proton determines the
property of every atom.
Isotopes are atoms with same atomic number but different atomic mass.
12
13
6C, 6C
and
14
6C
are isotopes of carbon
element. Atomic number of isotope is same. Means that numbers of protons are the same. Since protons assign the
property of atom, properties of isotopes of an element is same. All the isotopes of a given element reflect same
chemical properties due to chemical properties of atoms are controlled by number of electrons which are
determined by number of protons in the nucleus. Numbers of neutrons present in the nucleus have very little effect
on the chemical properties of an element.
The atomic mass is defined as the average mass of an atom of a particular element. Carbon has two stable isotopes
with masses of 12.0000 and 13.0036 amu, respectively. So why is the average mass 12.011 and not something
closer to 12.5? The answer is that when we take the average mass, we must account for the relative abundance of
each isotope. For carbon, the fact that we only need to consider two stable isotopes makes the calculation fairly
simple. We can multiply the mass by the fractional abundance to weight each isotope’s contribution to the atomic
mass.
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By Gizachew Berhanu
Carbon-12: 12.0000 × 0.9893 = 11.87
Carbon-13: 13.0036 × 0.0107 = 0.139
Weighted average mass = 11.87 + 0.139 = 12.01.
The value of 12.011 found in the periodic table is obtained using additional significant figures on the isotopic
abundance numbers.
Example 2.1
Calculate the number of protons, neutrons and electrons in 80
35𝐵𝑟
Solution
In the question we are given that
𝐴
𝑍𝑋
=
80
35𝐵𝑟
∴ Atomic number = Z = 35 and
Atomic mass = A = 80
Number of proton = 35
Since the given element, bromine is neutral
Number of electron = number of proton = 35
Atomic mass = number of proton + number of neutron
Number of neutron = atomic mass – number of proton
Number of neutron = 80 – 35 = 45.
Example 2.2
The number of electrons, protons and neutrons in species are equal to 18, 16 and 16 respectively. Assign the
proper symbol to the species.
Solution
From the question, we have
Number of electron = 18
Number of proton = 16 and
Number of neutron = 16
∴ Atomic number (Z) = number of proton = 16
∴ Atomic mass (A) = number of proton + number of neutron = 16 + 16 = 32
NOTE: 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 ≠ 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑟𝑜𝑡𝑜𝑛, 𝑡ℎ𝑒𝑛 𝑎𝑛 𝑎𝑡𝑜𝑚 𝑖𝑠 𝑐ℎ𝑎𝑟𝑔𝑒𝑑
Charge = number of proton – number of electron = 16 – 18 = -2
∴
32 −2
16𝑋
An element whose atomic mass is 32 and atomic number 16 is sulfur.
∴
32 −2
16𝑆
23
By Gizachew Berhanu
Example 2.3
a) Calculate the number of electrons which will together weight one gram and b) calculate the mass and charge
of one mole of electron.
Solution
a) We know that
Mass of one electron = 9.11x10-28gm
Mass of X electrons = 1gm
𝑋 𝑛𝑢𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑜𝑛𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑥 1𝑔𝑚
9.11𝑥10−28 𝑔𝑚
= 1.098𝑥1027
b) 1mole of electron = 6.022x1023electrons
9.11x10-28gm = X mass of one mole of electron
𝑋=
6.022𝑥1023 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑥 9.11𝑥10−28 𝑔𝑚
𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
= 5.48𝑥10−7 𝑘𝑔
𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑚𝑜𝑙𝑒 = 6.022𝑥1023 𝑥1.6𝑥10−19 𝐶 = 9.65𝑥104 𝐶
Example 2.4
Find a) the total number of neutrons and b) the total mass of neutrons in 7mg of
14
6𝐶 .
(Given that mass of a
-24
neutron = 1.67x10 gm).
Solution
a) First let us calculate number of neutrons present in one atom of carbon
14
6𝐶
Number of neutron = atomic mass – atomic number = 14-6 = 8 neutrons.
1mole of 146𝐶 = 14𝑔𝑚 𝑜𝑓 146𝐶 = 6.022𝑥1023 𝑎𝑡𝑜𝑚𝑠 𝑜𝑓 146𝐶
= 6.022𝑥1023
14
6𝐶
𝑥 8 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 = 4.802 𝑥1024 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
7mg =?
Total number of neutrons in 7mg =
7𝑚𝑔 𝑥 4.802 𝑥 1024
14000𝑚𝑔
= 2.409 𝑥 1021 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
b) 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑜𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 𝑥 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑖𝑛𝑔𝑙𝑒 𝑛𝑒𝑢𝑡𝑟𝑜𝑛
2.409 𝑥 1021 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 𝑥 1.67x 10−24 gm = 4.03 x 10−6 Kg
Example 2.5
Calculate the total number of electron present in one mole of methane.
Solution
One molecule of methane (C𝐻4 ) contains 4 electrons from hydrogen + 6 electrons from carbon = 10 electrons
One mole of methane = 6.022𝑥1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝐻4
1 electron = 6.022𝑥1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝐻4
One mole of methane = 6.022𝑥1023 𝑥 10 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 = 6.022𝑥1024 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
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By Gizachew Berhanu
Example 2.6
Find a) total number of proton and b) total mass of protons in 34mg of 𝑁𝐻3 at STP. Will the answer change if
temperature and pressure are change?
Solution
a) One molecule of 𝑁𝐻3 contains 3 proton from hydrogen + 7 proton from nitrogen = 10 protons.
1mole of 𝑁𝐻3 = 17gm of 𝑁𝐻3 = 6.022 𝑥 1023 𝑥 10 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 = 6.022 𝑥 1024 𝑝𝑟𝑜𝑡𝑜𝑛𝑠
Total number of proton in 34mg =
34𝑚𝑔 𝑥 6.02 𝑥1024
17000𝑚𝑔
= 1.204 𝑥1023 𝑝𝑟𝑜𝑡𝑜𝑛𝑠
b) Total mass = 1.67 𝑥 10−24 𝑔𝑚 𝑥 1.204 𝑥1023 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 = 2.017 𝑥10−5 𝐾𝑔
c) Will the answer change if temperature and pressure are change? No! Temperature and pressure doesn’t
matter.
Example 2.7
The chlorine present in PVC has two stable isotopes.
35
17𝐶𝑙
With a mass of 34.97 amu makes up 75.77% of the
natural chlorine found. The other isotope is 37
17𝐶𝑙 whose mass is 36.95 amu. What is the atomic mass of chlorine?
Solution
To determine the atomic mass, we must calculate the average mass weighted by the fractional abundance of
each chlorine isotope. Because there are only two stable isotopes, their abundances must add up to 100%. So we
35
can calculate the abundance of 37
17𝐶𝑙 from the given abundance of 17𝐶𝑙 .
First, we calculate the abundance of the chlorine-37 isotope:
𝐴𝑏𝑢𝑑𝑒𝑛𝑐𝑒 𝑜𝑓 37
17𝐶𝑙 = 100% − 75.77% = 24.23%
𝑤𝑒𝑖𝑔ℎ𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑚𝑎𝑠𝑠 = 35 𝑥 75.77% + 37 𝑥 24.23% = 35.45
Example 2.8
There are three naturally occurring isotopes of the element silicon, which is widely used in producing computer
chips. Given the masses and abundances below, calculate the atomic mass of silicon.
Isotope
𝟐𝟖
𝟏𝟒𝑺𝒊
Abundance
92.2%
Mass
27.977amu
𝟐𝟗
𝟏𝟒𝑺𝒊
4.67%
28.977amu
𝟑𝟎
𝟏𝟒𝑺𝒊
3.10%
29.974amu
Solution
The principle is the same as before we did in example 2.7.
𝐴𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑖𝑙𝑖𝑐𝑜𝑛 = 27.977 𝑥 92.2% + 28.977 𝑥 4.67% + 29.974 𝑥 3.10% = 28.077.
By Gizachew Berhanu
25
Example 2.9
Materials engineer has filed for a patent for a new alloy to be used in golf club heads. The composition by
mass ranges from 25 to 31% manganese, 6.3 to 7.8% aluminum, 0.65 to 0.85% carbon, and 5.5 to 9.0%
chromium, with the remainder being iron. What are the maximum and minimum percentages of iron possible
in this alloy?
Solution: - The summation of fraction components that built the alloy is equal to 100%
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒𝑠 𝑜𝑓 𝑖𝑟𝑜𝑛 = 100% − (25 + 6.3 + 0.65 + 5.5)% = 100% − 37.45% = 62.55%
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒𝑠 𝑜𝑓 𝑖𝑟𝑜𝑛 = 100% − (31 + 7.8 + 0.85 + 9.0 )% = 100% − 48.65% = 51.35%
Example 2.10
The element gallium, used in gallium arsenide semiconductors, has an atomic mass of 69.72 amu. There are
only two isotopes of gallium,
69
31𝐺𝑎
with a mass of 68.9257 amu and
71
31𝐺𝑎
with a mass of 70.9249 amu. What
are the isotopic abundances of gallium?
Solution
This problem is the inverse of those we solved in example 2.7 and 2.8 in which we used isotopic abundances to
find atomic masses. Here we must work from the atomic mass to find the isotopic abundances. Since there are
two isotopes, we have two unknowns. So we will need to write two equations that relate the percentages or
fractions of
69
31𝐺𝑎
and 71
31𝐺𝑎. We would also need to know the mass of each isotope, but presumably we could
look those up. The abundances of the two isotopes must add to 100%. This gives us a first equation:
69
𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 31
𝐺𝑎 + 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 71
31𝐺𝑎 = 100%
69
If we substitute 𝑋1 for fraction of 31
𝐺𝑎 and 𝑋2 for fraction of 71
31𝐺𝑎, then
𝑋1 + 𝑋2 = 100 …………. (1)
𝐴𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 = 68.9257𝑋1 + 70.9249𝑋2 = 69.72 ……….. (2)
But from equation (1) we have 𝑋2 = 100 − 𝑋1
Substitute this in equation (2) and multiply both sides by 100, we will get
{68.9257𝑋1 + 70.9249 (100 − 𝑋1 )}𝑥100 = 6972
Solution for the two variables is then:
𝑋1 = 60.26%
𝑋2 = 39.74%
26
By Gizachew Berhanu
2.2.2 Faraday Cathode Ray Tube
In 1850, Faraday began to study electrical discharge in partially evacuated tubes known as cathode ray discharge
tubes. Cathode ray tube is made of glass containing two thin pieces of metal, called electrodes sealed in it. The
electrical discharge through the gases could be observed only at very law pressures and at very high voltages. If
sufficient high voltage is applied across the electrodes, current starts flowing through a stream of particles moving
in the tube from the negative electrode (cathode) to the positive electrode (anode). The flow of current from cathode
to anode was further checked by making a hole in the anode and coating the tube behind anode with phosphorescent
material zinc sulfide. When these rays, after passing through anode, strike the zinc sulfide coating a bright spot on
the coating is developed.
Fig.2.1 Faraday cathode ray tube
Observations
The cathode rays start from cathode and move towards the anode. These rays themselves are not visible but their
behavior can be observed with the help of certain kind of materials (fluorescent or phosphorescent) which glow
when hit by them. In the absence of electric or magnetic field, these rays travel in straight lines. In the presence of
electric or magnetic field, the behaviors of cathode rays are similar to that expected from negatively charged
particles, suggesting that the cathode rays consist of negatively charged particles called electrons. The
characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas
present in the cathode ray tube.
2.3 Developments leading to Bohr Model
Dual character of the electromagnetic radiation and experimental results regarding atomic spectra were the two
important issues that Rutherford was unanswered. Before we discuss about Bohr model, we need to familiarize with
both those two issues.
27
By Gizachew Berhanu
2.3.1 Dual Character of the Electromagnetic Radiation
What do we mean by Dual character? Dual character means electromagnetic radiations possess both wave like and
particle like properties. Diffraction and interference are common properties of wave whereas black body radiation,
photoelectric effect, variation of heat capacity of solids as a function of temperature and line spectra of atoms with
special reference to hydrogen are particle nature of electromagnetic radiation.
2.3.1.1 Wave nature of Electromagnetic Radiation
James Maxwell (1870) suggested that when electrically charged particle moves under acceleration, alternating
electric and magnetic fields are produced and transmitted. These fields are transmitted in forms of waves called
electromagnetic or electromagnetic radiation. Example, Light is electromagnetic waves or electromagnetic
radiation. The electric and magnetic field components of an electromagnetic wave have the same wave length,
frequency, speed and amplitude, but they vibrate in two mutually perpendicular planes. It shows diffraction and
interference behavior. Unlike sound waves or water waves, electromagnetic waves do not require medium and can
move in vacuum. Units such as frequency (v) and wave length (λ) are used to represent electromagnetic radiation.
The SI unit for frequency is hertz. It is defined as the number of waves that pass a given point in one second.
Wavelength has the unit’s of length. SI unit of length is meter (m). Wave number (the number of wave lengths per
unit length) is also used to describe waves. There are many types of electromagnetic radiations, which one differ
from another in wavelength or frequency. These constitute what is called electromagnetic spectrum.
2.3.1.2 Particle Nature of Electromagnetic Radiation
 Variation of heat capacity of solids as a function of temperature
 Nature of emission of radiation from hot bodies (black body radiation)
 Ejection of electrons from metal surface when radiation strikes it (photo electric effect)
 Line spectra of atoms with special reference to hydrogen
Variation of heat capacity of solids as a function of temperature:
A phenomenon of black body radiation was given by Max Plank in 1900. When solids are heated, they emit
radiation over a wide range of wavelengths. When iron rod is heated in furnace, it turns to red. As it is heated
further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high. We
observed that the radiation emitted goes from a lower frequency to a higher frequency as the temperature increases.
Black Body Radiation
The ideal body which emits and absorbs all frequencies is called a black body and the radiation emitted by such a
body is called black body radiation. At a given temperature, intensity of emitted radiation increases with decrease
of wavelength, reaches maximum value at a given wavelength and then starts decreasing with further decreasing of
wavelength.
28
By Gizachew Berhanu
𝑇2
𝑇2 > 𝑇1
𝑇1
Wavelength
Fig. 2.2 intensity of emitted radiation increases with decrease of wavelength,
Planck gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of
electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency (v) and is
expressed by equation
E = nhv……………… (2.1), where v is frequency of the radiation and h is Planck’s constant 6.626x10-34Js.
Photoelectric effect
In 1887, Hertz performed an experiment in which electrons were ejected when certain metal (for example,
rubidium, cesium etc.) were exposed to a beam of light. The phenomenon is called photoelectric effect. The
electrons are ejected from the metal surface as soon as the beam of light strikes the surface. The number of
electrons ejected is proportional to the intensity (number of photons) or brightness of light. For each metal, there is
a characteristic minimum frequency Vo also known as threshold frequency below which photoelectric effect is not
observed. At a frequencyV > Vo , the ejected electrons come out with certain kinetic energy. The kinetic energies of
these electrons increase with the increases of frequency of light used. The kinetic energy will be generated doesn’t
depend on number of photons but depend on frequency of light used. For example, red light (V = 4.3 − 4.6) x
1014 Hz of any brightness (intensity) may shine on a piece of potassium metal for hours but no photoelectrons are
ejected due to the light frequency is less than threshold frequency. But as soon as even a very weak yellow light
(V = 5.1 − 5.2) x 1014 Hz shines on the potassium metal, the photoelectric effect is observed. The threshold
frequency (Vo ) for potassium metal is5.0 x 1014 Hz.
In 1905 Einstein was able to explain the photoelectric effect using Planck’s quantum theory of electromagnetic
radiation as starting point. Shining a beam of light on to a metal surface can therefore, be viewed as shooting a
beam of particles called photons. When a photon of sufficient energy strikes an electron in the atom of the metal, it
transfers its energy instantaneously to the electron during the collision and the electron is ejected without any time
lag or delay. Greater the energy possessed by the photon, greater will be the transfer of energy to the electron, and
greater the kinetic energy of the ejected electron. In other words, kinetic energy of ejected electron is proportional
to the frequency of the electromagnetic radiation. Since the striking photon has energy equal to hv and the
29
By Gizachew Berhanu
minimum energy required to eject the electron is hvo , then the difference in energy (hv − hvo ) is transferred as the
kinetic energy of the photoelectron.
1
mV 2
2
= h (V − Vo ) ……………………… (2.2)
1
𝑚𝑉 2
2
hv
v
Metal surface
a)
ℎ𝑣𝑜
b)
Fig. 2.3 photoelectric effect a) electrons are ejected from a surface of the metal when exposed to light of
frequency v in vacuum b) plot of the maximum kinetic energy of ejected electrons vs. frequency of the
incoming light
Line spectra
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. Atoms,
molecules or ions that have absorbed radiation are said to be excited. To produce an emission spectrum, energy is
supplied to a sample by heating it or irradiating it and the wavelength (or frequency) of the radiation emitted, as the
samples gives up the absorbed energy is recorded.
Conclusion
 Light has dual behavior. It behaves as waves as well as particles also.
 Microscopic particles like electrons also exhibit this particle-wave duality
Example 2.11
Calculate the energy of one mole of photons radiation whose frequency is5 𝑥 1014 𝐻𝑧.
Solution
In this section, we mentioned that energy of one molecule is given by
𝐸 = ℎ𝑣 , where h = Planck’s constant with value = 6.626 𝑥 10−34 𝐽. 𝑠 and 𝑣 is frequency of radiation
𝐸 = 6.626 𝑥 10−34 𝐽. 𝑠 𝑥 5.0 𝑥1014 𝑠 −1 = 3.313 𝑥 10−19 𝐽
From previous section we know that
1𝑚𝑜𝑙𝑒 = 6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
∴ 𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 1𝑚𝑜𝑙𝑒 = 6.022 𝑥 1023 ⁄𝑚𝑜𝑙𝑒 𝑥 3.313 𝑥 10−19 𝐽 = 199.51 𝐾𝐽⁄𝑚𝑜𝑙
30
By Gizachew Berhanu
Example 2.12
A 100wat bulb emits monochromatic light of wave length 400nm; calculate the number of photons emitted
per second by the bulb.
Solution
In the question we are given that
𝑃𝑜𝑤𝑒𝑟 = 100 𝐽⁄𝑠𝑒𝑐. = 100𝐽 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑠 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑖𝑛 𝑜𝑛𝑒 𝑠𝑒𝑐𝑜𝑛𝑑.
𝑊𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ = 400𝑛𝑚. And we are asked to calculate number of photons emitted (n).
From Planck’s quantum equation we know that
𝐸 = 𝑛ℎ𝑣
𝑛=
𝐸
ℎ𝑣
𝑛=
100𝐽 𝑥 4.0 𝑥10−7 𝑚
6.626 𝑥 10−34 𝐽𝑠−1 𝑥 3 𝑥 108 𝑚⁄𝑠
=
𝐸𝜆
ℎ𝑐
= 2.0 𝑥1020 𝑝ℎ𝑜𝑡𝑜𝑛𝑠/𝑠𝑒𝑐..
Example 2.13
When electromagnetic radiation of wavelength 300nm falls on the surface of sodium, electrons are emitted
with a kinetic energy of1.68 𝑥 105 𝐽⁄𝑚𝑜𝑙𝑒. a) What is the minimum energy needed to remove an electron
from sodium? b) What is the maximum wavelength that will cause a photoelectron to be emitted?
Solution
Given
𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ = 300𝑛𝑚
𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 =
1
𝑚𝑉 2
2
= 1.68 𝑥 105 𝐽⁄𝑚𝑜𝑙
We need to calculate
a) ℎ𝑉𝑜
b) 𝜆𝑚𝑎𝑥.
a) From equation (2.2), we have
1⁄ 𝑚𝑣 2 = ℎ(𝑣 − 𝑣 )
𝑜
2
1⁄ 𝑚𝑣 2 = ℎ𝑣 − ℎ𝑣
𝑜
2
ℎ𝑣𝑜 = ℎ𝑣 − 1⁄2 𝑚𝑣 2
From the given wavelength, we should calculate corresponding energy first
𝑐
ℎ𝑣 = ℎ 𝜆 = 6.626 𝑥 10−34 𝐽. 𝑠 𝑥
3.0 𝑥108 𝑚⁄𝑠𝑒𝑐.
3.0 𝑥 10−7 𝑚
= 6.626 𝑥10−19 𝐽
We need to convert this energy to joule per mole as we did in example 2.11 above since the kinetic energy
in which electron is emitted is given with this unit.
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By Gizachew Berhanu
ℎ𝑣 = 6.022 𝑥1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠⁄𝑚𝑜𝑙 𝑥 6.626 𝑥 10−19 𝐽⁄𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 = 3.99 𝑥 105 𝐽⁄𝑚𝑜𝑙𝑒
ℎ𝑣𝑜 = 3.99 𝑥 105 𝐽⁄𝑚𝑜𝑙 − 1.68 𝑥 105 𝐽⁄𝑚𝑜𝑙 = 2.31 𝑥 105 𝐽⁄𝑚𝑜𝑙
We often call such kind of minimum energy as work function.
b) 𝜆𝑚𝑎𝑥
Since wavelength is inversely proportional to the correspondence frequency, the maximum wavelength is
related with the minimum frequency which is a threshold frequency which again tied in its turn to minimum
energy required to remove electrons from the surface of the metal. But, here note that the unit per mol must
converted back to per molecule.
ℎ𝑣𝑜 ⁄𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 =
𝜆𝑚𝑎𝑥 =
2.31 𝑥 105 𝐽/𝑚𝑜𝑙
6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒/𝑚𝑜𝑙
3.84 𝑥 10−19 𝐽
ℎ𝑐
=
= 3.84 𝑥10−19 𝐽
3.84 𝑥 10−19 𝐽
3.0 𝑥 108 𝑚⁄𝑠𝑒𝑐.
6.626 𝑥 10−34 𝐽.𝑠 𝑥
= 517𝑛𝑚
Example 2.14
The threshold frequency 𝑣𝑜 , for a metal is 7.0 𝑥 1014 𝑠 −1 . Calculate the kinetic energy of an electron
emitted when radiation of frequency 𝑣 = 1.0 𝑥 1015 𝑠 −1 hits the metal.
Solution
From equation (2.2), we know that
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 = ℎ(𝑣 − 𝑣𝑜 ) Where,
𝑣 = 1.0 𝑥 1015 = 10.0 𝑥 1014 𝑠 −1
𝑣𝑜 = 7.0 𝑥 1014 𝑠 −1
𝐾. 𝐸 = 6.626 𝑥 10−34 𝐽. 𝑠(10.0 𝑥 1014 − 7.0 𝑥 1014 )𝑠 −1 = 1.988 𝑥 10−19 𝐽
Example 2.15
The Ethiopian Broadcasts Corporation (EBC) station from Addis Ababa is on frequency of 1,368 kHz.
Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the
electromagnetic spectrum does it belong to?
Solution
𝑣 = 1,368𝑘𝐻𝑧 = 1.368 𝑥 106 𝐻𝑧
𝜆=
𝑐
𝑣
=
3.0 𝑥 108 𝑚⁄𝑠
1.368 𝑥 106 /𝑠
= 219.3𝑚 , it belongs to radio wave.
32
By Gizachew Berhanu
Example 2.16
The wavelength range of the visible spectrum extends from violet (400nm) to red (750nm). Express these
wavelengths in frequencies.
Solution
𝑣1 =
𝑐
𝜆1
=
3.0 𝑥 108 𝑚⁄𝑠
4.0 𝑥 10−7 𝑚
= 7.5 𝑥 1014 𝐻𝑧
𝑣2 =
𝑐
𝜆2
=
3.0 𝑥 108 𝑚⁄𝑠
7.5 𝑥 10−7 𝑚
= 4.0 𝑥 1014 𝐻𝑧
Example 2.17
Calculate wave number and frequency of yellow radiation having wavelength 5800Å.
Solution
𝑤𝑎𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 𝑘 =
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑣 =
𝑐
𝜆
1
𝜆
=
=
1
5800Å
= 1.724 𝑥 106 𝑚−1
3.0 𝑥 108 𝑚⁄𝑠
5800Å
= 5.12 𝑥 1014 𝐻𝑧
Example 2.18
Calculate wavelength, frequency and wave number of light wave whose period is 2.0 𝑥10−10 𝑠𝑒.
Solution
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 =
1
𝑝𝑒𝑟𝑖𝑜𝑑
𝜆=
𝑐
𝑣
=
3.0 𝑥 108 𝑚⁄𝑠
5.0 𝑥 109 𝑠−1
𝑘=
1
𝜆
=
1
0.06𝑚
=
1
2.0 𝑥10−10 𝑠𝑒
= 5.0 𝑥 109 𝐻𝑧
= 6.0 𝑥 10−2 𝑚
= 16.67𝑚−1
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By Gizachew Berhanu
Example 2.19
A photon of wavelength 4.0 𝑥 10−7 𝑚 strikes on a metal surface and the work function of the metal
being2.23𝑒𝑉. Calculate a) the energy of the photon b) the kinetic energy of the emission and c) the velocity
of the photoelectron. (1Ev = 1.602 x 10-19J)
Solution
This example is same with example 2.13 with slightly deferent given quantities
𝑐
𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛 = 𝐸𝑝ℎ = ℎ𝑣 = ℎ 𝜆 = 6.626 𝑥 10−34 𝐽. 𝑠 𝑥
a)
3.0 108 𝑚⁄𝑠
4.0 𝑥 10−7 𝑚
= 4.9695 𝑥10−19 𝐽 =
3.102𝑒𝑉.
b) 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐾. 𝐸 = ℎ𝑣 − ℎ𝑣𝑜 = 3.102𝑒𝑉 − 2.23𝑒𝑉 = 0.97𝑒𝑉.
c) 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 =
2 𝑥 1.554 𝑥 10−19 𝐽
9.11 𝑥 10−31 𝐾𝑔
𝑉= √
1
𝑚𝑣 2
2
= 0.97 𝑥 1.602 𝑥 10−19 𝐽 = 1.554 𝑥 10−19 𝐽
= 5.84 𝑥 105 𝑚⁄𝑠
Self-test
1) Electromagnetic radiation of wavelength 242nm is just sufficient to ionize the sodium atom. Calculate the
ionization energy of sodium in KJ/mol. (𝑎𝑛𝑠𝑤𝑒𝑟 494.7𝐾𝐽/𝑚𝑜𝑙).
2) A 25watt bulb emits monochromatic yellow light of wavelength of 0. 57nm. Calculate the rate emission
of quanta per second. (Answer7.169 𝑥 1019 𝑝ℎ𝑜𝑡𝑜𝑛𝑠/𝑠𝑒𝑐.)
3) Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of
wavelength
6800Å.
Calculate
threshold
frequency
(𝑣𝑜 )
and
work
function
of
the
metal.
(Answer4.41 𝑥 1014 𝐻𝑧, 1.82𝐸𝑣)
2.4 Bohr atomic model
Bohr uses the emission spectrum of hydrogen to develop a model for hydrogen. He postulated that electron
moves around the nucleus in circular orbits. His model determines the radius of each of circular electron
orbits. Bohr calculated the energy of electron in various orbits and for each orbit predicted the distance
between the electron and nucleus. He offered a satisfactory model for explaining the spectra of hydrogen atom.
According to Bohr only certain orbits can exist and each orbit corresponds to specific energy. Electron energy
is quantized. Equivalently, it assumed that the orbiting electrons could take on only certain (quantized) value
of angular momentum L.
Fig. 2.4 Bohr’s model of atom
34
By Gizachew Berhanu
Photo emission
An excited atom relaxes from high energy to low energy by emitting a photon. We can determine the energy
c
difference (∆E) between levels by measuring the wavelength of the emitted photon. ∆E = h λ.
Continuous and quantized spectrum
When we say continuous spectrum ∆E has any value. But when we say quantized spectrum ∆E has only certain
values.
∆𝐸
∆𝐸
b)
a)
Fig. 2.5 continuous vs. quantized spectrum a) any ∆𝐸 is possible for continuous spectrum and b) Only
certain energies are allowed
For hydrogen atom whose atomic number is one, the mathematical expression for Bohr’s atomic postulate is
given by
Ln = mn vrn = nℏ … … … … … … … (2.3) where n = 1, 2, 3, 4, …
Ln = angular momentum of orbiting electron ,ℏ =
h
,
2π
mn = mass of n number of orbiting electrons ,
v = linear velocity of electron and
rn = radius of the electron from the nuclues for a given value of n
From equation (2.3) above, we will have
mo v =
nℏ
rn
……………………….. (2.4)
Since the electron orbits are assumed to be stable or at equilibrium, the centripetal force on the electron must
precisely balance the columbic attraction between the nucleus and the orbiting electron.
e2
4πϵo r2n
=
mo v2
rn
=
(mo v)2
mo
=
e2
4π∈o rn
…………….. (2.5)
Substituting equation (2.4) into equation (2.5) yields
(nℏ)2
mo rn
rn =
=
e2
4π∈o
4π∈o (nℏ)2
mo e2
………………… (2.6)
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By Gizachew Berhanu
If we again rearrange equation (2.5) and divided both sides by two, we will get
1
m v2
2 o
=
e2
… … … … (2.7)
8π∈o rn
Kinetic energy which is a part of total energy
And the second part of total energy is potential energy
P .E =
∞ drn
e2
∫
4π∈o 0 r2n
e2
= − 4π∈
o rn
………….. (2.8)
Therefore,
total energy = En = K. E + P. E =
e2
8π∈o ro
−
e2
4π∈o rn
=
e2
1
( −
4π∈o rn 2
1
e2
)
2 4π∈o rn
1) = − (
If we substitute for rn from equation (2.6), then we will have the total energy interims of n
1
e2
m e2
) ( o 2 2)
2 4π∈o
4π∈o n ℏ
En = − (
= −
1 mo e4
2 (4π∈o nℏ)2
= -13.6ev/n2 ………. (2.9)
Where
mo = Mass of stationary electron and
n = 1, 2, 3 … principal quantum number
The evidence for the quantized electronic energy level is atomic spectra. Bohr’s theory can also be applied to
the ions containing only one electron, similar to that present in hydrogen atom. For exampleHe+ , Li2+ , Be3+
and so on. The energies of the stationary states associated with these kinds of ions also known as hydrogen like
species are given by the expression stated in equation (2.9). It is possible to calculate the velocities of electrons
moving in these orbits. Magnitude of the velocity of electron increases with increase of the positive charge on
the nucleus and decreases with increase of principal quantum number. In Bohr model of hydrogen why the
total energy is negative?
In equation (2.9) above, the negative sign means that the energy of the electron in the atom is lower than the
energy of a free electron at rest. A free electron at rest is an electron that is infinitely far away from the nucleus
and is assigned the energy value of zero. When the electron is free from the influence of nucleus, the energy is
taken as zero. The electron in this situation is associated with the stationary state of principal quantum number
= n= ∞ and is called as ionized hydrogen atom. When the electron is attracted by the nucleus and is present in
orbit n, the photon is emitted and the electron loses the energy. That is the reason for the presence of negative
sign.
∆𝐸 = 𝐸𝑛 − 𝐸∞
Fig. 2.6 electron falling from infinity state to n orbital
36
By Gizachew Berhanu
Line spectrum of hydrogen
Further, each spectral line, whether in emission or absorption spectrum can be associated to the particular
transition in hydrogen atom. If you have large number of hydrogen atoms, different possible transitions can be
observed and thus leading to large number of spectral lines. The brightness or intensity of spectral lines
depends upon the number of photons of the same wavelength or frequency absorbed or emitted. According to
Bohr, radiation (energy) is absorbed if the electron moves from the orbit of smaller principal quantum number
to the orbit of higher quantum number, whereas the radiation (energy) is emitted if the electron moves from
higher orbit to lower orbit. The energy gap between the two orbits is given by equation
∆E = Ef − Ei . ..................... (2.10)
Stability of atom: - for stability of atoms Bohr gave explanation that in a particular orbit energy is conserved.
Only where electron jumps from one orbit to another, energy is either emitted or absorbed. No loss of energy
for particular orbit.
Bohr model of atom limitation
 Mainly for hydrogen
 Could not explain the spectra of multi-electron atoms
 In Bohr model, an electron is regarded as a charged particle moving in a well-defined circular orbit about the
nucleus. The wave character of the electron is ignored in Bohr’s theory. Does not follow De Broglie’s concept
of wave-particle duality for matter.
 Contradicts Heisenberg uncertainty principle
 Zeeman Effect (changes in spectral lines due to external magnetic fields) could not be explained.
 Stark effect (changes in spectral lines due to external electric fields) could not be explained
Example 2.20
How much energy is required to ionize a hydrogen atom if the electron occupies n = 5 orbits?
Compare your answer with the ionization enthalpy of hydrogen atom (amount of energy required to remove
the electron from n = 1 orbit)
Solution
It is given that the electron is present in n = 5 orbits currently and attracted from where it was free at infinity.
a) 𝐸∞ =
𝐸5 =
−13.6
∞2
−13.6
52
=0
= −0.544
∆𝐸 = 𝐸𝑓 − 𝐸𝑖 = 𝐸5 − 𝐸∞ = −0.544𝑒𝑣 − 0 = −0.544𝑒𝑣
b) ∆𝐸 = 𝐸1 − 𝐸∞ = −13.6𝑒𝑣
If we compare these tow energies it is more difficult to remove electron from n=1 orbit that at n=5 orbit.
37
By Gizachew Berhanu
Example 2.21
What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops
to the ground state?
Solution
If you see the electron may goes from
6 → 5, 6 → 4, 6 → 3, 6 → 2, 6 → 1 = 5 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑏𝑡𝑖𝑜𝑛𝑠.
5 → 4, 5 → 3, 5 → 2, 5 → 1 = 4 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑏𝑡𝑖𝑜𝑛𝑠
4 → 3, 4 → 2, 4 → 1 = 3 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑏𝑡𝑖𝑜𝑛𝑠
3 → 2, 3 → 1 = 2 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑏𝑡𝑖𝑜𝑛𝑠
2 → 1 = 1 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑏𝑡𝑖𝑜𝑛𝑠
𝑡𝑜𝑡𝑎𝑙 = 5 + 4 + 3 + 2 + 1 = 15 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑏𝑡𝑖𝑜𝑛𝑠
So the maximum emission lines is given by
𝑛(𝑛−1)
2
In our case 𝑛 = 6
∴
6(6−1)
2
= 15
Heisenberg uncertainty principle
Werner Heisenberg in 1927, stated that it is impossible to determine simultaneously, the exact position and exact
momentum (or velocity) of an electron.
∆X x ∆Px ≥
h
4π
. ………………. (2.11)
This principle is basically works for microparticles or small particles such as electrons or protons. It rules out
existence of definite paths or trajectories of electrons and other similar particles. It therefore, means that the precise
statements of the position and momentum of electrons have to be replaced by the statements of probability that the
electron has at a given position or momentum. The effect of Heisenberg uncertainty principle significant only for
motion of microscopic objects and is negligible for that of macroscopic objects.
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By Gizachew Berhanu
Example 2.22
A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1Å.
What is the uncertainty involved in the measurement of its velocity?
Solution
In the question we are given that
∆𝑋 = 0.1Å
𝑚𝑒 = 9.11 𝑥 10−31 𝐾𝑔
∆𝑋 𝑥 ∆𝑃𝑥 =
ℎ
4𝜋
∆𝑋𝑚𝑒 ∆𝑉 =
ℎ
4𝜋
∆𝑉 =
ℎ
4𝜋𝑚𝑒 ∆𝑋
, but ∆𝑃𝑥 = 𝑚𝑒 ∆𝑉
=
6.626 𝑥 10−34 𝐽.𝑠
4𝜋(9.11 𝑥 10−31 𝐾𝑔)(0.1 𝑥 10−11 𝑚)
= 5.79 𝑥 106 𝑚⁄𝑠𝑒𝑐.
Example 2.23
A golf ball has a mass of 40gm and speed of 45m/se. if the speed can be measured within accuracy of 2%,
calculate the uncertainty in the position.
Solution
In the question we are given that
𝑚 = 40𝑔𝑚 = 4.0 𝑥 10−2 𝐾𝑔
𝑉 = 45 𝑚⁄𝑠𝑒
∆𝑉 = 2% 𝑜𝑓 𝑠𝑝𝑒𝑒𝑑 =
∆𝑋𝑚∆𝑉 =
∆𝑋 =
2
100
𝑥 45 𝑚⁄𝑠𝑒 = 0.9 𝑚⁄𝑠𝑒
ℎ
4𝜋
ℎ
4𝜋𝑚∆𝑉
=
6.626 𝑥 10−34
4𝜋(4.0 𝑥 10−2 𝐾𝑔)(0.9𝑚⁄𝑠𝑒)
= 1.46 𝑥 10−33 𝑚
Wave-Particle duality for matter
De Broglie in 1924 proposed matter like radiation should also exhibit dual behavior. This means that just as the
photon has momentum as well as wavelength, electrons should also have momentum as well as wavelength. Then
De Broglie equated equation (2.1) with Albert Einstein equation of energy and solved for momentum and wave
length.
𝐸 = 𝑚𝑐 2 = ℎ𝑣
But we know that 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑣 =
𝑐
𝑚𝑐 2 = ℎ 𝜆 → 𝑚𝑐 =
𝐶
𝜆
ℎ
𝜆
Where C = velocity of light
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By Gizachew Berhanu
If we consider particle velocity instead of photon’s and replace C by V, then we will have
𝑚𝑣 =
ℎ
𝜆
→ 𝜆=
ℎ
𝑚𝑣
=
ℎ
𝑃
………………………… (2.12)
Where P = mv = momentum
The wavelengths associated with ordinary objects are so short due to their large mass that their wave properties
cannot be detected. The wavelengths associated with electrons and other subatomic particles with very small mass
can however be detected experimentally.
Example 2.24
What will be the wavelength of a ball of mass 0.1Kg moving with a velocity of 10𝑚⁄𝑠𝑒𝑐 ?
Solution
In the question it is given that
𝑚 = 0. 𝐾𝑔 and 𝑣 = 10 𝑚⁄𝑠𝑒
From equation (2.12), we have
𝜆=
ℎ
𝑚𝑣
𝜆=
6.626 𝑥 10−34 𝐽.𝑠
(0.1𝐾𝑔)(10𝑚⁄𝑠)
= 6.626 𝑥 10−34 𝑚
Example 2.25
The mass of an electron is 9.11 𝑥 10−31 𝐾𝑔. If its kinetic energy is 3.0 𝑥 10−25 𝐽, calculate its wavelength
Solution
Given
𝐾. 𝐸 = 3.0 𝑥 10−25 𝐽
𝑚 = 9.11 𝑥 10−31 𝐾𝑔
𝜆=?
𝜆=
ℎ
𝑚𝑣
We are given mass of the electron and we know the value of Planck’s constant and we need to calculate the
excepted velocity of electron from its kinetic energy
1
𝑚𝑣 2
2
𝐾. 𝐸 =
= 3.0 𝑥 10−25 𝐽
2𝐾.𝐸
𝑚
𝑉= √
𝜆=
ℎ
2𝐾.𝐸
𝑚
𝑚√
=
ℎ
√2𝐾.𝐸 𝑥 𝑚
=
6.626 𝑥 10−34
√2 𝑥 3.0 𝑥 10−25 𝑥 9.11 𝑥 10−31 𝐾𝑔
= 8.957 𝑥 10−7 𝑚
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By Gizachew Berhanu
Example 2.26
Calculate the mass of photon with wavelength 3.6Å.
Solution
𝑚𝑣 =
ℎ
𝜆
Since the issue we are discussing is about the photon the velocity is speed of light
𝑚=
ℎ
𝜆𝐶
=
6.626 𝑥 10−34 𝐽.𝑠
3.6Å 𝑥 3.0 𝑥 108 𝑚⁄𝑠
= 6.135 𝑥 10−29 𝐾𝑔
2.5 Quantum Mechanics
Classical mechanics, based on Newton’s law of motion, successfully describes the motion of macroscopic objects
such as a falling stone, orbiting planets etc, which have essentially a particle-like behavior. However, the principles
of classical mechanics do not provide the correct description of physical processes if very small length or energy
scales such as electrons, atoms, molecules, etc. are involved. This is mainly because of the fact that classical
mechanics ignores the concept of dual nature of matter especially for subatomic particles and the uncertainty
principle. Classical or Newtonian mechanics allows a continuous spectrum of energies (fig. 2.5) and allows
continuous spatial distribution of matter. For example, coffee is distributed homogeneously within a cup.
The branch of science that takes into account this dual behavior of matter is called quantum mechanics. Quantum
mechanical distributions are not continuous but discrete with respect to energy, angular momentum, and position.
For example, the bound electrons of an atom have discrete energies and the spatial distribution of the electrons has
distinct maxima and minima, that is, they are not homogeneously distributed. Quantum-mechanics does not
contradict Newtonian mechanics. When quantum mechanics is applied to macroscopic objects for which wave-like
properties are insignificant, the results are the same as those from classical mechanics. As will be seen, quantum
mechanics merges with classical mechanics as the energies involved in a physical process increase. In the classical
limit, the results obtained with quantum mechanics are identical to the results obtained with classical mechanics.
Quantum mechanics was developed independently 1926 by Werner Heisenberg and Erwin Schrödinger.
In classical or Newtonian mechanics the instantaneous state of a particle with mass m is fully described by the
particle’s position [x(t), y(t), z(t)] and its momentum [ Px (t), Py (t), Pz (t)]. For the sake of simplicity, we consider a
particle whose motion is restricted to the x-axis of a Cartesian coordinate system. The position and momentum of
the particle are then described by x(t) and P(t) = Px (t). The momentum P(t) is related to the particle’s velocity
v(t) by P(t) = m v(t) = m [dx(t) / dt] . It is desirable to know not only the instantaneous state-variables x(t)
and P(t), but also their functional evolution with time. Newton’s first and second law enables us to determine this
functional dependence.
Newton’s first law states that the momentum is a constant, if there are no forces acting on the particle, i. e.
P(t) = mv(t) = m
dx(t)
dt
= constant ………….. (2.13)
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By Gizachew Berhanu
Newton’s second law relates an external force, F, to the second derivative of the position x(t) with respect to t,
F=m
d2 x(t)
dt2
= ma ………………………… (2.14)
Where a is the acceleration of the particle.
Newton’s first and second law provide the state variables x(t) and P(t) in the presence of an external force.
Equation (2.14) is a form of second order differential equation in which we can solve with carry out two
integrations. You need to revise your previous mathematics course if you have forgotten how to solve ordinary
differential equation.
𝑑2 𝑥
𝑑𝑡 2
=𝑎
1) If we let 𝑈 =
𝑑𝑥
𝑑𝑡
= 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑡ℎ𝑒𝑛
𝑑𝑈
𝑑𝑡
=𝑎
𝑑𝑈 = 𝑎𝑑𝑡
∫ 𝑑𝑈 = 𝑎 ∫ 𝑑𝑡 → 𝑈(𝑡) = 𝑎𝑡 + 𝐶1
Where 𝐶1 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑝𝑎𝑟𝑡 𝑜𝑛𝑒
If we know that at 𝑡𝑜 the particle has velocity 𝑣𝑜 , then we have the additional relation
𝑈(𝑡𝑜 ) = 𝑎𝑡𝑜 + 𝐶1 = 𝑉𝑂
𝐶1 = 𝑉𝑂 − 𝑎𝑡𝑜
𝑈(𝑡) = 𝑉𝑜 + 𝑎(𝑡 − 𝑡𝑜 ) = 𝑉(𝑡)
2) From part one integration we know that
𝑈(𝑡) =
𝑑𝑥(𝑡)
𝑑𝑡
= 𝑉(𝑡)
𝑥(𝑡) = ∫ 𝑉(𝑡)𝑑𝑡
𝑥(𝑡) = ∫[𝑉𝑜 + 𝑎(𝑡 − 𝑡𝑜 )] 𝑑𝑡
𝑥(𝑡) = 𝑉𝑜 𝑡 +
1
𝑎𝑡(𝑡
2
− 2𝑡𝑂 ) + 𝐶2
Again if we know that at 𝑡𝑜 the particle has a position 𝑥𝑜 , then we have another additional relation
1
2
𝑥(𝑡𝑜 ) = 𝑉𝑜 𝑡𝑜 − 𝑎𝑡𝑜2 + 𝐶2 = 𝑥𝑜
𝐶2 = 𝑥𝑜 − 𝑉𝑜 𝑡𝑜 +
1
𝑎𝑡 2
2
Hence, 𝑥(𝑡) = 𝑥𝑜 + 𝑉𝑜 (𝑡 − 𝑡𝑜 ) +
1
𝑎(𝑡
2
− 𝑡𝑜 )2
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By Gizachew Berhanu
Example 2.27
Abenezar is a high roof public car driver continuously travelling from Adama to Addis Ababa on expressway.
His car which was at rest at Adama bus station set off on the road exactly at 12: 15 A.M and stop again at
Addis Ababa Kality bus station at 1: 00 P.M. Calculate the velocity of the car and the displacement between
Adama and Addis Ababa if the acceleration of the car is given by equation
𝑎 = 4.9 − 0.0325𝑉(𝑡).
Solution
In the question it is given that the car was continuously moving and never stop anywhere on the middle of the
way. Therefore, it was moving with accelerated velocity for 45 minutes which equal to 2700sec.
a) 𝑎 =
𝑑𝑉(𝑡)
𝑑𝑡
𝑑𝑉(𝑡)
𝑑𝑡
= 4.9 − 0.0325𝑉(𝑡)
= 0.0325(147.7 − 𝑉(𝑡))
𝑑𝑉(𝑡)
147.7−𝑉(𝑡)
= 0.0325𝑑𝑡
𝑡
𝑑𝑉(𝑡)
∫ 147.7−𝑉(𝑡) = ∫𝑡 0.0325𝑑𝑡
𝑜
−ln(147.7 − 𝑉(𝑡)) = 0.0325 (𝑡 − 𝑡𝑜 )
𝑉(𝑡) = 147.7 − 𝑒 −0.0325(𝑡− 𝑡𝑜 )
𝑉(2700𝑠𝑒𝑐. ) = 147.7 − 𝑒 −87.75 𝑚⁄𝑠𝑒𝑐. = 147.7 𝑚⁄𝑠𝑒.
b) 𝑥(𝑡) = 𝑥𝑜 + 𝑉𝑜 (𝑡 − 𝑡𝑜 ) +
1
𝑎(𝑡
2
− 𝑡𝑜 )2
𝑥𝑜 = 0, 𝑉𝑜 = 0, Since the vehicle starts from its stationary position
𝑥(𝑡) =
1
𝑎(𝑡
2
− 𝑡𝑜 )2 =
1
2
1
{(4.9 − 0.0325𝑉(𝑡))(𝑡 − 𝑡𝑜 )2 } = {2 (4.9 − 0.0325 ∗ 147.7)(2700)2 } =
363.6𝐾𝑚.
2.5.1 Energy
Newton’s second law is the basis for the introduction of work and energy. Work done by moving a particle along
the x axis from 0 to x by means of the force F(x) is defined as
x
W(x) = ∫0 F(x)dx ……………………. (2.15)
The energy of the particle increases by the (positive) value of the integral is given in Eq. (2.15). The total particle
energy, E, can be
(i) Purely potential,
(ii) Purely kinetic, or
(iii) A sum of potential and kinetic energy.
43
By Gizachew Berhanu
If the total energy of the particle is a purely potential energy, U(x), then W(x) = − U(x) and one obtains from Eq.
(2.15)
d
F = − dx U(x) ……………………………… (2.16)
If, on the other hand, the total energy is purely kinetic, Eq. (2.14) can be inserted in the energy equation, Eq. (2.15),
and one obtains
K. E =
1
mv 2
2
=
P2
2m
……………………………. (2.17)
If no external forces act on the particle, then the total energy of the particle is a constant and is the sum of potential
and kinetic energy
Etotal = K. E + P. E =
P2
+
2m
U(x) ………………………….. (2.18)
Consider a classical particle, e. g. a ball, positioned on one of the two slopes of the potential, as shown in Fig. 2.6.
Once the ball is released, it will move downhill with increasing velocity, reach the maximum velocity at the bottom,
and move up on the opposite slope until it comes to a momentary complete stop at the classical turning point. At
the turning point, the energy of the ball is purely potential. The ball then reverses its direction of motion and will
move again downhill. In the absence of friction, the ball will continue forever to oscillate between the two classical
turning points. The total energy of the ball, i. e. the sum of potential and kinetic energy remains constant during the
oscillatory motion as long as no external forces act on the object.
Fig.2.7 potential energy as a function of
spatial coordinate x. the total energy of the
particle shown is the sum of kinetic and
kinetic energies. The force acting on the
particle is the negative derivative of the
potential energy with respect to x.
2.5.2 Hamiltonian formulation of Newtonian mechanics
Equations (2.13) and (2.41) are known as the Newtonian formulation of classical mechanics. The Hamiltonian
formulation of classical mechanics has the same physical content as the Newtonian formulation. However, the
Hamiltonian formulation focuses on energy. The Hamiltonian function H(x, p) is defined as the total energy of a
system
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By Gizachew Berhanu
𝐻(𝑥, 𝑃) = 𝐸𝑡𝑜𝑡𝑎𝑙 =
𝑃2
2𝑚
+ 𝑈(𝑥) ………………………….. (2.19)
The partial derivatives of the Hamiltonian function with respect to x and P are given by
𝜕𝐻(𝑥,𝑃)
𝜕𝑥
=
𝑑
𝑈(𝑋)
𝑑𝑥
𝜕𝐻(𝑥,𝑃)
𝜕𝑃
=
𝑃
𝑚
Employing these partial derivatives and equations (2.13) and (2.16), one obtains two equations, which are known as
the Hamiltonian equations of motion:
d
 F = − dx U(x) = −
dP
dt
= −
∂H(x,P)
∂x
∂H(x,P)
∂x
dx
dt
=
∂H(x,P)
∂P
dP
dt
…………………………… (2.20)
 P = mv(x) → v(x) =
∴
=
dx
dt
=
P
m
=
∂H(x,P)
∂P
……………………………. (2.21)
Formally, the linear Equation (2.13) and the linear, second order differential Eq. (2.14) have been transformed into
the two linear, first order partial differential Equations (2.20) and (2.21). Despite this formal difference, the
physical content of the Newtonian and the Hamiltonian formulation is identical.
2.5.3 Schrödinger Equation
The formulation of quantum mechanics, also called wave mechanics focuses on the wave function, Ψ(x, y, z, t),
which depends on the spatial coordinates x, y, z, and the time t. In the following sections we shall restrict ourselves
to one spatial dimension x, so that the wave function depends solely on x. An extension to three spatial dimensions
can be done easily. The Schrödinger equation is the key equation of quantum mechanics. He provided a formulation
of wave mechanics to describe the motion of electrons, based on the principles of quanta and the wave-particle
duality. The one-dimensional Schrödinger equation is used when the particle of interest is confined to one spatial
dimension, for example the x axis. Due to the one-dimensional nature of many semiconductor heterostructures, the
one-dimensional Schrödinger equation is sufficient for most applications. To derive the one dimensional
Schrödinger equation, we start with the total-energy equation. From classical wave equation we have
𝑌(𝑥, 𝑡) = 𝐴𝑒 −𝑖(𝜔𝑡−𝑘𝑥) …………… (2.22)
Where
A = Amplitude
𝜔 = angular velocity
K=
𝜔
𝐶
x
= wave number and c = velocity of light
𝑖 = √−1 = 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑛𝑢𝑚𝑏𝑒𝑟
If ψ is introduced instead of Y to express quantum function and then given by
𝜓(𝑥, 𝑡) = 𝐴𝑒 −𝑖(𝜔𝑡−𝑘𝑥) ……………………….. (2.23)
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By Gizachew Berhanu
𝑘=
𝜔
𝐶
2𝜋𝑓
𝜆𝑓
=
=
2𝜋
𝜆
Equation (2.23) is state function or wave function and it depend on both time and position.
𝑥
𝜓(𝑥, 𝑡) = 𝐴𝑒 −2𝜋𝑖(𝑓𝑡− 𝜆)
Where again 𝑓 =
𝐸
ℎ
𝑎𝑛𝑑 𝜆 =
𝐸𝑡
𝜓(𝑥, 𝑡) = 𝐴𝑒 −2𝜋𝑖( ℎ −
Where ℏ =
𝑃𝑥
)
ℎ
ℎ
𝑃
2𝜋𝑖
= 𝐴𝑒 − ℎ
(𝐸𝑡−𝑃𝑥)
= 𝐴𝑒
−𝑖(𝐸𝑡−𝑃𝑥)
ℏ
……………………. (2.24)
ℎ
2𝜋
i) Derivate equation (2.24) twice with respect to position and keeping time constant yields
𝑑𝜓
𝑑𝑥
𝑖𝑃
𝜓
ℏ
=
−ℏ2 𝑑 2 𝜓
𝑑𝑥 2
and
𝑑2 𝜓
𝑑𝑥 2
=
−𝑃2
𝜓
ℏ2
…………. (Note: 𝑖 2 = −1)
= 𝑃2 𝜓 ………………………………… (2.25)
ii) Derivative equation (2.24) once again with respect to time and fix the position results
𝑑𝜓
𝑑𝑡
−𝐸𝑖𝜓
ℏ
=
→
−ℏ𝑑𝜓 𝑖
()
𝑖𝑑𝑡
𝑖
=
𝑖ℏ𝑑𝜓
𝑑𝑡
= 𝐸𝜓 ……………….. (2.26)
From equation (2.18) we know that the total energy is given by
𝐸=
𝑃2
2𝑚
+ 𝑈(𝑥)
Multiplying both sides of this equation by ψ gives
𝐸𝜓 =
𝑃2 𝜓
+
2𝑚
𝑈𝜓 …………………………………… (2.27)
Substituting equations (2.25) and (2.26) into equation (2.27) and simplify gives
ℏ2 𝑑 2 𝜓
2𝑚𝑑𝑥 2
+
𝑖ℏ𝑑𝜓
𝑑𝑡
− 𝑈𝜓 = 0 ………………………………… (2.28)
Equation (2.28) is known as Schrödinger equation. For a system such as an atom or molecule whose energy
does not change with time, the Schrödinger equation is written as
𝑃2
𝐸𝜓 = (2𝑚 + 𝑈) 𝜓
𝐸𝜓 = 𝐻𝜓 ……………………………………… (2.29), where 𝐻 =
𝑃2
2𝑚
+ 𝑈 a mathematical operator called
Hamiltonian, E is energy and ψ is a wave function that contains information of electron.
Schrodinger gave a recipe of constructing this operator from the expression for the total energy of the system. The
total energy of the system takes into account the kinetic energies of all sub-atomic particles (electrons-nuclei),
attractive potential between the electrons and nuclei and repulsive potential among the electrons and nuclei
individually. Solution of this equation gives E and ψ wave function pronounced as “Psi”. It is a mathematical
symbol to describe the properties of small particles.
46
By Gizachew Berhanu
Time Independent Schrodinger Equation
Schrodinger equation can be time independent but cannot be position independent. Let’s rewrite equation (2.24)
again and expand it.
−𝑖
𝜓(𝑥, 𝑡) = 𝐴𝑒 ℏ (𝐸𝑡−𝑃𝑥) = 𝐴𝑒 𝑃𝑥⁄ℏ . 𝑒 −𝑖𝐸𝑡⁄ℏ = 𝜓(𝑥)𝑒 −𝑖𝐸𝑡⁄ℏ
Equation (2.28) can be rewrite as same way with the above wave function
𝐸𝜓(𝑥). 𝑒 −𝑖𝐸𝑡⁄ℏ =
−ℏ2 𝑑 2 𝜓(𝑥) −𝑖𝐸𝑡⁄ℏ
.𝑒
2𝑚𝑥 2
+ 𝑈𝜓(𝑥). 𝑒 −𝑖𝐸𝑡⁄ℏ
Dividing both sides by 𝑒 −𝑖𝐸𝑡⁄ℏ results
𝐸𝜓(𝑥) =
𝑑 2 𝜓(𝑥)
𝑑𝑥 2
+
−ℏ2 𝑑 2 𝜓(𝑥)
2𝑚𝑑𝑥 2
2𝑚
(𝐸
ℏ2
+ 𝑈𝜓(𝑥)
− 𝑈)𝜓(𝑥) = 0 …………………………… (2.30)
Equation (2.30) is time independent Schrödinger equation.
Electron in free space: If there is no force acting on the particle, the potential function U(x) will be constant and E
> U(x). Assume U(x) = 0 for all x. If potential energy is zero, it can be simplified as the following
𝑑 2 𝜓(𝑥)
𝑑𝑥 2
+
2𝑚𝐸
𝜓(𝑥)
ℏ2
= 0 ……………………………… (2.31)
From equation (2.18) we know that
𝐸𝑡𝑜𝑡 =
𝑃2
2𝑚
+ 𝑈(𝑥)
Since 𝑈(𝑥) = 0 for electron in free space, then 𝐸𝑡𝑜𝑡 =
𝑃2
2𝑚
In equation (2.31) above
2𝑚
𝑃2
𝑃 2
4𝜋2
𝜆2
2𝜋 2
2𝑚𝐸
ℏ2
=
2𝑚𝐸
ℏ2
= 𝑘 2 ………………………………….. (2.32)
ℎ 2
( )
2𝜋
(2𝑚) = 4𝜋 2 (ℎ) =
= ( 𝜆 ) = 𝑘2
Now we can rewrite equation (2.31) again in the form of K as the following
𝑑2 𝜓
𝑑𝑥 2
+ 𝑘 2 𝜓(𝑥) = 0 …………………………………. (2.33)
The solution of this Schrodinger equation for U(x) = 0
Equation (2.33) is same form with second order differential equation and has the form of
𝑎𝜆2 + 𝑏𝜆 + 𝑐 = 0
Here in our case, 𝑏 = 0, 𝑎 = 1 𝑎𝑛𝑑 𝑐 = 𝑘 2 and reduced to the form
𝜆2 + 𝑘 2 = 0
𝜆2 = −𝑘 2
𝜆1 = 𝑖𝑘 𝑎𝑛𝑑 𝜆2 = −𝑖𝑘
General solution is then
47
By Gizachew Berhanu
𝜓(𝑥) = 𝐴𝑒 𝜆1 𝑥 + 𝐵𝑒 𝜆2 𝑥 = 𝐴𝑒 𝑖𝑘𝑥 + 𝐵𝑒 −𝑖𝑘𝑥
𝐸
You may recall that
𝑒 𝑖𝑘𝑥 = cos 𝑘𝑥 + 𝑖 sin 𝑘𝑥 and
𝑒 −𝑖𝑘𝑥 = cos 𝑘𝑥 − 𝑖 sin 𝑘𝑥
Question: what is the physical meaning of wave number (k)?
K
Let us look at equation (2.32) again
2𝑚𝐸
ℏ2
= 𝑘2
𝐸=
ℏ2 𝑘 2
2𝑚
=
Fig.2.8 energy vs. wave number graph
(ℏ𝑘)2
2𝑚
=
𝑃2
2𝑚
…………….. (2.34)
𝑃 = ℏ𝑘: 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑖𝑒𝑠 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 (𝑚𝑣)
𝑘=
2𝜋
𝜆
: Wave property of electron
A free particle with a well-defined energy has a well-defined wavelength and momentum
Free electron: continuous energy level → any energy allowed
2.5.4 Particle in the Box (Core electrons)
Consider a one dimensional box of length 𝑙 in which electron is assumed to being bounded in it. The potential
energy is zero in the box i.e. no any energy is there that protect an electron to move freely.
𝑈(𝑥)
𝑈(𝑥) = ∞
𝑈(𝑥) = ∞
𝑈(𝑥) = 0
𝑒−
0
𝑙
A particle in the infinite potential well is a classic example of a bound particle. The potential U(x) is infinite and
the particle is assumed to exist in the box (0 < x < l).
Boundary condition
𝑈(𝑥) = 0, 𝑓𝑜𝑟 0 < 𝑥 < 𝑙 , inside the box
𝑈(𝑥) = ∞, 𝑓𝑜𝑟 𝑥 < 0 𝑎𝑛𝑑 𝑥 > 𝑙 , outside of the box
Case 1
𝑈(𝑥) = ∞ , then Schrödinger equation is
𝑑 2 𝜓(𝑥)
𝑑𝑥 2
+
2𝑚
(𝐸
ℏ2
− 𝑈)𝜓(𝑥) = 0
𝑑 2 𝜓(𝑥)
𝑑𝑥 2
+
2𝑚
(𝐸
ℏ2
𝑑 2 𝜓(𝑥)
𝑑𝑥 2
−
2𝑚
(∞)𝜓(𝑥)
ℏ2
− ∞)𝜓(𝑥) = 0
= 0
48
By Gizachew Berhanu
𝜓(𝑥) =
ℏ2 𝑑 2 𝜓(𝑥)
2𝑚𝑑𝑥 2 (∞)
=0
This shows that there is no any particle out of the box.
Case 2
Inside the box where 𝑈(𝑥) = 0
Schrodinger equation is reduced to equation (2.33)
𝑑2 𝜓
𝑑𝑥 2
+ 𝑘 2 𝜓(𝑥) = 0
The general solution for this equation is
𝜓(𝑥) = 𝐴𝑒 𝑖𝑘𝑥 + 𝐵𝑒 −𝑖𝑘𝑥
We have two possible initial values on the boundaries at 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 𝑙
𝜓(0) = 𝐴 + 𝐵 = 0 ……………. (a)
𝜓(𝑙) = 𝐴𝑒 𝑖𝑘𝑙 + 𝐵𝑒 −𝑖𝑘𝑙 = 0 ……….. (b)
Substitute (a) into (b)
𝜓(𝑙) = 𝐴(𝑒 𝑖𝑘𝑙 − 𝑒 −𝑖𝑘𝑙 ) = 0
𝐴(cos 𝑘𝑙 + 𝑖 sin 𝑘𝑙 − (cos 𝑘𝑙 − 𝑖 sin 𝑘𝑙)) = 0
2𝑖𝐴 sin 𝑘𝑥 = 0
2𝑖𝐴 = 0 𝑜𝑟 sin 𝑘𝑙 = 0
But 2𝑖𝐴 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 ≠ 0, 𝑠𝑖𝑛𝑐𝑒 𝐴 = 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒, 𝑡ℎ𝑒 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑡ℎ𝑒𝑟𝑚 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑡ℎ𝑒𝑟𝑓𝑜𝑟𝑒, sin 𝑘𝑙 = 0
The zeros of the sine function occur at 0, ±𝜋, ±2𝜋, ±3𝜋, … … and hence
𝑘𝑙 = ±𝑛𝜋 ….. (n = 0, 1, 2, 3…..)
𝑘=
±𝑛𝜋
𝑙
………………… (2.35)
𝑛 = 𝑞𝑢𝑎𝑛𝑡𝑢𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑎𝑛𝑑 𝑞𝑢𝑎𝑛𝑡𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
𝑖𝑓 𝑛 = 0, 𝜓(𝑥) = 0 𝑛𝑜 𝑒𝑥𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑜𝑥
∴ 𝑛 = 0 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑒𝑔𝑛𝑜𝑟𝑑
If 𝜓 = 0, then nothing is in the universe with infinite energy; no particle present in the system. From equation
(2.34) we have
𝐸=
ℏ2 𝑘 2
2𝑚
=
ℏ2 (
±𝑛𝜋 2
)
𝑙
2𝑚
=
(
ℎ 2 2 2
) 𝜋 𝑛
2𝜋
2𝑚𝑙 2
=
𝑛2 ℎ 2
8𝑚𝑙 2
, where 𝑛 = 1, 2, 3, …
Conditions must be fulfilled: finite, ψ should be continuous, particle should have property at any time and
position, ψ should have single value.
49
By Gizachew Berhanu
Basic Concepts of Materials
Schrodinger equation for hydrogen
Solution of ψ, wave functions of hydrogen atom and hydrogen like species with one electron are called atomic
orbital. Probability of finding an electron at a point within an atom is proportional to |𝜓|2 at that point. When
Schrodinger equation is solved for hydrogen atom, the solution gives the possible energy levels that electron can
occupy and the corresponding wave function(s) ψ of the electron associated with each energy levels. The quantum
mechanical results of the hydrogen atom successfully predict all aspects of the hydrogen atom spectrum including
some phenomena that could not be explained by the Bohr model.
Application of Schrodinger equation to multi-electron atoms presents a difficulty. Schrodinger equation cannot be
solved exactly for multi-electron atom. This difficulty can be overcome by using approximate methods. Such
calculations with the aid of modern computers show that orbitals in atoms other than hydrogen do not differ in any
radical way from the hydrogen orbitals. The principal difference lies in the consequence of increased nuclear
charge. Because of this all the orbitals are somewhat contracted.
Quantum model feature
The energy of electrons in atoms is quantized. The existence quantized electronic energy levels is a direct result of
the wave like properties of electrons and are allowed solution of Schrodinger wave equation. Both the exact
position of and exact velocity of an electron in an atom cannot be determined simultaneously; the path of an
electron in an atom therefore, can never be determined or known accurately. An atomic orbital is the wave function
ψ for an electron in an atom. Since many such wave functions are possible for an electron, there are many atomic
orbitals in an atom. In each orbital, the electron has a definite energy and an orbital cannot contain more than two
electrons. In a multi-electron atom, the electrons are filled in various orbitals in the order of increasing energy.
All the information about the electron in an
atom is stored in its orbital wave function ψ
and quantum mechanics makes it possible to
extract this information out of ψ. The
probability of finding an electron at a point
within an atom is proportional to the square
of the orbital wave function i.e. |𝜓|2 at that
point. |𝜓|2 is known as probability density
and is always positive.
Fig.2.9 probability density of electrons
50
Gizachew Berhanu (student)
Basic Concepts of Materials
2.6 Orbitals and quantum numbers
A large number of orbitals are possible in an atom. 𝑠ℎ𝑒𝑙𝑙 → 𝑠𝑢𝑏𝑠ℎ𝑒𝑙𝑙 → 𝑜𝑟𝑏𝑖𝑡𝑎𝑙𝑠. Qualitatively, these orbitals
can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of
finding electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the
electron along certain directions than along others. Atomic orbitals are precisely distinguished by what are known
as quantum numbers. Each orbital is designated by three quantum numbers labeled as 𝑛, 𝑙 𝑎𝑛𝑑 𝑚𝑙 .
Table 2.1 Orbits and orbitals
Orbits (Bohr’s Model)
Orbitals (Quantum Model)
Doesn’t exist practically
Exists in real but no physical meaning
Well defined circular path around nucleus in which It is a quantum mechanical concept and refers to
electrons move
the one electron wave function 𝜓 in an atom. Three
dimensional space around which there is high
probability of finding electron
Contradicts dual nature of electron
Doesn’t contradicts dual nature of electron
Contradicts uncertainty principle
Doesn’t contradicts uncertainty principle
All orbits are circular like disc
They have various shape: s, p, d, f
Maximum number of electrons in an orbits is 2𝑛2
Maximum number of electrons in an orbital is 2
1) The principal quantum number (n)
The principal quantum number “n” is a positive integer with values of n = 1, 2, 3 …; the principal quantum number
determines the size and to large extent the energy of orbital. For hydrogen atom and hydrogen like species
(𝐻𝑒 + , 𝐿𝑖 2+ , … 𝑒𝑡𝑐) energy and size of the orbital depends only on ‘n. the principal quantum number also identifies
the shell. With the increase in the value ‘n’ the number of allowed orbital increases and are given by 𝑛2 . Maximum
number of electrons in a shell is given by 2𝑛2. All the orbitals of a given value of n constitute a single shell atom.
𝑛
= 1 2 3 4 . . . . . .
𝑠ℎ𝑒𝑙𝑙 = 𝐾 𝐿 𝑀 𝑁 . . . . . .
Size of orbital increases with increase of quantum number ‘n’. In other words the electron will be located away
from the nucleus. Since energy is required in shifting away the negatively charged electron from the positively
charged nucleus, the energy of the orbital will increase.
2) Azimuthal Quantum number (𝒍)
Azimuthal quantum number ‘𝑙’ is also known as angular quantum number or subsidiary quantum number. It defines
the three dimensional shape of the orbital. For a given value of 𝑛, 𝑙 can have 𝑛 values ranging from 0 𝑡𝑜 𝑛 − 1.
That is, for a given value of 𝑛 the possible value of 𝑙 are: 𝑙 = 0, 1, 2, … … . 𝑛 − 1
51
Gizachew Berhanu (student)
Basic Concepts of Materials
Each shell consists of one or more sub-shells or sub-levels. The number of sub-shells in principal shell is equal to
the values of 𝑛. Each sub-shell is designed an azitmuthal quantum number (𝑙).
𝑉𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑙:
0 1 2 3 4 5 . . . .
𝑛𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟: 𝑠 𝑝 𝑑 𝑓 𝑔 ℎ . . . .
𝑠𝑢𝑏 − 𝑠ℎ𝑒𝑙𝑙
Table 2.2 Notation for sub-shells
𝑛
1
2
𝑙
0
0
1
0
1
2
0
1
2
3
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
Sub-shell
3
4
notation
3) Magnetic orbital quantum number (𝒎𝒍 )
It gives information about the spatial orientation of the orbital with respect to standard set of coordinate axis. For
any sub-shell (defined by 𝑙 𝑣𝑎𝑙𝑢𝑒) 2𝑙 + 1 values of 𝑚𝑙 are possible and these values are given by: 𝑚𝑙 =
−𝑙, −(𝑙 − 1), −(𝑙 − 2) … , 0, 1 … , (𝑙 − 2), (𝑙 − 1), 𝑙
𝑙
0
1
2
3
4
5
Sub-shell notation
S
P
D
f
g
h
# of orbitals 2𝑙 + 1
1
3
5
7
9
11
4) Spin quantum number (𝒎𝒔 )
The three quantum numbers labeling an atomic orbital can be used equally well to define its energy, shape and
orientation. But all these quantum numbers are not enough to explain the line spectra observed in the case of
multi-electron atoms, that is some of the line actually occur in doublets (two lines closely spaced), triplets (three
lines closely spaced) etc. This suggests the presence of a few more energy levels than predicted by the three
quantum numbers.
52
Gizachew Berhanu (student)
Basic Concepts of Materials
In 1925, George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as
the electron spin quantum number (𝑚𝑠 ). An electron spins around its own axis, much in a similar way as earth
spins around its own axis. An electron has, besides charge and mass, intrinsic spin angular quantum number. Spin
angular momentum of the electron, a vector quantity, can have two orientations relative to the chosen axis. This
two chosen orientations are distinguished by the spin quantum numbers (𝑚𝑠 ) which can take the values of + 1⁄2
or − 1⁄2 . These are called the two spin states of the electron and are normally represented by two arrows ↑ spin
up and ↓ spin down. An orbital cannot hold more than 2 electrons and these 2 electros should have opposite spins.
Example 2.28
A certain one particle, one dimensional system has 𝜓 = 𝑎𝑒 −𝑖𝑏𝑡 𝑒 −𝑏𝑚𝑥
2 /ℏ
, where 𝑎 and 𝑏 are constants and 𝑚 is
mass of the particle. Find the potential energy function 𝑈. (Hint: Use the time dependent Schrodinger equation)
Solution
From equation (2.28) derived above, we know that time dependent Schrodinger equation is given by
ℏ2 𝑑 2 𝜓
2𝑚𝑑𝑥 2
+
𝑖ℏ𝑑𝜓
𝑑𝑡
− 𝑈𝜓 = 0 ………… (1)
And in the question, we are given that the one dimensional particle wave function equation is
𝜓(𝑥, 𝑡) = 𝑎𝑒 −𝑖𝑏𝑡 𝑒 −𝑏𝑚𝑥
2 /ℏ
…………… (2)
Derivate equation (2) two-times with respect to 𝑥 and once with respect to time to set up a similar equation
with Schrodinger equation form.
𝜓𝑥 =
𝜓𝑥𝑥 =
−2𝑏𝑚𝑥
𝜓
ℏ
−2𝑏𝑚
(𝑥 ′ 𝜓
ℏ
+ 𝑥𝜓 ′ ) =
−2𝑏𝑚
(𝜓
ℏ
+
−2𝑏𝑚𝑥 2
𝜓)
ℏ
4𝑏2 𝑚2 𝑥 2
ℏ2
= (
−
2𝑏𝑚
)𝜓
ℏ
……………. (3)
And derivative equation (2) w.r.t time gives
𝜓𝑡 = −𝑖𝑏𝜓 ……………….. (4)
Substitute equations (3) and (4) into equation (1)
ℏ2 4𝑏2 𝑚2 2
( 2 𝑥
2𝑚
ℏ
−
2𝑏𝑚
)𝜓
ℏ
+ 𝑖ℏ(−𝑖𝑏)𝜓 − 𝑈𝜓 = 0
𝑈 = 2𝑚𝑏 2 𝑥 2 − 𝑏ℏ + 𝑏ℏ
𝑈 = 2𝑚𝑏 2 𝑥 2
53
Gizachew Berhanu (student)
Basic Concepts of Materials
Example 2.29
A certain one particle, one dimensional system has the potential energy 𝑈 = 2𝐶 2 ℏ2 𝑥 2 ⁄𝑚 and is in stationary
2
state with 𝜓 = 𝑏𝑥𝑒 −𝑐𝑥 , where 𝑏 is a constant 𝑐 = 2.00𝑛𝑚−2 and 𝑚 = 1.00 𝑥 10−27 𝑔. Find the energy of the
particle.
Solution
From time-independent Schrodinger equation (2.30) we know that
𝑑 2 𝜓(𝑥)
𝑑𝑥 2
+
2𝑚
(𝐸
ℏ2
− 𝑈)𝜓(𝑥) = 0 ……......... (1)
And in the question we are given that
𝑈 = 2𝐶 2 ℏ2 𝑥 2 ⁄𝑚
𝜓 = 𝑏𝑥𝑒 −𝑐𝑥
2
𝑐 = 2.00𝑛𝑚−2
𝑚 = 1.00 𝑥 10−27 𝑔
Derivate twice the given function with respect to 𝑥
𝜓 = 𝑏𝑥𝑒 −𝑐𝑥
2
2
2
2
𝜓𝑥 = 𝑏(𝑒 −𝑐𝑥 − 2𝑐𝑥 2 𝑒 −𝑐𝑥 ) = 𝑏𝑒 −𝑐𝑥 − 2𝑐𝑥𝜓
′
2
2
𝜓𝑥𝑥 = (𝑏𝑒 −𝑐𝑥 − 2𝑐𝑥𝜓) = −2𝑐𝜓 − 2𝑐 (𝜓 + 𝑥(𝑏𝑒 −𝑐𝑥 − 2𝑐𝑥𝜓)) = −6𝑐𝜓 + 4𝑐 2 𝑥 2 𝜓 ………. (2)
Substitute (2) into (1)
−6𝑐𝜓 + 4𝑐 2 𝑥 2 𝜓 +
−6𝑐 + 4𝑐 2 𝑥 2 +
2𝑚𝐸
ℏ2
= 6𝑐
𝐸=
3𝑐ℏ2
𝑚
2𝑚
(𝐸
ℏ2
2𝑚𝐸
ℏ2
−
2𝐶 2 ℏ2 𝑥 2
)𝜓
𝑚
=0
− 4𝑐 2 𝑥 2 = 0
= 6.67 𝑥 10−20 𝐽
Example 2.30
What is the total number of orbitals associated with the principal quantum number 𝑛 = 3?
Solution
n
3
l
0
1
2
𝑚𝑙
0
-1 0 1
-2 -1 0 1 2
Total number of orbitals for the given value of quantum number 𝑛 is given by
𝑛 2 = 32 = 9
.
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Gizachew Berhanu (student)
Basic Concepts of Materials
2.6.1 Shapes of atomic orbitals
In previous section we have seen that the orbital wave functions or 𝜓 for an electron in an atom has no physical
meaning. It is simply a mathematical function of the coordinates of the electron. However, for different orbitals the
plots of corresponding wave functions as a function of 𝑟 (the distance from the nucleus) are different. Square of the
wave function (i.e.𝜓 2 ) at a point gives the probability density of the electron at that point. The region where this
probability density function reduces to zero is called nodal surfaces or simply nodes.
𝜓2
𝜓2
node
1s
nodes
+
2s
𝑟
+
𝑟
Fig.2.10 nodal surfaces (the point at which the probability of finding electron is almost zero)
Boundary surface diagram
Boundary surface diagrams constant probability densities for different orbitals give a fairly good representation of
the shapes of the orbitals. In these representations, a boundary surface or contour surface is drawn in space for an
orbital on which the value of probability density 𝜓 2 is constant. In principle many such boundary surfaces may be
possible. However, for a given orbital, only that boundary surface diagram of a constant probability density is taken
to be good representation of the shape of the orbital which encloses a region or volume in which the probability of
finding the electron is very high. Boundary surface diagram for 𝑆 orbital is actually sphere centered on the nucleus.
In two dimensions, these spheres looks like a circle as shown above.
Size of 𝑺 orbital
All 𝑆 orbitals are spherically symmetric. Size of the 𝑆 orbital increases with increase in principal quantum number
𝑛 that is, 4𝑆 > 3𝑆 > 2𝑆 > 1𝑆 and the electron is located further away from the nucleus as the principal quantum
number increases. Pictorially as shown above, a radial node appears to be spherical surface in which there is no
electron density and the sign of the wave function changes. Number of radial node: 𝑛 − 𝑙 − 1 radial nodes. For 𝑆
orbital 𝑙 = 0 and the radial node is equal to 𝑛 − 1. Number of angular node = 𝑙 nodes and for 𝑆 orbitals 𝑙 = 0
Size of 𝑷 orbitals
𝑃 orbital consists of two sections called lobes that are on either side of the plane that passes through the nucleus.
The probability density function is zero on plane where two lobes touch each other. The size, shape and energy of
the three orbitals are identical. They differ however, in the way the lobes are oriented. Since the lobes may be
considered to lie along 𝑥, 𝑦 and 𝑧 axis, they are given the designations 𝑃𝑥 , 𝑃𝑦 , and 𝑃𝑧 .
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Gizachew Berhanu (student)
Basic Concepts of Materials
Like 𝑆 orbitals, 𝑃 orbitals increase in size and energy with increase in the principal quantum number. The order of
the energy and size of various 𝑃 orbitals is 4𝑃 > 3𝑃 > 2𝑃. Like 𝑆 orbitals, the probability density functions for 𝑃
orbital also pass through value zero, besides at zero and infinite distance, as the distance from the nucleus increases.
The number of nodes are given by 𝑛 − 2 since value of 𝑙 is 1 for 𝑃 orbital. That is radial node is 0 for 2𝑃 orbital, 1
for 3𝑃 orbital, 2 for 4𝑃 orbital and so on. An angular node is usually a planar or conical surface in which there is no
electron density. The 𝑃 orbitals have one planar node through the center. Since the energy, 𝐸, of each orbital is a
function of only 𝑛 and then all the 𝑛 = 2 orbitals (2𝑆, 2𝑃𝑥 , 2𝑃𝑦 , 2𝑃𝑧 ) have the same energy.
𝒅 − orbitals
For 𝑙 = 2, the orbital is known as 𝑑 − 𝑜𝑟𝑏𝑖𝑡𝑎𝑙 and the minimum value of principal quantum number(𝑛) has to be 3
due to the value of 𝑙 cannot be greater than 𝑛 − 1. There are five 𝑚𝑙 values (-2, -1, 0, 1, 2) for 𝑙 = 2 and thus there
are five 𝑑 orbitals. All these five 3𝑑 orbitals are equivalent in energy. The 𝑑- orbitals for which 𝑛 is greater than 3
(4𝑑, 5𝑑 …) also have shapes similar to 3𝑑 orbital, but differ in energy and size. Besides the radial nodes, the
probability density functions for the 𝑛𝑃 and 𝑛𝑑 are zero at the plane(s) passing through the nucleus. There are 𝑙 or 2
angular nodes.
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑜𝑑𝑒𝑠 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑛𝑜𝑑𝑒(𝑙) + 𝑟𝑎𝑑𝑖𝑎𝑙 𝑛𝑜𝑑𝑒 (𝑛 − 𝑙 − 1) = 𝑛 − 1
Energies of orbitals: For hydrogen
The energy of an electron in a hydrogen atom is determined solely by principal quantum number. 1𝑠 < 2𝑠 = 2𝑝 <
3𝑠 = 3𝑝 = 3𝑑 < 4𝑠 = 4𝑝 = 4𝑑 = 4𝑓. The orbitals having the same energy are called degenerate. The 1𝑠 in a
hydrogen atom, as we mentioned earlier, corresponds to the most stable condition and is called the ground state
and an electron residing in this orbital is most strongly held by the nucleus. An electron in the 2𝑠, 2𝑝 or higher
orbitals in a hydrogen atom is in excited state. The only electrical interaction present in hydrogen atom is the
attraction between the negatively charged electron and the positively charged nucleus.
For multi-electron atom
The energy of an electron in a multi-electron atom depends not only on its principal quantum number (shell), but
also on its azimuthal quantum number (sub-shell). That is for a given principal quantum number, the orbitals
𝑠, 𝑝, 𝑑, 𝑓 … all have different energies. The main reason why sub-shells have different energies is the mutual
repulsion among the electrons in multi-electron atoms. Thus the stability of an electron in multi-electron is because
total attractive interactions are more than the repulsion interactions. The repulsive interactions of the electrons in
the outer shell with the electrons in the inner shell are more important. Attractive interactions of an electron
increases with increase of positive charge on the nucleus.
Shielding effect of out shell electrons
Due to the presence of electrons in the inner shells, the electron in the outer shell will not experience the full
positive charge on the nucleus(𝑍𝑒 ), but will be lowered due to the partial screening of positive charge on the
56
Gizachew Berhanu (student)
Basic Concepts of Materials
nucleus by the inner shell electrons. This is known as the shielding of the out shell electrons from the nucleus by
the inner shell electrons, and the net positive charge experienced by the electron from the nucleus is known as
effective nuclear charge(𝑍𝑒𝑓𝑓 ). Despite the shielding of the outer electrons from the nucleus by the inner shell
electrons, the attractive force experienced by the outer shell electrons increase with increase of nuclear charge.
Thus energy of interaction between, the nucleus and electron (that is orbital energy) decreases (that is more
negative) with the increase of atomic number (Z).
Effective nuclear charge (𝒁𝒆𝒇𝒇 ) and Energy: Azimutal quantum number
Both the attractive and repulsive interactions depend upon the shell and shape of the orbital in which the electron
is present. For example being spherical in shape, the 𝑠 orbital shields the electrons from the nucleus more
effectively as compared to 𝑝 orbital, 𝑝 orbitals shield the electrons from the nucleus more than the 𝑑 orbitals, even
though all these orbitals are present in the same shell. Further due to spherical shape, 𝑠 orbital electron spends
more time close to the nucleus in comparison to 𝑝 orbital and 𝑝 orbital spends more time in the vicinity of nucleus
in comparison to 𝑑 orbital. In other words, for a given shell (principal quantum number), the 𝑍𝑒𝑓𝑓 experienced by
the orbital decreases with increase of azimuthal quantum number(𝑙) that is the 𝑠 orbital will be more tightly
bound to the nucleus than 𝑝 orbital and 𝑝 orbital in turn will be better tightly bound than the 𝑑 orbital. The energy
of 𝑠 orbital will be lower (more negative) than that of 𝑝 orbital and that of 𝑝 orbital will be less than that of 𝑑
orbital and so on.
Energies of the orbitals on 𝒏 and 𝒍
Mathematically, the dependence of energies of the orbitals on 𝑛 and 𝑙 are quite complicated. But one simple rule is
that of combined value of 𝑛 and 𝑙. The lower the value of 𝑛 + 𝑙 for an orbital, the lower is its energy. If two orbitals
have the same value of 𝑛 + 𝑙, the orbital with lower value of 𝑛 will have the lower energy.
Note: Energies of orbitals in same sub-shell decrease with increase in atomic number.
𝐸2𝑠 (𝐻) > 𝐸2𝑠 (𝐿𝑖) > 𝐸2𝑠 (𝑁𝑎) > 𝐸2𝑠 (𝐾)
Since potassium has more protons than sodium, it can held electrons to tightly bind to the nucleus and has less
energy than sodium and sodium in its turn contains more protons than lithium and less energy than lithium and so
on.
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Gizachew Berhanu (student)
Basic Concepts of Materials
𝑛
Arranging orbitals in ascending of (𝒏 + 𝒍) rule
𝑙
𝑛 + 𝑙 orbital
From the right table one can see that 2𝑝 𝑎𝑛𝑑 3𝑠 have the same value of
1
0
1
1s
𝑛 + 𝑙 which equal to 3 for both. But the 𝑛 value for 2𝑝 is less than that
2
0
1
2
3
2s
2p
3
0
1
2
3
4
5
3s
3p
3d
4
0
1
2
4
5
6
4s
4p
4d
3
7
4f
of 3𝑠 and therefore from the rule, 2𝑝 will have lower energy than 3𝑠.
And also the value of 𝑛 + 𝑙 for 4𝑠 is less than that of 3𝑑 and therefore,
4𝑠 has lower energy than 3𝑑.
An ascending order is then
1𝑠, 2𝑠, 2𝑝, 3𝑠, 3𝑝, 4𝑠, 3𝑑, 4𝑝, 5𝑠, 4𝑑, 5𝑝, 4𝑓 ….
2.6.2 Filling of orbitals
It depends on four principles
 Aufbau principle
 Pauli’s exclusion principle
 Hund’s rule of maximum multiplicity
 Relative energies of the orbitals
Aufbau principle
This principle states that in the ground state of the atoms the orbitals are filled in order of their increasing energies
𝑜𝑟𝑑𝑒𝑟 = 1𝑠, 2𝑠, 2𝑝, 3𝑠, 3𝑝, 4𝑠, 3𝑑, 4𝑝, 5𝑠, 4𝑑, 5𝑝, 6𝑠, 4𝑓, 5𝑑, 6𝑝, 7𝑠 ….
Pauli’s exclusion principle
According to this principle no two electrons in an atom can have the same set of four quantum numbers. Pauli
Exclusion Principle can also be stated as: only two electrons may exist in the same orbital and these electrons must
have opposite spin (clock wise and anti-clockwise). This means that the two electrons can have the same value of
three quantum numbers 𝑛, 𝑙 𝑎𝑛𝑑 𝑚𝑙 , but must have the opposite spin quantum number. Maximum number of
electrons in shell with principal quantum number 𝑛 is equal to 2𝑛2.
Hund’s rule of maximum multiplicity
This rule deals with the filling of electrons into the orbitals belonging to the same sub-shell; that is orbitals of equal
energy called degenerate orbitals. For example 3𝑝 is filled after 3𝑠, but the principle is, in 3𝑝 sub-shell also there
are three orbitals 3𝑝𝑥 , 3𝑝𝑦 𝑎𝑛𝑑 3𝑝𝑧 and these all have equal energy. So which one will be filled first? We do not
know.
Hund’s rule of maximum multiplicity states that pairing of electrons in the orbitals belonging the same subshell
(𝑝, 𝑑, 𝑜𝑟 𝑓) does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly
occupied.
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Gizachew Berhanu (student)
Basic Concepts of Materials
Table 2.3 representation of arrangement of electrons
Atomic
Element
1s
2𝑝𝑦
2𝑝𝑥
2s
2𝑝𝑧
number
Number
oh
unpaired
electron
1
H
↑
1
2
He
↑↓
0
3
Li
↑↓
↑
1
4
Be
↑↓
↑↓
0
5
B
↑↓
↑↓
↑
6
C
↑↓
↑↓
↑
↑
7
N
↑↓
↑↓
↑
↑
↑
8
O
↑↓
↑↓
↑↓
↑
↑
2
9
F
↑↓
↑↓
↑↓
↑↓
↑
1
10
Ne
↑↓
↑↓
↑↓
↑↓
↑↓
0
1
2
3
Thus, if three electrons are to be filled in the p- level of any shell, one each will go into each of the three
(𝑝𝑥 , 𝑝𝑦 , 𝑝𝑧 ) orbitals. The fourth electron entering the p- level will go to 𝑝𝑥 orbital which now will have two
electrons with opposite spins (as shown above) and said to be paired. The unpaired electrons play an important part
in the formation of bonds. Since there are three 𝑝, five 𝑑 and seven 𝑓 orbitals, therefore, the pairing of electrons
will start in the 𝑝, 𝑑 𝑎𝑛𝑑 𝑓 orbitals with the entry of 4𝑡ℎ , 6𝑡ℎ 𝑎𝑛𝑑 8𝑡ℎ electron respectively. It has been observed
that half filled and full filled degenerate set of orbitals acquire extra stability due to their symmetry.
2.6.3 Electron configuration of atoms
The distribution of electrons into orbitals of an atom is called its electronic configuration. The electronic
configuration of different atoms can be represented in two ways.
i) 𝑠 𝑎 𝑝𝑏 𝑑𝑐 …………… notation and
ii) orbital diagram
𝑠
𝑝
𝑑
Advantage of second notation over first notation is that it represents all four quantum numbers.
Write electronic configuration of
a) Hydrogen
b) Helium
c) Lithium
d) Beryllium
e) Boron
f) Carbon
g) Nitrogen
h) Oxygen
i) Fluorine
j) Neon
k) Sodium
l) Magnesium m) Aluminum n) Silicon
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Gizachew Berhanu (student)
Basic Concepts of Materials
Solution
The following table summarizes the first way of electronic configuration for each above elements
Elements
Total # of electrons Electronic configuration
𝑠 𝑎 𝑝𝑏 𝑑𝑐 –notation
Hydrogen
1
1𝑠1
Helium
2
1𝑠 2 = most stable
Lithium
3
1𝑠 2 2𝑠1
Beryllium
4
1𝑠 2 2𝑠 2
Boron
5
1𝑠 2 2𝑠 2 2𝑝1
Carbon
6
1𝑠 2 2𝑠 2 2𝑝2
Nitrogen
7
1𝑠 2 2𝑠 2 2𝑝3
Oxygen
8
1𝑠 2 2𝑠 2 2𝑝4
Fluorine
9
1𝑠 2 2𝑠 2 2𝑝5
Neon
10
1𝑠 2 2𝑠 2 2𝑝6 = most stable
Sodium
11
1𝑠 2 2𝑠 2 2𝑝6 3𝑠1
Magnesium
12
1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2
Aluminum
13
1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝1
Silicon
14
1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝2
Simplified electronic configuration
The electronic configuration of the elements sodium (𝑁𝑎: 1𝑠 2 2𝑠 2 2𝑝6 3𝑠1 ) to argon (𝐴𝑟: 1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝6) follow
exactly the same pattern as the elements from lithium to neon with difference that the 3𝑠 and 3𝑝 orbitals are getting
filled now. i.e., 1𝑠 2 2𝑠 2 2𝑝6 is common. This process can be simplified if we represent the total number of electrons
in the first two shells by the name of element neon (𝑁𝑒). The electronic configuration of the elements from sodium
to argon can be written as 𝑁𝑎: [𝑁𝑒]3𝑠1 to 𝐴𝑟: [𝑁𝑒]3𝑠 2 3𝑝6 .
The electrons in the completely filled shells are known as core electrons and the electrons that are added to the
electronic shell with the highest principal quantum number are called valence electrons.
2.6.4 Stability of orbitals
According to Hund’s rule atoms having half-filled or completely filled orbitals are comparatively more stable and
hence more energy is needed to remove an electron from such atoms. The ionization potential or ionization
enthalpy of such atom is, therefore, relatively higher than expected normally from their position in the periodic
table.
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Gizachew Berhanu (student)
Basic Concepts of Materials
The extraordinary stability of half-filled and completely filled electron configuration can be explained in terms of
symmetry and exchange energy. The half-filled and completely filled electron configurations have symmetrical
distribution of electrons and this symmetry leads to stability. Moreover, in such configuration electron can
exchange their positions among themselves to maximum extent. This exchange leads to stabilization for example,
half-filled 2p orbital is Nitrogen and completely filled orbitals in Neon are given as follows.
Thus the 𝑝3 , 𝑝6 , 𝑑5 , 𝑑10 , 𝑓 7 and 𝑓 14 configuration which are either completely filled or exactly half-filled are more
stable.
Further, it may be noted that chromium and copper have five and ten electrons in 3d orbitals rather than four and
nine electrons respectively as expected. Therefore, to acquire more stability one of the 4s electron goes into 3d
orbitals so that 3d orbitals get half-filled or completely filled in chromium and copper respectively.
Special case
Chromium
Chromium and copper have five and ten electrons in 3𝑑 orbitals rather than four and nine as their position would
have indicated with two electrons with the 4𝑠 orbital.
1𝑠
2𝑠
2𝑝
3𝑠
3𝑝
4𝑠
3𝑑
Here both 4𝑠 and 3𝑑 are half filled which gives extra stability for chromium. The same is true for copper in which
4𝑠 is half filled and 3𝑑 is full filled.
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Gizachew Berhanu (student)
Basic Concepts of Materials
The reason is that fully filled orbitals and half-filled orbitals have extra stability. (That is lower energy). Electronic
configurations of elements are determined by spectroscopic methods.
Usage of electronic configuration
Explains chemical behavior

Why some elements are metals while others are non-metals?

Why elements like helium are not reactive whereas elements like halogens are reactive?
These questions have no answer in the Daltonian model of atom.
Half-filled sub-shell stability
 Symmetrical distribution of electrons
 Relatively small shielding
 Smaller columbic repulsion energy
 Large exchange energy
Example 2.31
1) Write the electronic configurations of the following ions. 𝑎 ) 𝐻 − 𝑏) 𝑁𝑎+ 𝑐) 𝑂+ 𝑑) 𝐹 −
2)
What
1
are
the
3
atomic
numbers
of
elements
whose
outermost
electrons
are
represented
5
by 𝑎) 3𝑠 𝑏) 2𝑝 𝑐) 3𝑝 ?
Solution
If the atom is negatively charged then it gains extra electrons and own more number of electrons than protons
and if positively charged it loses the electron and own less number of electron than protons. In question (1)
above hydrogen and fluorine have negative charge whereas sodium and oxygen have positive charge. In neutral
case hydrogen owns only one electron and here in ionic case it owns two electrons due to extra electron gained.
The same is true in case of fluorine
Also sodium atom has 11 numbers of electrons in neutral case and due to plus sign it loses 1 electron and
remains with 10 electrons. The same is true in case of oxygen also. The following table summarizes the
electronic configuration of those ions.
ions
𝐻−
𝑁𝑎+
𝑂+
𝐹−
Total number
of electron
2
10
7
10
Electronic
configuration
1𝑠 2
1𝑠 2 2𝑠 2 2𝑝6
1𝑠 2 2𝑠 2 2𝑝3
1𝑠 2 2𝑠 2 2𝑝6
2) a) the outermost is 3𝑠1 means the electronic configuration is something like 1𝑠 2 2𝑠 2 2𝑝6 3𝑠1 . Then the atomic
number is 2 + 2 + 6 + 1 = 11. The same is true for 𝑏 𝑎𝑛𝑑 𝑐.
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Gizachew Berhanu (student)
Basic Concepts of Materials
Example 2.32
Which atoms are indicated by the following configurations? 𝑎) [𝐻𝑒]2𝑠1 , 𝑏) [𝑁𝑒]3𝑠 2 3𝑝3 𝑐) [𝐴𝑟]4𝑠 2 3𝑑1
Solution
a) In the discussion section we mentioned that 𝐻𝑒 has 1𝑠 2 electronic configuration. Then [𝐻𝑒]2𝑠1 will
be 1𝑠 2 2𝑠1 . The total number of electron is then 2 + 1 = 3. Therefore the atom is Lithium.
b) In the same way as in (a) 𝑁𝑒 has 1𝑠 2 2𝑠 2 2𝑝6 electronic configuration. [𝑁𝑒]3𝑠 2 3𝑝3 will be
1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝3 . The total number of electron is 2 + 2 + 6 + 2 + 3 = 15 and an atom is phosphorus.
c) 𝐴𝑟 has 1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝6 electronic configuration. [𝐴𝑟]4𝑠 2 3𝑑1
will be 1𝑠 2 2𝑠 2 2𝑝6 3𝑠 2 3𝑝6 4𝑠 2 3𝑑1.
Therefore, total number of electron is 2 + 2 + 6 + 2 + 6 + 2 + 1 = 21. and an atom is scandium.
Example 2.33
What is the lowest value of 𝑛 that allows 𝑔 orbitals to exist?
Solution
n
1
2
3
4
5
l
0
0
1
0
1 2
0
1 2
3 0 1 2 3 4
orbitals s
s
p
s
p d
s
p d
f s p d f g
From this table we can see that “g” orbital appears at minimum value of 𝑙 = 4 or since 𝑙 is one less than 𝑛, the
lowest value that allows 𝑔 orbitals to exist is 𝑛 = 5
Example 2.34
An electron is in one of the 3𝑑 orbitals. Give possible values of 𝑛, 𝑙 𝑎𝑛𝑑 𝑚𝑙 for this electron.
Solution
The coefficient 3 in 3d indicates the value of principal quantum number (n) and for 𝑛 = 3 we have three
options of values of 𝑙 and five options of value of 𝑚𝑙 .
n
3
l
0 1
2
𝑚𝑙 0 -1 0 1 -2 -1 0 1 2
Here from the table 𝑙 = 0 𝑎𝑛𝑑 𝑙 = 1 represents 𝑠 𝑎𝑛𝑑 𝑝 orbitals respectively and the question is an electron
is one of the 3d orbitals that characterized by 𝑙 = 2. Therefore the possible values of 𝑛𝑙𝑚𝑙 in 3d orbitals are
{32(−2), 32(−1), 320, 321 𝑎𝑛𝑑 322}
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Gizachew Berhanu (student)
Basic Concepts of Materials
Self-test
1) An atomic orbital has 𝑛 = 3, a) what are the possible values of 𝑙 𝑎𝑛𝑑 𝑚𝑙 ? b) List the quantum numbers
(𝑙 𝑎𝑛𝑑 𝑚𝑙 ) of electrons for 3𝑑 orbital. c) Which of the following orbitals are possible? 1𝑝, 2𝑠, 2𝑝 and 3𝑠.
2) Using 𝑠, 𝑝, 𝑑 notations, describe the orbital with the following quantum numbers? (a) 𝑛 = 1, 𝑙 = 0 (b) 𝑛 = 3,
𝑙 = 1 (c) 𝑛 = 4, 𝑙 = 2 (d) 𝑛 = 4, 𝑙 = 3
3) Explain, giving reasons, which of the following sets of quantum numbers are not possible.
1
𝑛 = 0
𝑙=0
𝑚𝑙 = 0
𝑚𝑠 = + 2
𝑛=1
𝑙=0
𝑚𝑙 = 0
𝑚𝑠 = − 2
𝑛=1
𝑙=1
𝑚𝑙 = 0
𝑚𝑠 = +
1
2
𝑛=2
𝑙=1
𝑚𝑙 = 0
𝑚𝑠 = −
1
2
𝑛=3
𝑙=3
𝑚𝑙 = −3
𝑚𝑠 = + 2
𝑛=3
𝑙=1
𝑚𝑙 = 0
𝑚𝑠 = + 2
1
1
1
2.7 chemical Bonding
Until recently, we defined atom as a basic unit of matter that consists of a dense central nucleus surrounded by a
cloud of negatively charged electrons and except noble gases, under normal conditions no other elements exists as
an independent atom in nature due to every system tends to be more stable and bonding is nature’s way of lowering
the energy of the system to attain stability. The electron structure of the atoms plays the dominant role in
determining the nature of the chemical bond. When two atoms are brought together, the valence electrons interact
with each other and with the neighbor's positively charged nucleus. The result of this interaction is often the
formation of a bond between the two atoms, producing a molecule. Obviously there must be some force which
holds these constituent atoms together in the molecules. This attractive force that holds the constituting particles or
atoms with different chemical species together in a molecule is called chemical bond. Since the formation of
chemical compounds takes place as a result of combination of atoms of various elements in different ways, it raises
many questions. Why do atoms combine? Why are only certain combinations possible? Why do some atoms
combine while certain others do not? Why do molecules possess definite shapes? To answer such questions
different theories and concepts have been put forward from time to time. These are Kössel-Lewis approach,
Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB) Theory and Molecular Orbital (MO)
Theory. The evolution of various theories of valence and the interpretation of the nature of chemical bonds have
closely been related to the developments in the understanding of the structure of atom, the electronic configuration
of elements and the periodic table. Every system tends to be more stable and bonding is nature’s way of lowering
the energy of the system to attain stability.
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. 2.7.1
Theories on formation of molecules
Theory is a tested and testable concept that is used to explain an occurrence. These theories help us to explain and
predict behavior of atoms and molecules. They are right and wrong, they are useful and useless, depends upon what
you want to know.
Empirical Approach (by observation)

Kossel-Lewis approach (1916)

Valance Shell Electron Pair Repulsion (VSEPR) (1940)
Quantum Approach

Valance bond theory (1927)

Molecular orbital theory (1932)
Every system tends to be more stable and bonding is nature’s way of lowering the energy of the system to attain
stability.
Kossel-Lewis Approach to Chemical Bonding
In order to explain the formation of chemical bond in terms of electrons, a number of attempts were made, but it
was only in 1916 when Kössel and Lewis succeeded independently in giving a satisfactory explanation. They were
the first to provide some logical explanation of valence which was based on the inertness of noble gases. Lewis
pictured the atom in terms of a positively charged ‘Kernel’ (the nucleus plus the inner electrons) and the outer shell
that could accommodate a maximum of eight electrons. He further assumed that these eight electrons occupy the
corners of a cube which surround the ‘Kernel’. Thus the single outer shell electron of sodium would occupy one
corner of the cube, while in the case of a noble gas all the eight corners would be occupied. This octet of electrons,
represents a particularly stable electronic arrangement.
𝑂
𝐶
𝑁
𝐵
Fig.4.11 Lewis outer most electron representation. The centre sphere represents ‘KERNEL’ whereas most outer
𝐿𝑖
𝐵𝑒
most electrons represents at the corner of the cube
Lewis postulated that atoms achieve the stable octet when they are linked by chemical bonds. In the case of sodium
and chlorine, this can happen by the transfer of an electron from sodium to chlorine thereby giving the 𝑁𝑎+ and 𝐶𝑙 −
ions. In the case of other molecules like𝐶𝑙2 , 𝐻2 , 𝐹2 , etc., the bond is formed by the sharing of a pair of electrons
between the atoms. In the process each atom attains a stable outer octet of electrons.
Transfer of electrons: Electrovalent Bond
The formation of a negative ion from a halogen atom and a positive atom from an alkali metal atom is associated
with the gain and loss of an electron by the respective atoms; negative and positive ions thus formed attain stable
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noble gas electronic configurations. The negative and positive ions are stabilized by electrostatic attraction. The
bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as the
electrovalent bond. The electrovalence is thus equal to the number of unit charge(s) on the ion.
Example 𝑁𝑎𝐶𝑙 𝑎𝑛𝑑 𝐶𝑎𝐶𝑙2 .
Generally, there are four important mechanisms by which atoms are bonded in engineered materials. These are
1. Ionic bonds;
2. Covalent bonds;
3. Metallic bonds; and
4. Van der Waals bonds.
The first three bonds (ionic, covalent and metallic) are collectively known as primary bonds (relatively strong
bonds between adjacent atoms resulting from the transfer or sharing of outer orbital electrons). Whereas the fourth
one is a type of secondary bond and originate from a different mechanism and are relatively weaker. Let’s look at
each of these types of bonds one by one.
a) Ionic bond
If more than one type of atom is present, the outermost electrons can break free from atoms with excesses in their
valence shell, transforming them into positive ions. The electrons then transfer to atoms with deficiencies in their
outer shell, converting them into negative ions. The positive and negative ions have an electrostatic attraction for
each other, resulting in a strong bonding force.
Ionized atoms do not usually unite in simple pairs, however. All positively charged atoms attract all negatively
charged atoms. In sodium-chlorine bond, each sodium ion will attempt to surround itself with negative chlorine
ions, and each chlorine ion will attempt to surround itself with positive sodium ions. Since the attraction is equal in
all directions, the result will be a three-dimensional structure. Since charge neutrality must be maintained within the
structure, equal numbers of positive and negative charges must be present in each neighborhood. Formation of ionic
compounds would primarily depends upon:

The ease of formation of positive and negative ions from respective natural atoms;
𝑀(𝑔) → 𝑀+ (𝑔) + 𝑒 − ………………………. Ionization enthalpy
𝑋(𝑔) + 𝑒 − → 𝑋 − (𝑔) ………………………. Electron gain enthalpy

The electron gain process may be exothermic or endothermic. The ionization enthalpy, on the other is
always endothermic.

Ionic bonds will be formed more easily between elements with low ionization enthalpies and elements
with high negative value of electron gain enthalpies

The arrangement of the positive and negative ions in the solid that is the lattice of the crystalline
compound.
𝑀+ (𝑔) + 𝑋 − (𝑔) → 𝑀𝑋(𝑔)
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Lattice Energy in crystals: energy is released when requisite number of positive and negative ions combine to
form one mole of ionic compound is called lattice energy. Ionic compounds in the crystalline state consist of
orderly three-dimensional arrangements of cations and anions held together by coulombic interaction energies.
These compound crystallize in different crystal structures determined by the size of the ions, charge of ions, their
packing arrangements and other factors. Example, 𝑁𝑎𝐶𝑙
𝐹=
𝐾𝑞1 𝑞2
𝑟2
, where 𝐾 =
1
4𝜋𝜀𝑜
In ionic solids the sum of the electron gain enthalpy and the ionization enthalpy may be positive but still the crystal
structure gets stabilized due to the energy released in the formation of the crystal lattice.
Formation of 𝑁𝑎𝐶𝑙:

Ionic enthalpy for 𝑁𝑎+ (𝑔) formation from 𝑁𝑎 (𝑔) is

Electronic gain enthalpy for the change 𝐶𝑙(𝑔) + 𝑒 − → 𝐶𝑙 − 𝑖𝑠 − 148.7𝐾𝑔/𝑚𝑜𝑙

Enthalpy of lattice formation of 𝑁𝑎𝐶𝑙 (𝑠) is
495.8𝐾𝑔/𝑚𝑜𝑙
−788𝐾𝑔/𝑚𝑜𝑙
Qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not
simply by achieving octet of electrons around the ionic spices in gaseous state.
Properties of ionic solids:

Exist in solid state

Crystal structure: Arranged in regular pattern.

High melting and boiling point

Soluble in polar solvents and not soluble in non-polar solvents

Good conductor of electricity

Ionic reactions are very fast
General characteristics of materials joined by ionic bonds include moderate to high strength, high hardness,
brittleness, high melting point, and low electrical conductivity (since electrons are captive to atoms, charge
transport requires atom—or ion—movement).
b) Sharing of Electrons: Covalent Bond
A second type of primary bond is the covalent type. Langmuir (1919) refined the Lewis postulations by abandoning
idea of stationary cubical arrangement of octet and by introducing the term covalent bond. Here the atoms in the
assembly find it impossible to produce completed shells by electron transfer but achieve the same goal through
electron sharing. Adjacent atoms share outer-shell electrons so that each achieves a stable electron configuration.
The shared (negatively charged) electrons locate between the positive nuclei, forming a positive–negative–positive
bonding link. Figure 2-12 illustrates this type of bond for a pair of chlorine atoms, each of which contains seven
electrons in the valence shell. The result is a stable two-atom molecule 𝐶𝑙2 . The atoms in the assembly need not be
identical, the sharing does not have to be equal, and a single atom can share electrons with more than one other
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atom. For atoms such as carbon and silicon, with four electrons in the valence shell, one atom may share its valence
electrons with each of four neighboring atoms.
Like the ionic bond, the covalent bond tends to produce materials with high strength and high melting point. Since
atom movement within the three-dimensional structure (plastic deformation) requires the breaking of discrete
bonds, covalent materials are characteristically brittle. Electrical conductivity depends on bond strength, ranging
from conductive tin (weak covalent bonding), through semiconductive silicon and germanium, to insulating
diamond (carbon). Ionic or covalent bonds are commonly found in ceramic and polymeric materials.
Fig.2.12 Formation of a chlorine molecule by the electron sharing of a covalent bond
Note: both chlorine atoms attain outer shell octet of the nearest noble gas.
Each bond is formed as a result of sharing of an electron pair between the atoms. Each combining atom contributes
at least one electron to the share pair. The combining atoms attain the outer shell noble gas configurations as a
result of the sharing of electrons. Example, 𝐻2 𝑂, 𝐶𝐶𝑙4 .
Thus when two atoms share one electron pair they are said to be joined by a single covalent bond. Example,
𝐻2 𝑂, 𝐶𝐶𝑙4 , 𝐶𝐻4 , 𝑁𝐻3 . If two atoms share two pairs of electrons, the covalent bond between them is called a double
bond. 𝐶𝑂2 , 𝐶2 𝐻4 . And when combining atoms share three electron pairs, a triple bond is formed. Example,
𝑁2 , 𝐻2 𝐶2 .
2.7.2 Bond parameters
There are some parameters to define the bond.
1) Bond Length;- is defined as the equilibrium distance between the nuclei two bonded atoms in a molecule.
Fig.2.13, 𝑅 = 𝑟𝐴 + 𝑟𝐵 (𝑅 𝑖𝑠 𝑡ℎ𝑒 𝑏𝑜𝑛𝑑 𝑙𝑒𝑛𝑔𝑡ℎ 𝑎𝑛𝑑 𝑟𝐴 𝑎𝑛𝑑 𝑟𝐵 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑣𝑎𝑙𝑒𝑛𝑡
𝑟𝑎𝑑𝑖𝑖 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝐴 𝑎𝑛𝑑 𝐵)
Bond lengths are measured by spectroscopic, x-ray diffraction and electron-diffraction techniques.
Atomic radii: it is measured approximately, as radius of an atoms core which is contact with the core of an
adjacent atom in bonded situations.
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Van der waals Radii: - it represents the overall size of the atom which includes its valence shell in a nonbonded
situations.
Metallic Radii: - radius of atoms of element when joined to other atoms by metallic bonds.
2) Bond angle: - it is defined as the angle between the orbitals containing bonding electron pairs around the central
in a molecule/complex ion.
3) Bond Enthalpy: - it is defined as the amount of energy required to break one mole of bonds of a particular type
between two atoms in gaseous state. The unit of bond enthalpy is 𝐾𝐽𝑚𝑜𝑙 −1 . for example the 𝐻 − 𝐻 bond in
hydrogen molecule is 435.8𝐾𝐽𝑚𝑜𝑙 −1.
𝐻2 (𝑔) → 𝐻(𝑔) + 𝐻(𝑔), ∆𝐻 = 435.8𝐾𝑔𝑚𝑜𝑙 −1 .
In case of polyatomic molecules, the measurement of bond strength is more complicated. For example, in case of
𝐻2 𝑂 molecule, the enthalpy needed to break the two 𝑂 − 𝐻 is not the same. Therefore in polyatomic molecule the
term “mean” or average bond enthalpy is used.
𝐻2 𝑂(𝑔) → 𝐻(𝑔) + 𝑂𝐻(𝑔), ∆𝐻 = 502𝐾𝐽/𝑚𝑜𝑙
𝑂𝐻(𝑔) → 𝐻(𝑔) + 𝑂(𝑔),
∆𝐻 = 427𝐾𝐽/𝑚𝑜𝑙
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑏𝑜𝑛𝑑 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 =
502+427
2
= 464.5𝐾𝐽/𝑚𝑜𝑙
4) Bond Order:- in the Lewis description of covalent bond, the bond order is given by the number of bonds
between the two atoms in a molecule. Example, 𝐻2 (𝑠𝑖𝑛𝑔𝑙𝑒 𝑏𝑜𝑛𝑑), 𝑂2 (𝑑𝑜𝑢𝑏𝑙𝑒 𝑏𝑜𝑛𝑑), 𝑁2 (𝑡𝑟𝑖𝑝𝑙𝑒 𝑏𝑜𝑛𝑑). With
increase in bond order, bond enthalpy increases and bond length decreases.
5) Resonance Structures:- experimentally found structure of 𝑂3 cannot be represented by Lewis structures.
The concept of resonance was introduced to deal with the type of difficulty experienced in the depiction of accurate
structures of molecules like 𝑂3 . According to the concept of resonance, whenever the single Lewis structure cannot
describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and nonbonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule
accurately.
Resonance stabilizes the molecule as the energy of the resonance hybrid is less than the energy of any single
canonical structure. Resonance averages the bond characteristics as a whole. Thus the energy of the 𝑂3 (𝐼𝐼𝐼)
resonance hybrid is lower than either of the two canonical forms. There is no such equilibrium between the
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canonical forms. The molecule as such as a single structure which is the resonance hybrid of the canonical forms
and which cannot as such be depicted by a single Lewis structure.
The general principle of molecule formation is illustrated in figure below, showing two atoms brought together
from infinity.
2.7.2 Valence Bond Theory
As we know that Lewis approach helps in writing the structure of molecules but it fails to explain the formation of
chemical bond. It also does not give any reason for the difference in bond dissociation enthalpies and bond lengths
in molecules like H2 (435.8 kJ mol-1, 74 pm) and F2 (155 kJ mol-1, 144 pm), although in both the cases a single
covalent bond is formed by the sharing of an electron pair between the respective atoms. It also gives no idea about
the shapes of polyatomic molecules.
Similarly the VSEPR theory gives the geometry of simple molecules but theoretically, it does not explain them and
also it has limited applications. To overcome these limitations the two important theories based on quantum
mechanical principles are introduced. These are valence bond (VB) theory and molecular orbital (MO) theory.
Valence bond theory was introduced by Heitler and London (1927) and developed further by Pauling and others. A
discussion of the valence bond theory is based on the knowledge of atomic orbitals, electronic configurations of
elements (section 2.6), the overlap criteria of atomic orbitals, the hybridization of atomic orbitals and the principles
of variation and superposition. A rigorous treatment of the VB theory in terms of these aspects is beyond the scope
of this book. Therefore, for the sake of convenience, valence bond theory has been discussed in terms of qualitative
and non-mathematical treatment only. To start with, let us consider the formation of hydrogen molecule which is
the simplest of all molecules.
Consider two hydrogen atoms A and B approaching each other having nuclei 𝑁𝐴 and 𝑁𝐵 and electrons present in
them are represented by 𝑒𝐴 and 𝑒𝐵 . When the two atoms are at large distance from each other, there is no
interaction between them. As these two atoms approach each other, new attractive and repulsive forces begin to
operate. Attractive forces arise between:
(i) Nucleus of one atom and its own electron that is 𝑁𝐴 – 𝑒𝐴 and 𝑁𝐵 – 𝑒𝐵 .
(ii) Nucleus of one atom and electron of other atom i.e., 𝑁𝐴 – 𝑒𝐵 , 𝑁𝐵 – 𝑒𝐴 .
Similarly repulsive forces arise between
(i) Electrons of two atoms like 𝑒𝐴 – 𝑒𝐵 ,
(ii) Nuclei of two atoms 𝑁𝐴 – 𝑁𝐵 .
Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart
(Fig. 2.14).
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Fig.2.14 forces of attraction and repulsion during the formation of 𝐻2
Experimentally it has been found that the magnitude of new attractive force is more than the new repulsive forces.
As a result, two atoms approach each other and potential energy decreases. As the two atoms approach each other,
the atoms exert attractive and repulsive forces on each other as a result of mutual electrostatic interactions. Initially,
the attractive force 𝐹𝐴 dominates over the repulsive force 𝐹𝑅 . The net force 𝐹𝑁 between the two atoms is just the
sum of both attractive and repulsive components; that is,
𝐹𝑁 = 𝐹𝐴 + 𝐹𝑅
The potential energy 𝐸𝑟 of the two atoms can be found from by integrating the net force 𝐹𝑁 .
𝐹𝑁 =
𝑑𝐸
𝑑𝑟
Ultimately a stage is reached where the net force of attraction balances the force of repulsion and system acquires
minimum energy. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having
the bond length of 74 pm.
Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is
more stable than that of isolated hydrogen atoms. The energy so released is called as bond enthalpy, which is
corresponding to minimum in the curve depicted in Fig. 2.15. Conversely, 435.8 kJ of energy is required to
dissociate one mole of H2 molecule.
U
𝐻2 (𝑔) + 435.8𝐾𝐽 → 𝐻(𝑔) + 𝐻(𝑔)
𝐸𝑅
𝐸𝑁
F
𝐹𝐴
r
0
r
𝑟𝑜
𝐸𝐴
𝐹𝑁
𝐹𝑅
a)
b)
Figure 2.15a and b shows the variation of the net force 𝐹𝑁 (𝑟) and the overall potential energy 𝐸𝑟 with the
interatomic separation r as the two atoms are brought together from infinity. The lowering of energy
corresponds to an attractive interaction between the two atoms.
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The variations of 𝐹𝐴 and 𝐹𝑅 with distance are different. Force 𝐹𝐴 varies slowly, whereas 𝐹𝑅 varies strongly with
separation and is strongest when the two atoms are very close. When the atoms are so close that is, the individual
electron shells overlap, there is a very strong electron-to-electron shell repulsion and 𝐹𝑅 dominates.
An equilibrium will be reached when the attractive force just balances the repulsive force and the net force is zero,
or
𝐹𝑁 = 𝐹𝐴 + 𝐹𝐴 = 0
In this state of equilibrium, the atoms are separated by a certain distance 𝑟𝑜 , as shown in Figure 2.15. This distance
𝑑𝐸
is called the equilibrium separation and is effectively the bond length. On the energy diagram, 𝐹𝑁 = 0 means 𝑑𝑟 =
0, this means that the equilibrium of two atoms corresponds to the potential energy of the system acquiring its
minimum value. Consequently, the molecule will only be formed if the energy of the two atoms as they approach
each other can attain a minimum. This minimum energy also defines the bond energy of the molecule, as depicted
in Figure1.3b. Energy of 𝐸𝑜 is required to separate the two atoms, and this represents the bond energy.
Although we considered only two atoms, similar arguments also apply to bonding between many atoms, or between
millions of atoms as in a typical solid. Although the actual details of 𝐹𝐴 and 𝐹𝑅 will change from material to
material, the general principle that there is a bonding energy 𝐸𝑜 per atom and an equilibrium interatomic separation
𝑟𝑜 will still be valid. Even in a solid in the presence of many interacting atoms, we can still identify a general
potential energy curve 𝐸𝑟 per atom similar to the type shown in Figure 2.15b. We can also use the curve to
understand the properties of the solid, such as the thermal expansion coefficient and elastic and bulk moduli.
The magnitude of this bonding energy and the shape of the energy–versus–interatomic separation curve vary from
material to material, and they both depend on the type of atomic bonding. Furthermore, a number of material
properties depend on 𝐸𝑜 , the curve shape, and bonding type. For example, materials having large bonding energies
typically also have high melting temperatures; at room temperature, solid substances are formed for large bonding
energies, whereas for small energies, the gaseous state is favored; liquids prevail when the energies are of
intermediate magnitude.
2.7.3 Orbital Overlap Concept
In the formation of hydrogen molecule, there is a minimum energy state when two hydrogen atoms are so near that
their atomic orbitals undergo partial interpenetration. This partial merging of atomic orbitals is called overlapping
of atomic orbitals which results in the pairing of electrons. The extent of overlap decides the strength of a covalent
bond. In general, greater the overlap the stronger is the bond formed between two atoms. Therefore, according to
orbital overlap concept, the formation of a covalent bond between two atoms results by pairing of electrons present
in the valence shell having opposite spins.
Types of Overlapping and Nature of Covalent Bonds
The covalent bond may be classified into two types depending upon the types of overlapping:
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(i) Sigma (σ) bond, and (ii) pi (π) bond
(i) Sigma (σ) bond: This type of covalent bond is formed by the end to end (head-on) overlap of bonding orbitals
along the internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the
following types of combinations of atomic orbitals.
• s-s overlapping : In this case, there is overlap of two half-filled s-orbitals along the internuclear axis as shown
below :
• s-p overlapping: This type of overlap occurs between half-filled s-orbitals of one atom and half-filled p-orbitals
of another atom.

p–p overlapping : This type of overlap takes place between half-filled p-orbitals of the two approaching
atoms.

ii)𝑷𝒊 (𝝅)𝒃𝒐𝒏𝒅: in the formation of 𝜋 bond the atomic orbitals overlap in such a way that their axes remain parallel
to each other and perpendicular to the internuclear axis. The orbitals formed due to sidewise overlapping consists of
two saucer type charged clouds above and below the plane of the participating atoms.
Strength of Sigma and pi Bonds
Basically the strength of a bond depends upon the extent of overlapping. In case of sigma bond, the overlapping of
orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of
overlapping occurs to a smaller extent. Further, it is important to note that in the formation of multiple bonds
between two atoms of a molecule, pi bond(s) is formed in addition to a sigma bond.
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2.8 Bond hybridization in carbon
In order to explain the characteristic geometrical shapes of polyatomic molecules like CH 4, NH3 and H2O etc.,
Pauling introduced the concept of hybridization. According to him the atomic orbitals combine to form new set of
equivalent orbitals known as hybrid orbitals.
Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridization
which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute
their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.
For example when one 2s and three 2p-orbitals of carbon hybridize, there is the formation of four new sp3 hybrid
orbitals.
Bonding hybrid orbitals are directional in nature—that is, each extends to and overlaps the orbital of an adjacent
bonding atom. Furthermore, for carbon, each of its four sp3 hybrid orbitals is directed symmetrically from a carbon
atom to the vertex of a tetrahedron.
For diamond, its carbon atoms are bonded to one another with sp3 covalent hybrids—each atom is bonded to four
other carbon atoms. Diamond’s carbon–carbon bonds are extremely strong, which accounts for its high melting
temperature and ultrahigh hardness (it is the hardest of all materials).
c) Metallic bond
A third type of primary bond is possible when a complete outer shell cannot be formed by either electron transfer or
electron sharing. This bond is known as the metallic bond. As we have seen, the elements with the most
pronounced metallic characteristics are grouped on the left-hand side of the Periodic Table. In general, they have a
few valence electrons, outside the outermost closed shell, which are relatively easy to detach. In a metal, each ‘free’
valence electron is shared among all atoms, rather than associated with an individual atom, and forms part of the socalled ‘electron gas’, which circulates at random among the regular array of positively charged electron cores, or
cations. The metallic bond derives from the attraction between the cations and the free electrons and, as would be
expected, repulsive components of force develop when cations are brought into close proximity. However, the
bonding forces in metallic structures are spatially non-directed and we can readily simulate the packing and spacefilling characteristics of the atoms with modeling systems based on equal-sized spheres (polystyrene balls, even
soap bubbles). Other properties such as ductility, thermal conductivity and the transmittance of electromagnetic
radiation are also directly influenced by the non-directionality and high electron mobility of the metallic bond.
These highly mobile, free electrons account for the high electrical and thermal conductivity values as well as the
opaque (nontransparent) characteristic observed in metals (the free electrons are able to absorb the various discrete
energies of light radiation).They also provide the “cement” required for the positive–negative–positive attractions
that result in bonding. Bond strength, and therefore material strength and melting temperature, varies over a wide
range. More significant, however, is the observation that the positive ions can now move within the structure
without the breaking of discrete bonds. Materials bonded by metallic bonds can be deformed by atom-movement
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mechanisms and produce an altered shape that is every bit as strong as the original. This phenomenon is the basis of
metal plasticity, enabling the wide variety of forming processes used in the fabrication of metal products.
Secondary bonds
The existence of a hundred percent ionic or covalent bond represents an ideal situation. In reality no bond or a
compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there
is some ionic character.
When covalent bond is formed between two similar atoms, for example in H 2, O2, Cl2, N2 or F2, the shared pair of
electrons is equally attracted by the two atoms. As a result electron pair is situated exactly between the two
identical nuclei. The bond so formed is called nonpolar covalent bond. Contrary to this in case of a heteronuclear
molecule like HF, the shared electron pair between the two atoms gets displaced more towards fluorine since the
electronegativity of fluorine is far greater than that of hydrogen. The resultant covalent bond is a polar covalent
bond.
As a result of polarization, the molecule possesses the dipole moment (depicted below) which can be defined as the
product of the magnitude of the charge and the distance between the centers of positive and negative charge. It is
usually designated by a Greek letter ‘μ’. Mathematically, it is expressed as follows:
Dipole moment (μ) = charge (Q) × distance of separation (r)
Dipole moment is usually expressed in Debye units (D). The conversion factor is
1 D = 3.33564 × 10–30 C m where C is coulomb and m is meter.
Secondary bonding forces arise from atomic or molecular dipoles. In essence, an electric dipole exists whenever
there is some separation of positive and negative portions of an atom or molecule. The bonding results from the
coulombic attraction between the positive end of one dipole and the negative region of an adjacent one. Weak or
secondary bonds, known as van der Waals forces, can form between molecules that possess a nonsymmetrical
distribution of electrical charge. Some molecules, such as hydrogen fluoride and water, can be viewed as electric
dipoles. Certain portions of the molecule tend to be more positive or negative than others (an effect referred to as
polarization). The negative part of one molecule tends to attract the positive region of another, forming a weak
bond. Van der Waals forces contribute to the mechanical properties of a number of molecular polymers, such as
polyethylene and polyvinyl chloride (PVC).
Hydrogen Bonding
In many molecules, the centres of the positive and negative charges do not coincide i.e. they have permanent dipole
moment. Consider the example of water molecule (H2O). Since the electronegativity of oxygen is greater them
hydrogen, therefore, the oxygen atom pulls the bonding electrons (negative charge) to itself more strongly than
hydrogen does. These results in a net negative charge at the oxygen end and a net positive charge at the hydrogen
end of the molecule. Due to this imbalance in electrical charge, the water molecule possesses a permanent dipole
moment. The bond that is formed between water molecules due to attraction between the positively charged
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hydrogen end of a molecule and the negatively charged oxygen end of another molecule is called the hydrogen
bond.
REFERENCES
1) Concise Inorganic chemistry - J.D. Lee, 3rd Edn - 1977 and 5th Edn2002.
2) Theoretical Inorganic chemistry - M.C.Day and J.Sellbin, 2nd Edn. 1985.
3) Theorical princples of Inorganic chemistry - G.S.Manker, 9th Edn 1993.
4) Selected topics in Inorganic Chemistry - U.Mallik, G.D.Tute and
R.D. Madan, 6th Edn. - 1993c.
Problems
A. Choose the best answer
1. Atomic mass of an element is not necessarily a whole number because:
(a) It contains electrons, protons and neutrons
(b) It contains allotropic forms
(c) Atoms are no longer considered indivisible
(d) It contains isotopes
(e) None of these.
2. No two electrons in an atom will have all four quantum numbers equal. The statement is known as
(a) Exclusion principle (b) Uncertainity principle
(c) Hund’s rule (d) Aufbau principle
(e) Newlands law.
3. When the 3d orbital is complete, the new electron will enter the
(a) 4p orbital (b) 4f orbital (c) 4s orbital
(d) 4d orbital (e) 5s orbital.
4. The preference of three unpaired electrons in the nitrogen atom can be explained by:
(a) Pauling’s exclusion principle (b) Aufbau principle (c) Uncertainty principle (d) Hund’s rule (e) None
of these.
5. The number of orbitals in a p-sub-shell is
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(a) 1 (b) 2 (c) 3 (d) 6 (e) 5.
6. The nucleus of an atom contains:
(a) Electrons and protons (b) Neutrons and protons
(c) Electrons, protons and neutrons (d) Neutrons and electrons
(e) None of these.
7. Which is the lightest among the following? (a) An atom of hydrogen
(b) An electron (c) A neutron (d) A proton (e) An alpha particle.
8. Which of the following has no neutrons in the nucleus? (a) Deuterium
(b) Helium (c) Hydrogen (d) Tritium (e) An alpha particle.
9. When the value of the azimuthal quantum number is 3, the magnetic quantum number can have values:
(a) +1,-1 (b) +1, 0, 1 (c) +2, +1, 0, -1,-2
(d) +3, +2, +1, 0, -1,-2,-3 (e) +3,-3.
10. 2p orbitals have:
(a) n = 1,l = 2 (b) n=1, l = 0 (c) n = 2, l = 0
(d) n = 2, l =1 (e) n =1, l =1.
11. The atomic number of an element is 17 and its mass number is 37. The number of protons, electrons
and neutrons present in the neutral atom are:
(a) 17, 37, 20 (b) 20, 17, 37 (c) 17, 17, 20 (d) 17, 20, 17 (e) 37, 20, 17.
12. The maximum number of electrons that can be accommodated in the nth level is: (a) n2 (b) n+1 (c) n-1
(d) 2n2 (e) 2 + n.
13. The magnetic quantum number decides: (a) The distance of the orbital from the nucleus (b) The shape
of the orbital
(c) The orientation of the orbital in space (d) The spin of the electron
(e) None of these.
B. Write in one or two sentence
1. What is the charge of an electron, proton and a neutron?
2. What is atomic number?
3. What is the maximum number of electrons that an orbital can have?
4. How many orbitals are there in the second orbit? How are they designated?
5. Sketch the shape of s and p-orbital indicating the angular distribution of electrons.
6. What are the charge and mass of an electron?
7. What is an orbital?
8. Give the order of filling of electrons in the following orbitals 3p, 3d, 4p, 3d and 6s.
9. What is meant by principal quantum number?
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10. How many protons and neutrons are present in 188𝑂?
11. What are the particles generally present in the nuclei of atoms?
12. The atomic mass of an element is 24 and its atomic number is 12. Show how the atom of the element
is constituted?
13. How will you experimentally distinguish between a ray of neutron and ray of proton?
14. What is the principal defect of Bohr atom model?
15. Write the complete symbol for: (a) the nucleus with atomic number
56 and mass number 138; (b) the nucleus with atomic number 26 and mass number 55; (c) The nucleus
with atomic number 4 and mass number 9.
16. An atomic orbital has n = 3. What are the possible values of l?
17. An atomic orbital has l= 3. What are the possible values of m?
18. Give the electronic configuration of chromium. (Z=24).
19. Which energy level does not have p-orbital?
20. An atom of an element has 19 electrons. What is the total number of p-orbital?
21. How many electrons can have s+ ½ in a d-sub-shell?
22. Write the values of l and m for p-orbitals.
23. Which quantum accounts for the orientation of the electron orbital?
24. What is shape of the orbital with (i) n = 2 and l = 0; (ii) n = 2 and l = 1?
25. Give the values for all quantum numbers for 2p electrons in nitrogen (Z = 7).
26. Give the electronic configuration of Mn2+ and Cu. Atomic number of Cu = 29 and Mn = 25.
27. Explain why the electronic configuration of Cr and Cu are written as 3d5, 4s1 and 3d10 4s1 instead of
3d4 4s2 and 3d9 4s2?
C. Explain briefly on the following
1. Describe Aufbau principle. Explain its significance in the electronic buildup of atoms.
2. Using the s, p, d, notation, describe the orbital with the following quantum numbers? (a) n = 1, l=0; (b)
n = 2,l = 0; (c) n = 3, l = 1;
(d) n = 4, l =3.
3. Using the Aufbau principle, write the electronic configuration in the ground state of the following
atoms: Boron (Z = 5) Neon (Z = 10) and Aluminium (Z = 13).
4. What is Rutherford’s a- ray scattering experiment? What are its conclusions?
5. What are the postulates of Bohr Theory of atom?
6. Explain the various quantum numbers which completely specify the electron of an atom.
7. Who was Schrödinger?
8. Explain the formation of a chemical bond.
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9. Define octet rule. Write its significance and limitations.
10. Write the favorable factors for the formation of ionic bond.
11. Define the bond length.
12. Write the resonance structures for 𝑆𝑂3 , 𝑁𝑂2 and 𝑁𝑂3− .
13. Write the significance/application of dipole moment
14. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
15. Define electronegativity. How does it differ from electron gain enthalpy?
16. Explain with the help of suitable example polar covalent bond.
17. Distinguish between a sigma and a pi bond.
18. What is the total number of sigma and pi bonds in the following molecules?
(a) 𝐶2 𝐻2 (b) 𝐶2 𝐻4
19. What is meant by the term bond order?
20. Compare and contrast three primary bonds.
Problems
1. A mercury atom is initially in its lowest possible (or ground state) energy level. The atom absorbs a
photon with a wavelength of 185 nm and then emits a photon with a frequency of 6.88 × 1014 Hz. At the
end of this series of transitions, the atom will still be in an energy level above the ground state. Draw an
energy-level diagram for this process and find the energy of this resulting excited state, assuming that we
assign a value of E = 0 to the ground state. (This choice of E = 0 is not the usual convention, but it will
simplify the calculations you need to do here.)
2. When a photoelectric effect experiment was carried out using a metal “M” and light at wavelength 1,
electrons with a kinetic energy of 1.6 × 10−19 J were emitted. The wavelength was reduced to one-half its
original value and the experiment was repeated (still using the same metal target). This time electrons
with a kinetic energy of 6.4 × 10−19 J were emitted. Find the electron binding energy for metal M. (The
actual wavelengths used were not recorded, but it is still possible to find the binding energy.)
3. The photoelectric effect can be used in engineering designs for practical applications. For example,
infrared goggles used in night-vision applications have materials that give an electrical signal with
exposure to the relatively long wavelength IR light. If the energy
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needed for signal generation is 3.5×10−20 J, what is the minimum wavelength and frequency of light that
can be detected?
4. From the list of atoms and ions given, identify any pairs that have the same electron
configurations and write that configuration: Na+, S2−, Ne, Ca2+, Fe2+, Kr, I−
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Chapter Three: Thermodynamics of Materials
Thermodynamics of materials
3.1 Introduction
Thermodynamics involves work and heat. It is a branch of physics that deals with heat and temperature. It began in
the 19th century with the efforts of engineers to increase the efficiency of steam engines, but it has become the
general theory of the macroscopic behavior of matter at equilibrium. In another language it is a macroscopic
science. What is macroscopic? “Macro” means something which is big. So macroscopic is basically deals with
bulk systems. Therefore, thermodynamics never talk about very small properties of molecules or atoms. It talks
only about systems as a whole. It will not talk about the behavior or properties of each molecule or atoms which
constitute an object. So, macroscopic restriction is basically specified by macroscopic variables or quantities which
can be directly sense or measured such as volume and temperature.
Thermodynamics is based on empirical laws, as is classical mechanics. Although classical mechanics has been
superseded by relativistic mechanics and quantum mechanics, thermodynamics is an unchallenged theory. No
exceptions have been found to the laws of thermodynamics.
How thermodynamics differ from other branches of physics? Let’s compare thermodynamics with mechanics.
When we talk about mechanics, we talk only of motion of the system as a whole. Here note that both talk about a
system as a whole. Because they both deals with bulk systems. But in case of mechanics, we talk about the motion
of the system as a whole. For example if we look at a car which is moving, all what we are looking at is the motion
of the car. We never bother about the temperature of the car or its volume. We only bother about the velocity of the
car and how the car is accelerating. So, what we concentrate on as a whole is about the motion of the car. We never
care about the state of the car at that moment. But when we talk about thermodynamics, we will talk of only such
quantities which will tell us the state of an object.
Let us look another common example which relates thermodynamics with mechanics. When a bullet is fired from a
gun we only talk about the motion of this bullet. “With what velocity the bullet will be moving?” What the relative
velocity of the bullet is with compared to the gun?” and things like that. So, we are only worried about the motion
of the bullet. This is what mechanics deals with. But if the fired bullet hit the wall and embraces to the wall in front
of it, the kinetic energy at which the bullet was moving converted to heat energy. Whenever heat is generated, there
is some change in temperature. If there is some change in temperature there takes place some change in internal
energy of the bullet which deals in thermodynamics.
This chapter will concentrate on the changes in the world of thermodynamics of materials over the latter half of the
twentieth century. Within this field we have chosen to concentrate on a few basic concepts and cannot claim to
cover all of the major areas.
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Objectives of this chapter
At the end of this chapter, the reader should be able to

Understand core concepts of thermodynamics and compare it with other branches of physics

Describe systems, boundary system , characteristics of different systems

Differentiate the concept between system and its surroundings

Understand the important features of properties of materials either it is intensive or extensive.

Understand the concept behind the term thermodynamic equilibrium.

Explain the meaning of state variables (functions)

State the kinetic theory of ideal gases

Describe molecular nature of matter.

State Law of equipartition of Energy

Briefly explain three modes of energy transfer

Understand the physical meaning of Mean free path

Explain the phrase “specific heat capacity”.

Briefly describe Laws of thermodynamics

Define Enthalpy, Entropy and Gibbs free energy

Explain Engines and Refrigerators in thermodynamic language.
3.2 Thermodynamics
The name thermodynamics drives from the Greek words therme (heat) and dynamis (power), which is most
descriptive of the early efforts to convert heat into power. Today the same name is used to include all aspects of
energy and energy transformations, including power generation, refrigeration, and relationships among the
properties of matter. In a broader sense, thermodynamics studies the relationships between the macroscopic
properties of a system. A key property in thermodynamics is temperature. Temperature is an intensive property
associated with the hotness or coldness of a material. It determines the direction of spontaneous heat flow (always
from hot to cold).Thermodynamics is sometimes defined as the study of the relation of temperature to the
macroscopic properties of matter. This macroscopic property restriction of thermodynamics limit it to bulk
materials energy transfer in nature and does not care whether there are atoms or molecules.
3.2.1 Thermodynamic systems
Here, "system" refers to any set of physical interactions isolated from the rest of the universe. Anything outside of
the system, including all factors and forces irrelevant to a discussion of that system, is known as surrounding.
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System
System
boundary
Surrounding
Figure 3.1 Thermodynamics systems and its surrounding
Open system: is a system in which both matter and energy can transmit between system and its surroundings.
Closed system: is a system in which only energy is allowed to transmit between a system and its surroundings. No
transfer of matter between a system and its surroundings.
Isolated system: is also a type of system in which neither energy nor matter transfers between a system and its
surroundings. This type of system does not interact in any way with its surroundings. An isolated system is
obviously a closed system, but not every closed system is isolated.
System boundary: A system may be separated from its surroundings by different kinds of boundaries. The closed
surface that separates the system from its surroundings is known as the system boundary. In Fig. 2.1 above, the
system is separated from its surrounding by the red circumference of the sphere. A system boundary can be:

rigid or non-rigid (movable)

permeable or impermeable

adiabatic or non-adiabatic
By the term “permeable” we mean that a system boundary allows matter to pass through it and by the term
“impermeable” we mean that a system boundary does not allows matter to pass through it. Finally, when we say an
adiabatic system boundary, we mean that a boundary that does not conduct heat at all, whereas a non-adiabatic
system boundary does conduct heat. A system surrounded by a rigid, impermeable, adiabatic boundary cannot
interact with the surroundings and is isolated.
3.2.2 Thermodynamic Equilibrium
You may here the term equilibrium from your previous physics course or mechanics. In mechanics “equilibrium”
means if the net force or the net torque in a system is zero. Similarly in thermodynamics, a system is in equilibrium
when macroscopic variables such as temperature, pressure and volume do not change with time. Basically when we
talk of thermodynamic equilibrium, we talk about three types of equilibrium.

Thermal equilibrium

Chemical equilibrium and

Mechanical equilibrium
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Thermal equilibrium: a system is in thermal equilibrium if the temperature in the system is constant. If two
systems have same temperature and no flow of temperature occur the two systems are in thermodynamic thermal
equilibrium.
Chemical equilibrium: a system is in chemical equilibrium if the composition in the system is constant over the
time.
Mechanical equilibrium: a system is in mechanical equilibrium if the pressure on the system is uniformly
distributed.
Generally, if the system is in thermal equilibrium, chemical equilibrium as well as mechanical equilibrium, then we
say that the system is in thermodynamic equilibrium.
3.2.3 Thermodynamic state variables
In chapter one, we tried to see what we mean by the properties of materials. Here in thermodynamics, we are going
to correlate a property with thermodynamic state variables and define it as any characteristic of a system which can
be measured or calculated. Temperature and pressure are examples of properties that can be measured. Internal
energy, enthalpy and entropy are examples of properties that can be calculated. Properties can be either intensive
properties or extensive properties; the definitions of each are given below.
An extensive property is a property that changes when the size of the sample changes. The property is proportional
to the amount of material in the system. For example, both the mass and the volume of a diamond are directly
proportional to the amount that is left after cutting it from the raw mineral.
Examples of extensive properties include:
Length

Momentum

Mass

Number of moles

Particle number

Volume

Magnetic moment

Energy

Electrical charge

Entropy

Weight

Gibbs energy
An intensive property is a physical property of a system that does not depend on the system size or the amount of
material in the system. An intensive property doesn't change when you take away some of the sample. When a
diamond is cut, the pieces maintain their intrinsic hardness until their size reaches a few atoms thick.
Examples of intensive properties include:

Chemical potential

Pressure

Concentration

Specific energy

Density

Specific heat capacity

Ductility

Specific volume
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
Elasticity

Temperature

Electrical resistivity

Color

Hardness

Molecular weight

Magnetic field

Melting point and boiling point

magnetization

malleability
The ratio of two extensive properties of the same material or system is scale-invariant, and is therefore an intensive
property. For example, the ratio of the extensive properties mass and volume, the density, is an intensive
property.𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
= constant. If the system is divided in half, the mass and the volume change in the
identical ratio and the density remains unchanged.
3.3 Kinetic Molecular Theory
Many physics text books define energy as a capacity to cause a change. Energy forces the atoms or molecules to be
always in motion. There are different types of energy such as kinetic energy, potential energy, chemical energy,
heat energy, electrical energy, nuclear energy and more. We can categories these various forms of energy into two
major groups’ kinetic energy and potential energy. Kinetic energy is a form of energy possessed virtue of motion
whereas potential energy is an energy that an object possesses due to position or atoms and molecules possess due
to the bond associated with the intermolecular attractive forces.
At a temperature above absolute zero Kelvin, atoms or molecules are in state of random motions and hence possess
the kinetic energy. Three types of motions are identified during the movement of atoms or molecules, namely;
vibration motion, translation motion and rotational motion. The energy possessed due to the summation of those
modes of motions is kinetic energy.
Translation a movement of particle from one position to another is more convenient to monatomic molecules.
Solids, liquid and gases which comprise of two or more atoms can undergo internal vibration and rotation.
For moment discussion, let’s consider the situation occurs to gas molecules at temperature above absolute zero
Kelvin and applied pressure governed by the law of kinetic molecular theory (KMT).
What is kinetic theory? Kinetic theory explains the behavior of gases based on the idea that the gas consists of
rapidly moving atoms or molecules.
a) Solids
b) Liquids
c) Gases
Fig. 3.2 Demonstration of intermolecular distance between Solids, liquids and gases
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In solids, molecules are very closed each other and ordered as shown figure (a) above and in liquid state molecules
are slightly far apart from each other when compared with the intermolecular distance between solid whereas the
intermolecular distance between gas molecules is very far apart and molecules in gases freely move here and there
without any restriction. See figures (b) and (c) above respectively.
3.3.1 Success of kinetic theory
 Gives a molecular interpretation of pressure and temperature
 Consistent with gas law and Avogadro’s hypothesis
 Correctly explains specific heat capacities of many gases
3.3.2 Molecular nature of matters
There are many different scientists who gave their own ideas to prove molecular nature of matters.
Dalton’s atomic theory: this theory is also known as the molecular theory of matter. This theory was accepted
theory that suggested all objects around this universe are made up of small entities called atoms.
Gay Lussac’s theory: suggested that when gases combine chemically to yield another gas, their volumes are in
ratios of small integers.
Avogadro’s Law: equal volumes of all gases at equal temperature and pressure have same number of molecules.
Behavior of gases
Gases at low pressures and high temperatures much above that at which they liquefy or solidify approximately
satisfy the relation between their pressure, temperature and volume.
𝑷𝑽 = 𝑲𝑻…………. (3.1), where K = constant for a given sample of gas (varies with volume of gas). 𝑲 = 𝑵𝑲𝑩
Where N is number of molecules and KB is Boltzmann constant.
Then
𝑷𝑽 = 𝑵𝑲𝑩 𝑻……………… (3.2)
𝑷𝑽
 𝑵𝑻 = 𝑲𝑩 = constant (same for all gases)
If we have two different gases with gas one at (P1, T1, V1) and the other one at (P2, T2, V2), then we have the relation
𝑷𝟏 𝑽𝟏
𝑵𝟏 𝑻 𝟏
=
𝑷𝟐 𝑽𝟐
……………..
𝑵𝟐 𝑻 𝟐
(3.3)
This relationship will be true for all gases in the universe irrespective of whatever gases the relation satisfied for law
pressure and high temperature.
Justification of Avogadro’s hypothesis: if pressure, volume and temperature are the same, then N is also the
same for all gases i.e. if (P1, T1, V1) = (P2, T2, V2) then N1= N2. In plain language, number of molecules per unit
volume is the same at fixed pressure and temperature. (NA = Avogadro’s number = 6.02X 1023).
Perfect gas Equation
What is perfect gas equation? In behavior of gas equation above we have seen that 𝑷𝑽 = 𝑲𝑻. Perfect gas equation is
nothing different from that equation rather than it is basically representation of the same equation in different form.
Perfect gas equation deals with a behavior of gases at a particular substance or at a particular situation.
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𝑷𝑽 = 𝒏𝑹𝑻………….. (3.4)
Where n is number of mole,
𝒏=
𝑵
𝑵𝑨
=
𝒎
𝑴
= ………. (3.5)
𝒎 = 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇𝒎𝒂𝒔𝒔
M= molecular mass of the given gas.
R = ideal gas constant, = 𝑵𝑨 . 𝑲𝑩 ,
NA = Avogadro’s number and KB = Boltzmann constant
T = absolute temperature.
𝑷𝑽 = 𝒏𝑹𝑻 =
𝑵
. 𝑵𝑨 . 𝑲 𝑩 . 𝑻
𝑵𝑨
= 𝑵𝑲𝑩 𝑻, which is the same with equation (3.1)
Forms of perfect gas equation:
𝑷𝑽 = 𝒏𝑹𝑻
We have two possible values for n
1.𝒏 =
𝑵
𝑵𝑨
Then 𝑷𝑽 = 𝒏𝑹𝑻 =
𝑷 =
𝑹𝑻
𝝁𝑵
𝑨
Where 𝝁 𝒊𝒔
2. 𝒏 =
𝑵
𝑹𝑻,
𝑵𝑨
𝑷=
𝑵
𝑹𝑻
𝑽𝑵𝑨
𝑵
𝑹𝑻
𝑹𝑻
𝑨
𝑨
= (𝑽) . 𝑵 = 𝝁 𝑵
= 𝝁𝑲𝑻 ………………. (3.6)
𝑵
𝑽
= number density i.e. number of molecules per unit volume
𝒎𝒂𝒔𝒔
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
=
𝒎
𝑴
Then 𝑷𝑽 = 𝒏𝑹𝑻 =
𝒎
𝑹𝑻,
𝑴
𝑷=
𝒎
𝑹𝑻
𝑽𝑴
𝒎
= (𝑽) .
𝑹𝑻
𝑴
= 𝝆
𝑹𝑻
𝑴
𝑹𝑻
 𝑷 = 𝝆 𝑴 …………………. (3.7)
Where 𝝆 = mass density i.e. mass of the gas per unit volume.
 If the mass density or number density of a gas is given at a certain temperature, then we can calculate the total
exerted pressure by using either equation (3.6) or (3.7).
What is ideal gas?
A gas that satisfies the perfect gas equation exactly at all pressures and temperatures is an ideal gas. When we
started the topic behavior of gases we have said that nearly all gases satisfy the above equation at low pressure and
high temperature. But if there are some gases which satisfy this relation at all pressure and all temperature, then
these gases are ideal gases. But in reality no such gas exists. That is none of the gas is an ideal gas.
Even if the gas satisfies this equation at law pressure and high temperature, there is no gas that satisfies this
equation at all pressure and all temperature. So the concept of ideal gas is nothing but it is a theoretical concept
which is helpful for many other theoretical studies and it forms basics for theories but in reality there is no gas
87
By Gizachew Berhanu
Thermodynamics of materials
which is ideal. No real gas is truly ideal. So a gas which is not ideal is known as real gas. Real gases approach the
ideal gas behavior at low pressure and high temperatures.
Real gases deviation from ideal gas
Real gases approach the ideal gas behavior at low pressure and high temperatures.
𝑷𝑽 = 𝒏𝑹𝑻, For 1mole
n=1
𝑃𝑉
𝑇
𝑷𝑽 = 𝑹𝑻
𝑷𝑽
𝑻
Ideal gas
= 𝑹 = constant
𝒕𝒉𝒆𝒏 𝒕𝒉𝒆 𝒈𝒓𝒂𝒑𝒉 𝒊𝒔 𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒍𝒊𝒏𝒆
Fig. 3.3 the graph demonstrate that the real gas deviation from
ideal gas and approach to ideal gas at low pressure and high
P
temperature.
When the temperature increases, the molecules start getting even more random; so molecules are far away from
each other and the molecular interaction negligible. When the molecular interaction decreases we can see that the
gas behaves as ideal gas. From molecular structure we can also justify that real gas approach to ideal gas at low
pressure and high temperature.
Deducing Boyle’s law and charley’s law from perfect gas equation
𝑷𝑽 = 𝒏𝑹𝑻
1. Boyle’s Law
Consider the temperature and numbers of moles are constant. Then,
𝑷𝑽 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 …………. (3.8)
 Boyle’s law states that at constant temperature, pressure of a given mass of gas is inversely proportional to the
volume of the gas.
2. Charley’s Law
Consider again the above perfect gas equation at constant pressure and number of moles.
𝑷𝑽 = 𝒏𝑹𝑻 , P = constant and n = constant
𝑽
𝑻
=
𝒏𝑹
𝑷
= constant
𝑽 ∝ 𝑻……………………… (3.9)
 Charley’s law states that the volume of the gas is directly proportional to its absolute temperature at a fixed
pressure.
88
By Gizachew Berhanu
Thermodynamics of materials
Deducing Dalton’s Law of partial pressure from perfect gas equation
Total pressure of a mixture of ideal gases is the sum of partial pressure. That means if we have several ideal gases
mixed together in vessel, the total pressure of this mixture is equal to the sum of partial pressures. But what is
partial pressure? Partial pressure is basically the pressure exerted by a particular gas if only that gas is presented in
a vessel. For example if we have three types of gases let’s A, B and C, the partial pressure of A is the pressure
exerted by gas A if only gas A is presented in a vessel. The same is true for gases B and C.
From perfect gas equation we have
𝑷𝑽 = 𝒏𝑹𝑻
Since gases are mixed in one container, they have same volume and same temperature. Then the only thing that
gases differ is number of moles.
𝑷𝑽 = (𝒏𝟏 + 𝒏𝟐 + 𝒏𝟑 + ⋯ + 𝒏𝑵 )𝑹𝑻
𝑷=
(𝒏𝟏 + 𝒏𝟐 + 𝒏𝟑 +⋯+ 𝒏𝑵 )𝑹𝑻
𝑷=
𝒏𝟏 𝑹𝑻
𝒏 𝑹𝑻
𝒏 𝑹𝑻
+ 𝟐𝑽 + 𝟑𝑽
𝑽
𝑽
+ ….+
𝒏𝑵 𝑹𝑻
𝑽
𝑷 = 𝑷𝟏 + 𝑷𝟐 + 𝑷𝟑 + … . . + 𝑷𝑵 ……………… (3.10)
Example 3.1
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP if the
diameter of oxygen molecule is 3Å.
Solution
STP = standard temperature and standard pressure
= standard temperature = 273k
= standard pressure = 1atm
= diameter of O2 = 3Å = 3 x 10-10m = 3 x 10-8cm
= radius of O2 =
𝐷
2
=
3 𝑋10−8 𝑐𝑚
2
= 1.5 𝑋10−8 𝑐𝑚
= actual volume occupied by 1mole of any gas at STP = 22.4L = 22400cm3
Molecular volume = volume of 1molecule x number of molecules presented in O2 gas. =
4
𝜋𝑟 3 . 𝑁𝐴
3
NA = number of molecules in 1mole of O2 = Avogadro’s number
=
4
𝑥3.14𝑥(1.5𝑋10−8 𝑐𝑚)3 𝑥
3
∴ Ratio =
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒
𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
=
6.02𝑥1023 = 8.51𝑐𝑚3
8.51𝑐𝑚3
22400𝑐𝑚3
= 3.8𝑥10−4
89
By Gizachew Berhanu
Thermodynamics of materials
Example 3.2
A vessel contains two non-reactive gases neon (monatomic) and oxygen (diatomic). The ratio of their partial
pressures is 3:2. Estimate the ratio of:
a) Number of molecules and
b) Mass density of neon and oxygen in a vessel. Atomic mass of Ne = 20.2u and that of O2 = 32.0u.
Solution:
a) The first point noted here is the vessel contains non-reactive gases. That is ideal gases. Since
the mixture contains two ideal gases, it should satisfy Dalton’s partial pressure principle. So the Dalton’s partial law of
pressure states that the total pressure is equal to the sum of partial pressure of neon and oxygen.
𝑃 = 𝑃𝑁𝑒 + 𝑃𝑂2 , from equation (3.9)
In the question it is given that the ratio of partial pressure of neon to partial pressure of oxygen is equal to
𝑃
3:2, 𝑃𝑁𝑒
𝑂2
=
3
2
Since both gases are in the same vessel, their volume and temperature is the same.
𝑉𝑁𝑒 = 𝑉𝑂2 = 𝑉 𝑎𝑛𝑑 𝑇N𝑒 = 𝑇𝑂2 = 𝑇
𝑃𝑉 = 𝑛𝑅𝑇 ,
from equation (3.4)
𝑃𝑁𝑒 𝑉 = 𝑛𝑁𝑒 𝑅𝑇 …….. (1)
𝑃𝑂2 𝑉 = 𝑛𝑂2 𝑅𝑇 ………. (2)
Dividing (1) by (2)
𝑃𝑁𝑒 𝑉
𝑃𝑂2 𝑉
=
𝑛𝑁𝑒 𝑅𝑇
𝑛𝑂2 𝑅𝑇
𝑃𝑁𝑒
𝑃 𝑂2
,
=
𝑛𝑁𝑒
𝑛 𝑂2
𝑁
From equation (3.5) we have 𝑛 = 𝑁 ,
𝑛𝑁𝑒 =
𝐴
∴
𝑃𝑁𝑒
𝑃𝑂2
=
𝑛𝑁𝑒
𝑛 𝑂2
𝑁𝑁𝑒
𝑁𝑂2
=
=
𝑚𝑎𝑠𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦𝑜𝑓 𝑛𝑒𝑜𝑛
b) 𝑚𝑎𝑠𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑥𝑦𝑔𝑒𝑛 =
3
2
𝑁𝑁𝑒
𝑎𝑛𝑑𝑛𝑂2
𝑁𝐴
=
𝑁𝑂2
𝑁𝐴
= 1.5
𝜌𝑁𝑒
𝜌𝑂2
=?
From equation (3.7) we have
𝑃= 𝜌
𝑅𝑇
𝑀
𝑅𝑇
Then 𝑃𝑁𝑒 = ρ𝑁𝑒 𝑀
𝑁𝑒
𝑅𝑇
𝑎𝑛𝑑𝑃𝑂2 = 𝜌𝑂2 𝑀
𝑂2
By taking the ratio
𝑃𝑁𝑒
𝑃𝑂2
∴
𝑀𝑂
𝜌
𝜌
32
= (𝜌𝑁𝑒 ) . (𝑀 2 ) = (𝜌𝑁𝑒 ) . (20.2) =
𝜌𝑁𝑒
𝜌𝑂2
𝑂2
=
𝑁𝑒
3 20.2
𝑥 32
2
𝑂2
3
2
= 0.947
90
By Gizachew Berhanu
Thermodynamics of materials
Example 3.3
An air bubble of volume 1.0 cm3 rises from the bottom of Langano Lake 40m deep at a temperature of 12℃.
To what volume does it grow when it reaches the surface which has a temperature of 35℃?
Solution
Let’s suppose we have Langano Lake 40m deep as shown figure below.
Given: volume at the bottom 𝑉1 = 1.0𝑐𝑚3 = 1.0𝑥10−6 𝑚3
Temperature at the bottom 𝑇1 = 12℃ = 285𝐾
Temperature at the surface 𝑇2 = 35℃ = 308𝐾
Depth of the lake = 40m
Volume at the surface =?
Surface 35℃
40m
12℃
Fig 3.4 Animation picture represent Langano lake to show impact of ideal gas equation.
The pressure at the surface is nothing it is atmospheric pressure.
𝑃2 = 1𝑎𝑡𝑚 = 1.013𝑥105 𝑝𝑎
The pressure at the bottom
𝑃1 = 𝑃2 + 𝜌𝑔ℎ = 1𝑎𝑡𝑚 + 103 𝑥9.8𝑥40 = 493300𝑝𝑎
Where 𝜌 density of water 103kg/m3.
From ideal gas equation, we know that
𝑃1 𝑉1
𝑇1
𝑉2 =
=
𝑃2 𝑉2
𝑇2
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
=
493300𝑥1.0𝑥10−6 𝑥308
1.013𝑥105 𝑥285
= 5.263𝑐𝑚3
91
By Gizachew Berhanu
Thermodynamics of materials
3.3.3 Kinetic molecular theory of an ideal gas
The kinetic molecular theory of matter is a classical theory that can explain such seemingly divers as the pressure of
a gas, the heat capacity of metals, the average speed of electrons in semiconductors and electrical noise in resistors
are among many interesting phenomena.
Basics of kinetic theory:
Constant random motion: Gases are composed of a many particles that behave like hard spherical objects in a
state of constant, random motion.
Straight line motion: These particles move in a straight line until they collide with another particle or the walls of
the container.
Negligible volume: These particles are much smaller than the distance between particles, therefore the volume of a
gas is mostly empty space and the volume of the gas molecule themselves is negligible.
No forces of attraction: There is no force of attraction between gas particles or between the particles and the walls
of the container.
Elastic collisions: Collisions between gas particles or collisions with the walls of the container are elastic. That is,
none of the energy of the gas particle is lost in a collision.
The average kinetic energy of a collection of gas particles is dependent only upon the temperature of the gas. To
illustrate the significance of these postulates, consider the box containing a single molecule shown below. Start the
animation and observe the molecule, represented by the blue ball, bouncing and traveling (2L) back and forth
across the box. The collisions with the walls are perfectly elastic. Energy is neither gained nor lost from the
collision. Because the walls do not move, the molecule's speed is unaffected by the collision. The graph plots the
particle speed as a function of time. Observe that the speed has a constant value.
L
L
V
Vx
-Vx
L
L
t
Fig 3.5 illustration of oscillating particle throughout a box container and the particle oscillates with constant
velocity.
Again consider the box below in which several molecules exist. Start the animation and observe the blue molecule.
Unlike the system above, this system has multiple molecules and thus collisions between molecules occur.
92
By Gizachew Berhanu
Thermodynamics of materials
The speed is a constant in between collisions. At the moment of each collision there is an abrupt change in speed
and in direction. Thus the speed of a given molecule is not constant. From one collision to the next the molecule
speeds up or slows down.
Vz
Vy
Vx
Fig. 3.6 Illustration of multiple molecules oscillating across in a 3D container with different velocity
The resultant velocity is the magnitude of the summation of the three vector velocities.
V 2 = V2x + V2y + V2z … (3.11) Where V is the magnitude of average velocity.
Since the molecules are in random motion and collide randomly with each other, thereby exchanging kinetic
energy, the mean square velocity in the x-direction is the same with those in the y and z directions.
Vx = Vy = Vz …………… (3.12)
Assume the particle is vibrating in the x-direction and make a collision with a wall of the box length L and from
equation (3.11), we have
V 2 = 3V2x …………… (3.13)
The momentum of this collision is given by P = mv, in this case P = mvx, since we are only considering the
x dimension. The total momentum change for this collision is then given by
∆P = P2 – P1 = mv2 – mv1…… (3.14)
But V1 = V2 = Vx and opposite in direction.
∆P = mvx – ( - mvx) = 2mvx.........(3.15)
The total amount of time taken between collisions of the molecule with the wall for travelling forth and
bounce back is
L
t = t1 + t 2 = Vx +
L
Vx
=
2L
Vx
(3.16)
From Newton’s second law, the force exerted on the x-direction is given by
𝐹𝑥 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 =
𝑑𝑃
𝑑𝑡
=
2𝑚𝑉𝑥
2𝐿
𝑉𝑥
=
𝑚𝑉𝑋2
𝐿
(3.17)
The total force is the combinations of all forces contributed by particles in the system.
𝐹𝑥 = ∑ 𝐹𝑖 = 𝐹1 + 𝐹2 + 𝐹3 + ⋯ + 𝐹𝑁 =
𝑚
𝐿
( 𝑉12 + 𝑉22 + 𝑉32 + … . . + 𝑉𝑁2 ) …… (3.18)
93
By Gizachew Berhanu
Thermodynamics of materials
The mean square velocity is given by
V 2=
𝑉12 + 𝑉22 + 𝑉32 + …..+ 𝑉𝑁2
𝑁
= 3𝑣𝑥2 ……….. (3.19), where N is number of particles in the box.
N V 2 = 𝑉12 + 𝑉22 + 𝑉32 … + 𝑉𝑁2 = 3𝑣𝑥2 … . . (3.20)
Combining equations (3.13), (3.17), (3.18) and (3.20) gives
2
Nm V
𝐹=
…………… (3.21)
3𝐿
We can now solve for the total pressure exerted by the molecular collision across the area of the wall of
the box.
2
𝐹 = 𝑃𝐴 =
Nm V
2
= 𝑃𝑉 =
3𝐿
Nm V
….. (3.22), where A.L = V = volume
3
Example 3.4
Calculate the root mean square velocity of 0.5 moles of oxygen molecule in atmosphere at 2 atm if the
density of oxygen is 1.33kg/m3. Also calculate the root mean square velocity in x- direction.
Solution
1) From equation (3.22), we have
2
𝑃𝑉 =
Nm V
2
=𝑃=
3
Nm V
,
3𝑉
𝑚
𝑉
= 𝜌 and N = n.NA
Where NA is Avogadro’s number = 6.02x10-23 molecules.
2
2
3P = Nρ V , V
V= √
=
3P
Nρ
=
3P
ρnNA
3𝑥2𝑥1.013𝑥105 𝑁/𝑚2
3𝑃
𝑛𝜌𝑁𝐴
= √ 1.33𝑘𝑔
= 1.23𝑛 𝑚⁄𝑠𝑒𝑐.
6.02𝑥1023
(
)(0.5𝑚𝑜𝑙)(
)
𝑚3
𝑚𝑜𝑙
2) From equation (3.13) we have
2
V
= 3𝑉𝑋2
2
𝑉𝑟𝑚𝑠,𝑥 =
√V
3
=
√
(1.23 𝑥 10−9 )2
3
= 0.710𝑛 𝑚⁄𝑠𝑒𝑐.
94
By Gizachew Berhanu
Thermodynamics of materials
Temperature is proportional to average kinetic Energy.
We can obtain some insight relationship meaning of temperature by rewriting equation (3.21) and equating with
ideal gas law equation
2
Nm V
𝑃𝑉 =
3
2 1
( mV
3 2
1
mV
2
2
2
)=
=
Example 3.4
= 𝑛𝑅𝑇
𝑛𝑅𝑇
𝑁
3
KT
2
=
,
𝑅𝑇
𝑁𝐴
= 𝐾𝑇
K.E =
2
 𝑃𝑉 =
=
Nm V
3
2𝑁 3
( KT)
3 2
=
2𝑁 1
( mV
3 2
3
KT
2
2
)
= 𝑁𝐾𝑇, Where k is Boltzmann constant with value 1.382x10-23J/K
Example 3.5
Estimate the total number of air molecules in your class room at STP. Assume your class room is
a rectangular parallelepiped with area 15x15 and height 5m.
Solution: The number of air molecules in your classroom is given by the ideal-gas law. 𝑃𝑉 = 𝑁𝐾𝑇
𝑎𝑟𝑒𝑎 𝑜𝑓𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠 = 𝐴 = 𝑤. 𝑙 = 15𝑚𝑥15𝑚
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠 = ℎ 5𝑚
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠 = 𝑉 = 𝐴𝑥ℎ
𝑃𝑉 = 𝑁𝐾𝑇
𝑁=
𝑁=
𝑃𝑉
𝐾𝑇
,
𝑃𝑤𝑙ℎ
𝐾𝑇
but 𝑉 = 𝑤𝑙ℎ
𝑁=
1𝑎𝑡𝑚𝑋
101.325𝑘𝑝𝑎
𝑋15𝑚𝑋15𝑚𝑋5𝑚
𝑎𝑡𝑚
−23
1.382𝑋10
𝐽
(273𝐾)
𝐾
= 1.14 x 1031
Kinetic theory consistent with ideal gas equation and gas laws
1. consistent with ideal gas equation: from kinetic interpretation we found that
Average kinetic energy =
𝐸
𝑁
=
3
𝐾 𝑇
2 𝐵
=𝐸=
3
𝑁𝐾𝐵 𝑇
2
This shows that the total internal energy directly proportional to temperature. That is the internal energy of an ideal
gas depends only on temperature, not on pressure or volume. This is what exactly given by the ideal gas equation.
95
By Gizachew Berhanu
Thermodynamics of materials
2. consistent with Dalton’s law of partial pressure: for a mixture of ideal gases the total pressure is the sum of
each partial pressure of each constituent. From kinetic theory equation we have
2
Nm V
𝑃𝑉 =
2
𝑁𝑚 V
𝑃=
3
2 𝑁 1
3 𝑉 2
2
2
3
1
2
2
= ( ) 𝑚 V = 𝜇 𝑚 V where 𝜇 is number density.
3𝑉
2
𝜇. 𝑎𝑣𝑒. 𝑘. 𝐸
3
∴𝑃=
𝑃=
2
1
(𝜇1 2 𝑚1 V
3
𝑃=
1
(𝜇1 𝑚1 V
3
2
2
1
+ 𝜇2 2 𝑚2 V
1
+ 𝜇2 𝑚2 V
1
2
2
1
+ 𝜇3 2 𝑚3 V
2
+ 𝜇3 𝑚3 V
2
2
2
1
+ … . +𝜇𝑁 2 𝑚𝑁 V
3
+ … . + 𝜇𝑁 𝑚𝑁 V
3
2
𝑁
2
𝑁
)
)
In equilibrium, average kinetic energy of different gas will be equal.
1
𝑚 V
2 1
𝑚𝑁 V
𝑃=
2
=
1
2
𝑁
1
𝑚 V
2 2
2
2
=
3
𝐾 𝑇
2 𝐵
= 3𝐾𝐵 𝑇
1
(𝜇1 3𝐾𝐵 𝑇
3
+ 𝜇2 3𝐾𝐵 𝑇 + 𝜇3 3𝐾𝐵 𝑇 + ⋯ + 𝜇𝑁 3𝐾𝐵 𝑇)
= 𝐾𝐵 𝑇(𝜇1 + 𝜇2 + 𝜇3 + … + 𝜇𝑁 ) ,
=
𝑅
𝑇(𝜇1
𝑁𝐴
𝑅
+ 𝜇2 + 𝜇3 + … + 𝜇𝑁 )
𝑁
= 𝑁 𝑇( 𝑉1 +
𝐴
=
𝑅𝑇 𝑁1
(
𝑉 𝑁𝐴
=
+
𝑅𝑇
(𝑛1
𝑉
𝑁2
𝑉
𝑁2
𝑁𝐴
+
+
𝑁3
𝑉
𝑁3
𝑁𝐴
+ ⋯+
+ ⋯+
𝑁𝑁
),
𝑉
Do you remember 𝜇 =
𝑁
𝑉
?
𝑁𝑁
)
𝑁𝐴
+ 𝑛2 + 𝑛3 + … + 𝑛𝑁 ) , Do you remember 𝑛 =
𝑁
𝑁𝐴
?
𝑃1 + 𝑃2 + 𝑃3 + ….
This is the same with Dalton’s law of partial pressure. Therefore, the kinetic theory consists with ideal gas equation
as well as gas laws.
96
By Gizachew Berhanu
Thermodynamics of materials
Example 3.6
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the
root mean square speed of a helium gas atom at -20℃? (Atomic mass of Ar = 39.9u and that of He = 4.0u)
Solution
𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 ℎ𝑒𝑙𝑖𝑢𝑚 = 𝑀𝐻𝑒 = 4.0𝑢
𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑟𝑔𝑜𝑛 = 𝑀𝐴𝑟 = 39.9𝑢
𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 ℎ𝑒𝑙𝑖𝑢𝑚 = 𝑇𝐻𝑒 = −20℃ = 253𝐾
𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑎𝑟𝑔𝑜𝑛 = 𝑇𝐴𝑟 = ?
From kinetic theory we know that the mean average kinetic energy per molecules of a substance
is directly proportional to temperature
𝐸
𝑁
=
3
𝐾 𝑇
2 𝐵
What is this average kinetic energy? This average kinetic energy is nothing; but it is:
𝐸
𝑁
=
𝑚V
V
1
mV
2
2
𝑟𝑚𝑠
∴V
2
=
3
K T
2 B
= 3K B T V
3KB T
m
= √
𝐻𝑒
= √
2
3𝐾𝐵 𝑇
𝑚
=
3RT
M
= √
3𝑅𝑇𝐻𝑒
𝑀𝐻𝑒
𝑎𝑛𝑑 V
𝐴𝑟
3𝑅𝑇𝐴𝑟
𝑀𝐴𝑟
= √
In the question it is given that V
3𝑅𝑇𝐻𝑒
𝑀𝐻𝑒
√
𝑇𝐴𝑟 =
3𝑅𝑇𝐴𝑟
𝑀𝐴𝑟
= √
𝑇𝐻𝑒
. 𝑀𝐴𝑟
𝑀𝐻𝑒
=
=
𝑇𝐻𝑒
𝑀𝐻𝑒
=
𝐻𝑒
= V
𝐴𝑟
and then we have
𝑇𝐴𝑟
𝑀𝐴𝑟
253
. 39.9
4
= 2.52𝑥103 𝐾
∴This is the value of temperature at which root mean square speed of argon is equal to the root mean
square speed of helium.
97
By Gizachew Berhanu
Thermodynamics of materials
Example 3.7
A flask contains argon and chlorine in the ratio of 2:1 by mass. The temperature of the mixture is 27℃.
Obtain the ratio of:
1) Average kinetic energy per molecule and
2) Root mean square speed Vrms of the molecules of the two gases
(Atomic mass of Ar = 39.9u and that of Cl2 = 70.9u)
Solution
1) From kinetic theory we know that average kinetic energy =
3
𝐾 𝑇
2 𝐵
for all gases.
In the question it is given that the temperature of the mixture is 27℃ and constant for both gases, the two
gases are in thermal equilibrium.
The kinetic energy at 𝑇1 = 𝐸1 and at 𝑇2 = 𝐸2
Since 𝑇1 = 𝑇2 = 𝑇 𝑡ℎ𝑒𝑛, 𝐸1 = 𝐸2
∴ 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑠 𝐸1 : 𝐸2 = 1: 1
2) From example 3.6 above we have
V
𝑟𝑚𝑠
3KB T
m
= √
3RT
then,
M
= √
V
𝐴𝑟
= √
3𝑅𝑇𝐴𝑟
𝑀𝐴𝑟
𝑎𝑛𝑑 V
𝐶𝑙2
= √
3𝑅𝑇𝐶𝑙2
𝑀𝐶𝑙2
By taking the ratio Vrms of argon to Vrms of chlorine molecule we will get
V
V
3𝑅T
𝐴𝑟
𝐶𝑙2
=
√ 𝑀 𝐴𝑟
𝐴𝑟
3𝑅𝑇𝐶𝑙
√ 𝑀 2
𝐶𝑙2
𝑀𝐶𝑙2
= √𝑀
𝐴𝑟
70.9
= √39.9 = 1.33
3.3.4 Mean free paths
Mean free path is an important thing that we are going to discuss which the average distance between two
successive collisions. As have been already discussed above, inside the gas there are several molecules which are
randomly moving and colliding with each other. The distance that a particular molecule travels without colliding
with the other molecule is called mean free path. Any gas molecule travels some distance and collides with other
molecule in the gas. So the average distance it travels without collision is mean free path.
To derive mathematical expression for mean free path let’s assume each molecule of the gas is as sphere of
diameter “d”. Let us also assume the average speed of each molecule is ⟨𝑉⟩. Suppose this particular molecule
suffers to collision with any of molecules within a distance of d diameter as shown below.
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Thermodynamics of materials
𝑣̅ t
d
d
Fig. 3. 7 molecule randomly moving and makes a collision with other molecules. The arrow shows that
the free path the molecule can travel before collision.
The mean velocity at which molecules 𝐴 moves towards molecules 𝐵 is denoted by 𝑉𝐴 and has equal magnitude
with the velocity of molecule 𝐵 which denoted by 𝑉𝐵 . If the molecules collide each other by 90°, the resultant mean
velocity is
𝑉𝐴𝐵 = √𝑉𝐴2 + 𝑉𝐵2 = √2⟨𝑉⟩
The volume within which a molecule suffers collision = the total displacement (√2⟨𝑉⟩. ∆𝑡) times with area (𝜋𝑑2 )
𝑉 = √2⟨𝑉⟩∆𝑡𝜋𝑑2 = for one molecule.
If we have “𝜇” number of molecules per unit volume in the sample of the gas, then the total number of collisions
that will take place in time ∆𝑡 will be
= 𝜇⟨𝑉⟩∆𝑡𝜋𝑑2
Then the rate of collision, the number of collision per unit time will be
=
𝜇√2⟨𝑉⟩∆𝑡𝜋𝑑 2
∆𝑡
= 𝜇√2⟨𝑉⟩𝜋𝑑2 .
The time between any two collisions is given by
1
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
=𝜏=
1
𝜇√2⟨𝑉⟩𝜋𝑑 2
Then mean free path is the average distance travelled between two successive collisions.
𝑙 = 𝜏. ⟨𝑉⟩ =
∴𝑙=
1
. ⟨𝑉⟩
𝜇√2⟨𝑉⟩𝜋𝑑 2
1
…………..
√2𝜇𝜋𝑑 2
=
1
𝜇√2𝜋𝑑 2
(3.23)
Mean free path depends inversely on

Number density

Size of the molecules
99
By Gizachew Berhanu
Thermodynamics of materials
Example 3.8
Estimate the mean free path and collision frequency of a one mole of nitrogen molecule in a cylinder
containing nitrogen at 2.0atm and temperature 17℃. Take the radius of nitrogen molecule to be
roughly 1.0Å. Compare the collision time with the time the molecule moves freely between two
successive collisions. (Molecular mass of 𝑁2 = 28.0u).
Solution
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 𝑑 = 2𝑟 = 2𝑥1.0𝑥10−10 𝑚
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = 𝑀𝑁2 = 28.0𝑢
𝑃 = 2.0𝑎𝑡𝑚 = 2𝑥1.013𝑥105 𝑝𝑎
𝑇 = 17℃ = 290𝐾
𝑁 = 𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠𝑝𝑒𝑟𝑚𝑜𝑙𝑒 = 𝑛. 𝑁𝐴 = 1𝑥6.02𝑥1023 /𝑚𝑜𝑙𝑒
8.314𝐽
𝑅 = 𝑚𝑜𝑙𝑒.𝑘
a) From the expression we derived above, we have
𝑙=
1
√2𝜇𝜋𝑑 2
𝜇=
𝑁
𝑉
, 𝑤ℎ𝑒𝑟𝑒 d 𝑖𝑠 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑎𝑛𝑑 𝜇 𝑖𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡𝑣𝑜𝑙𝑢𝑚𝑒 ,
From perfect gas equation we know that
𝑃𝑉 = 𝑛𝑅𝑇,
𝜇=
𝑙=
𝑃𝑁
𝑅𝑇
=
𝑉=
𝑛𝑅𝑇
𝑃
=
𝑅𝑇
, 𝑓𝑜𝑟
𝑃
1𝑚𝑜𝑙𝑒
2.013 𝑥105 𝑝𝑎 𝑥 6.02𝑥1023 /𝑚𝑜l𝑒
8.314𝐽
𝑥
𝑚𝑜𝑙𝑒.𝐾
290𝐾
1
√2(3.14)(5.1 𝑥 1025 )(2 𝑥 10−10 )2
= 5.1𝑥1025 /𝑚3
= 1.0𝑥10−7 𝑚
b) This question asks to calculate the ratio between the time between collision and time taken for
collision. In order to calculate both the time between collision and time for collision, we need to
calculate the average velocity within which the molecule moves. In our previous light we know
that
𝑉𝑟𝑚𝑠 = √
3𝑅𝑇
𝑀
=√
3𝑥8.314𝑥290
28𝑥10−3
= 5.1𝑥102 𝑚/𝑠𝑒
𝑡𝑖𝑚𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 = 𝜏 =
𝑙
𝑉𝑟𝑚𝑠
=
1.0𝑥10−7 𝑚
5.1𝑥102 𝑚/𝑠𝑒𝑐.
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑠 𝑓𝑜𝑟 𝑎 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑡𝑜 𝑐𝑜𝑙𝑙𝑖𝑑𝑒 = 𝑡 =
𝑑
𝑉𝑟𝑚𝑠
= 2.0𝑥10−10 𝑠𝑒𝑐.
=
2.0𝑥10−10 𝑚
5.1𝑥102 𝑚/𝑠𝑒𝑐
= 4.0𝑥10−13 𝑠𝑒𝑐.
Since 𝑉𝑟𝑚𝑠 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
∴ 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 =
𝜏
𝑡
=
2.0𝑥10−10
4.0𝑥10−13
= 500
𝜏 = 500𝑡
100
By Gizachew Berhanu
Thermodynamics of materials
3.4 Law of Equipartition of Energy
Until now we have discussed about scope of thermodynamics, thermodynamic systems, thermodynamic
equilibrium, thermodynamic state variables and molecular kinetic theory. We will now start another new topic
called the law of equipartition of energy which is much related with molecular kinetic theory of gases. But before
we discuss the law of equipartition of energy, we will introduce a term called degrees of freedom.
3.4.1 Degree of freedom
What do we mean by degree of freedom?
Degree of freedom is independent displacements or rotations that specify the orientation of a system. In plain
language it is a degree of extent at which some particular object or system is free. To easily visualize the term
degree of freedom let’s consider a real life example. Suppose Chala and Yerosan are two brothers. Their mother
Adde Lello permits chala to go nearby place to play with his friend in the evening from 3:00 to 5:00. In the same
way she permits Yerosan to play in the evening from 3:00 to 8:00. As you can see, Chala has two hours free to play
whereas Yerosan has five hours to play. From this situation we can conclude that Yerosan is freer than Chala. There
is something which restricts Chala to play more. Therefore, Yerosan has more freedom than chala to spend his time
and play more.So degree of freedom will talk about the independent displacement or rotations that specify the
orientation of an object or systems. If we have an object, to what extent it can move whether it can translate, rotate
or it can vibrate to specify its orientation is known as degree of freedom.
Suppose we have a ball placed at the edge of the wall of the box and forced to move towards the opposite wall as
shown below.
Fig.3.8 the ball moving along horizontal
line represents one degree of freedom.
As you see from the above figure the ball can move only along horizontal line; no other option of moving. The
three arrows show the direction in which the only way the ball can move. Therefore, this ball will have one degree
of freedom. That is it can move in only one particular direction. So when a ball is restricted to move in a line, it can
move only in one way. Now let us consider another ball lying on a floor which will have two options to move.
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By Gizachew Berhanu
Thermodynamics of materials
Y
X
Fig. 3.9 the ball has two options to move and has two degrees of freedom
This floor is basically two dimensional. The ball can move either in horizontal arrow or vertical arrow. We can
conclude that this ball will have two degrees of freedom. If we consider the same ball is placed in space, it will
have three options to move; either in x-axis, y-axis or z-axis. Now we can say that it will have three degrees of
freedom. Till now we are talking about the movement or translation. So we will hope that you may have some idea
about what do we mean by the term degree of freedom. It tells you in how many ways a particular object can move
or rotate or vibrate.
Categories of degrees of freedom
At the beginning of this chapter we have tried to mention about three modes of motions. Translation motion,
rotation motion and vibrational motion. Therefore, we will have three categories of degrees of freedom.
 Translational degrees of freedom
 Rotational degrees of freedom
 Vibrational degrees of freedom
We will talk about each of them one by one as we go ahead.
1. Translational degree of freedom
What do we mean by translation? It is a motion of a particular object as a whole from one point to another. Till
now we have tried to explain the term degree of freedom as a ball example. At this moment we will try to see
degree of freedom interims of molecules. When we talk about an object, it is a group of molecules which again
molecules are group of atoms. Suppose we have oxygen molecules in which two oxygen molecules bonded
together. The oxygen molecule entire the bond is the body as a whole. When this body as a whole is moving from
one point to another that is called translation. It is not that one part of the body is moving and the other part is fixed
or not one atom is moving and another atom is at rest. It is not like that. Both of atoms as a whole is moving from
one point to another point. For a ball that we mention as example above basically talks about translation because it
was moving from one point to another as a whole. So in this case a molecule which is free to move in space will
need three coordinate to specify its location. Such a molecule will need three coordinate to specify its location that
is one coordinate along x-axis, another coordinate along y-axis and the third one is along z-axis. Therefore, it has
three degrees of freedom. Similarly a substance which is free to move in a plane that is two dimensional will need
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Thermodynamics of materials
two coordinate to specify its location. So it has two degrees of freedom. It is the same for a substance which is free
to move in a line will need one coordinate to specify its location and it has one degree of freedom.
Molecules of monatomic gas have only translational degrees of freedom. That is monatomic or gases which have
only one atom. For example if you consider helium atom it consists only helium atom and if you consider oxygen
molecule O2 that is diatomic; it consists two oxygen atoms. So for any monatomic gases we can have only
translational degrees of freedom. That is they can only move from one point to another; they cannot rotate because
there is no other molecule with respect to which it can do it. There is only one atom; so it cannot rotate neither it
can vibrate.
Each translational degree of freedom contributes a term that contains square of some variable motion. That is each
translational degree of freedom will contributes something to energy. Because when the object is moving from one
point to another, it is contributing something to energy. Variable of motion is a variable that determines motion
such as velocity. The velocity may be along Vx, Vy or Vz axis.
 when we talk of a motion along a line, we will talk about one of them Vx
 when we talk of motion along a plane, we will talk about two of them Vx and Vy
 When we talk of motion in space, we will talk about three of them Vx, Vy and Vz.
So each degree of freedom will contribute one term that has square of Vx, Vy or Vz variable of motion. It will
contribute a term
1
𝑚𝑉𝑥2
2
which will contribute to the energy. This is nothing but it is kinetic energy which involves
1
2
in the motion of molecule from one part to another. In thermal equilibrium, the average of each such term⟨ 𝑚𝑉𝑥2 ⟩
1
3
1
is equal to2 𝐾𝐵 𝑇. So when this is in 3D, it is equal to 2 𝐾𝐵 𝑇. Because for each term it contributes2 𝐾𝐵 𝑇 and for three
3
terms it is equal to 2 𝐾𝐵 𝑇.
2. Rotational degree of freedom
What do we mean by rotational degree of freedom? Rotational degree of freedom talks about independent rotations
that specify the orientation of a system. As we discussed in translational degree of freedom, it is only about the
motion of the whole body from one point to another. But, rotational degree is about the rotation of one part of the
body with respect to the other part to specify its orientation. This happens only in case of diatomic molecules. It is
not possible for monatomic molecules. Because in diatomic there are two atoms and the two atoms bonded together
so the rotation is possible for one atom with respect to the other one. Also in diatomic molecules there are
translational degrees of freedom too. Because of the entire molecule can move from one point to another point as a
whole. Therefore, they have translational degrees of freedom in addition to that they have rotational degrees of
freedom. Consider two oxygen atoms bonded together as shown in figure below.
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Thermodynamics of materials
Fig. 3.10 illustrates rotational possibilities in diatomic molecules and has five degrees of freedom
The two perpendicular lines to enter atomic distance are the two possibilities of rotations for diatomic molecules.
The above diagram shows that other than the three translational degrees of freedoms (along X, Y, Z axes), there are
extra two possible rotational degrees of freedoms for diatomic molecules. Therefore, similar to in translational,
each rotational degree of freedom contributes a term to the energy that contains square of a rotational variable of
motion. In translation, the variable of motion is linear velocity (Vx, Vy, Vz). But here in rotation, the variable of
motion is angular velocity (𝜔𝑥 , 𝜔𝑦 𝑎𝑛𝑑 𝜔𝑧 ). So the contribution for only rotational part along the two
perpendicular axes would be 𝐸𝑟 =
1
𝐼𝜔12
2
+
1
𝐼𝜔22 .
2
But if we consider the total energy contribution due to the
degrees of freedom for diatomic molecule, three would come from translational degrees of freedom
1
1
1
1
1
(2 𝑚𝑉𝑥2 , 2 𝑚𝑉𝑦2 , 2 𝑚𝑉𝑧2 ) and two from rotational degrees of freedom (2 𝐼𝜔12 , 2 𝐼𝜔22 ).
3. Vibrational degree of freedom
This vibrational degree of freedom vision come to picture for all kinds of molecules. It is observed in certain types
of molecules. For example some molecules like carbon monoxide have a mode of vibration even at moderate
temperature. That is their atoms oscillate along the inter-atomic axis like a one dimensional oscillator. This
vibration will not be observed in case of oxygen molecules. So it is not observed in all molecules; it is observed in
some molecules.
Fig.3.11 vibrational modes of motion of molecules in one dimension
So some molecules what will be happen are that the two atoms will vibrate about the inter-atomic axis. The
vibrational energy terms contain squared terms vibrational variables of motion as the same as translational and
rotational energy terms discussed above. What are vibrational variables of motion?
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Thermodynamics of materials
As discussed above, here we use the term one dimensional oscillator. When we talk of oscillator both kinetic as
well as potential energies are involved. So when we talk of contribution due to vibrational variables of motion two
things come into picture.
1. Kinetic energy due to the motion of the object. =
1
𝑑𝑦
𝑚(𝑑𝑥 )2
2
2. Potential energy due to force constant of oscillator =
𝐸𝑉 =
1
𝑑𝑦
𝑚( )2
2
𝑑𝑥
+
1
𝐾𝑦 2
2
1
𝐾𝑦 2
2
One remarkable thing we would care here is by translational and rotational degree of freedom contribute only one
square term of the variable of motion whereas vibrational degree of freedom contributes two square terms of the
variable of motion. That is because the vibrational degree of freedom consists of both kinetic energy and potential
energy.
Table 3.1 Comparison between three energy modes
Modes of motion
Translational
Energy
1
𝐸𝑡 = 𝑚𝑉𝑥2
Square term contributed
One square term
Rotational
𝐸𝑟 =
One square term
Vibrational
𝐸𝑣 =
2
1
𝐼𝜔2
2
1 𝑑𝑦 2
( )
2 𝑑𝑥
+
1
𝐾𝑦 2
2
Two square term
3.4.2 Law of Equipartition of Energy
This law states that in equilibrium, the total energy is equally distributed in all possible energy modes, with each
1
2
mode having an average energy equal to 𝐾𝐵 𝑇.What does this means that in equilibrium the total energy will be
equally distributed to each degree of freedom.
When we have been talking about translational mode, we said that it should have an average kinetic energy which
1
2
1
2
equal to 𝐾𝐵 𝑇 i.e. each translational degree of freedom should have 𝐾𝐵 𝑇, each rotational degree of freedom
1
1
should have 2 𝐾𝐵 𝑇, each vibrational degree of freedom should have 2x2 𝐾𝐵 𝑇 = 𝐾𝐵 𝑇. Because in each vibrational
degree of freedom there are two square terms of energy one is kinetic energy and the other is potential energy. In
the next section we will see how this law of equipartition of energy does help us to calculate the specific heat
capacity of various materials.
3.5 Laws of thermodynamics
In introduction section at the beginning of this chapter, we mentioned that thermodynamics is also based on
empirical laws, as is classical mechanics. There are four basic laws of thermodynamics.

Zeroth law of thermodynamics

Second law of thermodynamics and

First law of thermodynamics

Third law of thermodynamics
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3.5.1 Zeroth law of thermodynamics
Would you remember what do we mean by thermal equilibrium? Two systems will be in thermal equilibrium if
both of them are at the same temperature. What is zeroth law of thermodynamics? Zeroth law of thermodynamics
states that two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each
other. If we have two systems “system A” and “system B” such that “A” is in thermal equilibrium with a third
system “C” and “B” is also in thermal equilibrium with “C” , then “A” and “B” are in thermal equilibrium with
each other. The most significance of zeroth law of thermodynamics is that there should be a physical quantity
which should have the same value for the systems in thermal equilibrium. This physical quantity is defined as
temperature. So temperature is the quantity which determines whether or not the system is in thermal equilibrium
with the neighboring system.
Equation of state
In previous subsection of this chapter, we have discussed about thermodynamics state variables. When we talk of
state variables, it becomes very obvious to talk about equation of state. Equation of state is an equation which
relates state variables. Equation containing of any state variables such as pressure, volume, temperature or density
is an equation of state. All state variables are not independent i.e. you cannot say I have three state variables and all
of them are independent of each other; it doesn’t happen that way. If we take ideal gas as example and in case of
ideal gas, equation of state is:
𝑃𝑉 = 𝑛𝑅𝑇
Here we have three state variables P, V and T. For a fixed amount of mole we may have two independent variables
either (P, V), (T, V) or (P, T), but everything cannot be independent at the same time. This equation is an equation
of ideal gas that consists of state variables as example. Similarly, for any system you have an equation of state
which consists of state variables. But an important point to note here is that all state variables are not independent.
They cannot be independent all at a time.
Internal Energy
Internal energy is the sum of the kinetic energies and potential energies of the molecules constituting the system.
Do not get worried about the term molecules. You might be thinking that in thermodynamics we will deal about
macroscopic studies. You are right. We will have nothing to do with molecules or atoms in thermodynamics.
Please, note here that internal energy is a macroscopic quantity. It doesn’t see the kinetic energy or potential energy
of individual molecules or atoms. It is the total or the sum of kinetic energies and potential energies of all the
molecules constituting the system as a whole. So internal energy is basically the energy of the object as a whole.

Denoted by “U”.

Macroscopic variables of the system

Thermodynamic state variable
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Internal energy is an extensive state variable and depends on the size of the system and depends on its state
specified by specific values of pressure, volume and temperature. It doesn’t depend on how this particular state of
the system passed through, it only depend on that particular state. State variables by themselves tell us state of the
system at a particular instant. It doesn’t bother about the history how the system arrived to present state; it just
bothers about what is the state of the system at instant present time.
How can we change internal energy? Internal energy is something which we do not sense it from our self and
cannot visualize. Example, if you think of kinetic energy, you can immediately think of an object moving. When
we talk of potential energy, you can immediately think of something deformed from its shape or something being
released from its height. But if we talk of internal energy, you cannot visualize anything as such.
Because, it is the total energy of the molecules as the whole which is within the object. So how can we change the
internal energy? Basically, there are two ways by which we can change the internal energy. The two ways are heat
and work. Heat and work are the two different modes of changing the internal energy of the system.
The first way by we can change the internal energy of the system is transfer of heat. For instance,
assume we have a bottle in which we have tied a balloon on the top. This figure doesn’t tell exactly
how the balloon is tied to the bottle but if you try to do at home carefully, you will easily
understand the situation. You tied the balloon immediately at the neck of the bottle and then heat
the bottle. Finally, the bottle becomes in contact with an object which is at higher temperature.
Then due to the difference in temperature, heat flow takes place and then the balloon inflates or
blows up. The inflation of the balloon shows that the total internal energy increases. The kinetic
energy as well as potential energy of molecules inside the balloon is increasing. This means due
to transfer of heat there was a change in internal energy.
Similarly, the second mode which can change internal energy is work done. If we apply pressure
on the bottle from both sides, the air migrates to the balloon and inflates it as the first case. We conclude that
internal energy increases by two modes. One is when heat is absorbed and the second is when work is done on
the system. In contrast, internal energy decreases when the system loses heat and some work is done by
the system.
How internal energy different from heat and work? How these three quantities related to each
other?
In the previous light we saw that heat and work are two different ways to change the internal energy of the system.
Here we will see how to distinguish internal energy from heat and work. The first distinguish is that heat and work
are not state variables. Heat and work are just modes of energy transfer; they are not a property of any system.
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3.5.2 First law of thermodynamics
First law of thermodynamics mainly talks about the relation among heat, work and internal energy. This law states
that energy can neither be created nor destroyed; but transform from one form to another form. We have studied
conservation of energy everywhere. In all energy conservation it says that energy cannot be created or destroyed,
only transformed to other form. The same is true even in case of thermodynamics. If we take a ball falling from a
roof of a building as common examples, when the ball is at the roof of the building, it has only potential energy. As
it begins falling down the potential energy starts decreasing and kinetic energy starts increasing. This means
potential energy is not getting lost rather it is getting transform to kinetic energy.
Similarly, when we rub our hand palm together the kinetic energy or the mechanical
energy get transform into heat energy; because of which rubbing for quit long time we
tend to feel hot. The mechanical energy when we move our hands together get into heat
energy which giving us the feeling of warm. So first law of thermodynamics states that
sometimes it is appeared to us some kind of energy is getting lost somewhere, but
basically it is not getting lost somewhere it is getting transformed to some other form of energy. This was all about
the logical expression of first law of thermodynamics.
Now we will see mathematical expression, how is first law of thermodynamics expressed or defined
mathematically. First law of thermodynamics says that if an amount of heat ∆𝑄 is supplied to any system, then the
system is doing some amount of work ∆𝑤 on the surrounding.
Input
∆𝑄
System ∆𝑈
∆𝑊
Out put
∆𝑄 is the amount of input we give to system and ∆𝑤 is the amount of output of the system. It is observed that from
the amount of heat which is given to the system some part is lost and the remaining part is coming out as work done
on the surroundings. Where is the other part going? The lost part is used up in increasing the internal energy of the
system. So what happen when we supply an amount of heat to the system is the system uses small portion of this
heat in order to increase its own internal energy. Because, whenever heat is added to the system, there will be some
change in temperature of the system and the change in temperature will cause a change in internal energy. Since
internal energy is related to kinetic energy and potential energy, when temperature increases the movement of
molecules inside the system increases so the sum of kinetic and potential will change and the internal energy is also
change. To conclude, when you supply some amount of heat to the system, small portion of that heat is used up in
increasing or changing the internal energy of the system and the remaining amount of heat is given out as work
done on the surroundings. Then the first law thermodynamics can be summarized as
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∆𝑄 = ∆𝑈 + ∆𝑊………….. (3.24)
This is 1st law of thermodynamics
where
∆𝑄 = ℎ𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑏𝑦 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
∆𝑈 = 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
∆𝑊 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
Just before we started this topic on the 1st law of thermodynamics, we discussed the difference among heat, work
done and internal energy. Heat and work is not state variables whereas internal energy is state variables. The phrase
“state variable” is also mentioned as “state function” in many other thermodynamics text books. As mentioned
above it is something which is only depends on the state of the system; it doesn’t depend on how the state is
achieved. So, state variables are path independent. They do not depend up on which path was followed to reach that
particular state. Therefore, internal energy is path independent whereas heat and work are path dependents.
Conclusion of first law of thermodynamics
∆𝑄 = ∆𝑈 + ∆𝑊
∆𝑄 − ∆𝑊 = ∆𝑈
∆𝑄 − ∆𝑊 = 𝑝𝑎𝑡ℎ 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡
Table 3.2 path dependent and independent thermodynamic variables
∆𝑄
∆𝑊
∆𝑈
Non-state variable
Non-state variable
State variable
Path dependent
Path dependent
Path independent
∴The difference of two path dependent quantity is path independent quantity. This is the first conclusion which we
get from the 1st law of thermodynamics. Now let us consider certain spatial cases
Case 1: let us suppose a system undergoes a process such that ∆𝑈 = 0, no change in internal energy of the system.
1st law of thermodynamics: ∆𝑄 = ∆𝑈 + ∆𝑊
∆𝑄 = ∆𝑊
The heat supplied by surrounding to system should be equal to the work done by the system on the surroundings.
Whatever amount of heat is given to the system that much amount of work should be done on the surroundings.
Case 2: let us suppose that system is a gas in a cylinder with movable piston. Piston is a part of an engine that
consists of a short cylinder that fits inside a tube and moves up and down or backwards and forwards to make other
parts of the engine move. Since the piston is movable we can change the volume of the gas. If we push the piston
very much downwards, then the volume will decrease. Similarly if we move it upwards, the volume will increase.
So, as we move the piston the volume of the gas will change. As the piston is moved there is a work done involved
which is ∆𝑊.
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∆𝑊 = 𝐹𝑜𝑟𝑐𝑒 𝑥 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 but 𝐹 = 𝑃𝐴
𝑃 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑠𝑡𝑜𝑛
∆𝑊 = 𝑃𝐴∆ℎ = 𝑃∆𝑉 …………… (3.25)
1st law of thermodynamics
∆𝑄 = ∆𝑈 + 𝑃∆𝑉……………… (3.26)
This is the form of the 1st law of thermodynamics in case of a gas in a cylinder with movable piston.
3.5.3 Enthalpy
The enthalpy H of a thermodynamic system whose internal energy, pressure, and volume are U, P, and V is defined
as
𝐻 = 𝑈 + 𝑃𝑉 …….…………….. (3.27).
Since U, P, and V are state variables, H is also a state variable. Note from ∆𝑊 = 𝑃∆𝑉that the product of P and V
has the dimensions of work and hence of energy. Therefore it is legitimate to add U and PV. Naturally, H has units
of energy. Of course, we could take any dimensionally correct combination of state variables to define a new state
variable. The motivation for giving a special name to the state function U + PV is that this combination of U, P, and
V occurs often in thermodynamics. For example, let Q be the heat absorbed in at constant pressure process in a
closed system. The first law of thermodynamics ∆𝑄 = ∆𝑈 + 𝑃∆𝑉 gives
𝑄2 − 𝑄1 = 𝑈2 − 𝑈1 + 𝑃(𝑉2 − 𝑉1 )
= 𝑈2 − 𝑈1 + 𝑃𝑉2 − 𝑃𝑉1
= 𝑈2 + 𝑃𝑉2 − 𝑈1 − 𝑃𝑉1
= (𝑈2 + 𝑃𝑉2 ) − (𝑈1 + 𝑃𝑉1 )
𝑄2 − 𝑄1 = 𝐻2 − 𝐻1 → ∆𝑄 = ∆𝐻……………. (3.28) at constant pressure
Equation (3.28) says that for a closed system that can do only P-V work, the heat Q absorbed in a constant-pressure
process equals the system’s enthalpy change. For any change of state, the enthalpy change is
∆𝐻 = 𝐻2 − 𝐻1 = 𝑈2 − 𝑈1 + 𝑃𝑉2 − 𝑃𝑉1
∆𝐻 = ∆𝑈 + ∆(𝑃𝑉)………………… (3.29)
Where ∆(PV) = (PV)2 - (PV)1 =P2V2 - P1V1. For a constant-pressure process, P2 = P1 = P and ∆(PV) = PV2 - PV1 =
P ∆V. Therefore
∆𝐻 = ∆𝑈 + 𝑃∆𝑉 … … … … … .. (3.30)
3.5.3.1 Types of Enthalpy Change
Here we will focus on three major types of enthalpy change in which each type comprises its own subdivision
enthalpy change.

Enthalpy change for reaction
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
Enthalpy change for phase transformation

Bond enthalpy
3.5.3.1.1 Enthalpy change for reaction: The enthalpy change accompanying a reaction is called the reaction
enthalpy. The enthalpy change of a chemical reaction is given by the symbol ∆𝐻𝑟 . Mathematically:
∆𝐻𝑟 = (∑ 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 )– (∑ 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡)………………………. (3.31)
Example:
𝐶𝐻4 (𝑔) + 2𝑂2 → 𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙)
∆𝐻𝑟 = (∆𝐻𝐶𝑂2 + 2∆𝐻𝐻2 𝑂 ) − (∆𝐻𝐶𝐻4 + 2∆𝐻𝑂2 )
Enthalpy change for reaction consists:

Enthalpy change formation

Enthalpy of solution

Enthalpy change of combustion

Enthalpy change of hydration

Enthalpy change of neutralization
Standard enthalpy of reactions
Enthalpy of a reaction depends up on the conditions under which a reaction is carried out. It is therefore, necessary
that we must specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a
reaction when the participating substances are in their standard states. The standard state of the substance at
specified temperature is its pure form at 1bar pressure and 25℃ temperature. To identify it from other enthalpy of
reaction, standard conditions are denoted by adding superscript V to the symbol ∆𝐻 and written as ∆𝐻 𝑉 . Standard
enthalpy = sum of standard enthalpy of product – sum of standard enthalpy of reactant.
𝑉
𝑉
∆𝐻 𝑉 = ∑ ∆𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡
− ∑ ∆𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡
………………………………………. (3.32)
3.5.3.1.2 Enthalpy of formation
Enthalpy of formation is a form of enthalpy change for reaction which accompanying the formation of one mole of
a compound from its constituent elements.
Example: 𝐶 + 𝑂2 → 𝐶𝑂2 , ∆𝐻𝑓 = −394𝐾𝐽
Standard molar enthalpy of formation: when all the species of the chemical reaction are in their standard states,
the enthalpy of formation is called standard molar enthalpy of formation. It is the standard enthalpy change for the
formation of one mole of a compound from its elements in their most stable states of aggregation (also known as
reference states) and symbolized by ∆𝐻𝑓 .
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Table 3.3 standard enthalpy of formation
∆𝐻𝑓𝑜
Substance
𝐴𝑙2 𝑂3 (𝑠)
𝐵𝑎𝐶𝑂3 (𝑠)
𝐵𝑟2 (𝑙)
𝐵𝑟2 (𝑔)
𝐶𝑎𝐶𝑂3 (𝑠)
C(diamond)
C (graphite)
CaO(s)
𝐶𝐻4 (𝑔)
𝐶2 𝐻4 (𝑔)
𝐶𝐻3 𝑂𝐻(𝑙)
𝐶2 𝐻5 𝑂𝐻(𝑙)
𝐶6 𝐻6 (𝑙)
𝐶𝑂(𝑔)
𝐶𝑂2 (𝑔)
𝐶2 𝐻6 (𝑔)
𝐶𝑙2 (𝑔)
𝐾. 𝑚𝑙𝑒
-1675.7
-1216.3
0
+30.91
-1206.92
+1.89
0
-635.09
-74.81
+52.26
-238.86
-277.69
+49.0
-110.53
-393.51
-84.68
0
∆𝐻𝑓𝑜
Substance
𝐻𝐼(𝑔)
𝐾𝐶𝑙(𝑠)
𝐾𝐵𝑟(𝑠)
𝑀𝑔𝑂(𝑠)
𝑀𝑔(𝑂𝐻)2 (s)
𝑁𝑎𝐹(𝑠)
𝑁𝑎𝐶𝑙(𝑠)
𝑁𝑎𝐵𝑟(𝑠)
𝑁𝑎𝐼(𝑠)
𝑁𝐻3 (𝑔)
𝑁𝑂(𝑔)
𝑁𝑂2 (𝑔)
𝑃𝐶𝑙3 (𝑙)
𝑃𝐶𝑙5 (𝑠)
𝑆𝑖𝑂2 (𝑠)(𝑞𝑢𝑎𝑟𝑡𝑧)
𝑆𝑛𝐶𝑙2 (𝑠)
𝑆𝑛𝐶𝑙4 (𝑠)
𝐾. 𝑚𝑙𝑒
+26.48
-436.75
-393.8
-601.70
-924.54
-573.65
-411.15
-361.06
-287.78
-46.11
+90.25
+33.18
-319.70
-443.5
-910.94
-325.1
-511.3
Example 3.9
Calculate the standard enthalpy of formation of 𝐻2 𝐶4 for the thermochemical equation
𝑉
𝐶2 𝐻4 + 3𝑂2 → 2𝐶𝑂2 + 2𝐻2 𝑂 where ∆𝐻 𝑉 = −1323𝐾𝐽 , ∆ 𝐻𝐶𝑂
= −393.5 and ∆𝐻𝐻𝑉2 𝑂 = −249𝐾𝐽
2
per mole respectively.
Solution
𝐶2 𝐻4 + 3𝑂2 → 2𝐶𝑂2 + 2𝐻2 𝑂
In the question it is given that
∆𝐻𝑟𝑉 = −1323𝐾𝐽 𝑝𝑒𝑟 𝑚𝑜𝑙𝑒
𝑉
∆𝐻𝐶𝑂
= −393.5𝐾𝐽 𝑝𝑒𝑟 𝑚𝑜𝑙𝑒
2
∆𝐻𝐻𝑉2 𝑂 = −249𝐾𝐽 𝑝𝑒𝑟 𝑚𝑜𝑙𝑒
∆𝐻𝐻𝑉2 𝑂4 = ?
𝑉
𝑉
𝐻𝑟𝑉 = ∑ 𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡
− ∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡
−1323𝐾𝐽 = {2(−393.5) + 2(−249)}𝐾𝐽 − {∆𝐻𝐶𝑉2 𝐻4 + 3(∆𝐻𝑂𝑉2 = 0)}
−1323𝐾𝐽 = −1285𝐾𝐽 − ∆𝐻𝐶𝑉2 𝐻4
∆𝐻𝐶𝑉2 𝐻4 = 38𝐾𝐽 𝑝𝑒𝑟 𝑚𝑜𝑙𝑒
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3.5.3.1.3 Enthalpy of combustion
It is the enthalpy change accompanying the complete combustion of one mole of a substance in excess of
oxygen in air.
𝐶 + 𝑂2 → 𝐶𝑂2 , ∆𝐻 = −394𝐾𝐽
𝐶𝐻4 + 2𝑂2 → 𝐶𝑂2 + 2𝐻2 𝑂 , ∆𝐻𝐶 = −891𝐾𝐽
Enthalpy of combustion is always negative due to combustion is always exothermic process.
Example 3.10
A cylinder of cooking gas is assumed to contain 11.2kg of butane. The thermochemical equation for
combustion of butane is 𝐶4 𝐻10 (𝑔) + 13⁄2 𝑂2 (𝑔) → 4𝐶𝑂2 (𝑔) + 5𝐻2 𝑂(𝑙) , ∆𝐻𝐶 = −2658𝐾𝐽. If the family
needs 15,000KJ of energy per day for cooking, (a) how long would cylinder last? (b) assuming that
30% of gas is wasted due to incomplete combustion, how long the cylinder last?
Solution
𝐶4 𝐻10 (𝑔) + 13⁄2 𝑂2 (𝑔) → 4𝐶𝑂2 (𝑔) + 5𝐻2 𝑂 , ∆𝐻𝐶 = −2658𝐾𝐽
From this thermochemical equation we can see that 1mole of butane able to release 2658KJ energy.
Then we need to calculate the amount of energy expected from mass of 11.2kg.
Molar mass of butane = 4(12) + 1(10) = 58g/mol. Since we have 1mole of butane, we would have 58g
of mass of butane. If we have 2658KJ of released energy from 58g mass of butane, we will have
“X”KJ of energy from 11.2kg mass of butane. then 𝑋 =
11200𝑔 𝑋 2658𝐾𝐽
58𝑔
= 513268.97𝐾𝐽
a) In the question it is given that the family needs 15,000KJ amount of energy per day for cooking.
Then we need to calculate the number of days in which 513268.97KJ amount of energy is optimum.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 =
513268.97𝐾𝐽
15,000𝐾𝐽⁄𝑑𝑎𝑦
= 34 𝑑𝑎𝑦𝑠.
b) In question (a) above we assumed complete combustion of butane and set 100% of efficiency of the
cylinder and lasted 34 days. But here 30% of gas is wasted due to incomplete combustion means that
the cylinder of cooking is 70% efficient. Then the number of days the cylinder would last is
70%𝑥34𝑑𝑎𝑦𝑠
100%
= 23.8 ≈ 24 𝑑𝑎𝑦𝑠
113
By Gizachew Berhanu
Thermodynamics of materials
3.5.3.1.4 Enthalpy change of neutralization
It is the enthalpy change accompanying the complete neutralization of one gram equivalent of an acid by a base or
vice versa in a dilute aqueous solution
𝐻𝐶𝑙 + 𝑁𝑎𝑂𝐻 → 𝑁𝑎𝐶𝑙 + 𝐻2 𝑂 , ∆𝐻 = −57𝐾𝐽
Here note that enthalpy of neutralization for all strong acid and strong base is same = 57KJ.
3.5.3.1.5 Enthalpy change of solution
Enthalpy change of solution is an enthalpy accompanying the dissolution of one mole of substance in large excess
of a solvent so that further addition of solvent does not involve any more enthalpy change.
3.5.3.1.6 Enthalpy change of Hydration
It is the enthalpy change accompanying the hydration of one mole of anhydrous salt by combing with specific
number moles of water.
𝐶𝑢𝑆𝑂4 (𝑠) + 5𝐻2 𝑂 → 𝐶𝑢𝑆𝑂4 . 5𝐻2 𝑂 , ∆𝐻 = −78.3𝐾𝐽
3.5.3.2 Enthalpy change during phase transformation
Phase transformations also involve energy changes. For instance water exists in nature in the form of ice (solid
phase), water (liquid phase) and steam (gas phase) and one phase requires energy to transforms to another phase i.e.
ice requires heat to melt to water and water requires heat to become steam. The magnitude of energy change
depends on the strength of the intermolecular interactions in the substance undergoing the phase transformations.
This means that the energy required to vaporize one mole of acetone is less than it does to vaporize one mole of
water.
3.5.3.2.1 Enthalpy change of fusion
Enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard
enthalpy of fusion or molar enthalpy of fusion and symbolized by ∆𝐻𝑓𝑜 . Melting of a solid material is endothermic
and so all enthalpy of fusion are positive.
3.5.3.2.2 Enthalpy change of vaporization
Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure
𝑜
(1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization and symbolized by ∆𝐻𝑉𝑎𝑝
.
3.5.3.2.3 Enthalpy change of sublimation
𝑜
Standard enthalpy of sublimation ∆𝐻𝑆𝑢𝑏
is the change in enthalpy when one mole of a solid material sublimes at
constant temperature and standard pressure.
3.5.3.3 Bond Enthalpy Change
Energy is required to break a bond and energy is released when a bond is formed. With the reference to the
enthalpy associated with chemical bonds, two different terms are used in thermodynamics.
 Bond dissociation enthalpy
 Mean bond enthalpy
114
By Gizachew Berhanu
Thermodynamics of materials
Diatomic molecules = molecules dissociate to atoms
𝑜
𝐻2 (𝑔) → 2𝐻(𝑔); ∆𝐻𝐻−𝐻
= 435.0𝐾𝐽𝑚𝑜𝑙 −1
𝑜
𝐶𝑙2 (𝑔) → 2𝐶𝑙(𝑔); ∆𝐻𝐶𝑙−𝐶𝑙
= 242𝐾𝐽𝑚𝑜𝑙 −1
𝑜
𝑂2 (𝑔) → 2𝑂(𝑔); ∆𝐻𝑂−𝑂
= 428𝐾𝐽𝑚𝑜𝑙 −1
Polyatomic molecules = 𝐶𝐻4 → 𝐶(𝑔) + 4𝐻(𝑔); ∆𝐻 = 1664𝐾𝐽
3.5.3.3.1 l Lattice enthalpy
The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic
compound dissociates into its ions in gaseous state.
𝑁𝑎𝐶𝑙 → 𝑁𝑎+ + 𝐶𝑙 − ; ∆𝐻𝐿𝑜 = 788𝐾𝐽𝑚𝑜𝑙 −1
Reading assignment
It is impossible to determine lattice enthalpies directly by experiment; we use indirect method where we construct
an enthalpy diagram called a born- haber cycle. Please read this method from any chemistry book and try to
understand the sequences you will get.
Example 3. 11
The enthalpy of combustion of methane, graphite and dihydrogen at 298𝐾 are −890.3𝐾𝐽/𝑚𝑜𝑙,
−393.5𝐾𝐽/𝑚𝑜𝑙 and−285.8𝐾𝐽/𝑚𝑜𝑙 respectively. Calculate the enthalpy of formation of 𝐶𝐻4 (g).
Solution
The enthalpy of combustion of methane is given by
𝐶𝐻4 (𝑔) + 2𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙), ∆𝐻 = −890.3𝐾𝐽/𝑚𝑜𝑙
The enthalpy of combustion of graphite is given by
𝐶 + 𝑂2 → 𝐶𝑂2 ,
∆𝐻 = −393.5𝐾𝐽/𝑚𝑜𝑙
And the enthalpy of combustion of dihydrogen is given by
𝐻2 +
1
𝑂
2 2
→ 𝐻2 𝑂,
∆𝐻 = −285.8𝐾𝐽/𝑚𝑜𝑙
Then we have to form the reaction formation of 𝐶𝐻4 i.e.,
𝐶 + 2𝐻2 → 𝐶𝐻4
To do this we have to invert the first reaction and multiply the third reaction by 2.
𝐶𝑂2 + 2𝐻2 𝑂 → 𝐶𝐻4 + 2𝑂2 , ∆𝐻 = 890.3𝐾𝐽/𝑚𝑜𝑙
𝐶 + 𝑂2 → 𝐶𝑂2 , ∆𝐻 = −393.5𝐾𝐽/𝑚𝑜𝑙
2𝐻2 + 𝑂2 →
2𝐻2 𝑂, ∆𝐻 = −571.6𝐾𝐽/𝑚𝑜𝑙
𝑁𝑒𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 𝐶 + 2𝐻2 →
𝐶𝐻4 , ∆𝐻 = −74.8𝐾𝐽/𝑚𝑜𝑙
115
By Gizachew Berhanu
Thermodynamics of materials
Example 3.12
Enthalpies of formation of 𝑂, 𝐶𝑂2 , 𝑁2 𝑂 and 𝑁2 𝑂4 are −110, −393, 81 𝑎𝑛𝑑 9.7𝐾𝐽/𝑚𝑜𝑙. Find the value of
∆𝐻𝑟 if the reaction is: 𝑁2 𝑂4 + 3𝐶𝑂 → 𝑁2 𝑂 + 3𝐶𝑂2
Solution
∆𝐻𝑟 = ∑ 𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡 − ∑ 𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 = [81 + 3 ∗ −393]𝐾𝐽/𝑚𝑜𝑙 − [9.7 + 3 ∗ −110]𝐾𝐽/𝑚𝑜𝑙
= −777.7𝐾𝐽/𝑚𝑜𝑙
3.5.4 Specific Heat Capacity
Specific heat capacity is the amount of heat energy required to raise the temperature of a body per unit mass. In
order to raise the temperature of unit mass, whatever the amount of energy required is known as specific heat
capacity. Specific heat capacity depends on:

nature of the substance - it has no uniform value all the time

temperature
Specific heat capacity is denoted by “C” and has a unit𝐽(𝐾𝑔)−1 𝐾 −1. Mathematically expressed as
𝐶=
1 ∆𝑄
𝑚 ∆𝑇
…………….. (3.33)
Molar specific heat capacity
Molar specific heat capacity is the same as specific heat capacity with the only difference being in specific heat
capacity we talk about per unit mass whereas in molar specific heat capacity we talk about per unit mole.
𝐶𝑚 =
1 ∆𝑄
𝑛 ∆𝑇
…………… (3.34)
Depends on

Nature of the substance

Temperature and

Condition under which heat is supplied.
Unit = 𝐽𝑚𝑜𝑙 −1 𝐾 −1
Molar specific heat capacity at constant volume (𝑪𝑽 ) and Molar specific heat capacity at constant pressure
(𝑪𝑷 )
For ideal gas the relationship between molar specific heat capacity at constant pressure and molar specific heat
capacity at constant volume is
𝐶𝑃 − 𝐶𝑉 = 𝑅
Prove it!
116
By Gizachew Berhanu
Thermodynamics of materials
Solution
From 1st law of thermodynamics we know that
∆𝑄 = ∆𝑈 + ∆𝑊
∆𝑄 = ∆𝑈 + 𝑃∆𝑉…….. (1)
At constant volume ∆𝑉 = 0
∆𝑄 = ∆𝑈
Then molar specific heat capacity at constant volume (𝐶𝑉 )
1 ∆𝑄
1 ∆𝑈
( )
𝑛 ∆𝑇 𝑉
𝐶𝑉 = 𝑛 (∆𝑇 ) =
𝑉
∴ ∆𝑈 = 𝑛𝐶𝑉 ∆𝑇……… (2)
At constant pressure
𝐶𝑃 =
1 ∆𝑄
( )
𝑛 ∆𝑇 𝑃
∴ ∆𝑄 = 𝑛𝐶𝑃 ∆𝑇 …….. (3)
Combining equations (1), (2) and (3) we have
∆𝑄 = ∆𝑈 + 𝑃∆𝑉 = 𝑛𝐶𝑃 ∆𝑇 = 𝑛𝐶𝑉 ∆T + nR∆T
𝑛𝐶𝑃 ∆𝑇 = (𝐶𝑉 + 𝑅)𝑛∆𝑇
𝐶𝑃 = 𝐶𝑉 + 𝑅
𝐶𝑃 − 𝐶𝑉 = 𝑅
For ideal gas
3.5.4.1 Specific heat capacity ratio (𝜸) of gas molecules.
𝛾=
𝐶𝑃
𝐶𝑉
= Gamma
This gamma has specific values for different polyatomic gases
a) Monatomic gases: as we have discussed in the section of equipartition, for monatomic gases we can only have
translational degrees of freedom. They cannot have rotational or vibrational degrees of freedom. So the maximum
degrees of freedom they can have are three translational degrees of freedom.
1
In the same section we also discussed that each degree of freedom will contribute 2 𝐾𝐵 𝑇.Ttherefore, three degrees
3
of freedom will contribute2 𝐾𝐵 𝑇. Then the total internal energy of one gas from kinetic theory is
𝑈=
3
𝐾 𝑇𝑥𝑁𝐴
2 𝐵
=
3 𝑅
. 𝑇. 𝑁𝐴
2 𝑁𝐴
=
3
𝑅𝑇
2
From the above ideal gas theorem equation (2) we have ∆𝑈 = 𝑛𝐶𝑉 ∆𝑇 = 𝐶𝑉 ∆𝑇𝑓𝑜𝑟𝑜𝑛𝑒𝑚𝑜𝑙𝑒
∆𝑈
∆𝑇→0 ∆𝑇
𝐶𝑉 = lim
𝑑𝑈
= 𝑑𝑇 =
3
𝑅
2
𝐶𝑉 =
3
𝑅
2
117
By Gizachew Berhanu
Thermodynamics of materials
∴ 𝐹𝑜𝑟 𝑚𝑜𝑛𝑎𝑡𝑜𝑚𝑖𝑐 𝑔𝑎𝑠𝑒𝑠 𝑚𝑜𝑙𝑎𝑟 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑎𝑡 𝑐𝑜𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠
And 𝐶𝑃 = 𝐶𝑉 + 𝑅 =
3
𝑅
2
+𝑅 =
𝐶𝑃 =
5
𝑅
2
5
𝑅
2
∴ 𝐹𝑜𝑟 𝑚𝑜𝑛𝑎𝑡𝑜𝑚𝑖𝑐 𝑔𝑎𝑠𝑒𝑠 𝑚𝑜𝑙𝑎𝑟 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑠
𝐶𝑃
𝐶𝑉
∴ 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑖𝑜(𝛾) =
=
5
3
= 1.67
The ratio of the molar specific heat capacity remains the same value 1.67 for all monatomic gases.
b) Diatomic gases (rigid): rigid means the molecules we have has rotational degrees of freedom but they cannot
vibrate. They are rigid oscillator. Rigid oscillator means it can translate i.e. inter the molecule it can move from one
point to another. It can also rotate but cannot vibrate. Because, vibration is possible only when there is some
flexibility. There should not any rigid mass for vibration to take place. In our case we are considering that the
diatomic molecule is rigid. Therefore, for this molecule there will be three translational degrees of freedom plus
two rotational degrees of freedom and total five degrees of freedom. According to the law of equipartition of
1
2
energy, each degree of freedom will contributes 𝐾𝐵 𝑇 energy. Therefore five degrees of freedom will contribute
5
𝐾 𝑇
2 𝐵
𝑈=
energy. Then the total internal energy of one mole of gas will be
5
𝐾 𝑇. 𝑁𝐴
2 𝐵
5
𝑅𝑇
2
=
Then molar specific heat capacity at constant volume for diatomic molecule is
∴ 𝐶𝑉 =
𝑑𝑈
𝑑𝑇
=
For ideal gas
5
𝑅
2
𝐶𝑃 − 𝐶𝑉 = 𝑅
𝐶𝑃 = 𝐶𝑉 + 𝑅 =
5
𝑅
2
+𝑅 =
∴ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑖𝑜(𝛾) =
7
𝑅
2
𝐶𝑃
𝐶𝑉
=
7
5
= 1.4
For a rigid diatomic gas the ratio of molar specific capacity is 1.4.
Table 3.3 molar specific values at constant pressure and temperature
Gas
𝐶𝑃
𝐶𝑉
𝛾
Monatomic
5
𝑅
2
3
𝑅
2
1.67
Diatomic
7
𝑅
2
3
𝑅
2
1.4
118
By Gizachew Berhanu
Thermodynamics of materials
Example 3.13
What amount of heat must be supplied to 2.0x10-2kg of nitrogen at room temperature to raise its
temperature by 45℃ at constant pressure? (Molecular mass of𝑁2 = 28𝑢 and𝑅 = 8.314𝐽𝑚𝑜𝑙 −1 𝐾 −1)
Solution
The key point here is the constant pressure.
Mass of nitrogen = 2.0x10-2Kg = 20gm
Change in temperature = ∆𝑇 = 45℃ = 318𝐾
At constant pressure
∆𝑄 = 𝑛𝐶𝑃 ∆𝑇
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑛 =
Nitrogen is diatomic gas, then
=
𝐶𝑃 =
20
28
7
𝑅
2
= 0.714
=
7
𝑥8.314𝐽𝑚𝑜𝑙 −1 𝐾 −1
2
= 29.1𝐽𝑚𝑜𝑙 −1 𝐾 −1
∴ ∆𝑄 = 𝑛𝐶𝑃 ∆𝑇 = 0.714𝑥29.1𝑥318 = 6607.2𝐽 = 6.607𝐾𝐽
ii) Diatomic gases (non-rigid): let’s calculate specific heat capacity for diatomic gases which are non-rigid. Nonrigid means they will have translational degrees of freedom plus rotational degrees of freedom as well as vibrational
degrees of freedom. In this case we will have three translation degrees of freedom, two rotational degrees of
freedom and vibrational degree of freedom. The contribution from each degree of freedom is then
3 𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓𝑓𝑟𝑒𝑒𝑑𝑜𝑚 + 2 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓𝑓𝑟𝑒𝑒𝑑𝑜𝑚 + 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑚𝑜𝑑𝑒
1
1
3𝑥 2 𝐾𝐵 𝑇 + 2𝑥 2 𝐾𝐵 𝑇 + 𝐾𝐵 𝑇 =
7
𝐾 𝑇
2 𝐵
The total internal energy of one mole of the molecule is then 𝑈 =
7
𝑅𝑇.
2
∴ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 𝑡ℎ𝑒𝑛
𝐶𝑉 =
𝑑𝑈
𝑑𝑇
=
7
𝑅
2
∴ 𝐶𝑃 = 𝐶𝑉 + 𝑅 =
7
𝑅
2
+𝑅 =
Specific heat ratio = 𝛾 =
𝐶𝑃
𝐶𝑉
=
9
𝑅
2
9
7
= 1.3
c) Specific heat capacity of polyatomic gases: how many degrees of freedom will they have? They will have
3 𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓𝑓𝑟𝑒𝑒𝑑𝑜𝑚 + 3 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓𝑓𝑟𝑒𝑒𝑑𝑜𝑚
+ "n" 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓𝑓𝑟𝑒𝑒𝑑𝑜𝑚
In rotational degrees of freedom, we can also have along all three mutually perpendicular axes. Total internal
energy of one mole of gas will be equal to
3
𝑈 = (2 𝐾𝐵 𝑇 +
3
𝐾 𝑇
2 𝐵
+ 𝑛𝐾𝐵 𝑇) 𝑥𝑁𝐴 = (3 + 𝑛)𝑅𝑇
Where “n” is the number of vibrational mode.
119
By Gizachew Berhanu
Thermodynamics of materials
∴ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 𝑡ℎ𝑒𝑛
𝐶𝑉 =
𝑑𝑈
𝑑𝑇
= (3 + 𝑛)𝑅
𝐶𝑃 = (4 + 𝑛)𝑅
∴ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜(𝛾) =
4+𝑛
3+𝑛
3.5.4.2 Specific heat capacity of solids:
Till now we have been talking about specific heat capacity of gases; monatomic gases, diatomic gases and
polyatomic gases. Let’s now talk about specific heat capacities of solid materials. When we discuss about specific
heat capacity of solids, let’s consider the solid consists of “N” number of atoms. Each atom in the solid material can
oscillate about its main position and parts of vibrational degrees of freedom with energy contribution of𝐾𝐵 𝑇. In one
dimension the average energy will be𝐾𝐵 𝑇. But if we talk of 3-D, the average energy will be3𝐾𝐵 𝑇.Since we are
talking about solid materials, we need to talk about three dimensions. It is already mentioned that in case of solid
the atoms will oscillate about their main position. Therefore, the total internal energy of one mole of solid will be
𝑈 = 3𝐾𝐵 𝑇𝑥𝑁𝐴 = 3𝑅𝑇.
From 1st law thermodynamics we know that
∆𝑄 = ∆𝑈 + 𝑃∆𝑉
At constant pressure, the change in volume (∆𝑉) is negligible in case of solid materials since in solids the change in
volume is very less and solids are a very compacted molecules plus have a fixed shape.
∴ ∆𝑄 = ∆𝑈 = 3𝑅𝑇
𝐶𝑉
Molar specific heat capacity is then
𝑑𝑈
𝐶𝑉 = ( 𝑑𝑇 )
𝑉
3R
𝑑𝑄
And 𝐶𝑃 = (𝑑𝑇 ) since 𝑑𝑈 = 𝑑𝑄, 𝑡ℎ𝑒𝑛𝐶𝑉 = 𝐶𝑃
𝑃
∴ 𝐶𝑉 = 𝐶𝑃 = 3𝑅 This is the Dulong-Petit rule.
This is the relationship of specific heat capacity in solid materials.
T
3.5.4.3 Specific heat capacity of liquids (water)
In last subsection of specific heat capacity, we tried to discuss about both specific heat capacity of gases and solid
materials. Now let’s talk about specific heat capacity of water. Suppose let’s treat water as solid material. As
discussed before solid has “N” number of molecules and the same is true for water. For each atom we have the
average energy equal to3𝐾𝐵 𝑇 in three dimensions.For water (H2O), we have two hydrogen atoms and one oxygen
atom and we have a total three atoms. Therefore, the total internal energy will be
𝑈 = 3𝐾𝐵 𝑇𝑥3𝑥𝑁𝐴 = 9𝑅𝑇
∴ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑖𝑠 𝑡ℎ𝑒𝑛
𝐶𝑉 = 𝐶𝑃 =
𝑑𝑈
𝑑𝑇
= 9𝑅
∴This value 9R addresses approximately with experimental value.
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Conclusion on specific heat capacity discussion
We have been discussing all about specific heat capacity and its relation to degrees of freedom. What we were
studying so far was that classical mechanics tells us the predicted specific heat capacity which we calculated
depending on degrees of freedom should be independent of temperature since the term T never comes to picture
when we talk of specific heat capacity. However, if you see in reality, at temperature tends to zero, degree of
freedom doesn’t come into play. Degree of freedom becomes inefficient unfrozen at that time which is very
obvious. Degree of freedom basically deals with the motion of atoms and doesn’t come into picture at all when the
temperature of an object is zero. Therefore, how do we explain this paradox? This shows that the classical
mechanics was not enough to explain this contradiction. The latest class of physics quantum mechanics (chapter
two) able to overcome such paradox. Quantum mechanics states that minimum non-zero amount of energy is
required for degree of freedom to come into play.
3.5.5. Processes of thermodynamics
In the above discussion we tried to see the basics laws of thermodynamics (zero and first laws), enthalpy and
specific heat capacities of gases, liquids and solids. Now this is the time to discuss some of special processes of
thermodynamics. Before we go into those processes it is necessary to discuss about certain terminologies related to
thermodynamics.
Quasi static process
What is quasi static process? In a simple English static means something which is at rest and dynamic is something
which is in motion. Quasi static means something which is not totally at stationary or not totally moving. It is
partial or semi-static or intermediate process. It is hypothetical construct i.e. in reality there is nothing called quasi
static. In reality we do not see any such thermodynamic process which is semi-static, neither fully static nor fully
dynamic.
Now you may think that if it doesn’t exist in reality then what is the point of studying this? Or why was this process
ever evolved? That is because, when we go for our higher studying in science or when you go into your research
field, there are many things which you have to consider as hypothesis. Because using that concept you will reach to
any other things which are real life concept. Quasi static process is an infinitely slow process. It takes place very
slowly under way it departs from its original state. It is almost not observable. You do not realize its moving from
one state to another state. System changes its variables pressure, temperature and volume so slowly that it remains
in equilibrium with its surroundings throughout. Your variables do not change that much rapidly; you change your
state in such a way that the difference of the new variables to the initial variables is almost negligible. All the time
your system remains in equilibrium with the surroundings.
Which processes are approximated to quasi static processes? The first criterion is extremely slow process. Any fast
process cannot be quasi static process because a process which takes place in rapidly will not be in equilibrium with
surroundings due to map difference in macroscopic variables.
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The second criterion is processes not involving accelerating motion. Whenever, you will have accelerated motion,
you may have change in macroscopic variables and accelerated motion also signifies fast processes. The third
criterion is the process involving large temperature gradient. Let’s suppose you have a process in which initial
temperature is 𝑇1 and final temperature is 𝑇2. If the difference between 𝑇1 and 𝑇2 is very high, the system rapidly
changes from its initial state to its final state and cannot be quasi static process. Quasi static process should takes
place in such away there is very small temperature gradient.
Some special thermodynamic processes

Isothermal process

Isobaric process

Isochoric process

Adiabatic process
3.5.5.1 Isothermal process
You may have heard the term “iso” in any places such as isotopes, isomers in your chemistry class, isosceles
triangle in mathematics class; isotherms and isobars in physics class etc. Iso comes into picture whenever
something has same value. Isotope = elements with same atomic mass, isosceles triangle = triangle with same two
sides etc.
Whenever, the term “thermal” comes in any conversation we think something related to heat or temperature.
Similarly, when we say isothermal something is going to be the same. This something is temperature. Any process
in which temperature is constant is isothermal process. Other state variables change like pressure, volume etc. keep
changing but temperature remains constant throughout the process. For an ideal gas 𝑃𝑉 = 𝑛𝑅𝑇. If we keep
temperature constant throughout the process, then 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Isothermal P-V curve:
𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑐
𝑃=
𝐶
𝑉
∴ 𝑇ℎ𝑒 𝑃 − 𝑉 𝑔𝑟𝑎𝑝ℎ 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑙𝑖𝑐 𝑐𝑢𝑟𝑣𝑒 𝑤ℎ𝑖𝑐ℎ 𝑛𝑒𝑣𝑒𝑟 𝑐𝑟𝑜𝑠𝑠 𝑏𝑜𝑡ℎ 𝑉 − 𝑎𝑥𝑖𝑠 𝑎𝑛𝑑 𝑃 − 𝑎𝑥𝑖𝑠
P
Fig. 3. 9 P-V curve of isothermal process
V
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Since the graph shows inversely proportional, as the pressure keep increasing the volume keep decreasing and
the reverse is true.
Isothermal expansion of an ideal gas: is under constant temperature how does an ideal gas expand. Basically in
this portion what we deduce is how much work is done during isothermal expansion of ideal gas. If the
temperature is constant and gas expands, there is some amount of work done. Let’s suppose we have an ideal gas
whose initial state is given by (𝑃1 , 𝑉1 ) and isothermally expands to final state (𝑃2 , 𝑉2 ). This means that at constant
temperature T, the gas will expand from volume 𝑉1to volume𝑉2 . Note that all this expansion is quasi static
processes. At any intermediate stage the pressure is 𝑃 and the volume is 𝑉1 + ∆𝑉 i.e. the initial volume plus some
increasing volume. Then the work done is
∆𝑊 = 𝑃∆𝑉
We can find out the total work done by integrating the small amount of each work done
𝑉
2
∫ 𝑑𝑤 = ∫𝑉 𝑃𝑑𝑉
1
∆𝑊 =
𝑉2
∫𝑉 𝑃𝑑𝑉……………
1
(3.35)
But from ideal gas equation we know that
𝑃𝑉 = 𝑛𝑅𝑇 → 𝑃 =
𝑛𝑅𝑇
𝑉
Then
𝑉 𝑑𝑉
∆𝑊 = 𝑛𝑅𝑇 ∫𝑉 2
1
𝑉
𝑉
= 𝑛𝑅𝑇𝑙𝑛 (𝑉2 ) …………… (3.36)
1
In isothermal process there is no change in internal energy of an ideal gas. Because it is mentioned that
heat and work done are the two ways of changing internal energy.
In case of ideal gas change in internal energy depends only on temperature. If there is a change in
temperature, there is also a change in internal energy. In isothermal process temperature is constant and
change in temperature is zero. Since no change in temperature then no change internal energy during
isothermal process.
∆𝑇 = 0 → ∆𝑈 = 0
From first law of thermodynamics we know that
∆𝑄 = ∆𝑈 + ∆𝑊 → ∆𝑄 = ∆𝑊
∴ 𝑇ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑔𝑎𝑠 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑔𝑎𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠.
Isothermal contraction: the only difference from isothermal expansion is in case isothermal expansion work is
done by the gas and in case of the isothermal contraction on the gas by the surroundings. In general the work done
during isothermal process is given by
𝑉
𝑊 = 𝑅𝑇𝑙𝑛 (𝑉2 )
1
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Thermodynamics of materials
𝑪𝒂𝒔𝒆 𝟏
If 𝑉2 > 𝑉1 → 𝑊 > 0 → 𝐼𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛
Work is done by the gas; heat is absorbed by the gas and is endothermic process.
Case 2
If 𝑉2 < 𝑉1 → 𝑊 < 0 → 𝐼𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛
Work is done on the gas; heat is released from the gas and is exothermic process.
3.5.5.2 Adiabatic Process
Adiabatic process is thermodynamic process in which no any heat flow takes place between the system and its
surroundings. From first law of thermodynamics we have
∆𝑄 = ∆𝑈 + ∆𝑊 → 𝑏𝑢𝑡 𝑠𝑖𝑛𝑐𝑒 𝑛𝑜 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑜𝑐𝑐𝑢𝑟 𝑖𝑛 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 ∆𝑄 = 0 → 𝑑𝑈 = −𝑃𝑑𝑉
𝑛𝐶𝑉 𝑑𝑇 = −𝑃𝑑𝑉 …………. (1)
From ideal gas equation we know that 𝑃𝑉 = 𝑛𝑅𝑇. Since none of these state variables are fixed we have to
differentiate both sides of this equation and solve for change of temperature.
𝑑(𝑃𝑉) = 𝑑(𝑛𝑅𝑇) → 𝑑𝑃𝑉 + 𝑃𝑑𝑉 = 𝑛𝑅𝑑𝑇
𝑑𝑇 =
𝑑𝑃𝑉+𝑃𝑑𝑉
………….
𝑛𝑅
(2)
If we substitute (2) into (1), then we will have
𝑑𝑃𝑉+𝑃𝑑𝑉
)
𝑅
𝐶𝑉 (
= −𝑃𝑑𝑉
𝐶𝑉 𝑑𝑃𝑉 + 𝐶𝑉 𝑃𝑑𝑉 = −𝑅𝑃𝑑𝑉
𝐶𝑉 𝑑𝑃𝑉 = −(𝐶𝑉 + 𝑅)𝑃𝑑𝑉 → 𝑏𝑢𝑡 𝐶𝑉 + 𝑅 = 𝐶𝑃
𝐶𝑉 𝑑𝑃𝑉 = −𝐶𝑃 𝑃𝑑𝑉
𝑑𝑃𝑉 = −
𝐶𝑃
𝑃𝑑𝑉
𝐶𝑉
→ 𝑏𝑢𝑡
𝑑𝑃𝑉 = −𝛾𝑃𝑑𝑉 →
𝑃 𝑑𝑃
1 𝑃
∫𝑃 2
𝑉 𝑑𝑉
1 𝑉
= −𝛾 ∫𝑉 2
𝛾
𝑑𝑃
𝑃
𝐶𝑃
𝐶𝑉
= 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜 = 𝛾
= −𝛾
𝑑𝑉
𝑉
Integrate this equation from initial state to final state
= 𝑙𝑛𝑃2 − 𝑙𝑛𝑃1 = −𝛾𝑙𝑛𝑉2 + 𝛾𝑙𝑛𝑉1
𝛾
𝑙𝑛𝑃2 𝑉2 − 𝑙𝑛𝑃1 𝑉1 = 0
𝛾
𝛾
𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡………………….. (3.37)
∴ 𝑇ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑖𝑠 𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Adiabatic process P-V curve
Similar to isothermal process curve, we will also have adiabatic curve.
𝑃𝑉 𝛾 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑐
∴𝑃=
𝐶
𝑉𝛾
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Thermodynamics of materials
Similar to in isothermal process, pressure is inversely proportional to volume in adiabatic process with slightly
different (see example 3.14). The area under the curve represents the total work done.
P
𝑃2
𝑃1
𝑉1
V
𝑉2
Fig. 3. 10 P-V curve of adiabatic process
ocess
Adiabatic change of an ideal gas
In the section of isothermal process we discussed that either isothermal expansion or contraction there is some work
done. Similarly, during adiabatic change of an ideal gas also there is some work done. Let’s consider an ideal gas
whose initial state is (𝑃1 , 𝑉1 , 𝑇1 ), here in adiabatic case none of this three are constant and the final state is
(𝑃2 , 𝑉2 , 𝑇2 ). Then similar to isothermal process work done is:
𝑉
𝑊 = 𝑃 ∫𝑉 2 𝑑𝑉
1
Equation of state for adiabatic process is 𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑐 → 𝑃 =
𝑉 𝑑𝑉
𝐶
1−𝛾
𝑊 = 𝐶 ∫𝑉 2 𝑉 𝛾 = 1−𝛾 (𝑉2
1
𝛾
1−𝛾
− 𝑉1
)=
1
𝐶
(
1−𝛾 𝑉2𝛾−1
−
𝐶
𝛾−1
𝑉1
𝐶
𝑉𝛾
)
𝛾
𝑃2 𝑉2 = 𝑃1 𝑉1 = 𝐶
𝑊=
𝛾
𝑃 𝑉
1
(2 2
1−𝛾 𝑉2𝛾−1
−
𝛾
𝑃1 𝑉1
𝛾−1
𝑉1
)=
1
(𝑃2 𝑉2
1−𝛾
− 𝑃1 𝑉1 )
From ideal gas equation we have 𝑃𝑉 = 𝑅𝑇 for one mole.
𝑊=
𝑅
(𝑇
1−𝛾 2
∴𝑊=
− 𝑇1 ) =
𝑅
(𝑇
𝛾−1 1
𝑅∆𝑇
1−𝛾
− 𝑇2 )………………… (3.38)
Case 1: Work done is positive if initial temperature is greater than final temperature and work is done by the gas.
Therefore, in this case temperature of the gas decreases.
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Case 2: Work done is negative if initial temperature is less than final temperature and in this case temperature of
the gas increases.
Example 3.14
Compare the slopes of adiabatic and isothermal curves
Solution
In the discussion we tried to explain that an isothermal as well as adiabatic curve both looks similar.
The only difference is that adiabatic curve is steeper than isothermal curve due to the power of gamma.
For isothermal process:
𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑐 → 𝑃 =
𝑠𝑙𝑜𝑝𝑒 =
𝑑𝑃
𝑑𝑉
=
−𝐶
𝑉2
=
−𝑃𝑉
𝑉2
P
𝐶
𝑉
Isothermal
𝑃
= −𝑉
Adiabatic
For adiabatic process:
𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑐 → 𝑃 =
𝑠𝑙𝑜𝑝𝑒 =
𝑑𝑃
𝑑𝑉
=
−𝛾𝐶
𝑉 𝛾+1
=
−𝛾𝑃𝑉 𝛾
𝑉 𝛾+1
𝐶
𝑉𝛾
=
𝑃
𝛾 (− 𝑉)
V
= 𝛾. 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙.
∴ Slope of adiabatic curve is greater than slope of isothermal curve by a factor of gamma.
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Example 3.15
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure.
The walls of the cylinder are made of a heat insulator material and the piston is insulated by having
a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed quarter
of its original volume?
Solution
From the problem one can absolutely understood that since the walls of the cylinder as well as
the piston are insulated the process is adiabatic process. There is no any heat flow occur between
the system and its surroundings.
Initial state
final state
𝑃1
𝑃2
𝑉2 =
𝑉1
1
𝑉
4 1
The adiabatic gas equation for ideal gas is
𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, where gamma (𝛾) = 1.4 for diatomic molecule.
𝛾
𝛾
𝑃1 𝑉1 = 𝑃2 𝑉2
In the question it is given that the gas is compressed to quarter of its initial volume.
𝑃2
𝑃1
=
𝑉 𝛾
(𝑉1 )
2
𝑉1
𝛾
= ( 1 ) = 4𝛾
𝑉
4 1
𝑃2 = 4𝛾 𝑃1 = 41.4 𝑃1 = 6.964𝑃1
3.5.5.3 Isochoric Process
Isochoric process is a thermodynamic process in which volume remains constant throughout the process. In this
process no work is done on the system or by the system. From first law of thermodynamics we know that
∆𝑄 = ∆𝑈 + 𝑃∆𝑉
Here since volume is constant ∆𝑉 = 0 and then∆𝑄 = ∆𝑈 = 𝐶𝑉 ∆𝑇 𝑓𝑜𝑟 𝑢𝑛𝑖𝑡 𝑚𝑜𝑙𝑒. Absorbed
heat is completely used to change its internal energy and its temperature.
3.5.5.4 Isobaric Process
This process is also a thermodynamic process in which pressure remains constant throughout the process. Work
done in this case is ∆𝑊 = 𝑃∆𝑉
= 𝑃(𝑉2 − 𝑉1 ) = 𝑃𝑉2 − 𝑃𝑉1.
= 𝑛𝑅𝑇2 − 𝑛𝑅𝑇1 = 𝑛𝑅(𝑇2 − 𝑇1 ). Then first law of thermodynamics is
∆𝑄 = ∆𝑈 + 𝑃∆𝑉
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∴ Absorbed heat goes partly to increase internal energy and partly to do work.
3.5.5.5 Cyclic Process
As the name suggested the term “cyclic” is something related to circle. Cyclic process is the one when the system
returns to its initial state. In cyclic process the change in internal energy is zero.
P
Fig.3.11. cyclic process
V
Limitation of first law of thermodynamics: unable to predict direction of change or feasibility of process in
particular direction. For that we need to have another law that can reduce this limitation.
3.5.6 Second law of thermodynamics
We have very well discussed first law of thermodynamics and some of its applications. Now it is time to introduce
second law of thermodynamics. There are two statements for the second law of thermodynamics given by two
different scientists that they both mean the same. However, the student should remember those two statements
because many times he/she might be asked that give these particular statements of second law of thermodynamics.
1. Kelvin-Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and
the complete conversion of the heat into work. According to the first law of thermodynamics efficiency of an
engine can be 100%. Second law of thermodynamics put a restriction to this. It states that it is impossible that the
total amount of heat which is supplied to the system will get converted to work. There has to be some heat which
will be lost during the process. Complete conversion of heat to work is impossible.
2. clausius-statement: No process is possible whole sole result is the transfer of heat from a colder object to a
hotter object. This means any process in which transfer of heat takes place from a body at lower temperature to a
body at higher temperature is not possible. It is good if you very well remember both these statements even if they
both interpreted the same thing.
3.5.6.1 Reversible and irreversible process
a) Reversible (non- spontaneous) process: As the name indicate reversible means something which can be
reversed; process which has no natural tendency to occur and they do not occur at all naturally. A process that starts
from some initial state goes to some final state and can be reversed back in the same path to initial state. In general,
a thermodynamic process is reversible if the process can be turned back such that both the system and surroundings
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return to their original states with no other change anywhere else in the universe. Then a process is reversible if
only:
a) It is quasi static: a system which changes its state extremely slowly such that throughout the process the system
and surroundings remain in equilibrium.
b) No dissipative effects: includes things like frictional force, viscose force or any kinds of dissipative forces
where due to which you will lose some amount of energy. If there are dissipative effects in that case even if you try
to come back from final state to initial state, the system and surrounding will not able to completely reversed due to
some of the energy would be already lost.
So, if there are no dissipative effects and if the process is quasi- static only then the process can be reversible.
b) Irreversible process: the process which cannot be reversed back would be irreversible process. Irreversible
process is mostly known as spontaneous process. It is a process that takes place only in one direction without any
assistance once proper condition is set. May only be reversed by some external agency. Examples of irreversible
process are: falling of water from the hill, hot water cools down in room temperature, burning of wood, evaporation
of water and etc. an important note here is that spontaneous process doesn’t mean neither it occurs immediately nor
does it mean that it occurs fast. Once the condition is set no extra energy is required.
Doubts?
 What is the driving force of spontaneously occurring changes?
 What determines the direction of a spontaneous change?
The next section will answer such types of questions.
Enthalpy change and spontaneity
Daily observations

Water flows down the hill to lower or minimize its own potential energy.

Heat flows from hot body to cold body to minimize its energy.

Wood burns to lower its energy
These three processes shift from one position to another spontaneously or from one phase to another to lower its
own potential energy.
Conclusion: processes which are exothermic (Negative ∆𝐻) are spontaneous.
Let see some more spontaneous reaction
1
𝑁 (𝑔) +
2 2
𝑂2 (𝑔) → 𝑁𝑂2 (𝑔) ; ∆𝐻𝑟𝑜 = +33.2𝐾𝐽𝑚𝑜𝑙 −1
𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒, 𝑠) + 2𝑆(𝑙) → 𝐶𝑆2 (𝑙) ; ∆𝐻𝑟𝑜 = +128.5𝐾𝐽𝑚𝑜𝑙 −1
Final conclusion: decrease in enthalpy may be contributory factor for spontaneity, but it is not true for all cases.
Let’s examine another case in which ∆𝐻 = 0 i.e. there is no change in enthalpy, but still the process is spontaneous.
Consider diffusion of two gases into each other in a closed container as shown below.
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Fig. 3.12 spontaneous diffusion of two different gases and randomly distributed to each other
3.5.7 Entropy
Let us introduce another thermodynamic function entropy denoted by “S”. It is a measure of the degree of
randomness or disorder in the system. In chemical reaction entropy change can be attributed to rearrangment of
atoms or ions from one pattern in the reactants to another(the products). If the structure of products is very much
disorderd than that of the reactants there will be a resultant increase in entropy. Crystalline solid state is the state of
lowest entropy (most orderd) whereas the gaseous state is state of highest entropy. Increasing temperature, volume
or amount of substance increases the entropy. Note that entropy is state variable.
Entropy as a function of heat
Whenever heat is added to the system, it increases molecular motions causing increased randomness in the system.
Thus heat (Q) has randomizing influence on the system. Can we then equat entropy with heat? Expirience suggests
us that the distribution of heat also depends on the temperature at which heat is added to the system. System at
higher temperature randomness in it than one at lower temperature. Temperature is the measure of average kinetic
energy of particles in the system. This shows that entropy is directly proportional to the amount of heat added to the
system.
Heat added to a system at lower temperature cauces greater randomness than when the same quantity of heat is
added to it at higher temperature.This means that if we have two systems at different temperatures and add the same
amount of heat to both systems, the change in intropy at lower temperature is greater than the change in entropy at
higher temperature. Therefore, ∆𝑆 is related with 𝑄 and 𝑇 for reversible reaction as ∆𝑆 =
𝑄𝑟𝑒𝑣
.
𝑇
When a system is in
equilibrium, the entropy is maximum and change in entropy, ∆𝑆 = 0. Here one must note that entropy is directly
proportional with temperature whereas entropy change is inversely proportional with temperature.
∆𝑼 and ∆𝑺: Reversible and Irreversible Reaction
Both for reversible and irreversible expansion for an ideal gas under isothermal process ∆𝑈 = 0 whereas
∆𝑆𝑡𝑜𝑡 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟 ≠ 0. This shows that ∆𝑈 does not discriminate between reversible and irreversible
process whereas ∆𝑆 does.
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Second law of thermodynamics states that the entropy of the system is always increasing. To determine whether a
chemical reaction is spontaneous we must calculate the change in total entropy of the chemical system and its
surroundings.
∆𝑆𝑡𝑜𝑡 > 0 ∶ 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠
∆𝑆𝑡𝑜𝑡 = 0 ∶ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
∆𝑆𝑡𝑜𝑡 < 0 ∶ 𝑛𝑜𝑛 − 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠
Absolute Entropy
Disorder of a substance depends upon:Mass, molecular structure, temperature and Pressure
3.5.8 Third law of thermodynamics
Third law of thermodynamics suggustes that at absolute zero tempreature, the entropy of perfectly crystaline is
zero.
The entropy increases with increasing temperature. It is a measure of the thermal disorder introduced in
the solid, as it is heated above 0 K. The solid state is characterized by the random vibrations of atoms
about their mean positions. In the liquid state, the atoms have more freedom and can also move past one
another, resulting in greater disorder. Correspondingly, the entropy increases when a solid melts at
constant temperature.
𝑆=
𝑑𝑄
𝑇
= 𝐶𝑃
𝑑𝑇
𝑇
………….. (3.39)
3.5.8.1 Standard molar entropy
When one mole of substance is expressed at 298K and 1atm pressure, it is called standard molar entropy and is
denoted by 𝑆𝑚 .
Example: pridict in which of the following entropy increases or decreases.
a) a liquid crystallizes into a solid
b) temperature crystalline solid is raised from 0K to 115K.
Solution a) a liquid crystalizes to a solid means that from more disordering to ordering and then entropy decreases
b) since the temperature increases, the kinetic energy increase and the molecules become more random and entropy
encreases.
In addition to thermal entropy, a system may also possess configurational entropy, which is dependent on the
configurations of the system. Following Boltzmann’s definition, the configurational entropy can be written as
𝑆 = 𝑘 𝑙𝑛 𝑤 ……………….. (3.40)
Where k is Boltzmann’s constant and w is the number of different configurations of equal potential energy
in which the system can exist.
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3.5.8.2 The Statistical Nature of Entropy
The entropy of a system has been defined by Eqs. (3.39) and (3.40). The meaning of these equations can be
understood with reference to physical processes. Taking the configurational entropy first, consider the arrangement
of equal numbers of white and grey spheres on certain sites shown in Fig. 3.12. In Figs. 3.12, the arrangement is
disordered and random. It is easy to calculate the total number w of such distinguishable configurations that we can
have with these spheres. For generality, if we call the total number of sites as N and the number of white spheres as
n, the number of grey spheres is (N – n), and it is easily seen that
𝑤=
𝑁!
(𝑁−𝑛)!𝑛!
……………. (3.41)
The probability of the system existing in a disordered configuration is almost unity. Equation (3.40) states that the
configurational entropy increases as the logarithm of w. One mole of a solid contains more than 1023 atoms. If we
mix two different kinds of atoms randomly in a solid, we end up with an extremely large number of distinguishable
configurations and an appreciable amount of configurational entropy. Since w can never be less than one, the
configurational entropy is either zero or positive. It is zero for an absolutely pure solid, consisting of the same kind
of atoms on all its sites or for a perfectly ordered solid like a compound.
Consider a mole of atomic sites N in a solid. Let n atoms of B and (N – n) atoms of A be mixed randomly on these
sites. The configurational entropy is zero before mixing. The increase in entropy due to mixing is given by
𝑁!
∆𝑆 = 𝑆𝑓𝑖𝑛𝑎𝑙 − 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐾𝑙𝑛 ((𝑁−𝑛)!𝑛!) ……………… (3.42)
The following Stirling’s approximation is valid for small value of 𝑥.
𝑙𝑛𝑥! = 𝑥𝑙𝑛𝑥 − 𝑥 …………………….. (3.43)
Combining equations (3.42) and (3.43) we could have
∆𝑆 = 𝐾[𝑁𝑙𝑛𝑁 − (𝑁 − 𝑛) ln(𝑁 − 𝑛) − 𝑛𝑙𝑛(𝑛)] …………….. (3.44)
Equation (3.39) gives the thermal part of entropy as a function of temperature. At 0 K, we can visualize the atoms
of a solid to be at rest, so that there is no disorder due to temperature and, consequently, the entropy is zero. As the
temperature increases, the atoms begin to vibrate about their mean positions in the solid with increasing frequency
and amplitude. The atoms can be considered to oscillate in three modes (section 3.4.2) corresponding to the three
orthogonal directions. For many solids above room temperature, the frequency of these oscillations reaches a
constant value, about 1013 s–1. In contrast, the amplitude of the oscillations continues to increase with increasing
temperature. The average energy per atom per mode of oscillation is called the thermal energy and is equal to 𝑘𝑇,
where T is the temperature in kelvin and 𝑘 is Boltzmann’s constant.
All the atoms of a solid do not vibrate with the same amplitude and energy at any instant. The vibrating atoms
interact with their neighbours and, as a very large number of atoms are involved, the interaction effects are very
complex. Consequently, the vibrational energy of any particular atom fluctuates about the average value in a
random and irregular way. In other words, there is a statistical distribution of vibrational energies between the
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atoms. This distribution is the source of the thermal disorder. The entropy given by Eq. (3.39) should be associated
with this distribution. The thermal entropy arises from the vibrational energy distribution in a solid, while the
configurational entropy arises from the distribution in the configurational arrangements.
In the vibrational energy distribution, the fraction of atoms that possess energy equal to or greater than E is given
by the Maxwell-Boltzmann equation:
𝑁 = 𝑛𝑒𝑥𝑝 (– 𝐸/𝑘𝑇 ) ………………….. (3.45)
where n is the number of such atoms and N is the total number of atoms in the solid. The exponential term on the
right side, which is a fraction, also gives the probability that any given atom will have energy equal to or greater
than E. The Maxwell-Boltzmann equation is basic to the understanding of the equilibrium and kinetic processes in
materials.
3.5.8.3 Issue with total entropy
∆𝑆𝑡𝑜𝑡 = ∆𝑆𝑠𝑢𝑟𝑟 + ∆𝑆𝑠𝑦𝑠
We have eventhough find entropy of the system, it is very difficult to find entropy of surrounding. So how to find
entropy of surroundings?
The easiest way is∆𝑆𝑠𝑢𝑟𝑟 =
−𝑄
𝑇
=
−∆𝐻𝑠𝑦𝑠
𝑇
, where Q is released heat from the system and H is enthalpy of the system
at constant pressure.
∆𝑆𝑡𝑜𝑡 = ∆𝑆𝑠𝑢𝑟𝑟 + ∆𝑆𝑠𝑦𝑠
∆𝑆𝑡𝑜𝑡 =
−∆𝐻𝑠𝑦𝑠
𝑇
+ ∆𝑆𝑠𝑦𝑠
𝑇∆𝑆𝑡𝑜𝑡 = 𝑇∆𝑆 − ∆𝐻𝑠𝑦𝑠
−𝑇∆𝑆𝑡𝑜𝑡 = ∆𝐻𝑠𝑦𝑠 − 𝑇∆𝑆𝑠𝑦𝑠
∆𝐺𝑠𝑦𝑠 = ∆𝐻 − 𝑇∆𝑆𝑠𝑦𝑠 , where ∆𝐺𝑠𝑦𝑠 = −𝑇∆𝑆𝑡𝑜𝑡 , gibbs energy.
∆𝐺𝑠𝑦𝑠 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑓𝑜𝑟 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑝𝑟𝑜𝑐𝑒𝑠𝑠, 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑓𝑜𝑟 𝑛𝑜𝑛𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑎𝑛𝑑 𝑧𝑒𝑟𝑜 𝑓𝑜𝑟 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 .
3.5.9 Gibbs free Energy and spontaneity
Let’s define a new thermodynamic function Gibbs free energy or Gibbs function “G” as G = H – TS. Gibbs
function is an extensive property and state variable.
∆𝐺𝑠𝑦𝑠 = ∆𝐻𝑠𝑦𝑠 − 𝑇∆𝑆𝑠𝑦𝑠
Gibbs free energy is the amount of energy avaliable from a system at a given set of conditions that can be put into
useful work.
 If ∆𝐺𝑠𝑦𝑠 is negative (< 0), the process is spontaneous
 If ∆𝐺𝑠𝑦𝑠 is positive (> 0), the process is non-spontaneous and
 If ∆𝐺𝑠𝑦𝑠 is zero (= 0), the process is at equilibrium
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3.5.9.1 Gibbs free energy and feasibility of reaction
A process which is spontaneous at temperature 𝑇1 can be non-spontaneous at temperature 𝑇2 .
Endothermic process: Higher temperature favors spontanity
Exothermic process: Low temperature favors spontanity
∆𝐻 = 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒
∆𝐺 = ∆𝐻 − 𝑇∆𝑆
∆𝑆 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒
∆𝐻 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒
Always spontaneous
Spontaneous
at high temperature
∆𝑆 = 𝑁𝑒𝑔𝑎𝑡𝑖𝑣𝑒
Spontaneous
Always non spontaneous
at low temperature
Gibbs Free Energy change
𝐺 = 𝐻 − 𝑇𝑆
From equation (3.30), we know that 𝐻 = 𝑈 + 𝑃𝑉 and 𝑈 = 𝑇𝑆. Substitution and derivative of the above equation
will be reduced to
𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇 ………………………….. (3.46)
If temperature is kept constant the equation (3.46) will be reduced to
𝑑𝐺 = 𝑉𝑑𝑝 …………………………………… (3.47)
And from ideal gas equation we have
𝑃𝑉 = 𝑅𝑇
𝑉=
𝑅𝑇
𝑃
𝑑𝐺 =
substitude this into equation (3.47)
𝑅𝑇𝑑𝑃
𝑃
𝐺
𝑃
∫𝐺 𝑜 𝑑𝐺 = ∫𝑃 𝑅𝑇
𝑜
𝑑𝑃
𝑃
𝑃
→ 𝐺 = 𝐺 𝑜 + 𝑅𝑇𝑙𝑛 (𝑃 ) = 𝐺 𝑜 + 𝑅𝑇𝑙𝑛𝐾
𝑜
𝑜
Where 𝐺 standard gibbs free energy, 𝑃𝑜 standard pressure ,𝑃 intersted pressure and K reaction constant.
Gibbs free energy change and equilinrium
⃖ 𝐶+𝐼
𝐴+𝐵 →
∆𝐺𝑟 = 0
𝑃
∆𝐺 𝑜 + 𝑅𝑇𝑙𝑛 (𝑃 ) = 0
𝑂
𝑜
∆𝐺 = −𝑅𝑇𝑙𝑛𝐾 ……………………… (3.48)
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Thermodynamics of materials
Example 3.16
Reconsider fig.3.12 discussed in the section 3.5.6. If we assume white balls as 𝑁𝑖 atoms which are 16 in numbers
and grey balls as 𝐶𝑢 atoms which are 16 in numbers; in how many ways can we arrange if 32 sites are
available and calculate the possible configurational entropy.
Solutions
As it tried to mention in discussion section, if two atoms 𝐴 and 𝐵 are randomly mixed in available 𝑁 sites,
𝐴 type atoms randomly occupy 𝑛 and 𝐵 type atoms randomly occupy 𝑁 − 𝑛 sites. Therefore the number
of ways we can arrange them is given by
𝑤=
=
𝑁!
(𝑁−𝑛)!𝑛!
32!
(32−16)!16!
= 601080390 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠
∆𝑆 = 𝐾𝑙𝑛𝑤 = (1.38 𝑥 10−23 𝐽/𝑚𝑜𝑙. 𝐾) ln(601080390) = 2.8 𝑥 10−22 𝐽𝑚𝑜𝑙 −1 𝐾 −1
Example 3.17
Calculate the fraction of atoms with energies equal to or greater than 2 𝑒𝑉 at 300 𝐾 and 1500 𝐾.
Solution
The fraction of atoms that possess energy equal to or greater than E is given by the Maxwell-Boltzmann
equation:
𝑛 = 𝑁𝑒𝑥𝑝 (–
𝐸
)
𝑘𝑇
𝐴𝑡 𝑇 = 300𝐾 → 𝐾𝑇 =
𝑛
𝑁
= 0.026𝑒𝑣
= exp(− 2𝑒𝑣⁄0.026𝑒𝑣 ) = 3.92 𝑥 10−34
𝐴𝑡 𝑇 = 1500 → 𝐾𝑇 =
𝑛
𝑁
1.38 𝑥 10−23 𝑥 300
1.6 𝑥 10−19
1.38 𝑥 10−23 𝑥 1500𝐾
1.6 𝑥 10−19
= 0.13𝑒𝑣
= exp(− 2𝑒𝑣⁄0.13𝑒𝑣 ) = 1.93 𝑥 10−7
Example 3.18
Calculate ∆𝐺𝑟𝑜 for conversion of oxygen to ozone, 3⁄2 𝑂2 (𝑔) → 𝑂3 (𝑔) at 298K, if K for this
Conversion is 2.47 𝑥 10−29 .
Solution
∆𝐺𝑟𝑜 = −𝑅𝑇𝑙𝑛𝐾 = −8.314 𝑥 298 𝑥 ln(2.47 𝑥10−29 ) = 163 𝐾𝐽⁄𝑚𝑜𝑙
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Thermodynamics of materials
Example 3.19
At 60℃, dinitrogen tetra oxide is 50% dissociated. Calculate the standard free energy change at this
temperature and at one atmosphere.
Solution
𝑁2 𝑂4 ← 2𝑁𝑂2
𝑇 = 60℃ = 333𝐾
𝑃𝑁2 𝑂4 =
𝑃𝑁𝑂2 =
0.5
1.5
1
1.5
2𝑁2 𝑂
0 mole
1
total
1.5mole
1mole
= 1⁄3 𝑎𝑡𝑚.
= 2⁄3 𝑎𝑡𝑚.
2
𝐾𝑃 =
Start
At equilibrium
𝑁2 𝑂4
1mole
0.5mole
(𝑃𝑁𝑂2 )
𝑃𝑁2 𝑂4
=
(2⁄3)2
1⁄3
=
4
9
𝑥3=
4
3
4
∆𝐺 𝑜 = −𝑅𝑇𝑙𝑛𝐾𝑃 = −8.314 𝑥 333𝐾 ln 3 = −763.8𝐾𝐽/𝑚𝑜𝑙
3.6 Heat Engine
Till now whatever we studied about was zeroth law of thermodynamics, first law of thermodynamics, various terms
related to thermodynamics and second law of thermodynamics. Now we will look at application of thermodynamics
from which heat engine is one best example that was developed on the bases of thermodynamics. What is heat
engine? Heat engine is a device that converts heat into work. The name heat engine itself suggested an engine
related to heat. The train we have today and travelling in Addis Ababa from one sub-city to other are all modified
forms imported from china. The basic working of the heat engine in this train is that it converts heat into
mechanical work which causes the movement of the engine. We will not raise here the practical use of heat engine
and focus on the basic functioning of heat engine; how does heat engine basically works. No matter how
complicated the engine of the train looks like the basic functioning is still the same. Basic components which form
heat engine include:
1. Working substance: stem engine, diesel engine and gasoline engine are under category of heat engine. In this
case a working substance is stem for stem engine, mixture of diesel vapor, fuel vapor and water vapor in diesel
engine and fuel vapor and water in gasoline engine. The working substance is different in each of the cases but the
basic functioning remains the same.
2. External reservoir at higher temperature: external reservoir basically can be any object that is different for
different vehicles.
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3. External reservoir at lower temperature:
If you have these three things you will have heat engine. Basically in heat engine a body is configured to do work
when place in alternative contact with hot and cold bodies which are able to supply heat and absorb heat. Why do
we have external reservoir at higher temperature is to supply heat to the engine and why do we have another
external reservoir at lower temperature is to absorb heat from the engine. So one reservoir will supply heat to the
engine and the one will absorb heat from the engine. So what will the engine do is basically absorb heat from the
reservoir at higher temperature and do some work on the surrounding which converted to mechanical energy that
makes the vehicle move and the remaining amount of heat is given to the reservoir at lower temperature. And this
cycle keeps go on and on.
W
Hot reservoir
𝑄ℎ
Heat Engine
𝑄𝐶
Cold reservoir
Fig.3.13 diagram that illustrates how does heat engine work
The direction of the arrow shows that the heat engine takes 𝑄ℎ amount of heat from hot reservoir and does some
amount of work on the surroundings and the remaining amount of 𝑄𝐶 is returned to cold reservoir.
𝑄ℎ = 𝑊 + 𝑄𝐶 ……………………………. (3.49)
If you take stem engine or any other vehicle which you may see around you as example, you will see the work done
that the engine does on the surroundings is what you see in the form of mechanical energy that transfer to the
wheels and makes move them as a result the vehicle as a whole move. This is how heat engine converts heat energy
into work done which is very useful to us.
Efficiency of heat engine: when we talk of heat engine it is very important to talk about its efficiency. Efficiency
means how efficient the engine is. It indicates how much useful work is output for a given amount of heat energy
input and denoted by 𝜂.It is the ratio of output to input.
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝜂=
𝑊
𝑄ℎ
𝑜𝑢𝑡𝑝𝑢𝑡
𝑖𝑛𝑝𝑢𝑡
………………………. (3.50)
From equation (3.49) we know that 𝑊 = 𝑄ℎ − 𝑄𝐶
𝜂=
𝑄ℎ − 𝑄𝐶
𝑄ℎ
=1−
𝑄𝐶
𝑄ℎ
Is a 100% efficient heat engine possible? 100% efficiency means the heat engine properly use the given input and
did that amount of work without lose heat energy anywhere.
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𝜂 =1−
𝑄𝐶 = 0
𝑄𝐶
𝑄ℎ
Efficiency 100% means = 1 i.e.
𝑄𝐶
𝑄ℎ
=0,
No lose of heat.
First law of thermodynamics does not put any maximum efficiency of an engine whereas second law limits 100%
practical efficiency of any heat engine. Because there is always some amount of heat lost to the surroundings.
Example 3.20
A steam engine delivers 5.4x108J of work per minute and services 3.6x109J of heat per minute from its
boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Solution:
Work done = W = 5.4x108J per minute
Amount of heat the engine takes from the boiler = 𝑄1 = 3.6x109J per minute
Efficiency of the engine = η =?
Amount of heat wasted out of the engine = 𝑄2 =?
𝑄1
Boiler
Engine
𝑄2
Cold reservoir
W
Let’s suppose we have an engine as shown on the figure. This engine will take some amount of heat from
the boiler and do some amount of work on the surroundings and then give back the remaining heat to
cold reservoir. In this case boiler is the reservoir at higher temperature.
a) η =
𝑊
𝑄1
=
5.4𝑥108 𝐽
3.6𝑥109 𝐽
= 0.15
Q2 = wasted heat
Q1 = input
Engine
This means that the efficiency of an engine is 15%.
b) From equation (3.49) we know that 𝑄1 = 𝑊 + 𝑄2
W
𝑄2 = 𝑄1 − 𝑊
= 3.6x109J – 5.4x108J = 3.06x109J
3.06x109J per minute is wasted.
3.7 Refrigerators
In section 3.6 we have seen the basics of heat engine. Now let’s discuss about how a refrigerator function does. The
way of the functioning of a refrigerator is just the reverse of the heat engine. Heat engine used to take heat from the
hot reservoir and do some work on the environments and then return the remaining amount of heat to the cold
reservoir. But in case of refrigerator, it extracts heat from the cold reservoir and some work is done on the
refrigerator and then a certain amount of heat is transferred to hot reservoir.
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𝑄𝐶
Cold reservoir
Hot reservoir
Refrigerator
𝑄ℎ
W
Fig.3.14 diagram that illustrates principle functions of refrigerator.
𝑄ℎ = 𝑄𝑐 + 𝑊 ………………………………… (3.51)
Coefficient of performance (COP) of refrigerators (α): similar to efficiency of a heat engine, there is something
called COP of a refrigerator which tells how good the refrigerator function is. Mathematically it is the ratio of heat
extracted from cold reservoir to work done on the system and given by
𝛼=
𝑄𝐶
𝑊
………………………… (3.40)
It basically checks how nicely the system is able to extract heat from the cold reservoir. A point to be noted here is
that this COP value can be greater than one. Because, the value of heat which the system will take from the cold
reservoir can be greater than the amount of work which is done on the system. By energy conservation we can write
𝛼 in terms of 𝑄ℎ and 𝑄𝐶 .
𝛼=
𝑄𝐶
𝑊
=
𝑄𝐶
……………………………..
𝑄ℎ − 𝑄𝐶
(3.41)
Another important point must be noted here is the work done on the refrigerator can never be equal to zero. The
refrigerator will not function if there is no work done on it. In another language 𝑄ℎ − 𝑄𝐶 ≠ 0, which again implies
that 𝛼 will never be infinite. This means that the value of COP can be very well greater than one but cannot reach
infinity. There has to be a limit to its maximum value.
Example 3.19
A refrigerator is to maintain eatables kept inside at 9℃. If room temperature is 36℃, calculate the Coefficient of
performance.
Solution
In this problem what is given is that
Temperature of cold reservoir = 𝑇𝐶 = 9℃ = 282𝐾
Temperature of hot reservoir is room temperature = 𝑇ℎ = 36℃ = 309𝐾
𝛼=
𝑄𝐶
𝑄ℎ − 𝑄𝐶
=
𝐶𝑃 𝑇𝐶
𝐶𝑃 𝑇ℎ − 𝐶𝑃 𝑇𝐶
=
𝑇𝐶
𝑇ℎ − 𝑇𝐶
=
282𝐾
309𝐾−282𝐾
= 10.44
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Summary
 Thermodynamics involves work and heat. It is a branch of physics that deals with heat and temperature.
 Thermodynamics never talk about very small properties of molecules or atoms. It talks only about systems as a
whole. It will not talk about the behavior or properties of each molecule or atoms which constitute an object.
 A key property in thermodynamics is temperature. Temperature is an intensive property associated with the
hotness or coldness of a material. It determines the direction of spontaneous heat flow (always from hot to cold).
 System refers to any set of physical interactions isolated from the rest of the universe. Anything outside of the
system, including all factors and forces irrelevant to a discussion of that system, is known as surrounding.
 Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the
thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net
external force and torque on the system are zero.
 Properties can be either intensive properties or extensive properties; the definitions of each are given
below.
 An extensive property is a property that changes when the size of the sample changes. The property is
proportional to the amount of material in the system.
 An intensive property is a physical property of a system that does not depend on the system size or the
amount of material in the system. An intensive property doesn't change when you take away some of the sample.
 The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T ) is
𝑃𝑉 = 𝑛𝑅𝑇
 Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures.
 The law of equipartition of energy states that if a system is in equilibrium at absolute temperature T, the total
energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to
½ 𝑘𝐵 𝑇. Each translational and rotational degree of freedom corresponds to one energy mode of absorption and
has energy ½ 𝑘𝐵 𝑇. Each vibrational frequency has two modes of energy (kinetic and potential) with
corresponding energy equal to 2 × ½ 𝑘𝐵 𝑇 = 𝑘𝐵 𝑇.
 The mean free path l is the average distance covered by a molecule between two successive collisions.
 The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a third system are in
thermal equilibrium with each other’. The Zeroth Law leads to the concept of temperature.
 Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents
of the system. It does not include the over-all kinetic energy of the system. Heat and work are two modes of
energy transfer to the system. Heat is the energy transfer arising due to temperature difference between the
system and the surroundings.
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 Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing
the gas, by raising or lowering some weight connected to it.
 The first law of thermodynamics is the general law of conservation of energy applied to any system in
which energy transfer from or to the surroundings (through heat and work) is taken into account. It states
that
∆𝑄 = ∆𝑈 + ∆𝑊
where ΔQ is the heat supplied to the system, ΔW is the work done by the system and ΔU is the change in
internal energy of the system.
 Equilibrium states of a thermodynamic system are described by state variables. The value of a state
variable depends only on the particular state, not on the path used to arrive at that state. Examples of state
variables are pressure (P), volume (V), temperature (T), and mass (m). Heat and work are not state variables.
An Equation of State (like the ideal gas equation = 𝑛𝑅𝑇 ) is a relation connecting different state variables.
 A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical
equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the
environment can differ from those of the system only infinitesimally.
 The second law of thermodynamics disallows some processes consistent with the First Law of
Thermodynamics. Put simply, the Second Law implies that no heat engine can have efficiency η equal to 1
or no refrigerator can have co-efficient of performance α equal to infinity.
 A process is reversible if it can be reversed such that both the system and the surroundings return to their
original states, with no other change anywhere else in the universe. Spontaneous processes of nature are
irreversible. The idealized reversible process is a quasi-static process with no dissipative factors such as
friction, viscosity, etc.
 Heat engine is a device in which a system undergoes a cyclic process resulting in conversion of heat into
work.
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Further Reading
Materials Thermodynamics with Emphasis on Chemical Approach by HAE- GEON LEE Postech, Korea
Problems
A.
Choose the best answer
1. What is the name of the following statement: “When two systems are in thermal equilibrium with a
third system, then they are in thermal equilibrium with each other”?
(A) First Law of Thermodynamics
(B) Second Law of Thermodynamics
(C) Mechanical equivalent of heat
(D) Zeroth Law of Thermodynamics
(E) Thermal expansion of solids
2. An aluminum plate has a circular hole. If the temperature of the plate increases, what happens to the
size of the hole?
(A) Increases
(B) Decreases
(C) Stays the same
(D) Increases the top half of the hole
(E) More information is required
3. A bimetal plate consists of two materials of different coefficients of thermal expansion. The coefficient
of thermal expansion of the top part of the plate is less than the bottom part. If the temperature of the
entire plate increases, what happens to the plate?
(A) Expends
(B) Contracts
(C) Stays the same
(D) Bends down
(E) Bends up
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4. The state of an ideal gas was changed three times at three different temperatures. The diagram
represents three different isothermal curves. Which of the following is true about the temperature of the
gas?
(A) T1 > T2 > T3
(B) T1 > T2 < T3
(C) T1 < T2 < T3
(D) T1 > T2 = T3
(E) T1 = T2 > T3
5. A container filled with a sample of an ideal gas at the pressure of 1.5 atm. The gas is compressed
isothermally to one-fourth of its original volume. What is the new pressure of the gas?
(A) 2 atm (B) 3 atm (C) 4 atm (D) 5 atm (E) 6 atm
6. The state of an ideal gas was changed three times in a way that the pressure stays the same. The graph
represents three isobaric lines. Which of the following is true about the pressure of the gas?
(A) P1 > P2 > P3
(B) P1 > P2 < P3
(C) P1 < P2 < P3
(D) P1 = P2 > P3
(E) P1 > P2 = P3
7. The temperature of an ideal gas increases from 20 ̊C to 40 ̊C while the pressure stays the same. What
happens to the volume of the gas?
(A) It doubles
(B) It quadruples
(C) It is cut to one-half
(D) It is cut to one-fourth
(E) it slightly increases
8. The state of an ideal gas was changed three times in a way that the volume stays the same. The graph
represents three isobaric lines. Which of the following is true about the volume of the gas?
(A) V1 >V2 >V3
(B) V1 > V2 < V3
(C) V1 < V2 < V3
(D) V1 = V2 > V3
(E) V1 > V2 = V3
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9. A container with rigid walls filled with a sample of ideal gas. The absolute temperature of the gas is
doubled. What happens to the pressure of the gas?
(A) Doubles
(B) Quadruples
(C) Triples
(D) Decreased to one-half
(E) Decreased to one-fourth
10. The absolute temperature of an ideal diatomic gas is quadrupled. What happens to the average speed
of molecules?
(A) Quadruples
(B) Doubles
(C) Triples
(D) Increases by a factor of 1.41
(E) Stays the same
11. Two containers are filled with diatomic hydrogen gas and diatomic oxygen gas. The gases have the
same temperature. Compare the average speed of hydrogen molecules to the average speed of oxygen
molecules.
(A) 1/16
(B) 1/4
(C) 16/1
(D) 1/2
(E) 4/1
12. The average molecular kinetic energy of a gas depends on:
(A) Pressure
(B) Volume
(C) Temperature
(D) Number of moles
(E) None of the above
13. Kinetic Theory is based on an ideal gas model. The following statements about the ideal gas are true
EXCEPT:
(A) The average molecular kinetic energy is directly proportional to the absolute temperature
(B) All molecules move with the same speed
(C) All molecules make elastic collisions with each other and the walls of the container
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(D) The attractive force between the molecules can be ignored
(E) All molecules obey laws of classical mechanics
14. Internal energy of an ideal gas depends on:
i. the volume of the ideal gas
ii. The pressure of the ideal gas
iii. The absolute temperature of the ideal gas
(A) I (B) II (C) III (D) I and II (E) I, II and III
15. A sample of ideal gas has an internal energy U and is then compressed to one-half of its original volume
while the temperature stays the same. What is the new internal energy of the ideal gas in terms of U?
(A) U (B) 1/2U (C) 1/4U (D) 2U (E) 4U
16. An ideal gas with an internal energy U initially at 0 ̊C is heated to 273 ̊C. What is the new internal energy
in terms of U?
(A) U (B) 1/2U (C) 1/4U (D) 2U (E) 4U
17. Mechanical equivalent is associated with:
(A) Newton
(B) Kelvin
(C) Joule
(D) Boltzmann
(E) Avogadro
18. A sample of an ideal gas is taken through a closed cycle presented by P-V diagram. The process 1-2 is
perfectly isothermal. Which of the following is true about the change in internal energy and heat added to the
gas during the process 3-1?
(A) Δ U = 0 Q > 0
(B) Δ U > 0 Q > 0
(C) Δ U < 0 Q < 0
(D) Δ U = 0 Q = 0
(E) Δ U = 0 Q < 0
19. An ideal gas undergoes a cyclic process presented by the P-V diagram. Which of the following points has
the highest temperature?
(A) A (B) B (C) C (D) D (E) E
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20. An ideal heat engine operates between two temperatures 600 K and 900 K. What is the efficiency of the
engine?
(A) 50% (B) 80% (C) 100% (D) 10% (E) 33%
B. Write TRUE for true statement and write FALSE for false statement
1. The zeroth law of thermodynamics states that two objects in thermal equilibrium with each other must be
in thermal equilibrium with a third object.
2. The Fahrenheit and Celsius temperature scales differ only in the choice of the ice-point temperature.
3. The Celsius degree and the kelvin are the same size.
4. If the pressure of a fixed amount of gas increases, the temperature of the gas must increase.
5. Suppose that you compress an ideal gas to half its original volume, while also halving its absolute
temperature and then during this process, the pressure of the gas will be doubled.
6. The average translational kinetic energy of the molecules of a gas depends on the number of
moles and the temperature.
C. Write in one or two sentence
1. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at
standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The
stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T
surface?
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2. Liquid nitrogen is relatively cheap, while liquid helium is relatively expensive. One reason for
the difference in price is that while nitrogen is the most common constituent of the atmosphere, only
small traces of helium can be found in the atmosphere. Use ideas from this chapter to explain why it
is that only small traces of helium can be found in the atmosphere.
3. Imagine that you compress a gas while holding it at fixed temperature (perhaps by immersing the
container in cool water). Explain in terms of molecular motion why the pressure of the gas on the walls of its
container increases.
4. Imagine that you increase the temperature of a gas while holding its volume fixed. Explain in terms of
molecular motion why the pressure of the gas on the walls of its container increases.
5. Which speed is greater, the speed of sound in a gas or the rms speed of the molecules of the gas? Justify
your answer, using the appropriate formulas, and explain why your answer is intuitively plausible.
6. An astronomer claims that the temperature at the center of the Sun is about 107 degrees. Do you think that
this temperature is in kelvins, degrees Celsius, or doesn’t it matter?
7. Why might the Celsius and Fahrenheit scales be more convenient than the absolute scale for ordinary,
nonscientific purposes?
8. Two different gases are at the same temperature. What can be said about the average translational kinetic
energies of the molecules? What can be said about the rms speeds of the gas molecules?
9. By what factor must the absolute temperature of a gas be increased to double the rms speed of its
molecules?
10. “Yesterday I woke up and it was 20 ºF in my bedroom,” said JINENUS to her old friend DHUFERA
“That’s nothing, “replied DHUFERA. “My room was –5.0 ºC.” Who had the colder room, JINENUS or
DHUFERA?
Problems
1. A cylinder contains 7.0g of nitrogen gas. How much work must be done to compress the gas at a constant
temperature of 80℃ until the volume is halved?
2. A 50.0g aluminum disk at 300℃ is place in 200 𝑐𝑚3 of ethyl alcohol whose density is 0.79 𝑔⁄𝑐𝑚3 at10.0℃,
then quickly removed. The aluminum is found to have dropped 120℃ . What is the new temperature of the ethyl
alcohol?
3. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls
of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what
factor does the pressure of the gas increase if the gas is compressed to half its original volume?
4. 2 moles of a monatomic ideal gas is placed under a piston in a cylindrical container. The piston can move freely
up and down without friction and its diameter is 20 cm. A 50 kg weight is placed on the top of the piston. The
initial temperature of the gas is 0 ̊C.
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a. Calculate the absolute pressure in the container.
b. Calculate the volume of the gas.
The gas is heated to a temperature of a 100 ̊C at the constant pressure.
c. Calculate the new volume of the gas.
d. Calculate the amount of work done by the gas during its expansion.
e. Calculate the change in internal energy of the gas.
f. Calculate the amount of heat added to the gas.
5. One mole of an ideal gas is taken through a closed cycle A→B→C→A. Use the information
from the graph to answer the following questions:
a. Find the temperature at point A.
b. Is the net work done by the gas positive or negative? Explain.
c. How much work is done by the gas?
d. Is heat added to the gas or removed from the gas for the entire
cycle? Explain.
e. How much heat is added to the gas?
f. What is the change in internal energy for the entire cycle?
6. You are designing a vacuum chamber for fabricating reflective coatings. Inside this chamber, a small
sample of metal will be vaporized so that its atoms travel in straight lines (the effects of gravity are
negligible during the brief time of flight) to a surface where they land to form a very thin film. Assume
that the air in the chamber is at room temperature (300 K) and that the average diameter of an air
molecule is 4 × 10−10 m. The sample of metal is 30 cm from the surface to which the metal atoms will
adhere. How low must the pressure in the chamber be so that the metal atoms has diameter of only rarely
collide with air molecules before they land on the surface?
7. A steam engine delivers 5.4×108J of work per minute and services 3.6 × 109J of heat per minute from
its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
8 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75
joules per second. At what rate is the internal energy increasing?
9. Refrigerator is to maintain eatables kept inside at 9℃. If room temperature is 360C, calculate the
coefficient of performance.
10. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B,
an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process
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in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in
the latter case? (Take 1 cal = 4.19 J)
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Chapter Four: Crystal Structure
4.1 Introduction
In chapter three we have discussed that there are three states of matter; solids, liquids and gases. In the same
chapter under section 3.3, we tried to mention water as an example exists in three state forms; ice (solid), water
(liquid) and vapor (gas). If you see in this case the same matter is in three forms and the difference is only
temperature. Based on the pressure and temperature, a matter exists in various phases. Liquids and gases are called
fluids because of their ability to flow. The fluidity in both of these states is due to the fact that the molecules are
free to move around.
On the contrary, the constituent particles in solids have fixed positions and can only oscillate about their mean
positions. Under a given set of conditions of temperature and pressure, which of these would be the most stable
state of a given substance depends upon the net effect of two opposing factors. Intermolecular forces tend to keep
the molecules (or atoms or ions) closer, whereas thermal energy tends to keep them apart by making them move
faster. At sufficiently low temperature, the thermal energy is low and intermolecular forces bring them so close that
they cling to one another and occupy fixed positions. These can still oscillate about their mean positions and the
substance exists in solid state. The following are the characteristic properties of the solid state:
(i) They have definite mass, volume and shape.
(ii) Intermolecular distances are short.
(iii) Intermolecular forces are strong.
(iv) Their constituent particles (atoms, molecules or ions) have fixed positions and can only oscillate about their
mean positions.
(v) They are incompressible and rigid. This explains the rigidity in solids.
These solid materials have different applications based on their properties. These properties depend upon the nature
of constituent particles from which the material is composed and the binding forces operating between them. We
usually see different metals or any solid materials such as iron, plastics, glasses and many others. If we compare
these materials, all are solids but different properties. For instance, iron is conductor of electricity whereas plastics
and glasses are not. Plastics are soft while glasses are brittle. Therefore, study of the structure of solids is important.
The correlation between structure and properties discussed in chapter one helps in discovering new solid materials
with desired properties like high temperature superconductors, magnetic materials, biodegradable polymers for
packaging, biocompatible solids for surgical implants, etc.
In this Unit, we shall discuss different possible arrangements of particles resulting in several types of structures.
The correlation between the natures of interactions within the constituent particles and several properties of solids
will also be explored. How these properties get modified due to the structural imperfections or by the presence of
impurities in minute amounts would also be discussed.
By Gizachew Berhanu
Objectives of the Chapter
At the end of this chapter, the student should be able to

Describe general characteristics of solid state

Distinguish the difference between amorphous and crystalline solids

Understand space lattice, unit cell, 7 crystal structure and 14 Brarais lattices

Cite significant meaning of symmetry and symmetry operations

Explain close packing of particles in different unit cell

Derive point coordinate

Found out miller indices of crystallographic planes and direction

Describe different types of voids and close-packing structures

Calculate packing efficiency of different types of cubic and hexagonal unit cell

Correlate the density of the substance with its unit cell properties

Correlate some properties to crystal structure

Classify crystalline solids on the basis of the nature of binding forces

4.2 crystal
and amorphous structures
Solids can be classified as crystalline or amorphous on the basis of the nature of order present in the arrangement of
their constituent particles.
4.2.1 Crystal structure
Crystal is a three dimension translational periodic arrangement of atoms in space. Solid metals and most minerals
have a crystalline structure. A crystalline solid usually consists of a large number of small crystals, each of them
having a definite characteristic geometrical shape. In a crystal, the arrangement of constituent particles (atoms,
molecules or ions) is ordered. It has long range order which means that there is a regular pattern of arrangement of
particles which repeats itself periodically over the entire crystal. Sodium chloride and quartz are typical examples
of crystalline solids.
Crystalline solids are anisotropic in nature, that is, some of their physical properties like electrical resistance or
refractive index show different values when measured along different directions in the same crystals. This arises
from different arrangement of particles in different directions. This is illustrated in Fig. 4.2. Since the arrangement
of particles is different along different directions, the value of same physical property is found to be different along
each direction.
Note that the most of solid matter we see is crystalline. Nature by itself favors crystalline nature due to crystalline is
more symmetry and regular arrangement pattern. For example if you see your lips, nose, eyes, ears, arms and even
legs, all are symmetry by nature. So most of solids in nature are crystalline due to symmetry and the energy of
ordered arrangement is lower than the energy of irregular pattern. In chapter 2, we mentioned about energy in
By Gizachew Berhanu
which everybody (you, me, atom, molecule etc.) wants to be stable. The reason why you are learning is you want
stability live. The reason why a particular particle reacts with another substance is also same. Sodium reacts with
chlorine to become sodium ion which is more stable. Solid materials do the same thing and that is how the nature is.
In chapter 3, we discussed about lattice enthalpy and when crystal is formed energy is released due to crystal is
more stable and also difficult to break the bond between them.
Fig. 4.1 two dimensional crystalline structures
4.2.2 Amorphous structure
The term amorphous came from Greek word “amorphous” meaning no form. It consists of particles of irregular
shape and no long range order. In such structure, a regular and periodically repeating pattern is observed over short
distances only. Such portions are scattered and the arrangement of constituent particle is disordered. The structure
of amorphous solids is similar to that of liquids. Glass, rubber and plastics are typical examples of amorphous
solids.
Amorphous solids soften over a range of temperature and can be molded and blown into various shapes. On heating
they become crystalline at some temperature. Some glass objects from ancient civilizations are found to become
milky in appearance because of some crystallization. Like liquids, amorphous solids have a tendency to flow,
though very slowly. Therefore, sometimes these are called pseudo solids or super cooled liquids. Glass panes
fixed to windows or doors of old buildings are invariably found to be slightly thicker at the bottom than at the top.
This is because the glass flows down very slowly and makes the bottom portion slightly thicker.
Amorphous solids on the other hand are isotropic in nature. It is because there is no long range order in them and
arrangement is irregular along all the directions. Therefore, value of any physical property would be same along
any direction. These differences between crystalline and amorphous materials are summarized in Table 4.1.
Amorphous solids are useful materials. Glass, rubber and plastics find many applications in our daily lives.
Amorphous silicon is one of the best photovoltaic materials available for conversion of sunlight into electricity.
By Gizachew Berhanu
Table 4.1 differences between crystal and amorphous structure
Crystal
Amorphous
They have regular 3D arrangement of particles
They do not have regular arrangement
due to which they have well defined geometrical
of particles, therefore, do not have well defined
Shape.
geometrical shape.
They have long range order
They have short range order
They have sharp melting point
They do not have sharp melting point
They have high and fixed heat of fusion
They do not have fixed heat of fusion
They are anisotropic
They are isotropic
They are true solids
They are pseudo solids
They show two breaks in cooling curve
They show smooth cooling curve
More stable
Less stable
When cut with sharp tool, they split into two
When cut with a sharp edged tool, they cut into
pieces and newly generated surfaces are plain and
two pieces with irregular surfaces
smooth
4.2.3 Anisotropy and Isotropy
Crystalline substances differ from amorphous solids. The properties such as electrical conductivity, thermal
conductivity, mechanical strength and refractive index are different in different directions. Crystalline substances
are, therefore, said to be anisotropic. For example, the velocity of light passing through a crystal varies with the
direction in which it is measured. Thus, a ray of light entering such a crystal may split up into two components each
following a different path and travelling with a different velocity. This phenomenon is known as double
refraction. Thus, anisotropy in itself is a strong evidence for the existence of ordered molecular arrangements in
such materials. Amorphous solids, on the other hand, are isotropic, i.e., their physical properties are the same in all
directions. Liquids and gases are also isotropic.
If the properties are measured along the direction indicated by the slanting line CD in fig.4.2, they will be different
from those measured in the direction indicated by the vertical line AB. The reason is that while in the first case,
each row is made up of alternate type of atoms, in the second case; each row is made up of one type of atoms only.
In amorphous solids as well as in liquids and gases, atoms or molecules are arranged at random and in a disorderly
manner and, therefore, all directions are identical and all properties are alike in all directions.
By Gizachew Berhanu
B
C
D
A
a
b
Fig .4.2 Anisotropy (a) and isotropy properties of materials (b)
4.2.4 Crystal lattice
We have seen that the main feature of crystalline solid is the regular and repeating pattern of constituting particles.
Such regular three dimensional arrangements of points in space is called crystal lattice. A lattice is an infinite
periodic array of geometric points in space without any atoms which never formed in amorphous solids. A lattice
may be one-, two-, or three-dimensional. In one dimension, there is only one possible lattice: It is a line of points
with the points separated from each other by an equal distance, as shown in Figure 4-3.
Why we are studying crystal lattice? The physical properties we have discussed before such as melting point,
boiling point, malleability, ductility, electrical conductivity, thermal conductivity etc., of a material is actually
depends on this crystal lattice. Those physical properties are highlight view, but actually what we should
understand deep for a given compound is why this compound behaves like this manner, we have to understand
crystal lattice. It is very, very critical; this crystal lattice explains the physical properties of the crystal.
Crystal lattice characteristics

Each point in a lattice called lattice point or lattice site.

Each point in a crystal lattice represents one constituent which may be an atom, a molecules (group of
atoms) or ions.

Lattice points are joined by straight lines to bring out the geometry of the lattice as shown in fig. 4.3 below.
𝑟⃗
𝑏⃗⃗
𝑎⃗
Fig. 4.3 Two dimensional crystal lattice
By Gizachew Berhanu
Lattice point
Crystal
Lattice


A 3D translational periodic
arrangements of atoms in space.

Dials with what to repeat
A 3D translational periodic
arrangements of points in space.

Dials with how to repeat
There are two classes of lattices. Bravais and non-Bravais. In Bravais lattice, all lattice points are equivalent and
hence all atoms in the crystal are the same kind. That is Bravais lattice is an infinite array of discrete points
arranged and oriented in space, looks exactly the same, regardless of the point of which the array is chosen to be
viewed from. The lattice is invariant under translation. In contrary, in non-Bravais lattice, some of the lattice
points are nonequivalent. A non-Bravais lattice is sometimes referred to as a lattice with a basis; a group of one
or more atoms located in a particular way with respect to each other and associated with each lattice point also
known as motif. If we add atoms at each lattice point in an arrangement which may be one or more group of
atoms having the same spatial arrangement, we will get crystal structure.
Motif
Lattice
The basis must contain at least one atom, but it may contain many atoms of one or more types.
Crystal structure = lattice + basis
Fig. 4.4 crystal structure
Basis Vector
Consider the lattice shown in fig.4.3. If we choose the origin of coordinate at a certain lattice point, the position
vector of any lattice point can be written as
𝑟⃗ = 𝑝𝑎⃗ + 𝑞𝑏⃗⃗ , ………………………………… (4.1)
Where 𝑝 and 𝑞 are integers
The two vectors 𝑎⃗ and 𝑏⃗⃗ (which must be non-co-linear) form a set of basis vectors for the lattice, in terms of which
the positions of all lattice points can be conveniently expressed by equation (4.1). The set of all vectors expressed
by this equation is called basis vectors.
By Gizachew Berhanu
4.3 Unit Cell
Consider the lattice shown in fig. 4.3 once again. This lattice
is a big lattice that may be difficult to understand the
concept of entire crystal. The total area of the rectangle
is partitioned into 12 squares or one square is repeated to
form the whole crystal. This single square is the smallest
portion of the rectangle called unit cell.
Therefore, unit cell is the smallest portion of a crystal lattice that when
repeated in different directions, generates the entire crystal.
The unit cell is the smallest unit of the crystal structure that carries
Fig.4.5 A portion of a three dimensional
cubic lattice and its unit cell
the properties of the crystal. The case is same for one and three dimensions. Figure 4.5 given above indicates three
dimensional unit cells which is the smallest volume of the crystal.
This unit cell is not a unique entity. There is more than one way to choose a unit cell, so that it is better to select a
unit cell in the direction where three axes have the highest symmetry in 3-D. Generally, Unit cells can be broadly
divided into two categories, primitive and conventional unit cells. When constituent particles are present only on
the corner positions of a unit cell, it is called as primitive unit cell. It is the smallest structure of the crystal that
contains only one lattice point.
The second form of unit cell when a unit cell contains one or more constituent particles present at other positions
than corners in addition to those at corners, it is called a conventional or centered unit cell. Any non-primitive unit
cell that used because of more obvious relation with the point of symmetry operations is a conventional unit cell.
Fig.4.6 a unit cell of a lattice is NOT unique.
By Gizachew Berhanu
Characteristic parameters of unit cell
𝑧
(i) Its dimensions along the three edges, 𝑎, 𝑏 and 𝑐.
These edges may or may not be mutually perpendicular.
(ii) Angles between the edges, 𝛽 (between 𝑏 and 𝑐) 𝛼
𝑐
(between 𝑎and c) and γ (between a and b). Thus,
𝛽
a unit cell is characterized by six parameters,
𝛼
𝑎, 𝑏, 𝑐, α, β and γ. These parameters of a typical unit cell
𝑂
𝑎
are shown in Fig.4.7.
𝑥
𝛾
𝑏
𝑦
Fig. 4.7 Illustration of parameters of a unit cell
Unit cell shape CANNOT be the basis for classification of Lattices. Lattices are classified on the basis of their
symmetry.
4.3.1 Symmetry
Under section 4.2.1 it mentioned that nature by itself favors crystalline nature due to crystalline is more symmetry
and regular arrangement pattern. In ordinary life our first perception of symmetry is what is known as mirror
symmetry. Our bodies have, to a good approximation, symmetry in which our right side is matched by our left as if
a mirror passed along the central axis of our bodies. We can also see more complex symmetry in the patterns
around us. It can be found in wallpaper patterns, floor-tile arrays, churches, cloth designs, flowers, and mineral
crystals. There are many examples of other symmetry in ordinary life as shown below.
a
b
Fig.4.8 some kinds of symmetry exists in nature. (a) Wings of a bird are symmetry, (b) symmetry of
man with his own shadow
Symmetry plays an important role in crystallography. Our hands illustrate this most vividly; so that the image is
carried over to crystallography when one speaks of a molecule as being either “right”- or “left”- handed. The ways
in which atoms and molecules are arranged within a unit cell and unit cells repeat within a crystal are governed by
symmetry rules.
By Gizachew Berhanu
Such structure allows an object being divided by a point or line or plane or radiating lines or planes into two or
more parts exactly similar in size and shape and in position relative to the dividing point, etc., repetition of exactly
similar parts facing each other or a centre. This various constitute (lines, planes, or points) are called symmetry
elements. These are imaginary lines, planes and points.
The plane of symmetry (mirror plane, m) is an imaginary plane through the crystal that divides the crystal into two
equal parts such that one is the mirror image of the other. For instance if we consider a cubic crystal, it contains
nine planes of symmetry. In fig.4.9, it is seen that two planes passing through face diagonals 𝐴𝐵𝐺𝐻 and 𝐶𝐷𝐸𝐹 are
mirror planes.
b)
a)
c)
d)
Fig.4.9 Nine planes of symmetry elements of a cube
For each of the two pairs of opposite faces there are two such mirror planes, making a total of six planes of
symmetry which diagonally bisect pairs of opposite faces as shown in fig.4.9a, fig.4.9b and fig.4.9c.
IJKL in fig.4.9d and two other perpendicular planes which pass through the face centres of a pair of opposite faces
are also mirror planes. Therefore, there are 9 mirror planes for a cube (6 diagonal mirror plans + 3 perpendicular
mirror planes to the opposite faces).
An axis of symmetry (rotation axis) is an imaginary line through the crystal such that during rotation of the crystal
presents exactly the same appearance more than once during a rotation of 360°. The same appearance means, the
arrangement of faces and angles as in the starting position.
A crystal has an 𝑛 − 𝑓𝑜𝑙𝑑 axes of symmetry when a rotation through 360°⁄𝑛 about this axis, produces an
arrangement of indistinguishable from the first. The axis corresponding to 𝑛 = 2, 3, 4 𝑎𝑛𝑑 6 are called 𝑡𝑤𝑜 −
𝑓𝑜𝑙𝑑, 𝑡ℎ𝑟𝑒𝑒 − 𝑓𝑜𝑙𝑑, 𝑓𝑜𝑢𝑟 − 𝑓𝑜𝑙𝑑 𝑎𝑛𝑑 𝑠𝑖𝑥 − 𝑓𝑜𝑙𝑑 respectively. In fig.4.10 shown below an axis like 𝐿𝑀 passing
through face-centres of a pair of opposite faces is a 4-fold axis, since rotation about 𝐿𝑀 axis produces an
arrangement identical with the starting one, four times during a complete rotation. Two other 4-fold axis for the
other two pairs of opposite faces can be imagined. There are three 4-fold axis.
A 2-fold axis 𝐴𝐵, through mid-points of a pair of opposite edges, make a total of six 2-fold axis. An axis such as
𝑃𝑄 passing through diagonally opposite corners, there are three such 3-fold axes making a total of four 3-fold axes.
Thus, for a cube there are 13 (𝑠𝑖𝑥 2 − 𝑓𝑜𝑙𝑑 𝑎𝑥𝑒𝑠 + 𝑓𝑜𝑢𝑟 3 − 𝑓𝑜𝑙𝑑 𝑎𝑥𝑒𝑠 + 𝑡ℎ𝑟𝑒𝑒 4 − 𝑓𝑜𝑙𝑑 𝑎𝑥𝑒𝑠) rotation axes.
By Gizachew Berhanu
Fig.4.10 symmetry elements of a cube
The centre of symmetry is a point in a crystal such that any straight line through it cuts identical crystal surfaces at
equal distances on either side of the point. All the rotation axes pass through this point and are bisected by it. All
the planes of symmetry also, pass through this point. If the crystal is inverted about this point, the consequence
arrangement is indistinguishable from the original one. Thus the centre of symmetry is also an inversion centre (i).
For a cube the body centre is the centre of symmetry.
Totally, a cube has 23 elements of symmetry. (9 𝑚𝑖𝑟𝑟𝑜𝑟 𝑝𝑙𝑎𝑛𝑒𝑠 + 13 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑥𝑒𝑠 + 1 𝑖𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛 𝑐𝑒𝑛𝑡𝑟𝑒).
The symmetry word (somewhat distorted) itself shows 2-fold rotation symmetry (restored by 180° rotation). In the
picture below the plane looks identical after a 90° rotation. The plane has 4 fold rotation symmetry as it repeats
itself 4 times (shown by the red dot) in a full 360° rotation.
Fig.4.11 4-fold symmetry axes
4.3.2 Symmetry operations
From your high school Arithmetic math’s course, you may recall that two numbers are required to perform any of
addition, subtraction, multiplication or division; whereas to find the negative of a number we need only one
number. Also in logic you studied different connectives that need two statements and the negation connective needs
only one statement. We called then operations such as addition which need two numbers as binary operation and
operations which need only one number as unary operation.
If we define an operation ∆ on the set 𝑆; 𝑆 ≠ ∅. Then
𝑆 is called closed under ∆ if and only if 𝑥∆𝑦 𝜖 𝑆
∆ is called commutative if 𝑥∆𝑦 = 𝑦∆𝑥 for all 𝑥, 𝑦 𝜖 𝑆
∆ is called associative if (𝑥∆𝑦)∆𝑧 = 𝑥∆(𝑦∆𝑧) for all 𝑥, 𝑦, 𝑧 𝜖 𝑆
By Gizachew Berhanu
𝑒 𝜖 𝑆 is called the left identity if 𝑒∆𝑥 = 𝑥 for all 𝑥𝜖 𝑆
𝑒 𝜖 𝑆 is called the right identity 𝑥∆𝑒 = 𝑥 for all 𝑥 𝜖
𝑒 𝜖 𝑆 is called identity if 𝑒 is both right and left identity
𝑓 𝜖 𝑆 is called inverse of 𝑥 𝜖 𝑆 if 𝑥∆𝑓 = 𝑓∆𝑥 = 𝑒
The rules that govern symmetry are therefore, found in the mathematics of a set. Set addresses the way in which a
certain collection of mathematical “objects” are related to each other. For example, consider all the positive and
negative integers and zero. They can constitute element because under certain circumstances the relationships
between the integers obey the rules of the set:

There must be defined a procedure for combining two elements of the group to form a third. For the
integers one can choose the addition operation so that a + b = c is the operation to be performed and 𝑎, 𝑏,
and 𝑐 are always elements of the set.

There exists an element of the set, called the identity element and denoted 𝑒, that combines with any other
element to give the second one unchanged. In the case of the integers, the identity element is zero because
any integer plus zero gives that integer (a + 0 = a).

For every element of the set, there exists another element that combines with the first to give the identity
element; these are known as inverse elements. The negative integers constitute the inverses of the positive
integers because their pairwise sums all equal zero, the identity element (a + (–a) = 0).

Set operations in sequence obey the associative law. For addition of integers this means that (a + b) + c =
a+ (b + c). Notice that the commutative law, a + b = b + a, is not required even though it is true for this
particular set.

You might be tempted to say that the positive integers, when related by multiplication (a x b = c), also
constitute a group with the identity element now being one (a x 1 = a). In fact, the positive integers do not
constitute a group under these conditions because, to obey the group-theory rules, the no integer inverses
( 1 /a) as well as all the rational fractions (b/a) would have to be included. The expanded set of positive
rational numbers is a group under multiplication, and both it and the integer group already discussed are
examples of infinite groups because they each contain an infinite number of elements.
In the case of a symmetry group, an element is the operation needed to produce one object from another. For
example, a mirror operation takes an object in one location and produces another of the opposite hand located such
that the mirror doing the operation is equidistant between them (Fig.4. 12). These manipulations are usually called
symmetry operations. They are combined by applying them to an object sequentially.
The procedure in which an object is brought into a position indistinguishable from the initial by the use of symmetry element is
called a symmetry operation.
By Gizachew Berhanu
Fig.4. 12 A pair of left- and right-"footed” boots Illustrates
the mirror-plane symmetry operation. The right boot can be
positioned identically on the left boot by reflection through a
mirror between them and vice versa.
For example, doing a mirror operation twice on a right-handed object will, with the first operation, move it to the
left-handed position, and with the second operation, place it back on its original right-handed position. In fact,
applying a mirror operation twice in succession is equivalent to the identity operation, so that a mirror operation is
its own inverse. There are four types of symmetry operations in crystallography.

Translation

Reflection

Rotation

Inversion
Translation
The simplest type is the set of translation operations needed to fill a two-dimensional infinite plane or a threedimensional infinite space. These operations form a group by themselves and have essentially the same
characteristics as the example set of integers discussed above. The difference is that the translation group has two or
three sets of integers depending on whether a two dimensional plane or a three-dimensional space is filled. These
translation operations make the concept of a unit cell possible, because once the unit cell for a crystal is specified; it
takes only the right combination of translation operations to construct the full crystal lattice. There is also a type of
translation operation that relates objects within a unit cell so that the same objects are found at coordinates that are
half multiples of unit-cell distances along two or three of the axes. These last operations are, for example,
responsible for the face- centred and body-centered lattices found in three dimensions (Fig 4. 18). The possible
combinations of this full set of translations for plane- and space-filling arrays gives only five possible plane lattices
and fourteen possible space lattices (Fig.4. 13b).
By Gizachew Berhanu
a)
Fig.4.13 translational symmetry operation
(a) The first point is repeated at equal distances along a
line by a translation aT, where T is the translation vector
and a is an integer. Translation on a point with
coordinates xyz → x+a y+b z+c where, a, b and c are the
unit vectors in x, y and z directions respectively.
(b) Cobblestone forms a 2-D group by themselves and
has the same characteristics. A single cobblestone may
be regarded as a unit cell in a crystal.
b)
Rotation
The second type of crystallographic symmetry is rotation. A body shows n-fold rotation symmetry around an axis,
when a rotation by (360⁄𝑛)° 𝑜𝑟 2𝜋/𝑛 brings it into self-coincidence. A rotation can be applied on the translation
vector T in all directions, clock or anti-clock wise, through equal angles in the 2D space. If two rotation operations,
one each in clock and anti-clock direction, are applied on the translation vector T, it will create two more lattice
points. Because of the regular pattern, the translation between these two points will be some multiple of T (pT). An
n-fold rotation symmetry means rotation through an angle of 2𝜋/n will repeat the object or motif n times in a full
360° rotation. n =1 means no symmetry.
Table 4.2 rotational symmetry operations
Fig.4.14 rotation symmetry operations along different symmetry planes
By Gizachew Berhanu
Objects with 5, 7 and 8 or higher order symmetry do exist in nature, e.g. star fish (5-fold), flowers with 5 or 8-fold
symmetry.
However, these are not possible in crystallography as they cannot fill the space completely. Thus patterns with
pentagonal symmetry are non-repeating, non-periodic or incommensurate and consequently have, in the past, been
rather overlooked by crystallographers. There are gaps among pentagonal structures or unit cells are overlaps.
Fig.4.15 asymmetric crystal structures in 5-fold and 8-fold
Reflection or Mirror symmetry
An object with reflection symmetry will be a mirror image of itself across a plane called mirror plane (m).
Reflection operation: 𝑥𝑦𝑧 → −𝑥 𝑦 𝑧 (𝑥̅ 𝑦 𝑧) across a mirror plane perpendicular to x axis
Symmetry and Space lattice
Symmetry elements discussed so far define five types of 2D space lattices. When a translation is applied to the third
direction these lattices create a total of 7 crystal systems.
By Gizachew Berhanu
Fig.4.16 five types of 2D
space lattices
Inversion – Center of symmetry
A similar operation is inversion (Fig. 17) in which right- and
left-handed objects are arranged on opposite sides of a point,
called an inversion center. The presence of an inversion center
in a crystal is one of the primary classification features for
crystal structures: such crystal structures are centrosymmetric.
In this operation, every part of the object is reflected through
an inversion center called center of symmetry which is denoted as i.
Fig.4.17 inversion symmetry operation
The object is reproduced inverted from its original position.
Point and Space groups
Symmetry operations generate a variety of arrangements of lattice points in three dimensions. There are 32 unique
ways in which lattice points can be arranged in space. These non-translation elements are called point-groups.
A large number of 3D structures are generated when translations [linear translation, translation + reflection (glide
plane) and translation + rotation (screw axis)] are applied to the point groups. There are 230 unique shapes which
can be generated this way. These are called space groups.
Summary

Space lattice is arrangement of points with each point having exactly same surroundings.

Symmetry operations restore a body to its original position.

There are four symmetry operations –Translation, reflection, rotation and inversion.

There are 32 point groups and 230 space groups
By Gizachew Berhanu
4.4 The seven crystal systems and Fourteen Bravais Lattices
Above, we mentioned that crystal lattices are made of unit cell and unit cell has parameters𝑎, 𝑏, 𝑐, 𝛼, 𝛽, 𝑎𝑛𝑑 𝛾. If we
vary these parameters, we will get different varieties of unit cells. Based on these parameters, there are seven
crystal lattice systems which consist of 14 kinds of Bravais lattices(after the French mathematician who first
described them) namely: 1. Cubic, 2. Triclinic, 3. Monoclinic, 4. Orthorhombic, 5. Tetragonal, 6. Hexagonal and, 7.
Rhombohedra. That is, only seven different kinds of cells are necessary to cover all possible point lattices or all
crystals can be classified into one of these seven crystal systems. The symbols P, F, I, etc. in Table 4.2 are given on
the basis of the following rule. When a unit cell has only one lattice point, it is called a primitive (or simple) cell,
and usually represented by P. In addition, although the trigonal (rhombohedral) crystal system can also be classified
into primitive, we use R as the symbol. Other symbols are nonprimitive cells and more than one lattice point per
cell is included. Symbols I, C and F refer to body-centered, base centered and face centered cells, respectively.
Table 4.2 summary of seven crystal systems and Bravais lattice
Crystal system
Cubic
Axial length and angles
Three equal axes at right angle
𝑎 = 𝑏 = 𝑐, 𝛼 = 𝛽 = 𝛾 = 90°
Tetragonal
Three axes at right angles, two equals
𝑎 = 𝑏 ≠ 𝑐, 𝛼 = 𝛽 = 𝛾 = 90°
Three unequal axes at right angles
𝑎 ≠ 𝑏 ≠ 𝑐, 𝛼 = 𝛽 = 𝛾 = 90°
Orthorhombic
Trigonal
Hexagonal
Monoclinic
Triclinic
By Gizachew Berhanu
Three equal axes, equally inclined
𝑎 = 𝑏 = 𝑐, 𝛼 = 𝛽 = 𝛾 ≠ 90°
Two equal coplanar axes at 120°
Third axis at right angles
𝑎 = 𝑏 ≠ 𝑐, 𝛼 = 𝛽 = 90°, 𝛾 = 120°
Three unequal axes, one pair not at right
angles. 𝑎 ≠ 𝑏 ≠ 𝑐, 𝛼 ≠ 𝛽 ≠ 𝛾 = 90°
Three unequal axes, unequally inclined
and none at right angles. 𝑎 ≠ 𝑏 ≠
𝑐, 𝛼 ≠ 𝛽 ≠ 𝛾 ≠ 90°
Bravais lattice
Simple,
body-centered
Face-centered
Simple
Body-centered
Simple,
body-centered,
base-centered
face-centered
Simple
Lattice symbol
P
I
F
P
I
P
I
C
F
P
Simple
P
Simple
Base-centered
simple
P
C
P
Fig 4.18 The fourteen types of Bravais lattices grouped in seven crystal systems.
Table2.3 examples of seven crystal systems (reproduced from
http://www.youtube.com/watch?v=qh29mj6uXoM&feat e=relmfu )
Crystal system
Cubic
Tetragonal
Orthorhombic
Hexagonal
Rhombohedral
Monoclinic
Triclinic
Examples
𝐴𝑢, 𝑆𝑖, 𝐴𝑙, 𝐶𝑢, 𝐴𝑔, 𝐹𝑒, 𝑁𝑎𝐶𝑙
𝐼𝑛, 𝑇𝑖𝑂2 , 𝛽 − 𝑆𝑛
𝐺𝑎, 𝐹𝑒3 𝐶, 𝛼 − 𝑆,
𝑍𝑛, 𝐶𝑜, 𝐶𝑑, 𝑀𝑔, 𝑍𝑟, 𝑁𝑖𝐴𝑆
𝐻𝑔, 𝑆𝑏, 𝐴𝑠, 𝐵𝑖, 𝐶𝑎𝐶𝑂3 ,
𝐴𝑠4 𝑆4 , 𝐾𝑁𝑂2 , 𝐶𝑎𝑆𝑂4 . 2𝐻2 𝑂, 𝛽 − 𝑆
𝐾2 𝑆2 𝑂8 , 𝐾2 𝐶𝑟2 𝑂7
4.5 crystal structure of Metals
This sub-section will presents the principal crystal structures of elemental metals in details. Most elemental metals
(about 90 percent) crystallize upon solidification into three densely packed crystal structures: body-centered cubic
(BCC) (Fig. 4.21), face-centered cubic (FCC) (Fig. 4.23) and hexagonal close-packed (HCP)(Fig. 4.24). The HCP
structure is a denser modification of the simple hexagonal crystal structure shown in Fig. 4.18. Most metals
crystallize in these dense-packed structures because energy is released as the atoms come closer together and bond
By Gizachew Berhanu
more tightly with each other. Thus, the densely packed structures are in lower and more stable energy
arrangements. Of course, all structures cannot be covered here. The principal ionic and covalent crystal structures
such as zinc blend, diamond, etc. that occurs in ceramic materials would not be treated in detail here. Please, read them
from other sources mentioned at the end of this chapter.
i) Simple cubic Unit cell
The most simple unit cell is known as a simple cubic unit cell. This is where one atom occupies each of the eight
corners of a cube. The distance from atom to atom along the lattice is the same in every direction, and the angle
between each of axes is 90°. See table 4.2.
a
b
Fig.4.19 simple cubic crystal structure
a) Coordination Number
Coordination number is the number of nearest neighbor to a particular atom in the crystal. For the simple cubic
crystal structure the coordination number is six. This is demonstrated in the figure 4.19b above. Consider the atom
labeled A, which is located at the corner of the reduced-sphere simple cubic unit cell. It has three nearest neighbors
located in this unit cell—labeled 1, 3, and 5. In addition, the three shaded atoms, labeled 2, 4, and 6, are also
nearest neighbors that belong to adjacent unit cells.
b) Number of atoms per unit cell
Each unit cell contains a specific number of lattice points. When counting the number of lattice points belonging to
each unit cell, we must recognize that, like atoms, lattice points may be shared by more than one unit cell. A lattice
point at a corner of one unit cell is shared by seven adjacent unit cells (thus a total of eight cells); only one-eighth of
each corner belongs to one particular cell. Thus, the number of lattice points from all corner positions in one unit
cell is
𝑁 = 𝑁𝑖 +
𝑁𝑓
2
+
𝑁𝑐
8
………………………………………. (4.2)
Where
By Gizachew Berhanu
𝑁𝑖 = number of interior atoms
𝑁𝑓 = number of face atoms
𝑁𝑐 = number of corner atoms
For simple cubic, there are eight corner atoms and no face and no interior atoms i.e., N c = 8 and N i = N f = 0 .
N = Ni +
Nf
2
+
Nc
0 8
= 0 + + = 1 atom/per unit cell
8
2 8
c) Relationship between atomic radius and lattice parameters
If we refer to Figure 4.20, we find that atoms touch along the edge of the cube in SC structures.
The corner atoms are centered on the corners of the cube, so
𝑎
𝑟
= 2………………………………. (4.3)
a
Fig. 4.20 simple cubic parameter
d) Atomic packing Factor
In whatever way the constitute particles (atoms, molecules or ions) are packed, there is always some free space
in the form of voids. Packing efficiency is the percentage of total space filled by the particles. It is the sum of the
sphere volumes of all atoms within a unit cell (assuming the atomic hard-sphere model) divided by the unit cell
volume—that is,
𝐴𝑃𝐹 =
(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙) 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑖𝑛 𝑎 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝑡𝑜𝑡𝑎𝑙 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
………………… (4.4)
Therefore, the total sphere volume in terms of the atomic radius R is equal to
4
 4
NVS  (1 atom/unit cell)   R3    R3
3
 3
Because the unit cell is cubic, its volume is equal to the edge length (a) cubed—that is
VC = a3
And because a = 2R
VC = (2R)3 = 8R3
Finally, using Equation 4.4, the APF is computed as follows:
4
 R3
NVS 3
APF 

= 0.52
VC
8R3
By Gizachew Berhanu
ii) Body centered Cubic Unit cell
The next unit cell is known as the body-centered cubic. In this form of crystal, there is an atom at each corner of the
unit cell and also there is an additional atom in the center of the cube. This packing can fit more atoms into less
space than the simple cubic unit cell. Chromium, iron, tungsten, and several other metals listed in Table 4.4 exhibit
a BCC structure.
Table 4.4 Selected Metals that Have the BCC Crystal Structure at Room Temperature (20℃) and Their
Lattice Constants and Atomic Radii
Metal
Lattice constant 𝑎 (𝑛𝑚) Atomic radius 𝑅 (𝑛𝑚)
Chromium
0.289
0.125
0.287
0.124
Iron (𝛼)
Molybdenum
0.315
0.136
Potassium
0.533
0.231
Sodium
0.429
0.186
Tantalum
0.330
0.143
Tungsten
0.316
0.137
Vanadium
0.304
0.132
a) Coordination number
We see that each atom in the SC structure has a coordination number of six, while each atom in the BCC structure
has eight nearest neighbors.
1
2
3
4
5
6
8
7
Fig. 4.21 coordination number of BCC
b) Number of Atoms per Unit cell
In BCC unit cells, lattice points are located at the corners and the center of the cube: then from equation (4.2), we
have 𝑁𝑖 = 1, 𝑁𝑓 = 0 𝑎𝑛𝑑 𝑁𝐶 = 8
𝑁 = 𝑁𝑖 +
𝑁𝑓
2
+
𝑁𝑓
8
→ 𝑁 = 1+
By Gizachew Berhanu
0
2
+
8
8
=2
𝑎𝑡𝑜𝑚𝑠
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
c) Relationship between atomic radius and Lattice Parameters
In BCC structures, atoms touch along the body diagonal. There are two atomic radii from the center atom and one
atomic radius from each of the corner atoms on the body diagonal, so
Fig 4.22 BCC atomic radius
a
vs. lattice parameter
a
2
2
2
𝑎 + 𝑎 + 𝑎 = (𝑟 + 2𝑟 + 𝑟
)2
3𝑎2 = 16𝑟 2
𝑎
𝑟
=
4
√3
……………………………………. (4.5)
d) Atomic Packing Factor
From equation (4.4), we have
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓𝑎𝑡𝑜𝑚𝑠 𝑖𝑛 𝑎 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝐴𝑃𝐹 =
𝑡𝑜𝑡𝑎𝑙 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
4
In the BCC unit cell, there are 2 lattice points per cell; the volume of one atom is 3 𝜋𝑟 3 and the volume of the unit
cell is 𝑎3 where r is the radius of the atom and a is the lattice parameter. Therefore, the total sphere volume in terms
of the atomic radius r is equal to
4
 8
NVS  (2atom/unit cell)   R3    R3
3

 3
Because the unit cell is cubic, its volume is equal to the edge length (a) cubed—that is
VC = a3
And because a = R
4
√3
4𝑅 3
𝑉𝐶 = ( )
√3
𝐴𝑃𝐹 =
𝑁𝑉𝑆
𝑉𝐶
=
8 3
𝜋𝑟
3
4𝑟 3
(
√3
)
By Gizachew Berhanu
= 0.68
iii) The Face-Centered Cubic Crystal Structure
A slightly more tightly packed unit cell is the face-centered cubic unit cell. In this form of crystal, there is an atom
at each corner of the unit cell. And there is an atom at the center of each of the six faces of the cubic unit cell. This
crystal packing form has an even higher density than the body centered cubic unit cell. Atoms or ions are shared
between adjacent unit cells. The lattice position of the atom or ion determines the number of unit cells involved in
the share. There are four different lattice positions an atom or ion can occupy.
Body: Not shared
Face: Shared by two unit cells
Edge: Shared by four unit cells
𝑅
Corner: Shared by eight unit
cells.
𝑎
2𝑅
Fig.4.23 FCC
𝑅
a
b
𝑎
Some of the familiar metals having this crystal structure are copper; aluminum, silver, and gold (see also Table
4.5). Figure 4.23a shows a hard-sphere model for the FCC unit cell, whereas in Figure 4.23b the atom centers are
represented by small circles to provide a better perspective on atom positions. These spheres or ion cores touch one
another across a face diagonal.
Table 4.5 Selected Metals That have the FCC Crystal Structure at Room Temperature (20◦C) and Their
Lattice Constants and Atomic Radii
Metal
Lattice constant 𝑎 (𝑛𝑚)
Atomic radius 𝑅 (𝑛𝑚)
Almunium
0.405
0.143
Copper
0.3615
0.128
Gold
0.408
0.144
Lead
0.495
0.175
Nickel
0.352
0.125
Platinum
0.393
0.138
Silver
0.409
0.144
a) Coordination number
Each atom in the FCC structure has a coordination number of twelve, which is the maximum. This may be
confirmed by examination of Figure 4.23a; the front face atom has four corner nearest-neighbor atoms surrounding
it, four face atoms that are in contact from behind, and four other equivalent face atoms residing in the next unit cell
to the front (not shown).
By Gizachew Berhanu
b) Number of atoms per unit cell
In FCC form of crystal, there is an atom at each corner of the unit cell. And there is an atom at the center of each of
the six faces of the cubic unit cell.
For the FCC crystal structure, there are eight corner atoms
(𝑁𝐶 = 8), six face atoms (𝑁𝑓 = 6), and no interior atoms
(𝑁𝑖 = 0). Thus, from Equation 4.2,
𝑁 = 𝑁𝑖 +
𝑁=0+
6
2
𝑁𝑓
+
2
8
+
8
𝑁𝐶
8
= 4 𝑎𝑡𝑜𝑚𝑠⁄𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
c) Relationship between atomic radius and Lattice Parameters
The diagonal of the face is 𝑅 + 2𝑅 + 𝑅 = 4𝑅
Side of the face is = 𝑎 and the diagonal is = √𝑎2 + 𝑎2 = 𝑎√2
𝑎√2 = 4𝑅
𝑎
𝑅
= 2√2………………………………………………….. (4.6)
d) Atomic packing factor
𝐴𝑃𝐹 =
(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙)𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓𝑎𝑡𝑜𝑚
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
=
𝑁𝑥𝑉𝑠
𝑉𝐶
4
 16
NVS  (4atom/unit cell)   R3    R3
3
 3
Because the unit cell is cubic, its volume is equal to the edge length (a) cubed—that is VC = a3
And because 𝑎 = 2𝑅√2
𝑉𝐶 = (2𝑅 √2)
𝐴𝑃𝐹 =
3
𝑁𝑉𝑆
𝑉𝐶
=
16
𝜋𝑅 3
3
(2𝑅 √2)
By Gizachew Berhanu
3
= 0.74
iv) Hexagonal crystal structure
Not all metals have unit cells with cubic symmetry; the final common metallic crystal structure to be discussed has
a unit cell that is hexagonal. Figure 4.24 shows a reduced sphere unit cell for this structure, which is termed
hexagonal close-packed (HCP); an assemblage of several HCP unit cells is presented in Figure 4.24. The top and
bottom faces of the unit cell consist of six atoms that form regular hexagons and surround a single atom in the
center. Another plane that provides three additional atoms to the unit cell is situated between the top and bottom
planes.
Fig. 4.24 Hexagonal 2-D crystal structure 3-D hexagonal unit cell lattice
a) Coordination number
In hexagonal lattice the coordination number = 12. The center atom of the top face is in touch with six corner
atoms, three atoms of the mid layer and other three atoms of the mid layer of the unit cell above it. The HCP
metals include cadmium, magnesium, titanium, and zinc; some of these are listed in Table 4.6.
Table 4.6 Selected Metals That Have the HCP Crystal Structure at Room Temperature (20◦C) and Their
Lattice Constants, Atomic Radii, and c/a Ratios
Lattice
constants(nm)
Atomic radius
𝑅 (𝑛𝑚)
𝑐/𝑎 ratio
Metal
𝑎
Cadmium
Zinc
Ideal HCP
Magnesium
Cobalt
Zirconium
Titanium
Beryllium
By Gizachew Berhanu
𝑐
0.2973 0.5618 0.149
0.4947 0.4947 0.133
0.3209
0.2507
0.3231
0.2950
0.2286
% deviation
from ideality
0.5209
0.4069
0.5148
0.4683
0.3584
0.160
0.125
0.160
0.147
0.113
1.890
1.856
1.633
1.623
1.623
1.593
1.587
1.568
+15.7
+13.6
0
-0.66
-0.66
-2.45
-2.81
-3.98
b) Number of Atoms per Unit cell
The top and bottom faces of the unit cell consist of six atoms that form regular hexagons and surround a single
atom in the center. Another plane that provides three additional atoms to the unit cell is situated between the top
and bottom planes. The atoms in this mid-plane have as nearest neighbors atoms in both of the adjacent two planes.
In order to compute the number of atoms per unit cell for the HCP crystal structure, Equation 4.2 is modified to
read as follows:
𝑁 = 𝑁𝑖 +
𝑁𝑓
2
+
𝑁𝑐
6
…………………………………….. (4.7)
That is, one-sixth of each corner atom is assigned to a unit cell (instead of 8 as with the cubic structure). Because
for HCP there are 6 corner atoms in each of the top and bottom faces (for a total of 12 corner atoms), 2 face center
atoms (one from each of the top and bottom faces), and 3 mid-plane interior atoms, the value of N for HCP is
found, using Equation 4.7, to be
𝑁=3+
2
2
+
12
6
= 6.
Thus, 6 atoms are assigned to each unit cell.
c) Relationship between atomic radius and lattice constant
In the Hexagonal unit cell, number of atoms = 12 𝑐𝑜𝑟𝑛𝑒𝑟 𝑎𝑡𝑜𝑚𝑠 𝑥 1/6 (𝑠ℎ𝑎𝑟𝑒𝑑 𝑏𝑦 𝑠𝑖𝑥 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙𝑠) +
𝑇𝑤𝑜 𝑓𝑎𝑐𝑒 𝑎𝑡𝑜𝑚𝑠 𝑥 ½ + 3 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 = 6. From the HCP unit cell given in fig.4.24, we will deduce that 𝑎 = 2𝑟.
E
E
𝐶
2
a
D
G
F
A
G
B
Fig 4.25 Relationship between lattice parameters in HCP
𝐴𝐵 = 𝐵𝐷 = 𝐷𝐴 = 𝐴𝐸 = 𝐸𝐷 = 𝐸𝐵 = 𝑎 , 𝐵𝐹 = 𝐹𝐷 =
(𝐴𝐹 )2 + (𝐹𝐵)2 = (𝐴𝐵)2
𝑎 2
(𝐴𝐹 )2 = (𝐴𝐵)2 − (𝐹𝐵)2 = 𝑎2 − ( ) =
2
By Gizachew Berhanu
3𝑎 2
4
𝑎
2
𝑎𝑛𝑑 𝐸𝐺 =
𝐶
2
F
∴ 𝐴𝐹 =
(𝐸𝐺
)2
1
𝐸𝐺 =
𝐶
𝑐
𝑎
2
+ (𝐹𝐺)2 = (𝐹𝐸)2
𝐹𝐺 =
∴
𝑎√3
3
2
𝐴𝐹 =
𝑎
6
√3 𝑎𝑛𝑑 𝐹𝐸 = 𝐴𝐹 =
2
𝑎
2
√3
2
𝑎
𝑎
2
= √( √3) − ( √3) = 𝑎√
2
6
3
2
= 2√3 = 1.633…………………………………………. (4.8)
d) Atomic Packing Factor
𝐴𝑃𝐹 =
(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙)𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑡𝑜𝑚
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
=
𝑁 𝑥 𝑉𝑠
𝑉𝐶
In the HCP unit cell, there are 6 lattice points per cell; the volume of one atom is
4
3
𝜋𝑟 3and therefore, the total
sphere volume in terms of the atomic radius r is equal to
4

NVS  (6atom/unit cell)   r 3   8 r 3
3


The volume of the HCP unit cell can be obtained by determining the area of the base of the unit cell and then
multiplying this by its height (Fig. 4.25). The area of the base of the unit cell is area ABCDEF of Fig. 4.26a and b.
This total area consists of the areas of six equilateral triangles of area ABG of Fig. 4.26b.
Fig.4.26 Diagrams for calculating the volume of an HCP unit cell. (a) HCP unit cell. (b) Base of HCP unit
cell. (c) Triangle ABG removed from base of unit cell.
By Gizachew Berhanu
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐺 =
∴ 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐺 =
1
2
𝑎ℎ , where ℎ = 𝑎 sin 60° =
√3
𝑎
2
√3 2
𝑎
4
√3 2
𝑎
4
∴ 𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐻𝐶𝑃 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑏𝑎𝑠𝑎𝑙 𝑝𝑙𝑎𝑛𝑒 = 6 𝑥
𝑉𝐶 = 𝑏𝑎𝑠𝑒 𝑎𝑟𝑒𝑎 𝑥 ℎ𝑒𝑖𝑔ℎ𝑡 =
3√3
2
𝑎2 (𝑐) =
3√3
2
=
3√3
2
𝑎2
8
𝑎2 √ 𝑎 = 𝑎 3 3√2
3
And because a = 2r
𝑉𝐶 = (2𝑟)3 3√2 = 24√2𝑟 3
Finally, using Equation 4.4, the APF is computed as follows:
𝐴𝑃𝐹 =
𝑁𝑉𝑆
𝑉𝐶
=
8𝜋𝑟3
24√2𝑟 3
= 0.74
Table 4.7 crystal structure characteristics of some metals at room temperature.
Structure
𝑎 𝑣𝑒𝑟𝑠𝑢𝑠 𝑟
Atoms per cell
SC
BCC
𝑎 = 2𝑟
4𝑟
𝑎=
√3
𝑎 = 2𝑟√2
𝑎 = 2𝑟
FCC
HCP
Packing factor
1
2
Coordination
number
6
8
4
6
12
12
0.74
0.74
0.52
0.68
v) Density
A knowledge of the crystal structure of a metallic solid permits computation of its theoretical density. The
theoretical density of a material can be calculated using the properties of the crystal structure. The general formula
is:
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 (𝜌) =
(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓𝑎𝑡𝑜𝑚𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙)𝑥(𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠)
(𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙)𝑥(𝑎𝑣𝑜𝑔𝑎𝑑𝑟𝑜𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
=
𝑁𝑚
𝑉𝑁𝐴
… … … . (4.9)
4.6 Allotropic or Polymorphic Transformations
Materials that can have more than one crystal structure are called allotropic or polymorphic. The term allotropy is
normally reserved for this behavior in pure elements, while the term polymorphism is used for compounds. Some
metals, such as iron and titanium, have more than one crystal structure. At room temperature, iron has the BCC
structure, but at higher temperatures, iron transforms to an FCC structure. These transformations result in changes
in properties of materials and form the basis for the heat treatment of steels and many other alloys.
Iron has got the following three allotropic forms of crystal at different temperature. When it is heated from normal
temperature to high temperature (molten state), it undergoes all the allotropic forms.
a) Alpha iron: It occurs from normal temperature to 910℃ and has got BCC cubic lattice.

Ferromagnetic Alpha Iron: which occurs heating from normal temperature to 770℃ temperature.
By Gizachew Berhanu

Paramagnetic Alpha Iron: Which occurs from 770℃ to 910℃.
b) Gamma Iron: This occurs from 910℃ to 1400℃. It has got FCC crystal structure cubic lattice.
Fig.4.27 FCC iron transform to BCC crystal structure from1400℃ to 1540℃.
c) Delta Iron: this form occurs from 1400℃ to 1500℃ (molten state). The crystal structure is BCC cubic lattice as
shown in fig.4.15 above.
Fig.4.28 iron allotropic illustration diagram
Points of Arrests: in the above graph, there appear points of discontinuity which are called points of arrests or
simply A points. In order to distinguish between various A points, there are numbered as 𝐴1 , 𝐴2 , 𝐴3 , 𝐴4 . The change
from ferromagnetic alpha iron to paramagnetic alpha iron is called 𝐴2 and is occurred at 770℃. The change from
paramagnetic alpha iron to gamma iron which occurs at 910℃ is called 𝐴3 . When gamma iron changes to delta
iron at 1400℃, the change is called 𝐴4 .
By Gizachew Berhanu
Example 4.1
Prove that the volume of FCC primitive unit cell is one quarter of the volume of conventional unit cell.
Solution: In the section we discussed that primitive cell is the smallest area if the crystal is two dimensional
and the smallest volume if the crystal is three dimensional. First, let’s see volume of primitive unit cell
𝑉𝑃𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑒 = (𝑎⃗1 . 𝑎⃑2 𝑥𝑎⃑3 )
𝑎
𝑎
2
2
𝑎
𝑉′ = 0
𝑎
𝑎3
8
𝑎
2
2
𝑎
0
[2
𝑉′ =
0
+
=
𝑎 𝑎 𝑎
𝑎
2
2
( . −
2 2
. 0) −
𝑎
2
𝑎
(0. −
2
𝑎 𝑎
. )+ 0
2 2
2]
𝑎3
8
=
𝑎3
4
We inspect that the volume of conventional cell is given by 𝑎3
By taking the ratio, primitive to conventional cell
𝑉′
𝑉
=
𝑉′ =
𝑎3
4
𝑎3
1
4
=
1
4
𝑉
Example 4.2
Answer the following questions about gold (Au), which has FCC structure with the lattice parameter
𝑎 = 0.4070𝑛𝑚
(1) Obtain the nearest-neighbor distance, the second nearest-neighbor distance and
(2) Obtain the values of density
Solution
(1) When you look at the characteristic feature of the fcc lattice, the distance between the atoms which
occupy the corners of a unit cell is equal to the lattice parameter “𝑎,” and the distance between the atom
𝑎
located at a center position of the cell face and the atom which occupies a corner position is . Thus, the
√2
nearest-neighbor distance is estimated to be 𝑟1 =
seen that
𝑟1 =
𝑎
√2
(2) 𝜌 =
𝑎
√2
=
𝑎
and the second nearest-neighbor distance is 𝑟2 = 𝑎. It is
√2
< 𝑎.
0.4070
√2
= 0.2878𝑛𝑚, 𝑟2 = 𝑎 = 0.4070𝑛𝑚
4 𝑥 196.97
(0.4070 𝑥10−9)3 (6.022 𝑥1023 )
By Gizachew Berhanu
= 19.41𝑔/𝑐𝑚 3
Example 4.3
Calculate the theoretical volume change accompanying a polymorphic transformation in pure sodium from
the BCC to HCP crystal structure. Assume the hard-sphere atomic model and that there is no change in
atomic volume before and after the transformation. Use table 4.3 and ideal 𝑐⁄𝑎 ratio.
Solution
Sodium is a BCC crystal structure with 0.429nm lattice constant and then has a volume of
4𝑅 3
𝑎3
𝑉𝐶 =
= 0.5𝑎3 , 0.5 ( ) = 6.16𝑅3 since it has two atoms per unit cell,
2
√3
The volume per atom for the HCP crystal lattice, since it has six atoms per unit cell, is
𝑉𝐻𝐶𝑃 =
𝑎 33√2
6
= 0.707𝑎3 , 0.707(2𝑅)3 = 5.656𝑅3
The change in volume associated with the transformation from the BCC to HCP crystal structure, assuming
no change in atomic radius, is
∆𝑉
𝑉𝐻𝐶𝑃
=
𝑉𝐻𝐶𝑃 − 𝑉𝐵𝐶𝐶
𝑉𝐻𝐶𝑃
=
5.656𝑅 3 – 6.16𝑅 3
5.656𝑅 3
𝑥 100% = −8.9%
∴ The minus sign represents shrinkage in volume during transformation,i.e. atoms get more compact in HCP
than in BCC.
Example 4.4
Obtain the volume of void in BCC crystal structure with lattice parameter 𝑎 by supposing the case where
spherical atoms of radius 𝑟𝐴 are arranged in each lattice point. Also calculate the porosity.
Solution:
In body-centered cubic (BCC) structure, atoms contacting each other are seen on diagonals and then we
obtain the following relationship:
𝑟𝐴 =
√3
𝑎
4
1
It is discussed that in a unit cell of BCC structure, there are two atoms: one atom at eight corners (8 𝑥 8 = 1)
and one atom at the center. Therefore, the volume 𝑉𝐴 occupied by atom is described by
4
𝑉𝐴 = 2 𝑥 3 𝜋𝑟𝐴3 =
8
√3
3
𝜋 ( 4 ) 𝑎3 =
3
𝜋√3
8
𝑎3
The unit cell volume is expressed by 𝑎3 , so that the void volume 𝑉𝐻 is as follows:
𝑉𝐻 = 𝑉 − 𝑉𝐴 = 𝑎3 −
𝜋√3
8
𝑎3 = (1 −
𝜋√3
8
)𝑎3
The porosity in the BCC lattice is given in the following:
𝑉𝐻
𝑉
=
(1−
𝜋√3 3
)𝑎
8
3
𝑎
= 0.32
By Gizachew Berhanu
Example 4.5
If spheres of equal size are used to fill space, there are two ways for arranging spheres; in square form and in
hexagonal form. Compute the percentage of void of these two cases for two-dimensional array of spheres.
Solution
In two-dimensional array of spheres, each sphere is found to be in contact with four spheres in the square
form. On the other hand, each sphere contacts with six spheres in the hexagonal form, as illustrated in fig.
shown below. This implies that the coordination numbers of the nearest neighbors are 4 in the square form
and 6 in the hexagonal form, respectively.
2R
R
2R
2R
2R
b
a
When the radius of a sphere is given by 𝑅, the area produced by one set of sphere array in the square form is
expressed as 𝑎2 = 2𝑅 𝑥 2𝑅 = 4𝑅2 .
1
The area of the shaded part in the square unit cell is
4
𝜋𝑅2 𝑥 4 = 𝜋𝑅2 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑓𝑢𝑙𝑙 𝑐𝑒𝑟𝑐𝑖𝑙𝑒.
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑒𝑎𝑡𝑒𝑑 𝑣𝑜𝑖𝑑 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 − 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑝𝑎𝑟𝑡
= 4𝑅2 − 𝜋𝑅2 = 𝑅2 (4 − 𝜋) = 0.86𝑅2
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑣𝑜𝑖𝑑 𝑎𝑟𝑒𝑎 =
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑣𝑜𝑖𝑑
𝑥100% =
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
0.86𝑅 2
4𝑅 2
= 21.5%
In the hexagonal form, our notice focuses the area of an equilateral triangle with one edge-line of 2𝑅(see
1
Fig. above (b)). The area of the triangle is given by 2 𝑏ℎ =
1
2
(2𝑅)ℎ = 𝑅ℎ.
From the figure, we can see that ℎ = √4𝑅2 − 𝑅2 = √3𝑅
∴ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 = √3𝑅2
An (adjacent) interior angle of the equilateral triangle is 60°. This suggests that the area occupied by a
sphere is decreased to a factor of 2/3. then,
2
3
1
𝑥 𝜋𝑅2 𝑥3 = 1.57𝑅2 because it is equivalent to half of a
4
circle. Therefore, the percentage of void area in the hexagonal form is
By Gizachew Berhanu
√3𝑅 2−1.57𝑅 2
√3𝑅 2
𝑥100% = 9.3%
Example 4.6
Beryllium (Be) mineral is expressed by a chemical formula (3BeO. Al2O3. 6SiO2), and it is revealed that the
structure is hexagonal with the lattice parameters 𝑎 = 0.778𝑛𝑚 and density 2.68𝑔/𝑐𝑚3. Obtain the number of
molecules contained in a unit cell. The 𝑐⁄𝑎 𝑖𝑠 𝑖𝑑𝑒𝑎𝑙
Solution
At first, we obtain the molecular weight of beryllium mineral using a chemical formula from the molecular
weight per 1 mol of the individual oxide component:
𝐵𝑒𝑂 = 9 + 16 = 25𝑔/𝑚𝑜𝑙
𝐴𝑙2 𝑂3 = 2(27) + 3(16) = 102𝑔/𝑚𝑜𝑙
𝑆𝑖𝑂2 = 28 + 32 = 60𝑔/𝑚𝑜𝑙
3𝐵𝑒𝑂 + 𝐴𝑙2 𝑂3 + 6𝑆𝑖𝑂2 = 537𝑔/𝑚𝑜𝑙
Next, we estimate the volume of a unit cell for beryllium mineral from the given values of lattice parameters.
As readily seen in discussion section, in a unit cell of hexagonal system, the value of 𝑐 in ideal case is given by
𝑐 = 1.633𝑎. Therefore, the volume V of a unit cell of hexagonal system is given in the following equation
3√3𝑎 2(1.633𝑎)
3√3
𝑉𝐶 =
𝑎2 (𝑐) =
= 4.243𝑎3 = 4.243(0.778 𝑥10−9 𝑚 )3 = 1.998 𝑥 10−27 𝑚 3
2
2
The product of the volume of a unit cell and the density corresponds to the weight of one beryllium mineral
molecule, so that if this value is compared with the value calculated from molecular weight and Avogadro’s
number, the number of molecules in a unit cell will be obtained:
𝜌𝑉𝑁𝐴
𝑁=
𝑚
=
2.68 𝑥 106 𝑥 1.998 𝑥 10−27 𝑥 6.022 𝑥 1023
537
= 6 𝑎𝑡𝑚𝑜𝑠/𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
Example 4.7
The atomic weight per 1 mol of copper (Cu) with face-centered cubic (FCC) structure and the density at 298k
are 63.54𝑔 and 8.89𝑔/𝑐𝑚 3, respectively. Estimate the nearest-neighbor distance of Cu atoms.
Solution
In the FCC lattice, four atoms are known to be included in a unit cell.
From the relation =
𝑎3 =
𝑁𝑚
𝑉𝑁𝐴
4 𝑥 63.54
8.89 𝑥 6.022 𝑥 1023
, we have the equation 𝑎3 =
𝑁𝑚
𝜌𝑁𝐴
, where 𝑉 = 𝑎3
, 𝑎 = 3.621 𝑥 10−8 𝑐𝑚
The nearest-neighbor distance r of Cu atoms can be calculated since Cu atoms are in contact along the diagonal
line of a cell face in the FCC structure as shown in fig.4.23.
𝑟=
𝑎
√2
= 2.560 𝑥 10−8 = 0.2560𝑛𝑚
By Gizachew Berhanu
Example 4.8
At 278 K, iron (Fe) is found to show BCC structure with a lattice parameter of 0.2866 nm. Obtain the density of
iron from this information.
Solution
The BCC structure includes two atoms per unit cell and from equation (4.9) we have the relation
𝜌=
𝑁𝑚
𝑉𝑁𝐴
=
2 𝑥 56
(0.278𝑥10−9)3𝑥 6.022 𝑥1023
= 7.88𝑔/𝑐𝑚3
Self-test
1) Give the significant of lattice point
2) Name the parameters that characterizes a unit cell
3) Distinguish the differences between: a) Hexagonal and monoclinic unit cell and b) Face-centered and basecentered unit cell
4) Calculate the number of atoms contained in base-centered unit cell
5) Explain how much portion of an atom located at ( i) corner (ii) body-centered of a cubic unit cell is part of its
neighboring unit cell
6) X-ray diffraction studies show copper crystallizes in an FCC unit cell edge of 3.608 𝑥10−8 𝑐𝑚. In a separate
experiment, copper is determined to have a density of 8.91𝑔/𝑐𝑚3, calculate the atomic mass of copper.
7) Niobium crystallizes in body centered cubic structure. If the density is 8.55𝑔/𝑐𝑚 3, calculate atomic radius of
niobium using its atomic mass 93u.
By Gizachew Berhanu
4.7 Points, Directions and Planes in cubic Unit cell
4.7.1 Point Locations in Cubic Unit Cell
To locate atom positions in cubic unit cells, we use rectangular x, y, and z axes. In crystallography the positive x
axis is usually the direction coming out of the paper, the positive y axis is the direction to the right of the paper, and
the positive z axis is the direction to the top (Fig. 4.29). Negative directions are opposite to those just described.
Atom positions in unit cells are located by using unit distances along the x, y, and z axes, as indicated in Fig. 4.29a.
For example, the position coordinates for the atoms in the BCC unit cell are shown in Fig. 4.29b. The atom
positions for the eight corner atoms of the BCC unit cell are
(0,0,0), (1,0,0), (0,1,0), (0,0,1) (1,1,1), (1,1,0), (0,1,1), (1,0,1)
a)
b)
Fig.4. 29 (a) Rectangular x, y, and z axes for locating atom positions in cubic unit cells. (b) Atom
positions in a BCC unit cell.
The center atom in the BCC unit cell has the position coordinates(1/2, 1/2, 1/2). For simplicity sometimes only
two atom positions in the BCC unit cell are specified which are (0, 0, 0) and (1/2, 1/2, 1/2). The remaining atom
positions of the BCC unit cell are assumed to be understood. In the same way the atom positions in the FCC unit
cell can be located.
4.7.2 Directions in cubic unit cells
Often it is necessary to refer to specific directions in crystal lattices. This is especially important for metals and
alloys with properties that vary with crystallographic orientation. In explaining crystal properties, we must
frequently specify a direction in a crystal, or a particular plane of atoms. Many properties, for example: The elastic
modulus, electrical resistivity, magnetic susceptibility, magnetization etc, are depends on the crystallographic
By Gizachew Berhanu
direction within the crystal. For cubic crystals the crystallographic direction indices are the vector components of
the direction resolved along each of the coordinate axes and reduced to the smallest integers.
To specify the crystallographic direction, let us consider fig.4.30. A point P on the vector can be expressed by the
coordinates xo, yo, and zo. Where xo, yo, and zo are projections from Point P on to the x, y, and z axes.
1. Identify the coordinates of P (xo, yo, zo) from the projection on x, y and z axes.
2. Express the coordinates (xo, yo, zo) in terms of lattice parameters a, b & c
𝒙𝒐 = 𝒂⁄𝟐 , 𝒚𝒐 = 𝒃, 𝒛𝒐 = 𝒄⁄𝟐
3. Specify these intercepts in fractional co-ordinates. Fractional intercepts. 𝑥1 = 𝑎⁄2𝑎 , 𝑦1 = 𝑏⁄𝑏 , 𝑧1 = 𝑐⁄2𝑐 i.e,
𝑥1 = 1⁄2 , 𝑦1 = 1, 𝑧1 = 1⁄2
4. Multiply or divide these numbers to obtain the smallest integers, 𝑥1 = 1, 𝑦1 = 2, 𝑧1 = 1 =
5. Put these integers in square bracket without any comma. [121].
Fig.4.30 direction location in cubic unit cell
Fig.4.31 Family of <111> directions equivalent
directions
These integers are denoted by u, v and w and put in the format [u v w]. If any integer is negative number, we place
̅ 2] where integer in the y-direction is negative number. If some directions
a bar on top of that integer. Example:[11
are equivalent then the indices are put in a curly bracket, i.e. <1 1 1> and <1 0 0>. Therefore, <1 0 0> represent:
̅ 00], [01
̅ 0], [001
̅]
[100], [010], [001], [1
By Gizachew Berhanu
Directions and their multiples are identical. Vectors and multiples of vectors have the same number of lattice
points/length.
Fig.4.32 Directions and their multiples are identical
4.7.2 Planes in Cubic Unit Cell
The geometry of a crystal may be completely defined with the help of coordinate axes all meeting at a point
(origin). The number and inclination of this crystal intercept them at definite distances from the origin or are
parallel to some of the axes, i.e., intercepting at infinity. The law of rational indices or intercepts states that it is
possible to choose along the three coordinate axes unit distances (𝑎, 𝑏, 𝑐) not necessarily of the same length such
that the ratio of the intercepts of any plane in the crystal is given by (𝑙𝑎 ∶ 𝑚𝑏 ∶ 𝑛𝑐) where 𝑙, 𝑚 𝑎𝑛𝑑 𝑛 are simple
integers like 1, 2, 3 or fractions of whole numbers. For example consider plane LMN in the crystal shown in Fig.
4.33.
This plane has intercepts 𝑂𝐿, 𝑂𝑀 and 𝑂𝑁 along the 𝑥, 𝑦 and z-axes at distances 2𝑎, 4𝑏 and 3𝑐 respectively. These
intercepts are in the ratio of 2𝑎 ∶ 4𝑏 ∶ 3𝑐 whereas 2, 4, 3 are simple integral whole numbers.
The coefficients of a, b and c (2, 4 and 3 in this case) are known as the Weiss indices of a plane. It may be borne in
mind that the Weiss indices are not always simple integral whole numbers as in this case. They may have fractional
values as well as infinity (an indefinite quantity). Weiss indices are, therefore, rather awkward in use and have
consequently been replaced by miller indices.
By Gizachew Berhanu
z𝑁
N
Fig.4.33. the intercepts of a crystallographic plane
c
𝑂O
a
b
x
M
𝑀 y
𝐿
L
Taking the reciprocals of Weiss indices and multiplying throughout by the smallest number (least common
multiplication) in order to make all reciprocals as integers obtain the Miller indices of a plane. Miller indices are
defined as the reciprocals of the fractional intercepts which the plane makes with the crystallographic axes.
Consider a plane which in Weiss notation is given by 2 a: 4b: 3c. Taking reciprocals of coefficients of a, b and c,
we get the ratio 1/2, 1/4, 1/3. Multiplying by 12 in order to convert them into whole numbers, we get 6, 3, 4. These
numbers are called the Miller indices of the plane, and the plane is designated as the (634) plane.
For example, if a plane is described by the Miller indices of (h k l), the plane makes fractional intercepts of 1⁄ℎ,
1⁄ , and 1⁄ with the axes a, b, and c, respectively. This reciprocal symbolism enables us to give the Miller indices
𝑘
𝑙
being zero, when a plane is parallel to an axis. For example, the center position of a body-centered cubic lattice is
expressed by 1⁄2 1⁄2 1⁄2 and the position of face-centered lattice as1⁄2 1⁄2 0; 1⁄2 0 1⁄2 ; 0 1⁄2 1⁄2.
Similarly the Miller’s indices for the plane which the Weiss notation is given by ∞𝑎 ∶ 2𝑏 ∶ 𝑐. Taking reciprocals
of coefficients of a, b and c, we get the ratio 1/∞, 1/2, 1/1, i.e., 0, 1/2, 1. Multiplying by 2 in order to convert
them into whole numbers, we get 0, 1, 2. The plane is designated as the (012) plane in which, ℎ = 0, 𝑘 =
1 𝑎𝑛𝑑 𝑙 = 2.
The Miller indices of a face of a crystal are inversely proportional to the intercepts of that face on the various axes.
The Miller indices of this particular family of planes are given by the reciprocals of the fractional intercepts along
each of the cell directions. The procedure for determining the miller indices for a plane is as follows:
(1) The distance from the origin to the intersection of the desired plane with each crystal axis is determined from
the basis of unit length, such as a lattice parameter. As shown in Fig. 4.34, the a-axis intersects at the unit
length of 1⁄ℎ .
(2) The reciprocals of three numbers are taken and let the minimum integer ratio (h k l) be the index of the
corresponding plane.
By Gizachew Berhanu
(3) If the desired plane is parallel to a certain axis, the distance from the origin in the axis to the intersection
becomes infinite. In that case, the index is expressed by zero.
For example, (h 0 0) represents a plane parallel to b-axis and c-axis.
(4) Although a set of planes parallel to it can be found for every plane, Miller indices usually refer to that plane in
the set which is nearest to the origin.
(5) When a plane intercepts at the negative side in any axis, such negative value is represented by writing a bar over
the Miller indices, for example, ( hkl ).
𝑐
𝑐 ⁄𝑙
(ℎ𝑘𝑙 )
𝑏
𝑏⁄𝑘
𝑎
𝑎 ⁄ℎ
Fig. 4.34 Example of Miller indices for plane
(6) There are sets of equivalent lattice planes related by symmetry, for example, the planes of a cube, (100), (010),
( 1 00), (0 1 0), (001), and (00 1 ).They are called “planes of a form” and the expression of {001} is used. The number
of the equivalent lattice planes in one plane of a form corresponds to the multiplicity factor and they are given for
seven crystal systems.
(7). Prepare a three-column table with the unit cell axes at the tops of the columns.
(8). Enter in each column the intercept (expressed as a multiple of
a, b or c) of the plane with these axes. (9). Invert all numbers.
(9). Clear fractions to obtain ℎ, 𝑘 and 𝑙.
Consider the x-, y-, z- axes in the
figure 4.35 with the dots representing atoms in a single crystal lattice. To
determine the Miller indices, one finds the intercepts on the three axes. The intercepts are: x = 4, y = 2 and z = 3.
Then the reciprocals are taken, i.e.,1⁄4,1⁄2,1⁄3and finally these fractions are reduced to the smallest integers, i.e.,
3, 6, 4 if multiplied by 12. Then the Miller indices represented as (364).
By Gizachew Berhanu
z
4
3
2
Fig.4. 35. 𝑥, 𝑦, 𝑧Intercepts
1
1
1
2
3
4
y
2
3
x
4
Let us look at the most common planes in a cube, shown below in figure.4.36. As an example the front crystal face
shown here intersects the x-axis but does not intersect the y- or z -axes but parallel to y and z axis. The front crystal
face intersects only one of the crystallographic axes (x-axes).So the miller index for the plane is (100). The side
plane has Intercepts x = ∞, y = 1, z = ∞ because the plane is parallel to the x- and z-axes, forming the Miller
indices gives (010). The top plane has intercepts x = ∞, y = ∞ , z = 1 because the plane is parallel to the x- and yaxes , forming the Miller indices gives (001). The (110) plane intercepts x=1, y=1 and z= ∞ which is parallel to zaxis. Similarly the other two planes are (101) and (011). The (111) plane intercepts all the three axes x=1, y=1 and
z=1.
4.36 Cubic unit cell planes
By Gizachew Berhanu
Example 4.9
Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c) (ii) (a, b, c) (iii)
(6a, 3b, 3c) and (iv) (2a, -3b, -3c).
Solution
Following the procedure given above, we prepare the tables as follows:
b
c
Remarks
b
c
Remarks
𝑖a
𝑖𝑖a
2
3
1
1
1
1
Intercepts
Intercepts
1⁄
1/3
1
1
1
1
Reciprocals
Reciprocals
2
3
2
6
1
1
1
clear fractions
clear fractions
Hence, the Miller indices are (326).
Hence, the Miller indices are (111).
b
c
Remarks
𝑖𝑖𝑖a
6
3
3
Intercepts
1⁄
1/3
1/3 Reciprocals
6
1
2
2
clear fractions
Hence, the Miller indices are (122).
𝑖𝑣a
2
1/2
3
b
-3
-1/3
c
-3
-1/3
Remarks
Intercepts
Reciprocals
-2
-2
clear fractions
Hence, the Miller indices are (3 2 2 ).
Example 4.10
The vectors of FCC unit cell connect atoms at the vertices of the FCC cube with the face centered atoms. Write
these vectors in the standard Miller notation and calculate the angle between any two of them.
Solution
The question says that the vectors of the primitive FCC unit cell connect atoms at the vertices of the FCC cube
with the face centered atoms. The following figure illustrates this statement.
𝑎1 =
𝑎
2
(𝑥̂ + 𝑦̂), 𝑎2 =
𝑎
2
(𝑦̂ + 𝑧̂ ), 𝑎3 =
𝑎
2
(𝑥̂ + 𝑧̂ )
[111] and [222] are standard direction miller indices and
111
{222} for parallel planes, because (2 2 2) are intercepts.
𝑎1 . 𝑎2 = |𝑎1 ||𝑎2 | cos 𝜃
2
2
𝑎
𝑎
|𝑎1 | = |𝑎2 | = √( ) + ( ) =
2
2
cos 𝜃 =
𝑎1.𝑎2
|𝑎1 ||𝑎2|
=
𝑎⁄ 𝑥 𝑎⁄
2
2
𝑎
𝑎
⁄ 𝑥 ⁄
√2
√2
𝜃 = cos −1 (1⁄2) = 60°
By Gizachew Berhanu
=
𝑎
√2
𝑎 2⁄
4
𝑎 2⁄
2
= 1⁄2
If the crystal plane being considered passes through the origin so that one or more intercepts are zero, the plane
must be moved to an equivalent position in the same unit cell and the plane must remain parallel to the original
plane. This is possible because all equispaced parallel planes are indicated by the same Miller indices. If sets of
equivalent lattice planes are related by the symmetry of the crystal system, they are called planes of a family or
form, and the indices of one plane of the family are enclosed in braces as {hkl} to represent the indices of a family
of symmetrical planes. For example, the Miller indices of the cubic surface planes (100), (010), and (001) are
designated collectively as a family or form by the notation {100}.
Example 4.11
Draw the following crystallographic planes in cubic unit cells:
̅ 0)
a) (11
b) (632)
̅ 11)
c) (1
d) (221)
̅ 10)
e) (1
Solution
I will show (a) and (b) whereas (c), (d) and (e) left as exercises.
̅ 0) plane. These are 1, -1, ∞. The (11
̅ 0) plane
a) First determine the reciprocals of the Miller indices of the (11
must pass through a unit cube at intercepts x = 1 and y = -1 and be parallel to the z axis.
̅ 0)
Same planes (11
b) (632), with same manner as in (a), first determine the reciprocals of the plane indices (632) .
1⁄ , 1⁄ 𝑎𝑛𝑑 1⁄
6 3
2
To reduce its lowest factor multiply all by 2
(1⁄6 , 1⁄3 , 1⁄2) 𝑥 2 = (1⁄3 , 2⁄3 , 1)
These are intercept points. Then connect the
three points with straight lines. The figure
on right is the final answer.
By Gizachew Berhanu
Planes and their multiples are not identical unlike in direction
Fig.4.37 planes and their multiples are not identical
Planes of a Zone and Interplanar Spacing
As shown earlier, there are sets of equivalent planes by symmetry in any crystal lattice and they are called planes of
a form. Atoms in crystals can be arranged not only on a lattice plane but also on a group of straight lines which are
mutually parallel. This straight line is called “zone axis” and all the planes parallel to the direction of this line are
called “planes of a zone.” Such planes have quite different indices and spacings, but their parallelism to a line is
satisfied. For example, the plane of a zone which belongs to the zone axis [001] in a cubic system is shown in
Fig.4.38. If a plane belongs to a zone whose indices are (h k l) and the indices of zone axis are [u v w], the
following relation is satisfied:
(ℎ𝑘𝑙 ) . [𝑢𝑣𝑤] = 0
ℎ𝑢 + 𝑘𝑣 + 𝑙𝑤 = 0 ………. (4.10)
Fig. 4.38 planes and zone axis
Let us consider any two planes to be planes of a zone, when they are both parallel to their line of intersection. If
these two planes are denoted by (ℎ1 𝑘1 𝑙1 ) and (ℎ2 𝑘2 𝑙2 ), the indices of their zone axis [u v w] are given by the crossproduct relations:
[𝑢𝑣𝑤] = (ℎ1 𝑘1 𝑙1 )𝑋(ℎ2 𝑘2 𝑙2 )
[𝑢𝑣𝑤] = [
ℎ1
ℎ2
𝑘1
𝑘2
𝑙1
]
𝑙2
𝑢 = 𝑘1 𝑙2 − 𝑘2 𝑙1 ; 𝑣 = 𝑙1 ℎ2 − 𝑙2 ℎ1 ; 𝑤 = ℎ1 𝑘2 − ℎ2 𝑘1 : …………… (4.11)
The value of the interplanar spacing d is a function of both the plane indices (ℎ 𝑘 𝑙) and the lattice parameters
(𝑎, 𝑏, 𝑐, 𝛼, 𝛽, 𝑎𝑛𝑑 𝛾). The relationship between the plane indices (ℎ 𝑘 𝑙) and the interplanar spacing 𝑑 depends on
By Gizachew Berhanu
crystal systems. For example, the interplanar spacing d of the plane of (h k l) for the cubic and tetragonal systems is
given in the following equations:
𝑑ℎ𝑘𝑙 =
𝑑=
𝑎
√ℎ 2+ 𝑘 2+ 𝑙 2
, …………….. (4.12)For cubic
𝑎
2
√ℎ 2+ 𝑘 2 + 𝑙 2 (𝑎 ⁄
𝑐2
)
… ……….. (4.13) for (tetragonal)
It may be worth mentioning that lower indices the plane has, larger the value of interplanar spacing becomes, and
the density of the lattice points in the corresponding plane also becomes large.
Example 4.12
Calculate the angle between ( 010 ) and ( 110 ) planes in fig.4.35 given above.
Solution
In the discussion section, we mentioned that the zone axis of certain planes is equal to the cross-product of the
̅ 0) and ( 110 ). Then the zone axis of these
corresponding planes. In the question we are given two planes (01
planes is given by:
[𝑢𝑣𝑤] = [0 −1 0] = [001]
1 −1 0
And if we are given two vectors say 𝐴 and 𝐵, then the magnitude of the cross-product of these two vectors is
given by
|𝐴 𝑋 𝐵| = |𝐴||𝐵| sin 𝜃
̅ 0) and = (11
̅ 0) , then |𝐴| = 1 𝑎𝑛𝑑 |𝐵| = √2
In our case let 𝐴 = (01
|𝐴 𝑋 𝐵 | = 1
|𝐴 𝑋 𝐵|
1
𝜃 = sin−1 ( |𝐴||𝐵| ) = sin (1.√2) = 45°
Example 4.13
Copper has an FCC crystal structure and a unit cell with a lattice constant of 0.361 nm. What is its interplanar
spacing 𝑑220 ?
Solution
𝑑ℎ𝑘𝑙 =
𝑑220 =
𝑎
√ℎ 2+ 𝑘 2+ 𝑙 2
0.361𝑛𝑚
√4+4+0
=
= 0.128𝑛𝑚
By Gizachew Berhanu
4.7.3 Hexagonal Planes and Directions
4.8.3.1 Indices for Crystal Planes in HCP Unit Cells
With respect to the hexagonal system, a slightly different method for plane indexing is employed: the so called
Miller–Bravais indices refer to plane indices with four axes such as (ℎ 𝑘 𝑖 𝑙), instead of Miller indices. The unit cell
of a hexagonal lattice is given by two equal and coplanar vectors of 𝑎1 and 𝑎2 with 120° to one another, and a
fourth axis 𝑐 at right angle, as shown in Fig. 4.39. In addition, the third axis 𝑎3 , lying on the basal plane of the
hexagonal prism, is symmetrically related to 𝑎1 and 𝑎2 and then it is often used with the other two. The complete
hexagonal lattice is obtained by repeated translations of the points at the unit cell corners by the vectors 𝑎1 , 𝑎2 , and
𝑐. It is noted for the Miller–Bravais indices that the relation of 𝑖 = −(ℎ + 𝑘) between h and k is always satisfied.
This is because the value of i depends on the h and k values, since the intercepts of a plane on 𝑎1 and 𝑎2 determine
its intercept on 𝑎3 .
Fig.4.39 The four coordinates axes (a1, a2, a3, and c) of the HCP crystal structure unit cell.
The procedure for finding the indices of planes is exactly the same as before, but four intercepts are required, giving
indices of the form (hkil). Because of the redundancy of the a3 axis and the special geometry of the system, the first
three integers in the designation, corresponding to the a1, a2, and a3 intercepts, are related by h + k = -i.
The indices of the four shaded planes (A, B, C and D) are summarized in the
table 4.8 given below.
By Gizachew Berhanu
Table 4.8 hexagonal indices
Planes
Intercepts
𝑎1
𝑎2
𝑎3
𝑐
indices
A
B
C
1
1
1
-1
-1
0
0
0
-1
0
0
0
D
0
1
-1
0
̅ 00)
(11
̅ 00)
(11
̅ 0)
(101
̅ 0)
(011
Why A and B Planes have equal indices?
Fig.4.40 indices of planes in Hexagonal structure unit cell
4.8 Planer and linear density
4.8.1 Planar density
Planar density (PD) refers to density of atomic packing on a particular plane.
𝑃𝑙𝑎𝑛𝑎𝑟 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑝𝑙𝑎𝑛𝑒
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒
……………………. (4.15)
Assume you want to know the planar density of (101) plane of aluminum element at room temperature.
At room temperature aluminum has FCC crystal structure and you need to a particular given plane as
discussed before.
Fig.4.41 Plane density along (101) plane
As shown on the fig.4.41 above the one corner of the plane contribute 1/4 atom and one side face contribute
1/2 atom to the total area. The plane owns four corners.
1
∴ 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑐𝑜𝑟𝑛𝑒𝑟 = 4 𝑥 4 = 1𝑎𝑡𝑜𝑚
1
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑖𝑑𝑒 𝑓𝑎𝑐𝑒 = 𝑥 2 = 1𝑎𝑡𝑜𝑚
2
Totally there are 2atoms on the (101) plane.
By Gizachew Berhanu
The diagonal side plane is equal to 𝑎√2 and the area of the plane is
𝑎 𝑥 𝑎√2 = 𝑎2 √2
∴ 𝑃𝑙𝑎𝑛𝑎𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒
=
2
𝑎 2√2
=
√2
𝑎2
𝐼𝑛 𝑐𝑎𝑠𝑒 𝑜𝑓 𝑎𝑙𝑚𝑢𝑛𝑖𝑢𝑚 𝑎 = 0.405𝑛𝑚
∴ 𝑃𝑙𝑎𝑛𝑎𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝐴𝑙 =
√2 𝑎𝑡𝑜𝑚𝑠
(0.405 𝑥 10 −9𝑚)2
= 8.62 𝑥1018 𝑎𝑡𝑜𝑚𝑠/𝑚 2
In the {111} planes of the FCC lattice there are 2 𝑎𝑡𝑜𝑚𝑠 (1/6 𝑥 3 corner atoms + 1/2 𝑥 3 side atoms).
Fig.4.42 Planar density of {111} planes in the FCC crystal
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 = 1⁄2 (𝑎𝑏 sin 𝜃)
= 1⁄2 (𝑎 √2)(𝑎√2) sin 60° =
𝑃𝐷(111) =
2𝑎𝑡𝑜𝑚𝑠
√3 2
𝑎
2
=
√3 2
𝑎
2
4𝑎𝑡𝑜𝑚𝑠
√3𝑎2
This is higher than {110} and any other plane. Therefore, {111} planes are most densely packed planes in the FCC
Crystal.
4.8.2 Linear Density
Linear density (LD) is the number of atoms per unit length along a particular direction
𝐿𝑖𝑛𝑒𝑎𝑟 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟
………………………. (4.16)
< 110 > directions in the FCC lattice have 2 atoms (1/2 𝑥 2 𝑐𝑜𝑟𝑛𝑒𝑟 𝑎𝑡𝑜𝑚𝑠 + 1 𝑐𝑒𝑛𝑡𝑒𝑟 𝑎𝑡𝑜𝑚) and the length is
2a .
𝐿𝐷[110] =
2
𝑎√2
=
√2
𝑎
This is the most densely packed direction in the FCC lattice.
By Gizachew Berhanu
4.9 close packed structure
In solids, the constituent particles are close-packed, leaving the minimum vacant space. Let us consider the
constituent particles as identical hard spheres and build up the three dimensional structure in three steps.
(a) Close Packing in One Dimension:
There is only one way of arranging spheres in a one dimensional close packed structure, that is to arrange them in a
row and touching each other (Fig. 4.43)
Fig.4.43 close packing of spheres in one dimension
In this arrangement, each sphere is in contact with two of its neighbours. The number of nearest neighbours of a
particle is called its coordination number. Thus, in one dimensional close packed arrangement, the coordination
number is 2.
(b) Close Packing in Two Dimensions
Two dimensional close packed structure can be generated by stacking (placing) the rows of close packed spheres.
This can be done in two different ways.
(i) The second row may be placed in contact with the first one such that the spheres of the second row are exactly
above those of the first row. The spheres of the two rows are aligned horizontally as well as vertically. If we call the
first row as ‘A’ type row, the second row being exactly the same as the first one, is also of ‘A’ type. Similarly, we
may place more rows to obtain AAA type of arrangement as shown in Fig. 4.44 (a).
a)
b)
Fig. 4.44: (a) Square close packing (b) hexagonal close packing of spheres in two dimensions
By Gizachew Berhanu
In this arrangement, each sphere is in contact with four of its neighbours. Thus, the two dimensional coordination
number is 4. Also, if the centers of these 4 immediate neighbouring spheres are joined, a square is formed. Hence
this packing is called square close packing in two dimensions.
(ii) The second row may be placed above the first one in a staggered manner such that its spheres fit in the
depressions of the first row.
If the arrangement of spheres in the first row is called ‘A’ type, the one in the second row is different and may be
called ‘B’ type. When the third row is placed adjacent to the second in staggered manner, its spheres are aligned
with those of the first layer. Hence this layer is also of ‘A’ type. The spheres of similarly placed fourth row will be
aligned with those of the second row (‘B’ type). Hence this arrangement is of ABAB type. In this arrangement there
is less free space and this packing is more efficient than the square close packing. Each sphere is in contact with six
of its neighbours and the two dimensional coordination number is 6. The centres of these six spheres are at the
corners of a regular hexagon (Fig. 4.44b) hence this packing is called two dimensional hexagonal close-packing. It
can be seen in Figure 4.44 (b) that in this layer there are some voids (empty spaces). These are triangular in shape.
The triangular voids are of two different types. In one row, the apex of the triangles is pointing upwards and in the
next layer downwards.
(c) Close Packing in Three Dimensions
All real structures are three dimensional structures. They can be obtained
by stacking two dimensional layers one above the other. In the last
Section, we discussed close packing in two dimensions which can be of
two types; square close-packed and hexagonal close-packed.
Let us see what types of three dimensional close packing can be obtained
from these.
(i) Three dimensional close packing from two dimensional square closepacked layers: While placing the second square close-packed layer above
the first we follow the same rule that was followed when one row was
placed adjacent to the other.
Fig. 4.45 Simple cubic lattice formed
by A A A .... arrangement
The second layer is placed over the first layer such that the spheres of the upper layer are exactly above those of the
first layer. In this arrangement spheres of both the layers are perfectly aligned horizontally as well as vertically as
shown in Fig. 4.45. Similarly, we may place more layers one above the other. If the arrangement of spheres in the
first layer is called ‘A’ type, all the layers have the same arrangement. Thus this lattice has AAA.... type pattern.
The lattice thus generated is the simple cubic lattice, and its unit cell is the primitive cubic unit cell (See Fig. 4.19).
By Gizachew Berhanu
(ii) Three dimensional close packing from two dimensional hexagonal close packed layers: Three dimensional close
packed structure can be generated by placing layers one over the other.
(a) Placing second layer over the first layer
Let us take a two dimensional hexagonal close packed layer ‘A’ and place a similar layer above it such that the
spheres of the second layer are placed in the depressions of the first layer. Since the spheres of the two layers are
aligned differently, let us call the second layer as B. It can be observed from Fig. 4.46 that not all the triangular
voids of the first layer are covered by the spheres of the second layer. This gives rise to different arrangements.
Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is
formed. These voids are called tetrahedral voids because a tetrahedron is formed when the centres of these four
spheres are joined.
Fig. 4.46 A stack of two layers of close packed spheres and voids generated in them. Tetrahedral void and
Octahedral void
At other places, the triangular voids in the second layer are above the triangular voids in the first layer, and the
triangular shapes of these do not overlap. One of them has the apex of the triangle pointing upwards and the other
downwards. Such voids are surrounded by six spheres and are called octahedral voids. One such void has been
shown separately in Fig. 4.47.
The number of these two types of voids depends upon the
number of close packed spheres.
Let the number of close packed spheres be N, then:
The number of octahedral voids generated = N
The number of tetrahedral voids generated = 2N
Fig 4.47 Tetrahedral and octahedral voids top view
By Gizachew Berhanu
(b) Placing third layer over the second layer
When third layer is placed over the second, there are two possibilities.
(i) Covering Tetrahedral Voids: Tetrahedral voids of the second layer may be covered by the spheres of the third
layer. In this case, the spheres of the third layer are exactly aligned with those of the first layer. Thus, the pattern of
spheres is repeated in alternate layers.
This pattern is often written as ABAB ....... pattern. This structure is called hexagonal close packed (HCP) structure.
This sort of arrangement of atoms is found in many metals like magnesium and zinc.
(ii)Covering Octahedral Voids: The third layer may be placed above the second layer in a manner such that its
spheres cover the octahedral voids. When placed in this manner, the spheres of the third layer are not aligned with
those of either the first or the second layer. This arrangement is called “C’ type. Only when fourth layer is placed,
its spheres are aligned with those of the first layer. This pattern of layers is often written as ABCABC ........... This
structure is called cubic close packed face-centred cubic (FCC) structure. Metals such as copper and silver
crystallize in this structure.
Both these types of close packing are highly efficient and 74% space in the crystal is filled. In either of them, each
sphere is in contact with twelve spheres. Thus, the coordination number is 12 in either of these two structures.
4.10 Structure-Property correlation
Aluminium (Al) is ductile while iron (Fe) and magnesium (Mg) are not. This can be explained from their crystal
structures. Al is FCC whereas Fe is BCC and Mg is HCP. Plastic deformation in metals takes place mainly by a
process called slip. Slip can broadly be visualized as sliding of crystal planes over one another. Slip occurs on most
densely packed planes in the most closely packed directions lying on that plane. The slip plane and the direction
together is called a Slip system

In FCC, {111} planes are close-packed and there are four unique {111} planes. Each of these planes
contains three closely packed <110> directions. Therefore, there are 4 x 3 = 12 slip systems

In HCP, the basal plane, (0001) is the close-packed and it contains three <1 12 0> directions. Hence,
number of slip system = 1 x 3 = 3

Slip in more number of slip systems allows greater plastic deformation before fracture imparting ductility
to FCC materials

Close-packed planes are also planes with greatest interplanar spacing and this allows slip to take place
easily on these planes.

BCC structure on the other hand has 48 possible slip systems. However, there is no close-packed plane.
Hence, plastic deformation before fracture is not significant. Slip might occur in {110}, {112} and {123}
planes in the <111> directions. (SEE CHAPTER SIX!)
By Gizachew Berhanu
Fig.4. 48. Slip planes and directions
4.11 Crystal Structure of Ionic Materials
In chapter 2, we tried to mention that crystals can be broadly classified into three categories from the point of view
of bonding: “metallic,” “ionic,” and “covalent.” In metallic crystals, a large number of electrons (conduction or
valence electrons) are free to move the inside of the system, without belonging to specific atoms but shared by the
whole system. This bonding arising from a conduction electron is not very strong. For example, the interatomic
distances of alkali metals are relatively large, because the kinetic energy of conduction electrons is relatively low at
the large interatomic distances (equation 2.7). This leads to weak binding and simple structure. On the other hand,
ionic crystals consist of positive and negative ions, and the ionic bond results from the electrostatic interaction of
oppositely charged ions in the solid state. These solids are hard and brittle in nature. They have high melting and
boiling points. Since the ions are not free to move about, they are electrical insulators in the solid state. However,
in the molten state or when dissolved in water, the ions become free to move about and they conduct electricity.
Typical examples are metal–halogen compounds, and two typical structures found for ionic crystals are sodium
chloride and cesium chloride structures.
In ionic crystals, ionic arrangements that minimize electrostatic repulsion and maximize electrostatic attraction are
preferred. In many cases, the negative ions (anions) of large size are densely arranged so as to avoid their direct
contact, and the positive ions (cations) of small size occupy the positions equivalent to the vacant space produced
by anions.
A wide variety of crystalline solids of non-metals result from the formation of covalent bonds between adjacent
atoms throughout the crystal. They are also called giant molecules. Covalent bonds are strong and directional in
nature, therefore atoms are held very strongly at their positions. Such solids are very hard and brittle. They have
extremely high melting points and may even decompose before melting.
By Gizachew Berhanu
They are insulators and do not conduct electricity. Diamond and silicon carbide are typical examples of such solids.
Graphite is soft and a conductor of electricity. Its exceptional properties are due to its typical structure. Carbon
atoms are arranged in different layers and each atom is covalently bonded to three of its neighbouring atoms in the
same layer. The fourth valence electron of each atom is present between different layers and is freeto move about.
These free electrons make graphite a good conductor of electricity. Different layers can slide one over the other.
This makes graphite a soft solid and a good solid lubricant.
a) Sodium chloride
Space lattice of sodium chloride is known to consist of a face centered cubic lattice of 𝑁𝑎+ ions interlocked with a
similar lattice of 𝐶𝑙 − ions. A unit cell of this combined lattice is shown in figure 4.49. This unit cell repeats itself in
three dimensions throughout the entire crystal. The yellow spheres indicate chloride ions and red spheres represent
sodium ions. The lattices are constituted entirely by ions are known as ionic lattices. All electrovalent compounds
show such lattices.
𝑁𝑎+
𝐶𝑙 −
Fig. 4.49 sodium chloride structure
There are four units of 𝑁𝑎𝐶𝑙 in each unit cube with atoms in the positions
𝐶𝑙: 0 0 0 ; 1⁄2 1⁄2 0 ;
𝑁𝑎: 1⁄2 1⁄2 1⁄2 ; 0 0
1⁄ 0 1⁄ ; 0 1⁄ 1⁄
2
2
2 2
1⁄ ; 0 1⁄ 0 ; 1⁄ 0 0
2
2
2
As shown in the figure 4.49, the unit cell of sodium chloride consists of 14 chloride ions and 13 sodium ions. Each
chloride ion is surrounded by 6 sodium ions and similarly, each sodium ion is surrounded by 6 chloride ions. Notice
that the particles at corners, edges and faces do not lie wholly within the unit cell. Instead these particles are shared
by other unit cells. A particle at a corner is shared by eight unit cells, one at the centre of face is shared by two and
one at the edge is shared by four. The unit cell of sodium chloride has 4 sodium ions and 4 chloride ions as shown
below;
# 𝑜𝑓 𝑠𝑜𝑑𝑖𝑢𝑚 𝑖𝑜𝑛𝑠 = 12 (𝑎𝑡 𝑒𝑑𝑔𝑒 𝑐𝑒𝑛𝑡𝑒𝑟𝑠) 𝑥
𝑁𝑒𝑑
4
+
𝑁𝑓
2
+
𝑁𝑏
1
+
𝑁𝐶
8
1
4
+ 1 (𝑎𝑡 𝑏𝑜𝑑𝑦 𝑐𝑒𝑛𝑡𝑒𝑟)𝑥 1
…………………………………….. (4.17)
In case of sodium ion 𝑁𝑒𝑑 = 12, 𝑁𝑓 = 0, 𝑁𝑏 = 1 𝑎𝑛𝑑 𝑁𝑐 = 0. If we substitute these values in equation
(4.10), we will get
# 𝑜𝑓 𝑠𝑜𝑑𝑖𝑢𝑚 𝑖𝑜𝑛 =
12
4
+
0
2
+
1
1
+
0
8
= 3+1= 4
In case of chlorine ion, 𝑁𝑒𝑑 = 0, 𝑁𝑓 = 6, 𝑁𝑏 = 0 𝑎𝑛𝑑 𝑁𝐶 = 8
By Gizachew Berhanu
# 𝑜𝑓 𝑐ℎ𝑙𝑜𝑟𝑖𝑛𝑒 𝑖𝑜𝑛 =
0
4
+
6
2
+
0
1
+
8
8
= 3+1= 4
Thus, number of NaCl units per unit cell is 4. The sodium chloride structure is also called rock-salt structure.
Representative crystals having the NaCl arrangements includes: LiH, NaI, KCl, RbF, RbI, PbS etc.
b) Cesium Chloride structure
The cesium chloride, CsCl, structure has body-centred cubic system and is shown in figure. The body-centred cubic
arrangement of atoms is not a close packed structure. There is one molecule per primitive cell, with atoms at the
corners (000) and body-centred positions 1/2 1/2 1/2 of the simple cubic space lattice.
𝐶𝑠 +
𝐶𝑙 −
Fig. 4.50 Cesium Chloride structure
1. The 𝐶𝑙 − ions are at the corners of a cube whereas 𝐶𝑠 + ion is at the centre of the cube or vice versa.
2. Each 𝐶𝑠 + ion is connected eight 𝐶𝑙 − ions and each 𝐶𝑙 − ion is connected eight 𝐶𝑠 + ions i.e., 8:8 coordination.
Thus each atom is at the center of a cube of atoms of the opposite kind, so that the coordination number is eight.The
unit cell of cesium chloride has one 𝐶𝑠 + ion and one 𝐶𝑙 − ion as shown below.
1
# 𝑜𝑓𝐶𝑙 − 𝑖𝑜𝑛𝑠 = 8 (𝑎𝑡𝑐𝑜𝑟𝑛𝑒𝑟𝑠)𝑥 (𝑐𝑜𝑚𝑚𝑜𝑛𝑡𝑜𝑒𝑖𝑔ℎ𝑡𝑢𝑛𝑖𝑡𝑐𝑒𝑙𝑙)
8
1
8𝑥8 = 1
# 𝑜𝑓𝐶𝑠 + 𝑖𝑜𝑛𝑠 = 1 𝑥 1 = 1
Thus number of CsCl units per unit cell is 1. Representative crystals having the CsCl arrangements include: CsBr,
CsI, TlBr, TlI, NH4Cl etc.
Location of tetrahedral void and octahedral void in FCC
The tetrahedral void of the regular tetrahedron surrounded by four
spheres is shown in Fig.4.51. Such tetrahedral voids form a simple cubic
lattice with the lattice parameter 𝑎⁄2. This tetrahedral void has eight
equivalent positions per unit cell. Consider fig.4.49 sodium chloride
structures in which one 𝑁𝑎+ ion is positioned at the center and extra 𝑁𝑎+
ions are place on the 12 edges of the cub whereas 𝐶𝑙 − are placed at the
corner and faces of the cub.
By Gizachew Berhanu
Fig.4.51 Tetrahedral void
As shown in the figure, there are 8 𝐶𝑙 − ions located at each corner and there is one tetrahedral void for each corner
ion and therefore, 8 tetrahedral voids are available. If 𝑁 is number of atoms per unit cell and there are 2𝑁
tetrahedral voids. Hence, here in FCC crystal structure 𝑁 = 4 and 2 𝑥 4 = 8 number of tetrahedral voids.
Let us consider the case where the hard sphere with the atomic radius 𝑟 =
𝑎
√2
is arranged in the FCC lattice. As
shown in Fig.4.52, the void is found at the center of a unit cell as well as the midpoint of each edge-line. The void
at the center of unit cell is surrounded by six spheres and constitutes the octahedral interstitial site.
Fig.4.52 Octahedral void
Since such void position forms the FCC lattice with the lattice parameter “a,” the octahedral void has four
equivalent positions per unit cell.
1
1 (𝑜𝑐𝑡𝑎ℎ𝑒𝑑𝑟𝑎𝑙𝑣𝑜𝑖𝑑𝑎𝑡 𝑐𝑒𝑛𝑡𝑒𝑟) + 12 𝑥 4 (𝑜𝑐𝑡𝑎ℎ𝑒𝑑𝑟𝑎𝑙𝑣𝑜𝑖𝑑𝑎𝑡𝑒𝑑𝑔𝑒𝑠) = 1 + 3 = 4
∴ 𝑁𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑜𝑐𝑡𝑎ℎ𝑒𝑑𝑟𝑎𝑙𝑣𝑜𝑖𝑑 = 𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑎𝑡𝑜𝑚𝑠𝑝𝑒𝑟𝑢𝑛𝑖𝑡𝑐𝑒𝑙𝑙 =
1
2
𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑡𝑒𝑡𝑟𝑎ℎ𝑒𝑑𝑟𝑎𝑙𝑣𝑜𝑖𝑑
Next, we will estimate the maximum radius of the sphere which is just fit to the octahedral void, 𝑟𝑜 , and tetrahedral
void, 𝑟𝑡 , in fact structure. With the help of the relationships found in Fig. above geometry of the octahedral void is
illustrated and we obtain the value of 𝑟𝑜 in the following way: First let’s consider a void that surrounded by three
atoms or molecules or ions which is same as coordination number three.
The two ends of the triangle rest at the center of the two
atoms whereas the third end rests in the center of the void
created among three atoms as shown in the fig.4.53 on the
right. Since these atoms are regarded as the same atom, they
have same radii and the two lines drawn to the void center
have equal length which is 𝑅 + 𝑟𝑣 . Therefore, the triangle is
isosceles and the angle subscribed between the two sides is
60°. The size of the longer side is 2𝑅.
By Gizachew Berhanu
Fig.4. 53 coordination number three
To calculate the ratio of radius of an atom to radius of void let’s apply cosine law of triangle.
(𝑅 + 𝑟𝑣 )2 + (𝑅 + 𝑟𝑣 )2 + 2(𝑅 + 𝑟𝑣 )2 cos 60° = 4𝑅2
3(𝑅 + 𝑟𝑣 )2 = 4𝑅2
𝑟𝑣
=
𝑅
2
√3
− 1 = 0.155
If a fourth atom is added to the top of the three atoms figure above, there will be a tetrahedral void discussed above.
Let points 𝐴, 𝐵, 𝐷 𝑎𝑛𝑑 𝐸 are location of the four atoms and point 𝐹 is location of tetrahedral void as shown fig. 4.54
given below. Then, letter 𝑋represents 𝐸𝐹 = 𝐴𝐹, the distance from the center of the atom to the center of the void.
From Pythagoras theorem, we know that:
(𝐴𝐶 )2 + (𝐶𝐵)2 = (𝐴𝐵)2
𝐴𝐶 = √(𝐴𝐵)2 − (𝐶𝐵)2 = √4𝑅2 − 𝑅2 = √3𝑅 and
(𝐴𝐺 )2 + (𝐺𝐸 )2 = (𝐴𝐸 )2
𝐴𝐺 =
2
3
𝐴𝐶 =
2
√3
𝑅 , since point 𝐺 is centroid of triangle 𝐴𝐵𝐷.
𝐺𝐸 = √(𝐴𝐸)2 − (𝐴𝐺 )2 = √4𝑅2 −
4
3
8
𝑅2 = √ 𝑅
3
Another triangle we will consider again is ∆𝐴𝐺𝐹 which is
a right triangle.
(𝐴𝐹 )2 = 𝑋 2 = (𝐴𝐺 )2 + (𝐺𝐹 )2 , where 𝑋 = 𝑅 + 𝑟𝑣
𝑏𝑢𝑡, 𝐺𝐹 = 𝐺𝐸 − 𝑋
Fig. 4.54 tetrahedral void
2
𝑋 = (𝐴𝐺 )2 + (𝐺𝐸 − 𝑋)2 =
8
𝑋
4𝑅2 = 2√3 𝑅𝑋 , 𝑅 =
𝑅+ 𝑟𝑣
𝑅
4
3
2
2
8
8
𝑅 + (√3 𝑅 − 𝑋) = 4𝑅2 − 2√3 𝑅𝑋 + 𝑋 2
= √3⁄2,
𝑟𝑣
𝑅
= √3⁄2 − 1 = 0.225
Example 4.14
A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How
many of these are tetrahedral voids?
Solution
In the discussion we mentioned that 𝑁 molecules have 𝑁 octahedral voids and 2𝑁 tetrahedral voids. Then if
we find number of molecules, we can easily find number of octahedral voids. There is 0.5 numbers of moles,
means:
0.5 𝑥 6.022 𝑥1022 𝑎𝑡𝑜𝑚𝑠 = 3.011 𝑥1023 𝑎𝑡𝑜𝑚𝑠.
Number of octahedral = 3.011 𝑥1023 voids
Number of tetrahedral= 2 𝑥 3.011 𝑥1023 = 6.022 𝑥1023 𝑣𝑜𝑖𝑑𝑠
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑖𝑑 = 9.033 𝑥1023
By Gizachew Berhanu
Example 4.15
1
A compound is formed by two elements 𝑀𝑎𝑛𝑑 𝑁. The element 𝑁 forms FCC and atoms of 𝑀 occupy 𝑟𝑑 of
3
tetrahedral voids. What is the formula of the compound?
Solution
The question is say the compound is formed from the elements 𝑀 and 𝑁 where atom 𝑁 forms FCC and 𝑀 is
1
located at some where 3 𝑟𝑑 of tetrahedral void.
If 𝑁 = 𝑋 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠, 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑡𝑟𝑎ℎ𝑒𝑑𝑟𝑎𝑙 𝑣𝑜𝑖𝑑𝑠 = 2𝑋,
𝑀=
1
3
𝑜𝑓 𝑡𝑒𝑡𝑟𝑎ℎ𝑒𝑑𝑟𝑎𝑙 𝑣𝑜𝑖𝑑𝑠 =
𝑀to𝑁 ratio =
𝑀
𝑁
=
2𝑋
3𝑋
=
1
3
𝑥 2𝑋 =
2
3
𝑋
2
3
∴ Then the formula of the compound is 𝑀2 𝑁3.
Example 4.16
An element with molar mass 2.7 𝑥10−2 𝑘𝑔/𝑚𝑜𝑙 forms a cubic unit cell with edge length 405pm. If its density
is 2.7 𝑥103 𝑘𝑔/𝑚3 . What is the nature of the cubic unit cell?
Solution
From equation (4.4), we know that
𝜌=
𝑁𝑀
𝑉𝑁𝐴
→ 𝑁=
𝜌𝑉𝑁𝐴
𝑀
=
2.7 𝑥 103 𝑥 6.022 𝑥 1023 𝑥 (405 𝑥 10−12)
2.7 𝑥 103
3
= 3.84 ≈ 4
∴ The nature of this cubic unit cell is FCC
Example 4.17
Calculate the bond angle 𝜃, of O–Si–O in silica (SiO2) assuming that Si and O atoms form regular tetrahedron
with Si at the center.
Solution
Let us consider the case where the peak positions of a cube characterized by the length of 𝑎 are connected to
form regular tetrahedron as shown in Fig. shown below.
The position of O in this figure corresponds to a center of both cube and
regular tetrahedron. The positions denoted by A and C represent corners
of regular tetrahedron and B gives the midpoint of regular tetrahedron
which is characterized by the length of AC. The angle AOC is equivalent
to the desired angle, 𝜃, of the O–Si–O bond. Therefore, the following
relationship is readily found in triangle AOB:
By Gizachew Berhanu
tan(𝜃⁄2) =
𝐴𝐵
𝑂𝐵
tan(𝜃⁄2) =
𝐴𝐵 =
,
𝑎 √2
𝑎
22
𝐴𝐶
2
=
𝑎√2
2
, 𝑎𝑛𝑑 𝑂𝐵 =
𝑎
2
= √2, → 𝜃⁄2 = tan−1(√2) = 54.74°
→ 𝜃 = 109.48°
Example 4.18
The position of the maximum void in the body-centered cubic lattice is known to be corresponding to the
tetrahedral site (1/2, 1/4, 0), and to equivalent position. Obtain the radius of maximum sphere that fits to this
space if the BCC crystal structure with lattice parameter 𝑎 by supposing the case where spherical atoms of
radius 𝑟𝐴 are arranged in each lattice point.
Solution
If the radius of the sphere which fits to void is 𝑟𝑥 , the following relationship is obtained by geometric
conditions.
Tetrahedral voids in the bcc lattice. (Filled circle) Metal atoms, (open circle) Tetrahedral Void
𝑎 2
𝑎 2
(𝑟𝑥 + 𝑟𝐴 )2 = ( ) + ( ) =
4
2
5𝑎 2
16
𝑟𝑥 + 𝑟𝐴 = 𝑎⁄4 √5
𝐹𝑜𝑟 𝐵𝐶𝐶 𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒, 𝑟𝐴 = 𝑎⁄4 √3
√5 − √3
𝑟𝑥 = 𝑎⁄4 √5 − 𝑟𝐴 = 𝑎⁄4 √5 − 𝑎⁄4 √3 = 𝑎 ( 4 )
If we compare the ratio radius of void to radius of atom,
𝑟𝑥
𝑟𝐴
=
(
√5−√3
)𝑎
4
√3
𝑎
4
= 0.29 ,
𝑟𝑥 = 0.29𝑟𝐴
The maximum radius which fits to void in the BCC lattice is about 30% of the radius of the constituent atom.
By Gizachew Berhanu
Example 4.19
In ionic crystals, anions of the relatively larger size are densely arranged so as to avoid their direct contact,
whereas cations of relatively smaller size occupy the positions equivalent to the vacant space produced by
anions. For this reason, if the radii of cation and anion are described by 𝑟𝑐 and 𝑟𝑎 , respectively, some
correlations are recognized between the coordination numbers and the size ratio of 𝑟𝑐⁄𝑟𝑎 . Estimate the specific
values of 𝑟𝑐⁄𝑟𝑎 for cases that cations are surround by anions with the coordination numbers of 3, 4, 6, and 8.
Solution
a) 3-Coordination Number
A stable structure is obtains if the central ion is in contact with three anions. If the central ion is sufficiently
large, the three anions can be in contact with the central ion without mutual contact. However, as the central
ion becomes smaller, stage is reached when the three ions are in contact with each other. This is the limiting
condition for three coordination number, i.e. above this value 3-coordination is possible.
sec 30° =
𝑟𝐶
𝑟𝐴 + 𝑟𝐶
= 2⁄
𝑟𝐴
√3
⁄𝑟𝐴 = 2⁄ − 1 = 0.155
√3
b) 4-Coordination Number
The four anions may be considered to be tetrahedral arrangement around a central cation.
In example 4.17 we calculated that the bisected tetrahedral angle is 54.74°
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒,
sin(54.74°) =
𝑟𝐶
𝑟𝐴
−
1−0.81654
0.81654
𝑟𝐴
𝑟𝐴 + 𝑟𝐶
= 0.81654
= 0.225
6- Coordination number and 8- coordination number are left as exercises.
By Gizachew Berhanu
Self-test
1) If the radius of octahedral void is 𝑟 and radius of the atoms in close packing is 𝑅, then derive the relation
between 𝑟 and 𝑅.
2) A cubic solid is made of two elements 𝑃 and 𝑄. Atoms of 𝑄 are at the corners of the cube and 𝑃 at the body
center. What is the formula of the compound? What are the coordination numbers of 𝑃 and 𝑄?
3) Explain the basic similarities and differences between metallic and ionic crystal.
4) Ferric Oxide crystallizes in a hexagonal close-packed array of oxide ions with two out of every three
octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
5) Gold (atomic radius = 0.144nm) crystallizes in a face centered unit cell. What is the length of a side of the
cell?
6) Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is 125pm.
a) What is the length of the side of the unit cell?
b) How many unit cells are there in 1.00𝑐𝑚 3 aluminium?
7) A solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high
temperature. What type of solid is it? Ionic, Covalent, or metallic?
8) Ionic solids conduct electricity in molten state but not in solid state. Explain why.
9) What type of solids are electrical conductors, malleable and ductile?
10) Which of the following lattice has the highest packing efficiency? (i) Simple Cubic, (ii) body centered
cubic and (iii) hexagonal close-packed lattice.
11) How can you determine the atomic mass of unknown metal if you know its density and the dimension of its
unit cell? Explain.
4.12 covalent structure
Carbon is one of the most abundant elements on earth. Carbon is an essential component of all living organisms,
and it has enormous technological significance with a wide range of applications. For example, carbon dating is a
process by which scientists measure the amount of a radioactive isotope of carbon that is present in fossils to
determine their age, and now some of the most cutting-edge technologies exploit one of the world’s strongest
materials: carbon nanotubes. And of course, a small amount of carbon (e.g., 0.5 wt%) converts iron into steel.
Pure carbon exists as several allotropes, meaning that pure carbon exists in different forms (or has different
arrangements of its atoms) depending on the temperature and pressure. We discussed more about allotropes in
section 4.6. Two allotropes of carbon are very familiar to us: diamond and graphite.
By Gizachew Berhanu
a) Diamond
In a crystal structure of diamond each carbon atom is covalently
bonded to four other atoms; tetrahedrally arranged and this
arrangement extended throughout the crystal (fig.4.55). The
whole crystal can thus be regarded as macromolecule. The
carbon-carbon bonds result from 𝑠𝑝3 hybrid orbitals of each
carbon overlapping with four such hybrid orbitals from four other
carbon atoms. The hardness of diamond can be attributed to the
large bond energy of the carbon-carbon single bond. A crystal can
be readily broken if it has a plane where cleavage can occur. The
absence of such cleavage planes and the enormous number of
carbon-carbon bonds that have to be broken in order to break the
Fig.4.55 Diamond structure
crystal are also responsible for its hardness.
The carbon-carbon covalent bonds being strong, thermal vibrations do not affect the bond distance. This results in a
low coefficient of expansion. Since a large amount of energy is required to break the carbon-carbon bonds in
macromolecule, the melting point of diamond is high. Since all the four electrons of each carbon atom in the
macromolecules are utilized for bond formation and are practically localized, these are not available for conduction
and thus diamond is an insulator.
Another allotrope of carbon, graphite has a
completely different crystal structure, which is
responsible for the difference in the properties of
diamond and graphite. It crystallizes in the
hexagonal system. The atoms are arranged in
parallel layers. In every layer each carbon atom
is covalently bound to three other carbon atoms
to give an extended hexagonal pattern. The
carbon atoms in a plane are obviously 𝑠𝑝2
hybridized, the carbon-carbon distance in plane
being 3.41Å, is too large for any orbital overlap.
The planes are held by weak Van der Waals
Fig.4.56 The structure of graphite. The carbon atoms are
arranged in layers, and in each layer, the carbon atoms are
arranged in a hexagonal pattern.
By Gizachew Berhanu
forces. Such crystals as graphite are said to form
layer lattices.
As a result of the weak Va der Waals forces between the planes, graphite is able to cleave in layers and a plane is
able to cleave slide or glide over another plane. This may be responsible to a certain extent for the lubricating action
of graphite. In graphite, since 𝑠𝑝2 hybrid bonds are used, each carbon also has one 𝜋-electron. The mobile 𝜋electron system is responsible for the electrical conductivity of graphite. Graphite is able to form intercalation
(insertion between) compounds as result of the availability of the 𝜋-electrons. This 𝜋-electrons are used in forming
a compound (𝐶𝐹 )𝑛 with fluorine, in which the color and electrical conductivity of graphite are lost.
Example 4.20
Describe the diamond cubic structure as a lattice and a basis and determine its packing factor.
Solution
The diamond cubic structure is a face-centered cubic lattice with a basis of two atoms of the same type located
at (0, 0, 0) and (1/ 4, 1/ 4, 1/ 4). The basis atom located at (0, 0, 0) accounts for the atoms located at the FCC
lattice points, which are (0, 0, 0), (0, 1/ 2, 1/ 2), (1/ 2, 0, 1/ 2), and (1/ 2, 1 /2, 0) in terms of the coordinates
of the unit cell. By adding the vector [1/ 4 1/ 4 1/ 4] to each of these points, the four additional atomic
coordinates in the interior of the unit cell are determined to be (1/ 4, 1/ 4, 1/ 4), (1/ 4, 3/ 4, 3/ 4),
(3/ 4, 1/ 4, 3/ 4), and (3/ 4, 3/ 4, 1/ 4). There are eight atoms per unit cell in the diamond cubic structure:
4 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 𝑝𝑜𝑖𝑛𝑡𝑠/ 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 ∗ 2 𝑎𝑡𝑜𝑚𝑠/𝑙𝑎𝑡𝑡𝑖𝑐𝑒 𝑝𝑜𝑖𝑛𝑡 = 8 𝑎𝑡𝑜𝑚𝑠/ 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝑎 = 8𝑟/√3
The packing factor is the ratio of the volume of space
occupied by the atoms in the unit cell to the volume of the
unit cell:
𝑃𝑎𝑐𝑘𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟 =
By Gizachew Berhanu
𝑁𝑉𝑆
𝑉𝐶
=
4
3
8𝑟 3
( )
√3
8 𝑥 𝜋𝑟3
= 0.34
4.13 crystal structure analysis
Our present knowledge of crystal structures has been obtained mainly by x-ray diffraction techniques that use xrays about the same wavelength as the distance between crystal lattice planes. However, before discussing the
manner in which x-rays are diffracted in crystals, let us consider how x-rays are produced for experimental
purposes.
Diffraction patterns are produced when light passes through a recurrent structure such as a diffraction grating. From
the wavelength of the incident radiation and the angle of diffraction, it is possible to calculate the repeat distance of
the recurrent structure.
We discussed at the begging as crystal is a 3D periodic arrangement of atoms structure and so attempts were made
to measure the interplanar distance with the help of radiation of known wavelength. For obtaining a diffraction
pattern the wavelength of the probing radiation should be less than or equal to distance between atoms. Von Laue
(1912) suggested that x-rays could produce diffraction effects by using a crystal as a 3D grating. A diffraction
pattern was obtained when Friedrich and Knipping passed a beam of polychromatic x-rays (consisting of x-rays of
different wavelengths) through a crystal of 𝑍𝑛𝑆 and allowed the emergent beam to fall on a photographic plate. The
diffraction pattern called Laue Pattern is rather difficult to interpret.
W.H.Bragg (1913) argued that the diffraction effects observed for x-ray could be interpreted in terms of
interference between x-ray beams reflected from the various lattice planes in the crystal. In other words the crystal
can be looked upon as a reflection grating. The Bragg was also suggested the use of homogeneous x-rays, i.e. of
uniform wavelength.
4.13.1 Bragg’s Law
Consider a parallel beam of monochromatic x-rays of wavelength 𝜆 incident on the face of the crystal, which can be
considered to be a series of parallel lattice planes (PP’, QQ’, RR’ etc.) at repeat or interplanar distance of 𝑑
(Fig.4.57). The glancing angle is 𝜃. A part of the beam gets reflected at point 𝐵 along 𝐵𝐶, while another part of the
beam penetrates the crystal and gets reflected from another atom-bearing lattice plane along 𝐸𝐻. If the path
difference between rays is a whole number of whole number of wavelengths, the two reflected beams interfere
constructively giving rise to a diffraction maximum.
Fig.4.57 Derivation of Bragg’s Law
By Gizachew Berhanu
𝑃𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝐹𝐸 + 𝐸𝐺
𝐴𝑛𝑔𝑙𝑒 𝐹𝐵𝐸 = 𝐴𝑛𝑔𝑙𝑒 𝐸𝐵𝐺 = 𝜃
𝐹𝐸 = 𝐸𝐺 = 𝑑 sin 𝜃
∴ 𝑃𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝐹𝐸 + 𝐹𝐺 = 𝑑 sin 𝜃 + 𝑑 sin 𝜃 = 2𝑑 sin 𝜃
The condition for constructive interference is then
𝑛𝜆 = 2𝑑 sin 𝜃 …………………….. (4.18)
In equation (4.18) known as Bragg’s Law, 𝑛 is called order of reflection. For homogeneous x-rays, 𝜆 is fixed and
for a given family of lattice planes, i.e. of fixed ℎ𝑘𝑙 values, 𝑑 is also fixed. Hence the condition for obtaining a
diffraction maximum depends on 𝜃. As 𝜃 is increased gradually, diffraction maximum occur only for 𝑛 =
1, 2, 3 𝑒𝑡𝑐.
For other non-integral values of 𝑛 the path difference is such that there is destructive interference. The reflection
maximum corresponding to integral values of 𝑛 are separated by regions of low intensity. The successive reflection
maxima correspond to the first, second, etc. orders of reflection. As 𝑛 increases the intensity decreases. Higherorder reflections are weak for identical atoms as lattice constituents. It is usual to assume that all reflections are first
order (𝑛 = 1). From Fig.4.54 it is seen that 𝑑200 =
𝑑100⁄
2, where the 𝑑200 and 𝑑100 are the distances between the
(200) and (100) planes respectively. If we know the wavelength of the x-rays, we can determine the interplanar
spacings and, eventually, the identity of the planes that cause the diffraction. To identify the crystal structure of a
cubic material, we note the pattern of the diffracted lines—typically by creating a table of 𝑠𝑖𝑛2 𝜃 values. By
combining Equation 4-12 with Equation 4-18 and squaring both sides for the interplanar spacing, we find that:
𝑑ℎ𝑘𝑙 =
𝑎
√ℎ 2 + 𝑘 2 + 𝑙 2
…………. (4.12)
𝑛𝜆 = 2𝑑ℎ𝑘𝑙 sin 𝜃 ……………. (4.18)
𝑠𝑖𝑛2 𝜃 =
𝜆2
4𝑎 2
(ℎ2 + 𝑘 2 + 𝑙2 ) …………………….. (4.19)
Fig.4.58 Schematic diagramof an x-ray diffractometer;
(T) x-ray source, (S) specimen, (C) detector, and (O)
the axis around which the specimen and detector
rotate.
By Gizachew Berhanu
Fig.4.59 Record of the diffraction angles for a tungsten sample obtained by the use of a diffractometer with copper
radiation.(After A. G. Guy and J. J. Hren, “Elements of Physical Metallurgy,” 3d ed., Addison-Wesley, 1974, p.
208.)
In simple cubic metals, all possible planes will diffract, giving an ℎ2 + 𝑘 2 + 𝑙2 pattern of 1, 2, 3, 4, 5, 6, 8, . . . . In
body-centered cubic metals, diffraction occurs only from planes having an even ℎ2 + 𝑘 2 + 𝑙2 sum of 2, 4, 6, 8, 10,
12, 14, 16, . . . . For face-centered cubic metals, more destructive interference occurs, and planes having ℎ2 + 𝑘 2 +
𝑙2 sums of 3, 4, 8, 11, 12, 16, . . . will diffract. By calculating the values of 𝑠𝑖𝑛2 𝜃 and then finding the appropriate
pattern, the crystal structure can be determined for metals having one of these simple structures, as illustrated in
Example 4.21.
Table 4.9 X-Ray Diffraction Reflection Rules and Reflection Indices for Body-Centered Cubic, FaceCentered Cubic, and Simple Cubic Crystal Structures
Crystal structure
Reflections present
Simple cubic
all
Reflection indices for first six
pattern
100, 110, 111, 200, 210, 211
BCC
(ℎ + 𝑘 + 𝑙) even
110,200, 211, 220.310, 222
FCC
ℎ, 𝑘 𝑎𝑛𝑑 𝑙 either all add or all even 111, 200, 220, 311, 222, 400
By Gizachew Berhanu
Table 4.10 Miller Indices of the Diffracting Planes for BCC and FCC Lattices
Cubic planes
𝒉 𝟐 + 𝒌𝟐 + 𝒍 𝟐
Sum
∑(𝒉𝟐 + 𝒌𝟐 + 𝒍𝟐 )
{100}
{110}
{111}
{200}
{210}
{211}
…
{220}
{221}
{310}
12 +
12 +
12 +
22 +
22 +
22 +
1
2
3
4
5
6
7
8
9
10
02 +
12 +
12 +
02 +
12 +
12 +
02
02
12
02
02
12
22 + 22 + 02
22 + 22 + 12
32 + 12 + 02
Cubic diffracting
Planes {𝒉𝒌𝒍}
BCC
FCC
110
200
…
111
200
211
220
200
310
…
Interpreting Experimental X-Ray Diffraction Data for Metals with Cubic Crystal Structures
We can use x-ray diffractometer data to determine crystal structures. As it is mentioned earlier, for homogeneous xrays, 𝜆 is fixed and for a given family of lattice planes, i.e. of fixed ℎ𝑘𝑙 values, 𝑑 is also fixed. Hence the condition
for obtaining a diffraction maximum depends on 𝜃. A simple case to illustrate how this analysis can be used is to
distinguish between the BCC and FCC crystal structures of a cubic metal. Let us assume that we have a metal with
either a BCC or an FCC crystal structure and that we can identify the principal diffracting planes and their
corresponding 2θ values. From equation (4.19) we have
𝑠𝑖𝑛2 𝜃 =
𝜆2
4𝑎 2
(ℎ2 + 𝑘 2 + 𝑙2 )
From x-ray diffraction data we can obtain experimental values of 2θ for a series of principal diffracting {hkl}
planes. Since the wavelength of the incoming radiation and the lattice constant a are both constants, we can
eliminate these quantities by forming the ratio of two sin2 θ values as
𝑠𝑖𝑛2 𝜃1
𝑠𝑖𝑛2 𝜃2
=
ℎ12+ 𝑘12+ 𝑙12
ℎ22+ 𝑘22+ 𝑙22
…………………….. (4.20)
Where 𝜃1 and 𝜃2 are two diffracting angles associated with the principal diffracting planes {ℎ 1 𝑘1 𝑙1 } and {ℎ 2 𝑘2 𝑙2 },
respectively.
Using Eq. 4.20 and the Miller indices of the first two sets of principal diffracting planes listed in Table 3.7 for BCC
and FCC crystal structures, we can determine values for the sin2 θ ratios for both BCC and FCC structures. For the
BCC crystal structure the first two sets of principal diffracting planes are the {110} and {200} planes (Table 4.10).
Substitution of the Miller {hkl} indices of these planes into Eq. 4.20 gives
𝑠𝑖𝑛2 𝜃1
𝑠𝑖𝑛2 𝜃2
=
ℎ12+ 𝑘12+ 𝑙12
ℎ22+ 𝑘22+ 𝑙22
=
12 + 12 + 02
22 + 02 + 02
By Gizachew Berhanu
=
2
4
= 0.5
Thus, if the crystal structure of the unknown cubic metal is BCC, the ratio of the 𝑠𝑖𝑛2 𝜃 values that correspond to
the first two principal diffracting planes will be 0.5.
For the FCC crystal structure the first two sets of principal diffracting planes are the {111} and {200} planes (Table
4.10). Substitution of the Miller {hkl} indices of these planes into Eq. 4.20 gives
𝑠𝑖𝑛2 𝜃1
𝑠𝑖𝑛2 𝜃2
ℎ12+ 𝑘12+ 𝑙12
=
ℎ22+ 𝑘22+ 𝑙22
=
12 + 12 + 12
22 + 02 + 02
=
3
4
= 0.75
Thus, if the crystal structure of the unknown cubic metal is FCC, the ratio of the sin2θ values that correspond to the
first two principal diffracting planes will be 0.75.
Example Problem 4.23 uses Eq. 4.20 and experimental x-ray diffraction data for the 2θ values for the principal
diffracting planes to determine whether an unknown cubic metal is BCC or FCC. X-ray diffraction analysis is
usually much more complicated than Example Problem 4.23, but the principles used are the same. Both
experimental and theoretical x-ray diffraction analysis has been and continues to be used for the determination of
the crystal structure of materials.
Example 4.21
Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC
Chromium when monochromatic radiation of wavelength 0.0711 nm is used. Given that atomic radius of 𝐶𝑟 is
0.1249nm.
Solution
We already discussed that there are 2atoms per unit cell in BCC crystal structure and lattice constant is related to
atomic radius by 𝑎 =
4𝑅
√3
Diffraction of plane (310) and x-ray wavelength at 𝑛 = 1 (𝜆𝑛=1 ) 0.0711nm
From equation (4.12) we have
𝑑ℎ𝑘𝑙 =
𝑑310 =
𝑎
√ℎ 2+ 𝑘 2+ 𝑙 2
4𝑅
=
4 𝑥 (0.1249𝑛𝑚)
√3 𝑥 √32+ 12 + 02
√3 .√ℎ 2+ 𝑘 2+ 𝑙 2
= 0.0912𝑛𝑚
𝑛𝜆 = 2𝑑 sin 𝜃
sin 𝜃 =
𝑛𝜆
2𝑑
=
1 𝑥 0.0711𝑛𝑚
2 𝑥 0.0912𝑛𝑚
= 0.39
𝜃 = sin−1(0.39) = 22.9°
An expected diffraction angle is then
2𝜃 = 2 𝑥 22.9° = 45.8°
By Gizachew Berhanu
Example 4.22
For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of 44.53°
for FCC nickel when monochromatic radiation having a wavelength of 0.1542nm is used. Given that atomic
radius of 𝑁𝑖 is 0.1246nm.
Solution
This question is about determining the plane indices from which a first-order diffraction is obtained with a
glancing angle of 44.53°. The given crystal structure is FCC in which 4 atoms per unit cell are available and its
lattice constant is equal to
𝑎 = 2√2𝑅 where 𝑅 is atomic radius.
X-ray 𝜆𝑛=1 = 0.1542𝑛𝑚 and 2𝜃 = 44.53° → 𝜃 =
44.53°
2
= 22.265°
𝑛𝜆 = 2𝑑ℎ𝑘𝑙 sin 𝜃
𝑑ℎ𝑘𝑙 =
𝑛𝜆
2 sin 𝜃
=
1 𝑥 0.1542𝑛𝑚
2 sin(22.265°)
= 0.2035𝑛𝑚
And from equation (4.12) we also have
𝑑ℎ𝑘𝑙 =
𝑎
√ℎ 2+ 𝑘 2+ 𝑙 2
𝑑ℎ𝑘𝑙 = 0.2035𝑛𝑚 =
√ℎ 2 + 𝑘 2 + 𝑙 2 =
𝑎
√ℎ 2+ 𝑘 2+ 𝑙 2
2√2𝑅
0.2035𝑛𝑚
=
=
2√2𝑅
√ℎ 2+ 𝑘 2+ 𝑙 2
2√2 (0.1246𝑛𝑚)
0.2035𝑛𝑚
= 1.732
ℎ 2 + 𝑘 2 + 𝑙 2 = (1.732)2 = 3
∴ 𝑇ℎ𝑒 𝑜𝑛𝑙𝑦 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 (111)
Example 4.23
An x-ray diffractometer recorder chart for an element that has either the BCC or the FCC crystal structure shows
diffraction peaks at the following 2θ angles: 40, 58, 73, 86.8, 100.4, and 114.7. The wavelength of the incoming
x-ray used was 0.154 nm. Determine:
(a) The crystal structure of the metal;
(b) The indices of the planes that produce each of the peaks; and
(c) The lattice parameter of the metal.
By Gizachew Berhanu
Solution
a) Determination of the crystal structure of the element. First, the sin2θ values are calculated from the 2θ
diffraction angles.
2𝜃 (𝑑𝑒𝑔. )
40
58
73
86.8
100.4
114.7
Peak
1
2
3
4
5
6
𝜃 (𝑑𝑒𝑔. )
20
29
36.5
43.4
50.2
57.35
𝑠𝑖𝑛𝜃
0.3420
0.4848
0.5948
0.6871
0.7683
0.8420
𝑠𝑖𝑛2 𝜃
0.1170
0.2350
0.3538
0.4721
0.5803
0.7090
Next the ratio of the sin2θ values of the first and second angles is calculated:
𝑠𝑖𝑛2 𝜃1
𝑠𝑖𝑛2 𝜃2
=
0.1170
0.2350
= 0.498 ≈ 0.5
The crystal structure is BCC since this ratio is ≈ 0.5. If the ratio had been ≈ 0.75, the structure would have
been FCC.
(b) The indices of the planes that produce each of the peaks
We can first determine the 𝑠𝑖𝑛2 𝜃 value for each peak, then divide through by the lowest denominator,
0.02185 and reconstruct the above table
Peak
2𝜃 (𝑑𝑒𝑔. )
𝜃 (𝑑𝑒𝑔. )
𝑠𝑖𝑛𝜃
𝑠𝑖𝑛2 𝜃
1
2
3
4
5
6
40
58
73
86.8
100.4
114.7
20
29
36.5
43.4
50.2
57.35
0.3420
0.4848
0.5948
0.6871
0.7683
0.8420
0.1170
0.2350
0.3538
0.4721
0.5803
0.7090
𝑠𝑖𝑛2 𝜃 ⁄
0.1170
1
2
3
4
5
6
ℎ2 + 𝑘2 + 𝑙2
2
4
6
8
10
12
(ℎ𝑘𝑙 )
(110)
(200)
(211)
(220)
(321)
(222)
(c) The lattice parameter of the metal.
𝑑ℎ𝑘𝑙 =
𝑎
√ℎ 2+ 𝑘 2+ 𝑙 2
Substituting into h = 1, k = 1, and l = 0 for the h, k, l Miller indices of the first set of principal diffracting
planes for the BCC crystal structure, which are the {110} planes, the corresponding value for sin2θ , which is
0.117, and 0.154 nm for λ, the incoming radiation, gives
𝑑110 =
𝑎=
𝑎
√2
√2𝜆
2 sin 𝜃
=
=
𝜆
2 sin 𝜃
(√2)𝑥 0.154𝑛𝑚
2 𝑥 0.3420
= 0.318𝑛𝑚
This value is much close to the lattice constant of tungsten which is 0.316nm.
By Gizachew Berhanu
Summary
Solids have definite mass, volume and shape. This is due to the fixed position of their constituent particles,
short distances and strong interactions between them. In amorphous solids, the arrangement of constituent
particles has only short range order and consequently they behave like super cooled liquids, do not have sharp
melting points and are isotropic in nature. In crystalline solids there is long range order in the arrangement of
their constituent particles. They have sharp melting points, are anisotropic in nature and their particles have
characteristic shapes. Properties of crystalline solids depend upon the nature of interactions between their
constituent particles. On this basis, they can be divided into four categories, namely: molecular, ionic, metallic
and covalent solids. They differ widely in their properties. The constituent particles in crystalline solids are
arranged in a regular pattern which extends throughout the crystal. This arrangement is often depicted in the
form of a three dimensional array of points which is called crystal lattice. Each lattice point gives the location
of one particle in space. In all, fourteen different types of lattices are possible which are called Bravais lattices.
Each lattice can be generated by repeating its small characteristic portion called unit cell. A unit cell is
characterized by its edge lengths and three angles between these edges. Unit cells can be either primitive
which have particles only at their corner positions or centred. The centred unit cells have additional particles
at their body centre (body-centred), at the centre of each face (face-centred) or at the centre of two opposite
faces (base-centred). There are seven types of primitive unit cells. Taking centred unit cells also into account,
there are fourteen types of unit cells in all, which result in fourteen Bravais lattices. Close-packing of particles
result in two highly efficient lattices, hexagonal close-packed (hcp) and cubic close-packed (ccp). The latter is
also called face-centred cubic (fcc) lattice. In both of these packings 74% space is filled. The remaining space
is present in the form of two types of voids-octahedral voids and tetrahedral voids. Other types of packing are
not close-packings and have less efficient packing of particles. While in body-centred cubic lattice (bcc) 68%
space is filled, in simple cubic lattice only 52.4 % space is filled.
Using the hard-sphere model for atoms, calculations can be made for the volume, planar, and linear density of
atoms in unit cells. Planes in which atoms are packed as tightly as possible are called close-packed planes, and
directions in which atoms are in closest contact are called close-packed directions. Atomic packing factors for
different crystal structures can also be determined by assuming the hard-sphere atomic model.
Some metals have different crystal structures at different ranges of temperature and pressure, a phenomenon
called polymorphism.
Crystal structures of crystalline solids can be determined by using x-ray diffraction analysis techniques. Xrays are diffracted in crystals when the Bragg’s law (nλ = 2d sin θ) conditions are satisfied. By using the x-ray
diffractometer and the powder method, the crystal structure of many crystalline solids can be determined.
By Gizachew Berhanu
References
1) Materials science and Engineering an introduction; William. D. Callister, Jr. David.G. Rethwisch 8th ed. (2012).
2) The science and Engineering of Materials; Donald R. Askeland, Pradeep P.Fulay, Wendelin J.Wright. 6th ed
(2010).
3) Mechanical behavior of materials, MareMeyers and Krishan Chawla 2nd ed. (2009).
4) Symmetry by Hermann Weyl Princeton University press, 1952 Princeton, new Jerse
5) An introduction to materials Engineering and science; Brain S. Mitchell,3rd,(2004)
6) Physical Metallurgy and Advanced Materials, R.E Smallman and A.H.W. Ngan, 7th (2007)
7) DeGarmo’s, Materials and Processes in Manufacturing, J T. Black and Ronald A. Kohser, 10th (2008)
8) Materials Engineering, Science, Processing and Design Michael Ashby, Hugh Shercliff and David Cebon (2007)
9) Principles of Electronic Materials and Devices, S.O Kasap,3rd (2006)
10) L.V.Azaroff, Introduction to Solids, TMH edition, Tata MCGraw-Hill, New Delhi.
11. C.Kittel, Introduction to Solid State Physics, Third edition, John Wiley, 1966.
12. A.F.Wells, Structural Inorganic Chemistry, Oxford University Press, 1962.
13. Anthony R. West Solid State Chemistry John Wiley & Sons, New York, 1989.
Web References
http://www.youtube.com/watch?v=qh29mj6uXoM&feature=relmfu
http://www.youtube.com/watch?v=8zWySdeXB0M&feature=relmfu
http://www.youtube.com/watch?v=Rm-i1c7zr6Q&feature=related
http://www.youtube.com/watch?v=PWQ89UoxOK8&feature=related
http://www.youtube.com/watch?v=mQy2CdKYqX0&feature=related
By Gizachew Berhanu
Problems
A. Choose the best answer
B. Fill in the Blanks:
1. The structure of sodium chloride crystal is:
1. In NaCl ionic crystal each 𝑁𝑎+ ion is surrounded
(a) Body centred cubic lattice
by -------- 𝐶𝑙 − ions and each 𝐶𝑙 − ion is surrounded
(b) Face centred cubic lattice
by -------- Na+ ions.
(c) Octahedral
2. The coordination number of Cs+ in CsCl crystal
(d) Square planar
is -----------
2. The number of atoms in a face centred cubic unit
3. ---------- solids do not possess sharp melting
cell is:
points and can be considered as ------ liquids.
(a) 4 (b) 3 (c) 2 (d) 1
4. A body centred unit cell has an atom at the each
3. The 8:8 type of packing is present in:
vertex and at --------- of the unit cell.
(a) CsCl (b) KCl (c) NaCl (d) MgF2
5. The three types of cubic unit cells are --------, ----
4. In a simple cubic cell, each point on a corner is
-------- and -------
shared by
6. A crystal may have a number of planes or axes of
(a) 2 unit cells (b) 1 unit cells (c) 8 unit cells (d) 4
symmetry but it possesses only one ----- of
unit cells
symmetry.
5. An amorphous solid is:
7. Amorphous solids that exhibit same physical
(a) NaCl (b) CaF2 (c) glass (d) CsCl
properties in all the directions are called ---------.
6. Each unit cell of NaCl consists of 4 chlorine ions
8. Crystalline solids that exhibit different physical
and:
properties in all directions are called ---------.
(a) 13 Na atoms (b) 4 Na atoms
9. The number of atoms in a single unit cell of cubic
(c) 6 Na atoms (d) 8 Na atoms
close packed sphere is ---------
7. In a body centred cubic cell, an atom at the body
10. In a bcc, an atom of the body centre is shared by
of centre is shared by:
----------- unit cell.
(a) 1 unit cell (b) 2 unit cells (c) 3 unit cells (d) 4
11. The Weiss indices of a plane are 1/2, 1/2, 1/2.
unit cells
Its miller indices will be -----and the plane is
8. In the sodium chloride structure, formula per unit
designated as ---------.
cell is equal to
12. A plane is parallel to x & z axes and makes unit
(a) 2 (b) 8 (c) 3 (d) 4
intercepts along y-axis. Its Weiss indices are --------
9. In a face centred cubic cell, an atom at the face
. Its Miller indices are --------. The plane is
centre is shared by:
designated as -------.
(a) 4 unit cell (b) 2 unit cells (c) 1 unit cells (d) 6
unit cells
By Gizachew Berhanu
C. Write in one or two sentence:
8. How many lattice points are there in one unit cell of
1. What governs the packing of particles in
each of the following lattice?
crystals?
(i) Face-centred cubic
2. What is meant by ‘unit cell’ in crystallography?
(ii) Face-centred tetragonal
3. How many types of cubic unit cell exits?
(iii) Body-centred
4. What are Miller Indices?
9. What makes a glass different from a solid such as
5. Mention the number of sodium and chloride ions
quartz? Under what conditions could quartz be
in each unit cell of NaCl
converted into glass?
6. Mention the number of cesium and chloride ions
10. What is meant by the term 'coordination number'?
in each unit cell of CsCl
(i) What is the coordination number of atoms:
D. Explain briefly on the following:
(a) In a cubic close-packed structure?
1. Define and explain the following terms
(b) In a body-centred cubic structure?
a) Crystalline solids b) Amorphous solids c) Unit
11. How can you determine the atomic mass of an
cell
unknown metal if you know its density and the
2. Give the distinguishing features of crystalline
dimension of its unit cell? Explain.
solids and amorphous solids.
Problems
3. Explain the terms Isotropy and Anisotropy.
1. How many atoms are there per unit cell in (i) simple
4. What is the difference between body centred
cubic arrangement of atoms, (ii) body centred cubic
cubic and face centred cubic?
arrangement of atoms, and (iii) face-centred cubic
5. Five-fold rotation with and without mirror lines
arrangement of atoms?
are certainly possible and are widely represented in
2. Molybdenum at 20℃ is BCC and has an atomic
nature. However, the 5-fold rotation is idle in
radius of 0.140 nm. Calculate a value for its lattice
regarding to symmetry. Explain the reason behind.
constant a in nanometers.
5. Draw a neat diagram for sodium chloride
3. Niobium at 20℃ is BCC and has an atomic radius
structure and describe it accordingly.
of 0.143 nm. Calculate a value for its lattice constant a
6. Draw a neat diagram for Cesium chloride
in nanometers.
structure and describe it accordingly.
4. Lithium at 20℃ is BCC and has a lattice constant of
7. How will you distinguish between the following
0.35092 nm. Calculate a value for the atomic radius of
pairs of terms:
a lithium atom in nanometers.
(i) Hexagonal close-packing and cubic close-
5. Sodium at 20℃ is BCC and has a lattice constant of
packing?
0.42906 nm. Calculate a value for the atomic radius of
(ii) Crystal lattice and unit cell?
a sodium atom in nanometers.
(iii) Tetrahedral void and octahedral void?
6. How many atoms per unit cell are there in the
FCC crystal structure?
By Gizachew Berhanu
7. What is the coordination number for the atoms in
Calculate a value for its lattice constant a in
the FCC crystal structure?
nanometers.
8. Gold is FCC and has a lattice constant of 0.40788
20. Osmium at 20℃ is HCP. Using a value of 0.135
nm. Calculate a value for the atomic radius of a
nm for the atomic radius of osmium atoms, calculate a
gold atom in nanometers.
value for its unit-cell volume. Assume a packing factor
9. Platinum is FCC and has a lattice constant of
of 0.74.
0.39239 nm. Calculate a value for the atomic radius
21. Draw the following directions in a BCC unit cell
of a platinum atom in nanometers.
and list the position coordinates of the atoms whose
10. Palladium is FCC and has an atomic radius of
centers are intersected by the direction vector:
0.137 nm. Calculate a value for its lattice constant a
(a) [100] (b) [110] (c) [111]
in nanometers.
22. Draw direction vectors in unit cubes for the
11. Strontium is FCC and has an atomic radius of
following cubic directions:
0.215 nm. Calculate a value for its lattice constant a
̅1
̅1
̅ ] (b) [11
̅ 0] (c) [1
̅ 21
̅ ] (d) [1
̅1
̅ 3]
(a) [1
in nanometers.
23. Draw direction vectors in unit cubes for the
12. Calculate the atomic packing factor for the FCC
following cubic directions:
structure.
̅2
̅ ] (c) [3
̅ 31] (e) [21
̅ 2] (g) [1
̅ 01] (i) [321] (k)
(a) [11
13. How many atoms per unit cell are there in the
[122]
HCP crystal structure?
(b) [123] (d) [0.21] (f) [233] (h) [121] (j) [103] (l)
14. What is the coordination number for the atoms
[223]
in the HCP crystal structure?
24. What is the notation used to indicate a family or
15. What is the ideal c/a ratio for HCP metals?
form of cubic crystallographic planes?
16. Of the following HCP metals, which have
25. What are the {100} family of planes of the cubic
higher or lower c/a ratios than the ideal ratio: Zr, Ti,
system?
Zn, Mg, Co, Cd, and Be?
26. Draw the following crystallographic planes in a
17. Copper crystallizes into a fcc lattice with edge
BCC unit cell and list the position of the atoms whose
–8
length 3.61 × 10 cm. Show that the calculated
centers are intersected by each of the planes:
density is in agreement with its measured value of
(a) (100) (b) (110) (c) (111)
–3
8.92 g cm
27. Draw the following crystallographic planes in an
18. Calculate the volume in cubic nanometers of the
FCC unit cell and list the position coordinates of the
titanium crystal structure unit cell. Titanium is HCP
atoms whose centers are intersected by each of the
at 20℃ with a = 0.29504 nm and c = 0.46833 nm.
planes:
19. Rhenium at 20℃ is HCP. The height c of its
(a) (100) (b) (110) (c) (111)
unit cell is 0.44583 nm and its c/a ratio is 1.633.
By Gizachew Berhanu
28. A cubic plane has the following axial intercepts:
39. Draw a schematic diagram of an x-ray tube used
a = 1/3, b = −2/3, c = 1/2. What are the Miller
for x-ray diffraction, and indicate on it the path of the
indices of this plane?
electrons and x-rays.
29. A cubic plane has the following axial intercepts:
40. What is the characteristic x-ray radiation? What is
a = −1/2, b = −1/2, c = 2/3. What are the Miller
its origin?
indices of this plane?
41. Distinguish between destructive interference and
30 A cubic plane has the following axial intercepts:
constructive interference of reflected x-ray beams
a = 1, b = 2/3, c = −1/2 . What are the Miller indices
through crystals.
of this plane?
42. Derive Bragg’s law by using the simple case of
31. What is polymorphism with respect to metals?
incident x-ray beams being diffracted by parallel
32. How do the spacings of the three planes (100),
planes in a crystal.
(101) and (111) of simple cubic lattice vary?
43. A sample of BCC metal was placed in an x-ray
33. How do the spacings of the three planes (001),
diffractometer using x-rays with a wavelength of λ =
(011) and (111) of bcc lattice vary?
0.1541 nm. Diffraction from the {221} planes was
34. How do the spacings of the three planes (010),
obtained at2θ = 88.838◦. Calculate a value for the
(110) and (111) of fcc lattice vary?
lattice constant a for this BCC elemental metal.
35. The d422 interplanar spacing in an FCC metal is
(Assume first-order diffraction, n = 1.)
0.083397 nm. (a) What is its lattice constant a? (b)
44. X-rays of an unknown wavelength are diffracted by
What is the atomic radius of the metal? (c) What
a gold sample. The 2θ angle was 64.582° for the {220}
could this metal be?
planes. What is the wavelength of the x-rays used?
36. Titanium goes through a polymorphic change
(The lattice constant of gold = 0.40788 nm; assume
from BCC to HCP crystal structure upon cooling
first-order diffraction, n = 1.)
through 882℃. Calculate the percentage change in
45.
volume when the crystal structure changes from
element that has either the BCC or the FCC crystal
BCC to HCP. The lattice constant a of the BCC unit
structure showed diffraction peaks at the following 2θ
cell at 882℃ is 0.332 nm, and the HCP unit cell has
angles: 41.069°, 47.782°, 69.879°, and 84.396°. (The
a = 0.2950 nm and c = 0.4683 nm.
wavelength of the incoming radiation was 0.15405
37. Pure iron goes through a polymorphic change
nm.)
from BCC to FCC upon heating through 912℃.
(a) Determine the crystal structure of the element.
Calculate the volume change associated with the
(b) The indices of the planes that produce each
change in crystal structure from BCC to FCC if at
of the peaks;
912℃ the BCC unit cell has a lattice constant a =
(c) Determine the lattice constant of the element.
0.293 nm and the FCC unit cell a = 0.363 nm.
(d) Identify the element.
38. What are x-rays, and how are they produced?
By Gizachew Berhanu
An x-ray diffractometer recorder chart for an
46. An x-ray diffractometer recorder chart for an
(a) Determine the crystal structure of the element.
element that has either the BCC or the FCC crystal
(b) The indices of the planes that produce each of the
structure showed diffraction peaks at the following
peaks;
2θ angles: 38.60°, 55.71°, 69.70°, 82.55°, 95.00°,
(c) Determine the lattice constant of the element.
and 107.67°. (Wavelength λ of the incoming
(d) Identify the element.
radiation was 0.15405 nm)
48.
(a) Determine the crystal structure of the element.
element that has either the BCC or the FCC crystal
(b) The indices of the planes that produce each of
structure showed diffraction peaks at the following 2θ
the peaks;
angles: 40.663°, 47.314◦, 69.144°, and 83.448°. (The
(c) Determine the lattice constant of the element.
wavelength λ of the incoming radiation was 0.15405
(d) Identify the element.
nm.)
47. An x-ray diffractometer recorder chart for an
(a) Determine the crystal structure of the element.
element that has either the BCC or the FCC crystal
(b) The indices of the planes that produce each of the
structure showed diffraction peaks at the following
peaks;
2θ angles: 36.191°, 51.974°, 64.982°, and 76.663°.
(c) Determine the lattice constant of the element.
(The wavelength of the incoming radiation was
(d) Identify the element.
0.15405 nm.)
By Gizachew Berhanu
An x-ray diffractometer recorder chart for an
Chapter Five: Imperfection and Movements of atoms in Solids
5.1 Introduction
From our earlier studies, we know that different possible arrangements of particles resulting in several types of
structures. The correlation between the nature of interactions within the constituent particles and several properties
of solids were also being explored. We were overall talking about crystal structure of solids as orderly arranged
and considering it as a perfect form. By bringing all the atoms together to try to form a perfect crystal, we lower the
total potential energy of the atoms as much as possible for that particular structure. But nothing is perfect in this
world. Everybody has some imperfection and so are the crystals. One solid has large number of crystals and defects
in crystals happen during crystallization process. If the crystallization is very fast it will become amorphous itself
and if it is very slow it will be almost perfect crystal which is very rare. So for very good quality of crystals, the
crystallization should occur with very, very slow speed. Even if you do with a very slow speed some defects will be
occurs. This defect takes place with irregularity in arrangement. Slightly deviation from ideal arrangement will be
observed.
In this unit, we will deal how these properties get modified due to the structural imperfections or by the presence of
impurities in minute amounts. We will introduce the five basic types of imperfections: point defects, line defects (or
dislocations), surface defects, Volume (bulk) defects and Vibrations defects.
In many applications, the presence of such defects is useful. There are a few applications, though, where we strive
to minimize a particular type of defect. For example, defects known as dislocations are useful for increasing the
strength of metals and alloys; however, in single crystal silicon, used for manufacturing computer chips, the
presence of dislocations is undesirable. Often the “defects” may be created intentionally to produce a desired set of
electronic, magnetic, optical, or mechanical properties. For example, pure iron is relatively soft, yet, when we add a
small amount of carbon, we create defects in the crystalline arrangement of iron and transform it into plain carbon
steel that exhibits considerably higher strength. Similarly, a crystal of pure alumina is transparent and colorless, but,
when we add a small amount of chromium, it creates a special defect, resulting in a beautiful red ruby crystal. In the
processing of Si crystals for microelectronics, we add very small concentrations of P or B atoms to Si. These
additions create defects in the arrangement of atoms in silicon that impart special electrical properties to different
parts of the silicon crystal.
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By Gizachew Berhanu
Objectives of this chapter
At the end of this chapter, the student should be able to

Understand about the five types imperfections could occur in crystals

Describe the differences between points defect and line defects

Understand the physical meaning of imperfections

Understand the important features of the impurity atoms in crystals

Understand nature of stability, metastability and instability

Explain the differences among five types of imperfections in crystalline materials

State how tension and compression occurs during dislocation

Define edge, screw and mixed dislocations separately

State the principle and why atoms moves from one location to another location in crystalline solids

Write and interpret the equations in atomic diffusion
5.2 Imperfections
What happens when the crystal is grown from a liquid or vapor; do you always get a perfect crystal? What happens
when the temperature is raised? What happens when impurities are added to the solid? Nothing is perfect.
There is no such thing as a perfect crystal. We must therefore understand the type of defects that can exist in a
given crystal structure. Quite often, key mechanical and electrical properties are controlled by these defects. The
crystalline structures that we have looked at all have imperfections. We will quantify these imperfections here.
Crystalline Defects
A crystalline defect is a lattice irregularity having one or more of its dimensions on the order of an atomic
dimension. There are 5 major categories of crystalline defects:
 Zero dimensional: Point defects
 One dimensional: Linear defects (dislocations)
 Two dimensional: Planar (surface) defects
 Three dimensional: Volume (bulk) defects
 Vibrations
5.2.1 Point Defects
Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline
substance. Point defects allow for diffusion to occur in the solid state.
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By Gizachew Berhanu
Point defect
Line defect
Fig.5.1 Point and Line defects
5.2.1.1 Types of Point Defects
Point defects can be classified into three types: (i) stoichiometric defects
(ii) Impurity defects and (iii) non-stoichiometric defects.
(a) Stoichiometric Defects
These are the point defects that do not disturb the stoichiometry of the solid. They are also called intrinsic or
thermodynamic defects. Basically these are of two types, vacancy defects and interstitial defects.
i)
Vacancy Defect: In kinetic molecular theory we discussed that all the atoms in a crystal vibrate about
their equilibrium positions with a distribution of energies, a distribution that closely resembles the
Boltzmann distribution. At some instant, there may be one atom with sufficient energy to break its bonds
and jump to an adjoining site on the surface, as depicted in Figure 5.1. This leaves a vacancy behind, just
below the surface. This vacancy can then diffuse into the bulk of the crystal, because a neighboring atom
can diffuse into it. These are produced by thermal vibrations of the crystal lattice and/or from need to
maintain charge neutrality. This defect can develop when a substance is heated. When some of the lattice
sites are vacant, the crystal is said to have vacancy defect (Fig. 5.2). This results decrease the density of the
substance. From the figure below two lattice sites are dismissed and this minus the number of atoms from
the same volume crystal. This kind of defect is non-ionic.
This latter process of vacancy creation has been shown to be a sequence of events in Figure 5.2. Suppose that Ev
is the average energy required to create such a vacancy. Then only a fraction, of all the atoms in the crystal can
have sufficient energy to create vacancies. If the number of atoms per unit volume in the crystal is N, then the
vacancy concentration 𝑛𝑣 is given by
𝑄
𝑛𝑣 = 𝑁𝑒𝑥𝑝(− 𝑅𝑇)
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By Gizachew Berhanu
𝑛𝑣 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑐𝑎𝑛𝑐𝑖𝑒𝑠 𝑝𝑒𝑟 𝑐𝑚3
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠
𝑄 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
Vacant site
𝑐𝑎𝑙
8.314𝐽
𝑅 = 𝑔𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 1.987 𝑚𝑜𝑙.𝐾 𝑂𝑟 𝑚𝑜𝑙.𝐾
𝑇 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑖𝑛 𝑘𝑒𝑙𝑣𝑖𝑛
Fig. 5.2 Vacancy defect in crystal
(ii) Interstitial Defect: When some constituent particles (atoms or molecules) occupy an interstitial site, the
crystal is said to have interstitial defect (Fig.5.3). If it is occupied by the same atomic species it is a self-interstitial
or self-interstitialcy. It creates large distortions in the crystal lattice and so the concentrations are small. In the
following figure the green balls are anions and the red colored small balls represent cation which the same structure
with sodium chloride discussed in chapter 4. The cation has small size and occupies interstitial position. This form
of imperfection against the vacancy imperfection discussed above. In this case the crystal contains extra constituent
particles within the same volume and the density of the crystal will increase. It is also non-ionic.
Fig.5.3 Interstitial imperfection
Vacancy and interstitial defects as explained above can be shown by non-ionic solids. Ionic solids must always
maintain electrical neutrality. Rather than simple vacancy or interstitial defects, they show these defects as Frenkel
and Schottky defects.
iii) Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its
normal site to an interstitial site (Fig.5.4). It creates a vacancy defect at its original site and an interstitial defect at
its new location. Frenkel defect is also called dislocation defect. It does not change the density of the solid. Frenkel
defect is shown by ionic substance in which there is a large difference in the size of ions, for example, ZnS, AgCl,
AgBr and AgI due to small size of Zn2+and Ag+ ions.
228
By Gizachew Berhanu
Fig.5.4 Frenkel defect
Schottky Defect: In an ionic crystal, such as NaCl, which consists of anions (Cl-) and cations (Na+), one
common type of defect is called a Schottky defect. This involves a missing cation-anion pair (which may have
migrated to the surface), so the neutrality is maintained, as indicated in Figure 5.5. These Schottky defects are
responsible for the major optical and electrical properties of alkali halide crystals. It is basically a vacancy defect
in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal
(Fig.5.5). Like simple vacancy defect, Schottky defect also decreases the density of the substance. Number of
such defects in ionic solids is quite significant. For example, in NaCl there are approximately 10 6Schottky pairs
per cm3 at room temperature. In 1 cm3 there are about 1022 ion. Thus, there is one Schottky defect per 1016 ions.
Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes. For
example, NaCl, KCl, CsCl and AgBr. It may be noted that AgBr shows both, Frenkel as well as Schottky
defects.
Fig.5.5 Schottky defect.
b) Impurity Defect: Vacancies are only one type of point defect in a crystal structure. Point defect generally
involve lattice changes or distortions of a few atomic distances. If molten NaCl containing a little amount of CaCl2
is crystallized, some of the sites of Na+ ions are occupied by Ca2+ (Fig.5.6). Each Ca2+ replaces two Na+ ions. It
229
By Gizachew Berhanu
occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in
number to that of Ca2+ions. Another similar example is the solid solution of CdCl2 and AgCl.
𝐶𝑎𝐶𝑙2 (Little)
𝑁𝑎𝐶𝑙 (Molten)
𝐶𝑎2+ 𝐶𝑙 −
𝑁𝑎+
𝐶𝑙 −
𝐶𝑙 −
𝐶𝑎2+ → 𝑁𝑎+ + 𝑁𝑎+
Fig.5.6 The charge of one calcium ion is equivalent to the
charge of two sodium ion and one calcium ion kicked out
two sodium ion; occupies the position of one sodium and
left vacant space for the second one.
The crystal structure may contain impurities, either naturally or as a consequence of intentional addition, as in the
case of silicon crystals grown for microelectronics. If the impurity atom substitutes directly for the host atom, the
result is called a substitutional impurity and the resulting crystal structure is that of a substitutional solid solution.
When a Si crystal is "doped" with small amounts of arsenic (As) atoms, the As atoms substitute directly for the Si
atoms in the Si crystal; that is, the arsenic atoms are substitutional impurities. In general, the impurity atom will
have both a different valency and a different size. It will therefore distort the lattice around it. For example, if a
substitutional impurity atom is larger than the host atom, the neighboring host atoms will be pushed away. The
crystal region around an impurity is therefore distorted from the perfect periodicity and the lattice is said to be
strained around a point defect. A smaller substitutional impurity atom will pull in the neighboring atoms.
c) Non-Stoichiometric Defects
The defects discussed so far do not disturb the stoichiometry of the crystalline substance. However, a large number
of non-stoichiometric inorganic solids are known which contain the constituent elements in non-stoichiometric ratio
due to defects in their crystal structures. These defects are of two types: (i) metal excess defect and (ii) metal
deficiency defect.
(i) Metal Excess Defect
Metal excess defect due to anionic vacancies: Alkali halides like NaCl and KCl show this type of defect. When
crystals of NaCl are heated in an atmosphere of sodium vapor, the sodium atoms are deposited on the surface of the
crystal. The Cl– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by
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By Gizachew Berhanu
loss of electron by sodium atoms to form Na+ ions. The released electrons diffuse into the crystal and occupy
anionic sites. As a result the crystal now has an excess of sodium. The anionic sites occupied by unpaired electrons
are called F-centres (from the German word Farbenzenter for colour centre). They impart yellow colour to the
crystals of NaCl. The colour results by excitation of these electrons when they absorb energy from the visible light
falling on the crystals. Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCl
crystals violet (or lilac).
Metal excess defect due to the presence of extra cations at interstitial sites: Zinc oxide is white in colour at room
temperature. On heating it loses oxygen and turns yellow.
Now there is excess of zinc in the crystal and its formula becomes Zn1+xO. The excess Zn2+ions move to interstitial
sites and the electronsto neighboring interstitial sites.
(ii) Metal Deficiency Defect
There are many solids which are difficult to prepare in the stoichiometric composition and contain less amount of
the metal as compared to the stoichiometric proportion. A typical example of this type is FeO which is mostly
found with a composition of Fe0.95O.
It may actually range from Fe0.93O to Fe0.96O. In crystals of FeO, some Fe2+cations are missing and the loss of
positive charge is made up by the presence of required number of Fe3+ions.
Example 5.1
What type of defect can arise when a solid is heated? Which physical property is affected by it and in what
way?
Solution
On heating, the constituent particle of the solid will get energy that enable them to exits and leave the lattice. If
the particles leave the lattice, then vacancy will be created in the lattice. A vacancy created is vacancy defect.
Since some of the particles leave the lattice, it decreases the density of the solid. Then an affected physical
property is a density.
Example 5.2
What type of stoichiometric defect is shown by (i) ZnS (ii) (AgBr)
Solution
(i) Frenkel defect
(ii) Both Frenkel and Schottky defect
231
By Gizachew Berhanu
Example 5.3
Explain how vacancies are introduced in an ionic solid when a cation of a higher valence is added as an
impurity in it?
Solution
As discussed in the section, we have first ionic solid let say sodium chloride which contains𝐶𝑙 − 𝑁𝑎+ . If a higher
valence let say 𝑆𝑟 2+ is added, it kick out two 𝑁𝑎+ and occupies the position of one 𝑁𝑎+ and left empty to rest
one.
𝑆𝑟 2+ → 𝑁𝑎+ + 𝑁𝑎+ (See fig.5.6).
Example 5.4
Stability of a crystal is reflected in the magnitude of its melting points. Comment.
Solution
Obviously melting point is a good indication of the intermolecular farce of attraction. This means that higher
melting point implies higher intermolecular force of attraction. This again implies greater stability.
Example 5.5
Analysis shows that nickel Oxide has the formula 𝑁𝑖0.98 𝑂1.00 . What fractions of nickel exist as 𝑁𝑖 2+ and 𝑁𝑎3+
ions?
Solution
Assume there are 100 oxygen atoms. Then there will be 98 nickel atoms. This nickel contains both 𝑁𝑖 2+
and 𝑁𝑖 3+. Let’s represent number of 𝑁𝑖 2+ by 𝑥 and then 𝑁𝑖 3+ = 98 − 𝑥.
Now net charge has to be zero i.e. 𝑁𝑒𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑛𝑖𝑐𝑘𝑒𝑙 = 𝑁𝑒𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑂𝑥𝑦𝑔𝑒𝑛
2𝑥 + 3(98 − 𝑥) − 2(100) = 0
2𝑥 + 294 − 3𝑥 = 200
𝑥 = 94
𝑁𝑖 2+ = 94
𝑁𝑖 3+ = 98 − 𝑥 = 98 − 94 = 4
∴ 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑁𝑖 2+ =
𝑁𝑖 2+
𝑁𝑖 2+ + 𝑁𝑖 3+
𝑥 100% =
94
98
𝑥 100% = 96%
𝑁𝑖 3+ = 100 − 96 = 4%
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By Gizachew Berhanu
Example 5.6
a) Calculate the concentration of vacancies in copper at room temperature (25°C). b) What temperature will be
needed to heat treat copper such that the concentration of vacancies produced will be 1000 times more than the
equilibrium concentration of vacancies at room temperature? Assume that 20,000 cal are required to produce a
mole of vacancies in copper
Solution
Copper has a FCC crystal structure with lattice parameter 0.3615 nm from table 4.5. There are four atoms per
unit cell in FCC crystal structure; therefore, the number of copper atoms per 𝑐𝑚3 is
𝑛=
4𝑎𝑡𝑜𝑚𝑠/𝑐𝑒𝑙𝑙
(3.615 𝑥 10−8 𝑐𝑚)3
= 8.467 𝑥 1022 𝑐𝑜𝑝𝑝𝑒𝑟 𝑎𝑡𝑜𝑚𝑠/𝑐𝑚3
a) 𝑉𝑎𝑐𝑎𝑛𝑐𝑦 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑛𝑣 = 𝑛𝑒𝑥𝑝(−𝑄⁄𝑅𝑇)
At room temperature 𝑇 = 25 + 273 = 298𝐾
22
𝑛𝑣 = 8.467 𝑥 10 𝑒𝑥𝑝 (
20,000𝑐𝑎𝑙
𝑚𝑜𝑙
1.987𝑐𝑎𝑙
𝑥 298𝐾
𝑚𝑜𝑙.𝐾
−
) = 1.814 𝑥 108 𝑣𝑎𝑐𝑎𝑛𝑐𝑖𝑒𝑠/𝑐𝑚3
b) We wish to find a heat treatment temperature that will lead to a concentration of vacancies that is
1000 times higher than that calculated in (a)
If we assume the room temperature as 𝑇1 = 298𝑘 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑐𝑎𝑛𝑐𝑦
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑖𝑠 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑠 𝑎𝑠 𝑛1𝑣 = 1.814 𝑥
108 𝑣𝑎𝑐𝑎𝑛𝑐𝑖𝑒𝑠
𝑐𝑚3
𝑤𝑖𝑡ℎ 𝑠𝑎𝑚𝑒 ℎ𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑦.
The temperature desired to produce vacancy concentration 1000 times than at 𝑡1 is definitely 𝑇2
From the above vacancy concentration with temperature and heat, we have
𝑛1𝑣 = 𝑛𝑒𝑥𝑝(−𝑄/𝑅𝑇1 ) and
𝑛2𝑣 = 𝑛𝑒𝑥𝑝(−
𝑄
)
𝑅𝑇2
= 1000𝑛1𝑣
By taking the ratio of both relations we have
𝑛2𝑣
𝑛1𝑣
=
1000𝑛1𝑣
𝑛1𝑣
𝑄
1
= exp(𝑅 (𝑇 −
1
1
))
𝑇2
After some arrangement and simplification we could have
𝑇2 =
𝑄𝑇1
𝑄−3𝑅𝑇1 ln(10)
=
20,000𝑐𝑎𝑙
𝑥 298𝐾)
𝑚𝑜𝑙
1.987𝑐𝑎𝑙
20,000−3(
)298𝐾 𝑥
𝑚𝑜𝑙.𝐾
(
ln(10)
= 374.6𝐾 𝑂𝑟 101.6℃
By heating the copper slightly above 100°C, waiting until equilibrium is reached, and then rapidly cooling the
copper back to room temperature, the number of vacancies trapped in the structure may be one thousand times
greater than the equilibrium number of vacancies at room temperature. Thus, vacancy concentrations
encountered in materials are often dictated by both thermodynamic and kinetic factors.
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By Gizachew Berhanu
Example 5.7
If 𝑁𝑎𝐶𝑙 is introduced with 10−3mol% of 𝑆𝑟𝐶𝑙2 , what is the concentration of cation vacancies?
Solution
This problem is same with problem on example 5.3. Here 𝑆𝑟𝐶𝑙2 is added to 𝑁𝑎𝐶𝑙 in which 𝑆𝑟 is a higher
valence element than 𝑁𝑎. If a higher valence let 𝑆𝑟 2+ is added, it kicks out two 𝑁𝑎 + and occupies the
position of one 𝑁𝑎+ and left empty to the rest one. Now 10−3 𝑚𝑜𝑙% means: 100 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 →
𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑤𝑖𝑡ℎ 10−3 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑟𝐶𝑙2
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑁𝑎𝐶𝑙 =
10−3
𝑚𝑜𝑙𝑒𝑠
100
𝑜𝑓 𝑆𝑟𝐶𝑙2 = 10−5 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑟𝐶𝑙2
The cation vacancy created by one 𝑆𝑟 2+ is one because of the above reason.
10−5 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑆𝑟𝐶𝑙2 Contains 10−5 𝑥 6.022 𝑥 1023 𝑖𝑜𝑛𝑠 𝑜𝑓 𝑆𝑟 = 6.022 𝑥 1018 𝑆𝑟 2+
∴ 6.022 𝑥 1018 𝑐𝑎𝑡𝑖𝑜𝑛 𝑣𝑎𝑐𝑎𝑛𝑐𝑖𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑐𝑟𝑒𝑎𝑡𝑒𝑑
5.2.2 Linear Defects (Dislocations)
A line defect is formed in a crystal when an atomic plane terminates within the crystal instead of passing all the
way to the end of the crystal, as depicted in Figure 5.7a. The edge of this short plane of atoms is therefore like a line
running inside the crystal. The planes neighboring (i.e., above) this short plane are dislocated (displaced) with
respect to those below the line. We therefore call this type of defect an edge dislocation and use an inverted T
symbol. The vertical line corresponds to the half-plane of atoms in the crystal, as illustrated in Figure 5.7a. It is
clear that the atoms around the dislocation line have been effectively displaced from their perfect-crystal
equilibrium positions, which results in atoms being out of registry above and below the dislocation. The atoms
above the dislocation line are pushed together, whereas those below it are pulled apart, so there are regions of
compression and tension above and below the dislocation line, respectively, as depicted by the shaded region
around the dislocation line in Figure 5.7a. Therefore, around a dislocation line, we have a strain field due to the
stretching or compressing of bonds. The energy required to create a dislocation is typically in the order of 100 eV
per nm of dislocation line. On the other hand, it takes only a few eV to form a point defect, which is a few
nanometers in dimension. In other words, forming a number of point defects is energetically more favorable than
forming a dislocation. Dislocations are not equilibrium defects. They normally arise when the crystal is deformed
by stress, or when the crystal is actually being grown. A line defect is a lattice distortion created about a line formed
by the solidification process, plastic deformation, vacancy condensation or atomic mismatch in solid solutions.
Dislocations explain the observation of plastic deformation at lower stress than would be required in a perfect
lattice. They also explain the phenomenon of work hardening. There are three categories of line defects:
 Edge dislocation
o
b is perpendicular to dislocation line
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By Gizachew Berhanu
o
there will be tensile & compressive strain fields in the crystal lattice in their vicinity
 Screw dislocation
Another type of dislocation is the screw dislocation, which is essentially a shearing of one portion of the crystal
with respect to another, by one atomic distance, as illustrated in Figure 5.7b. The displacement occurs on either side
of the screw dislocation line. The circular arrow around the line symbolizes the screw dislocation. As we move
away from the dislocation line, the atoms in the upper portion become more out of registry with those below; at the
edge of the crystal, this displacement is one atomic distance, as illustrated in Figure 5.7b.
o
b is parallel to dislocation line
o
there will be shear strain fields in the crystal lattice in their vicinity
 Mixed dislocation
Both edge and screw dislocations are generally created by stresses resulting from thermal and mechanical
processing. A line defect is not necessarily either a pure edge or a pure screw dislocation; it can be a mixture, as
depicted in Figure 5.7c. Screw dislocations frequently occur during crystal growth, which involves atomic stacking
on the surface of a crystal. Such dislocations aid crystallization by providing an additional "edge" to which the
incoming atoms can attach, as illustrated in Figure 5.7c. To explain, if an atom arrives at the surface of a perfect
crystal, it can only attach to one atom in the plane below. However, if there is a screw dislocation, the incoming
atom can attach to an edge and thereby form more bonds; hence, it can lower its potential energy more than
anywhere else on the surface. With incoming atoms attaching to the edges, the growth occurs spirally around the
screw dislocation, and the final crystal surface reflects this spiral growth geometry.
o
b is at an oblique angle to dislocation line
o
strain fields will be T, C and S
The Burgers Vector, b, quantifies the magnitude and direction of the structural defect. It is the displacement
vector necessary to close a step-wise loop around the defect. For common metals this is the repeat distance along
the highest atomic density direction. Equivalently, the Burgers vector denotes the direction and magnitude of the
atomic displacement that occurs when a dislocation moves.
The phenomenon of plastic or permanent deformation of a metal depends totally on the presence and motions of
dislocations, as will be discussed in chapter 6 on the mechanical properties of materials. In the case of electrical
properties of metals, we will see in volume 2 that dislocations increase the resistivity of materials, cause significant
leakage current in a pn junction, and give rise to unwanted noise in various semiconductor devices. Fortunately, the
occurrence of dislocations in semiconductor crystals can be controlled and nearly eliminated. In a metal
interconnection line on a chip, there may be an average of 104-105 dislocation lines per mm2 of crystal, whereas a
silicon crystal wafer that is carefully grown may typically have only 1 dislocation line per mm2 of crystal.
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By Gizachew Berhanu
Fig.5.7 dislocation defects
5.2.3 Planar (Interfacial) Defects
External Surfaces
Many materials are polycrystalline; that is, they are composed of many small crystals oriented in different
directions. In fact, the growth of a flawless single crystal from what is called the melt (liquid) requires special
skills, in addition to scientific knowledge. When a liquid is cooled to below its freezing temperature, solidification
does not occur at every point; rather, it occurs at certain sites called nuclei, which are small crystal-like structures
containing perhaps 50 to 100 atoms. The liquid atoms adjacent to a nucleus diffuse into the nucleus, thereby
causing it to grow in size to become a small crystal, or a crystallite, called a grain. Since the nuclei are randomly
oriented when they are formed, the grains have random crystallographic orientations during crystallite growth. As
the liquid between the grains is consumed, some grains meet and obstruct each other. At the end of solidification,
therefore, the whole structure has grains with irregular shapes and orientations.
In contrast to a single crystal, polycrystalline material has grain boundaries where differently oriented crystals meet.
The atoms at the grain boundaries obviously cannot follow their natural bonding habits, because the crystal
orientation suddenly changes across the boundary. Therefore, there are both voids at the grain boundary and
stretched and broken bonds. In addition, in this region, there are misplaced atoms that do not follow the crystalline
pattern on either side of the boundary. Consequently, the grain boundary represents a high-energy region per atom
with respect to the energy per atom within the bulk of the grains themselves. The atoms can diffuse more easily
along a grain boundary because (a) less bonds need to be broken due to the presence of voids and (b) the bonds are
strained and easily broken anyway. In many polycrystalline materials, impurities therefore tend to congregate in the
grain boundary region. We generally refer to the atomic arrangement in the grain boundary region as being
disordered due to the presence of the voids and misplaced atoms.
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Atoms at any surface are not in their perfect crystal positions because they will have different CN than the atoms
inside the volume of the material. Hence these atoms will be at higher energies. During solidification materials try
to minimize this.
Grain Boundaries
Since the energy of an atom at the grain boundary is greater than that of an atom within the grain,
these grain boundaries are nonequilibrium defects; consequently, they try to reduce in size to give the
whole structure a lower potential energy.
At or around room temperature, the atomic diffusion process is slow; thus, the reduction in the grain
boundary is insignificant. At elevated temperatures, however, atomic diffusion allows big grains to
grow, at the expense of small grains, which leads to grain coarsening (grain growth) and hence to a
reduction in the grain boundary area.
Mechanical engineers have learned to control the grain size, and hence the mechanical properties of
metals to suit their needs, through various thermal treatment cycles.
For electrical engineers, the grain boundaries become important when designing electronic devices
based on polysilicon or any polycrystalline semiconductor. For example, in highly polycrystalline
materials, particularly thin-film semiconductors (e.g., polysilicon), the resistivity is invariably
determined by polycrystallinity, or grain size, of the material, as discussed in Chapter 6.
Grains are formed during the solidification process. A grain boundary is the area of mismatch between volumes of
material that have a common orientation of the crystallographic axes. The atoms at grain boundaries are not in their
perfect crystal positions and hence the grain boundary is less dense. These atoms are at higher energies than the
atoms inside the volume of a grain. The thickness in on the order of 2-5 atoms wide.
There are different categories of grain boundaries:
 Low angle tilt boundary – a few isolated edge dislocations
 High angle tilt boundary – more complex

Coincident site lattice (CSL) – there is registry between the two adjacent crystal lattices in the vicinity of the
boundary region.

These boundaries will tend to consist of regions of good correspondence separated by grain boundary
dislocations (GBD)

GBDs are linear defects within the boundary plane. They might be secondary in that they have Burgers
vectors different from those found in the bulk material (primary dislocations)
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By Gizachew Berhanu
At ambient temperatures, grain boundaries give strength to a material. So in general, fine grained materials are
stronger than coarse grained ones because they have more grain boundaries per unit volume. However, at higher
temperatures, grain boundaries act to weaken a material due to corrosion and other factors.
Grain size can be quantified by

ASTM grain size number

Average grain diameter

Grain density
The size and shape of grains are determined by a number of factors during solidification:

Lots of nucleation sites  fine grains

Fewer nucleation sites  coarse grains

Equal growth in all directions  equiaxed grains

Thermal gradients  elongated or columnar grains

Slow growth  coarse grains

Rapid growth  fine grains
Most materials are polycrystalline (or polygranular). There are some applications where the expense and
time to produce single crystal materials (and hence no grain boundaries) is justified:

Turbine blades

Silicon wafer chips
 Twin boundaries – a special type of grain boundary across which there exists a mirror image of the
crystal lattice. It is produced by mechanical shear stresses and/or annealing some materials
 Stacking faults – the interruption of the stacking sequence of close packed planes.
 Phase boundaries-The surface area between the grains in multiphase materials
 Ferromagnetic domain walls – The boundary between regions that have a different orientation of
magnetic dipoles.
5.2.4 Volume (Bulk) Defects
These are introduced during processing and fabrication.
 Pores
 Cracks
 Foreign inclusions
Noncrystalline solids
 The Zachariasen model is the visual
definition of the random network theory
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By Gizachew Berhanu
of glass structure (which is analogous to
the point lattice associated with crystal
structure).
It is “perfectly” random,
although it does have short-range order
(SRO) as the building block of the crystal
is retained.
(e.g. AlO33- triangles)
 Synonyms for noncrystalline: amorphous, vitreous, glassy, short-range order
 Materials processed to avoid crystallization can have desirable unique props.
o
Amorphous Metals (Metallic glasses)
have high strengths and corrosion
resistance due to lack of grain
boundaries (but the process is
expensive). The Bernal Model is
helpful in visualizing this structure.
o
o
Amorphous semiconductors – economical (but complex electronic props).
Medium-range order refers to a tendency for some structural order to occur in the medium ranges of a few
nanometers. This can occur in common glasses containing significant amounts of modifiers such as Na 2O and
CaO. For example CaO-SiO2 glass has CaO6 octahedra share edges in a regular arrangement. This is in
contrast to the random distribution of Na+ ions in the random network model that describes vitreous SiO2
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By Gizachew Berhanu
5.2.5 Atomic Vibrations
At any temperature above absolute zero, the atoms in a solid are vibrating very rapidly about their lattice positions.
These vibrations are considered to be defects in the material. At room temperature a typical vibration frequency is
1013 vibrations/sec and the amplitude is 0.003 nm.
Temperature is a measure of the average vibrational energy of the atoms in a material. In other words, an average
energy, Eave, is associated with the atoms in a material at any given temperature. The higher the temperature, the
higher is this average energy.
However, not all of the atoms have the average energy associated with the given temperature. Some will have an
energy that is lower; others will have an energy that is higher. There will be a statistical distribution of the energies
of all the atoms about the average energy for that temperature. Furthermore, an individual atom’s vibrational
energy will vary randomly with time.
The percentage of atoms with energies above the average for a given temperature can be found with the MaxwellBoltzmann distribution:
P  e - (E* - Eave) / kT
P is the probability of finding an atom with energy E*. (E* is above Eave, the average energy characteristic of a
particular temperature.)
For a given E*, the probability of finding atoms with that energy will increase as the temperature increases. And of
course for a given temperature, the probability of finding an atom with energy E* will be high if E* is close to Eave.
5.3 Movement of atoms in solids
5.3.1 Metastability and stability
The concept of stability is easily understood by considering a mechanical analog. In Fig. 5.11, a rectangular block
of square cross-section is shown in various tilted positions. In position 1, the block is resting on the square base; the
arrow from the centre of mass indicates the line along which the weight acts. In position 2, the block is resting with
the wider plane, along which the weight acts is still within the square base. In position 3, the tilt is increased to such
an extent that the line of force just falls on the periphery of the base. The centre of mass is now at the maximum
possible height from the base. Further tilting lowers it. The line of force now falls outside the square base but
within the rectangular base, position 2. On coming to rest on the rectangular face, position 2, the centre of mass is at
the lowest possible position for all configurations of the block.
The potential energy of the block is measured by the height of the centre of the mass from the base. Position 5
corresponds to the lowest potential energy for all configurations and is correspondingly described as the most stable
state or simply the stable state. A system always tends to go towards the most stable state. Position 3 has the
240
By Gizachew Berhanu
maximum potential energy and is called an unstable state. Positions 2 and 4 are also unstable states but do not have
the maximum energy. Position 1 is called a metastable state.
α
Fig.5.8 rectangular box positioned with different sites to illustrate stability, metastability and unstability
We can use this mechanical analogy to illustrate various equilibrium configurations of a system. Figure 2.2 shows
the potential energy of a system as a function of configuration. The potential energy curve has two valleys and a
peak. At these positions, the curve has zero slope, that is, the energy does not
Stability and Metastability vary as a function of configuration for infinitesimally small perturbations. Such
configurations are called equilibrium configurations. Corresponding to the terminology used for the tilting block,
we have stable equilibrium, unstable equilibrium and metastable equilibrium, see Fig.5.8. Even though the potential
energy is a minimum in the metastable state, it is not the lowest for all configurations of the system. Due to the
valley position, after small perturbations, the original configuration is restored in both stable and metastable
equilibrium. Such restoration does not occur in the case of unstable equilibrium.
A metastable state of existence is very common in materials. For example, most metals at room temperature are
stable only in the form of an oxide. Oxygen is easily available in the surrounding air. Yet a metal may not combine
with oxygen at room temperature (except for a very thin film on the surface). It may exist in the metastable metallic
state for an indefinite period. This period could be centuries, as is borne out by the unchanging state of some
ancient metallic statues and pillars. This fortunate set of circumstances enables us to use metals in many
engineering applications.
In the mechanical analogy we used, the block can be tilted and brought to the most stable configuration starting
from a metastable state by an external supply of energy in the form of a hand push or a jiggling base. In materials,
the most common source of such energy is the thermal energy. As the temperature of a solid is increased from 0 K,
the atoms in the solid vibrate about their mean positions with increasing amplitude. When the temperature is
sufficiently high, rotation of atoms (or small groups of atoms) also becomes possible in some solids. The
translational motion of atoms past one another is more characteristic of liquids and gases. The energy associated
with the vibrations, rotations and translations aids in taking a material from a metastable state to a stable state. In
the above example of the oxidation of metals, at sufficiently high temperatures, thermal energy will aid the
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By Gizachew Berhanu
chemical combination of the metal with the surrounding oxygen, taking it from the metastable to the stable state.
This, of course, poses the problem of protecting metals against oxidation, when used at high temperatures.
5.3.2 Atomic Diffusion and the diffusion Coefficient
Consider the motion of the impurity atom in Figure 5.12. For simplicity, assume a two dimensional crystal in the
plane of the paper, as in Figure 1.30. The impurity atom has four neighboring voids into which it can jump. If 0 is
the angle with respect to the x axis, then these voids are at directions 0 = 0°, 90°, 180°, and 270°; as depicted in
Figure 5.13. Each jump is in a random direction along one of these four angles. As the impurity atom jumps from
void to void, it leaves its original location at O, and after N jumps, after time t, it has been displaced from O to O'.
Let 𝑎 be the closest void-to-void separation. Each jump results in a displacement along x which is equal to a cos 0,
with 0 = 0°, 90°, 180°, or 270°. Thus, each jump results in a displacement along x which can be a, 0, -a, or 0,
corresponding to the four possibilities.
E
EA*
A*
EA
EA = EB
A
B
Fig.5.9 Diffusion of an interstitial impurity atom in a crystal from one void to a neighboring void. The
impurity atom at position A must possess an energy Ea to push the host atoms away and move into the
neighboring void at B.
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By Gizachew Berhanu
O
90
Y
L
x
O
180
O
0
O
270
Fig.5.10 An impurity atom has four site choices for diffusion to a neighboring interstitial vacancy.
After N jumps, the impurity atom would have been displaced from the original position at O.
After N jumps, the mean displacement along x will be close to zero, just as the mean voltage of the ac voltage from
a power outlet is zero, even though it has an rms value of 120 V. We therefore consider the square of the
displacements. The total square displacement, denoted X2, is
𝑋 2 = 𝑎2 𝑐𝑜𝑠 2 𝜃1 + 𝑎2 𝑐𝑜𝑠 2 𝜃2 + 𝑎2 𝑐𝑜𝑠 2 𝜃3 + 𝑎2 𝑐𝑜𝑠 2 𝜃4 + ⋯ + 𝑎2 𝑐𝑜𝑠 2 𝜃𝑁
Clearly, 𝜃 = 90° and 270° give 𝐶𝑂𝑆 2 𝜃 = 0. Of all N jumps, are 𝜃 = 0 and 180°, each of which gives 𝑐𝑜𝑠 2 𝜃 = 1.
Thus,
1
𝑋 2 = 2 𝑎2 𝑁
There will be a similar expression for Y2, which means that after N jumps, the total square distance L2 from O to O'
in Fig.5.10 is
L2 = X2 + Y2 = a2N
The rate of jumping (frequency of jumps) is given by Equation
𝑣 = 𝑣𝑜 𝐴𝑒𝑥𝑝(− 𝐸 ⁄𝐾𝑇)
So the time per jump is 1 /𝑣. Time t for N jumps is 𝑁/𝑣. Thus, 𝑁 = 𝑣𝑡 and
𝐿2 = 𝑎2 𝑣𝑡
If we introduce 𝐷 for 𝑎2 𝑣, then
𝐿2 = 𝐷𝑡 and
𝐿 = √𝐷𝑡
Where, 𝐷 = 𝑎2 𝑣 = 𝑎2 𝑣𝑜 𝐴𝑒𝑥𝑝 (− 𝐸 ⁄𝐾𝑇) = 𝐷𝑜 𝑒𝑥𝑝(− 𝐸 ⁄𝐾𝑇)
Where D0 is a constant. The root square displacement L in time t, is given by L = [Dt]1/2. Since L2 is evaluated from
X2 and y2, L is known as the root mean square (rms) displacement.
The preceding specific example considered the diffusion of an impurity in a void between atoms in a crystal; this is
a simple way to visualize the diffusion process. An impurity, indeed any atom, at a regular atomic site in the crystal
can also diffuse around by various other mechanisms. For example, such an impurity can simultaneously exchange
243
By Gizachew Berhanu
places with a neighbor. But, more significantly, if a neighboring atomic site has a vacancy that has been left by a
missing host atom, then the impurity can simply jump into this vacancy. The activation energy is a measure of the
difficulty of the diffusion process. It may be as simple as the energy (or work) required for an impurity atom to
deform (or strain) the crystal around it as it jumps from one interstitial site to a neighboring interstitial site, as in
Figure 5.9; or it may be more complicated, for example, involving vacancy creation.
Various Si semiconductor devices are fabricated by doping a single Si crystal with impurities (dopants) at high
temperatures. For example, doping the Si crystal with phosphorus (P) gives the crystal a higher electrical
conductivity. The P atoms substitute directly for Si atoms in the crystal. These dopants migrate from high to low
dopant concentration regions in the crystal by diffusion, which occurs efficiently only at sufficiently high
temperatures.
Example 5.8
The diffusion coefficient of phosphorus atoms in the Si crystal with D0 = 10.5 cm2 s-1 and EA = 3.69
eV. What is the diffusion coefficient at a temperature of 1100℃ at which dopants such as P are diffused
into Si to fabricate various devices? What is the rms distance diffused by P atoms in 5 minutes? Solution
From diffusion equation we have
Solution
𝐷 = 𝐷𝑂 exp(− 𝐸𝐴 ⁄𝐾𝑇)
2 −1
𝐷 = 10.5𝑐𝑚 𝑠
exp (−
(3.69𝑒𝑣)𝑥 (1.6 𝑥
(1.38
10−19 𝐽
)
𝑒𝑣
𝑥10−23 𝐽
)𝑥 (1100+273)𝐾
𝐾
) = 3.0 𝑥10−13 𝑐𝑚2 𝑠 −1
The rms distance L diffused in a time t = 5 min = 5 x 60 seconds is
𝐿 = √𝐷𝑡 = [√(3.0 𝑥 10−13 𝑐𝑚2 𝑠 −1 ) 𝑥 60 𝑠)] = 6.5𝑥10−6 𝑐𝑚
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By Gizachew Berhanu
Chapter Six: Mechanical Behavior of Materials
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6.1 Introduction
In chapter one, we tried to group materials properties into six main classes. The properties of metals are the
characteristics by which it can be accurately identified from polymers and ceramics or by which its range of
usefulness can be determined.
Among the properties of materials discussed there, mechanical properties of solids will be smoothly presented in
this chapter. Mechanical properties of solids describe characteristics such as their strength, elasticity, plasticity,
hardness, ductility, malleability and resistance to deformation. Basically, the chapter focuses on the strength of
metals. Strength is the ability of a material to withstand the applied stress without failure. When you give the stress
to the material, to what extent it can bear to that stress is strength of that material.
Elasticity is the mechanical property of a material to regain its original shape and size when the external force is
removed. If you stretch a helical spring by gently pulling its ends, the length of the spring increases slightly. When
you release the ends of the spring, it regains its original size and shape. The property of a material, by virtue of
which it tends to regain its original size and shape when the applied force is removed, is known as elasticity and the
deformation caused is known as elastic deformation. On the other hand when we talk of plasticity we mean that
once we change the shape of the material, it never comes back to its original shape even when the external force is
removed. It remains permanently deformed.
When we talk of resistance to deformation, we mean that at certain time we want to change the shape of the
material. So it talks about how resistance that particular material is to the change of the shape. At certain times we
see some materials which change their shape very easily; for instance, if you think of certain material from clay or
mud, it is very easy to change their shape into your desire form. This means their resistance to deformation is less
whereas if you think of certain materials like iron, it is not at all possible to change their shape easily as before. Iron
changes its shape only when it is heated to a very high temperature. Because, their resistance to deformation is very
high. So different materials, depending up on their properties, they have different resistance to deformation.
The elastic behavior of materials plays an important role in engineering design. For example, while designing a
building, knowledge of elastic properties of materials like steel, concrete etc. is essential. The same is true in the
design of bridges, automobiles, ropeways etc. One could also ask — can we design an aeroplane which is very light
but sufficiently strong? Can we design an artificial limb which is lighter but stronger? Why does a railway track
have a particular shape like I? Why is glass brittle while brass is not? Answers to such questions begin with the
study of how relatively simple kinds of loads or forces act to deform different solids bodies. In this chapter, we
shall study the elastic behavior and mechanical properties of solids which would answer many such questions.
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6.2 Elastic Behaviors of Solids
In chapter four, we studied that in a solid, each atom or molecule is surrounded by neighboring atoms or molecules
called coordination numbers. These coordination numbers are bonded together by interatomic or intermolecular
forces and stay in a stable equilibrium position. When a solid is deformed, the atoms or molecules are displaced
from their equilibrium positions causing a change in the interatomic (or intermolecular) distances. When the
deforming force is removed, the interatomic forces tend to drive them back to their original positions. Thus the
body regains its original shape and size. The restoring mechanism can be visualized by taking a model of springball system shown in the Fig. 6.1. Here the balls represent atoms and springs represent interatomic forces.
Fig.6.1 Spring-ball model for the illustration of elastic behavior of solids
If you try to displace any ball from its equilibrium position, the spring system tries to restore the ball back to its
original position. Thus elastic behavior of solids can be explained in terms of microscopic nature of the solid.
Robert Hooke, an English physicist (1635 - 1703 A.D) performed experiments on springs and found that the
elongation (change in the length) produced in a body is proportional to the applied force or load. In 1676, he
presented his law of elasticity, now called Hooke’s law. We shall study about it in Section 6.4. This law, like
Boyle’s law studied in chapter three, is one of the earliest quantitative relationships in science. It is very important
to know the behavior of the materials under various kinds of load from the context of engineering materials design.
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6.3 Stress and strain
Whenever a force is applied on a certain material, it is deformed to a small or large extent depending upon the
nature of the material and the magnitude of the deforming force. The deformation may not be noticeable visually in
many materials but it is there. When a material is subjected to a deforming force, a restoring force is developed
inside of the material. This restoring force is equal in magnitude but opposite in direction to the applied force. The
restoring force per unit area is known as stress (𝝈). If F is the force applied and A is the area of cross section of the
material, then:
𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 (𝜎) =
𝐹
𝐴
……………………… (6.1)
The SI unit of stress is N m–2 or Pascal (Pa) and its dimensional formula is [ML–1T–2].
Types of stresses:
There are three ways in which a solid may change its dimensions when an external force acts on it.
1. Longitudinal stress: is a restoring force per unit area when the applied force is normal to the cross-sectional
area of cylindrical object.
Fig.6.2 Cylinder subjected to tensile stress stretches it by an amount ΔL.
When you apply the force perpendicular to the area of the cylinder as shown on fig.6.2, there is a restoring force
develops inside the cylinder in the opposite direction to normal force with equal magnitude. This restoring force per
unit area is known as longitudinal stress. Due to the effect of this stress, there is a change in length of the cylinder.
If we assume the cylinder is made of material that easily stretches such as rubber and tied with small thread on it to
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hung a heavy box containing a lot of things to the cylinder, then the cylinder gets stretched down and increases in
length. Under longitudinal stress we have two types of stresses; tensile stress and compressive stress.
Tensile stress: In Fig.6.3, a cylinder is stretched by two equal forces applied normal to its cross-sectional area. The
restoring force per unit area in this case is called tensile stress
Fig.6.3 Cylinder subjected to tensile stress
Compressive stress: If the cylinder is compressed under the action of applied forces, the restoring force per unit
area is known as compressive stress.
Fig.6.4 compressive stress
2. Tangential or shear stress (𝝉): if two equal and opposite deforming forces are applied parallel to the crosssectional area of the cylinder, as shown in Fig. 6.4, there is relative displacement between the opposite faces of the
cylinder. The restoring force per unit area developed due to the applied tangential force is known as tangential or
shearing stress.
Fig.6.5 shear stress
tan 𝛾 = 𝑎⁄ℎ
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3. Hydraulic stress: a solid sphere placed in the fluid under high pressure is compressed uniformly on all
sides. The force applied by the fluid acts in perpendicular direction at each point of the surface and the body is
said to be under hydraulic compression. This leads to decrease in its volume without any change of its
geometrical shape. The body develops internal restoring forces that are equal and opposite to the forces applied
by the fluid (the body restores its original shape and size when taken out from the fluid). The internal restoring
force per unit area in this case is known as hydraulic stress and in magnitude is equal to the hydraulic pressure
(applied force per unit area).
Fig.6.6 A solid sphere subjected to a uniform hydraulic stress shrinks in volume by an amount ΔV.
Strain (𝜺)
Strain is a measure of deformation representing the displacement between particles in the body relative to a
reference length. In stress section we mentioned that due to applied load per unit area there was some deformation
in an object. It is either there is a change in length or there is a relative motion between the opposite faces of the
material or there will be some contraction of the material. Refer fig 6.3 – 6.6.
Strain is nothing but a measure of the deformation or how much the change in the shape takes place. Therefore, it is
basically measures what it was initially and what is it know. Since it is the ratio of two same quantities, strain is
dimensionless.
Fig.6.7 strained material up on applied load (tensile stress)
𝑙 = ∆𝑙 + 𝑙𝑜
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𝑙𝑜 is initial length
∆𝑙 is additional length due to stretched of material
𝑙 is length after elongation
𝑠𝑡𝑟𝑎𝑖𝑛 (𝜀) =
∆𝑙
𝑙𝑜
=
𝑙− 𝑙𝑜
𝑙𝑜
…………………………………… (6.2)
Types of strain
Similar to in the case of stress, there are three types of strains forms
1. Longitudinal strain: change in length to the original length of the body due to the longitudinal stress.
𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =
∆𝑙
𝑙𝑜
2. Shearing strain: is measure of the relative displacement of the opposite faces of the body as a result of
shearing stress. Consider fig.6.5 once again. The cylinder is shift 𝑎 amount of displacement from its original
position with an angle 𝛾.
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 =
∆𝑙
𝑙𝑜
=
𝑎
ℎ
= tan 𝛾
Where 𝛾 is the angular displacement of the cylinder from the vertical (original position of the cylinder). Usually
𝛾 is very small, tan𝛾 is nearly equal to angle, (if 𝜃 = 10°, for example, there is only 1% difference between 𝛾 and
tan𝛾).
𝑎
∴ Shear strain (𝛾) = ℎ …………………….. (6.3)
3. Volume strain: ratio of change in volume to the original volume as a result of the hydraulic stress.
𝐻𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑠𝑡𝑟𝑒𝑠𝑠 =
∆𝑉
𝑉
……………………. (6.4)
6.4 Hook’s Law
This law was named after scientist Robert Hook who established it and stated that within elastic limit, stress
developed is directly proportional to the strain produced in a material. Stress and strain take different forms in the
situations depicted in the Fig. (6.3- 6.5). For small deformations the stress and strain are proportional to each other.
This is known as Hooke’s law.
Thus, stress ∝ strain
Stress = E × strain = 𝜎 = 𝐸𝜀 …………………… (6.5)
Where E is the proportionality constant and is known as modulus of elasticity.
Hooke’s law is an empirical law and is found to be valid for most materials. However, there are some materials
which do not exhibit this linear relationship.
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6.5 Stress-Strain Curve
The relation between the stress and the strain for a given material under tensile stress can be found experimentally.
In a standard test of tensile properties, a test cylinder or a wire is stretched by an applied force. The fractional
change in length (the strain) and the applied force needed to cause the strain are recorded.
The applied force is gradually increased in steps and the change in length is noted. A graph is plotted between the
stress (which is equal in magnitude to the applied force per unit area) and the strain produced. A typical graph for a
metal is shown in Fig. 6.8. Analogous graphs for compression and shear stress may also be obtained. The stressstrain curves vary from material to material. These curves help us to understand how a given material deforms with
increasing loads. From the graph, we can see that in the region between O to A, the curve is linear. In this region,
Hooke’s law is obeyed. The body regains its original dimensions when the applied force is removed. In this region,
the solid behaves as an elastic body.
Fig.6.8 A typical stress-strain curve for a metal.
A) Proportional Limit:
In Figure 6-8 it can be noted that the initial response is linear. Up to point (A), the stress and strain are directly
proportional to one another. The stress at which this proportionality ceases is known as the proportional limit.
Below this value, the material obeys Hooke’s law, which states that the strain is directly proportional to the stress.
The proportionality constant, or ratio of stress to strain, is known as Young’s modulus or the modulus of elasticity.
This is an inherent property of a given material and is of considerable engineering importance. As a measure of
stiffness, it indicates the ability of a material to resist deflection or stretching when loaded and is commonly
designated by the symbol E.
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B) Elastic Limit:
Up to a certain stress, if the load is removed, the specimen will return to its original length. The response is elastic
or recoverable, like the stretching and relaxation of a rubber band. The uppermost stress for which this behavior is
observed is known as the elastic limit. For most materials the elastic limit and proportional limit are almost
identical, with the elastic limit being slightly higher. Neither quantity should be assigned great engineering
significance, however, because the determined values are often dependent on the sensitivity and precision of the
test equipment.
C) Yield Point:
Whenever the elastic limit is exceeded, increases in strain no longer require proportionate increases in stress. For
some materials, a stress value may be reached where additional strain occurs without any further increase in stress.
This stress is known as the yield point and the corresponding stress is yield-point stress 𝑺𝒚 . For low-carbon steels,
with curves like that in Figure 6-8, two distinct points are significant. The highest stress preceding extensive strain
is known as the upper yield point, and the lower, relatively constant, “run-out” value is known as the lower yield
point. The lower value is the one that usually appears in tabulated data.
Plastic Behavior:
If the load is increased further, the stress developed exceeds the yield strength and strain increases rapidly even for
a small change in the stress. The portion of the curve between B and D shows this. When the load is removed, say
at some point C between B and D, the body does not regain its original dimension. In this case, even when the
stress is zero, the strain is not zero. The material is said to have a permanent set. The deformation is said to be
plastic deformation.
Most materials, however, do not have a well-defined yield point and exhibit stress–strain curves more like that
shown in Figure 6-9. For these materials, the elastic-to-plastic transition is not distinct, and detection of plastic
deformation would be dependent upon machine sensitivity. To solve this dilemma, we elect to define a useful and
easily determined property known as the offset yield strength. Offset yield strength does not describe the onset of
plastic deformation but instead defines the stress required to produce a given, but tolerable, amount of permanent
strain. By setting this strain, or “offset,” to 0.2% (a common value), we can determine the stress required to
plastically deform a 1mm length to a final length of 1.002mm (a 0.2% strain). If the applied stresses are then kept
below the 0.2% offset yield strength of the material, the user can be guaranteed that any resulting plastic
deformation will be less than 0.2% of the original dimension. While 0.2% is a common offset for many mechanical
products, applications that cannot tolerate that amount of deformation may specify offset values of 0.1% or even
0.02%.
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Fig.6.9 Stress–strain diagram for a material not having a
well-defined yield point, showing the offset method for
determining yield strength. is the 0.1% offset yield
strength; is the 0.2% offset yield strength.
Offset yield strength is determined by drawing a line parallel to the elastic line, but displaced by the offset strain,
and reporting the stress where the constructed line intersects the actual stress–strain curve. Figure 6-9 shows the
determination of both 0.1% offset and 0.2% offset values, and, respectively. The intersection values are
reproducible and independent of equipment sensitivity. Offset yield values are meaningless unless they are reported
in conjunction with the amount of offset strain used in their determination. The 0.2% value is most common and is
generally assumed unless another number is specified.
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