Materials Science and Engineering Introduction

By: Gizachew Berhanu
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
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].
Chapter one: Basic Concepts of Materials
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.

Natural rock
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.
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
b) Seat table
Aggregation of rocks for concrete materials

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].

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.

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 fast-
paced world, the discovery of silicon single crystals and an understanding of their properties have enabled the
information age.

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].
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.
Figure 1.3 Representative strengths of various categories of materials

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 materials
groups and categories of applications.
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].
Window glass
Ceramic floor
Reinforcement metals
Polymer tire

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].
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

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,
ceramic
Glass-ceramic
Glass bottles

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.

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)
Figure 1.8 all the preceding materials metals, ceramics, polymers, composites, and semiconductors may
be used as biomaterials.

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,
polymer, 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.
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
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

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 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.
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”.
Summary

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, PradeepP.Fulay, WendelinJ.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 KrishanChawla 2nd
ed. (2009).
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.

1) Identify whether it is engineering materials or not from the following alternatives.
a) Beach sand
b) Laser
c) Cotton
2. Define the following terminologies with less than 25 words.
3. Compare and contrast the twins given below.
a) Metals, ceramics b) glass, glass-ceramics c) ceramics, polymers
d) Materials science, materials engineering 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 under construction at Adama Science
and Technology University exactly when this material was being written. The construction company charged to
construct this Research Park was Tekleberhan Abaye 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?
Problems
a) Engineering materials
b) Properties
c) Synthesis
d) Composition
e) Structure
f) Metal
g) Sensor
h) Piezoelectric material
i) Actuator
j) Memory shape
d) Gel
e) Forklift
f) Pin

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 working 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.

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 21st
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.
Chapter Two: Structure of Atoms

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.
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
Objectives of this chapter

 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 × 10-19
C, 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.

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.
𝐀𝐭𝐨𝐦𝐢𝐜𝐧𝐮𝐦𝐛𝐞𝐫( 𝐙) = 𝐍𝐮𝐦𝐛𝐞𝐫𝐨𝐟𝐩𝐫𝐨𝐭𝐨𝐧𝐬𝐢𝐧𝐭𝐡𝐞𝐧𝐮𝐜𝐥𝐞𝐮𝐬𝐨𝐟𝐭𝐡𝐞𝐚𝐭𝐨𝐦
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 NAof atoms. The quantity NA= 6.022 X 1023
atoms per mol is the
number of atoms or moleculesin a mole. Therefore, the atomic mass has units of g/mol. An alternative unit
foratomic mass is the atomic mass unit, or amu, which is 1/ 12 the mass of carbon 12 (i.e., thecarbon atom with
twelve nucleons—six protons and six neutrons). As an example, one moleof iron contains 6.022 X 1023
atoms and
has a mass of 55.847 g, or 55.847 amu. Calculationsincluding a material’s atomic mass and the Avogadro’s
constant are helpful to understandingmore about the structure of a material.
𝐌𝐚𝐬𝐬𝐧𝐮𝐦𝐛𝐞𝐫( 𝐀) = 𝐌𝐚𝐬𝐬𝐨𝐟𝐩𝐫𝐨𝐭𝐨𝐧 + 𝐦𝐚𝐬𝐬𝐨𝐟𝐧𝐞𝐮𝐭𝐫𝐨𝐧
Atom representation = 𝐗𝐙
𝐀
where X represent an element
Isobars and isotopes
Isobars are the atoms with same mass number but different atomic number. For instance C6
14
and N7
14
. 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. C6
12
, C6
13
and C6
14
are isotopes of carbon
element. Atomic number of isotope is same. Means that number of protons is 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.

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 𝐵𝑟35
80
Solution
In the question we are given that
𝑋𝑍
𝐴
= 𝐵𝑟35
80
∴ 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: 𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 ≠ 𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑝𝑟𝑜𝑡𝑜𝑛, 𝑡ℎ𝑒𝑛𝑎𝑛𝑎𝑡𝑜𝑚𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒𝑑
Charge = number of proton – number of electron = 16 – 18 = -2
∴ 𝑋−2
16
32
An element whose atomic mass is 32 and atomic number 16 is sulfur.
∴ 𝑆−2
16
32

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 = 9.11x10-28
gm
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 = 6.022x1023
electrons
9.11x10-28
gm = 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 2.5
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
Example 2.4
Find a) the total number of neutrons and b) the total mass of neutrons in 7mg of 𝐶6
14
. (Given that mass of a
neutron = 1.67x10-24
gm).
Solution
a) First let us calculate number of neutrons present in one atom of carbon 𝐶6
14
Number of neutron = atomic mass – atomic number = 14-6 = 8 neutrons.
1mole of 𝐶 =6
14
14𝑔𝑚𝑜𝑓 𝐶6
14
= 6.022𝑥1023
𝑎𝑡𝑜𝑚𝑠𝑜𝑓 𝐶6
14
= 6.022𝑥1023
𝐶6
14
𝑥 8 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 = 4.802 𝑥1024
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
7mg =?
Total number of neutrons in 7mg =
7𝑚𝑔𝑥 4.802 𝑥1024
14000𝑚𝑔
= 2.409 𝑥1021
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
b) 𝑡𝑜𝑡𝑎𝑙𝑚𝑎𝑠𝑠𝑜𝑓𝑛𝑒𝑢𝑡𝑟𝑜𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠𝑥𝑚𝑎𝑠𝑠𝑜𝑓𝑠𝑖𝑛𝑔𝑙𝑒𝑛𝑒𝑢𝑡𝑟𝑜𝑛
2.409 𝑥1021
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠𝑥1.67x 10−24
gm = 4.03 x 10−6
Kg

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 𝑥1023
𝑥 10 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 = 6.022 𝑥1024
𝑝𝑟𝑜𝑡𝑜𝑛𝑠
Total number of proton in 34mg =
34𝑚𝑔𝑥 6.02 𝑥1024
17000𝑚𝑔
= 1.204 𝑥1023
𝑝𝑟𝑜𝑡𝑜𝑛𝑠
b) Total mass = 1.67 𝑥10−24
𝑔𝑚𝑥 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. 𝐶𝑙17
35
With a mass of 34.97 amu makes up 75.77% of the
natural chlorine found. The other isotope is 𝐶𝑙17
37
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
can calculate the abundance of 𝐶𝑙17
37
from the given abundance of 𝐶𝑙17
35
.
First, we calculate the abundance of the chlorine-37 isotope:
𝐴𝑏𝑢𝑑𝑒𝑛𝑐𝑒 𝑜𝑓 𝐶𝑙17
37
= 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 Abundance Mass
𝑺𝒊𝟏𝟒
𝟐𝟖 92.2% 27.977amu
𝑺𝒊𝟏𝟒
𝟐𝟗 4.67% 28.977amu
𝑺𝒊𝟏𝟒
𝟑𝟎 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.

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, 𝐺𝑎31
69
with a mass of 68.9257 amu and 𝐺𝑎31
71
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 𝐺𝑎31
69
and 𝐺𝑎31
71
. 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:
𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺𝑎31
69
+ 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐺𝑎31
71
= 100%
If we substitute 𝑋1 for fraction of 𝐺𝑎31
69
and 𝑋2 for fraction of 𝐺𝑎31
71
, 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%

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.
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.
Fig.2.1 Faraday cathode ray tube

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.

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 6.626x10-34
Js.
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 Voalso 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) x1014
Hz shines on the potassium metal, the photoelectric effect is observed. The threshold
frequency (Vo) for potassium metal is5.0 x1014
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
𝑇1
Wavelength
𝑇2
𝑇2>𝑇1
Fig. 2.2 intensity of emitted radiation increases with decrease of wavelength,

to the frequency of the electromagnetic radiation. Since the striking photon has energy equal to hv and the
minimum energy required to eject the electron ishvo, then the difference in energy (hv − hvo) is transferred as the
kinetic energy of the photoelectron.
1
2
mV2
= h (V − Vo) ……………………… (2.2)
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 𝑥1014
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⁄
Metal surface
hv
1
2
𝑚𝑉2
ℎ𝑣𝑜
v
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
a) b)

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
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
=
𝐸𝜆
ℎ𝑐
𝑛 =
100𝐽 𝑥 4.0 𝑥10−7 𝑚
6.626 𝑥 10−34 𝐽𝑠−1 𝑥 3 𝑥 108 𝑚 𝑠⁄
= 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 𝑥105
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
2
𝑚𝑉2
= 1.68 𝑥 105
𝐽 𝑚𝑜𝑙⁄
We need to calculate
a) ℎ𝑉𝑜
b) 𝜆 𝑚𝑎𝑥.
a) From equation (2.2), we have
1
2⁄ 𝑚𝑣2
= ℎ(𝑣 − 𝑣𝑜)
1
2⁄ 𝑚𝑣2
= ℎ𝑣 − ℎ𝑣𝑜
From the given wavelength, we should calculate corresponding energy first
ℎ𝑣 = ℎ
𝑐
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.
ℎ𝑣𝑜 = ℎ𝑣 − 1
2⁄ 𝑚𝑣2

ℎ𝑣 = 6.022 𝑥1023
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑚𝑜𝑙⁄ 𝑥 6.626 𝑥10−19
𝐽 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒⁄ = 3.99 𝑥105
𝐽 𝑚𝑜𝑙𝑒⁄
ℎ𝑣𝑜 = 3.99 𝑥105
𝐽 𝑚𝑜𝑙⁄ − 1.68 𝑥105
𝐽 𝑚𝑜𝑙⁄ = 2.31 𝑥105
𝐽 𝑚𝑜𝑙⁄
We often call such kind of minimum energy as work function.
b) 𝜆 𝑚𝑎𝑥
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⁄ =
2.31 𝑥105 𝐽/𝑚𝑜𝑙
6.022 𝑥1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒/𝑚𝑜𝑙
= 3.84 𝑥10−19
𝐽
𝜆 𝑚𝑎𝑥 =
3.84 𝑥10−19 𝐽
ℎ𝑐
=
3.84 𝑥10−19 𝐽
6.626 𝑥10−34 𝐽.𝑠𝑥 3.0 𝑥108 𝑚 𝑠𝑒𝑐.⁄
= 517𝑛𝑚
Example 2.14
The threshold frequency 𝑣𝑜, for a metal is7.0 𝑥1014
𝑠−1
. Calculate the kinetic energy of an electron emitted
when radiation of frequency 𝑣 = 1.0 𝑥1015
𝑠−1
hits the metal.
Solution
From equation (2.2), we know that
𝐾𝑖𝑛𝑒𝑡𝑖𝑐𝑒𝑛𝑒𝑟𝑔𝑦 = ℎ(𝑣 − 𝑣𝑜) Where,
𝑣 = 1.0 𝑥1015
= 10.0 𝑥1014
𝑠−1
𝑣𝑜 = 7.0 𝑥1014
𝑠−1
𝐾. 𝐸 = 6.626 𝑥10−34
𝐽. 𝑠(10.0 𝑥1014
− 7.0 𝑥1014) 𝑠−1
= 1.988 𝑥10−19
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.

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 =
1
𝜆
=
1
5800Å
= 1.724 𝑥 106
𝑚−1
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑣 =
𝑐
𝜆
=
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
=
1
2.0 𝑥10−10 𝑠𝑒
= 5.0 𝑥 109
𝐻𝑧
𝜆 =
𝑐
𝑣
=
3.0 𝑥 108 𝑚 𝑠⁄
5.0 𝑥 109 𝑠−1 = 6.0 𝑥 10−2
𝑚
𝑘 =
1
𝜆
=
1
0.06𝑚
= 16.67𝑚−1

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.
Example 2.19
A photon of wavelength 4.0 𝑥10−7
𝑚strikes 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 10-19
J)
Solution
This example is same with example 2.13 with slightly deferent given quantities
a)𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑛 = 𝐸 𝑝ℎ = ℎ𝑣 = ℎ
𝑐
𝜆
= 6.626 𝑥 10−34
𝐽. 𝑠 𝑥
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 =
1
2
𝑚𝑣2
= 0.97 𝑥 1.602 𝑥 10−19
𝐽 = 1.554 𝑥 10−19
𝐽
𝑉 = √
2 𝑥 1.554 𝑥 10−19 𝐽
9.11 𝑥 10−31 𝐾𝑔
= 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 𝑥1019
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 𝑥1014
𝐻𝑧, 1.82𝐸𝑣)
Fig. 2.4 Bohr’s model of atom

Photo emission
An excited atom relaxes from high energy to low energy by emitting a photon. We can determine the energy
difference (∆E) between levels by measuring the wavelength of the emitted photon. ∆E = h
c
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.
For hydrogen atom whose atomic number is one, the mathematical expression for Bohr’s atomic postulate is given
by
Ln = mnvrn = nℏ … … … … … … … (2.3) where n = 1, 2, 3, 4, …
Ln = angular momentum of orbiting electron, ℏ =
h
2π
,
mn = mass of 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
mov =
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
4πϵorn
2 =
mov2
rn
=
(mov)2
mo
=
e2
4π∈orn
…………….. (2.5)
Substituting equation (2.4) into equation (2.5) yields
(nℏ)2
morn
=
e2
4π∈o
rn =
4π∈o(nℏ)2
moe2 ………………… (2.6)
∆𝐸
∆𝐸
a) b)
Fig. 2.5 continuous vs. quantized spectrum a) any ∆𝐸 is possible for continuous spectrum and b) Only
certain energies are allowed

If we again rearrange equation (2.5) and divided both sides by two, we will get
1
2
mov2
=
e2
8π∈orn
… … … … (2.7) Kinetic energy which is a part of total energy
And the second part of total energy is potential energy
P . E =
e2
4π∈o
∫
drn
rn
2
∞
0
= −
e2
4π∈orn
………….. (2.8)
Therefore,
totalenergy = En = K. E + P. E =
e2
8π∈oro
−
e2
4π∈orn
=
e2
4π∈orn
(
1
2
− 1) = −
1
2
(
e2
4π∈orn
)
If we substitute for rn from equation (2.6), then we will have the total energy interims of n
En = −
1
2
(
e2
4π∈o
)(
moe2
4π∈on2ℏ2
) = −
1
2
moe4
(4π∈onℏ)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.
Fig. 2.6 electron falling from infinity state to n orbital
∆𝐸 = 𝐸 𝑛 − 𝐸∞

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 lose 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∞ =
−13.6
∞2 = 0
𝐸5 =
−13.6
52 = −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.

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.
∆Xx∆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.
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

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 = 𝑚𝑐2
= ℎ𝑣
But we know that 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑣 =
𝐶
𝜆
𝑚𝑐2
= ℎ
𝑐
𝜆
→ 𝑚𝑐 =
ℎ
𝜆
Where C = velocity of light
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
∆𝑋 = 0.1Å
𝑚 𝑒 = 9.11 𝑥10−31
𝐾𝑔
∆𝑋𝑥∆𝑃𝑥 =
ℎ
4𝜋
, but ∆𝑃𝑥 = 𝑚 𝑒∆𝑉
∆𝑋𝑚 𝑒∆𝑉 =
ℎ
4𝜋
∆𝑉 =
ℎ
4𝜋𝑚 𝑒∆𝑋
=
6.626 𝑥10−34 𝐽.𝑠
4𝜋(9.11 𝑥10−31 𝐾𝑔)(0.1 𝑥10−11 𝑚)
= 5.79 𝑥106
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
𝑚 = 40𝑔𝑚 = 4.0 𝑥 10−2
𝐾𝑔
𝑉 = 45 𝑚 𝑠𝑒⁄
∆𝑉 = 2% 𝑜𝑓 𝑠𝑝𝑒𝑒𝑑 =
2
100
𝑥 45 𝑚 𝑠𝑒⁄ = 0.9 𝑚 𝑠𝑒⁄
∆𝑋𝑚∆𝑉 =
ℎ
4𝜋
∆𝑋 =
ℎ
4𝜋𝑚∆𝑉
=
6.626 𝑥 10−34
4𝜋(4.0 𝑥 10−2 𝐾𝑔)(0.9𝑚 𝑠𝑒⁄ )
= 1.46 𝑥 10−33
𝑚

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. 𝐾𝑔and𝑣 = 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 𝑥10−31
𝐾𝑔. If its kinetic energy is 3.0 𝑥10−25
𝐽, calculate its wavelength
Solution
Given
𝐾. 𝐸 = 3.0 𝑥10−25
𝐽
𝑚 = 9.11 𝑥10−31
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
𝐾. 𝐸 =
1
2
𝑚𝑣2
= 3.0 𝑥10−25
𝐽
𝑉 = √
2𝐾.𝐸
𝑚
𝜆 =
ℎ
𝑚√
2𝐾.𝐸
𝑚
=
ℎ
√2𝐾.𝐸𝑥𝑚
=
6.626 𝑥10−34
√2 𝑥 3.0 𝑥10−25 𝑥 9.11 𝑥10−31 𝐾𝑔
= 8.957 𝑥10−7
𝑚

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)
andP(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)
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

Newton’s second law relates an external force, F, to the second derivative of the position x(t) with respect to t,
F = m
d2x(t)
dt2 = ma ………………………… (2.14)
Where ais 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
𝑑𝑡2 = 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 𝑡 𝑜 the particle has velocity 𝑣𝑜, then we have the additional relation
𝑈( 𝑡 𝑜) = 𝑎𝑡 𝑜 + 𝐶1 = 𝑉𝑂
𝐶1 = 𝑉𝑂 − 𝑎𝑡 𝑜
𝑈( 𝑡) = 𝑉𝑜 + 𝑎( 𝑡 − 𝑡 𝑜) = 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 − 𝑡 𝑜)] 𝑑𝑡
𝑥( 𝑡) = 𝑉𝑜 𝑡 +
1
2
𝑎𝑡(𝑡 − 2𝑡 𝑂) + 𝐶2
Again if we know that at 𝑡 𝑜 the particle has a position 𝑥 𝑜, then we have another additional relation
𝑥( 𝑡 𝑜) = 𝑉𝑜 𝑡 𝑜 −
1
2
𝑎𝑡 𝑜
2
+ 𝐶2 = 𝑥 𝑜
𝐶2 = 𝑥 𝑜 − 𝑉𝑜 𝑡 𝑜 +
1
2
𝑎𝑡2
Hence, 𝑥( 𝑡) = 𝑥 𝑜 + 𝑉𝑜( 𝑡 − 𝑡 𝑜) +
1
2
𝑎( 𝑡 − 𝑡 𝑜)2

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
W(x) = ∫ F(x)dx
x
0
……………………. (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
Example 2.27
Abenezar is a high roof public car driver continuously travelling from Addis Ababa to Bahir Dar on straight
way. His car which was at rest at Atobistera (Addis Ababa main bus station) set off on the road exactly at 12:
15 A.M and stop again at Bahir Dar bus station at 1: 00 P.M. Calculate the velocity of the car and the
displacement between Addis Ababa and Bahir Dar 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
= 4.9 − 0.0325𝑉(𝑡)
𝑑𝑉(𝑡)
𝑑𝑡
= 0.0325(147.7 − 𝑉(𝑡))
𝑑𝑉(𝑡)
147.7−𝑉(𝑡)
= 0.0325𝑑𝑡
∫
𝑑𝑉(𝑡)
147.7−𝑉(𝑡)
= ∫ 0.0325𝑑𝑡
𝑡
𝑡 𝑜
−ln(147.7 − 𝑉( 𝑡)) = 0.0325 (𝑡 − 𝑡 𝑜)
𝑉( 𝑡) = 147.7 − 𝑒−0.0325(𝑡− 𝑡 𝑜)
𝑉(2700𝑠𝑒𝑐. ) = 147.7 − 𝑒−87.75
𝑚 𝑠𝑒𝑐.⁄ = 147.7 𝑚 𝑠𝑒.⁄
b) 𝑥( 𝑡) = 𝑥 𝑜 + 𝑉𝑜( 𝑡 − 𝑡 𝑜) +
1
2
𝑎( 𝑡 − 𝑡 𝑜)2
𝑥 𝑜 = 0, 𝑉𝑜 = 0, Since the vehicle starts from its stationary position
𝑥( 𝑡) =
1
2
𝑎( 𝑡 − 𝑡 𝑜)2
=
1
2
{(4.9 − 0.0325𝑉( 𝑡))( 𝑡 − 𝑡 𝑜)2} = {
1
2
(4.9 − 0.0325 ∗ 147.7)(2700)2} =
363.6𝐾𝑚. Currently such type of vehicle which able to cover this much distance within 45min. never appear
in Ethiopia. The next generation has a chance to win the design of such type material.

(iii) A sum of potential and kinetic energy.
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)
F = −
d
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
2
mv2
=
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.
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
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.

Hamiltonian formulation focuses on energy. The Hamiltonian function H(x, p) is defined as the total energy of a
system
𝐻( 𝑥, 𝑃) = 𝐸𝑡𝑜𝑡𝑎𝑙 =
𝑃2
2𝑚
+ 𝑈(𝑥) ………………………….. (2.19)
The partial derivatives of the Hamiltonian function with respect to xand Pare given by
𝜕𝐻(𝑥,𝑃)
𝜕𝑥
=
𝑑
𝑑𝑥
𝑈(𝑋)
𝜕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:
 F = −
d
dx
U(x) = −
∂H(x,P)
∂x
=
dP
dt
dP
dt
= −
∂H(x,P)
∂x
…………………………… (2.20)
 P = mv(x) → v(x) =
dx
dt
=
P
m
=
∂H(x,P)
∂P
∴
dx
dt
=
∂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
= wave number and c = velocity of light
𝑖 = √−1 = 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑛𝑢𝑚𝑏𝑒𝑟
x

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