KENYATTA UNIVERSITY
INSTITUTE OF OPEN LEARNING
SPH 202
MODERN PHYSICS
LAWRENCE OTIENO OCHOO
DEPARTMENT OF PHYSICS
2005
TABLE OF CONTENTS
Page
General Overview………………………………………………….
1
Lesson 1
1.0 Atomic Structure and Properties of Matter …………………………………
3
1.1 Introduction…………………………………………………………
1.2 Composition of Matter……………………………………………...
1.2.1 Thomson’s Model………………………………………
1.2.2 Rutherford’s Experiment……………………………….
1.2.3 Rutherford’s Model…………………………………….
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3
5
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Lesson 2
2.0 Rutherford’s Model and the Electromagnetic Theory..…………………..
2.1 Introduction………………………………………………………..
2.2 Electromagnetic theory about accelerated charged particles………
2.3 Atomic Spectra…………………………………………………….
2.4 Bohr’s theory and Atomic Model………………………………….
2.4.1 Bohr’s Atomic Model………………………………….
2.4.2 The Bohr’s Postulates…………………………………
2.4.3 Implications of Bohr’s postulates……………………..
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Lesson 3
3.0 Atomic radius and Orbital energy…………………………………………
16
3.1 Introduction……………………………………………………….
3.2 Atomic radius……………………………………………………..
3.3 Orbital Energy…………………………………………………….
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Lesson 4
4.0 Shortfalls of the Bohr’s Theory…………………………………………..
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4.1 Introduction……………………………………………………….
4.2 Rotational Motion of the nucleus…………………………………
4.3 de Broglie’s Wave Theory………………………………………..
4.4 Heisenberg’s Uncertainty Principle………………………………
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Lesson 5
5.0 Photoelectric Effect………………………………………………………
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5.1 Introduction………………………………………………………
5.2 Absorption of Light Energy in quanta……………………………
5.3 Black body radiation ……………………………………………..
5.4 Millikan’s Photoelectric Effect Experiment……………………...
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27
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31
Lesson 6
6.0 Effect of Light Intensity in photoelectric Effect……………………………
6.1 Introduction…………………………………………………………
6.2 Effect of Light Intensity on Kinetic Energy and Number of electrons
6.3 Characteristic Features of the light Intensity graph…………………
6.4 Davison-Germer Experiment………………………………………..
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Lesson 7
7.0 X-rays………………………………………………………………….
7.1 Introduction…………………………………………………….
7.2 Generation of X-rays………………………..…………………
7.3 Application of X-rays…………………………………………..
7.4 Compton Effect…………………………………………………
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40
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Lesson 8
8.0 Introduction to Nuclear Physics………………………………………
48
8.1 Introduction…………………………………………………...
8.2 Composition of the Nucleus and Nuclear forces……………...
8.3 Exchange force theory………………………………………..
8.4 The Stability of a nucleus…………………………………….
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51
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Lesson 9
9.0 Radioactivity…………………………………………………………
54
9.1 Introduction………………………………………………….
9.2 Process of Radioactivity…………………………………….
9.3 Magnetic field Effect on emitted radiations…………………
9.4 Application of the emitted radiations……………………….
9.5 Decay Law………………………………………………….
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Lesson 10
10.0
Radioactivity Model and Other nuclear reactions……….……
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10.1 Introduction…………………………………………….
10.2 Radioactivity Model……………………………………
10.3 Other Nuclear reactions………………………………..
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10.4 Introduction to the Theory of Relativity…………………………
10..1 Introduction………………………………………………
10.2 Effect of Relative on basic physical quantities ………….
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Lesson 11
11.0 Relativistic equations ………………………………….
11.1 Introduction …………………………………………...
11.2 Newtonian Transformation…………………………….
11.3 Effect on length…………………………………………
11.4 Lorentz Transformation and its implications ………….
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Lesson 12
12.0 Relativistic Effect on Time and wave nature of Light…………
12.1 Introduction……………………………………………
12.2 Effect of relativity on time…………………………….
12.3 Effect of relativity on mass……………………………
12.4 Light waves and the theory of relativity………………
12.5 Michelson –Morley Experiment………………………
Useful Constants and Answers to odd-numbered questions
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General Overview
Modern physics, as the name suggests, is about new ways of interpreting the causes of various
properties of matter and aspects of the behaviour of matter when it interacts with things in the
environment. It teaches us that we may not be able to comprehend or justify causes of certain
properties of matter if we don’t relate property to the modern ideas which purport to give the most
accurate description of the characteristics of the particles which make up matter. The new ideas
present the particles of matter as extraordinary particles which do not behave as simple particles but
as particles which also have wave nature, depending on the prevailing conditions. A number of
scientists have tried to justify this wave nature of such particles. In a nutshell, particles in matter
exhibit properties which are a characteristics of particle and wave, hence, laws of physics which
apply to them are also those which apply to both:
•
•
Particle properties and
Wave properties
By systematically going through original beliefs and contributions of different scientists about the
true nature of matter and factors behind some of their properties, modern physics helps a student to
identify factors which led to the idea of wave aspects of the particles of matter. It also sets a firm
foundation about wave behaviour of tiny particles, which is crucial for advanced areas of physics
and technology. Through this course students also notice and learn that almost every theory used in
explaining certain properties of matter may still have some things to be corrected, hence, scientists
in search for truth are better off not looking at every idea or law of physics as complete.
For the purpose of outlining areas where the nature of matter proved complex for the ordinary
minds and the kind of new ideas that have helped shape the current level of understanding, this
course is built around three main areas, namely:
•
•
•
Introduction to the atomic theory of matter
Introduction to nuclear physics and
Introduction to theory of relativity
Introduction to the atomic theory of matter
In this topic, we go through what the early scientists had speculated and how they scrutinised
various aspects of the particles that make up matter, we look at how attempts have been made to
present accurate picture about the nature, structure and properties of particles in atom. It outlines
how atomic spectra contributed greatly in the advent of new theories which have helped correct and
refine information about the structure of atom and wave aspect of electron as a stable particle
outside the nucleus of an atom.
1
Introduction to nuclear physics
In this part, we look at the other particles of atom which reside in the nucleus. We start by looking
at the composition of the nucleus in order to gain valuable information concerning the manner in
which the nuclear particles co-exist for the stability of both the nucleus and the atom in general. It
will highlight how the need for satisfactory answers about what keeps the nuclear particles
(nucleons) together led to the speculation and discovery of new elementary other particles originally
not known.
Introduction to theory of relativity
Around this subject of relativity, a total revolution that threatened to tear physics apart is
considered. It points out how difference in velocity and high speed motion can make particles of
matter change their behaviour and identity to observers, hence, need for a new set of laws which
express what happens to basic quantities such as length, mass and time when subjected to very high
speed motion past an observer. The study of this subject shows how scientists laboured to establish
a clear relationship which may be used to standardise results reported by observers at different
relative velocities with one another.
2
Lesson 1
1.0 Atomic Structure and Properties of Matter
1.1 Introduction
Let me start by mentioning that the collection of ideas about matter, which have been
crystallized into modern physics, started becoming clear in the late 1800 and early 1900 when it
became evident that there was an increasing failure by scientists to accurately account for properties
of matter which had proved complex for the understanding of ordinary minds. Some of the reasons
for their failure were attributed to:
a) lack of clear understanding of the correct nature of matter, i.e composition, arrangement and
pattern of motion of the particles which make up matter.
b) Lack of appropriate laws of physics which would apply to particles so small to be seen by
naked eyes
In this course, therefore, in order to gain a clear understanding of how to justify certain properties of
matter we need to analyse causes of the past failures.
Objectives
By the end of this lesson, you should be able to:
• Describe the correct nature of matter
• Describe the composition and name the smallest particles of matter
• State the atomic theory of matter and identify discrepancy in the theory and
Explain what led to the discovery of the discrepancy in the atomic theory
• Compare Thomson’s model with Rutherford’s model
• Describe Rutherford’s experiment, aim, results and conclusion
• Discuss features of the Rutherford experiment
• Discuss challenges brought about by Rutherford’s model
1.2 Composition of Matter
In simple description and understanding, matter has been described as anything which occupies
space and has mass and for a long time such things have been perceived to be existing as a
continuous state in the following forms; solid, liquid and gas. Even at the start of the nineteenth
century no body had any defensible idea about the exact nature, composition and structure of matter
although there had been many speculations about a possibility of smaller and discrete particles in
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matter. Then the least expected person, a teacher without any formal training in science, John
Dalton, advanced a simple theory about matter, now known as atomic theory of matter.
According to the theory;
Matter is made up of smallest particles called atoms which can exist as separate
entities.
While Dalton’s atomic theory was the first bold view about the nature of matter, it was not a
valid theory as far as the word “smallest” would want to imply. Later in the century, the idea that
atom is the smallest particle of matter was challenged when a much smaller particle, electron, was
discovered when testing if gas could conduct electricity. Dalton’s theory was then threatened with
being invalidated, however, scientists still accepted the existence of atom as separate entities in
matter and the only correction made was that atoms are not the smallest particles of matter.
Note:
•
Dalton’s atomic theory asserted that atom is the smallest particle of matter,
however, it is not the smallest particle since much smaller particles, electrons,
were discovered in gas. This correction, however, did not invalidate the
existence of atom.
Although the discovery of electrons resolved one issue about the fundamental nature of
matter, it generated another problem. No one had any faintest idea about how the electrons and
atom fit in the matter and this triggered another speculation that electrons and other particles smaller
than atoms could be housed in the atom itself. A number of scientists made attempts, with the help
of models, to describe the possible ways of how the smaller particles fit in the atom.
Activities
1. What was originally known to be the smallest particle in matter?
2. State the Dalton’s atomic theory of matter
3. Explain how you would demonstrate that each of the following is made up of
small particles. Water, air and metal
4. State some of the reasons believed to have made it difficult for the early
scientists to accurately account for some of the observed properties of matter?
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1.2.1 Thomson Model
In 1898 J.J Thomson speculated that atom is simply a uniform positively charged sphere in
which electrons are embedded, just like ‘plum pudding’, figure 1.
Figure 1: The J.J Thomson’s plum pudding model
Positively charged sphere
Negative charges
At the time, Thomson’s hypothesis seemed perfectly reasonable and almost thirteen years
passed before a definite experimental test by Rutherford invalidated this model. Rutherford
designed the first experimental test to verify Thomson’s hypothesis. Following the Rutherford’s
experiment, the Thomson’s hypothesis, although plausible was abandoned and in its place a
structure incomprehensible in the light of classical physics (for large particle) was introduced.
1.2.2 Rutherford’s Experiment
This is the experiment which compelled the abandonment of the Thomson’s hypothesis. It
was the first experimental test to probe the structure of atom to establish the distribution of positive
charges (protons) in the atom, as alleged by J.J Thomson. Rutherford was a scientist with a good
background, talented, hard working and had already been awarded a Nobel Prize in 1908 for some
other investigations, so, by any standard, his experimental results should have had influence, respect
and little challenge if any.
In 1911, Rutherford, bombarded a thin gold foil with a narrow beam of fast moving
positively charged helium particles (alpha) from a radioactive element and analyzed the scattering
behaviour of the alpha particles as caused by the repulsion from the protons in the atoms in a thin
gold foil, figure 2.
Figure 2: Rutherford’s experiment
alpha particles
Luminescent Screen
thin metallic foil
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It was anticipated that, because like-charges repel one another, alpha particles would suffer
deflection from positive charges in the atom and the magnitude of deflection would depend on the
distribution of the positive charges in the atom.
Note
Rutherford chose helium particles because:
1.they are positively charged particles and, hence, suitable to probe the presence of
other positive charges due to repulsion expected between them.
2. they are heavy and, therefore, cannot be deterred from penetrating deep into the
atom by the repulsive force of the positive charges in the atom
Most alpha particles passed straight through the atoms undeflected while at some regions of
the atoms alpha particles were scattered through very large angles - right angle and even back in the
direction from which they had come. It was noticed that the region of greatest deflection appeared
to be a very small central point as illustrated in Figures 3. This suggested that forces which acted
upon the heavy alpha particles to cause such large deflections must have been very strong.
Figure 3: Observed deflection of the alpha particles
1.2.3 Rutherford’s Model
The surprising results of alpha particles ploughing straight through the atom undeterred,
except for a very small region where there were greatest deflections, led Rutherford to a belief that
atom is just an empty space with the positive charges concentrated in a tiny nucleus at the center,
Figure 4. He compared electrons in atom to planets and said they must be:
•
•
at a greater distance from the nucleus, beyond the reach of the nuclear forces
in motion, circling the nucleus at appropriate constant speeds for them to remain
outside there, otherwise they can be drawn into the nucleus by the protons, which
positive charges considered to be concentrated in the nucleus.
He then speculated the possible existence of a neutral particle, neutron. The presence of neutron,
however, was later confirmed by other scientists in other experiments.
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Figure 4: The Rutherford model of the atom.
Nucleus is at the centre with electrons around it
While the results of Rutherford, were so much welcome, suggestion by the 1908 Nobel Prize
winner that electrons are in circular motion around the nucleus, however, generated controversy and
doubt. This was because of the implication it had on some classical laws of physics, which were
known to be very successful that no one expected any idea to challenge. This issue will be
considered in the next lesson.
Summary
We have learned that matter is made up of atoms and atom itself is composed of much smaller
particles, electrons, protons and neutrons. Rutherford, in his experiment, confirmed that positively
charged particles are concentrated in a very small space, at the centre of the atom, called nucleus.
He speculated that electrons must be outside the nucleus, circling the nucleus at some appropriate
speed enough to maintain them outside the nucleus the same way planets do around the sun.
Exercise 1
1. What was J.J Thomson’s concept of atomic structure?
2. What was the disputed aspect of the Rutherford’s idea of atom and what was the reason behind
the dispute?
3. Describe the Rutherford’s experiment and the model it led to?
4. From your understanding what was the main reason behind the Rutherford’s experiment?
5. Name any properties of any form of matter whose causes you have no defensible explanation
for.
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Lesson 2
2.0 Rutherford’s model and the Electromagnetic Theory
2.1 Introduction
In lesson 1 we ended with the Rutherford’s model and stated that, although the results of his
experiment were helpful in identifying the location of protons in the atom, his idea of how electrons
exist in the atom generated controversy because of the conflict it would have with some well known
law of physics. In this lesson we’ll look at its conflict with electromagnetic theory, which is a theory
about what should happen to the energy of moving charge if it experiences any form of acceleration.
We will then explore how this conflict is resolved in the case of electron.
Objectives
By the end of this lesson, you should be able to:
• State the electromagnetic theory about effect of accelerated motion on a charge
• Explain the prediction of the combined consequence of Newton’s second law of
motion and the electromagnetic theory, if electrons are in motion as speculated by
Rutherford
• Explain how the controversy between the Rutherford’s idea and the Laws of physics
is resolved
• State some of the phenomena which support Rutherford model
• Describe features of atomic spectrum
• Explain how they contradict the prediction of electromagnetic theory
• Explain some of the implications of the features of the spectrum on the stability of
atom
• Explain Bohr’s view about electromagnetic theory
• State how Balmer formula assisted Bohr in his contribution
• State Bohr’s postulates
• State the implication about angular momentum of electron being integral multiple of
h/2π
2.2 Electromagnetic theory about accelerated charged particle
According to the electromagnetic theory;
Any charged particle in accelerated motion should radiate energy
continuously as long as the acceleration persists.
Now, electron itself is a negatively charged particle and if, as stated by Rutherford, electron
is really in circular motion around the nucleus then it must be expected to have centripetal
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acceleration according to Newton’s second law of motion for particles in circular motion, whose
equation is expressed in the form;
mv2
F
= ma =
r
By electromagnetic theory, therefore, electron should be expected to radiate energy
continuously so long as it remains moving in the circular path, this is because there will also be
continuous centripetal acceleration. The continuous energy emission, however, would imply that
electron would gradually exhaust its total energy and this may lead to continuous decrease in its
orbital radii. A decrease in the orbital radius, on the other hand, should be expected to lead to
increase in the centripetal acceleration of electron as it gets closer to the nucleus. Electron should
then approach and land in the nucleus in a spiralling path as shown in figure 5a. Also, the energy
emitted by electrons would then be expected to be radiation of all wavelengths which should appear
as a continuous spectrum, figure 5b.
Figure 5: The expected implication of Electromagnetic theory
continuous bright region
Electron
-
nucleus
a)
spiralling path of electron
b) continuous energy spectrum
If the electromagnetic theory really holds then estimates put it that an electron would take a
time (lifetime) of the order of 10-10 sec only to exhaust all its energy and land in the nucleus. This is
so fast and no electron would be expected to be existing outside the nucleus in any atom. That is to
say, by obeying the electromagnetic theory no electron would exist outside the nucleus as
Rutherford is trying to imply. The big challenge, therefore, was to either accept the electromagnetic
theory and reject the Rutherford assertion or to accept Rutherford’s idea and reject the
electromagnetic theory.
Note
It is evident that the Rutherford’s model triggered real challenge to electromagnetic
theory
• Accepting Rutherford’s idea meant that electron, although a charged particle, would
not be radiating energy as expected from the electromagnetic theory or, even if it
radiates energy, it would neither exhaust it nor spiral towards the nucleus? Then,
what would be the source of the renewable energy to maintain electron in motion
outside there?
• Accepting electromagnetic theory would suggest that an atom is a very unstable
particle of matter, expected to be uncontrollably emitting energy of all frequencies.
Thus, a continuous spectrum should be expected from the emission of an atom, and
matter.
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2.3 Atomic Spectra
There is more evidence which give more weight to Rutherford’s model over the
electromagnetic theory. Known phenomena such as chemical reactions, static and current
electricity, in which electrons are easily transferred, prove that to some extent electrons must be free
in the atom. If they were in the nucleus where other particles are strongly and permanently bound
by strong forces they could not be that free.
Also, attempts to verify if atoms emit continuous spectrum of energy, as would be expected
by the electromagnetic theory, produced negative results. Atomic spectra of the energy emitted by
hydrogen gas consisted of well separated and non-uniformly distributed spectral lines, figures 6, as
opposed to the expected continuous spectrum.
Figure 6 : Features of atomic spectra
bright regions
Discrete energy spectrum
The emission of discrete spectral lines by atom, therefore, removed the possibility that electron
emits energy continuously, hence, suggesting that;
i) in terms of energy, atom has some limits of stability and
ii) if it has to emit energy, only certain frequencies of energy are emitted, not all.
Although the discrete energy observation seemed to have faulted electromagnetic theory and thus
gives weight to Rutherford’s idea, it all the same created another question.
Why should atom emit only certain well separated frequencies (discrete or
quantized) of energy?
The most likely cause of the discrete energy emission by atom was suggested in 1913 by a relatively
little known physicist, Niels Bohr, and this is considered next.
Activities
1. State the electromagnetic theory about accelerated charged particles
2. Discuss challenges triggered by Rutherford’s idea about how electron exists in
atom?
3. Discuss factors which seem to favour Rutherford’s idea over the
electromagnetic theory?
4. Discuss features of the observed atomic spectra and explain how it strengthens
Rutherford speculation
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2.4 Bohr’s theory and Atomic Model
We recall that, while the discrete nature of energy emitted by electrons favoured the
Rutherford’s theory against the electromagnetic theory, no reason was given as to why electrons
only emit discrete energy and even Rutherford never seemed to have had any clue. Here we discuss
how Bohr, in favour of the Rutherford’s model, tried to explain why electrons would not emit
continuous energy as had been expected by electromagnetic theory, even though they are charged
particles in accelerated circular motion in the atom.
Bohr, a Danish physicist at the age of twenty eight, was at that time working in Rutherford's
laboratory when he attempted to give the correct picture about electrons in atom. He adopted
Rutherford’s model but with a radical theory to overcome the conflict with electromagnetic theory.
He suggested that even though electrons would be in continuous circular motion with centripetal
acceleration they may be having a unique behaviour which may not be understood from the
electromagnetic theory and the electromagnetic theory needed to be modified when considering
phenomena of atomic scale. He asserted that not all the classical laws would apply to atomic
systems and that, instead, a different set of laws may have to be identified, and Bohr had his
suggestions.
While our interest is on Bohr’s contribution about the picture of atom, it should be noted that
he was not alone in the Search for the origin of the spectral Lines and that he benefited from the
contributions of other scientists to map his direction. We would like to look at one case upon which
Bohr might have benefited.
In attempt to identify mechanisms behind features of the atomic spectra, most views were
influenced by the belief that light emission could have been caused by vibration of atoms of a
material. As a result, frequencies of such vibrations were expected to be related in the same way
that harmonics occur in sound vibrations, in which other frequencies are multiples of the lowest
frequency value as follows.
fundamental
fo
1st overtone
2fo
2nd overtone
3fo
It became necessary to investigate the existence of such a relationship between the
wavelengths of the spectral lines of the observed atomic spectra. In the case of numbers which form
a series, it is easy to notice relationship and formulate general expression between them. For
example, a series of the following numbers
2, 4, 6, 8 ...
form 2n patter, where n is 1st , 2nd , 3rd,… term in the series
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Johann Balmer, a Swiss schoolmaster, analysed and noticed a pattern of relationship in the
frequencies and wavelengths of the spectral lines from a discharge of hydrogen gas. Balmer
discovered that the wavelengths of emissions fitted the description of the expression
1
λ
1
=
R
22
1
-
n2
Where R is the Rydberg’s constant whose experimental value is given as 1.0968 x 107 meter-1.
While Balmer expression succeeded in fitting a few spectral lines well, the electronic
behaviour which could be associated with the pattern of the spectral lines, however, was not known
and it became Bohr’s main interest.
Balmer’s assumption was that electrons in atom always jump to the second orbit (hence, the
22 in the equation) in the atom in order to emit energy. He later suggested that additional series of
lines might be found in which electrons jumped to the first, third, fourth orbits and so on. This led
many scientists into the search for similar expressions which would describe other series of lines.
Lyman discovered a pattern for a jump to the first orbit while Paschen discovered for the third orbit.
It soon turned out that these predictions did not fit all the emitted wavelengths and the relationships
that were discovered were disproved when more accurate wavelength measurements became
available. Thus, the Balmer contribution, although significant, was in itself not comprehensive and
merely opened avenues for piecemeal contributions to the subject of the cause of discrete energy
emission. A comprehensive account became necessary although it was not known who would come
up with one.
Activities
1. What was Bohr’s declaration about the electromagnetic theory and micro
particles?
2. Give Balmer equation and explain what conclusion it led Balmer in making
3. State any weaknesses noticed with the Balmer’s account of cause for discrete
emission.
2.4.1 Bohr’s Atomic Model
It is interesting to note that as soon as Bohr saw the Balmer formula he immediately
developed a similar idea to explain the cause of discrete energy spectrum. He adopted the Planck’s
quantum theory in which atoms emit energy in packets rather than like a continuous stream of
water. He then made a simple assumption that electrons are found on specific orbits in atom where
they don’t just emit energy, he called them non-radiating orbits. He then asserted that the emission
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of energy maybe as a result of electron making a sudden jump from any one of the non-radiating
orbits to the other of a lower energy and, hence, emitting energy equal to the energy difference
between the two orbits.
According to Bohr, different non radiating orbits have different amounts of energy because
of differences in distance from the nucleus. He perceived the spacing between the spectral lines as
proportional to the spacing between different orbits. It can be noted that Bohr’s idea of electron
jumping from one orbit to the other is in agreement with the Balmer’s idea except that for Balmer
the jump must be to the second orbit while for Bohr the jump is between any two orbits.
Figure 6: Energy emission as electrons jump from one orbit to the other
Emitted
Energy
discrete energy
Note
•
•
•
Bohr speculated that atomic systems have unique characteristics that may only be
described by a different set of laws, asserting that not all the classical laws would
apply to atomic systems and, as a result, it is the electromagnetic theory which
needed to be modified when considering phenomena of atomic scale.
According to Bohr, atoms are stable particles which should not be expected to radiate
energy under normal conditions but only when its electron takes a jump from one
orbit to the other, that of higher energy to that of lower energy.
While in Bohr’s case the jump taken by electron can be between any two orbits, for
Balmer it is from any orbit to the second orbit.
Bohr’s idea about what leads to discrete spectral lines are summarised in three statements
known as postulates. Postulates are theories, not based on any experimental confirmation for their
validity, put forward to explain a phenomena.
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2.4.2 The Bohr’s postulates
•
•
•
In atom, electrons are found on certain selected orbits, where they are stable and never
radiate energy. Such orbits are known as non-radiating orbits.
Emission of energy occurs only when an electron jumps from one non-radiating orbit to
the other and that the emit energy equals to the energy difference between the orbits.
hf =E1 –E2
The non-radiating orbits where electrons are found are those in which the angular
momentum of electron about the nucleus is integral multiple of h/2π
i.e
mvr = nh/2π where h is Planck’s constant and n is an integer
2.4.3 Implications of the Bohr’s Postulates
The postulate, especially the third, have far reaching implications,
•
•
•
that given many orbits, electron would select an orbit where to sit, mainly where its
angular momentum is an integral multiple of h/2π. How does electron get to know
where this is true?
that angular momentum of electron is quantized, because n is a quantum number and
that knowing angular momentum is like knowing the position of electron in an atom.
Consider the case when electricity is switched on, electrons are assumed to be in circulation
in the circuit. Now, when switched off what happens? It would be expected that electrons find
atoms nearby to fit themselves. So, by this third postulate, they will not stop just on any orbit but
will go to specific ones. How do they identify those orbits?
This question suggests that there are still issues to clarify if Bohr’s theory holds. This
notwithstanding, quantization of angular momentum has been confirmed to be true and Bohr’s
derived wavelength equation was found to predict accurate spectra of hydrogen atoms.
Activities
1. State the Bohr’s postulates
2. Explain how Balmer series helped Bohr in his attempt to explain the
causes of discrete lines observed in the atomic spectra
3. Explain the consequences of Bohr’s postulate about angular momentum of
electron and the orbits where electrons are found
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Summary
Bohr, in his contribution about the structure of atom, accepted Rutherford’s theory about the
location of electron and dismissed the electromagnetic theory. He asserted that not all the classical
laws would apply to atomic systems and suggested that it is the electromagnetic theory which
needed to be modified when considering phenomena of atomic scale. To account for the discrete
energy spectrum of an atom, he assumed that electrons occupy certain specific orbits only, where
they circle the nucleus without emitting energy and that energy is only to be emitted when electron
jumps from one allowed orbit to the other. That electron only absorbs or emits energy equal to the
energy difference between the two orbits that it jumps between. He described such orbits as only
those in which the angular momentum of an electron would be integral multiple of h/2π. It is noted,
however, that Bohr had no justification as to why electrons would be found only on such orbit but
his idea sounded like that of Balmer and led to accurate prediction of the spectral lines for simple
atoms such as hydrogen.
Exercise 2
1. What is a postulate?
2. One may argue that it is the features of the atomic spectra that contributed significantly in
support of Rutherford’s view about electrons, what is your view?
3. Bohr tried to explain features of the atomic spectra, explain his view about the cause of
discrete lines?
4. What were some of the weaknesses identified in the Bohr’s postulates?
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Lesson 3
3.0 Atomic radius and Orbital energy
3.1 Introduction
Every valid postulate or theory would be expected to lead to accurate values of measurable
quantities of the system whose structure or bahaviour it expresses. In this lesson we look at how
well Bohr’s postulates can predict atomic radius and energy emitted by hydrogen whose atom is the
simplest of all elements.
Objectives
At the end of this lesson you will be able to
• Derive expression for Bohr’s orbital radius of atom
• Show that the orbits spacing is proportional to n2 and, hence, not uniformly spaced
• Apply Bohr’s postulates to determine expression for energy gap between two orbits
• Show that Bohr’s postulate lead to Balmer formula with accurate Rydberg constant
• What are non-radiating orbits, what is their main characteristic, i.e how are they
determined according to Bohr?
3.2 Atomic radius
The radius r is contained in the expression for the electrostatic force (F), which is the
centripetal force keeping electron in motion within the orbit of any radius r, i.e from;
e2
F
=
4πε0 r2
And, according to Newton’s second law of motion, F is also expressed in terms of r and mass of
electron as
mv2
F =
r
hence,
mv2
e2
=
r
4πε0 r2
e2
Leading to
r
=
4πεo (mv2)
e2
=
1
4πεom v2
16
Now, on applying the Bohr’s postulate, by taking expression for v from the Bohr’s angular
momentum expression,
mvr = nh/2π
i.e
nh
v
=
2πmr
e2
r
=
4π εo m
r
r
=
=
⇒
4π2m2r2
e2
=
nh
εo n2 h2
εo
2
x
e2 π m r 2
thus
2πmr
1/ v2
and substituting for
4π εo m
x
=
n2 h2
εo n2 h2
πmr2
e2
εo
x
n2 h2
r2
=
e2 π m
r
n2 h2
πme
2
n = 1, 2, 3, …….
Orbital radius depends on n and it can be expressed as
where ro = εo h2/πme2
r = n2 r o
where
εo = 8.85 x 10-12 , h = 6.62 x 10-34 Js, m = 9.11 x 10-31 kg , e = 1.6 x 10-19 C, π = 3.14
The first orbit is n = 1, hence, r = n2 ro = 5.296 x 10-11 m
The Bohr’s orbital radius for n =1was found to be in good agreement with atomic diameter as
estimated by other methods.
Since the radii of the possible orbits are directly proportional to n2 i.e (1, 4, 9, 16..), it would
be expected that the spacing between any two orbits and, hence, that of the spectral lines would not
be equal but should vary as displayed in figure (7).
Figure 7: Orbital radius is Proportional to n2
r = n2ro
n=1
n =2
n =3
17
n=4
Activities
1. Derive expression for the Bohr’s atomic radius
2. Show that the radii of the Bohr’s orbits are proportional to the square of the
orbital number n2
3. Calculate radius of the first, second and third Bohr’s orbital radii
3.3 Orbital Energy
Orbital energy is the total energy electron would have in a particular orbit. It is given as the
sum of the kinetic energy due to its motion plus potential energy due to the electrostatic force it
experiences in that orbit. i.e
E = ½mv2 + P.E
It should be noted that potential energy of a particle exists only where there is force field
acting on it. Consider where an object of mass m is raised to some height, h, above the ground.
Potential energy,
P.E = mgh
where mg is gravitational pull on the object.
If force is zero then
mg = 0 and P.E = 0
electron
m
h
r
The potential energy of an electron on a given orbit of radius r is given by
e2
P.E
= -
Therefore,
4πεor
1
E
=
2
mv +
2
e2
4πεor
18
Again, from the Bohr’s third postulate for angular momentum (mvr = nh/2π), v takes the form
nh
v =
2mrπ
Hence,
1
nh 2
e2
E= m
+
2
2mrπ
4πεor
Recall, r had been obtained as
n2 h2
r = εo
πme2
On substituting for r,
1
E=
πmoe2
nh
m
2
2
-
2πm εo n h
2 2
πme2
e2
4πεo εo n2 h2
me4
=
me4
8εo n h
2
2
me4
=
2
1
8
εo2 n2 h2
-
4εo2 n2 h2
1
4
me4
En
= -
8εo2 n2 h2
From the equation, all quantities are constant except n, implying that energy of electron depends on
n (the orbital number). For the first orbit n =1 implying the first energy E1 is given
me4
E1
= -
8εo2 h2
9.1 x 10-31 x (1.6 x 10-19)4
=
-
8 x (8.85 x 10-12)2 x (6.62 x 10-34)2
= - 13.6 eV
Therefore, energy for any other orbit (n =2,3,4…) can be expressed in terms of E1 and n as
En
-
E1
n
2
- 13.6
=
2
eV
n
Substituting values for n = 1,2,3,4,……∞ and representing them on an energy diagram the
following energy values and spacing pattern are obtained, figure 8.
19
Figure 8: Atomic Energy Levels
n=∞
0 eV
n=3
- 1.51 eV
n=2
- 3.4 eV
n=1
- 13.6 eV
If electron takes a transition from n = 1 to n =∞, the energy emitted is ionisation energy of
hydrogen and is equal to the energy difference between the two levels.
For two different orbits n and l , therefore, energy difference between them, as suggested by Bohr,
is,
∆E = En – El
∆E =
me4
1
8εo h
2
2
l
1
2
-
n2
Where ∆E is equal to the energy that an electron would emit when it jumps from n to l. This
expression has similar features with the Balmer equation. In terms of frequency ∆E= hf,
me4
f =
1
1
-
8εo h
l
n2
and, from f = c/ λ , the expression in terms of wavelength would be given as
2
1
λ
=
3
2
me4
1
8εo2 h3c
1
l2
-
n2
The quantity outside the bracket, me4 /8εo2 h3c, seems to represent the Rydberg constant (R) in the
Balmer equation, its theoretical value can be calculated from the expression. It gives 1.0974 x 107
meter-1 while the experimental is 1.0968 x 107 meter-1.
Note
It may be important to note that through the Bohr’s postulates
• the expression resembling the Balmer formula is obtained
• the predicted theoretical value of the Rydberg constant approximates well to the
experimental value.
• Bohr’s theory solved the problem of piecemeal discoveries witnessed in the
contributions of Balmer, Paschen and Lyman, which would have continued.
20
For Bohr, this must have been a major triumph and a vital piece of evidence for justifying
the correctness of his postulates. However, as pointed out earlier, Bohr’s postulates also had its
weaknesses as will be discussed in the next lesson.
Activities
1. What is orbital energy?
2. Give one example to prove that where there is no force potential energy is
zero
3. What is the expression for the potential energy associated with electron in an
orbit?
4. Derive expression for the orbital energy associated with an electron?
5. Determine the orbital energy of the 3rd orbit
6. Show that emitted energy between two orbits leads to Balmer type of
expression
Summary
It has been observed that Bohr’s theory leads to accurate predication of the Balmer type of
equation which accurately predicts wavelengths of the spectral lines for the hydrogen spectra. The
calculated Rydberg constant also approximates well to the experimental value. These, coupled with
the fact that Bohr’s postulate solved the problem of piecemeal discoveries of orbital jumps which
had been witnessed in the case of Balmer, Paschen and Lyman, Bohr’s contribution was a major
triumph. However, it is also noted that Bohr had not offered justification for some of the
implications of his postulates, such as why electrons are only found in non-radiating orbits whose
angular momenta are integral multiples of h/2π.
Exercise 3
1. Determine the Bohr’s radii and energy for the orbits n = 3, 4
2. Show that Bohr’s theory about energy emitted between two orbits lead to similar expression as
the Balmer equation for wavelength ( λ )
3. Calculate wavelength of light emitted by electron jumping from third orbit to Bohr’s second
orbit
4. Determine the Rydberg constant R from the derived Bohr’s expression for λ
5. In your view, do you think Bohr’s contribution was anything different from what Balmer had
offered? Explain
21
Lesson 4
4.0 Shortfalls of the Bohr’s Theory
4.1 Introduction
From the highlighted successes of Bohr’s postulates, it may appear insignificant and petty to
say that the small difference between the experimental and the theoretical values of the Rydberg
constant is an indication of some inadequacy in the Bohr’s theory. The truth is that in science such
small discrepancies turn out to be due to some important unrealistic assumptions in the formulation
of theories. In this lesson, we will learn that there are factors which may have been overlooked by
Bohr and whose consequences could have truly led to the difference between the calculated and the
experimental values of the Rydberg constant. As a result we shall have learned that in science any
small difference between a theoretical and experimental values may signify important omission or
assumption by a theory.
Objectives
At the end of this lesson you will be able to
• Explain how shortfalls in the Bohr’s theory came to light
• State some of the important factors Bohr overlooked in his theory about atom
• State de Broglies idea about energy emitted by electron
• Point out differences and similarities between de Broglie’s idea and Bohr’s ideas
• Show how wave theory can be used to justify the cause of discrete energy emission
• State Heisenberg’s Uncertainty Principle
• State the Bohr’s theory which is in contradiction to the Heisenberg Uncertainty
Principle
• Express and apply the equations form for the Heisenberg’s uncertainty principle
4.2 Rotational motion of the nucleus
Bohr’s model started becoming suspect when it failed to satisfactorily account for the
features of the spectra of atoms with many electrons and for those of molecules. The Bohr’s model
assumed that an electron revolves about a stationary nucleus, however, this can only be true if the
central nucleus is infinitely heavy. It was, therefore, found necessary to take both the electron and
proton as a system of connected bodies rotating about a common center of mass, Figure 9.
Figure 9: A system of electron and proton rotating about an axis
nucleus
M
electron
m
axis at centre of mass
22
This consideration alters the effective mass of the electron to
m
mo =
where M is mass of nucleus, m is mass of electron
1 + m/M
This expression shows that effective mass will be less than when nucleus is assumed to be
stationary and it is referred to as reduced mass equation, leading to expression for the emitted
frequency being;
m
f =
e4
1 +m/M 8εo2h3
1
l2
1
-
n2
If the nucleus is very heavy, m/M ≈ 0 and the equation reduces to the original form.
Also, since 1 + m/M > 1, it is expected that the calculated mass reduces, hence,
lower theoretical Rydberg constant, which now approaches the experimental value.
It was also realized that there was need to have a satisfactory explanation as to why angular
momentum associated with electron should be quantized as implied by the Bohr’s postulate. The
next development which attempted to improve the weaknesses of the Bohr’s postulate followed
about ten years later, from a scientist known as de Broglie. He likened the nature of electron to
nature of light and introduced wave aspect to explain quantization of both energy and momentum.
Activities
1. Write the expression for the Bohr’s wavelength equation, for the reduced mass
of electron
2. From a suitable expression, demonstrate that inclusion of the mass of nucleus
would lead to lower Rydberg constant
4.3 de Broglie’s Wave Theory
The de Broglie wave theory was a way of overcoming problems associated with Bohr’s
postulates. According to de Broglie, electrons do not need to jump from one orbit to the other in
order to emit discrete energy. Instead, he asserted that electrons form standing wave patterns in their
orbits and they emit energy when the standing wave patterns change from a higher mode of
vibration to a lower mode, on the same orbit, figure 10. Thus, electrons absorb and emit energy in
amounts corresponding to the energy difference between any two different wave patterns.
23
Figure 10: Various modes of standing waves
A higher energy mode
Low energy modes
de Broglie further asserted that electrons exist only on orbits which fit a whole number of
wavelengths of different modes (patterns) (λ, 2λ, 3λ, .. nλ), figure 11. i.e for orbit of radius r, whose
circumference is 2πr, the full length of the wave should be equal to the orbital circumference.
Figure 11: Electronic wave on orbit
λ
2λ
If a given orbit accommodates n wavelengths of the electronic wave then the whole length of the
wave is nλ, leading to a statement
2πr
= nλ
n =1,2,3,4……
According to de Broglie, because a photon of light wave of frequency υ has a momentum given as
hυ
p =
c
where h is Planck’s constant and c is speed of light
and frequency υ = c / λ
h
leading to
λ=
where λ is the wavelength of light
p
any material particle with wave characteristics and momentum p = mv should also be expected to
have the wavelength of its wave expressed in a similar form as
h
λ =
h
=
p
where
p = mv is linear momentum
mv
This was later verified through experiment to be true. If this is substituted, it implies that
nh
2πr = nλ =
mv
24
Leading to the Bohr equation, on exchanging mv with 2π
nh
mv r =
2π
On its part, the de Broglie’s wave theory appears more realistic than the Bohr’s theory
because it offers some form of justification for the expression for momentum. This led to the
adoption of the wave theory as a way of explaining features of the atomic spectra. The introduction
of wave theory created a turning point in the manner in which behaviour and properties of microparticles should be looked at. One other scientist whose contribution strengthened the wave aspect
in the behaviour of electron and whose contribution brought serious implication to our desire to gain
accurate knowledge of certain quantities associated with electron was Heisenberg.
Note
• Bohr’s theory failed to account well for many- electron elements and this prompted
need for better reasons to explain cause for discrete spectral lines
• de Bgroglie introduced wave aspect as a way of explaining the discreteness of the
atomic spectra.
• According to de Broglie, electron has dual nature of behaviour (wave-particle nature)
• According to de Broglie, electrons remain within the same orbit but change their
wave pattern and would emit energy on changing this wave pattern for that of higher
energy state to that of low energy state
• Atoms absorb and emit energy only in amounts exactly corresponding to the energy
difference between the energy levels
4.4 Heisenberg Uncertainty Principle
In this section, we look at the implication of wave aspect of electron on our ability to
accurately determine important quantities associated with electron and other particles of the same
size. The aim is to understand whether wave behaviour may or may not interfere with accurate
determination of quantities which are important for the accurate description of the state and
structure of an atom.
According to Heisenberg, for very small particle, it would be difficult to accurately
determine the following pairs of quantities in a simultaneous measurement, momentum and position
or energy and time. That is, if measurements are to be taken simultaneously, high accuracy in the
measurement of any one quantity of a given pair perturbs the other quantity leading to high
inaccuracy in its measurement.
The Heisenberg’s uncertainty principle states that:
For micro-particles it is difficult to determine the following pairs of observable quantities,
momentum and position or energy and time, with the highest degree of accuracy in a
simultaneous measurement.
25
This principle is expressed in the form;
h
∆x ∆p ≥
h
∆x
or
4π.
≥
4π∆p
where ∆x represents error made in the measurement of position and ∆p represents error made in the
measurement of momentum in a simultaneous measurement.
It showing that because ∆x is inversely proportional to ∆p, if the error in one quantity reduces
through accurate measurement, error in the other increases. We are informed that this impossibility
is not due to some kind of inaccuracy or inadequacy of apparatus that one may use but a phenomena
that exists because of wave behaviour, which is unpredictable when something alters any state of a
particle.
Note
• Bohr’s postulate that electrons are found in orbits where their angular momenta are
integral multiples of h/2π seems to be in contradiction with the Heisenberg’s
Uncertainty principle.
The Bohr’s statement implies that our accurate knowledge of angular momentum is
like accurate knowledge of the position of an electron in an atom.
Summary
We have seen that the higher value of the Rydberg constant from the Bohr’s theory could be
due to unrealistic assumptions made by Bohr. That is, if the nucleus and electrons can move as a
system then the calculated Rydberg constant reduces towards the experimental value. Also, the de
Broglie’s idea of wave is more realistic because it has tried to justify why angular momentum is
quantized as had been expressed by Bohr. We have also heard that the Bohr’s postulate that
electrons are found in orbits where their angular momenta are integral multiples of h/2π seems to
be in contradiction with the Heisenberg’s Uncertainty principle
Exercise 4
1. Point out some of the factors believed to be contributing to the features of atomic spectra but
which Bohr could have assumed.
2. Discuss de Broglie’s ideas about the cause of discrete energy lines in atomic spectra and
quantization of angular momentum
3. State Heisenberg’s uncertainty principle
4. State the Bohr’s postulate which is in contradiction to the Heisenberg’s principle and explain
how.
5. Wave aspect puts limitation to our accurate knowledge of certain observable quantities in a
simultaneous measurements. Comment on the validity of this statement
26
Lesson 5
5.0 Photoelectric effect
5.1 Introduction
This lesson looks at how electrons behave when they interact with light of different
frequencies or wavelengths. Through this, the manner in which electrons absorb light energy will
enhance our understanding of its wave aspect as well as applications to which photoelectric effect
has been put.
Objectives
At the end of this lesson you will be able to
•
•
•
•
•
•
Describe photoelectric effect
Explain the meaning of threshold frequency
Explain the significance of workfunction of a material
Express Einstein equation relating the kinetic energy of
liberated electrons to the absorbed energy and
workfunction
State Einstein’s postulates about existence of light in
quanta
Describe Millikan’s experiment and its results
5.2 Absorption of light energy in quanta
Photoelectric effect is a phenomenon demonstrating consequences of the interaction
between light energy and electrons in different materials. We have already seen, through features of
atomic spectra, that atoms emit energy in discrete form (quanta). Photoelectric effect is one means
by which the mode of absorption of light energy by atom can be understood.
At one time it was imagined that electrons would have to absorb energy continuously and
cumulatively up to a certain level before it can be dislodged from an atom and, therefore, light of
any frequency (energy) could dislodge electron from a material. However, it became surprising that
only light of certain frequencies dislodge electrons from the surface of a material and the emission
of electrons start instantly when light of the right frequency falls on a material. This suggested that
there is no accumulation of energy and, hence, nothing like continuous absorption of energy by
electron.
Even before the photoelectric effect, Planck had already advanced a theory that atoms either
emit or absorb light in quanta (smaller units or packets of same amount of energy) and not
continuously but many scientists found Planck’s theory too radical to accept and for sometime it
was ignored. However, in 1905, the same Planck’s idea helped Albert Einstein formulate theories
that helped in strengthening the idea of light being absorbed in quanta and, hence, the understanding
27
of the mode of absorption of light energy. Planck’s idea came through his attempt to form equation
which could be used to describe the nature of radiation emitted by hot bodies ( classified as “black
body” radiation ). We need to look at what black body radiation is and how it came to be the source
of idea about the quantization of energy absorbed or emitted by materials.
5.3 Black body radiation
In 1900, one of interesting problems in physics was about the nature of radiation observed
from hot materials, whose characteristics seemed to vary with the variation of temperature. The first
clue to the idea about the quantization of energy came from the attempt to identify relationship
which could be relied on for accurate prediction of wavelength or frequencies emitted by any hot
body at any temperature. It is a matter of common experience that as an object gets hotter, not only
does it radiate more strongly but its colour changes perceptibly. E.g when electric heater is first put
on you can feel the heat radiated by the element, although you cannot see visible radiation. This is
because nearly all the energy is radiated in the low frequency or long wavelength, infrared, to which
our eyes are not sensitive. As the temperature increases, the element glows a dull red then a brighter
red or even orange.
To be able to understand the issues related to radiation of a material, an idealized “black body” was
assumed, one which can absorb perfectly at all wavelengths. A black body is also expected to be an
ideal radiator, that is one which radiates all wavelengths at any temperature. This was to be the most
suitable kind of body for analysing the form in which hot bodies emit energy because the use of any
common body may give inadequate results, since absorption and emission by a material may
depend on the nature of the material, the state of its surface and so on.
By definition, a black body is one which absorbs all radiation incident upon it, regardless of
wavelength or frequency.
There is no real body which completely satisfies this ideal definition, but a close approximation to a
black body consists of a small hole in a large enclosure, or cavity. If the hole is sufficiently small,
entering radiation has a negligible chance of getting out through the hole before being absorbed, due
to multiple reflections. Thus the hole is “black” and is an ideal model of a black body and can be
used to analyse the kind of radiation from black bodies. The kind of radiation from such an ideal
body is called “cavity radiation” or black body radiation.
The experimental approach, therefore, would be to measure radiation from a cavity in a hot
material, formed by drilling a very small hole in a hollow tungsten cylinder or sphere for example.
From the definition of a black body, it may not appear that there could be anything unique about the
spectrum of radiation emitted.
Usually, the radiation emitted by a hot, and black body for that matter, is not confined to one
wavelength or narrow spectral region but spans the entire electromagnetic spectrum. That is, at any
temperature a body emits radiation of all wavelengths. However, the proportion (distribution) of
each wavelength varies with the temperature. At low temperature much of the energy appear as
infrared radiation but as the temperature increases much of the energy is radiated in the high
frequency or shorter wavelength, i.e distribution is a curve whose peak (most abundant
wavelength/intense) shifts to shorter wavelengths as the temperature increases., figure 12. The
position of the peak signifies the most intense wavelength or energy in the mixture and the
28
frequency at the peak was found to be directly proportional to the absolute temperature of the body
and the radiated intensity goes smoothly to zero as the wavelength approaches zero or becomes
infinite (∞).
Figure 12: Black body radiation
1500K
Intensity
1200K
900K
700K
0
wavelength λ
It became necessary that people understand and be able to explain these simple features of
the radiation. The problem that occupied the attention of Max Planck and a few other theorists was
of finding a satisfactory formula for this spectral distribution. In 1874, J. Stefan showed that the
power emitted per unit area of a body, at absolute temperature T , could be expressed as
P = εσT4
where ε is emissivity (≈ 1 for good emitter) and σ = 5.67 x 10-8 W m-2 K-4 is known as Stefan’s
constant, it is independent of the nature of the radiating surface.
Then in 1884, Boltzmann derived the same relation from thermodynamics for the use of a
black body (ε =1). It is now known as the Stefan-Boltzmann law and it suggests that the energy
radiated by a real black body(ε =1) depends only on the temperature.
Towards the end of 1900, Planck , in much the same way that Balmer had done, proposed a
formula, based on a revolutionary hypothesis in which the emission of energy by a material is
quantized. He constructed a mechanical model of oscillators or resonators, as he called them, able to
absorb and emit radiant energy and postulated that the energy of an oscillator of a given frequency
cannot take arbitrary values between zero and infinity, but can only take on the discrete values in
the form
∆E = hv
29
This idea was later supported by Einstein. Instead of Einstein disagreeing with the Planck’s idea he
even compounded it by postulating that
•
•
•
Light is not only emitted or absorbed in quanta as speculated by Planck but it also travels
in quanta each of value E = hf.
at absorption of light, a single photon gives up the whole of its energy to a single electron.
out of the energy absorbed by electron from a single photon, part is used to dislodge the
electron from the material and the remaining part is used as kinetic energy by the electron to
enable it move after it is dislodged. This is expressed in equation form as
½mv2 = hf - φ
where
hf - is the energy of the absorbed photon
φ =hfo -is called work function of a material. The amount of energy required to dislodge
electron from a material.
Note
• The idea of quantization of the energy emitted or absorbed by matter first came through
the study of black body radiation, following Plack’s hypothesis.
According to Einstein
•
•
a ray of light is supposed to be made up of several packets of energy (photons) and
these packets of energy move as distinct entities through space figure 13
electron absorbs a single photon only, and only if it is of the right frequency and, if
the absorbed photon has energy higher than what is needed to just dislodge electron
from the surface of a material, excess energy provides kinetic energy to the electron.
Figure 13: Photons travelling in space
A single photon absorbed
Photons
Atom
The first experimental test of photoelectric effect, to determine the mode of absorption of
light energy by electron in a material and to verify Einstein’s equation, was carried out by Millikan.
30
Activities
1. Explain the difference between the Plancks’ postulate and the Einstein’s
postulates about packets of energy in light.
2. Discuss characteristics of a blackbody radiation and state factors believed to
be influencing the most occurring frequency/energy emitted.
3. State Einstein’s postulate about the manner in which an electron absorbs
energy from light
4. Explain the Einstein’s idea of how the absorbed energy is utilized by the
electron
5. Discuss how radiation from just any hot body would compare to black body
radiation
5.4 Millikan’s Photoelectric effect Experiment
By observing photoelectric effect in different materials, Millikan investigated how the
energy absorbed from light by electron is utilized and how it depends on the material used. By
measuring how the kinetic energy of the liberated electrons varied with frequency of light and the
material used, he investigated Einstein’s idea about mode of energy absorption. The apparatus and
the set up are as shown in figure 14.
Figure 14: Experimental setup for photoelectric effect
Light (hf)
Cathode
anode
electrons
V
Voltmeter
A
Ammeter
Battery connected in reverse
Light of the selected frequency or wavelength is allowed to fall on the cathode material to
liberate electrons from its surface. Some of the liberated electrons may have enough kinetic energy
to cross and reach the anode, through the evacuated space, and conduct current whose value
depends on the number of electrons able to cross. The current reading is taken from the
ammeter (A).
The kinetic energy of liberated electrons can be determined by applying opposing
potential (V) to stop them from reaching the anode. When the opposing energy due to the applied
opposing potential equals the kinetic energy of the electrons, no electron would be expected to
31
reach the anode and the anode current will be zero. By adjusting the reverse voltage (V) until the
ammeter reading goes to zero, the kinetic energy of the electrons is calculated from the product of
the applied reverse voltage and the electric charge on electron (i.e E = eV). This reverse voltage (V)
which stops the liberated electrons from reaching the anode is stopping potential. For reverse
voltage of value V, the opposing energy E is given by
E = eV
= 1.6 x 10-19 x V Joules
where e = 1.6 x 10-19 Coulomb
Then, since the kinetic energy of electron equals the applied opposing energy,
½mv2 = eV
By changing frequency or wavelength of light, one is changing energy of the absorbed
photon (hf). If the material remains the same, the work function remains the same even as light
frequency is changed. Therefore, if light of higher frequency is used, excess energy and, hence, the
kinetic energy of the electrons would be high because it is equal to the excess energy of the photon
after the work function part is used. Thus, the stopping potential (V) is expected to increase with
increased light frequency. Therefore, by recording the reverse potential (V) for different frequencies
or wavelength used, one can determine variation of the kinetic energy (eV) with frequency (f = c/
λ). From this a graph of eV against f can be plotted, with kinetic energy (eV) along y-axis and f
along x-axis.
Millikan’s graph for different materials appeared as represented in figure 15, it revealed that;
•
irrespective of the materials used, the graphs were linear and all had the same
slope (gradient)
• for each material, its graph starts at some minimum frequency value called threshold
frequency (fo), signifying no emission of electrons by frequencies less than this value
Figure 15: Kinetic energy versus frequency of light
Kinetic Energy/
½mv2
A
B
C
½mv2
0
fo
fo
fo
f
frequency
The minimum frequency implies each material requires some minimum energy to dislodge
an electron from it, this depends on the strength of the force holding an electron in the material.
Threshold frequency, therefore, is an intrinsic property of a material, hence, it has to vary from one
material to the other.
32
Definition: Threshold frequency is the minimum frequency of light that is capable of dislodging
electron from a material. It varies for different materials.
By determining gradient of any of the lines, say C, the expected expression is as follows
½mv2 - 0
Change in kinetic energy
grad =
=
change in frequency
f - fo
Millikan found the gradients of his lines to be equal to 6.56 x 10-34 Js, which is closely equal to the
Plancks constant (h), leading to an expression of the form
½mv2 - 0
h
=
f - fo
leading to
½mv2 = hf - hfo
which is the same form as the Einstein’s equation
½mv2 = hf - φ
where
φ = hfo
Example
A photon in the light of frequency 2 x 1015 Hz falls on a material whose work function is 2.28 eV.
Calculate
i)
Energy in the photon
ii)
Maximum kinetic energy of the liberated electrons
iii)
The threshold frequency of the material used
iv)
Stopping potential required to stop the electrons from reaching the anode
(i) Energy of the light of frequency f is the energy of each photon in it, given by
E = hf = 6.62 x 10-34 x 2 x 10 15 = 13.24 x 10-19 J
In electron volts
13.24 x 10-19
=
= 8.28eV
1.6 x 10-19
(ii) Maximum energy is the maximum possible energy any liberated electron would have. It is equal
to the excess energy after the work function part is removed from the energy of the absorbed
photon.
i.e excess energy ( Kinetic energy) = hf - φ
i.e K.E = 8.28 – 2.28 = 6.0 eV
33
(iii) threshold frequency is the lowest frequency of light that can dislodge electron without any
excess energy left as kinetic energy, i.e kinetic energy equals to zero.
From
= hf - φ
= hf - φ
K.E
0
hf = φ
φ
So
f0 = f =
2.28 x 1.6 x 10-19
6.62 x 10-34
h
f0 = 5.51 x 1014 Hz
iv) the stopping potential (V), as the name suggests, potential which stops the electrons with kinetic
energy from reaching the anode, it provides energy (eV) equal to the kinetic energy (K.E) of the
emitted electrons
from
K.E = eV
K.E
V =
for the light used, K.E was found to be 6.0 eV
e
6.0 x 1.6 x 10-19
hence, V =
1.6 x 10
-19
= 6.0 V
Summary
When light of frequency greater than the threshold frequency falls on a material electrons
are liberated instantly not after sometime. This confirms that electrons do not accumulate energy by
continuous absorption in order to be liberated but they absorb energy in quanta. According to
Einstein, Planck’s theory that atoms absorb light energy in quanta is valid and he added that it also
travels in quanta. At absorption, a single photon gives up all its energy to a single electron out of
which part is used to dislodge electron from the material and the remaining portion is utilized by the
electron as kinetic energy.
34
Exercise 5
1.
2.
3.
4.
Discuss Millikan’s experiment and its results
What is threshold frequency?
What is the meaning of work function?
In the photoelectric effect, what is the relation between the threshold frequency f0 and the
work function φ ?
5. Calculate the energy in a photon of ultraviolet light whose frequency is 2 x 1016 Hz.
6. An energy of 5 x 10-19 joule is required to remove an electron from the surface of a particular
metal.
i)
What is the frequency of light that will just dislodge photoelectrons from the
surface?
ii)
What is the maximum energy of photoelectrons emitted through the action of
light of wavelength 3 x 10-7 m?
7. The photoelectric threshold wavelength of the tungsten is 2.73 x 10-5 cm. Calculate the maximum
kinetic energy of the electrons ejected from a tungsten surface by ultraviolet radiation of
wavelength 1.80 x 10-5 cm.
8. A photoelectric surface has a work function of 4.00 electron volts. What is the maximum velocity
of photoelectrons emitted by light of frequency 3 x 1015 sec-1 ?
9. The photoelectric work function of potassium is 3.0 eV. If light having a wavelength of 3600 Å
falls on potassium, find (a) the stopping potential, (b) the kinetic energy of the most
energetic electrons ejected, and (c) the velocities of these electrons.
10. When a certain photoelectric surface is illuminated with light of different wavelengths, the
stopping potentials in the table below are observed.
Wavelength,
angstroms
Stopping
Potential/ volts
3660
4050
4360
4920
5460
5790
1.48
1.15
0.93
0.62
0.36
0.24
Plot the stopping potential as ordinate against the frequency of the light as abscissa.
Determine
(a) the threshold frequency,
(b) the threshold wavelength,
(c) the photoelectric work of the material, and
(d) the value of Planck’s constant h (the value of e being 1.6 x 10-19 J).
35
Lesson 6
6.0 Effect of Light intensity on Photoelectric effect
6.1 Introduction
This lesson intends to look at how the Einstein’s postulate of a “single photon energy
transfer to a single electron” can be justified. The aspect of photoelectric effect which may seem to
justify this is the light intensity effect. We’ll take the light intensity to mean the number of photons
falling on unit area of a material. This lesson will also discuss the significance of the Davisson and
Germer experiment which tried to demonstrate that electrons do exhibit wave properties such as
diffraction and interference.
Objectives
At the end of this lesson you will be able to
•
•
•
•
•
•
Explain the aspects of photoelectric effect influenced by light intensity
Describe the Einstein’s postulate about photon absorption which is justified by effect
of intensity
Use illustrative diagram to explain how variation in light intensity influences the
number of electrons emitted
Describe the Davisson and Germer experiment
Discuss results of the Davisson-Germer experiment
Explain what the behaviour of electrons in this experiment demonstrates about its
property
6.2 Effect of Light Intensity on the Kinetic Energy and Number of electrons
Classically, if light intensity is increased an electron may absorb more than one photon, as a
result its kinetic energy is expected to increase with the increased intensity or proportionately to the
number of photons absorbed since the excess energy retained by electron above the work function
would be high. Any change in the kinetic energy of the emitted electrons, therefore, would be a
better indicator for monitoring the of effect of light intensity and, hence, if one electron can absorb
more than one photon from light.
Photoelectric experiments have shown that provided the frequency of light used remains
constant, the kinetic energy of the emitted electron do not vary with the light intensity, however, the
number of electrons emitted per second (current) from the surface of any material is directly
proportional to the intensity of the incident light.
36
It was also observed that liberation of electrons from the surface of a material does not
depend on the intensity of the light used but only on the frequency. That is, any light of frequency
less than the threshold frequency of a material (f < fo) does not dislodge electrons from the material
even at very high intensity. These observations confirm that intensity does not increase the energy
absorbed by electron but merely increases the number of emitted electrons. This is in agreement
with the Einstein’s postulate that electron absorbs energy only from a single photon, therefore, if the
number of photons is increased by increased intensity then the number of emitted electrons increase
and not the kinetic energy.
The number of electrons produced by light of a given intensity can be monitored by
measuring the current produced by a given light intensity. Recall, current level is proportional to the
number of electrons flowing in a circuit. The same apparatus for photoelectric effect are used,
however, with the voltage in the forward direction to sweep any produced electron to anode to be
counted by the ammeter.
6.3 Characteristic Features of the Light intensity graph
It is worth noting that even when voltage equals to zero (V=0), at the origin, current value is
not zero. This point shows that electrons still reach anode even without the applied voltage,
confirming that some electrons are dislodged with enough kinetic energy to reach the anode. As the
voltage (potential difference) between cathode and anode is increased, more electrons are helped to
cross to anode and the ammeter reading is seen to increase to maximum value (saturation point) as
shown by the curves in figure 16.
Figure 16: The Variation of current with Light Intensity
Current (I)
High intensity
Low intensity
-V
0
p.d
volts
The saturation point is a sign that all electrons which are liberated by a given light intensity
are reaching the anode, those with enough energy and those aided by the applied voltage, hence, no
more increase in the current even as voltage is increased.
It is also noted that, irrespective of the intensity, the curves start at the same point on the
negative side of the x-axis. At this point current is zero although a negative voltage is applied. This
voltage is the stopping potential.
37
Note
•
•
•
Light intensity increases the number of electrons emitted from the surface of a
material,
Intensity does not alter kinetic energy of electron, kinetic energy depends mainly on
the wavelength of the light used.
Kinetic energy of the liberated electrons can be determined by applying
reverse/negative potential, K.E = eV
This behaviour seems to confirm that an electron absorbs energy from a single photon (hf)
because its Kinetic energy does not increase with increased number of photons falling on a material
(intensity).
Activities
1. Describe the effect light intensity has on photoelectrons
2. Draw the curves showing the effect of light intensity on the current and
discuss features of the curves
3. Explain how the results of the effect of light intensity justifies the Einstein’s
postulate about energy transfer between electrons and photons
6.4 Davisson-Germer Experiment
While the particle-wave nature of electron had been accepted and supported by many
scientists, one significant problem remaining was the question of whether other characteristic
properties of wave such as diffraction, interference and reflection can be observed in electrons. The
argument being, a wave should exhibit the three properties, therefore, if electron has wave
behaviour it should be expected to show the three properties. In order to investigate this
characteristic in electron, Davisson and Germer tried to observe scattering of electrons on
crystalline surfaces.
Davison conducted the first experiment alone and observed that electrons elastically
reflected from the surface of a Platinum undergo interference with strong maxima in intensity,
figure 17. He suggested that the observed maxima are a sign of interference phenomena which is in
line with the de Broglie idea of wave aspect.
38
Figure 17: Interference pattern with strong maxima
incident beam of electrons
interference Maxima
Scattered beam of electron
Crystal
Later, in 1927, Davison and Germer tried scattering slow moving electrons from the surface
of a single nickel crystal and observed that electrons accelerated through some potential difference
(p.d) are scattered in a similar way to x-rays of the same wavelength, clearly suggesting existence of
wave characteristic in electrons. From the diffraction results, they found that the wavelength,
determined in terms of Bragg’s law, were consistent with wavelength calculated from de Broglie
equation.
According to the de Broglie wave equation
h
λ =
mv
and from the kinetic energy expression for electrons accelerated by a potential difference V
½mv2
= eV
mv2 = 2 eV
On introducing ‘m’ to both sides leads to
(mv)2 = 2meV and
mv = 2meV
Hence, the de Broglie wavelength of the beam of used electrons is
h
λ =
2meV
By manipulating V, electronic waves of different wavelength comparerable to x-rays could
be obtained and used in the experiment and, as reported earlier, the diffraction and interference
behaviour of the electrons were found to compare well to x-rays of the same wavelengths.
39
This property, however, is not limited to electrons alone. T.H Johnson later carried out a similar
experiment with hydrogen atoms and showed that wave-like property exists. Elsewhere, the same
was observed with helium atoms. This experiment, therefore, confirmed the wave aspect of
electron.
Example
An electron is accelerated through a potential difference of 40,000 Volts, calculate its de Broglie
wavelength.
the de Broglie wavelength has been derived as
λ =
h
2meV
λ =
6.62 x 10-34
2 x 9.1 x 10-31 x 1.6 x 10-19 x 40,000
λ = 6.13 x 10-12m = 0.0613Å where 1 Å = 10-10 m
Summary
It has been noted that light intensity increases the number of electrons emitted from the
surface of a material but does not alter kinetic energy of the electrons, kinetic energy depends
mainly on the wavelength of the light used. This behaviour confirms that an electron absorbs energy
from a single photon (hf) because its Kinetic energy does not increase with increased number of
photons falling on a material (intensity). Davisson and Germer experiment confirmed the wave-like
behaviour of electron through diffraction of electrons on nickel crystals. This is a clear
demonstration of the diffraction and interference behaviour of electrons. This behaviour, however,
is not limited to electrons only but it is a behaviour of all particles of the size of atom and its
constituents.
Exercise 6
1. Discuss the Davisson-Germer experiment, its results and implication.
2. State the Einstein’s postulate about energy transfer process between photon and electron and
describe an experiment which can be used to demonstrate the validity of the postulate.
3. Calculate the voltage through which electrons should be accelerated if their wavelength is to
be equal to that of x-rays of wavelength 0.2 Å
40
Lesson 7
7.0 X-Rays
7.1 Introduction
The same way the energy of photon can be transferred to electron and be transformed into
kinetic energy, it was speculated that the reverse could also be true. That is, transforming the kinetic
energy of electron back into photon when accelerated electron gives up all its kinetic energy to a
single electron that will absorb it and re-emit it in the electromagnetic form known as x-rays. This
lesson looks at the process that would lead to this reverse process.
Objectives
At the end of this lesson you will be able to
• Describe how x-rays are generated using illustrative diagram
• Discuss factors influencing the quality of x-rays
• Calculate wavelength of x-rays generated under different p.ds
• Describe the Compton effect
• Discuss the results of the Compton effect
• Relate the scattering behaviour of photon to the Einstein’s idea about photons in light
• Explain what the property of photon that the Compton effect demonstrates
• Apply Compton equation to determine shift in the wavelength of photon after
collision
7.2 Generation of x-rays
X-Rays are electromagnetic waves of very short wavelengths generated when very fast
moving electrons are abruptly brought to rest by a metal target of high melting point. X-rays were
first discovered by Roentgen in 1895, they were called X-rays because their nature could not be
identified. It is a tendency in mathematics to let something not known be x. The kinetic energy of
the moving electrons is transferred to the atoms of the metal which then re-emits the energy in the
form of photons of very short wavelengths. The emission of x-rays may be seen as the reverse of
photoelectric effect. This is because, in photoelectric emission the energy of a photon is transformed
into kinetic energy of an electron while in this case it is the kinetic energy of the electron that is
converted to photon. X-rays are usually produced in a specially designed tube, modern one being
that designed by Coolidge, figure 18.
41
Figure 18: x-ray production tube
Cooling
filaments
cathode
anode
x-rays
heating source
p.d
In the x-ray tube, electrons are produced at the cathode by electrical heating. The liberated
electrons are accelerated by very high p.d, about 40,000V or more, towards the anode where they
are brought to rest by the target material, tungsten or other metal plate embedded in copper head.
The accelerated electrons transfer their kinetic energy to electrons in the atom which later re-emit
the energy as x-rays. It has been estimated that only a small amount, about 0.2%, of the energy of
the bombarding electrons is transformed into x-rays. The remainder is dissipated as heat in the
anode, which therefore gets very hot. This is why a target of high melting point is usually used and
cooling filaments added to enhance fast cooling. The cooling fins are air cooled for small x-ray
tubes and water or oil cooled for large tubes.
The p.d across the tube governs the quality of the x-rays. There is relationship between the
voltage and the kinetic energy of the electrons. Even the early workers with x-rays noted that
increasing the voltage –which means increasing speed of the bombarding electrons led to the
production of x-rays of greater penetrating power (hard X-rays). Reducing the voltage led to x-rays
of low penetrating power (soft X-rays). It was shown that x-rays of high penetrating power have
very short wavelengths or high frequency and the wavelength can be expressed in terms of the
voltage.
Since kinetic energy is to convert to photon energy,
hc
K.E = hf
=
λ
and
½mv2
K.E =
= eV
hc
Therefore,
eV =
λ
hc
hence
λ
=
eV
42
Example
Electrons are accelerated to the anode of the x-ray tube at a potential difference of 20,000V,
calculate the wavelength of the generated x-rays.
Kinetic energy gathered by the accelerated electrons is
K.E = eV
= 1.6 x 10-19 x 20,000
hc
This energy is converted to x-rays of energy
hence
λ
λ
-34
hf
=
8
hc
6.62 x 10
eV
1.6 x 10-19 x 20,000
=
λ
x 3 x 10
= 6.21 x 10-12 m
7.3 Applications of x-rays
X-rays are widely known to be used in hospitals and X-ray tubes are now an essential part of
the equipment of hospitals throughout the country. They are used to monitor inner structures of
solid materials, e.g broken limbs, congestions in lungs for signs of TB etc. Although x-rays
penetrate flesh, they are stopped by the bone and other solid objects. The medical application was
recognized shortly after the discovery of X-rays by Roentgen.
There are extensive industrial applications of x-rays such as for the detection of flaws in
metal casts, cracks invisible to eye and defects in motor tyres from x-ray photographs (radiographs),
see figure 19
Figure 19: Detection of cracks in metal pipes by x-rays
Crack
43
Activities
1. Describe how x-rays are produced in the x-ray tube
2. Discuss factors influencing the quality of x-rays
3. x-ray tube is operated at 30000V, calculate wavelength of the generated xrays.
7.4 Compton Effect
Einstein’s idea that light energy is carried across space in packet-like smaller units (photons)
led to the impression that photons move like a stream of particles which could be scattered the way
particles are scattered when they collide with other particles. Many cases of particle scattering in
physics demonstrate elastic collision property of colliding particles in which energy and momentum
are conserved. This lesson looks at the experiment by Compton which demonstrates that photons
have a similar scattering behaviour, in which energy and momentum are conserved in their collision
with free particles such as electrons.
In 1923, Compton investigated the scattering of X-rays by passing a monochromatic ( single
wavelength) beam through a thin target material. He observed that photons are scattered to different
angles θ from the main path and the scattered photons have increased wavelengths compared to the
incident photons. He observed that the difference between the wavelength of the scattered photon
(λ’ ) and that of the undeflected photon (λo) increased with the angle of scattering, θ, figure 20
Figure 20: Compton scattering effect
θ = 0°
λo
θ = 45°
θ = 90°
λo λ’
λo
λ’
θ = 135°
λo
λ’
To address the cause of the increase in wavelength, he assumed that photon gives up some
of its energy in the collision but the total energy before and after collision remains conserved. He
visualised radiation to be travelling in photons, as stated by Einstein, and that the scattering process
is the result of a ball-like elastic collision between a photon and a free electron, figure 21.
44
Figure 21: Elastic collision between photo and electron
λ’
scattered photon
λo
photon
θ
φ
electron
A drop in the energy of a photon is expected to lead to an increase in the wavelength since
the two are related inversely from
hc
E =
λ
Using the law of conservation of both momentum and energy, Compton derived a satisfactory
expression which accurately predicts a shift in the wavelength of photons scattered by different
angles, i.e
h
λ’ - λo =
( 1 - Cos θ )
mec
where h is Planck’s constant, m is mass of electron and c is the speed of light.
The value h/mec is called Compton Shift, it is the change in wavelength for θ = 90º
Definition:
Compton effect is a phenomenon in which a shift occurs in the wavelength of a photon when it
collides with a free electron as a result of transfer of part of it energy to the electron.
Note
The Compton’s success, based on the assumptions, provided a further justification for ;
•
•
the Einstein’s idea that light travels in photons
elastic collision between photon and free electrons, which demonstrates the photon’s
particle-like property
45
Example
A photons of light whose wavelength is 2 x 10-7 m collides with a free electron and is scattered at an
angle 60º. Calculate shift in the wavelength and the energy transferred to the electron.
i) Change in wavelength
From
h
λ’ - λo =
thus
λ’ - λo
( 1 - Cos θ )
mec
6.62 x 10-34
=
9.1 x 10-31 x 3 x 108
( 1 - Cos 60º) but
=1.21 x 10-12 m = 0.00121 nm where
Cos 60º = 0.5 = ½
1nm = 10-9 m
ii)Energy lost to the electron
Energy lost to the electron is equal to change in the energy of the photon.Energy of any photon is
calculated from
E = hf
or, in terms of wavelength
E = hc/λ
where
since f = c/λ
from c = fλ
c = 3 x 108 m/s speed of light
Energy before collision depends on the wavelength before collision, λ = 2 x 10-7m
6.62 x 10-34 x 3 x 108
E1 = hf = h c/λ
=
=
E1 =
2 x 10-7
9.93 x 10-19 J
OR, since 1.6 x 10-19 J = 1eV
6.20625 eV
Energy after collision depends on the wavelength after collision, λ’ = 2.0000121 x 10-7m
6.62 x 10-34 x 3 x 108
E2 = hf = h c/λ
=
=
Thus
E2 =
2.0000121 x 10-7
9.92994 x 10-19 J OR, since 1.6 x 10-19 J = 1eV
6.20621 eV
Therefore, energy lost by photon = E1 - E2
= 0.00004 eV
46
Activities
1. What are the meaning of elastic and inelastic collisions
2. Describe the Compton’s experiment and what its results demonstrates.
3. State the assumptions made by Compton in his attempt to explain the
results of his experiment
4. A photon whose energy is 2 x 105 eV strikes a free electron which acquires
an energy of 1.5 x 105 ev in the collision. Find the frequency of the
scattered photon.
Summary
The same way energy of a photon can be converted to kinetic energy of electron in
photoelectric effect, the reverse is possible in x-rays production. Very fast moving electrons can
transfer their kinetic energy to atom on impact and the transferred energy is transformed into x-rays.
The X-rays can be classified as soft or hard X-rays depending on the range of the wavelength,
which depends on the potential difference used in the x-ray tube. Increasing the voltage leads to the
production of highly penetrating x-rays (hard X-rays).
Also, from the Compton effect, when photons are involved in elastic collision with free
electrons part of their energy is transferred to the electron, leading to scattering at different angles
and a shift in the wavelength. This kind of behaviour demonstrates the particle-like behaviour of
photons showing that photons also have the wave-particle duality nature.
Exercise 7
1. Sketch and labelled an x-ray tube and discuss its features
2. What is Compton effect and what does it demonstrate about the nature of photon?
3. A photon of undergoes a 90º scattering on its collision with electron in an atom, determine
the shift in its wavelength
4. A photon experiences a wavelength shift of 0.002nm, determine its angle of scattering.
47
Lesson 8
8.0 Introduction to nuclear Physics
8.1 Introduction
The study of atom now moves from electron to the other particles in the atom. Rutherford’s
experimental results which revealed that protons reside in the nucleus brought the desire to know
much about the nucleus of an atom. This lesson will provide valuable information concerning the
manner in which the nuclear particles co-exist for the stability of both the nucleus and the atom in
general. It will highlight how the need for satisfactory answers about what keeps the nuclear
particles (nucleons) together led to the discovery of new elementary particles originally not
observed. We start by looking at the composition of the nucleus.
Objectives
At the end of this lesson you will be able to
• Discuss sources of the strong nuclear forces.
• Define nuclear binding energy
• Define nuclear mass defect
• Explain the significance of high or low nuclear binding energy
• state theory about factors behind the instability of nuclei of some atoms
• Name new elementary particles believed to be in the nucleus.
8.2 Composition of the Nucleus and Nuclear forces
Atom, from the time of Rutherford, has been known to contain electrons, protons and
neutrons. The existence of neutrons as neutrally charged particles was confirmed by Chadwick in an
experiment. While electrons are said to be maintained in circular paths by the electrostatic force
from protons in the nucleus, the nature of force maintaining protons and neutrons together in the
nucleus of the atom appears mysterious because protons are expected to repel one another while
neutrons are neutral and cannot be held by protons.
The fact that these particles have remained together in the nucleus confirms that whatever force is
holding them together, it must be
(i)
(ii)
a very strong force attractive, otherwise repulsive force due to protons would scatter the
protons
It is also short ranged within the limits of the nucleus, otherwise it may bring even the
electrons into the nucleus if it can reach far.
48
(iii)
It appears charge-independent i.e it acts between proton-proton, neutron- neutron, and
proton-neutron.
Note
•
•
The force that holds the nuclear particles must be a very strong attractive force,
otherwise, protons should not be held together against repulsive force between them
The mystery about the source of such a strong force is that it cannot be electrostatic
since neutrons are neutral particles and cannot be involved in electrostatic force of
attraction, neither is it gravitational in nature because calculation reveals that it is
much stronger than even the repulsive force between the protons.
The strength of the nuclear force can be determined in terms of the amount of energy that it
would require to break it up. It has been observed that when nuclei are formed there is loss in mass
and this, according to Einstein, is an equivalent measure of the energy involved in the formation of
the nucleus. This loss of mass causes nuclear mass defect, where the mass of the nucleus is less than
the total mass of the individual nucleons forming the nucleus.
Definition
Nuclear mass defect is the difference between the mass of the nucleus and the combined
mass of the nucleons (protons and neutrons) forming it.
According to Einstein, the higher the mass defect the higher the energy lost and, by
extension, the stronger and stable the nucleus. The mass-energy conversion is expressed as
E = ∆mc2
The energy equivalent of this mass is known as nuclear binding energy. The nucleus that is
particularly stable is characterised by a high nuclear binding energy.
Definition
Nuclear binding energy is the energy equivalent of the mass lost in the formation of the
nucleus. It is mostly defined as the amount of energy that would be required to break up the
nucleus into its constituent particles.
Worked Example
Consider the formation of Helium nucleus whose atomic number is 2 and mass number is 4 (2He4)
using 2 protons and 2 neutrons. Determine nuclear mass defect and the binding energy.
49
The masses are as follows
1proton ( p ) = 1.007277 a.m.u (atomic mass unit)
1 neutron ( n ) = 1.008666 a.m.u
Actual mass of Helium = 4.001509 a.m.u
1a.m.u = 1.66 x 10-27 kg
It would be expected that mass of Helium, from the masses of its nucleons, be
2p + 2n = (2 x 1.007277) + (2 x 1.008666)
= 4.031886 a.m.u
So, mass lost in the formation of helium is
∆m = 4.03186 – 4.00159 = 0.030377 a.m.u
∴ 0.030377 a.m.u = 1.66 x 10-27 x 0.030377 kg
thus, lost mass ∆m = 5.043 x 10-29kg
∆E = ∆mc2
From
But ∆m = 5.043 x 10-29 x (3 x 108)2 J
so
∆E =
=
45.387 x 10-13J
45.387 x 10-13 eV
1.6 x10-19
∆E = 28.365 x 106 eV
= 28.4 MeV
Activities
1. Workout the nuclear mass defect (∆m) and nuclear binding energy in eV for
Lithium 37Li which is made up of 3 protons and 4 neutrons. Actual mass of
Lithium is 7.016005 a.m.u. proton = 1.007277 a.m.u, 1neutron = 1.008666
a.m.u.
1a.m.u = 1.66 x 10-27 kg
Ans: ∆m = 0.032490 a.m.u,
E = 30.2 MeV
50
8.3 Exchange force theory
While the strength of the nuclear force can be calculated from the nuclear mass defect, the
big question is still about the origin of this strong short ranged force. The source of the strong
nuclear forces was speculated as due to the existence of a third elementary particle in the nucleus,
other than the protons and neutrons, which is being exchanged between the nucleons. It is believed
that, like children playing by exchanging ball, this particle exchange is expected to keep the
nucleons together.
In 1935 a Japanese physicist Hideki Yukawa, was seeking the explanation of the then still
unknown nature of the nuclear forces. In the course of this search, he was led to speculate that there
existed a previously unobserved particle which was being exchanged by the nucleons, figure 22.
That this particle acted as a “glue” which held them together as long as the exchange continued,
thus introducing what would be referred to as exchange force.
Figure 22: Exchange of a third Particle between neutron and proton
nucleus
This new particle was named π-meson or pion. Yukawa postulated its mass to be ≈ 300
times the mass of electron and that π-meson may be positively charged, negatively charged or
neutral and the charge is equal to that of electron. The existence of this particle, hence, confirmation
of Yukawa’s theory, came in 1946. The very reason for which the idea of meson was predicted
required that it interact very strongly with atomic nuclei. Yukawa’s work earned him the 1949
Nobel Prize of physics. Mesons have so far been detected in cosmic rays.
Activities
1. What are nucleons?
2. Describe the theory of what is believed to be the source of the strong
nuclear forces
3. In your view do you think it is justified to accept the idea of π-meson, as
the glue for the nucleons, to be a good scientific idea? Explain your
reason.
51
8.4 The Stability of a Nucleus
This section looks at factors contributing to the instability in the nuclei of some elements.
By calculating nuclear binding energies for many elements, physicists have confirmed that the
binding energy of a nucleus is proportional to the number of nucleons it contains, i.e. the number of
protons plus neutrons combined. If the nuclear energy is divided by the number of nucleons, the
binding energy per nucleon obtained shows that elements with low and high mass numbers have
lower binding energies per nucleons than those of intermediate mass number. Therefore, because
the magnitude of the nuclear binding energy is proportional to the strength of the nuclear force, it is
expected that the nuclei of the lightest and heaviest atoms are less stable than the nuclei of elements
of intermediate atomic mass.
A plot of nuclear binding energy against atomic mass yields a graph in which the binding
energy per nucleon reaches peak then starts decreasing with increasing protons, confirming that
stability start dropping for higher atomic numbers, figure 23 shows roughly the variation of the
binding energy per nucleon among elements.
Figure 23: Variation of the Binding Energy per nucleon with mass number
Binding energy
per nuclei
mass number (No. of protons)
Example
Calculate the binding energy per nucleon for helium nucleus (refer to earlier example for
calculating nuclear binding energy for helium)
The nuclear binding energy for helium was found to be = 28.4 MeV
Helium has 2 protons and 2 neutrons
=
4 nucleons
28.4
Therefore, binding energy per nucleon =
= 7.1 MeV
4
The great majority have a value of about 8 MeV per nucleon. In spite of considerable
binding energy, elements with high mass numbers have a tendency to disintegrate. This may not be
surprising because many protons and neutrons are packed into a tiny volume.
52
It has been observed that all the naturally occurring elements with atomic numbers greater
than 83 have unstable nucleus and would spontaneously decay by emitting radiation which have
effect on photographic plate. A few elements with atomic numbers less than 83 are also known to
be unstable.
The major factors believed to be having influence on the stability of atomic nuclei, other
than mass, which are in agreement with the exchange force theory are;
•
•
the ratio of neutrons to protons in the atom.
The even – odd number ratio relationship between the number of neutrons to the number
of protons.
Of atoms having equal masses, the most stable nuclei are those whose ratio of protons to
neutrons is 1:1. This supports the exchange force theory of nuclear bonding.
Summary
We have heard that in the formation of the nucleus of an atom some mass is lost by
conversion into energy of formation, as a result there is a difference between the mass of the
nucleus and the total mass of the nucleons which come together to form it. The higher the nuclear
binding energy the more stable the nucleus is expected to be, however there are other factors
believed to be behind the stability of a nucleus. It is believed that the strong nuclear forces originate
from the exchange force brought about by particle called π-meson being exchanged between the
nucleons in nucleus and this requires that the ratio of the protons to neutrons be 1:1 for better
exchange and more stability.
Exercise 8
Where necessary use the following values for mass
1 proton = 1.007277 amu, 1 neutron = 1.008666 amu, 1 amu = 931 MeV and
the atomic mass (gram) contains 6.02 x 1023 atoms.
1.
2.
3.
4.
5.
6.
What is nuclear mass defect?
What is the meaning of nuclear binding energy and how does it relate to the mass defect?
How does the magnitude of the nuclear binding energy relate to stability of the nucleus?
Discuss factors believed to be influencing the stability of a nucleus?
Discuss the characteristics/nature of the nuclear forces
Draw a rough sketch of the curve showing variation between the binding energy per nucleon
against mass number s of elements and discuss its features
7. Calculate the binding energy per nucleon in MeV for Sulfure (S=32) which consists of 16
protons, 16 neutrons if the atomic mass is 31.972064 amu
8. Calculate the binding energy per nucleon in MeV for carbon 12 which consists of 6 protons
and 6 neutrons if its atomic mass is 12.000 amu
53
Lesson 9
9.0 Radioactivity
9.1 Introduction
This lesson will look at the system and rate of disintegration of the unstable nuclei. It has
been observed that the nuclei of some elements disintegrate spontaneously by emitting some
particles and energy. This spontaneous disintegration, which is known as radioactivity, leads to
transformation of the atom of an element into a new element, and not all the atoms transform into
the new element at once, a few may be transformed today while others remain for several year
before they become new atoms. This implies that, with time a radioactive substance will exist with a
mixture of atoms of different elements in different proportions. It is necessary to know how, after
some time, one can get the proportions of different atoms in a mixture which starts as a pure sample
of one type of atoms only.
Objectives
At the end of this lesson you will be able to
• Define radioactivity
• Name products of radioactivity.
• define and calculate half life
• distinguish between fusion and fission
• name some of the applications for the products of radioactive decay
• state factors influencing rate of radioactivity.
9.2 Process of Radioactivity
Radioactivity is the spontaneous breakdown of the nuclei of atoms by emitting some
particles and gamma rays so as to become stable. In radioactivity an element transforms itself into
another element or isotope when it emits particles which lead to a change in the atomic number. The
transformation, however, is gradual, not all the existing atoms in a given mass of a radioactive
substance change simultaneously into new atoms.
A typical example of transformation reaction is
88Ra
226
86Rn
222
+ 2He4 + energy
Radium emits helium and in the process a new element, radon, is formed. The superscript is
the mass number of an element while the subscript is the atomic number. The equation follows law
of conservation of mass and charge, in which the total mass of products equals mass of the
54
radioactive element and the total charge(atomic number) of the products add up to charge on the
radioactive element.
Example
Find x and y in the nuclear reactions represented by
226
xP
=
y
86Z
+ 2He4 + γ
i) by law of conservation of mass, the superscripts are used as follows
226 = y + 4
y = 226 – 4
y = 222
thus
so, element Z can be expressed as
222
86Z
i) by law of conservation of charge, subscripts are used as follows
x = 86 + 2
x = 88
thus
Hence, element P can be expressed as
226
88P
Activities
1. An atom is represented as 84X218, what do the numbers 218 and 84 represent?
2. State the laws obeyed by equation of nuclear reaction
3. Find x and y in the nuclear reactions represented by
y
226
+ 2He4 + γ
88Ra
xRn
xRn
y
84Po
218
+
2He
4
+ γ
4. How much energy in MeV is evolved in the decay of an 86Rn222 nucleus? Atomic
masses of the isotopes: 86Rn222 = 222.0165amu; 84Po218 = 28.0079amu; 2He4 4.0026amu.
The common radiations emitted by radioactive elements are;
•
•
•
Alpha particles (α) – these are positively charged heavy particles composed of 2 protons
and 2 neutrons.
Beta (β) particles – they are negatively charged particles, mass equivalent to that of
electron.
Gamma (γ) rays – these are high energy electromagnetic waves, the same kind of
radiation as visible-light but of much shorter wavelength, they are not charged.
55
These radiations can be distinguished from one another by passing them through magnetic field or
electric field, which would separate them by deflection in different directions according to charge.
Charged particles are always deflected in both electric and magnetic fields.
9.3 Magnetic field effect on emitted radiation
In the case of magnetic field being used, the deflection of the charged particles take place
according to Fleming’s left hand rule which requires the thumb, forefinger and middle finger of the
left hand to be set at right angles to each other according to figure 24. The forefinger is set to point
in the direction of the magnetic field (into page), the middle finger in the direction of the
conventional current- which can be the direction of the positive charge or direction opposite the
flow of negative charge-the thumb then shows the direction of the deflection of the charge in
question. Gamma rays are not charged and, therefore, are not deflected.
Figure 24: Fleming’s left hand rule
current
field
Force
Note
The Middle finger which points to the direction of the current should be set
•
•
To point in the same direction as that of the movement of the positive charges – for
positive
To point in the opposite direction to that of the movement of negative charges-for
negative charges
Consider a case where magnetic field is directed into the page of a book while the emitted
particles are directed along the y-axis as shown in figure 25, then the deflection of the positive and
the negative charges, according to Fleming’s left hand rule, will be as illustrated in the figure.
56
Figure 25: Deflections of radiations by a Magnetic field
α (alpha) current (+)
Field
y-axis Magnetic Field
α (alpha)
deflection
α (alpha)
positive
γ gamma
β (beta)
negatively charged
magnetic field
Field
Lead
Container
β (beta)
deflection
current (+) for case of
β (beta)
Beta (β) – due to their negative charge, they are deflected to the right in this setup. Also, because
they are very light particles, they are deflected so much out of the y-axis.
Alpha (α) – being positively charged, they are deflected to the left and, because they are heavy
particles, they are deflected slightly out of the path.
In addition to the behaviour observed in the magnetic field, the speed of beta (β) is near that
of light and, therefore, are highly penetrating than the alpha (α) which are stopped by even a few
centimetres of air. A thin sheet of paper stops the alpha particles because they are heavy and slow in
speed. The gamma rays are high energy rays, as a result, they even penetrate through a few
centimetres of steel metal.
57
9.4 Applications of Emitted radiations
Radiation from natural radioactive sources can kill plant seeds, bacteria and even small
animals. Controlled radiation is used to destroy diseased tissues in the treatment of cancer and
certain skin diseases.
The emission of any of these particles from a nucleus, however, is a haphazard event. It is
not possible to predict the moment when an unstable nucleus is likely to emit them. However,
radioactivity has been observed to be a process which follows a certain law of decay from which the
number of nuclei that will have undergone decay after a certain time interval can be predicted. The
rate of radioactivity does not depend on any other factor apart from the existing number of
undecayed nuclei i.e things such as temperature or pressure do not influence the rate of decay.
9.5 Decay law
If N is the number of undecayed nuclei at a given time then this number will reduce at a rate
proportional to the number N. As the number of the undecayed nuclei reduces the rate of decay also
decreases and this can be expressed in equation form as follows;
dN
∝
-
N
negative shows that rate of decay decrease with time
dt
The equation leads to
dN
= λN
where λ is called decay constant
dt.
This equation can be integrated to find expression-relating remaining undecayed nuclei (N)
to starting number (No).
-
dN
= - λdt
expressing as
N
The limits of integration become No to N
N
thus
1
- λ dt =
dN
N
No
N
Leading to
and
loge N
= - λt
No
loge N – log No = - λt
58
N
loge
= - λt
No
N = e- λt
No
leading to
giving
N = No e- λt
From this equation, it is clear that radioactivity is not a linear process but an exponential
decay, due to e- λt, in which N decreases with time. This can be represented by a decay curve of the
type shown in figure 26.
Figure 26: Decay Curve
No. of undecayed
nucleii (N)
N’
t’
time
From such a decay curve, one can determine the number of remaining undecayed nuclei
after some time (t), e.g. after time t’ number remaining is read as N’ . The quantity of undecayed
nuclei can be expressed in grams or number of atoms as can be calculated from the Avogadro’s
constant 6.02 x 1023 . The number of undecayed nuclei can also be determined by calculation, using
the exponetial decay equation and it can be expressed in terms of the number of undecayed nuclei or
in mass.
Example 1 (using nuclei )
Radium 88Ra226 has a decay constant λ = 1.36 x 10-11 s-1, determine how long it can take 2g to
reduce to 1g .
Converting mass to the number of nuclei undergoing decay, from Avogadro’s number,
atomic mass of radium 226g has 6.02 x 1023 nuclei
2 x 6.02 x 1023
so,
and
No = 2g has
N = 1g has
= 5.32 x 1021 nuclei
226
2.66 x 1021 nuclei
59
N = No e- λt
From
21
2.66 x 10
= 5.32 x 10
21
-1.36 x 10-11 t
e
-1.36 x 10-11 t
1
=2 e
take natural logarithm of both sides, to help eliminate ‘ e’
- 1.36 x 10-11 t
loge 1 = loge 2 e
- 1.36 x 10-11 t
loge 1 = loge 2 + loge e
loge 1 = loge 2 - 1.36 x 10-11 t loge e but loge e = 1
loge 1
1.36 x 10-11 t
= loge 2 -
1.36 x 10-11 t = loge 2 - loge 1
but loge 1 = 0
⇒ 1.36 x 10-11 t = loge 2
⇒
∴
t =
t =
loge 2
1.36 x 10-11
0.693
1.36 x 10-11
but
loge 2 = 0.693
= 5.1 x 1010 sec = 1620 years, to the nearest thousand
Example 2 ( using mass)
A radioactive element has a decay constant λ = 1.36 x 10-11 s-1, determine how long it takes for its
mass to reduce from 2g to 1g .
From
N = No e- λt
No = 2g,
N = 1g
-1.36 x 10-11 t
⇒
1=2e
take natural logarithm of both sides, to help eliminate ‘ e’
- 1.36 x 10-11 t
loge 1 = loge 2 e
- 1.36 x 10-11 t
loge 1 = loge 2 + loge e
loge 1 = loge 2 - 1.36 x 10-11 t loge e but loge e = 1
loge 1 = loge 2 -
1.36 x 10-11 t
60
1.36 x 10-11 t = loge 2 - loge 1
⇒ 1.36 x 10-11 t = loge 2
⇒
t = loge 2
1.36 x 10-11
∴ t = 0.693
since loge 2 = 0.693
-11
= 5.1 x 1010 sec = 1620 years
1.36 x 10
We note that it take 1620 years for 2g of radium to reduce to 1g, whether calculated by mass or in
terms of number of nuclei.
Activities
Work out time it would take the following masses of radium (λ = 1.36 x 10-11 s-1 ) to
reduce to the shown values, do by number of nuclei and by mass and compare the
obtained values.
i)
ii)
iii)
From 1g to 0.5g
from 0.5g to 0.25g
from 100g to 50g.
You probably noticed that it take the same period to reduce 1g to 0.5g, 0.5g to 0.25g or 100g
to 50g. That is, the same period to reduce any given mass or number of the nuclei of a radioactive
substance to half the original value. This time is known as half life, It varies for different
radioactive elements and ranges from fractions of seconds to years. An important way of
conceptualising and comparing the rate of radioactivity of any radioactive substances is by half life.
Definition:
half life is the time taken by a given number of the nuclei of a radioactive substance
to decay by half
The half life of radium 88Ra226 which is about 1620 years means that radium decays at such a
rate that of 2g of radium which existed 1620 years ago, only 1g remains today, the rest have been
transformed into another element; and 1620 years from now, only half of this present amount
(0.5 g) will remain. Half life of a radioactive element can be read from the decay curve of a
radioactive substance or it can be calculated from
N = No e- λt
61
Let time taken to reduce the number of nuclei No to ½No be T½,
½No = No e- λT½
Implying
Leading to
1 = 2 e- λT½
loge 1 = loge 2 - λT½ loge e
and
λT½ = loge 2 - loge 1
T½
=
loge 2
but
loge 2 = 0.693
λ
In general, half life of any substance is given by
0.693
T½ =
λ
Thus half life of any radioactive substance can be calculated from the expression so long as
its decay constant (λ ) is known.
Activities
1. In your opinion, is it possible for all the atoms in 2g of radium to transform
themselves by radioactivity process so as to have a completely pure new
element?
2. Name some activities/events you know whose change with time follows the
exponential decay law
Summary
Unstable nuclei of atoms can spontaneously breakdown by emitting some particles and gamma rays
so as to become stable in a process called radioactivity. In radioactivity an element transforms itself
into another element or isotope when it emits particles which lead to a change in the atomic number.
The rate of radioactivity does not depend of any other factor apart from the existing number of
undecayed nuclei.
62
Exercise 9
1.
2.
3.
4.
5.
What is the meaning of half life of a radioactive substance?
Write the expression for determining half life and identify terms in it
What is the decay constant of the isotope 86Rn222 which has a half life of 3.82 days?
Determine the decay constant of 92U238 if its half life is 4.49 x 109 years.
Find x and y in the nuclear reactions represented by
x
i) 82Pb214
+ -1e0 + γ
yBi
238
92U
6.
+
1
0n
x
yU
Calculate the energy evolved when a 82 Rn214 nucleus decays. Atomic masses of the
particles involved in this decay: 82 Rn214; 83Bi214 = 213.9972amu; -1e0 = 0.0005amu.
63
Lesson 10
10.0 Radioactivity Model and Other nuclear reactions
10.1 Introduction
In this lesson we’ll complete our study of the introduction to nuclear physics, by looking at
the model of radioactivity and other forms of nuclear reaction. Our study will then move to the
theory of relativity in which we’ll be concerned about the effect of high speed relative motion on
the measurement of physical quantities and how to relate laws of physics to any differences caused
by relative motion.
10.2 Radioactivity Model
Water from a Leaking container behaves so much like a decaying radioactive substance. The
rate of leakage depends on the water level above the leakage point.
It can be noted that at start of leakage, much like start of radioactivity, the rate of drop of
water level is high and the rate decreases as the water level drops. If one measures the rate of drop
of water level in a container with a hole made at near the bottom, by measuring water level after
some time interval, it can be observed that it follows the decay curve.
Water level /cm
Time taken /sec
64
Activities
Project
•
•
•
•
•
•
•
•
•
•
•
Take transparent plastic bottles, about 1 litre
Make a hole at the side, near bottom of the bottle, use nails, different
diameters for different bottles (2 inch , 3 inch or 4 inch nails)
Have a 1ft ruler for checking the change in water level
Leave the bottle top open throughout this experiment
Fix ruler, using cello tape, the 0 cm mark down and 30 cm mark up, let
the 28 cm or any convenient mark be the starting point for reading
Prepare a table for recording values as shown in table 1 below
Set your watch /clock to zero (0:00) for readiness
Pour water to a level above the starting mark, as shown
Start timing once the level drops to your chosen starting point
Record time in seconds and water level after every drop of level by
some convenient uniform amount, e.g 4 cm or 5cm (what you can cope
with)
Plot a graph of water level against time
i)
Discuss the shape of the graph and compare it to normal radioactivity curves
1
ii)Table From
the graph determine time taken for water level to get to half the stating
point
Time
seconds
Reading
iii)
From the
graph
still, determine time taken
for various water levels to reduce to
/
cm
half their values and compare them
iv)
What do0 you conclude about the time taken28for any level to drop to half its
5 is this time called.
value, what
10 hole sizes, use half life to determine decay constant using the
v)
For different
general half life equation.
Repeat this with bottles of different holes and compare them to find if shape of the
graph and time for half decay depend on the size of the hole
10.3 Other Nuclear Reactions
It is recognised that apart from radioactivity there are other ways by which nuclear energy
can be released, with the consequence of increased stability. A process in which light-weight nuclei
can combine to form one heavier nuclei is one means of liberating energy to improve stability. This
process, often referred as fusion, is believed to be the source of sun’s energy.
There is also an opposite reaction to this, in which one heavy nucleus is split into smaller
nuclei on bombardment with neutrons. This process, in which a heavy nucleus splits into smaller
stable nuclei, is called fission. In 1939 it was discovered that nuclei of the uranium 92U235
65
sometimes split into two smaller nuclei when bombarded with neutrons. The most striking aspects
of fusion and fission reactions is the magnitude of the energy liberated, which gives the impression
that nuclear reactions are potential means of generating energy. In whatever reaction, the equation
of reaction obeys the law of conservation of mass and charge as stated earlier. The two terms, fusion
and fission, are usually confusing, however, they can be differentiated from the ‘S’ and ‘SS’.
Note
•
•
Fusion is a nuclear reaction in which two or more nuclei of light weight are joined
together to produce one heavy nucleus. hence, its single ‘S’
Fission is a nuclear reaction in which one heavy nucleus splits into two light weight
nuclei. hence, its double ‘SS’
Activity
1. Describe different nuclear reactions by which energy liberation and nuclear
stability can be realised
2. Explain the difference between radioactivity and these other nuclear reactions
10.4 Introduction to the theory of Relativity
As stated under section 10.1, the subject of relativity is concerned with the effect that
relative motion and high speed would have on measurements of physical quantities and laws of
physics as would be reported by different observers. It is believed that if different people were to
describe their experience of an event going on in a frame in relative motion to one of them then
their descriptions would be quite different.
Example
The trajectory of an object dropped from an aeroplane flying horizontally at some height above the
ground appears different to the observer on earth from how it appears to an observer in the same
aeroplane, figure 27.
66
Figure 27: Trajectory of an object dropped from aeroplane
To observer on ground
To the observer in the aeroplane
a) Parabolic curve
b) linear vertical motion
Note
•
An object dropped from a moving frame continues moving in the same direction as
the frame and with the same speed as the frame. This is why if one tries to alight
from a moving vehicle he falls towards the vehicle’s direction of the motion. This
applies to an object dropped from an aeroplane.
The practical trajectory of a projectile fired from earth’s surface in the north-south direction
is a parabola to observer on earth, both the observer and the projectile are in motion at the same
speed in the west-east direction, figure 28.
Figure 28: Trajectory of a projectile fired in the north-south, as seen from earth
67
It is clear that results of events depend on the relative motion between frames and, hence,
our perception is expected to be influenced by the magnitude of the relative motion between our
frames and that of the event.
Activity
Describe trajectory of the projectile fired on earth in the north-south direction as it
would appear to an observer outside the earth.
To scientists, the effect of relative motion should be worrying since it would appear to
present different results and, hence, different laws for same scientific events. The study of relativity
is aimed at establishing the effect of high speed relative motion on the measurement of quantities of
interest and how to relate laws of physics to any differences caused by relative motion.
10.5 Effect of relativity on Basic physical quantities
For over 200 years, the Newton’s equations of motion were believed to be so accurate in
their description of nature and laws that govern results of measurements of various quantities,
especially the basic ones, mass, length and time and it was never imagined such measurements
would be influenced by difference in speed between an observer and the frame of the quantity.
When it became clear that relative motion between frames at very high velocities have influence on
our observation of various quantities, it became essential to try to either identify a universal frame
of reference in which experimental results would not vary or identify the relationships which can be
used to standardize the results. It emerged that there was no such frame of reference.
To Einstein, a universal frame of reference in which laws of physics would be the same was
not necessary, he believed that laws of physics would be the same between all frames of reference
so long as there is no relative motion between them. And, if there is relative motion, influence
would depend on whether the frames are at constant relative speed or at accelerated relative motion
In 1905, at 26 years only, Einstein developed a special theory of relativity applicable only to
inertial frames of reference. These are frames of reference moving at constant velocity with respect
to one another.
He stated that;
All laws of physics take the same form in all inertial frames of reference
That is, according to Einstein, our measurements of quantities and events in any frame of
reference should be the same unless relative motion between the frame of the measured quantity and
that of the observer somehow changes the characteristics of the measuring instruments. e.g the
perceived decrease in the length of a meter rule at a high relative velocity with respect to an
observer is pronounced and there may be need for correcting the value obtained with it. Thus,
68
transformation expressions for events observed from different frames of reference at relative motion
were deemed necessary.
Summary
This lesson has looked at other forms of nuclear reaction, fusion and fission, which involve a
large magnitude of the energy liberated and making nuclear reactions a potential means of
generating energy. These forms of reaction also obey the law of conservation of mass and charge.
We have also discussed how effect of relativity on perception of various physical quantities depend
on the relative velocity between our frame of reference and that of the event under observation
Exercise 10
1. What is the difference between fusion and fusion?
2. Discuss the main concern of the theory of relativity and explain whether this concern is
justified
3. Discuss Einstein’s theory of special relativity and what it expects the laws of physics
4. What are inertial frames of reference?
69
Lesson 11
11.0 Relativistic equations
11.1 Introduction
In the previous lesson it was mentioned that Einstein asserted that all laws of physics remain
the same in all frames of reference unless there is relative motion between the measured event and
the observer taking the measurement. That is, the laws are independent of the nature of the frame
except that it is our perception that will not be the same if the frame of the event is moving at a very
high velocity relative to our frame. In this lesson we’ll look at how the laws of physics should be
corrected to conform with our perception when there is relative velocity between frames of the
event and the observer. We’ll compare two systems of transformation, in which one treats
transformation system as independent of the magnitude of velocity, low or high while the other
takes the system of transformation as dependent on the magnitude of relative velocity.
11.2 Newtonian Transformation
Under this system of transformation the effect of relative motion on our perception was
assumed to be independent of the magnitude of the relative velocity and the transformation
equations were generalised. Consider two observers whose co-ordinate axes are parallel to one
another. Let the two frames be S and S’, in which S’ is moving at a constant relative velocity with
respect to S in the positive x-direction, figure 29. Assume that at time t =0 the two frames have yaxes at the same starting point.
Figure 29: The two frames of reference at time t = 0
y
S
y’
S’
x’
x
O
z
z’
If after some time t a particular event is recorded by the observer in S as it occurs at point P
in S’. The observer in S will see the location of the event, with respect to his co-ordinates,
as (x,y,z t). To the observer in S’ the event will be occurring at (x’,y’,z’ t) with respect to his coordinate system.
70
Figure 30: The two frames of reference after time t
y
y’
S
S’
A
B
P (x,y,z,t) and (x’,y’,z’,t)
vt
x’
x’
O
O’
x
z’
z
Here, the event is taking place in frame S’ and x,y,z,t are co-ordinates as seen in the frame S.
The co-ordinates x’,y’,z’,t’ are those taken in the frame ( S’), same frame of the event. The observer
in S’ is moving at the same speed as the event while to the observer in S the event is moving faster
by v, i.e relative velocity between the two is v
The location of P in the frame S, after time t, can be seen as
x =AB + BP
AB = distance moved by S’ in the time t = relative speed( S’) x time = vt
BP is the distance of P from y’-axis of frame S’
In terms of x-co-ordinate on the S’, i.e BP = x’
So,
x = vt + x’
or conveniently expressed in the form
x’ = x - vt
This shows that along the direction of motion co-ordinates reported will be dependent on the
relative velocity (v). For other axes, there is no motion along y-axis or z-axis, therefore, readings of
the corresponding co-ordinates are;
x = vt + x’
y = y’
z = z’
t = t’
or
or
x’ = x - vt
y‘ = y
z’ = z
t’ = t
The expressions obtained are called ‘Newtonian transformation’ equations between the coordinate systems of the two frames of reference. The equations suggest that apart from the quantity
along the direction of motion which requires a correction factor (vt), quantities along other axes
perpendicular to that of motion and the time-scale are not affected by the relative motion between
the observers.
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Activity
Derive the Newtonian transformation co-ordinate system and discuss the implication
of the results in terms of the how the values observed by different observers relate to
one another
From the predicted effect of relativity on the co-ordinates of an event expected between
different observers, the prediction of Newtonian transformation about the effect of relativity on
lengths and time as basic physical quantities can be determined.
11.3 Effect on Length
Consider an object lying between two points in a moving frame S’, whose co-ordinates are as
shown in the moving frame of figure 31.
Figure 31: Object moving in the direction parallel to its length
S
S’
x1
x1
x2
x2
If the time of reading the co-ordinates of the ends of the object is the same, simultaneous, the
transformation will be as follows;
x’1 = x1 – vt
x’2 = x2 – vt
So, the length between the ends is
x’2 - x’1 = ( x2 – vt) – (x1 – vt)
x’2 - x’1 = x2 - x1
The results show that the left hand side equals right hand side. Thus, according to the Newtonian
transformation, the length of an object will neither be dependent on the relative velocity between
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the frame of the object and that of the observer nor its magnitude. Meaning that the two observers
will report equal length after using their own co-ordinates..
This implies that under the Newtonian transformation, the reported length of an object
should not be influenced by relative motion between two frames, as a result, observers in different
frames with a constant relative velocity between them are expected to report same length. It is also
noted that the magnitude of velocity does not matter in this case. According to Newtonian
transformation, the following quantities are not affected by relative motion between the frames
i.
ii.
iii.
Length along the direction of motion – x-axis
Length perpendicular to the direction of motion –y and z-axes
Time
Note
•
According to Einstein, laws of physics take the same form in all inertial frames of
reference and the Newtonian transformation agrees with this for the behaviour of
length since two observers at constant relative velocity report same length.
Einstein on the other hand postulated that for very high relative speed the prediction of the
Newtonian transformation fails to give correct co-ordinates and that there can’t be equality between
the time co-ordinates t and t’ as in the Newtonian transformation. That is, the magnitude of the
relative velocity is a factor to consider. He found that appropriate transformation was one which had
already been derived by Lorentz (Dutch physicist), now known as Lorentz transformation which
takes the form
x – vt
x’ =
1 – v2 / c 2
y’ =
y
z’ =
z
t – vx/c2
t’ =
1 – v2/ c2
It is noted that a new factor, (1 – v2/ c2 )½, is introduced to modify the Newtonian transformation
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11.4 Lorentz Transformation and its implications
First, it is noted that if v is very small compared to the velocity of light c, the Lorentz
transformation approximate to the Newtonian transformation. i.e if velocity is much less than that
of light ( v << c), the quantity v2/ c2 will be so small ( ≈ 0 ) because c is so big.
x – vt
so,
x’
=
1– 0
x – vt
=
1
x’
= x – vt
The agreement between the Lorentz transformation and the Newtonian transformation for
velocities far much less than the speed of light is necessary because the Newtonian equations hold
to high degree of precision for low relative velocities that all other theories should agree with their
predictions at those low velocities.
To determine the effect of relative motion on the length of an object, by Lorentz
transformation, let the co-ordinates as observed in the different frames be as shown, figure 32
Figure 32: Object moving in the direction parallel to its length
S
S’
x1’
x1
lo
x2’
x2
By Lorentz transformation
x’1
=
x1 – vt1
1 – v2/ c2
x’2
=
x2 – vt2
1 – v2/ c2
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So, the length between the ends is
x’2 - x’1
= (x2 – vt2)
1 – v2/ c2
x’2 - x’1
–
(x1 – vt1)
1 – v2/ c2
= (x2 – x1) – vt2 + vt1
1 – v2/ c2
If the readings of the co-ordinates are taken simultaneously.
t2 = t1
So,
– vt2 + vt1 = 0
Hence,
x’2 - x’1 =
(x2 – x1)
1 – v2/ c2
The observer in the same frame with the object is the one whose perception of the length of the
object is not affected by the motion, the actual length of the object can be taken to be
lo = x’2 - x’1
So,
x’2 - x’1 = lo =
(x2 – x1)
1 – v2/ c2
Thus, according to the observer who records from the outer frame, the co-ordinates of his frame
would show him that the length of the object in relation to the actual length is
(x2 – x1) =
lo
1 – v2/ c2
Let it be l
Thus
l = lo
1 – v2/ c2
i.e if v ≈ c the term v2/ c2 ≈ 1 and, hence, 1 – v2/ c2 ≈ 0
The new length, as noted by the observer outside the frame of the object, would be less then
the actual length, i.e the object appears shorter to the observer outside that frame. That is, for
extraordinary velocities, the Lorentz transformation predicts ‘length contraction’ according to an
observer outside the frame in motion. Thus, l
0 (tends to zero) as v increases towards velocity
of light (c).
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Example
Show that the length of a 1m stick would appear to reduce to near zero if it moves at a velocity
approximately equal to that of light past a stationary observer.
Since, the observer is stationary, relative velocity (difference in velocity) is just equal to the velocity
of the stick.
v ≈ c
Let velocity be
Then from
l = lo
1 – v2/ c2
v2/ c2 ≈ c2/ c2 ≈ 1
hence,
l ≈ lo
1–1
l ≈ 0
Activity
i) Derive the Lorentz transformation expression for the length of an object moving in
the direction parallel to its length at a velocity v past a stationary observer
ii) Explain conditions under which the Lorentz transformation would reduce to
Newtonian transformation
ii) Explain what would happen to our perception of moving buses if the speed of light
would suddenly drop to the speed of a bus?
Summary
Under the Newtonian transformation, the length of an object should not be influenced by
relative motion between two frames with a constant relative velocity between them. It is also noted
that the magnitude of velocity does not matter in this case. According to Newtonian transformation,
the following quantities are not affected by relative motion between the frames.
i.
ii.
iii.
Length along the direction of motion – x-axis
Length perpendicular to the direction of motion –y and z-axes
Time
It has been stated that the Newtonian transformation does not agree well with the observed
values for the physical quantities observed under very high velocities, as a result a new system of
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transformation called Lorentz transformation is introduced for high velocity. For extraordinary
velocities, the Lorentz transformation predicts ‘length contraction’ according to an observer outside
the frame in motion. The agreement between the Lorentz transformation and the Newtonian
transformation occurs for velocities far much less than the speed of light.
Exercise 11
1. A stick of length 2m is moving past an observer at a relative velocity of 2.7 x 108 m/s.
What is the length detected by the observer.
2. Write expressions comparing the Newtonian and Lorentz transformations of co-ordinates of
one frame of reference to the other.
3. Show that for Newtonian transformation the length of an object reported by observers in
different frames at constant relative velocity v, one outside the frame of the object and the
other moving with the object in the same frame, is the same and independent of the relative
velocity between them.
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Lesson 12
12.0 Relativistic effect on time and wave nature of light
12.1 Introduction
We recall Einstein’s postulate that there can’t be equality in the time taken by event as may
be reported by different observers whose frames of reference are at high relative velocity with one
another, which is a contradiction of the Newtonian transformation. In this lesson we’ll look at the
Einstein’s postulate, through the Lorentz transformation. We’ll also look at how the subject of
relativity opened new analysis on the behaviour of light wave, this will be considered through
Michelson and Morley experiment which became so much helpful in proving the nature of light as a
wave which does not require any medium for its propagation.
12.2 Effect of relativity on time
From the Lorentz transformation, it can be shown that if observers in different frames at
relative motion with one another are recording time taken by an event in one of the frames, their
perceived time taken will be different and are related by
t
=
to
1 – v2/ c2
where t - is time noted by observer outside the frame of the event
to - is time noted by observer in the frame of the event
According to the observer outside the frame of the event, time taken by a given event
increases as the speed of the frame of the event increases. That is, to the observer outside the frame
of the event, the observed event is seen to be taking long , i.e it appears to be slow. This perceived
slowness of the event is known as ‘time dilation’
Activity
•
Derive the expression for the relationship between the time taken by any event
recorded by two observers in different frames, one being in the frame where the
event is happening, as expected by the Lorentz transformation of co-ordinates.
Say t1 is the start and t2 is end of the recorded event.
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12.3 Effect of relativity on Mass
The theory of relativity revolutionised many areas of physics. For the case of mass, which is
a third basic quantity, it had been assumed that it is a constant quantity and the laws of physics
which depend on mass have been expressed with that assumption. For example the Newton’s
second law has been and is being expressed, without regard for the effect of high speed, as
F = ma
However, in the Einstein’s theory of relativity, mass like length and time has a value which depends
on the relative velocity (v) between observers and, for particles whose velocities approach that of
light, it takes the form
mo
m =
1 – v2/ c2
where
mo - is the mass when the object is at a rest (rest mass) or when there is no relative motion with
respect to an observer.
v – is the relative velocity between the observer and the object.
c – is the speed of light 3 x 108 m/s.
From the expression above, If v << c such that v2/ c2 ≈ 0 then m = mo and if v is increased to
near c, mass increases and the ratio of m/mo increases.
Activity
•
•
•
•
Write the Einstein’s equation relating mass to relative velocity?
Identify each of the terms in the equation and state their units of measurement
Discuss implication of the equation about the influence of relative velocity on
mass
What is the ratio of m/mo of a body whose velocity is 40% of the velocity of
light?
Example
If electron is accelerated at a velocity 1.5 x 108 m/s what would be its observed mass. Speed of
light is 3 x 108 m/s and mass of electron is 9.1 x 10-31 Kg
From
m =
mo
1 – v2/ c2
9.1 x 10-31 Kg
m =
1 - (1.5 x 108/3 x 108)2
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= 10.5 x 10 –31 Kg
12.4 Light wave and the theory of relativity
The issue of relativity also revived interest in the wave theory of light. Initially, it was
believed that the speed of light would depend on the properties of the medium in which it travels,
liquid or gas. Because every wave was associated with some form of medium for propagation, every
physicist firmly believed that light required a medium for its propagation and that there must have
been some medium assisting it. A medium called ether was proposed to be for light waves.
This medium had to be transparent and completely at rest, although if earth moves then, like
breeze in opposite direction, ether should be expected to have effect on the light wave. Thus, the
relative motion between the earth and the ether, like any other medium that affects any wave which
depends on it, was expected to have effect on light.
12.5 Michelson-Morley Experiment
In 1887 Michelson and Morley used an interferometer, of the type originally designed by
Michelson, to attempt to detect effect of relative motion between earth and ether on light as a wave.
The interferometer was to detect any slowing down or speeding up of the light wave by ether. This
instrument, in effect was to compare the velocities of light along two equal paths set at right angle
with each other, SM1 and SM2. A ray of light is split into two by a ray splitter (S) and allowed to
move to two reflecting mirrors, M1 and M2, to reflect and send them back to be analysed for
differences in speed, figure 33.
Figure 33: Setup of the Michelson –Morley experiment
M1
M2
s
Continuous spectrum
It was anticipated that for the direction parallel to the earth rotation the velocity would
undergo changes in its velocity due to effect of ether. By interference principle, such a difference
should make waves along SM1 and SM2 be out of phase with one another when they arrive back,
due to difference in time taken. Any such effect was expected to reveal itself through appearance of
fringes (alternating bright and dark regions). However, no fringes were observed even after rotating
the mirrors in such a way that the ray parallel to the direction of rotation of the earth is sent against
the direction of rotation of the earth. The no-fringes results indicated that there was no shift in the
velocity of light caused by relative motion between earth and the ether. The negative results
suggested that there was no effect of ether on the light waves.
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At first it did not occur to physicists to give up the notion of the ether. However, the baffling
negative results was significant and Einstein realized that the negative results would be expected if
the relativistic equation is introduced to the Newtonian mechanics. It followed from this relativistic
theory that the speed of light is the same whether it is moving in ether, against ether or
perpendicular to the ether. Here again Einstein postulated that velocity of light is independent of
relative motion between its frame of reference and any other. It also implied that light does not need
a medium such as ether for its propagation. This postulate was summarised as follows;
The speed of light in a vacuum, relative to any inertial frame of reference, is constant.
This was a major triumph for both the theory of relativity and the wave theory of light.
Summary
It has been noted that under the Newtonian transformation time is not supposed to be
influenced by relative motion between frames of reference, however, Einstein, through the Lorentz
transformation, postulated that time is also affected by high relative velocity, leading to time
dilation. He also gave expression for the change that would be reported if a given mass moves at a
very high relative velocity past an observer. The idea of relativity also led to assessment of the
nature of light waves which had for sometime been believed to be like any other mechanical waves
whose propagation should have depended on ether as a medium. In the Michelson and Morley
experiment, it was confirmed that light waves did not depend on any medium for its propagation, as
its speed remained the same even in a vacuum.
Exercise 12
1. What is time dilation?
2. Discuss the ideas or reasons behind the medium called ether.
3. Describe the Michelson-Morley experiment, their findings and implication of their findings.
4. Calculate the mass of electron whose velocity is 80% that of light.
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APPENDIX
Important physical quantities and constants that may be useful
Electron mass me
Electron charge
Mass of proton
Permittivity of free space
Planck’s constant
Avogadro’s Number
1eV
Speed of light
1 atomic mass unit (a.m.u)
= 9.1 x 10-31 kg
= 1.6 x 10-19 C
= 1.7 x 10-27 kg
= 8.85 x 10-12 F m-1
= 6.62 x 10-34 Js
= 6.02 x 1023 mol-1
= 1.6 x 10-19 J
= 3 x 108 m/s
= 1.66 x 10-27 kg = 931 MeV
ANSWERS
Answers to Odd-numbered Problems
LESSON 3: Exercise 3
1) For n = 3;
For n = 4;
radius = 4.77 x 10-10 m, Energy = - 1.51 eV
radius = 8.47 x 10-10 m, Energy = - 0.85 eV
3) Wavelength = 6.58 x 10-7 m
LESSON 5: Exercise 5
LESSON 8: Exercise 8
5) 0.8275 eV
7) 2.35 eV
9 (i) Stopping potential = 0.45 volts.
(ii) Kinetic energy = 0.45eV
(iii) velocity of electron v = 3.98 x 105 m/s
7) Energy = 263.5 MeV
LESSON 6: Exercise 6
LESSON 9: Exercise 9
3) p.d = 3760 volts
3) decay constant λ = 2.1 x 10-6 s-1
5) (i) x = 214,
y = 83
LESSON 7: Exercise 7
LESSON 11: Exercise 11
3) Change in wavelength = 0.0024 nm
1) Observed length L = 0.872 m
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