Cambridge
International AS and A Level
Sciences
2014
SAM
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A Level
17 Selection and evolution
Charles Darwin and Alfred Russel Wallace proposed a theory of
natural selection to account for the evolution of species in 1858.
A year later, Darwin published On the Origin of Species providing
evidence for the way in which aspects of the environment act
The variation that exists
within a species is
categorised as continuous
and discontinuous.
The environment has
considerable influence on
the expression of features
that show continuous (or
quantitative) variation.
as agents of selection and determine which variants survive and
which do not. The individuals best adapted to the prevailing conditions succeed in the ‘struggle for existence’.
By the end of this topic you should be able to:
a) describe the differences between continuous and discontinuous variation and explain the genetic
basis of continuous (many, additive genes control a characteristic) and discontinuous variation
(one or few genes control a characteristic)
b) explain, with examples, how the environment may affect the phenotype of plants and animals
c) use the t-test to compare the variation of two different populations
d) explain why genetic variation is important in selection
17.1 Variation
Introducing variation
Individuals of a species are strikingly similar, which is how we may identify them, whether humans
buttercups or houseflies, for example. But individuals also show many differences, although we
may have to look carefully in members of species other than our own. Within families there are
remarkable similarities between parents and their offspring, but no two members of a family are
identical, apart from identical twins. About these differences we can say:
● some differences may be controlled by genes – such as human blood groups
● other differences between individuals may be due to the effect of our environment, such as the
fur colour of the Arctic hare or the Siamese cat
● other differences between individuals may be due to both genetics and environment, such as our
body height and weight.
Another important point about variation is that it is of two types (Figure 17.1).
● Discontinuous variation arises when the characteristic concerned is one of two or more
discrete types with no intermediate forms. Examples include the garden pea plant (tall or dwarf)
and human ABO blood grouping (group A, B, AB or O). These are genetically determined.
● Continuous variation results in a continuous distribution of values. Height in humans is a good
example. Continuous variation may be genetically determined, or it may be due to environmental
and genetic factors working together.
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17.1 Variation
between the two data sets. An example will illustrate the method. However, you should note that
you are not expected to calculate values of t.
Applying the t-test
An ecologist was investigating woodland microhabitats, contrasting the communities in a shaded
position with those in full sunlight. One of the plants was ivy (Hedra helix), but relatively few
occurred at the locations under investigations. The issue arose: were the leaves in the shade
actually larger than those in the sunlight?
Leaf widths were measured, but because the size of the leaves varied with the position on the
plant, only the fourth leaf from each stem tip was measured. The results from the plants available
are shown in Table 17.1.
Table 17.1 Sizes of sun and shade leaves of Hedera helix.
Size-class / mm
Sun leaves (A)
20–24
24
Shade leaves (B)
25–29
26, 26
26
30–34
30, 31, 31, 32, 32, 33
33, 34
35–39
37, 38
35, 35, 36, 36, 36, 37
40–44
43
41, 42
45–49
45
Steps to the t-test:
1 The null hypothesis (negative hypothesis) assumes the difference under investigation has arisen
by chance; in this example the null hypothesis is:
‘There is no difference in size between sun and shade leaves.’
The role of the statistical test is to determine whether to accept or reject the null hypothesis. If it
is rejected here, we can have confidence that the difference in the leaf sizes of the two samples is
statistically significant.
Next, check that the data are normally distributed. This is done by arranging the data for sun
leaves and shade leaves as in Table 17.1 (and plotting a histogram, if necessary).
2 You are not necessarily expected to calculate values of t. This is a statistic which when required can be
found using by using a scientific or statistics calculator, or by means of a spreadsheet incorporating
formulae. Actually, a formula for the t-test for unmatched samples (data sets a versus b) is :
xa – xb
t=
sa2 sb2
na nb
where:
xa 5 mean of data set a
xb 5 mean of data set b
sa2 5 standard deviation for data set a, squared
sb2 5 standard deviation for data set b, squared
na 5 number of data in set a
nb 5 number of data in set b
√ 5 square root of
3 Once a value of t has been calculated (the value of t here is 2.10) we determine the degrees of
freedom (df) for the two samples, using the formula:
df (total number of values in both samples) 2
na nb 2
In this case, df 11 11 22.
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17 Selection and evolution
Allele frequency: the
commonness of the
occurrence of any particular
allele in a population.
on, from generation to generation. Remember, an allele is one of a number of alternative forms of
a gene that can occupy a given locus on a chromosome. The frequency with which any particular
allele occurs in a given population will vary.
When allele frequencies of a particular population are investigated they may turn out to be
static and unchanging. Alternatively, we may find allele frequencies changing. They might do so
quite rapidly with succeeding generations, for example.
When the allele frequencies of a gene pool remain more or less unchanged, then we know that
population is static as regards its inherited characteristics. We can say that the population is not
evolving.
However, if the allele frequencies of gene pool of a population is changing (i.e. the proportions
of particular alleles is altered, we say ‘disturbed’ in some way), then we may assume that evolution
is going on. For example, some alleles may be increasing in frequency because of an advantage
they confer to the individuals carrying them. With possession of those alleles the organism is more
successful. It may produce more offspring, for example. If we can detect change in a gene pool we
may detect evolution happening, possibly even well before a new species is observed.
How can we detect change or constancy in gene pools?
The answer is, by a mathematical formula called the Hardy–Weinberg formula. Independently
this principle was discovered by two people, in the process of explaining why dominant
characteristics don’t take over in populations, driving out the recessive form of that characteristic.
(For example, at that time people thought [wrongly] that human eye colour was controlled by
a single gene, and that an allele for blue eyes was dominant to the allele for brown eyes. ‘Why
doesn‘t the population become blue-eyed?’, was the issue.)
Let the frequency of the dominant allele (G) be p, and the frequency of the recessive allele (g) be q.
The frequency of alleles must add up to 1, so p + q = 1.
This means in a cross, a proportion (p) of the gametes carries the G allele, and a proportion (q) of the gametes
carries the g allele.
The offspring of each generation are given by the Punnett grid.
gamete frequency
G
p
g
q
Gg
pq
G
p
g
q
GG
p2
Gg
pq
So the progeny are respectively:
p2 = frequency of GG homozygote
2pq = frequency of Gg heterozygote
q2 = frequency of gg homozygote
gg
q2
Hardy–Weinberg formula
If the frequency of one allele (G) is p, and the frequency of the other allele (g) is q then the frequencies of the three
possible genotypes GG, Gg and gg are respectively p2, 2pq and q2.
In this way, Hardy and Weinberg developed the following equation to describe stable gene pools:
p2
frequency of
homozygous
dominant
individuals
+
2pq
frequency of
heterozygous
individuals
+
q2
frequency of
homozygous
recessive
individuals
=
1
total
Figure 17.12 Deriving the Hardy–Weinberg formula
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17 Selection and evolution
The steps to domestication of wild animals
Question
6 Explain the key
difference between
natural and artificial
selection.
1 A wild population of a ‘species’ useful as a source of hides, meat, etc. is identified and people
learn to distinguish it from related species. Herd animals (such as sheep and cattle) are
naturally sociable and lend themselves to this.
2 Once a wild herd has been trapped inside a compound or field, there is a selective killing
(culling) of the least suitable members of this herd in order to meet immediate needs for food
and materials for living.
3 Breeding is encouraged among members of the herd showing the desired feature and they are
protected from attack by wild predators.
4 From the offspring the individual with the most useful features is selected and used as the future
breeding stock.
5 The breeding stock is maintained during unfavourable conditions, such as drought and floods.
6 After a long period of artificial selection, a domesticated herd, dependent on the herds people
rather than living wild, is established. This leads to the possibility of trading individuals of a
breeding stock with other groups of herds people living nearby who have similar stocks. This
introduces a wider gene pool into the herd – and may introduce fresh characteristics.
Wild sheep or mouflon (Ovis musimon) occur today
on Sardinia and Corsica
Soay sheep of the outer Hebrides
suggest to us what the earliest
domesticated sheep looked like
Modern selective breeding has produced
shorter animals with a woolly fleece in
place of coarse hair and with muscle of
higher fat content. Many breeds have lost
their horns
Figure 17.13 From wild to domesticated species and the origins of selective breeding skills
Selective breeding of dairy cattle for improved
milk yield
In highly developed agricultural systems, principally the USA and Western Europe, milk yield per
cow has more than double in the past 50 years. Holstein Friesian cattle are today’s most productive.
This breed originated in Europe, in Holland and Germany, and proved to be the animals that could
best exploit the abundant grass resources available (Figure 17.14).
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17 Selection and evolution
maternal inheritance of mtDNA
outer membrane
egg cell
(egg and sperm
not to scale)
crista
inner membrane
rings of DNA
(mtDNA)
mitochondria
with mtDNA
mitochondrion (enlarged)
sperm
nucleus with
DNA of the
chromosomes
speciation, that is,
two closely related
species have
evolved from a
common ancestor
A
only the sperm
nucleus enters
the egg cell at
fertilisation.
A1
A2
passage of time
*
A is one mtDNA
gene (of known
base sequence)
ring of mtDNA
with many
genes
*
accumulation of genetic differences in mtDNA
In mtDNA, mutations involve about 1–2 base changes in every
100 nucleotides per million years.
Studying change in the base pairs of mtDNA genes allows us to detect
evolutionary changes occurring over several hundred thousand years.
*
*
*
chance mutations
accumulate at an
approximately
constant rate but at
different locations
*
*
* random mutations
Figure 17.20 The use of mitochondrial DNA in measuring evolutionary divergence
Speciation
Species: a group of
organisms that are
reproductively isolated,
interbreeding to produce
fertile offspring. Organisms
belonging to a species have
morphological (structural)
similarities, which are often
used to identify to which
species they belong.
Present-day plants and animals have arisen by change from pre-existing forms of life. This process
has been called ‘descent with modification’ and ‘organic evolution’, but perhaps ‘speciation’ is
better because it emphasises that species change. So, what is a species?
When Linnaeus devised the binomial system of nomenclature in the 18th century there was
no problem in defining species. It was believed that each species was derived from the original
pair of animals created by God. Since species had been created in this way they were fixed and
unchanging.
In fact the fossil record provides evidence that changes do occur in living things – human fossils
alone illustrate this point. Today, as many different characteristics as possible are used in order to
define and identify a species. The three main characteristics used are:
● morphology and anatomy (external and internal structure)
● cell structure (whether cells are eukaryotic or prokaryotic)
● physiology (blood composition, renal function) and chemical composition (comparisons of
nucleic acids and proteins, and the similarities in proteins between organisms, for example).
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17 Selection and evolution
Many organisms (e.g. insects and birds) may have flown or been carried on wind currents to the
Galapagos from the mainland. Mammals are most unlikely to have survived drifting there on a natural
raft over this distance, but many large reptiles can survive long periods without food or water.
immigrant travel to the Galapagos
The Galapagos Islands
Today the tortoise population
of each island is distinctive
and identifiable.
Equator
Pinta
Genovesa
Galapagos
Marchena
0°
Santiago
Santa Cruz
Fernandina
1°S
The giant iguana lizards on the Galapagos Islands became
dominant vertebrates, and today are two distinct species,
one still terrestrial, the other marine, with webbed feet and
a laterally flattened tail (like the tail fin of a fish).
Isabela
San Cristóbal
Santa Fé
~
Espanola
Floreana
91°W
90°W
Figure 17.23 The Galapagos Islands and species divergence there.
Another example of behavioural and ecological separation leading to speciation has been
demonstrated in the fauna of the Galapagos islands. Today there are 14–15 species of finch,
and they have all been derived from a common ancestor, and have evolved, living in the same
environment. These birds were noted by Charles Darwin (but they failed to attract his detailed
attention) whilst on his visits to the islands. At a later date, detailed study of these birds was begun
by the ornithologist, David Lack.
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17.3 Evolution
Summary
●
●
(continued)
Natural selection may work to keep the characteristics of a species constant (stabilising
selection), but if the environment changes then new forms may emerge (directional and
disruptive selection). Balancing selection is a process which actively maintains multiple alleles in
the gene pool of a population.
Humans have obtained the animals and plants used in today’s agriculture, transport and leisure
pursuits by a process of domestication of wild organisms by means of artificial selection.
Selective breeding is carried out by careful selection of the parents in breeding crosses and
the selection of progeny with the required features.
Examination style questions
1 a) Explain what is meant by artificial selection.
[4]
b) In a plant breeding programme, corn, Zea mays, was bred in
an attempt to produce a high yield of protein in the grain.
2 a) Explain how changes in the nucleotide sequence of DNA
may affect the amino acid sequence in a protein.
[7]
b) Explain how natural selection may bring about evolution. [8]
[Total: 15]
The results of this programme are shown in Fig. 1.1.
(Cambridge AS and A Level Biology 9700, Paper 04 Q9 June 2009)
24
3 Spartina species (cord grass) are common plants of
estuaries and salt marshes in many parts of the world. Two
species, S. maritima (60 chromosomes – AA genome) and
S. alterniflora (62 chromosomes – BB genome), once grew
apart in different waters of the northern hemisphere. Now
they occur together in many habitats. Today they have
been joined by a new species of cord grass, S. angelica (122
chromosomes – AABB genome). This latter cord grass is a larger
plant, and grows vigorously.
a) It is assumed that S. angelica has evolved by a particular
mechanism, involving the other two species. Describe this
type of change, how it may have come about, and the
steps that would have been involved.
[12]
b) Identify another plant species that has evolved by this
mechanism.
[2]
mass of protein / g per 100 g of grain
20
16
12
8
4
[Total: 14]
0
0
5
10
15
20
25 30 35
generations
40
45
50
55
60
Fig. 1.1
i) With reference to Fig. 8.1, calculate the percentage
increase in grain protein by the end of the experiment.
Show your working.
[2]
ii) Suggest why the protein yield does not increase
steadily in each generation.
[2]
[Total: 8]
(Cambridge AS and A Level Biology 9700, Paper 04 Q8
June 2007)
4 a) Define the terms ‘gene pool’ and ‘differential mortality’. [2]
b) Illustrate what you understand by stabilising selection by
means of an example, and explain the suggestion that
stabilising selection does not lead to evolution.
[6]
c) What is directional selection? By means of an example,
show how directional selection may lead to new varieties
of organism.
[6]
d) Disruptive or diversifying selection is said to result in
balanced polymorphism. Elaborate this idea by means
of a named example and by specifying the selection
forces operating.
[6]
[Total: 20]
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Contents
Introduction ............................................................................................................... . . . . . . . . . . . . . . . . . . . . vi
Structure of syllabus .......................... ........................................................................... . . . . . . . . . . . . . . . . . . . vii
AS Level
Physical Chemistry
Topic 1 Chemical formulae and moles .......................................................................................................................... 1
Topic 2 The structure of the atom ............................................................................................................................. 20
Topic 3 Chemical bonding in simple molecules ........................................................................................................... 44
Topic 4 Solids, liquids and gases ................................................................................................................................ 69
Topic 5 Energy changes in chemistry .......................................................................................................................... 97
Topic 6 Acids and bases ........................................................................................................................................... 118
Topic 7 Oxidation and reduction .............................................................................................................................. 136
Topic 8 Rates of reaction ......................................................................................................................................... 154
Topic 9 Equilibria .................................................................................................................................................... 172
Inorganic chemistry
Topic 10 Periodicity ......................... ............................................................................ . . . . . . . . . . . . . . . . . 186
Topic 11 Group 17 ........................... ........................................................................... . . . . . . . . . . . . . . . . . . 202
Organic chemistry
Topic 12 Introduction to organic chemistry .............................................................................................................. 209
Topic 13 Alkanes ..................................................................................................................................................... 234
Topic 14 Alkenes ..................................................................................................................................................... 251
Topic 15 Halogenoalkanes ....................................................................................................................................... 267
Topic 16 Alcohols .................................................................................................................................................... 281
Topic 17 Aldehydes and ketones.............................................................................................................................. 294
Topic 18 Carboxylic acids and their derivatives ........................................................................................................ 305
Advanced practical skills (Paper 3)
Topic 19 Practical work ..................... ............................................................................ . . . . . . . . . . . . . . . . . 316
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PHYSICAL CHEMISTRY
Contents
A Level
Physical Chemistry
Topic 20 Further energy changes....................................................................................... . . . . . . . . . . . . . . . . 329
Topic 21 Quantitative kinetics ......................................................................................... . . . . . . . . . . . . . . . . . 347
Topic 22 Quantitative equilibria............ ........................................................................... . . . . . . . . . . . . . . . . . 365
Topic 23 Electrochemistry ............................................................................................... . . . . . . . . . . . . . . . . 379
Inorganic chemistry
Topic 24 The 3d block ........................ ........................................................................... . . . . . . . . . . . . . . . . . 395
Organic chemistry
Topic 25 Arenes and phenols .......................................................................................... . . . . . . . . . . . . . . . . 413
Topic 26 Further reactions of carboxylic acids and ketones ...................................................... . . . . . . . . . . . . . . . . . 438
Topic 27 Amines, amides and amino acids........................................................................... . . . . . . . . . . . . . . . . . 447
Topic 28 Addition and condensation polymers ..................................................................... . . . . . . . . . . . . . . . . . 467
Topic 29 Techniques of analysis ............ ............................................................................ . . . . . . . . . . . . . . . . 493
Topic 30 Organic Synthesis and analysis .. ........................................................................... . . . . . . . . . . . . . . . . . 517
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2
The structure of the atom
Worked example 1
Chlorine consists of two isotopes, with mass numbers 35 and 37, and with relative abundances
76% and 24% respectively. Calculate the average relative atomic mass of chlorine.
Answer
The percentages tell us that if we took 100 chlorine atoms at random, 76 of them would
have a mass of 35 units, and 24 of them would have a mass of 37 units.
Now try this
1 Chromium has four stable isotopes,
with mass numbers 50, 52, 53 and 54,
and relative abundances 4.3%, 83.8%,
9.5% and 2.4% respectively. Calculate
the average relative atomic mass of
chromium.
2 Use the figures in Table 2.2 (page 22)
to calculate the average relative atomic
masses of boron, neon and magnesium
to 1 decimal place.
3 (Harder) Iridium has two isotopes, with
mass numbers 191 and 193, and its
average relative atomic mass is 192.23.
Calculate the relative abundances of
the two isotopes.
total mass of the 100 random atoms = (35 × 76) + (37 × 24)
= 3548 amu
so
3548
100
= 35.5 amu
average mass of one atom =
that is,
Ar = 35.5
Worked example 2
Calculate the average relative atomic mass of krypton from the table in Figure 2.6.
Answer
We can extend the 100-random-atom idea from Worked example 1 to include fractions of
atoms. Thus the average mass of one atom of krypton
(78 × 0.3) + (80 × 2.3) + (82 × 11.6) + (83 × 11.5) + (84 × 56.9) + (86 × 17.4)
100
= 83.9
=
2.6 Chemical energy
The concept of energy is central to our understanding of how changes come
about in the physical world. Our study of chemical reactions depends on energy
concepts.
Chemical energy is made up of two components – kinetic energy, which is a
measure of the motion of atoms, molecules and ions in a chemical substance, and
potential energy, which is a measure of how strongly these particles attract one
another.
Kinetic energy
Kinetic energy increases as the temperature increases. Chemists use a scale of
temperature called the absolute temperature scale, and on this scale the kinetic
energy is directly proportional to the temperature (see section 4.13).
Kinetic energy can be of three different types. The simplest is energy due to
translation, that is, movement from place to place. For monatomic gases (gases
made up of single atoms, for example helium and the other noble gases), all
the kinetic energy is in the form of translational kinetic energy. For molecules
containing two or more atoms, however, there is the possibility of vibration
and rotation as well (see Figure 2.7). Both these forms of energy involve the
movement of atoms, even though the molecule as a whole may stay still. In
diatomic gases, the principal form of kinetic energy is translation, but in more
complex molecules, such as ethane, vibration and rotation become the more
important factors.
Figure 2.7 The two atoms in a diatomic gas
molecule, such as nitrogen, behave as though
they are joined by a spring, which lets them
vibrate in and out. The two atoms can also
rotate about the centre of the bond.
25
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PHYSICAL CHEMISTRY
The energy E of a photon of light is related to its frequency, f, by Planck’s equation:
E = hf
where h is the Planck constant.
If photons of a particular frequency are being emitted by an atom, this means that
the atom is losing a particular amount of energy. This energy represents the difference
between two states of the atom, one more energetic than the other (see Figure 2.13).
Figure 2.13 A photon is emitted as the
atom moves between state E1 and state E2.
E2
photon emitted, with energy
energy
hf = ΔE
ΔE = E2 – E1
E1
The spectrum of the simplest atom, hydrogen, shows a series of lines at different
2.14). This suggests that the hydrogen atom can lose different
amounts of energy, which in turn suggests that it can exist in a range of energy states.
Barking
Dog Art between the various energy states cause photons to be emitted at various
Transitions
frequencies. These energy states can be identified with situations where the single hydrogen
electron is in certain orbitals, at specific distances from the nucleus, as we shall see.
(see Figure
02_13frequencies
Cam/Chem AS&A2
Figure 2.14 Part of the hydrogen
spectrum, showing the corresponding
energy transitions
E6
n = • (ionisation)
n=7
n=6
E5
n=5
E4
n=4
excited electron
falls to n = 2
n=3
E3
energy, E
a photon is released
E2
n=2
E1
n=1
f1 =
(E3 – E2)
f2 =
h
5
(E4 – E2)
f3 =
h
6
(E5 – E2)
h
f4 =
(E6 – E2)
h
7
14
frequency, f/10
Hz
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Barking Dog Art
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5
Energy changes in chemistry
Alternative energy sources
Alternative energy sources do not rely on the
combustion of fuels. The following are some of
those in current use.
• Hydroelectric power – there are many places
in the world where rivers have been dammed
and the resulting difference in water level
used to provide energy to generate electricity,
for example at Kariba on the Zambezi river
between Zimbabwe and Zambia, at Itaipu on
the Parana river between Brazil and Paraguay,
and the Three Gorges dam on the Yangtze river
in China. The last two examples alone generate
over 6 × 1017 J of energy per year, the equivalent
of burning 6 × 107 tonnes of coal in a coal-fired
power station. This saves pumping millions of
tonnes of carbon dioxide into the atmosphere.
However, there can be resultant damage to the
Figure 5.5 Wind farms provide pollution-free energy, but have a visual effect on
environment, which needs to be considered as it
the landscape.
may be unacceptable.
• Tidal and wave power – these are little used, even though the large-scale tidal
power generator at La Rance in northern France first produced electricity 50 years ago.
• Wind power – the use of wind turbines is increasing, although there is some
objection to the construction of large windmills in isolated regions as they have a
dramatic effect on the landscape. The wind speed is not always high enough to turn
the blades in summer, but it usually is in winter, when the demand for electricity
is highest. Another difficulty is that the places which are the most windy (and
therefore the most suitable for the production of wind power) are often not the
most populous, so the power has to be transported to where it is needed or power
lines constructed to link to a national grid.
• Geothermal power – potentially this could be an easy way to provide energy for
home heating, though it can only be carried out at sites with particular geological
features. Water is pumped from deep under the ground and comes to the surface
at a temperature near to boiling point. It is not usually possible to use this hot
water to produce electricity, because superheated steam is needed in order to
obtain a reasonable energy conversion. However, the Philippines, New Zealand
and Costa Rica obtain a reasonable proportion of their energy from this source.
• Solar panels – the direct conversion of sunlight into electricity using solar cells
used to be quite expensive, but the application of conducting polymers has
reduced the cost of these, and in sunny places, for example in California, Spain
and India, solar energy is now economically viable.
• Nuclear power – the use of nuclear reactors based on fission reactions, in
which large nuclei split into two smaller nuclei and give out nuclear energy, has
sharply declined as a result of accidents such as that at Chernobyl in 1986. There
is much debate over the long-term implications of the disposal of nuclear waste,
and most countries had cut back on their nuclear programmes until a few years
ago, when concern for the rise in atmospheric carbon dioxide caused a rethink.
There is the possibility that reactors using fusion reactions, in which lighter
nuclei join to form a heavier one, may be developed in the next few decades.
These provide the best hope for a long-term solution to the energy problem,
because they are ‘cleaner’ and do not produce radioactive waste.
Renewable fuels
Most of the alternative energy sources listed above are renewable because they are
powered by the Sun, and so are always available. Some fuels are made out of resources
that can be replaced quickly. For example, tropical countries such as Brazil can grow
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7
Oxidation and reduction
Titrations using potassium manganate(VII)
as the oxidant
Usually potassium manganate(VII) titrations are carried out in acid solution. Under
these conditions, the following half-equation is relevant:
MnO4− + 5e− + 8H+ → Mn2+ + 4H2O
You will meet this equation often in redox reactions, and it is a good idea to learn it.
The deep purple manganate(VII) ion is used as its own indicator. This solution
is run into the flask from the burette, and in the course of the reaction the colour
changes from purple to colourless as nearly colourless manganese(II) ions are
produced. At the end-point, the first extra drop of manganate(VII) ions makes the
solution turn pink. The following experimental details should be noted.
●
●
●
Figure 7.3 The top of the meniscus is read
when using potassium manganate(VII) in a
titration. Compare the readings with those in
Figure 6.5 (page 127).
●
Potassium manganate(VII) is not very soluble, and the highest concentration used is
0.02 mol dm−3.
The reaction is carried out in the presence of sulfuric acid at approximately 1 mol dm−3.
It is usual to read the top, rather than the bottom, of the meniscus of potassium
manganate(VII) in the burette because the deep purple colour obscures the reading
at the bottom (see Figure 7.3). Because the volume delivered is the difference
between the readings, it does not matter whether we read the top or bottom of the
meniscus, as long as we do the same for both readings.
If the titration is carried out too quickly, the solution may turn brown (see
Figure 7.4). This is due to the formation of manganese(IV) oxide, MnO2. This can
be avoided either by increasing the acidity of the solution or by warming it.
Figure 7.4 To avoid the solution turning
brown, warm the solution or use plenty
of acid.
Worked example 1
In a titration, 25.0 cm3 of a solution of iron( II ) sulfate required 22.4 cm3 of 0.0200 mol dm −3
potassium manganate(VII ) solution at the end-point. What is the concentration of the
solution of iron( II ) sulfate?
Answer
n(KMnO4 ) = c ×
v
22.4
= 4.48 × 10 −4 mol
= 0.0200 ×
1000
1000
oxidation: Fe2+ → Fe3+ + e−
reduction: MnO4− + 5e− + 8H+ → Mn2+ + 4H2O
To balance the electrons, 5 mol of Fe2+ react with 1 mol of KMnO4. Therefore:
n(Fe2+) used is = 5 × 4.48 × 10−4 = 2.24 × 10−3 mol
1000
c(Fe2+) =
×n
v
1000
× 2.24 × 10−3
=
25.0
= 0.0896 mol dm-3
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AS Level
Organic chemistry
13 Alkanes
Functional group:
C¬C¬H
In this topic the properties and
reactions of the simplest of the
homologous series, the alkanes, are
discussed. Their extraction from crude
oil, and how they are transformed
into fuels and feedstocks for the
chemical industry are described,
and their reactions with oxygen,
chlorine and bromine are studied.
Their characteristic reaction of radical
substitution is explained in detail.
Learning outcomes
By the end of this topic you should be able to:
1.5b)
14.1a)
14.2a)
15.1a)
15.1b)
15.1c)
15.1d)
15.1e)
15.3a)
15.3b)
perform calculations, including use of the mole concept, involving volumes of
gases (part, see also Topic 1)
interpret, and use, the general, structural, displayed and skeletal formulae of the
alkanes (part, see also Topic 12)
interpret, and use, the following terminology associated with organic reactions:
homolytic fission, free radical, initiation, propagation, termination (part, see also
Topic 12)
show awareness of the general unreactivity of alkanes, including towards polar
reagents
describe the chemistry of alkanes as exemplified by the following reactions of
ethane: combustion, and substitution by chlorine and by bromine
describe the mechanism of free-radical substitution at methyl groups with
particular reference to the initiation, propagation and termination reactions
explain the use of crude oil as a source of both aliphatic and aromatic
hydrocarbons
suggest how ‘cracking’ can be used to obtain more useful alkanes and alkenes
of lower Mr from larger hydrocarbon molecules
describe and explain how the combustion reactions of alkanes lead to their use
as fuels in industry, in the home and in transport
recognise the environmental consequences of:
● carbon monoxide, oxides of nitrogen and unburnt hydrocarbons arising from
the internal combustion engine and of their catalytic removal
● gases that contribute to the enhanced greenhouse effect.
13.1 Introduction
The alkanes are the simplest of the homologous series found in organic chemistry.
They contain only two types of bonds, both of which are strong and fairly non-polar
(see Table 13.1). (See section 3.10 for an explanation of bond polarity.)
Table 13.1 Carbon–carbon and carbon–
hydrogen bonds are strong and non-polar.
Bond
Bond enthalpy/kJ mol−1
Polarity (difference in
electronegativity)
C¬C
346
0
C¬H
413
0.35 (Hδ+¬Cδ−)
Consequently, the alkanes are not particularly reactive. Their old name (the
‘paraffins’) reflected this ‘little (parum) affinity’. What few reactions they have,
however, are of immense importance to us. They constitute the major part of crude
oil, and so are the starting point of virtually every other organic compound made
industrially. Apart from their utility as a feedstock for the chemical industry, their
major use is as fuels in internal combustion engines, jet engines and power stations.
We shall be looking at their role in combustion reactions in detail later in this topic.
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13
Alkanes
The straight-chain alkanes form a homologous series with the general formula
CnH2n+2. As the number of carbon atoms in the molecule increases, the van der Waals
attractions between the molecules become stronger (see section 3.17). This has a
clear effect on the boiling points and the viscosities of members of the series (see
Figure 13.1 and Table 14.1, page 440).
Figure 13.1 The boiling points and viscosities
of the alkanes
150
250
boiling point
100
150
100
viscosity
50
viscosity/poise
boiling point/C
200
50
0
0
5
6
7
Figure 13.2 Space-filling and ball-and-stick a
models of ethane, C2H6 (a and b) and butane,
C4H10 (c to f). The space-filling models show
13_01 Cam/Chem AS&A2
the overall shape of the molecule but the
ball-and-stick models show the positions of
Barking Dog Art
the atoms more clearly. Note that e and f are
identical to c and d, as they are formed by free
rotation of the central C¬C bond.
c
d
8
9
number of carbon atoms
10
11
12
b
e
f
13.2 Isomerism and nomenclature
As was mentioned in section 12.5, the first step in naming an alkane is to identify
the longest carbon chain, which gives the stem. Side branches are then identified
by prefixes. Each prefix is derived from the stem name denoting the number of
carbon atoms the side chain contains, and is preceded by a number denoting the
position it occupies on the longest carbon chain. Two examples will make this
clear.
CH3
CH3
CH
CH3
●
●
CH2
The longest carbon chain contains four carbon atoms, so the alkane is a derivative
of butane.
There is just one side branch, containing one carbon atom (and hence called
methyl).
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ORGANIC CHEMISTRY
●
By the oxidation of aryl side chains. When treated with hot alkaline potassium
manganate(VII), aryl hydrocarbons produce benzoic acids by oxidation:
CH3
CO2H
heat with KMnO4 OH¯(aq)
Summary
●
●
●
RCO2H + NaOH → RCO2−Na+ + H2O
2RCO2H + Na2CO3 → 2RCO2−Na+ + CO2 + H2O
RCO2H + SOCl2 → RCOCl + SO2 + HCl
RCO2H + PCl5 → RCOCl + POCl3 + HCl
The carboxylic acids, RCO2H, are weak acids, ionising to the
extent of 1% or less in water.
Carboxylic acids react with alcohols to form esters, and with
phosphorus(V) chloride to form acyl chlorides.
Esters undergo acid- or base-catalysed hydrolysis.
Key reactions you should know
● Carboxylic acids:
1
RCO2H + Na → RCO2−Na+ + H2
●
RCO2H + R′OH
Esters:
RCO2R′
2
heat with conc. H+
RCO2R′ + H2O
heat with H3O+ or OH−(aq)
RCO2H + R′OH
Examination-style questions
1 Isomerism occurs in many organic compounds. The two
main forms of isomerism are structural isomerism and
stereoisomerism. Many organic compounds that occur
naturally have molecules that can show stereoisomerism,
that is cis–trans or optical isomerism.
a i Explain what is meant by structural isomerism.
ii State two different features of molecules that can give
rise to stereoisomerism.
Unripe fruit often contains polycarboxylic acids, that is
acids with more than one carboxylic acid group in their
molecule.
One of these acids is commonly known as tartaric acid,
HO2CCH(OH)CH(OH)CO2H.
b Give the structural formula of the organic compound
produced when tartaric acid is reacted with an excess of
NaHCO3.
Another acid present in unripe fruit is citric acid,
neutralisation.
i Use these data to deduce the number of carboxylic
acid groups present in one molecule of W.
ii Suggest the displayed formula of W.
[Cambridge International AS & A Level Chemistry 9701,
Paper 21 Q5 Summer 2010]
2 a Complete the following reaction scheme which starts
with ethanal.
In each empty box, write the structural formula
of the organic compound that would be formed.
e
HCN
CH3CHO
Tollens’
reagent
B
E
dilute H2SO4
heat
reduction
OH
HO2CCH2CCH2CO2H
D
CO2H
c Does citric acid show optical isomerism? Explain your
answer.
A third polycarboxylic acid present in unripe fruit is a
colourless crystalline solid, W, which has the following
composition by mass: C, 35.8%; H, 4.5%; O, 59.7%.
d i Show by calculation that the empirical formula of W is
C4H6O5.
ii The Mr of W is 134. Use this value to determine the
molecular formula of W.
A sample of W of mass 1.97 g was dissolved in water
and the resulting solution titrated with 1.00 mol dm−3
NaOH. 29.4 cm3 were required for complete
314
conc. H2SO4
heat
CH2
CHCO2H
cold dilute MnO−4/H+
2−
Cr2O7 /H+
heat under reflux
C
18_08 Cam/Chem AS&A2
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Contents
AS Level
Topic 1
1.1
1.2
1.3
Physical quantities and units ................................................................................ . . . . . . . . . . . . . . . . . . 1
Physical quantities .................. ........................................................................... . . . . . . . . . . . . . . . . . . 2
SI quantities and base units ................................................................................... . . . . . . . . . . . . . . . . . 2
Scalars and vectors ............................................................................................ . . . . . . . . . . . . . . . . . . 7
Topic 2 Measurement techniques ...................................................................................... . . . . . . . . . . . . . . . 15
2.1 Measurements .................................................................................................. . . . . . . . . . . . . . . . . 15
2.2 Errors and uncertainties ........... ............................................................................ . . . . . . . . . . . . . . . 31
Topic 3 Kinematics ........................... ............................................................................ . . . . . . . . . . . . . . . 40
3.1 Speed, displacement, velocity and acceleration ......................................................... . . . . . . . . . . . . . . . . 40
Topic 4
4.1
4.2
4.3
Dynamics ............................. ............................................................................ . . . . . . . . . . . . . . .
Relationships involving force and mass .................................................................... . . . . . . . . . . . . . . . .
Weight ................................ ........................................................................... . . . . . . . . . . . . . . . .
The principle of conservation of momentum .............................................................. . . . . . . . . . . . . . . .
55
55
58
61
Topic 5
5.1
5.2
5.3
5.4
Forces, density and pressure .................................................................................. . . . . . . . . . . . . . . .
Types of force ........................ ............................................................................ . . . . . . . . . . . . . . .
Moment of a force .................. ........................................................................... . . . . . . . . . . . . . . . .
Equilibrium of forces ............... ........................................................................... . . . . . . . . . . . . . . . .
Density and pressure ............... ............................................................................ . . . . . . . . . . . . . . .
71
71
71
73
76
Topic 6
6.2
6.1
6.3
6.4
Work, energy, power ............... ............................................................................ . . . . . . . . . . . . . . .
Work ................................... ........................................................................... . . . . . . . . . . . . . . . .
Energy............................................................................................................. . . . . . . . . . . . . . . .
Kinetic energy ................................................................................................... . . . . . . . . . . . . . . .
Power ................................. ............................................................................ . . . . . . . . . . . . . . .
80
80
83
85
89
Topic 9 Deformation of solids ......................................................................................... . . . . . . . . . . . . . . . . 94
9.1 & 9.2 Force and deformation ................................................................................. . . . . . . . . . . . . . . . . 94
Topic 14
14.1
14.2
14.3
14.4
14.5
Waves ............................................................................................................. . . . . . . . . . . . . . . 101
Wave motion .................................................................................................... . . . . . . . . . . . . . . . 101
Graphical representation of waves ......................................................................... . . . . . . . . . . . . . . . 102
The determination of the frequency of sound using a calibrated c.r.o. .............................. . . . . . . . . . . . . . . . 107
Doppler effect ................................................................................................. . . . . . . . . . . . . . . . 108
The electromagnetic spectrum ... ............................................................................ . . . . . . . . . . . . . . 109
Topic 15 Superposition ................................................................................................... . . . . . . . . . . . . . . . 114
15.3 Interference ..................................................................................................... . . . . . . . . . . . . . . . 114
15.1 Stationary waves ............................................................................................... . . . . . . . . . . . . . . . 119
14.3 Measuring the speed of sound using stationary waves ................................................. . . . . . . . . . . . . . . . 124
15.2 & 15.4 Diffraction ....................................................................................................................................126
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Contents
Topic 17 Electric fields. ........................ ............................................................................ . . . . . . . . . . . . . . 136
17.1 Electric forces and fields.. .......... ............................................................................. . . . . . . . . . . . . . 136
17.2 Electric field strength. .......................................................................................... . . . . . . . . . . . . . . 139
Topic 19
19.1
19.2
19.3
Current of electricity. ............... ............................................................................. . . . . . . . . . . . . . 145
Charge and current. ................. ............................................................................. . . . . . . . . . . . . . 145
Potential difference. ................ ............................................................................ . . . . . . . . . . . . . . 148
Resistance. ............................ ............................................................................ . . . . . . . . . . . . . . 149
Topic 20
20.1
20.2
20.3
D.C. circuits. ...................................................................................................... . . . . . . . . . . . . . . 156
Electrical circuits. .................... ............................................................................ . . . . . . . . . . . . . . 156
Kirchhoff’s first and second laws. ............................................................................. . . . . . . . . . . . . . 159
Potential dividers and potentiometers....................................................................... . . . . . . . . . . . . . . 162
Topic 26 Particle physics....................... ............................................................................ . . . . . . . . . . . . . . 168
26.1 Atomic structure and radioactivity........................................................................... . . . . . . . . . . . . . . 168
26.2 Fundamental particles. .......................................................................................... . . . . . . . . . . . . . 178
A Level
Topic 7 Motion in a circle................................................................................................ . . . . . . . . . . . . . . 184
7.1 Radian measure and angular displacement.................................................................. . . . . . . . . . . . . . 184
7.2 Centripetal acceleration and centripetal force. ............................................................ . . . . . . . . . . . . . . 185
Topic 8 Gravitational fields. ................. ............................................................................. . . . . . . . . . . . . . 191
8.1 Gravitational field. .................. ............................................................................ . . . . . . . . . . . . . . 191
8.2 & 8.3 Gravitational field strength...............................................................................................................192
8.4 Gravitational potential and gravitational potential energy............................................... . . . . . . . . . . . . . 198
Topic 10 Ideal gases. ........................................................................................................ . . . . . . . . . . . . . 202
10.1 & 10.3 Equation of state of an ideal gas ................................................................................................. 202
10.2 A microscopic model of a gas................................................................................... . . . . . . . . . . . . . 205
Topic 11
11.1
11.2
11.3
Temperature. ...................................................................................................... . . . . . . . . . . . . . 211
Temperature. ...................................................................................................... . . . . . . . . . . . . . 211
Temperature scales. ................. ............................................................................. . . . . . . . . . . . . . 212
Thermometers........................ ............................................................................ . . . . . . . . . . . . . . 215
Topic 12 Thermal properties of materials... ............................................................................. . . . . . . . . . . . . . 217
12.1 Solids, liquids and gases, and thermal (heat) energy . . .................................................... . . . . . . . . . . . . . 217
12.2 Internal energy . .................................................................................................. . . . . . . . . . . . . . 225
Topic 13
13.1
13.2
13.3
Oscillations......................................................................................................... . . . . . . . . . . . . . 230
Oscillations......................................................................................................... . . . . . . . . . . . . . 230
Energy changes in simple harmonic motion................................................................. . . . . . . . . . . . . . . 235
Free and damped oscillations................................................................................... . . . . . . . . . . . . . 237
Topic 14 Ultrasound............................. ............................................................................. . . . . . . . . . . . . . 247
14.6 The generation and use of ultrasound. . ...................................................................... . . . . . . . . . . . . . . 247
Topic 16 Communication....................... ............................................................................. . . . . . . . . . . . . . 255
16.1 Communication channels. ...................................................................................... . . . . . . . . . . . . . . 255
16.2 Modulation............................ ............................................................................ . . . . . . . . . . . . . . 257
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Contents
16.3 Analogue and digital signals.................................................................................. . . . . . . . . . . . . . . . 262
16.4 Relative merits of channels of communication. ........................................................... . . . . . . . . . . . . . . . 265
16.5 Signal attenuation. .................. ........................................................................... . . . . . . . . . . . . . . . 268
Topic 17
17.3
17.4
17.5
Electric fields. ........................ ............................................................................ . . . . . . . . . . . . . . 274
Point charges. .................................................................................................... . . . . . . . . . . . . . . 274
Electric field strength due to a point charge.. ............................................................. . . . . . . . . . . . . . . . 276
Electric potential energy and electric potential. ........................................................... . . . . . . . . . . . . . . 276
Topic 18 Capacitance. ...................................................................................................... . . . . . . . . . . . . . . 281
18.1 Capacitors and capacitance.................................................................................... . . . . . . . . . . . . . . 281
18.2 Energy stored in a capacitor. ...... ........................................................................... . . . . . . . . . . . . . . . 285
Topic 19 & 20 Electronic sensors. ........... ........................................................................... . . . . . . . . . . . . . . . 292
19.4 Sensing devices...................... ........................................................................... . . . . . . . . . . . . . . . 292
20.3 The use of potential dividers. ................................................................................ . . . . . . . . . . . . . . . 295
Topic 21
21.1
21.2
21.3
Electronics........................................................................................................ . . . . . . . . . . . . . . . 298
The ideal operational amplifier (op-amp).. ................................................................. . . . . . . . . . . . . . . . 298
Operational amplifier circuits...... ............................................................................ . . . . . . . . . . . . . . 299
Output devices. ...................... ........................................................................... . . . . . . . . . . . . . . . 305
Topic 22
22.1
22.2
22.3
22.4
22.5
Magnetic fields.................................................................................................. . . . . . . . . . . . . . . . 311
Concept of magnetic field. ......... ............................................................................ . . . . . . . . . . . . . . 311
Force on a current-carrying conductor. ...................................................................... . . . . . . . . . . . . . . 314
Force on a moving charged particle in a magnetic field.................................................. . . . . . . . . . . . . . . 318
Magnetic fields due to currents. ............................................................................. . . . . . . . . . . . . . . . 324
The use of (nuclear) magnetic resonance imaging. ........................................................ . . . . . . . . . . . . . . 326
Topic 23 Electromagnetic induction. ........ ............................................................................ . . . . . . . . . . . . . . 332
23.1 Magnetic flux and electromagnetic induction . ............................................................ . . . . . . . . . . . . . . 332
Topic 24
24.1
24.2
24.3
24.4
Alternating currents. ............................................................................................ . . . . . . . . . . . . . . 341
Characteristics of alternating currents....................................................................... . . . . . . . . . . . . . . 341
Transformers. ......................... ........................................................................... . . . . . . . . . . . . . . . 343
Transmission of electrical energy. ............................................................................ . . . . . . . . . . . . . . 344
Rectification. ......................... ........................................................................... . . . . . . . . . . . . . . . 345
Topic 25 Quantum physics. ............................................................................................... . . . . . . . . . . . . . . . 350
25.1 & 25.2 Photoelectric emission of electrons and energy of a photon .......................................................... 350
25.3 Wave–particle duality............... ........................................................................... . . . . . . . . . . . . . . . 355
25.4 Energy levels in atoms and line spectra. ................................................................... . . . . . . . . . . . . . . . 356
25.5 Band theory........................... ............................................................................ . . . . . . . . . . . . . . 360
25.6 The production and use of X-rays. ............................................................................ . . . . . . . . . . . . . . 363
Topic 26 Nuclear physics. ................................................................................................. . . . . . . . . . . . . . . . 376
26.3 Mass defect and nuclear binding energy. ................................................................... . . . . . . . . . . . . . . 376
26.4 The spontaneous and random nature of radioactive decay............................................. . . . . . . . . . . . . . . . 381
AS Level Answers to Now it’s your turn and Examination style questions.......................................... . . . . . . . . . . . . . . 388
A Level Answers to Now it’s your turn and Examination style questions.......................................... . . . . . . . . . . . . . . . 394
Index . . . . . . . . . ....................................... ........................................................................... . . . . . . . . . . . . . . . 399
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2.1 Measurements
Methods of measuring mass
The method of measuring mass is with a balance. In fact, balances compare the
weight of the unknown mass with the weight of a standard mass. But because weight
is proportional to mass, equality between the unknown weight and the weight of the
standard mass means that the unknown mass is equal to the standard mass.
In your laboratory, you may have access to a number of different types of balance,
including the top-pan balance, the lever balance and the spring balance. It is important
that you should familiarise yourself with the use of all types that are available to you,
so that you do not restrict your choice to one particular type. Note also that some
types of spring balance may be calibrated in force units (that is, in newton) rather than
in mass units (kilogram).
The top-pan balance
Figure 2.15 Top-pan balance
The top-pan balance (Figure 2.15) is a direct-reading instrument, based on a pressure
sensor, or sometimes a spring. The unknown mass is placed on the pan, and its weight
applies a force to the sensor. The mass corresponding to this force is displayed on a
digital read-out.
When using the balance, ensure that the initial (unloaded) reading is zero. There is
a control for adjusting the zero reading. The balance may have a tare facility, for use in
backing off the mass of an empty container so that the mass of material added to the
container is obtained directly. This works in the same way as adjusting the balance for
zero error.
The uncertainty in the reading of a particular top-pan balance will be quoted in the
manufacturer’s manual. As with other digital instruments, it is likely to be expressed
as a percentage uncertainty of the reading shown on the scale, together with the
uncertainty in the final figure of the display.
The spring balance and the lever balance
Spring balances (Figure 2.16) are based on Hooke’s law (see Topic 9): the extension of
a loaded spring is proportional to the load. The extension is measured directly, by a
marker moving along a straight scale, or by a pointer moving over a circular scale. As
with any instrument using a scale and pointer, you should take care not to introduce
a parallax error when you take readings. Position yourself so that your line of sight is
perpendicular to the scale. Before placing the object of unknown mass on the pan,
check for zero error. There is likely to be a zero-error adjustment screw on the balance.
Lever balances are based on the principle of moments. In one common type
(Figure 2.17), the unknown mass is placed on a pan, and balance is achieved by sliding
a mass along a bar, calibrated in mass units, until the bar is horizontal. This represents
the condition in which the moment of the load is equal and opposite to the moment
of the sliding mass and the bar. A reading is taken from the edge of the sliding mass
on the divisions marked on the bar. In this case, parallax error is less likely to be
serious. Again, check for zero error before taking a reading.
Figure 2.17 Lever balance
Figure 2.16 Spring balance
Figure 2.18 Lever balance with
circular scale
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14 Waves
14.4 Doppler effect
The whistle of a train or the siren of a police car appears to increase in frequency
as it moves towards a stationary observer. The frequency change due to the relative
motion between a source of sound or light and an observer is known as the
Doppler effect.
When the observer and source of sound are both stationary, the number of waves per
second reaching the observer will be the same frequency as the source (see Figure 14.13).
stationary
λ
λ
λ
λ
source
stationary
observer
Figure 14.13 The source emits waves of
wavelength λ. The observer is stationary and
receives waves with the same wavelength λ.
When the source moves towards the observer the effect is to shorten the wavelength of
the waves reaching the observer (see Figure 14.14).
P
moving
source
stationary
observer
vs
Figure 14.14 Source of sound moving
towards a stationary observer
Let v be the speed of sound in air. A source of sound has a frequency fs and
wavelength λ. The source moves towards an observer at a speed vs.
The period of oscillation of the source of sound is T (= 1/fs). In the time of one
oscillation the source moves towards the observer a distance vsT. Hence the
wavelength is shortened by this distance. The wavelength of the sound received by the
observer is λ − vsT.
Hence the frequency observed fo = v/(λ − vsT) = v/(v/fs − vs/fs)
fo = fsv/(v − vs)
The source would move away from a stationary observer at position P on the left-hand
side of Figure 14.14. The observed wavelengths would lengthen.
For a source of sound moving away from an observer the observed frequency can
be shown to be
fo = fsv/(v + vs)
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26.2 Fundamental particles
The particles in an atom must experience forces in order to maintain its structure.
The forces were the gravitational force that acts between all masses (see Topic 8)
and the electrostatic force that acts between charged objects (see Topic 17). The
electrostatic force of repulsion is approximately 1036 times greater than the gravitational
force of attraction between protons. Another attractive force must keep the protons
together in the nucleus. This force is known as the strong force and acts between
nucleons. The force does not seem to have any effect outside the nucleus and is,
therefore, considered to be very short range (a little more than the diameter of nuclei,
10−14 m). There appears to be a limiting spacing between nucleons which is similar in
different nuclei and this suggests that the force is repulsive as soon as the nucleons
come close together. The strong force does not act on electrons.
The strong force acts on protons and neutrons but not on electrons.
Example
Figure 26.13 illustrates a hydrogen atom with an electron orbiting the nucleus.
(a) State, for the forces acting on the electron and the proton,
(i) their nature,
(ii) their direction.
(b) Explain why a strong force does not act on the electron or proton.
proton (+e)
+
(a)
(i)
gravitational force (due to the mass of the electron and proton), electrostatic
force (due to the charge on the electron and proton)
(ii) both forces are attractive and, therefore, directed from the one particle
towards the other particle.
(b) The electron is not a nucleon and, hence, is not affected by the strong force. There
is only one nucleon and the strong force acts between nucleons.
electron (−e)
Figure 26.13 Hydrogen atom
Now it’s your turn
4
State the forces acting on the nucleons of a helium nucleus.
Hadrons and leptons
The discovery of antimatter in cosmic radiation supported the theory developed from
the special theory of relativity and quantum theory that all fundamental particles
have a corresponding antimatter particle. The matter and antimatter particles have the
same mass but opposite charge. The following particles were required to support the
theory: the antiproton, the antineutron and the antielectron. The symbols used for the
–
–
antiparticle are p
for the antiproton, n
for the antineutron and e– for the antielectron.
The antielectron or positive electron was introduced in β-particle decay on page 172.
It is also known as the positron.
Many other particles were discovered in cosmic radiation throughout the twentieth
century, giving support for the idea that the electron, proton and neutron were not the
only fundamental particles.
The numerous types of particles are placed into two main categories. Those affected
by the strong force are called hadrons, for example protons and neutrons, and
those not affected by the strong force are called leptons, for example electrons and
positrons.
The many different particles discovered in cosmic radiation have been reproduced
in high-energy collisions using accelerators such as those at Stanford in California and
CERN in Switzerland during the second half of the twentieth century. A vast number of
collisions were carried out and a large number of hadrons were produced. Two of the
conclusions to these reactions were:
●
●
the total electrical charge remains constant
the total number of nucleons normally remains constant.
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26 Particle physics
The quark model of hadrons
The problem of what were considered to be fundamental particles was resolved by
the quark model for hadrons. In this model, the hadrons are made up of three smaller
particles called quarks. The types of quark, called flavours of quark, are up (u),
down (d) and strange (s). The quark flavours have charge and strangeness as shown
in Table 26.4.
Table 26.4 Charge and strangeness values for the three quarks
flavour
charge
strangeness
up (u)
+2
3
1
−
3
0
−1
3
−1
down (d)
strange (s)
0
–
There are three antiquarks, u–, d and s– : these have the opposite values of charge and
strangeness.
Protons and neutrons consist of three quarks.
proton:
charge
+1
strangeness
0
u
u
d
+2
3
0
+2
3
0
−1
3
0
neutron:
charge
0
strangeness
0
u
d
d
+2
3
0
−1
3
0
−1
3
0
In strong interactions, the quark flavour is conserved.
Example
–
State the values of charge and strangeness for the antiquarks u−, and d.
charge − 2
strangeness
0
u−
3
–
charge + 1
strangeness
0
and d.
3
Now it’s your turn
5
Show whether the following reaction can occur.
p + p → p + p– + n
Leptons
Leptons are particles that are not affected by the strong force. The electron and
neutrino and their antimatter partners, the positron and antineutrino, are examples of
leptons. These types of particle do not appear to be composed of any smaller particles
and are, therefore, considered to be fundamental particles.
The emission of electrons or positrons from nuclei was discussed earlier in this topic
(β-decay page 172). During the decay of a neutron in the nucleus, a proton is formed
and an electron and antineutrino emitted. In terms of the fundamental particles,
quarks, the reaction can be shown as follows:
1
0n
→ 11p + –10e + 00v–
u
u
d
u
d
d
The quark flavour is not conserved as a down quark has changed to an up quark. The
reaction cannot be due to the strong force. The β-decay must be due to another force.
This force is called the weak force or weak interaction.
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9781471809217
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Mike Crundell & Geoff Goodwin
9781471809255
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