PHYSICS TUTORIAL NOTES
From
Quanta to Quarks
SYLLABUS
TOPIC 9.8
These notes are meant as a
guide only and are designed to focus your thoughts on the dot points mentioned
in the syllabus. They give a very brief overview of the topic and should be used
in conjunction with your class notes, your textbooks and your research from
other areas such as the library and the Internet.
Notes compiled by:
CARESA EDUCATION SERVICES
FROM QUANTA TO QUARKS
“1. Problems with the
“Students learn to discuss the structure
of the
Under
Most going through undeflected
indicated that most of the atom was empty space. The alpha particles that
bounced back indicated that they were deflected by something massive and
positively charged.
“Students learn to analyse the
significance of the hydrogen spectrum in the development of Bohr’s model of the
atom” (syllabus)
- Electrons orbiting the nucleus are oscillating electric charges and should be continuously emitting electromagnetic radiation. This does not occur.
With the emission of electromagnetic radiation the electrons should gradually lose energy and spiralinto the nucleus. However atoms are stable.
A spiralling electron should emit radiation of all frequencies, producing a continuous spectrum. This is in contrast to the line spectrum of elements that is observed.
- Bohr proposed that electrons could orbit only in certain orbits or energy levels. Under normal conditions it will follow the innermost orbit, which has the least energy and is known as the ground state.
While it is folowing an allowed orbit the electron will not emit radiation or lose energy.
An electron can absorb energy and jump up to a higher energy level. In this energy level it is unstable and will return to its former energy level in one or more steps. As it drops back to a lower energy level it will give out a photon of electromagnetic radiation. The difference between the two energy levels will determine the frequency and hence the colour of the radiation which is governed by Planck’s equation; E = hf.
“Students learn to define Bohr’s
postulates” (syllabus)
Bohr proposed that electrons could orbit only in certain orbits or energy levels. Under normal conditions it will follow the innermost orbit, which has the least energy and is known as the ground state.
While it is following an allowed orbit the electron will not emit radiation or lose energy.
An electron can absorb energy and jump up to a higher energy level. In this energy level it is unstable and will return to its former energy level in one or more steps. As it drops back to a lower energy level it will give out a photon of electromagnetic radiation. The difference between the two energy levels will determine the frequency and hence the colour of the radiation which is governed by Planck’s equation; E = hf.
“Students learn to discuss Planck’s
contribution to the concept of quantised energy” (syllabus)
Planck’s contribution to the concept of quantised energy came from his study of black bodies. A black body is one that will absorb all radiation that hits it. At equilibrium the amount of incident radiation equals the amount of emitted radiation so the temperature remains constant.
Planck
investigated the spectra emitted by cavity radiators and observed that the peak
wavelength was dependent on the temperature of the object and not on the
material that the cavity radiator was made of.
Cavity Radiator
Planck proposed that the atoms inside the cavity oscillated back and forth, emitting radiation, similar to the way a radio antenna works. Energy would be absorbed by the walls of the cavity radiator and re-emitted. Oscillating atoms inside the cavity could receive electromagnetic radiation, but only at certain wavelengths. He proposed that atoms could only receive and emit energy in fixed amounts consistent with the equation E = hf = hc/l, i.e. energy was quantised.
Emitted radiation comes in whole number multiples of a specific quantity; no fraction values are allowed.
“Students learn to describe how Bohr’s postulates
led to the development of a mathematical model to account for the existence of
the hydrogen spectrum:
1/l = R(1/nf2
– 1/ni2 ) ” (syllabus)
The wavelengths of the four visible lines of the hydrogen spectrum were first determined by Anders Angstrom in the mid nineteenth century. Johann Balmer found an equation that would fit these wavelengths and his equation was later modified by Janne Rydberg to 1/l = R(1/22 – 1/n2 ) where n is an integer (3,4,5 or 6) and R is a constant (Rydberg’s constant = 1.097 x 107 m-1). Both Balmer and Rydberg used empirical methods to develop the equation i.e. they found an equation that would fit the numbers with no explanation as to why it worked.
Niels Bohr proposed that electrons could only orbit in certain stable orbits of fixed energy levels. He began by equating the angular momentum of the electron to multiples of h/2p and proceded to develop a mathematical model for the allowed electron orbits and ended up with the Rydberg equation, 1/l = R(1/nf2 – 1/ni2 ). Bohr’s analysis predicted spectral lines outside the visible region and these were later discovered by Lyman, Paschen, Brackett and Pfund.
A site that shows how the Bohr model can be used to derive the Rydberg equation is
http://www.launc.tased.edu.au/online/sciences/Physics/brhydrogen.html
“Students learn to discuss the
limitations of the Bohr model of the hydrogen atom” (syllabus)
The Bohr model does not allow the theoretical calculation of spectral lines other than hydrogen.
It has trouble explaining what happens to atoms with more than one valence electron.
It does not explain why spectral lines are not of equal intensity.
It does not explain the Zeeman effect or hyperfine splitting.
“Students perform a first-hand
investigation to observe the visible components of the hydrogen spectrum”
(syllabus)
To complete this section of the syllabus properly your teacher should set up a hydrogen discharge tube and you should observe it through a spectroscope. The following illustration of the hydrogen spectrum is taken from
http://www.physchem.co.za/Atomic/Hydrogen%20Spectrum.htm
Some other websites that show the hydrogen spectrum are listed below.
http://230nsc1.phy-astr.gsu.edu/hbase/hyde.html
http://cat.middlebury.edu/~chem/chemistry/class/general/ch103/hatom/index.html
http://csep10.phys.utk.edu/astr162/lect/light/absorption.html
“Students process and present
diagrammatic information to illustrate Bohr’s explanation of the Balmer series” (syllabus)
The Balmer series represents the situation where electrons drop down to the second energy level i.e. where nf = 2. It was widely studied because its spectral lines are in the visible region of the spectrum. It is represented in the diagram below.
Balmer Series
Similar diagrams can be drawn for the Lyman
series (nf = 1, the ground state), the
Paschen series (nf = 3), the Brackett
series (nf = 4), and the Pfund series (nf = 5).
Series
for Electron Energy Levels
The energy levels can also be represented in a diagram like the one below. In each case the amount of energy stated is the energy released when ni = T. This is the energy required to completely release an electron and is known as the ionisation energy.
For example, the energy required to completely remove a hydrogen electron in its ground state can be found by:
(i) Calculating the wavelength of the photon required
1/l = R(1/nf2 – 1/ni2 )
1/l = 1.097 x 107 (1/12 – 1/T2)
1/l = 1.097 x 107 (1– 0) = 1.097 x 107
l = 9.1158 x 10-8m (Note that this is in the U.V. region)
(ii) Calculating the energy of the photon in joules.
E = hf = hc/l
E = 6.626 x 10-34 x 3 x 108/ 9.1158 x 10-8 = 2.181 x 10-18 Joules
(iii) Convert to electron volts.
An electron Volt is the energy required to accelerate an electron through a potential difference of 1 volt. W = qV = 1.6 x 10-19 x 1 1 eV = 1.6 x 10-19 J
2.181 x 10-18 Joules = 2.181 x 10-18 / 1.6 x 10-19 eV = 13.6 eV
Electron Energy
Levels for Hydrogen
“Students solve problems and analyse
information using: 1/l = R(1/nf2 – 1/ni2 ) ”
(syllabus)
- The Balmer series of spectral lines occurs when nf = 2. It includes the visible lines of the hydrogen spectrum. It is not the ground state. Calculate the wavelengths of the four visible lines of the hydrogen spectrum.
A. 1/l = R {(1/nf)2 – (1/ni)2}
1/l = 1.097 x 107{(1/2)2 – (1/3)2} l = 6.563 x 10-7 m.
1/l = 1.097 x 107{(1/2)2 – (1/4)2} l = 4.862 x 10-7 m.
1/l = 1.097 x 107{(1/2)2 – (1/5)2} l = 4.341 x 10-7 m.
1/l = 1.097 x 107{(1/2)2 – (1/6)2} l = 4.102 x 10-7 m.
“Students analyse secondary information
to identify the difficulties with the Rutherford-Bohr model, including its inability
to completely explain:
-
the spectra of larger atoms
-
the relative intensity of
spectral lines
-
the existence of hyperfine
spectral lines
-
the Zeeman
effect” (syllabus)
Larger atoms:
While the mathematical model developed by Bohr works well in predicting hydrogen spectral lines it has not been successful with larger atoms. It has had limited success with helium and group I metals but has been unable to predict spectral lines of other atoms. Bohr made no provision for restricting the number of electrons in each orbit. The maximum number of electrons in each orbit is 2n2 and no two electrons can have exactly the same energy levels (Pauli exclusion principle). Bohr’s mathematical model did not allow for these restrictions and consequently does not fit larger atoms.
Intensity:
Electrons drop from a high energy level to a lower energy level and emit a photon of light as they do so. The number of electrons dropping between different energy levels varies and so too does the intensity of the different wavelengths.
Hyperfine
spectral lines:
Hyperfine splitting of spectral lines results from energy level changes in atoms of different isotopes and also from the interaction of nuclear spin with electron spin.
Zeeman effect:
The Zeeman effect refers to the splitting of spectral lines when they are placed in a strong magnetic field. It occurs due the interaction of the electron spin and the magnetic field.
“2. The limitations of classical physics
gave birth to quantum physics.” (syllabus)
“Students learn to describe the impact
of de Broglie’s proposal that any kind of particle
has both wave and particle properties” (syllabus)
The implications of De Broglie’s wave-particle duality theory allow the use of the wave nature of things considered as particles. It explained the stability of electron orbits in atoms by proposing that the orbit of these particles was made up of a whole number of wavelengths. He said that the electron could rise to a higher energy level by increasing its circumference by multiples of its wavelength. Electrons (and all particles) have both wave and particle properties.Electron diffraction allowed the structure of crystals to be determined with greater accuracy while the electron microscope allows such things as cell components to be studied in much greater detail than is possible with the light microscope.
“Students learn to define diffraction
and identify that interference occurs between waves that have been diffracted”
(syllabus)
Diffraction is the spreading out or bending of waves as they pass around a barrier or through an aperture. It occurs when water waves go past a rock or post and join up again on the other side. It also occurs with other types of waves including electromagnetic waves.
In 1665 Francesco Grimaldi showed that when sunlight passed through a very small hole, it spread out and showed coloured fringes at the edge if it was projected onto a screen. He had demonstrated the diffraction of light.
Diffraction
was further demonstrated in 1704 by
In 1803 Thomas Young further demonstrated the diffraction of light with his double slit experiment. He shone light first through a single slit and then through two slits and observed interference patterns on a screen. The single slit ensured coherent sources at the two slits since they had both come from the same wave rather than from two separate waves.
These diffraction and interference effects demonstrated the wave nature of light. This was extended to other electromagnetic waves in 1912 when Max von Laue demonstrated the diffraction of x-rays as they passed through crystal lattices and in 1913 when William Henry Bragg and William Laurence Bragg used x-ray diffraction to determine the crystal structure of diamond and several minerals.
In 1924 Louis de Broglie proposed that waves and matter were inter-related and that matter could be represented as waves where l = h / mv. This phenomenon is known as the wave-particle duality where waves have some matter properties and matter has some wave properties. This was demonstrated by Americans Davisson and Germer in 1927 and by the Englishman George Thomson in 1928 when they demonstrated the diffraction of electrons by bouncing a beam of cathode rays off a crystal of nickel.
While x-rays have the advantage of having wavelengths that approximate the separation of atoms in crystals and hence show maximum diffraction, electrons have the advantage of having a charge and so can be focused. This has enabled the development of the electron microscope.
It is interesting that Joseph John Thomson received the Nobel Prize in 1906 for his work in showing that electrons were particles and his son George Thomson received the Nobel Prize in 1937 (jointly with Clinton Davisson) for his work in showing that electrons are waves.
“Students learn to describe the
confirmation of de Broglie’s proposal by Davisson and Germer” (syllabus)
Davisson and Germer demonstrated electron
diffraction patterns and thus showed that electrons have a wave nature. They
reflected electrons off the surface of a nickel crystal and found that certain
angles produced diffraction patterns. While Davisson
and Germer were demonstrating electron diffraction in
“Students learn to explain the stability
of the electron orbits in the Bohr atom using de Broglie’s
hypothesis” (syllabus)
De Broglie proposed that the circumference of the electron’s orbit was a whole number of wavelengths. He said that the electron could rise to a higher energy level by increasing its circumference by multiples of its wavelength.
“Students solve problems using: l = h/mv” (syllabus)
What is the wavelength of (i) an electron travelling at 106 ms-1?
(ii) a 65 kg physics teacher running at 5 ms-1?
(i) l = h / mv = 6.63 x 10-34 / 9.1 x 10-31 x 106 = 7.3 x 10-10 m.
(ii) l = h / mv = 6.63 x 10-34 / 65 x 5 = 2.0 x 10-36 m.
“Students gather, process, analyse and
present information and use available evidence to assess the contributions made
by Heisenberg and Pauli to the development of atomic
theory” (syllabus)
- Heisenberg’s uncertainty principle states that it is impossible to measure both the momentum and the position of a particle at the same time. The observation of one property disturbs the system such that the other changes.
- Pauli’s exclusion principle states that no two electrons in an atom can have the same set of quantum numbers.
- Pauli predicted the neutrino on the basis of energy loss in beta decay.
“3. The work of Chadwick and Fermi in
producing artificial transmutations led to practical applications of nuclear
physics.” (syllabus)
“Students learn to define components of
the nucleus (protons and neutrons) as nucleons and contrast their properties. ”
(syllabus)
The major nuclear particles, i.e. protons and
neutrons, are known as nucleons.
Protons
and neutrons have similar mass with the neutron being slightly heavier. (Proton
= 1.673 x 10-7 kg, Neutron = 1.675 x 10-7 kg)
The
proton has an electrical charge equal in magnitude and opposite in sign to that
of the electron. (Charge on proton = + 1.602 x 10-19 Coulomb).
The
neutron has no electric charge.
Both
protons and neutrons are composed of three quarks.
Particles
composed of quarks are known as “hadrons” and those composed of three quarks
are known as “baryons”. So protons and neutrons are both hadrons and baryons
(as opposed to mesons which are composed of two quarks – these are hadrons but
not baryons).
Protons
and neutrons are composed of up quarks which have an electric charge of +2/3 e- and of down quarks that
have a charge of – 1/3 e-.
Protons
consist of two up quarks and a down quark while neutrons consist of two down
quarks and an up quark.
“Students learn to discuss the
importance of conservation laws to Chadwick’s discovery of the neutron. ”
(syllabus)
Chadwick’s discovery of the neutron was based on his interpretation of the work of Bothe and Becker and the subsequent work of Frederic and Irene Joliot.
Bothe and Becker bombarded beryllium with alpha rays and noticed that it gave off a penetrating radiation that was undeflected by electric and magnetic fields. They incorrectly thought that these rays were high-energy gamma rays.
Frederic and Irene Joliot repeated Bothe and Becker’s experiment and placed a paraffin sheet in front of the penetrating radiation. They found that protons were emitted from the paraffin. They calculated the energy required by gamma rays to knock out a proton and found it to be unacceptably high. It seemed that either the laws of conservation did not hold or that the radiation was not gamma rays.
James Chadwick read the Joliots’ research and repeated their experiment as well as bombarding several light elements such as helium, lithium and nitrogen with the penetrating radiation and measuring the recoil of their nuclei. Chadwick examined the energy of the penetrating radiation and the energy of the emitted protons and found that if it was gamma radiation then the energy would be insufficient to knock protons out of the paraffin. He also studied the energy of recoil of the several light elements that he bombarded with the penetrating radiation. He was faced with the dilema of either rejecting conservation laws or coming up with an alternative explanation. He found that the conservation laws of energy and momentum would hold if the radiation consisted of neutral particles of about the same mass as the proton.
“Students learn to define the term
“transmutation”. ” (syllabus)
Transmutation is the changing of one element into another. It can occur naturally by the element emitting an alpha particle or a beta particle, or artificially by bombarding the nucleus with another particle.
“Students learn to describe nuclear
transmutations due to natural radioactivity” (syllabus)
Radioactivity
was first discovered by Henri Becquerel in 1896. Further investigation by
Earnest Rutherford showed that there were two types of radiation given off;
alpha rays and beta rays. ( Paul Villard later found a third type of radiation
which he called gamma rays)
The
emission of an alpha particle (2 protons + 2 neutrons) or a beta particle
(electron) changes the atom to a different substance i.e. the atom undergoes
nuclear transmutation.
e.g. 
à 
+ alpha
particle
à 
+ beta
particle
+ 
à 
+ 
+
or
+ à 
+
In 1925 Blackett used a cloud chamber to show
that only two particles were produced. This showed that the reaction
represented by the second equation took place and nitrogen underwent
transmutation into oxygen.
“Students learn to describe Fermi’s
initial experimental observation of nuclear fission. ” (syllabus)
Following the discovery by Curie and Joliot of artificial radioactivity (1934), he demonstrated that nuclear transformation occurs in almost every element subjected to neutron bombardment. This work resulted in the discovery of slow neutrons that same year, leading to the discovery of nuclear fission and the production of elements lying beyond what was until then the Periodic Table. Copied from http://www.nobel.se/physics/laureates/1938/fermi-bio.html
. In 1934, while at the
“Students learn to discuss Pauli’s suggestion of the existence of neutrino and relate
it to the need to account for the energy distribution of electrons emitted in b-decay. ”
(syllabus)
Beta decay seemed to result in some energy loss. While some scientists, such as Bohr, suggested that conservation laws might not hold for sub-atomic particles, Pauli suggested that the energy was carried away by another sub-atomic particle. He predicted that the particle had no charge, had high penetrating power and very little mass.
Enrico Fermi constructed a theory of beta decay based on Pauli’s predictions and studied the shape of the spectrum produced by beta decay. He concluded that the mass of the particle was very close to zero. He named the particle the”neutrino” (little neutral one in Italian).
“Students learn to evaluate the relative
contributions of electrostatic and gravitational forces between nucleons. ”
(syllabus)
The relative contributions of the electrostatic and gravitational forces between nucleons is best understood by doing a numerical example.
1. The separation of nucleons is in the order of 10-15m. Calculate the force of repulsion between two protons at this distance.
F = kq1q2 / r2 = 9.0 x 109 (1.6 x 10-19)2 / (1.0 x 10-15)2 = 230N
2. What is the gravitational attraction between two protons that are 10-15 m apart?
F = Gm1m2/ r2 = 6.67 x 10-11 (1.673 x 10-27)2 /(1.0 x 10-15)2 = 1.87 x 10-34N
3. What is the ratio of the electrostatic force to the gravitational force of two protons that are 10-15 m apart?
230 / 1.87 x 10-34 = 1.2 x 1036 : 1
Note that
the electrostatic force is huge compared with the gravitational force.
“Students learn to account for the need
for the strong nuclear force and describe its properties.” (syllabus)
Two
protons in a nucleus are separated by around 10-15 metres and have
an electrostatic force of repulsion of around 230 N between. Since nuclei are
stable there has to be a stronger force than this to hold the nucleus together.
This is the strong nuclear force and acts only over very small distances. It is
a force between nucleons and acts between protons and protons, neutrons and
neutrons as well as protons and neutrons. At very close distances nucleons
repel each other but the repulsive force turns to one of attraction at
separations of just under 10-15 m. At separations of about 3 x 10-15
m the strong nuclear force drops to zero.
“Students learn to explain the concept of
a mass defect using Einstein’s equivalence between mass and energy” (syllabus)
In a
nuclear reaction the sum of the masses of the reactants is usually greater than
the sum of the masses of the products. (There are some endothermic nuclear
reactions where the opposite is true) The missing mass has been converted to
energy. The amount of energy can be determined by Einstein’s equation: E = mc2
This can
be illustrated by the fusion of two deuterium nuclei to form an isotope of
helium and a neutron. The masses in atomic mass units are:
= 2.01410u
= 3.01603u
= 1.00867u
The
reaction is represented by the following equation.
+ à 
+
The sum
of the masses of the reactants is 2.01410u + 2.01410u = 4.02820u
The sum
of the masses of the products is 3.01603u + 1.00867u = 4.02470u
The loss
of mass (mass defect) is 4.02820u - 4.02470u = 0.00350u
The
energy released = 0.00350 x 931.5 = 3.26 Mev = 5.22 x
10-13 J
“Students learn to describe Fermi’s
demonstration of a controlled nuclear chain reaction in 1942” (syllabus)
Upon the discovery of fission, by Hahn and Strassmann early in 1939, he immediately saw the
possibility of emission of secondary neutrons and of a chain reaction. He
proceeded to work with tremendous enthusiasm, and directed a classical series
of experiments which ultimately led to the atomic pile and the first controlled
nuclear chain reaction. This took place in
Fermi’s own account is given on http://hep.uchicago.edu/cp1.html
“Students learn to compare requirements
for controlled and uncontrolled nuclear chain reactions. ” (syllabus)
Nuclear
reactions produce a lot of energy. This can be released all at once such as in
a nuclear bomb or over an extended period of time such as in a nuclear
reactor. The secret to controlling the rate
of a nuclear reaction is to regulate the rate at which free neutrons collide
with fissionable nuclei. This is done by using a material such as cadmium to
absorb excess neutrons.
In a nuclear reactor the neutron absorbing material is made into control rods that are inserted in between the fuel rods to regulate the rate of the nuclear reaction. To speed the reaction up some control rods are removed from in between the fuel rods while to slow it down, more control rods are inserted between the fuel rods.
“Students perform a first-hand
investigation or gather secondary information to observe radiation emitted from
a nucleus using Wilson Cloud Chamber or similar detection device” (syllabus)
historical intro;
http://www.sciencemuseum.org.uk/on-line/electron/section3/1911b.asp
cloud chamber for cosmic rays
http://www.ulg.ac.be/masc/cloudchamber.htm
“Students solve problems and analyse
information to calculate the mass defect and energy released in natural
transmutation and fission reactions.” (syllabus)
Q.
Tritium, an isotope of hydrogen with two neutrons, emits a beta particle to
become helium.
à + 
Masses
are: = 3.01605u CHECK DATA
= 3.01603u
= 0.00055u
Calculate
the mass defect and binding energy.
A. Mass defect = 3.01605 – 3.01603 – 0.00055
= -0.00053u
Energy = 0.00053 x 931.5 = 0.494 Mev = 7.9 x 10-14 J absorbed
Q.An example of a nuclear fission reaction is shown below.
92U235 + 0n1
92U236 57La148 + 35Br85
+ 3 0n1
The masses of the atoms are;
92U235 = 235.044u
57La148 = 147.915u
35Br85 = 84.911u
0n1 = 1.009u
1 u = 1.66 x 10-27 kg.
(i) What is the mass defect of the reaction in atomic mass units per fission?
(ii) What is the mass defect of the reaction in kilograms per fission?
(iii) What is the energy released, in joules, by the fission of one atom of 92U235?
(iv) What is the energy released, in MeV, by the fission of one atom of 92U235?
A.(i) Mass on left = 235.044 + 1.009 = 236.053u
Mass on right = 147.915 + 84.911 + 3(1.009) =235.853u
Mass defect = 236.053 – 235.853 = 0.200u
(ii) 0.200 x 1.66 x 10-27 = 3.32 x 10-28 kg.
(iii) E = mc2 = 3.32 x 10-28 (3x108)2 = 2.988 x 10-11 J
(iv) 0.200 x 931.5 = 186.3 Mev.
“4. An understanding of the nucleus has
led to large science projects and many applications.” (syllabus)
“Students learn to explain the basic
principles of a fission reactor” (syllabus)
A nuclear
fission reactor enables a nuclear fuel such as uranium or plutonium to undergo
fission (splitting of the nucleus) in a controlled manner with the release of
energy. A diagram illustrating the core of a nuclear reactor is shown below.
The core of a nuclear
reactor
A suitable fuel for the reactor would be enriched uranium.
Enriched uranium is uranium with a higher percentage than normal of the U-235 isotope. Naturally occurring uranium consists of about 99.3% U-238 and about 0.7% U-235 and traces of other uranium isotopes. This low percentage of U-235 is too low to sustain a chain reaction so physical separation techniques are used to remove some of the U-238 and leave a residue with about 3% U-235 and 97% U-238.
The uranium-235 will undergo fission in the reactor core and release energy.
Several equations are possible since U-235 does not always fission in the same way. Here are some examples;
92U235 + 0n1 à 92U236 à 57La148 + 35Br85 + 3 0n1
92U235 + 0n1 à 92U236 à 56Ba141 + 36Kr92 + 3 0n1
92U235 + 0n1 à 92U236 à 50Sn127 + 42Mo104 + 5 0n1
1. A chain reaction is a self-sustaining nuclear reaction in which enough neutrons are released to fission more atoms and keeps the reaction going.
2. The moderator slows the neutron so that it can be captured by the nucleus of the fissile material rather than bouncing off.
3. Heavy water or graphite are suitable moderators.
4. Control rods regulate the rate of the nuclear reaction by absorbing excess neutrons. To speed the reaction up some control rods are removed from in between the fuel rods while to slow it down, more control rods are inserted between the fuel rods.
5. Cadmium is a suitable material for control rods.
6. The coolant lowers the temperature of the reactor core and transfers heat from the core to where it can do useful work.
7. Light water or heavy water are suitable coolants. Heavy water is usually preferred as it can also be used as the moderator.
8. Workers are protected from radiation by a thick layer of reinforced concrete around the reactor core. This is often surrounded by lead to provide further protection.
“Students learn to describe some medical
and industrial applications of radio-isotopes.” (syllabus)
Medicine:
Technetium-99 is used in diagnostic medicine. It is either injected into the bloodstream or consumed. It emits low-energy gamma rays that are easily detected outside the body.
Cobalt-60 emits high-energy gamma rays that are used to treat some forms of cancer. The cancer is bombarded with gamma rays from different directions so that there is maximum exposure of the cancerous cells to the radiation and less exposure of the healthy tissue surrounding the cancer.
Industry:
Cobalt-60 can be used to ensure uniform thickness of sheet metal. A cobalt-60 source is situated above the sheet metal as it rolls along the mill. It emits gamma rays that penetrate the metal and are detected by a geiger counter below. Variations in the thickness of the metal are detected as variations in the radiation count.
“Students learn to describe how neutron
scattering is used as a probe by referring to the properties of neutrons”
(syllabus)
Neutron scattering involves the interaction of neutrons with matter, and investigation of their subsequent path. Scattering can be elastic, in which case diffractometers are used to study the structure of matter, or inelastic, in which case spectrometers are used to analyse the motion and energy of matter.
Neutrons have no electric charge so are unaffected by the charge on the nucleus or its surrounding electrons. Also their small deBroglie wavelength makes them ideal for studying the structure of matter.
“Students learn to identify ways by
which physicists continue to develop their understanding of matter, using
accelerators as a probe to investigate the structure of matter” (syllabus)
The linear accelerator consists of a series of tubes (called drift tubes) in a vacuum. The tubes are connected to high voltage, high frequency, alternating current; alternate tubes to opposite terminals of the A.C. When a charged particle enters the first tube it drifts through it because the electric field strength inside a hollow conductor is zero. When it reaches the gap between the tubes, it is subjected to a high potential difference and accelerates across the gap. It then enters the second tube where it drifts until it reaches the gap between the next two tubes. While it is drifting, the polarity of the tubes changes so that the potential difference between the tubes is always accelerating the particles in the one direction. As the particle moves from one tube to the next it gets faster. To synchronise with the A.C. it must spend the same time in each tube. This is achieved by making the tubes longer as the particle progresses along the linear accelerator.
The cyclotron is more compact than the linear accelerator. It consists of two D shaped chambers connected to a high voltage, high frequency source of A.C. The chambers are in a vacuum and a strong magnetic field. The charged particle enters near the centre of the cyclotron and is accelerated across the gap by the potential difference between the chambers. The magnetic field then deflects the charged particle in a semi-circular path inside the chamber until it again reaches the gap between the chambers. While the particle is in the chamber the polarity of the A.C. between the chambers changes so that the charged particle is always accelerated (not decelerated) between the gaps. It accelerates across the gap twice with each revolution and after many revolutions it emerges with a very high energy.
“Students learn to discuss the key
features and components of the standard model of matter, including quarks and
leptons” (syllabus)
The first
half of the 20th century saw the model of the atom arise whereby
atoms consisted of a nucleus of protons and neutrons surrounded by orbiting
electrons. To make things more complicated, positrons and neutrinos were
discovered. Then with the development of particle accelerators a whole myriad
of new particles was discovered.
Particles
can be divided into three groups: (i) hadrons which
experience a strong nuclear force between them (ii) leptons that do not
experience the strong nuclear force and (iii) bosons which carry the force
fields between other particles.
Hadrons are composed of quarks. There are two types;
(i)
Baryons that have a half integer spin and obey the Pauli
exclusion principle. They consist of three quarks. Protons and neutrons are baryons.
(ii)
(ii) Mesons that have zero or integer spin. They are composed of two
quarks (actually a quark and an antiquark). They are
unstable and decay in a small fraction of a second.
Leptons do not experience the strong nuclear force.
They have half integer spins and obey the Pauli
exclusion principle. Electrons and electron neutrinos are the most common
leptons.
Bosons are particles that carry the force fields.
The strong nuclear force that holds quarks together is facilitated by an exchange
of gluons between the quarks. The electroweak force includes electric fields,
magnetic fields and the weak nuclear force responsible for beta decay. This
force is facilitated by an exchange of photons. The third force field is
gravity and this is believed to be due to an exchange of gravitons. Gravitons
are yet to be discovered.
The Pauli
Exclusion Principle states that two particles cannot exist in the same quantum state. In the
case of electrons, only two electrons can have the same energy level and if
they do, they have opposite spins. Particles that obey the Pauli
exclusion principle have half integer spins and are called fermions.
“Students gather, process and analyse
information to assess the significance of the
The Manhattan Project was the project that developed the atomic bomb during World War II.
The most profound effect on society was the
death of several hundred thousand people in
After the war
The Manhattan Project led to the increased knowledge and understanding of nuclear processes. This has enabled the technology to be used for peaceful purposes such as the production of electricity by nuclear power stations but with this comes the danger of nuclear accidents such as that at Chernobyl and the problem of disposing of nuclear wastes. It has also facilitated the development of nuclear isotopes used in medicine, industry and agriculture.
“Students identify data sources, and
gather, process, and analyse information to describe the use of
-
a named isotope in medicine
- a named isotope in agriculture
- a
named isotope in industry” (syllabus)
Medicine:
Technetium-99 is used in diagnostic medicine. It is either injected into the bloodstream or consumed. It emits low-energy gamma rays that are easily detected outside the body.
Cobalt-60 emits high-energy gamma rays that are used to treat some forms of cancer. The cancer is bombarded with gamma rays from different directions so that there is maximum exposure of the cancerous cells to the radiation and less exposure of the healthy tissue surrounding the cancer.
Agriculture:
Carbon-14 can be used to trace the pathways of biological reactions such as photosynthesis in agriculture. The plant is exposed to carbon dioxide made up of the carbon-14 isotope. By analysing different parts of the plant for carbon-14 it is possible to determine the pathway of carbon dioxide through the plant and to develop more efficient growth patterns.
Industry:
Cobalt-60 can be used to ensure uniform thickness of sheet metal. A cobalt-60 source is situated above the sheet metal as it rolls along the mill. It emits gamma rays that penetrate the metal and are detected by a Geiger counter below. Variations in the thickness of the metal are detected as variations in the radiation count.
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