Category Archives: HSC Physics – Quanta to Quarks

Applications of Nuclear Physics

Explain the basic principles of a fission reactor

A fission reactor uses a nuclear reaction to generate electricity. As with all generators, this involves producing rotation to turn a generator. In a nuclear reactor, heat from the nuclear reaction is used to produce steam which turns a turbine, in the same way that burning coal generates steam in a coal power plant. As outlined before, there are several requirements for a controlled reaction. These must be met in a fission reactor to ensure that firstly a reaction takes place, and secondly that the reaction doesn’t go out of control and produce an explosion. In addition to this, there are several key components to a fission reactor. Fuel rods consisting of enriched uranium are placed inside the reactor to provide the critical mass required. Control rods consisting of cadmium or boron are also placed in the reactor, such that they can be moved in and out to control the reaction. The control rods absorb excess neutrons to prevent the reaction from taking place too quickly. When they are lowered, more neutrons are absorbed and the reaction slows, and when pulled out the reaction rate increases. The entire reactor is immersed or surrounded by a moderator to slow down neutrons and thus increase the rate of reaction. The moderator consists of either heavy water, graphite, or various other organic compounds. A coolant is required to extract heat from the reaction and to prevent the reactor from melting. The coolant flows through the reactor then out into a heat exchanger that takes heat extracted from the coolant and uses it to boil water. Spent fuel rods that have been depleted in the reactor are extracted and processed or stored. They are extremely radioactive, making them very difficult to dispose of. Finally, the reactor is surrounded by multiple layers of shielding. There is a graphite shield that reflects neutrons back into the core, followed by a thermal shield to prevent unwanted heat loss from the core, a pressure vessel surrounding the core to isolate and contain everything inside the core, and lastly a biological shield of about 3 metres of concrete mixed with lead pellets, to absorb gamma rays and neutrons.

Figure 1

Remember- A nuclear reactor has a reactor core with fuel rods, control rods and a moderator. Coolant is heated inside the core and pumped out where it boils water. The steam produced turns a turbine, and is then condensed back into water. Shielding is used inside the reactor to prevent radiation and heat from escaping.

Gather, process and analyse information to assess the significance of the Manhattan Project to society

The Manhattan Project was one of the most significant scientific undertakings of the 20th century because of the dramatic impacts it had on society. It consisted of American efforts to produce nuclear weaponry, which were eventually successful and resulted in the deployment of nuclear weapons over Japan in 1945. In terms of impact on society, there were direct scientific impacts, namely the development of nuclear power offering a possible solution to the depletion of fossil fuels and a way of reducing greenhouse gas emissions from power generation. Much more significant however, were the social impacts that atomic weaponry had on global politics. To begin with, nuclear power had terrified the world with its incredible destructive power as witnessed in Japan. As a result, countries with nuclear weapons, primarily the USA and Russia in the period following WW2, became very reluctant to use them, firstly because of their long term destructive power, but secondly because of fear that retaliation would take the form of nuclear reprisal. As a result, although significant political tension built between Russia and USA, it never broke out into conflict, as either side was concerned that aggressive action would result in nuclear warfare, resulting in mutually assured destruction. Where a conventional war would have broken out previously, peace was maintained due to the development of nuclear weaponry. In modern times however, nuclear power is proving to be a dangerous bargaining chip for rogue states such as North Korea and Iran which are using nuclear weapons as leverage in negotiations with the Western world. It has led to a situation where small nations with comparatively weak conventional forces can use the threat of nuclear warfare to negotiate equally with large nations such as the USA. This has led to significant problems in terms of global politics and the power balance between nations necessary to maintain peace. Arguably however, even in these situations the threat of mutually assured destruction is preventing warfare. As a result of the nuclear threat, the UN and the USA are focussing on a diplomatic, sanctions-based approach to resolving conflict rather than an aggressive military approach. Overall, although the Manhattan Project led to the deaths of many Japanese people in Hiroshima and Nagasaki, and although it resulted in a build-up of nuclear arsenals across many nations providing a constant threat to global security, in the end the resulting nuclear stalemate has prevented several wars and therefore averted many possible deaths.

Remember- Although the Manhattan project led to many deaths at the end of WW2, the threat of nuclear war has prevented conflict in the decades after.

Describe some medical and industrial applications of radioisotopes

There are many applications of radioisotopes in medical and industrial fields. In the medical field, radioisotopes are mainly used for imaging/diagnosis and for treatment. In imaging, the transmission of radiation through the body and the degree to which radiation is absorbed can be used to remotely examine the body. They are often used to examine brain activity (using Positron Emission Tomog- raphy). By injecting radioisotopes into the body and examining where they end up (made possible because the radioisotopes are emitting radiation), the circulatory system can be investigated. Fi- nally, radioisotopes are frequently used to kill cancer cells, the radiation destroying them. In industry, they are used to examine stress fractures in metals such as in aircraft wings (because although the fractures may not be visible, radiation can pass through them), detecting leaks in pipes that may be otherwise difficult to find (since radiation will escape from a leaking pipe), and to irradiate medical supplies and food to kill bacteria.

Remember- Radioisotopes are in medicine used for imaging and cancer treatment, while in industry they are used to examine stress fractures and to sterilise objects and food.

Identify data sources and gather, process and analyse information to describe the use of a named isotope in medicine, agriculture and engineering

Iridium-192 is used in medicine to kill cancerous tumours. Iridium-192 pellets are implanted into the tumour where gamma emissions kill the cancer cells. Because the cancer cells are directly exposed to the radiation since they surround the pellet, damage to healthy cells is minimised (as opposed to external irradiation). Because it has a half-life of around 80 days the iridium must be surgically removed after treatment is complete to prevent over-exposure to radiation. The iridium undergoes beta decay, and transmutates to turns to inert platinum that poses no health risk. This makes iridium implants an extremely effective way to treat cancer.

In engineering, cobalt-60 is used to detect stress fractures in metals, particularly in aircraft. Stress fractures occur when metals are repeatedly exposed to strong forces, such as those experienced by the wings of an aircraft. Small fractures can form in the metal, which can eventually result in a catastrophic failure (e.g. there were several cases where early jet aircraft had fuselage explosions because the metals used to construct the aircraft eventually broke apart due to stress). These fractures are the precursors to actual breaks in the metal, but they are extremely hard to detect. By placing cobalt-60 on one side of the metal, and a gamma detector on the other side (often photographic film), the cracks can be identified easily and non-destructively because the gamma radiation only penetrates in areas where stress fractures have formed.

In agriculture, the elements that plants require can be substituted with radioisotopes of the same element. This allows the path of the material to be tracked through the plant’s structure. For example, replacing the phosphorus in soil with a mix containing radioactive phosphorus-32 will allow the path of phosphorus to be tracked into plants. By measuring how radioactive the plants are, how much phosphorus was used by the plant can be determined, as well as the areas in the plant where the phosphorus is concentrated. This has benefits in terms of better understanding the conditions favourable for plant growth, thereby maximising yield and increasing efficiency of the farming process.

Remember- Iridium-192 is implanted to kill cancer cells, cobalt-60 is used is used to detect fractures in metal, and phosphorus-32 is used to trace element flow in plants.

Describe how neutron scattering is used as a probe by referring to the prop- erties of neutrons

In the same way that an electron microscope uses electrons to probe materials, neutrons too can be used in microscopes. However, unlike electrons, neutrons do not carry charge and are therefore not affected by the nuclei of atoms which would deflect electrons. They are therefore extremely useful for imaging crystal structures, as well as substances containing light atoms such as hydrogen. While electrons do pass through crystals and diffract, they are deflected by charges in the crystal, resulting in errors. Neutrons penetrate crystals very effectively, allowing for a clearer and more accurate interference pattern to be produced. According to the de Broglie equation, neutrons have a wavelength shorter that light, similar to the wavelength of an electron.

Remember- Neutrons can be used as effective probes because they have wavelengths similar to electrons. However, because they are not charged they can image objects where charge interferes with electrons.

Identify ways in which physicists continue to develop their understanding of matter, using accelerators as a probe to investigate the structure of matter

Physicists now develop their understanding of matter by examining the component of atoms to better understand them. This necessitates separating the atom into its components, requiring large inputs of energy and sophisticated equipment that was previously unavailable. However, modern particle accelerators are used to break atoms into their components, which are then examined, developing our understanding of matter.

All particle accelerators use magnetic fields to accelerate charged atoms or particles to very high velocities. The three most common types are the linear accelerator, the cyclotron and the synchrotron. A linear accelerator is simply a very long track down which an atom is propelled. It is simple, but its main constraint is size, and energy input is dependant on the length of the accelerator. A cyclotron uses high-frequency AC current to generate a magnetic field that causes the electron to accelerate in a spiral. This reduces the size of the reactor, while at the same time using relatively simple equipment. A synchrotron is a complete circle in which a particle travels. While the particle can be accelerated indefinitely, powerful computers are required to manipulate the magnetic field in the synchrotron in order to propel the particle. Particle accelerators also usually have the ability to generate collisions between high-energy accelerated particles, allowing scientists to examine the properties of matter from the collision. So through using particle accelerators, scientists are able to develop their understanding of matter.

Remember- Scientists are now trying to understand the components of atoms. Accelerators provide the high energies required to break atoms into their components.

Discuss the key features and components of the standard model of matter, including quarks and leptons

The standard model of matter is a theory that states all matter is composed of small elementary particles that exist by themselves or group together to form subatomic particles and to transmit force (because under quantum theory, forces that result due to a field are caused by particles travelling between the objects). There are broadly 3 types of particles. Bosons are force-carrying particles, examples of which include photons that carry electric and magnetic force, gluons that carry the strong nuclear force, and gravitons that cause gravity. Leptons are single elementary particles that exist by themselves and are not affected by the strong nuclear force. They include electrons, muons and taus as well as their neutrino subsidiaries (the electron, muon and tau neutrinos respectively). Quarks are the building blocks of hadrons, which are groups of quarks.

There are 6 types of quarks- up, down, top, bottom, strange and charm, each with different properties. Baryons are groups with 3 quarks, such as protons and neutrons, while mesons are pairings of a quark and an antiquark. Because quarks each have a half-integer spin value (spin being one of Pauli’s quantum numbers), combining two gives a whole integer, while combining three gives a half-integer. Therefore, baryons have half-integer spin values while mesons have whole integer spin values. Thus baryons are also fermions, as fermions are particles that have half-integer spins and therefore obey Pauli’s exclusion principle. Bosons are not fermions.

Figure 2

 

Nuclear Physics and Nuclear Energy

Define the components of the nucleus (protons and neutrons) as nucleons and contrast their properties

Protons and neutrons are both nucleons- particles found in the nucleus, and are slightly different. Both have masses on the same order (measured in amu) but the neutron is slightly heavier than the proton. In terms of charge, the proton has the same charge as an electron only positive, while the neutron has no charge at all. Protons are therefore affected by magnetic and electric fields, while neutrons are not.

Discuss the importance of conservation laws to Chadwick’s discovery of the neutron

Don’t forget that the focus of this dotpoint is the use of conservation laws in regard to discovering the neutron. In an exam make sure you don’t waste time by going into too much detail about the experiment itself.

Chadwick predicted the existence of the neutron based on an experiment that otherwise had no other explanation. When a beryllium atom was bombarded by alpha particles, it emitted a form of radiation. This radiation could not be detected in a cloud chamber and didn’t appear to be a particle- in fact it was initially thought to be gamma radiation. The radiation was capable of knocking protons out of a block of paraffin wax, with the protons travelling away with considerable momentum. In terms of conservation laws there were two applicable to this- conservation of atomic mass and number, and conservation of momentum/energy. Chadwick found that the energy required to eject the proton with the observed momentum could not have been produced by EMR as the energy required would be insufficient (and conservation of momentum would be violated as the photon would not contain enough momentum). However, he realised that a neutral particle would be capable of colliding with a proton and imparting the observed momentum without violating conservation laws. So conservation of momentum was vital in terms of discovery of the neutron. Secondly, the nuclear reaction was \frac{9}{4} Be + \frac{4}{2} He \frac {12}{6} C + ? . By adding mass numbers, according to conservation of atomic mass there would have to be an unknown particle with \frac{1}{0} ? to explain the reaction- so through conservation of mass Chadwick was able to prove the existence of the neutron (and show that the initially observed radiation was in fact a particle).

Figure 1

Remember- Chadwick used conservation of energy to determine the radiation was a particle, and conservation of mass to determine its mass and charge.

Define the term transmutation

Transmutations are nuclear reactions where one element is transformed into another because the number of protons in the nucleus changes- this can occur either due to alpha or beta decay.

Describe nuclear transmutations due to natural radioactivity

Some atoms are inherently unstable because their nuclei exist outside the zone of stability in terms of proton-neutron ratio, or because they have too many protons. This can cause natural radioactive decay to occur, resulting in nuclear transmutation. There are two forms of natural radioactive decay that result in transmutations- alpha and beta decay. In alpha decay, the nucleus emits an alpha particle consisting of two protons and two neutrons, in the process reducing its mass by 4 and its atomic number by 2. A common example is the alpha decay of uranium-238, which occurs according to

 

\frac{238}{92}U -> \frac {234}{90}Th + \frac{4}{2}He

More accurately, the alpha particle doesn’t have any electrons, thus a more complete equation would be

\frac{238}{92}U -> \frac {234}{90}Th^2- + \frac{4}{2}He^2+

However, the alpha particle rapidly gains electrons from surrounding atoms (hence why alpha radiation is the least penetrative form of nuclear radiation) and the Thorium ion formed rapidly loses its extra electrons. Therefore, the charges are typically omitted since they are so short lived.

In beta decay, a neutron decays into a proton (which stays in the nucleus raising the atomic number by one), an electron (which is emitted), and an antineutrino (which is also emitted). In the typical case of beta decay

\frac{1}{0}n -> \frac {1}{1}p + \frac{0}{-1}e+ \overline{v}

This form of beta decay is known as ‘beta-minus’. There is another form of beta decay, ‘beta-plus’, where the proton decays into a neutron, a neutrino, and a positron (antielectron). However, beta-plus decay is not included in the HSC- only beta-minus is considered.

Remember- Alpha decay releases 2 protons and 2 neutrons (an alpha particle) while beta decay releases an electron (a beta particle), an antineutrino, and converts a neutron into a proton.

Describe Fermi’s initial experimental observation of nuclear fission

Fermi initially joined the many other scientists who were using neutron bombardment of heavy nuclei in order to investigate their properties. Fermi was trying to cause uranium to undergo beta decay to produce transuranic elements heavier than uranium. What he initially found was that slow neutrons (slowed by a paraffin wax block) were far more effective than fast neutrons, because they had a greater chance of being captured by the nuclei (since slow neutrons spend more time close to the nucleus, because they travel slower). But most importantly, what he observed was that when he bombarded the nuclei with neutrons, instead of producing a single heavy radioisotope he found 4 separate products each with different half lives. This was his first observation of fission, although he did not realise what was happening in his experiment.

Remember- Fermi was the first to observe nuclear fission when he realised that following a nuclear reaction there was more than one product.

Perform a first-hand investigation or gather secondary information to observe radiation emitted from a nucleus using a Wilson Cloud Chamber or similar detection device

The shape of the trails can be directly linked to the properties of alpha and beta particles. Alpha particles form strong trails because they are more highly charged, and therefore ionise more air as they travel (resulting in more condensation), while beta particles form less intense trails because their ionisation strength is not as great. However, alpha particles trails are shorter, because their strong charge causes them to attract electrons rapidly, so before they travel a long distance they get converted to neutral helium, and can therefore no longer ionise the air. In fact, this is the same reason that alpha radiation is both not very penetrative, yet highly dangerous inside the body. Beta particles react less with their surroundings, which is why they travel a longer distance. Finally, the alpha trails are relatively straight, because the large mass of the alpha particle means that it is deflected less by other particles as it travels. On the other hand, beta particles have a very low mass, and so are very susceptible to having their path changed through interactions with other particles. However, their path is still relatively straight because their high velocity and low charge means that they are less likely to have their path changed.

In this experiment, we constructed a cloud chamber by filling a transparent glass container with a supersaturated vapour. We did this by placing filter paper soaked in methylated spirits inside the container, then cooling the container with dry ice placed below the container. When we placed a radiation source next to the chamber, we were able to observe trails left by alpha and beta particles. This is because when the charged particles travel through the chamber, they ionise the surrounding air. The ions created served as points for the vapour to condense, leaving a trail. The trails from alpha particles were straight, relatively short, and thick/well-defined, while the trails from beta particles were longer and thinner, though they too were straight. As the gamma radiation did not create a stream of ions for condensation, gamma emissions were not visible.

Figure 2

Remember- Alpha particles produce short, straight trails, beta particles produce longer, less straight trails, and gamma radiation doesn’t produce any trail.

Discuss Pauli’s suggestion of the existence of the neutrino and relate it to the need to account for the energy distribution of electrons emitted in beta decay

During beta decay, initially scientists thought only beta particles were emitted. When they evaluated the energies involved, they came up with a figure for the maximum kinetic energy that a beta particle should have. All beta particles should have been emitted with this velocity, but this wasn’t the case. Instead, almost none were emitted with the full amount of kinetic energy, and most of them were emitted with significantly less. This meant that the slow beta particles were missing kinetic energy, leading to a violation of conservation of energy. Also, the sum of the momentums before and after beta decay was not equal- assuming the nucleus starts off stationary, the sum of momentums should be zero. However, when the momentums of the beta particle and the remainder of the nucleus were added, it was not zero, so conservation of momentum was being violated.

Pauli realised that conservation of energy (and momentum) could be resolved if there was an addi- tional particle being emitted in beta decay- the neutrino, so named as it means little neutral one. The sum of all 3 momentums was equal, thereby maintaining conservation of momentum, although the neutrino couldn’t be detected at the time. So the proposal of the neutrino explained the variable kinetic energies of beta particles, and resolved conservation of momentum. It was an excellent idea in this regard, but critically it lacked evidence because scientists at the time could not detect neutrinos. They were only detected 20 years later using more advanced techniques (Later they were shown to be antineutrinos, rather than neutrinos).

 

Figure 3

Remember- Pauli suggested the existence of the neutrino to account for variable beta particle veloc- ities and to fulfil conservation of momentum.

Evaluate the relative contributions of electrostatic and gravitational forces between nucleons

Electrostatic repulsion between like-charged positive protons and gravitation attraction between masses in the nucleus are two of the forces that act between nucleons. However, in terms of relative contributions i.e. relative strength, electrostatic repulsion is far stronger than gravitational attrac- tion. Indeed, the force of gravitational attraction is so insignificant that it can be disregarded in most calculations regarding forces acting between nucleons. The end result of this is that if electrostatic repulsion is forcing the nucleus apart, and gravitational attraction cannot hold it together, then there must be another strong force acting to hold the nucleus together, to prevent it from disintegrating.

Remember- Electrostatic repulsion is far stronger than gravitational attraction within the nucleus.

 

Account for the need for the strong nuclear force and describe its properties

The strong nuclear force is required to hold the nucleus together (as observed countless times because nuclei don’t just fly apart) given that the only significant other force is electrostatic repulsion in the nucleus. Therefore, the strong nuclear force is needed to be an attractive force that opposes electrostatic repulsion and holds the nucleus together. The strong nuclear force is experienced only over very short distances- at extreme short distances it is repulsive, then it becomes attractive as distance increases, then increasingly weaker at large distances (while electrostatic repulsion remains relatively strong). This means that there is a balance of separation where at a particular point, the two forces are balanced and the nucleus is stable. Incidentally, the strong nuclear force is only repulsive at extremely small distances- at reasonably small distances it is attractive, and far stronger than the electrostatic force in terms of magnitude. In fact, the strong nuclear force is the strongest known force in the universe. The force is independent of charge and only acts on neighbouring nucleons, not on the entire nucleus.

Figure 4

Remember- The strong nuclear force is required to hold together the nucleus given that gravitational attraction is so much weaker than electrostatic repulsion. It repels at extremely short distances, then attracts with decreasing strength at increasing distances.

 

Explain the concept of a mass defect using Einstein’s equivalence between mass and energy

See the Extra Content chapter for an explanation of the apparent contradiction described in this dotpoint, where both fusion and fission release energy. Also note that the correct unit for ‘atomic mass unit’ is simply ‘u’, not ‘amu’.

The actual mass of a nucleus is always less then the sum of the masses of the constituents of the nucleus. This means that a helium nucleus with 2 neutrons and 2 protons has less mass than the combined mass of 2 neutrons and 2 protons measured separately. This implies that there is missing mass- this missing mass is called mass defect. It is related to the need for nucleons to lose energy in order to bind together, a stable bond being representative of a low energy state. In order for the nucleons to bond together, they need to lose energy. They do this by losing mass, as according to Einstein mass and energy are equivalent. The mass defect is calculated by simply taking the difference between the mass of the nucleus and the sum of its constituents, usually all carried out in amu. Further, this mass loss can be expressed in energy terms, as MeV, according to 1u = 931.5MeV. The mass defect in terms of energy is also known as the binding energy- the energy required to completely separate out all the parts of the nucleus by breaking bonds. Binding energy is the energy input required to restore the nucleons to their original energy states, thereby breaking the bonds that hold them together in the nucleus. This is also related to the release of energy in fission- the binding energy of a single atom is less than the binding energy of the two atoms produced when the single atom is split. If the total binding energy has increased, that means that more energy is now required to break the bonds, and therefore some energy must have been emitted in the splitting process- this is the energy release from fission (similarly, when split the total mass defect of two smaller nuclei is more than the mass defect of just one nucleus)

 

Figure 5

Remember- Binding energy is the energy required to break a nucleus into its constituents, and mass defect is binding energy expressed in amu by using Einstein’s equivalence between mass and energy.

Describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942

Fermi realised that since the fission of a uranium atom released 3 neutrons, and that since only 1 neutron is required to cause fission in a uranium nucleus, a chain reaction of nuclear fission could be produced that would release a great deal of energy. If neutrons were absorbed such that not all of them produced additional fission, a controlled chain reaction could be produced to release power. This is exactly what Fermi demonstrated in 1942 in a squash court in Chicago at Stagg Field, when he took 50 tonnes of natural uranium in 20000 slugs, in a reactor with 400 tonnes of graphite as a moderator. He used cadmium control rods to prevent the reaction from going out of control. His reaction was successful and was able to generate 0.5 watts in a self-sustaining reaction.

Remember- Fermi built a nuclear reactor at Stagg Field in Chicago with a graphite moderator and cadmium control rods.

Compare requirements for controlled and uncontrolled nuclear chain reactions

To produce an uncontrolled nuclear chain reaction, all that is required is a mass of fissionable material such as Uranium-235 greater than the critical mass specified for that material. The critical mass for a material is the minimum amount of material required so that the neutrons emitted from fission go on to cause further fission reactions in a chain reaction, sustaining the reaction. So for an uncontrolled reaction, all that is required is a source of neutrons, a means of slowing them down, and a super- critical mass of fissionable material. A large lump of fissionable material will generally meet all 3 criteria, as the material itself is a super-critical mass, a source of neutrons, and a means for slowing down neutrons.

To produce a controlled reaction, a reactor is required with a mechanism to capture neutrons so that the overall number of neutrons that cause fission is constant. Normally uranium releases 3 neutrons- the control rods in a reactor capture two of the neutrons so that only one goes on to cause another fission reaction. If all 3 neutrons caused fission, each stage of fission would have triple the number of reactions, resulting in exponential growth of energy release and an uncontrolled reaction. By absorbing excess neutrons, the reaction is controlled and doesnt spiral into an explosion. Also, a moderator is used to slow fast neutrons in the reactor. Fast neutrons travel past nuclei rapidly and have a low chance of being absorbed, causing fission. Slow neutrons spend much longer in the vicinity of atomic nuclei (since they are travelling slower) and so have a much greater chance of being captured by the nucleus. So a controlled nuclear reaction needs a super-critical mass of fissionable material, a source of neutrons, a way to slow those neutrons down, and a mechanism to absorb excess neutrons from the reaction.

Remember- An uncontrolled reaction requires a critical mass of fissionable material, along with a moderator and source of neutrons (which is often the material itself). A controlled reaction also needs a control mechanism to capture excess neutrons.

Matter Waves and the Quantum Atom

Describe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties

The immediate and most important impact that de Broglie’s proposal had was to provide a model to accompany Bohr’s first assertion that there were stable orbits where electrons did not emit energy. Under the first postulate, Bohr simply claimed they would not emit energy, directly contravening Maxwell’s theories without explanation. This, without a model to explain it, deprived the Bohr atom from receiving scientific credibility, and as such it was rejected by the scientific community. De Broglie’s proposal gave a workable solution to explain stable orbits that don’t emit energy, and this gave the Bohr model the credibility it required to be accepted and developed upon by the scientific community, which proved vital in terms of understanding the structure of the atom. Later, de Broglie’s proposal was used to exploit the wave nature of electrons in electron microscopes which could be used to image objects at far greater resolutions than was possible with light due to the smaller wavelength of electrons. De Broglie’s proposal also reconciled Einstein’s theory of light with classical physics by showing that light could have both wave and particle nature.

Remember- de Broglie’s proposal provided a model to solve one of the biggest problems with the Bohr atom, namely the stability of orbits. Later it was used as the foundation for electron micro- scopes.

Define diffraction and identify that interference occurs between waves that have been diffracted

The diagram for this dotpoint has been deliberately oversimplified to show the obstruction of the original wave, and the generation of point-source waves at the corners of the object. In reality, there will be actual interference between the original and the diffracted waves (not shown in this diagram)

Diffraction is the bending of waves around obstructions. It is solely a wave property, and is observed when the passage of a wave is obstructed by an object. The wave can bend around the object and exist where there should be a shadow from the object- this effect is strongest when the size of the object is of the same order as the wavelength of the wave. The corner of the object acts as a point source for the wave, resulting in a curved wave that radiates outward. There are now two waves- the point source and the main wave, and because they exist in the same location interference occurs between the two waves. This means that the process of diffraction results in an interference pattern. This is because at some points the waves interfere destructively and at others they interfere constructively. This results in lines of light and dark, or light and dark rings if it is a circular obstruction.

Remember- Diffraction is when waves bend around objects, and since the corners of the object act as point sources interference occurs between the original wave and the new point source waves.

Describe the confirmation of de Broglie’s proposal by Davisson and Germer

Davisson and Germer were studying the surface of nickel with an electron beam, expecting that even the smoothest surface would appear rough to the electrons. In their experiment an accident occurred and the nickel oxidised when it was exposed to air. To remove the oxide film, they heated the nickel to near its melting point, resulting in the formation of crystals larger than the width of their electron beam. Then, when they fired the beam at the nickel and reflected it to a detector, they observed an interference pattern very similar to an x-ray diffraction pattern, confirming the wave nature of electrons and confirming their wavelength as being very close to what de Broglie predicted.

Remember- Davisson and Germer observed the electron diffraction de Broglie had predicted.

Explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis

Electrons under the de Broglie hypothesis exist as standing waves around the nucleus. This means that they have a closed orbit with no movement of energy, and therefore the orbit is stable with no energy emission. This was the model to explain the stability of electron orbits. The electron orbits must have a circumference equal to a multiple of the wavelength of the electron, as this allows for a standing wave. Therefore,

which was Bohr’s third postulate. Therefore, using de Broglie’s theory of matter waves not only could there be an explanation for Bohr’s first postulate, but it was also possible to mathematically derive Bohr’s third postulate that had initially been nothing more than an assertion.

Remember- Stable orbits in the Bohr atom exist at radii where the circumference of the orbit is a multiple of the wavelength of an electron.

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

Heisenberg and Pauli both made very significant contributions to quantum theory, Heisenberg through his uncertainty principle and Pauli through his exclusion principle.

Heisenberg firstly devised matrix mechanics to explain the atom in terms of quantum probabilities, rather than mixing classical and quantum theory as Bohr had done. This led to an entirely quantum theory of the atom, helping to mathematically understand its nature. Secondly, he devised the uncertainty principle which essentially stated that the more was known about the momentum of a particle, the less could be known about its position in space and vice versa. This changed the way science viewed atomic structure, and is perhaps one of the most important central principles of quantum mechanics, that knowledge of one thing can be mutually exclusive to knowledge of another. This isn’t just due to measurements changing quantities- it’s a fundamental property of quantum mechanics. Heisenberg’s work greatly changed the way in which scientists approached quantum physics.

Eventually, it was realised that the position and properties of an electron could be described in terms of 4 quantum numbers. Pauli’s exclusion principle stated that no two electrons could have all 4 numbers exactly the same- this explained the maximum number of electrons in each shell, and provided a quantum explanation for the position of the first 20 elements in the periodic table. Further, Pauli was able to use his work with quantum numbers to explain the Zeeman effect. Pauli also proposed the existence of the neutrino, another significant subatomic particle.

Remember- Heisenberg devised matrix mechanics and the uncertainty principle, and Pauli developed the exclusion principle and the neutrino.

Atomic Structure

Discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits

The Rutherford atom consisted of a small positive nucleus with negatively charged electrons orbiting the nucleus. The Rutherford atom was devised following Rutherford’s experiments in which he fired positively charged alpha particles at thin gold foil. Rutherford found that while most alpha particles passed straight through the foil, a small proportion of them were reflected back. He hypothesised that they had encountered very dense areas of positive charge. The fact that most alpha particles passed through the gold foil led Rutherford to model the atom with a great deal of empty space. Rutherford modelled the atom with a dense, positively charged nucleus, negatively charged electrons that orbited the nucleus, and free space between the nucleus and the electrons. The model was essentially a simplified version of what we use today- it was groundbreaking at the time as it was a step in the right direction for other scientists to build on, but it lacked a description of where the electrons were and failed to address how atoms had stability without energy emission from accelerating electrons.

figure 14

Remember- The Rutherford atom had a positive nucleus, negative electrons orbiting the nucleus, and empty space between the electrons and the nucleus.

Analyse the significance of the hydrogen spectrum in the development of Bohr’s model of the atom.

Bohr’s model of the atom was quite similar to Rutherford’s, but with two important differences- firstly, it assigned positions to the electrons, but secondly the electron energy levels were quantised. This was radically new, the idea that electrons had energy states and could absorb and emit energy to change states, and had no evidence. Bohr realised that if his model was correct, each atom would have a spectral fingerprint related to the differences between electron energy levels in that atom. The Rydberg equation, otherwise known as the Balmer equation, gave him evidence for the quantised emission of energy from the hydrogen atom, leading to him going on to further his model and define his postulates. So the hydrogen spectrum was very significant to the development of Bohr’s model of the atom, because without an understanding of it Bohr may not have continued to work on his model.

Remember- The hydrogen spectrum was extremely significant because it provided the only evidence at the time for an otherwise purely theoretical model.

Perform a first-hand investigation to observe the visible components of the hydrogen spectrum

In our experiment, we had a discharge tube (vacuum tube with a cathode and anode, powered by a high-voltage induction coil) with low-pressure hydrogen inside it. When high-voltage current was passed through the tube, the hydrogen fluoresced, emitting light that was visible in our darkened room. We observed the visible components of the spectrum with handheld spectrometers that used a diffraction grating to split the light. Using the spectrometer, we could clearly observe the red and blue/violet hydrogen emission lines, although the violet lines were very hard to observe. The red line was very clear and intense compared to the other observed lines.

Remember- Hydrogen discharge tube observed with a spectrometer.

Discuss Planck’s contribution to the concept of quantised energy

The concept of quantised energy is that energy can only occur in small packets of fixed amounts, and distinguished between energy increases due to increased intensity (bigger packets) and energy increases due to greater intensity (more packets). This was developed entirely by Planck in his work on black body radiation, and although Einstein significantly improved upon Planck’s ideas, the underlying idea was Planck’s alone, and so Planck made a huge contribution to the underlying concept of quantised energy. However, his involvement was limited to developing the mere concept- others developed it into a functional model.

Remember- Planck developed the concept of quantised energy, but not a functioning model.

Define Bohr’s postulates

Bohr had 3 postulates. The first was that electrons in an atom exist in stationary states of stability and emit no energy when in these states. The second was that energy is only lost or gained by an electron when it moves from state to state, and when it moves from a high energy state to a low energy state it releases a photon with energy equal to the difference between the states (and therefore a characteristic frequency). His third postulate was that electron angular momentum in a

Describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum (the Rydberg equation)

Balmer originally devised the equation empirically by examining the lines in the hydrogen spectrum and creating a formula to fit them. Rydberg used Bohr’s postulates and manipulated them (especially the third) to create the same formula (derived from calculating differences in energy states). Essentially, there were two paths to the Rydberg equation and one of them used Bohr’s postulates to arrive at the equation, while the other didn’t.

Process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series

In this diagram, the energy levels described by Bohr are clearly marked. According to Bohr, the Balmer series (shown on the top of the diagram as the hydrogen spectrum) was caused by electrons changing energy levels. The electron makes a transition from a higher energy level to a lower energy level, in the process releasing light. As shown, larger energy changes produce more energetic photons, as seen in the Balmer series, and further, this diagram shows how the Balmer series is formed by successive electron transitions to the 2nd shell (transitions to other shells produce additional lines named after their discoverers).

figure 15

Remember- Bohr explained the Balmer series as being the result of successive electron transitions down to the 2nd shell.

Discuss the limitations of the Bohr model of the hydrogen atom (including “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, and the Zeeman Effect”)

For all the questions the Bohr model answered, it posed still more. There was still no explanation for there being no energy emission from accelerating electrons as Maxwell predicted- instead it was simply an assumption. Further, there was no evidence for the Bohr model to give it scientific credibility. Finally, in terms of explaining spectral lines there were observed effects that simply could not be explained. These were

  • Relative intensity of spectral lines- When observing spectra, some lines were much brighter than The Bohr model could not explain why some lines were more intense than others (i.e. why some electron transitions were preferred to others)
  • Hyperfine splitting- When the spectral lines were examined closely, it was observed that each line actually consisted of many small lines, the existence of which the Bohr model could not explain as it only predicted one clear line for each transition
  • Larger atoms- The Bohr model could not explain the spectra of larger atoms with more than one electron, a problem that Bohr tried unsuccessfully to
  • Zeeman effect- The Zeeman effect occurs when a magnetic field is passed through the discharge The magnetic field increases the hyperfine splitting of spectral lines, further breaking them up. Again, the Bohr model was unable to explain the experimental evidence

Although the Bohr model lay down the framework for the quantum model of the atom, which ended in a scientific revolution out of which quantum mechanics (a vital part of modern physics) emerged, it was left to future scientists such as Pauli and Heisenberg to fully explain these phenomena.

Remember- Relative intensity, hyperfine splitting, larger atoms, Zeeman effect.