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).

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.

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).

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.

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)

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.