What is covered in this post?
In this post, we are going to look into two more main topics with AQA A-Level Nuclear Physics: Nuclear Radius, Mass and Energy. We are going to start by discussing the topic of Nuclear Radius and will work our way through to the other topics. I have included an image below of what the AQA specification says regarding these two sub-topics below:
Nuclear Radius
Firstly, it is important to understand a very accurate method of determining the diameter of a nucleus. This can be done through the use of high-energy electrons, which is produced by accelerating electrons through a potential difference. A beam of high-energy electrons that are directed at a thin solid sample of an element will diffract by the nuclei of the atoms in the foil. The electrons diffract as the de Broglie wavelength of the electrons is in the same order of magnitude as the average diameter of a nucleus. A detector is used to measure the number of electrons per second that are diffracted through different angles. I have included an image of the set-up below:
Below is a typical graph for the detector reading on the y-axis against the angle of diffraction on the x-axis.
As can be seen on the graph, as the angle θ increases, the number of electrons detected per second (otherwise known as the intensity of the beam) decreases, then slightly increases, and finally decreases again. This scattering of the electron beam occurs due to the fact that the negatively charged electrons are attracted by the positively charged nucleus. Thus, as the angle θ increases, the intensity of the beam decreases.
Just like the pattern of fringes observed when carrying out Young's Double Slit Experiment (if you don't know what it is, check out this amazing explanation: https://www.onlyphysics.org/post/the-two-slits-that-changed-everything), the same thing essentially happens when the electron beam is scattered - just on a much smaller scale. Diffraction of the electron beam by each nucleus causes a pattern of maximas and minimas in the intensity - just as we see in the double slit experiment. The angle of the first minimum (which is highlighted on the graph) from the centre is measured and used to calculate the diameter of the nucleus - remember that you can calculate the de Broglie wavelength of an electron.
How does nuclear radius depend on nucleon number?
Oftentimes, in Physics, we plot our data on graphs in order to find a mathematical relationship between two variables. In this case, you need to know the relationship between the radius R of different nuclides and their respective mass numbers A. By doing this, physicists were able to deduce the following relationship:
Therefore, if you are to plot the radius R of different nuclides on the y-axis and the value of A^1/3 (so the mass number of each nuclide raised to a power of a third) on the x-axis, you should get a straight line through the origin. The gradient will be equal to r - this is essentially the constant of proportionality between R and A^1/3 and has a value of 1.05 femtometres. I have included an image of this below:
You could also plot ln(R) against ln(A). This should also result in a straight line with a gradient of 1/3. The y-intercept of the graph will be ln(r). This relationship can be shown by taking the natural logarithm of both sides of the equation and comparing it to a standard y = mx+c relationship.
Nuclear Density
Another interesting fact that you must know about nucleons is the fact that the density of a nucleus is constant (irrespective of its size). This can be shown by analysing a few equations.
Firstly, if we assume that the nucleus is spherical, we can write the following equation for the volume of the nucleus. I have included an image below:
From this, we can deduce that the volume of a nucleus is proportional to the mass of the nucleus. We also know that density = mass/volume and so if the volume of a nucleus depends on mass, A, and so does obviously the mass itself, the density of the nucleus is a constant as the masses cancel out. Take a look at the image below for a mathematical explanation:
Note how every term (3, u, 4, π, and r) is a constant and therefore so is the density of a nucleus. So, what does it mean if the density of a nucleus is constant, independent of its radius, and uniform throughout a nucleus? Well, we can conclude that nucleons are separated by the same distance regardless of the size of the nucleus and are thus evenly distributed throughout the nucleus.
We can actually calculate the density of a nucleus. We know that the mass of a nucleus is its atomic mass A multiplied by the value of one atomic mass unit, u, which has a value of 1.661x10^-27 kg. From this, we just sub in the numbers to calculate the density of a nucleus.
That is it for this first sub-topic; now we move onto Mass and Energy!
Nuclear Energy: Energy and Mass
Let's remind ourselves of what the AQA specification requires you to know about the follow topic.
In 1905, Einstein published his theory of special relativity, which was instrumental for the development of modern-day physics. However, all you need to know is the following infamous equation:
You should have seen this equation when learning about Particle Physics: in particular, when you studied about what pair production and annihilation is. This equation essentially means that every quantity of mass has an equivalent amount of energy and vice versa. The mass m actually refers to an objects change in mass (more on this later) and c is, as we know, the speed of light in a vacuum. This equation applies to any modern day event; however, the changes are so minuscule that are they are negligible. It is only when we consider these changes in nuclear reactions that they begin to have significant effects.
Energy changes in reactions
Reactions on a nuclear scale involve minute changes of mass; however, these changes result in immense releases of energy on a macro scale. We can calculate, using the equation below, the energy released Q from the difference in the total mass before and after the nuclear reaction, Δm. Note how the equation is pretty much the same as the E=mc^2 that we have seen before.
In any nuclear change where energy is released, the total mass after the change is always less than the total mass before the change. This is due to the fact that, in the change, some of the mass is converted into energy, which is then released. Below, I have included three particle interactions that you should be familiar with and how the Energy-Mass principle can be applied to them.
Alpha Decay
In alpha (α) decay, the nucleus recoils when an alpha particle is emitted. This is to conserve momentum. But, the energy released in this reaction is shared between the α particle and the nucleus. The energy released that is shared between the α particle and the nucleus is done so in inverse proportion to their masses - you must remember this fact as it has come up in previous exam questions! You can prove this using conservation of momentum, but it is much easier to just remember it as fact.
Beta Decay
In beta (β) decay, the energy released is also shared in different proportions between the β particle, the nucleus, the neutrino or antineutrino released in the decay. When the β particle has maximum kinetic energy, the neutrino/antineutrino has negligible kinetic energy. The maximum kinetic energy of the β particle is less than the energy released in the decay: this is due to the recoil of the nucleus. Note that in calculations we often assume that the mass of a neutrino is negligible (in other words, we ignore its mass).
Electron Capture
In electron capture, energy is released in the decay. This energy is carried away by the neutrino that is emitted from the nucleus. An x-ray photon is also emitted. This is due to the fact that there is an inner-shell vacancy due to the electron that interacts with a proton in electron capture; this vacancy is then filled by a de-exciting electron, resulting in a release of energy through an x-ray photon.
The Strong Nuclear Force
You have probably learnt about the strong nuclear force in AS/Year 12 Physics but we need to recap before we can continue with the Nuclear Physics specification.
The strength of the strong nuclear force can be estimated by working out the force of repulsion between two protons that are distance of 1 fm apart (this is the approximate size of the nucleus), using Coulomb's law of force. The strong nuclear force must be of similar magnitude as it prevents protons from fully repelling from each other.
As we know from before, when we looked at how high-energy electrons can be used to deduce the diameter of a nucleus, nucleons are evenly spaced throughout the nucleus. This, therefore means that the strong nuclear force acts only between neighbouring nucleons (as otherwise the nucleons would not be equally distributed throughout the nucleus).
Another aspect of the strong nuclear force that you could be asked to calculate is the work done by the force to prevent two nucleons from being fully repelled away from each other. In order to calculate this, you just need to understand the fact that the force does work against the force of repulsion between two nucleons (such as two protons). Thus, as work done, W, is equal to Fs (where F is the magnitude of the force and s is the displacement of the object in the direction of the force), we can use this equation to calculate the work done by the force. We can work out the magnitude of the strong nuclear force by considering the force of repulsion between two protons. We can then multiply this force between the distance over which the strong nuclear force acts (about 3 fm), to get the work done by the force. I have included an image of the relevant calculations below:
Now, you may not be asked to do this in a question, and, if you do, the variables may have different values. However, the same principle applies and it is an interesting calculation nonetheless.
Binding Energy
Binding energy is a key part of Nuclear Physics and it is quite difficult to get your head around. The definition of binding energy is: 'the binding energy of the nucleus is the work that must be done to separate a nucleus into its constituent neutrons and protons'.
If you think about it, work must be done to overcome the strong nuclear force and to thus separate each nucleon from the others. As a result, the potential energy of every nucleon is increased as it is removed from the nucleus.
On the flip side, when separate protons and neutrons come together and form a nucleus, energy is released as the potential energy of each nucleon is decreasing. The energy realised is equal to the binding energy of the nucleus. But, if energy is being released in this process, the mass of the nucleus must be less than the mass of the separated nucleons. This is known as the mass defect.
The definition that you need to remember is that 'the mass defect Δm of a nucleus is defined as the difference between the mass of the separated nucleons and the mass of the nucleus'.
The binding energy of a nucleus - so the energy that is released when a nucleus is formed - is equal to the mass defect multiplied by the speed of light squared (c^2). Essentially, we are just using the E = mc^2 equation.
Calculating the binding energy or the energy released in a nuclear reaction is a topic that is tested pretty much all the time. So, I have included an example question below and have included the mark scheme at the bottom of the blog post.
We are almost at the end of this blog post; we just have one more small topic to go over.
Nuclear Stability
As you may expect, the binding energy of each nuclide is different. The binding energy per nucleon of a nucleus is the average work that is required per nucleon to remove all of the nucleons from the nucleus. The more energy required to do so, the more stable the nucleus is. The image below shows the binding energy per nucleon (in MeV) against the mass number (A). You need to have a pretty good mental image of how the graph looks like as questions do come up about it.
In nuclear fission, a large unstable nucleus splits into two fragments which are more stable than the original nucleus. This causes the binding energy per nucleon to increase. Whereas, in nuclear fusion, small nuclei are forced to fuse together to form a larger nucleus. The binding energy per nucleon of the product nucleus is greater than the smaller nuclei. Nuclear fusion can occur provided that the mass number of the product nucleus is not greater than about 56 (Iron's atomic mass number) as this is where nuclei become their most stable and thus fusion stops (too much energy is required for fusion to take place).
Image References
Thank you for reading! If you have any questions about this post, feel free to comment down below or Contact Us via email. Have a look at Part 4 of this series to learn about the final part of the specification - induced fission and safety aspects of nuclear fission reactors.
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