Let's look at nuclear power. Part 2: Physics again

Door mux op woensdag 24 mei 2017 20:26 - Reacties (4)
Categorie: Techniek, Views: 976

Nuclear power is a contentious subject, for a bunch of reasons. But really, what are those reasons? Even if you have an opinion on the subject, more often than not that will mostly be based on qualitative arguments, not an in-depth understanding of the subject. Today I would like to start to change that.

Nuclear Physics 102

https://tweakers.net/ext/f/6xbw7tM7buAl4eg0Muz9HOSd/full.png

In the previous part of this blog series, we had a look at elementary nuclear physics and how to conceptualize the atom. We learned that atoms consist of nuclei and electrons, where those nuclei again consist of protons and neutrons, which in turn are made of quarks. All matter is held together by four fundamental forces - strong and weak nuclear force, electromagnetism and gravity - but some are more important than others in specific cases. Atomic nuclei mostly experience the nuclear strong force. All particles have associated energies, and the Pauli Exclusion Principle says that two particles cannot exist in the same (energy/other properties) state, so every additional particle in an atom exists at different states. We visualized this as a bucket where each nucleon sat at a different height in the bucket, and that that bucket can either overflow or particles can tunnel through the walls, shedding some of that energy (which we call binding energy).

In this part of the series, we will finalize your nuclear physics course with all you need to know about actual nuclear reactions, radioactivity, reaction chains and conserved quantities.

Conserved quantities
You have probably had chemistry in your secondary education. For instance, you know that you can make hydrogen (H2) and oxygen (O2) from water (H2O). However, water does not contain equal numbers of hydrogen and oxygen atoms - it's got twice as many H as it has O. So if we want to write down a correct reaction, we say:

2 H2O --> 2 H2 + O2

Two water molecules go in, producing two hydrogen and two oxygen molecules. So in total there are four hydrogen atoms and 2 oxygen atoms in this reaction, and the reaction conserves this. It just moves around how these atoms are bonded to each other. We say that the quantity of each element is conserved.

Nuclear reactions also conserve certain quantities. Obviously not atoms - nuclear reactions can alter atoms to be completely different elements. But they still need to make sure that all of its elementary properties, also called quantum numbers are conserved. These properties are:
  • Mass-energy (as well as (angular) momentum)
  • Charge
  • Baryon minus lepton number
  • Weak isospin
There is a little more to it depending on the kind of nuclear reaction (for instance, the nuclear strong force conserves all quark flavours), but this is all you need to know for now. So let's look at a couple of particles and their quantum numbers:

ParticlesymbolRest mass (eV/c2)ChargeBaryon numberLepton numberIsospin
Protonp938 272 081+1+10+1/2
Neutronn939 565 4130+10-1/2
Electrone-510 999-101-1/2 or 0


In this table, rest mass simply means the mass without taking into account linear or angular momentum, which is energy and can, as you know, also be expressed as mass. Also note that the electron has this interesting property that its isospin can be -1/2 or 0, depending on its handedness. I won't go into this here, but it's a fascinating asymmetry and one of those enigmatic facts of this universe. The only thing you need to know is that a particle without isospin cannot undergo weak nuclear interactions.

Alright, let's try to do a nuclear reaction now! Hey, why don't we transform a proton into a neutron? They only differ a little bit by mass and electric charge, and it seems like if you just smash an electron into a proton, it should make a neutron, right? So, let's do this:

p + e- --> n

Let's see if we conserved everything. Well, first of all, it seems like we're missing a little bit of mass-energy. The neutron is about 1.3MeV/c2 heavier than the proton, yet an electron only has a rest mass of 0.5MeV/c2. That's fine, we can just add some momentum to the electron. Charge is obviously conserved, but then we run into trouble. The baryon number of the proton and neutron are 1, but the electron has a lepton number of 1. So B-L on the left-hand side of the equation is 0, while on the right hand side it's 1. Oh, and the isospin is a bit of a troublemaker as well; it's 0 or +1/2 on the left hand side, but -1/2 on the right hand side.

This means we have to do one of two things:
  • Either find a particle with -1 lepton number and -1/2 or -1 isospin to add to the left-hand side
  • Or find a particle with +1 lepton number and +1/2 or +1 isospin to add to the right-hand side of the equation
This particle can have no charge, has to be a lepton and ideally it shouldn't be too big, otherwise we get a big energy imbalance. So let's dig through our lepton drawers to find what we can find:

ParticleSymbolRest mass (eV/c2)ChargeBaryon numberLepton numberIsospin
Electrone-510 999-101-1/2 or 0
Electron neutrinoνeVery little001+1/2
Muonμ-105 658 375-101-1/2 or 0
Muon neutrinoνμVery little001+1/2
Tauτ-1 776 820 000-101-1/2 or 0
Tau neutrinoντVery little001+1/2


Hey, awesome, we actually have a couple of particles that fit! We can put in an electron neutrono on the right hand side. It has the required +1 lepton number and +1/2 isospin. But wait a second, we can just as well add in a muon neutrino or tau neutrino. They all have the same... everything! Well, I can tell you that the electron neutrino is the one you really want, because we're dealing with an interaction involving an electron. Why is this? Well, there are more conserved quantities - every type of interaction has its own little family of quantities. They're not too important for what we're learning today, so I'm ignoring them for now, but you can see how this can quickly lead to trouble reconciling interactions.

Regardless, we now have a working interaction:

p+ + e- --> n + νe

This reaction is called Electron capture and is an important part of the mechanism by which heavy elements are created in supernovae, as well as a decay mode in certain isotopes.

Energy in reactions
So, we've now successfully converted a proton into a neutron - or did we? I already alluded to it - there is an excess of energy in the 'reactant' particles compared to the 'reaction products'. In other words, this reaction is 'exothermic'. I'm using chemistry terms here because I assume many readers are familiar with them. The real reaction should be something like this:

p+ + e- --> n + νe + ~1MeV

This reaction energy isn't a discrete thing - it is distributed over the reaction products in such a way that momentum is conserved. Momentum is energy, but depending on the rest mass of a particle it can result in wildly different speeds. For instance, when a decaying atomic nucleus ejects a neutron spontaneously, usually in the order of 10 MeV of energy is released. This is equally distributed over the neutron and nucleus (assuming no other particles are involved in the process) - 5MeV for the neutron and 5MeV for the nucleus. However, that nucleus is probably over 200 times heavier than the neutron, meaning that if we consider that E = 1/2 mv2, the neutron will be accelerated to about 31 000 km/s, or about 10% of the speed of light. The nucleus, on the other hand, only gets kicked up to about 150 km/s, a comparative snail's pace.

It's also important to note that any reaction that requires energy to be put in, in chemistry terms 'endothermic', will not happen until at least that amount of energy is somehow supplied. That energy needs to be supplied in a way that actually interacts with the nucleus as well - for this reason photons (light, lasers) are often completely unusable, because all they really do is interact with the electrons, not the nucleus.

Cross sections
Say you have a nice deuterium atom (a proton and neutron) and you want to make it into a helium-3 atom (2 protons and a neutron).

21H + 11H --> 32He ( + γ + 5.49MeV)

(By the way, this is probably a good time to introduce the notation we use for nuclear reactions. The H and He are the atoms hydrogen and helium, and to the left of it are two numbers. The top one is the sum of neutrons and protons, while the bottom one is the number of protons. That's all, really. Also, we use a + or - in superscript after the element or particle to denote its electric charge)

So you take the deuterium atom and fire a regular hydrogen atom (proton + electron) at it. This reaction liberates a LOT of energy - 5.49MeV - so you'd imagine that, because the reaction products are much more stable, this is a really easy thing to achieve, right? Unfortunately, the coulomb (electromagnetic) force between the nuclear proton and the proton you're firing at it is very strong, so even if you fire it almost exactly head-on, it is usually deflected. Even if you fire the proton head-on into the deuterium nucleus, it still has a big chance of not getting over the nuclear strong force 'barrier'. Then if you try to accelerate the proton even faster to hit it harder, the entire nucleus falls apart because you add more energy than the binding energy of the nucleus. There is just no winning here.

You can imagine the nucleus as a tiny little target, both in spatial and energy dimensions. If you hit it just in the right spot and just with the right energy, it sticks. The apparent size of the target you need to hit is called the reaction cross-section, and sometimes this also includes the energy window you need to hit. The deuterium-hydrogen reaction we just discussed has an absolutely tiny cross-section. You have a much better chance of fusing deuterium and tritium together, because this has a reaction cross-section (and required energy level) many orders of magnitude wider.

This concept of reaction cross-sections is especially important for nuclear reactions because of the central role this plays in getting these reactions to work at all. But, obviously, it would apply just as much to any other reaction, even those not involving any nucleons or atoms at all. Like the interactions between photons.

Atomic nuclei

Binding energy
We have touched on this before in the first part of the series: Atomic nuclei have less mass than you would expect based on the rest mass of the sum of its particles. Let's take an atom of iron for instance, specifically the isotope iron-56, the most abundant isotope in nature. Iron has the atomic number 26, so it has 26 protons and 30 neutrons, as well as 26 electrons. A neutron weighs 1.674929 x 10-27 kg, a proton 1.672623 x 10-27 and an electron 9.109390 x 10-31. Also, trust me on this one, there are 6.0221309 x 1023 atoms in one mole of material.

So how heavy should iron be per mole? Well:
  • a mole of 26 protons weigh 26 x 1.6726 x 10-27 x 6.022 x 1023 = 0.0261892 kg
  • a mole of 30 protons weigh 30 x 1.6749 x 10-27 x 6.022 x 1023 = 0.0302599 kg
  • a mole of 26 electrons weighs 26 x 9.1040 x 10-31 x 6.022 x 1023 = 0.0000143 kg
So our mole of iron-56 atoms should weigh a nice and even 56.4634 grams. So let's verify our result by going to the Wikipedia page and see that... it actually weighs 55.9349 u. That u, short for the atomic mass unit, means grams per mole. There is about half a gram missing! This mass, or more accurately mass-energy, is called the binding energy. And as you can tell, it's quite a lot - about a percent of the average atom's mass is missing.

Not all atoms have the same binding energy, and generally speaking - as everything in nature - atomic nuclei want to be more stable, i.e. have a higher binding energy. If you graph the binding energy of all elements by weight, you find this:

https://upload.wikimedia.org/wikipedia/commons/thumb/5/53/Binding_energy_curve_-_common_isotopes.svg/500px-Binding_energy_curve_-_common_isotopes.svg.png

This is probably the most important graph in this entire article, because it shows you at a glance why nuclear reactors work the way they do. You see, the graph first goes up, and then it goes down. In other words, somewhere in the middle, near Iron-56, is the 'most stable' atomic nucleus, the one with the highest binding energy. If you have anything lighter than iron, you can fuse them together to create not just a heavier atom, but release a ton of energy in the process. Likewise, take something heavier, break it apart (fission it) and you ALSO release energy. This is why both fusion and fission can release energy, but only with the right kind of 'starter' atoms and 'reaction products'. It clearly wouldn't make sense to fission uranium into tons of hydrogen.

A couple of things in this graph are odd. I said it goes up, then down, but that's a simplification - it goes up and down a lot, especially at the beginning. Most notably, helium-4 - 2 protons and 2 neutrons - is incredibly tightly bound. This is what I was talking about earlier; the helium-4 nucleus sort of exists 'on its own' within a lot of heavier cores, exactly because this spatial arrangement of neutrons and protons is so tightly bound. Likewise, C-12 and O-16 are especially tightly bound.

Decay
As you know from part 1 of this series, atomic nuclei are a careful balancing act of just the right amount of neutrons to protons. This ratio is called the N/Z ratio, where N is the amount of neutrons and Z is the amount of protons. Stable atomic nuclei have an N/Z ratio near 1 for the very smallest atoms (excluding hydrogen, just a proton) to 1.537 for lead-208, the highest N/Z ratio for stable nuclei.

https://upload.wikimedia.org/wikipedia/commons/thumb/8/80/Isotopes_and_half-life.svg/1200px-Isotopes_and_half-life.svg.png

This means that there are always quite a bit more neutrons in an atomic nucleus than protons, and this is the natural state of things. But what happens when this ratio is not obeyed? In other words; what does an isotope do when it has 'too many' or 'too few' neutrons? Well, as you can imagine, it will always try to achieve stability, so too many neutrons means it either wants to shed neutrons, convert neutrons into protons or add protons. Conversely... well, you get the picture.

So we can imagine a couple of ways in which an atomic nucleus can rearrange itself, given the constraints we know already. Let's say we have a nucleus with too many neutrons as compared to protons. What can we do to fix this?

Problem: too many neutrons. Solutions?
  • Convert a neutron into a proton. This increases the atomic number by 1 and causes an electron to be emitted (we've done this before). This is called beta decay (an electron is a beta particle)
  • Eject a neutron. This keeps the atomic number the same but reduces N. This is called neutron emission.
That seems pretty obvious, to be honest. Now, the other option:

Problem: too many protons. Solutions?
  • Convert a proton into a neutron. This decreases the atomic number by 1 and causes an electron from one of the inner shells of an atom to be 'captured' by a proton in the nucleus. That's why we call this electron capture**.
  • Eject a proton. Guess what this is called. Proton emission
  • Eject a Helium-4 nucleus. What?! Well, think about it. There are way more neutrons than protons in a nucleus, so if you eject the same number of protons as neutrons, you proportionally reduce the share of protons in the remaining nucleus. He-4 is extremely tightly bound, so it is the most convenient way to do this trick. But theoretically, you can eject any number of neutrons and protons together that improve the N/Z situation. Ejecting a He-4 nucleus, also known as an alpha particle, is called alpha decay. Ejecting more than one particle at once is called cluster decay.
Which type of decay occurs mostly just depends on what is most energetically favourable. Ejecting an alpha particle generally doesn't happen to light elements - the internal binding energy in the nucleus of lighter elements is too small to accomodate this process. But even in many heavier atoms, it's energetically generally more favourable to do electron capture. You see, if you think back to the bucket model of a nucleus - small nuclei generally sit pretty deep in the bucket, so it's really unlikely for one nucleon to spontaneously pop out, let alone more than one. In order for something like that to happen, you need a lot of energy to push multiple particles over the edge of the bucket.

This can happen, though! For instance when for some reason you put a lot of energy into an atomic nucleus, which puts the nucleus into an excited state. This is often denoted by a little star symbol behind the isotope, e.g. 5626Fe*. This extra energy in the core can cause other types of decay to happen.

There are also other ways to get rid of this extra energy. If a nucleus has extra energy (for instance, it just decayed but couldn't get rid of the energy in its decay products and is left with an excited nucleus) it can emit a photon. This nicely conserves all quantum numbers (a photon has B-L of 0, no isospin, no rest mass, so all you do is get rid of energy with this method) and gets rid of an arbitrary amount of energy for the nucleus. Because these photons often have massive amounts of energy - millions of times as much energy as visible light photons - they're very dangerous and in a class of their own, called gamma rays. This type of 'decay', almost always accompanied by other decay methods, is called gamma decay.

So, with all of this information in hand, we can look again at the picture at the start of this subchapter and show what kind of decay modes are most common for the isotopes. This should be no surprise anymore:

https://upload.wikimedia.org/wikipedia/commons/7/79/NuclideMap_stitched_small_preview.png

Light blue is beta decay, pink is electron capture, yellow is alpha decay and - these are hard to see - dark purple (at the bottom of the picture) is neutron emission while orange (mostly at the extreme top) is proton emission. If it's not clear why all of this is obvious, please re-read this chapter. This is important to understand the rest of this series!

Decay chains
Some elements just can't take a break. Take for instance uranium-238, the most abundant isotope of uranium and one of the very few naturally occurring radioactive isotopes. 23892U, like most very heavy isotopes, is predominantly an alpha emitter. When it ejects an alpha particle, it loses 2 protons and 2 neutrons, becoming 23490Th (Thorium-234). But wait, thorium is ALSO not stable. This is a very heavy isotope if thorium, so it experiences beta decay, becoming one element higher: Protactinium-234. And guess what, Pa-234 is also unstable.

I could go on, but it would get really boring. It takes at least 14 steps for U-238 to decay into a stable isotope, Lead-206:

https://upload.wikimedia.org/wikipedia/commons/thumb/6/65/Decay_chain%284n%2B2%2C_Uranium_series%29.svg/350px-Decay_chain%284n%2B2%2C_Uranium_series%29.svg.png

But, and this is crucial: the entire series of decay products, called the decay chain, is predictable if you follow the rules outlined above and calculate which transitions are allowed based on energy. We also know the half-lives of various products - the time it takes, on average, for half of the atoms in a batch to decay. The bandwidth of half-lives is gigantic - some isotopes decay in less than a nanosecond, others live on for many times the age of the universe. Things that have a short half-life are generally rare, whereas things that stick around - well, they stick around. This makes it so that we can calculate the amount of certain radioactive decay products that we expect given a certain starting situation.

Decay chains also release massive amounts of energy. For instance, even though the initial uranium-to-thorium decay only releases about 5MeV, if you add up all the energy released in nuclear decay from U-238 to Pb-206, you end up with over 50MeV. With the vast majority of decay products only having half-lives of seconds to minutes, we really shouldn't say that nuclear reactors run on Uranium or Thorium - they run just as much on the many decay products in between. Note though - a decent chunk of energy is lost to neutrino's, particles which are extremely hard to interact with, so these generally just shoot off into the cosmos.

Prompt decay
Now, the last piece of the puzzle you need to understand why we build nuclear reactors the way we do is the concept of prompt decay. This hooks back into the explanation on nuclear decay in general: you can cause some isotopes to decay in ways different to natural decay if you blast them with an extra bit of energy. This happens a lot in decay chains in general; a nuclear decay releases energy, but sometimes this energy is not released with the reaction products. This energy then causes the nucleus to exist in an excited state, which is either resolved through a process called [b]internal conversion[/i] (where it for instance ionizes itself by shooting off an electron with the excess energy - note that this is fundamentally different from beta decay, even though it also releases an electron), or it can go into different decay modes that are not possible with the nucleus in a ground state. Most commonly, excited nuclei undergo cluster decay, becoming alpha emitters or even emitting heavier nuclei (sometimes called spontaneous fission). This is especially common with superheavy elements.

The cause of the excited nucleus can also be the decay product of a nearby atom. For instance, a neutron emitted from a nearby nucleus can hit another nucleus, causing it not just to gain energy but an extra neutron as well. In a few very special circumstances, for instance with U-235 and Pu-239, on average each fission reaction produces more neutrons than it took to happen. These isotopes can cause rapidly accelerated nuclear decay, because the decay of one nucleus will cause a chain reaction of subsequent neighboring decays, releasing a lot of energy as it goes on. If left unchecked, you have a nuclear bomb at your hands. If you moderate this process, you can get a lot of energy out of it, even when the fuel ore could never produce that much energy by itself.

https://upload.wikimedia.org/wikipedia/commons/thumb/9/9a/Fission_chain_reaction.svg/497px-Fission_chain_reaction.svg.png
Hey, what about fusion?
We've been primarily talking about nuclear fission, but nuclear fusion is of course just as important for the discussion of energy production from nuclear reactions. However, all the information you just learned can just as well be applied to fusion.

The major difference between fusion and fission is the fact that fission occurs spontaneously at standard conditions - room temperature, typical earth surface pressure, that kind of stuff. In a nuclear reactor, you really have to keep the nuclear fuel from releasing its energy too quickly. Fusion is very much the opposite.

Although - if you look at the graph of binding energies - there is MUCH more energy to be gained by fusing atoms together (especially if you consider that the energy from fused atoms comes from about 100x smaller atoms), the big issue keeping nature from spontaneous fusion is the reaction cross-section. Overcoming the Coulomb and Strong nuclear force requires in the order of 0.1 MeV, which is quite a lot of energy, but the reaction yields neutrons and beta particles with energies in the order of 10MeV. That's good, right? Those can go on and fuse with other nuclei!

Well, no - reaction cross-sections are really reaction windows: too much energy (specifically: more than the binding energy of a nucleus), and you end up destroying your target, consuming the momentum as binding energy and rendering the whole endeavour fruitless. Considering that these are already really tiny targets to hit physically, and it should come as no surprise that you need incredibly high pressures and temperatures all at the same time to initiate a fusion reaction.

Otherwise, there is nothing in this article that can't just as well be applied to both fission and fusion.

Conclusion

This concludes the physics tutorial for this series. Next time we will be looking at some of the simpler reactor designs and understand exactly why they work the way they do. It's going to be exciting, although not nearly as much as Tc-99m.

Let's look at nuclear power. Part 1: Physics

Door mux op zondag 14 mei 2017 15:13 - Reacties (15)
Categorie: Techniek, Views: 2.780

Nuclear power is a contentious subject, for a bunch of reasons. But really, what are those reasons? Even if you have an opinion on the subject, more often than not that will mostly be based on qualitative arguments, not an in-depth understanding of the subject. Today I would like to start to change that.

We're not just talking about nuclear power

Before we start the technical discussion proper, I'd like to say a few words about the motivation behind and aims of this blog series.

There is quite a vocal - but small - group of vehemently pro-nuclear and massively anti-nuclear people on the internet. There is no real discussion between these groups; like most pro/anti-debates, the different camps tend to make straw men out of each other and fight those - or keep repeating the same, debunked or un-nuanced, points over and over again. Instead of indulging this catharsis, I think this energy is much better spent (2 intentional puns there) learning how nuclear power works. This also strongly aids in making sense of the torrent of information present in media streams.

This is a series of blog articles with in-depth treatment of the technology behind nuclear fission and nuclear fusion reactors. I believe there is no good way to understand the political and societal issues without having at least an elementary understanding of how nuclear reactors work. Don't worry though - there are almost no equations and it's generally enough to just follow along with the broad strokes of the articles. However, if you are interested in this subject, I highly recommend using this blog series as a springboard for further reading. I do employ a fair amount of simplifications and generalizations.

I also welcome any constructive criticism, corrections and so on.

Nuclear physics 101

Like rocket science, nuclear physics has a popular image of being fiendishly difficult. But as a qualified rocket scientist, I can tell you that nuclear physics is easy. Anyone can understand it in principle.

Nuclear physics revolve around the nucleus of atoms. Atoms consist of three types of subatomic particles: electrons (which we'll ignore for the time being), neutrons and protons. Neutrons and protons are very similar; they have very similar masses and respond to the same kinds of forces. The major difference is that protons carry a positive electric charge and neutrons do not have an electrical charge. Protons therefore like to repulse each other and need the neutrons as 'glue' to keep them all together and stable. This means that for a given number of protons (which determines what kind of atom you're dealing with - oxygen, carbon, iron) you need a certain number of neutrons to go with it. Too few or too many and the nucleus isn't stable.

Fundamental forces
So far for the TL;DR. Let's dive in! In order to really understand nuclear physics, you need to at least dip your toes in fundamental forces. You can think of fundamental forces as ways of interacting; seeing, hearing, smelling, etc. Not all particles can see, hear or smell the same way. Some can't see at all, or be seen for that matter. The four fundamental forces are:

- The strong nuclear force
- The weak nuclear force
- The electromagnetic force
- Gravity

https://imgs.xkcd.com/comics/fundamental_forces.png

There are no other ways for any two things in the universe, regardless of its composition, to interact with anything else. Now, we have to understand what interacting means. When you see something, what actually happens is that the thing you're seeing has emitted a photon and your eye has detected that photon. Many photons actually. This *is* the electromagnetic force. We call them forces, but in layman terms that conjures up the wrong idea. They're really just ways of interacting, and seeing literally leverages the electromagnetic force.

The strong and weak nuclear force are very different. On the scale of humans, you don't experience these forces. Because whereas the electromagnetic force can reach all the way across the universe (we can literally see the edge of the observable universe - it's called the CMB), the nuclear strong and weak forces - as the name implies - only really works on the scale of atomic nuclei. The further away particles are, the weaker its influence becomes. And I'm not saying that the influence becomes half at twice the distance - it becomes one-millionth. At the scale of molecules, the influence of these forces is already infinitesimally small - let alone at human scales. But, much like with light (which is mediated by photons), the nuclear forces are also transferred by fundamental particles: bosons and mesons.

https://www-tc.pbs.org/wgbh/nova/education/activities/images/3012_elegant_fonffpart.gif

Gravity is a bit of a weird one. We're pretty sure it is a fundamental force, but it has some confusing properties. For instance, it is incredibly weak; you need to mass together millions of trillions of tons of rock to make the gravitational interactions even come close to the strength of interactions within nuclei. Also, we're not really sure if there are mediating particles that actually 'transfer' gravitational interactions between particles. There might be a 'graviton', but we haven't definitively discovered it yet. Also, contrary to all other fundamental forces, gravity acts on *everything*, not just a subset of all particles. Nothing escapes gravity.

Fundamental particles
Much like fundamental forces, there are fundamental particles. You might have heard of them: they are part of the Standard Model. These are particles that cannot be broken up into smaller parts. Each group of particles can interact with a certain subset of fundamental forces.

https://upload.wikimedia.org/wikipedia/commons/thumb/0/00/Standard_Model_of_Elementary_Particles.svg/1200px-Standard_Model_of_Elementary_Particles.svg.png
Standard Model - Wikipedia

All matter is made from these fundamental particles. For instance, neutrons and protons are made from quarks, held together by, primarily, the strong nuclear force mediated by gluons. This is often depicted as the nucleons being made from quarks AND gluons, but that is a little bit of a misnomer; even though gluons mediate the strong force, you shouldn't think of them as literal bits of glue physically existing between the quarks, holding them together. For all practical purposes, these particles are virtual, i.e. they are a mathematical way to represent interactions, not actual physical things.

The valley (or bucket) model of nuclei
So, atoms are what we're really talking about in this article. Atomic nuclei, their cores, are made from protons and neutrons. Protons carry an electric charge and are thus subject to the electromagnetic force. Because they have like charges, they are repelled from each other. On the other hand, the strong nuclear force binds the nucleons together on very short distances. You can imagine that this is a little tug of war between fundamental forces.

The strong force really, REALLY wants to pull nucleons together. Imagine the strong force like a valley between two mountains. In order to pull two nucleons apart, you have to push them up the mountain. Let go of them, and they roll into the valley again. This energy you put in is actual energy and you can measure this! When you make an atom larger, for instance by adding more neutrons or protons, the nucleons will necessariy have to sit further apart (due to the Pauli Exclusion Principle or PEP that says that no two particles can occupy the same exact place and state). Them being further apart means they have to sit higher up the valley, and thus contain more energy and - because energy equals mass - they are heavier then you'd expect! This is called the binding energy. This leads to the interesting situation that free neutrons and protons actually weigh more than neutrons and protons within an atomic nucleus.

https://c1.staticflickr.com/4/3521/3772864128_5ce7c80f86.jpg
In this valley, you can put protons and neutrons. Put more in, and they sit at higher energy levels (denoted by the horizontal lines, each line represents an energy level

The mountains surrounding the valley aren't infinitely tall. They have summits and if you manage to roll a nucleon up the mountain far enough, it will pass the summit and... roll down the other side! It's loose now. The height of the mountain is an analogy for the total binding energy, or conversely, the amount of energy you need to add to a nucleus to break it apart.

Of course, part of this energy exists in the form of that repelling force between protons. For small atoms, for instance hydrogen, the mountains really aren't much more than hills and just adding a second proton to its single proton nucleus already adds so much electromagnetic energy to the mix that the proton will spontaneously pop over the hill; it doesn't want to stay. You have to add at least one neutron to those 2 protons to increase the valley depth and push down those protons. You've just made Helium-3. 2 protons and 1 neutron.

http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/imgnuc/nucpot.gif
The potential well (i.e. valley) is 'pushed down' (from the clear outline to the red outline) by adding neutrons

This gets us into the territory of isotopes. We tend to give different names to atoms if they differ in the amount of protons: Hydrogen has one proton, Helium 2, Lithium 3, etc. We do this because regardless of the amount of neutrons, the chemical properties - what we're most interested in in everyday life - of atoms are purely determined by the amount of electrons. Because electrons naturally occur in equal numbers to protons in atoms, that's how we order atoms. But what about the neutrons? And what about the electrons for that matter?

Most types of atoms (which we usually call elements or species) can have two or more numbers of associated neutrons. We call these different configurations isotopes. For instance, Helium - atomic number 2, so 2 protons and 2 electrons - is stable with either 1 or 2 neutrons. Iron, atomic number 26, is stable with 28, 30, 31 and 32 neutrons. Notice that it's not stable with 29, even though that number is surrounded by stable isotopes. This really is where the bucket analogy fails, and where you should stop imagining atoms as buckets.

http://web.sahra.arizona.edu/programs/isotopes/images/iron.gif

By the way, aside from differing amounts of neutrons, there is also the phenomenon of unequal numbers of protons and electrons. Yes, this does exist; these are called ions. While the vast, vast majority of atoms at any time are neutrally charged - i.e., they have the same number of protons as they have electrons - chemical reactions can cause atoms to lose or gain a few electrons. Not too many, usually 1-6 even if the atom has many more protons. The process of stripping or adding electrons is called ionization. The field of studying interactions that have to do with electron exchange is called chemistry. It is therefore outside the scope of nuclear physics, that only has to do with the nucleus.

A better model for the nucleus
So, we're familiar and comfortable with the model that atoms are wiggling balls in a valley, or bucket, that - if you give them enough energy - they will be able to jump out. This model omits a couple of very important properties of atomic nuclei, though.

http://people.uwplatt.edu/~sundin/114/image/l1423a.gif

First of all, there is tunnelling. You see, there is this law called the Heisenberg Uncertainty Principle (HUP), which says that it's impossible to know with infinite precision where a particle is AND how much momentum it has. Or in other words, you HAVE to trade off positional accuracy with information about momentum. Try to pinpoint exactly where a particle is, and it can have any amount of momentum (energy). Try to determine its exact energy, and you can't know where the particle is.

Now, the HUP only really applies to incredibly tiny things like fundamental particles. The scales that we're talking about are infinitesimal to normal human eyes. But in a nucleus, they can mean very significant things. It should be quite intuitive to understand that particles aren't fixed points - they are more of a cloud of possible locations and energies. That cloud does not have a sharp boundary; the possibility that a particle is further away from its 'average' position simply drops off, but never to zero. This also means that particles have a tiny chance to suddenly find themselves outside of the bucket. They have overcome the potential barrier of the bucket walls not by climbing the walls, but by 'tunnelling' through them. This is not some wishy-washy mathematical thing; there is a measurable (hell, pretty high!) tunnelling ('leakage') current in the processor that powers the device you're reading this on. Electrons tunnel across atoms in your CPU all the time! You can even build microscopes that leverage tunnelling (scanning tunnelling microscopes, or STMs). And you can build nuclear power plants that use tunnelling as their main power source. But I'm getting ahead of myself.

https://qph.ec.quoracdn.net/main-qimg-674c5917220f56bbaa2d611bb8e1c78f-c

The second important nuance of the bucket model is the fact that nucleons aren't independent from each other. First of all, they don't have a super well-defined position within the nucleus; they are all little possibility clouds phasing through each other. But they don't just do this individually; you can imagine that certain arrangements of nucleons just don't work as well as others. You wouldn't expect lots of protons near each other surrounded by a shell of neutrons; the charge would rip apart such an arrangement. Really, the nucleus behaves like a lot of loosely interconnected groups of particles. Some, for instance, are really stable by themselves - the alpha particle (2 protons and 2 neutrons) is a very special case. It isn't just a helium atom, it also exists as a pretty solid subgroup of many other atomic nuclei. And a lot of atoms, when they fall apart, don't just randomly lose mass: they lose this specific type of particle all at once. That is called alpha radiation.

Energy
Whohoo, we're finally going to talk about nuclear energy! Well, sorry, no. We first need to understand, fundamentally, what energy means. As you might know, energy comes in many different forms; macroscopically energy is for instance the momentum of mass, the temperature of a thing, light, etc. This goes just as well for subatomic particles, but with a twist. By Einstein's famous equation E2 = m2c4 + p2c2, energy and mass are the same thing. So when we talk about the mass of some particle, we can say, with full equivalence, that that particle has an energy equivalent to its mass times the speed of light squared. We often express this in units of eV/c2 (electronvolt per light speed squared), which is a measure of energy (trust me on this). And this is an extremely profound thing.

Why? Because this determines exactly what kind of interactions you can expect. You see, like everything in the universe, energy (and thus mass) is always conserved. Every interaction that happens on the subatomic level has to observe this law of conservation of energy. Sort of, I'll get into that later. And interactions on subatomic levels all have associated energies and masses with them.

As a general rule, an interaction can only take place if the associated energy is greater than the rest mass of its gauge particle. Those are those fundamental particles I showed in the picture earlier. For instance, the weak nuclear force has the W and Z gauge bosons mediating its effects. These have a rest mass of 80-90 GeV/c^2. This means that in order to cause a weak nuclear interaction, you have to somehow scrounge together that amount of energy to create this particle.

So what happens if you have less energy than that? Well, you create other kinds of particles. There are rules to this. For instance, energy is not the only thing that is conserved; so are a few other fundamental properties of elementary particles like charge, spin, color (in the case of quarks) and so on. As long as your 'ingredients' of your interaction have the exact amount of these properties AND enough energy to surpass the rest mass of the particle you want to produce, there is a chance you'll create that particle!

A chance eh? What kind of lame excuse is that. See, the kinds of interactions that are possible are constrained mostly by the conserved properties, but not necessarily the energy. You can never produce a particle from just energy - you need its constituent properties at least - but you can produce a particle with just its fundamental properties and much less energy than its rest mass. What? You just said the opposite! Well, this is the HUP again: the location and energy of a particle can't be known with infinite precision, it's a trade-off. So if you have an interaction where a particle only exists for an incredibly short amount of time, i.e. its spacetime location is very well-defined, it can have any energy. So the actual energy you have to put in externally can be arbitrarily small, depending on your point of view the rest of the energy is either 'borrowed' from the void (for a small amount of time) or simply never existed in the first place.

The fact that this is how nature works on the quantum scale doesn't mean that nature is without rules. As I said, you still need to conserve fundamental properties. Also, perhaps most importantly, very out-of-whack interactions don't happen with the same probability as 'well-behaved' particle interactions. The more energy you need to borrow, the shorter the interaction time, the less likely it is to happen.



* Sidebar: This is for instance the reason why the weak nuclear force is so weak - the nominal rest mass of the W and Z bosons are very high, so either the chance of that amount of energy to be available all at once or the chance of a much lower-energy (mass) W/Z boson to be created is extremely small. Small chances of the mediating particle to arise means it doesn't happen very often, that means its effects on the whole are pretty weak. Hence, we call it a weak force.

Energy states
We should also really talk about energy states in a more formal manner. I've talked about the PEP already - it says that no two particles can be in the same state (within certain bounds, but that's another discussion entirely). For instance, a very often-used model for electrons around the nucleus of an atom is to imagine them as a little solar system, with different electrons all orbiting around the atom at different distances. Those differences in distance (=energy) are not arbitrary - they exist at well-defined intervals. We call this quantization, and it's what gives quantum physics its name. Further away means a higher energy state. Back in the day we even assigned names to each of those 'orbitals' - r, s, p, etc. Because electrons have more than just energy differentiating themselves (they also have spin), you can generally fit 2 electrons in one orbital, each with opposite spin.

This model is initially useful to get your head around the concept, but it falls apart when you really look into it. Like everything on this scale, you can't know the exact position and trajectory of an electron; it exists as a cloud of possible locations. This cloud does have a reasonably well-defined shape, though. We can see them here:

https://qph.ec.quoracdn.net/main-qimg-11d8cbaa18d2d5d23e78a080e9f7a667

So all electrons exist in a discrete energy state. The orbital model is sort-of accurate in one way: higher-energy electrons do tend to exist further away from the nucleus, and are thus more exposed to the 'outside world'. The innermost (sometimes called ground state or inner orbital) electrons are usually well-shielded from interaction with other particles. You can imagine where this is going: nucleons exhibit the same kind of energy quantization. Parts of the nucleus are 'more on the outside' (i.e. have a probability density function, or pdf, with a higher mean diameter) and others stay more on the inside.

All this is to say: there is, again, a method to the madness. Things are generally well-ordered and predictable to a large degree. For instance, if you fire a high-energy photon at an atom, in the overwhelming majority of cases the 'outermost' electron will take the hit and get excited. If the energy the photon imparts (plus the intrinsic energy of its orbital location) is more than the binding energy of that electron, the electron will become free from the atom and the atom has been ionized. Whatever excess energy was imparted beyond the binding energy will become the momentum of that electron (and the rest of the atom).

However, it's really more interesting to see what happens when the same thing happens, but the incoming energy source is not strong enough to ionize the atom. Instead of the electron flying away completely, it will 'ascend' an orbital (or rarely two or more). If more energy is available, it is either re-radiated as a lower-energy photon or converted to heat (momentum of the atom). However, the buck doesn't stop here - now we have an electron in an excited state. This configuration is not stable in the long term, and eventually the electron will fall back to its 'rest' state again, emitting a photon. Because the energy difference between orbitals is well-defined, the exact photon energy is always the same for the same transition type. You can imagine that these transitions depend on the type of atom this happens to. Conversely, you can look for photons with this exact energy (=wavelength, =color) and determine with fairly high accuracy that if you see a lot of light of that particular wavelength, you are looking at a specific element in the periodic table. This is how we know what stars are made of - they exhibit anomalous emissions at these particular wavelengths.

http://www.cas.miamioh.edu/~yarrisjm/F21_03.jpg

Nucleons work much the same way, although nucleons have much, much larger binding energies and excitation energies. So just firing a photon at a bare nucleon probably won't cause a proton or neutron to shoot away - but firing a fast neutron or other 'heavy' particle at it could.

Conclusion

We've just laid all the groundwork you need to discuss actual nuclear reactions in the next part of this blog series. Next time we'll talk about nuclear stability, reactions (transmutations), reaction chains and some important quantities in nuclear physics. But don't worry, after that we are all set to talk about nuclear reactor designs!