An Introduction to Neutrons Part I, Neutron Cross Sections and the Mean Free Path
The 1930's were a wild time.
So, at long last the edging has finished. We made it, neutrons, what they are and how they interact with matter. So as always we have three atomic particles at this scale. Electrons exist around the nucleus, protons and neutrons exist in the nucleus. Electrons are negatively charged while protons have a positive charge. Neutrons have no charge, hence the neutral name. Protons and electrons were discovered in the late 19th and early 20th centuries by using electrostatic forces to ionize elements. The first to be discovered was the electron, found by J.J. Thompson using a Crookes Tube. A Crookes tube is a glass tube that has had the air pumped out of it. High voltage electrodes are set up inside it and when a voltage is placed across them a ‘ray’ shoots out from the cathode of the tube. This ‘cathode ray’ was found to be susceptible to deflection from electric and magnetic fields and was realized to be composed of particles. These particles were given the name electrons and were the first sub atomic particle discovered. The proton was discovered later on by a very important figure in the history of nuclear science, Ernest Rutherford. Rutherford was a titan and his work can (and does) fill volumes. For now just know that he was able to produce ionized hydrogen (that is a hydrogen nucleus without any electrons orbiting it) without using any hydrogen to begin with. To do this he bombarded nitrogen (specifically nitrogen-14) with alpha particles produced by what I think was radium. Regardless these alpha particles hit the nitrogen and produced the following reaction.
Oxygen-13 was produced along with a positively charged particle with an atomic mass of 1. This was found to be identical to the hydrogen ion and was given the name of proton in 1920.
There was a question remaining though. It was found that the atomic number (the number of protons in the nucleus) corresponded to the electron number but the mass of nuclei was not linear. As the atomic number went up, the mass did not track in a linear fashion. Helium-4 has an atomic number of two and an atomic mass of 4 but uranium has an atomic number of 92 and an atomic mass of 238 (u-238 does ant that’s the vast majority of available uranium). And of course there was the isotope problem, that some fractions of elements had the same atomic number but different mass numbers.
There were a number of theories about this question. Rutherford himself theorized that some sort of ‘neutral proton’ might exist in the nucleus. Another related model assumed the existence of electron-proton pairs that existed in the nucleus which neutralized each other and a surplus of protons corresponded to the atomic number. In the case of helium-4 for example the nucleus would be composed of four protons and two nuclear electrons which would be close enough in proximity to neutralize two of the protons.
There were problems with this and the question went unanswered definitively for over a decade until some strange apparently gamma-ray like emissions were noticed in certain nuclear reactions beginning in 1930. First noticed by a team in Germany under Bothe and expanded upon by the Curies in France, the reaction of beryllium-9 with an alpha particle produced what was taken to be the following.
Where h is planck’s constant and f is the gamma ray frequency. The alpha particle produced was from polonium with an energy of ~5.4 MeV. The radiation, called beryllium-radiation, was assumed to be a gamma-ray because it was not charged; it could not be deflected by electric or magnetic fields like electrons, protons or ions could be. As the Curies demonstrated in their results, the radiation produced in a reaction was such that if it was pointed at a brick of paraffin wax (or other material containing a lot of hydrogen) the rays would cause protons to be ejected from the material. It was assumed this was a mechanism analogous to Compton scattering where electrons are scattered by photons. In this case, protons would be scattered by the so-called beryllium radiation.
There was a problem though. Let’s do a calculation to see where the problem might lie. We’ll assume that the interaction between the gamma ray photon and the hydrogen proton is analogous to that of Compton scattering with electrons. The ‘photon particle’ hits the proton and scatters like a billiard ball being hit. So if we know the final energy of the proton, and we know it’s initial velocity is at rest, what is the energy required for the incoming photon? We can use conservation of momentum to solve the problem. But I can hear your objection, the photon is mass-less, how does this work? Thanks to our old chum matter-energy equivalence, we can do what Compton did and evoke “Associated Mass”. Behold!
Since momentum is just p = mv and the velocity of all light is the speed of light, the momentum of a photon is simply.
Just to be clear, E is the energy, m mass, c is the speed of light, h is Plack’s constant and f is the frequency of the photon. From experiments bombarding paraffin wax with the ‘beryllium radiation’ it was found that the energy of the scattered protons was around 5.7 MeV. That is much smaller than the 938.2 MeV rest mass of the proton, and so we can thankfully avoid relativistic effects in this calculation. How this final energy value was determined is due to something called the Bragg curve, which is for another time. Basically the distance the protons traveled in air was measured, and from that distance velocity and energy were determined. Now we need to do some edging, Compton style. The energy transfer in a collision between a photon and a particle is:
Where Ef is the final energy of the gamma-ray. E is the initial energy of the gamma-ray. m is the mass of the target, in this case a proton and c is as always the speed of light. This is a bit confusing but the smaller the value of the final gamma ray energy the more energy of proton receives in the collision, since the most energy was transferred. The maximum transfer of energy occurs when the scattering angle, θ is 180 degrees or π. At that point, cos(θ) = -1 so we get a factor of 2 in the denominator. This maximum energy transfer is called the Compton edge in gamma ray spectroscopy. Now after a messy derivation we find this simplified form for the final energy.
Ef is also the amount of energy transferred to the target, in this case our proton because we’re assuming a total photon absorption by the target.
But we already know that value; We need the initial energy, and by using our old friend the quadratic formula we can find a function for it.
Running the numbers assuming a final energy proton energy of 5.7 MeV we find a needed initial gamma ray energy of ~54 MeV to impart that amount of energy to the proton. This on its face seems suspect because of the implications of conservation of energy violation. Where does a gamma-ray with that energy come from? But an even more troubling effect was observed when the beryllium rays were pointed at a nitrogen gas target. It was found that nitrogen ions had an energy of ~1.2 MeV. This in turn corresponds to a gamma-ray energy of 90 MeV. So as the target got larger, somehow the gamma-ray got stronger. There was some tomfoolery about!
An experimental physicist named James Chadwick who was a college of Rutherford’s at Cambridge decided to repeat the earlier experiments. He got similar results but had a different interpretation. He reasoned that, instead of a gamma-ray what if there was a neutral particle, with roughly the same mass as the proton but with no electric charge? As it was neutral, it would be highly penetrating, meaning it would be able to travel far into a material before being stopped as only direct collisions with nuclei would effect it and since it had a mass like a proton it would transfer energy much more efficiently than a gamma-ray would. Using this assumption and his experimental data Chadwick was able to demonstrate that such a neutral particle would resolve the odd features of the beryllium radiation problem. Let’s look at the energy required for such a particle to impart 5.7 MeV to a proton. We can start with a conservation of momentum relationship.
In this case we can assume that mp = mn because the mass of the proton and neutron are so similar. The p subscripts refer to protons and the n subscripts to neutrons. i and f refer to initial and final, and v is velocity of course. Next we consider the conservation of kinetic energy relation.
We assume the proton is at rest in the target so vpi = 0. Doing a bit of algebra with two equations and two unknowns (we don’t know either the final neutron or proton velocity) gives us the values for the final proton velocity.
As we assumed our two masses were equal the above function becomes.
In short it is an elastic collision and the initial neutron energy would be around the final proton energy of 5.7 MeV. This is of course an oversimplification when one considers what happens in reality but it demonstrates Chadwick’s insight that a large neutral particle, rather than a high energy gamma-ray explained the experimental results much more readily. Chadwick published his findings in 1932 and received the Nobel Prize in 1935 for the discovery of the neutron but the most important ramification of his work went unnoticed at the time.
A Hungarian theoretical physicist named Leo Szilard working in London read about Chadwick’s work in the London press and a report about other research by Cockroft and Walton that involved splitting lithium into two alpha particles (We will have more to say on the latter experiment and others in a later post). The 1933 paper quoted Lord Rutherford (who was the director of the Cavendish Laboratory at Cambridge where the discoveries had been made) Rutherford had said that anyone talking about the near term use of the energies within atomic nuclei was “talking moonshine”; The neutron was primarily of interest as an academic discovery and a tool for future research. Szilard disagreed. To him the neutron threw the possibility wide open with terrifying geopolitical implications.
The barrier to atomic power at that time was the electrostatic columb barrier. Protons and heavier ions have a positive charge and so they are repelled by positively charged nuclei. To get them to interact requires particle accelerators to push them into energies where they have a chance of overcoming the barrier and interacting. The neutron, being neutral has no such limitation. With barely any energy to its name a neutron can simply slide into a nucleus as there is nothing to repel it electromagnetically. Once it gets close enough the strong force takes over and *pop* it’s in the nucleus.
This principle was put to use almost immediately by physicists to synthesize new isotopes and new elements, most famously by (future Szilard collaborator) Enrico Fermi and his research team. Szilard reasoned one step beyond. If there was an element which when hit with a neutron gave off two neutrons and had a positive Q value (energy, think exothermic reaction) then those two neutrons could each interact with two more nuclei and cause them to give off two more neutrons each. One becomes two, two becomes four, four becomes eight, eight becomes sixteen… A chain reaction and the possibility of a massive explosion. Szilard had no idea at the time what, if any isotope could sustain such a reaction, but the possibility seemed more likely to him than not. It would be six years before such a reaction was to be found. Before we get there though I want to explain the concept of neutron cross sections.
In the wild and wooly world of alchem- I mean physics a cross section is a way of describing a probability. The probability that a certain reaction will happen. For example, let’s say a neutron hits hits a nucleus and is absorbed or captured to use the technical lingo. The cross section describes the probability of that reaction occurring. The best way to think about neutron cross sections or any kind of nuclear cross sections generally is to realize they’re described as an area. The greater the chance of an interaction happening, the larger the cross section. Recall our old friend the nuclear radius formula.
From this we can calculate the surface area of a nucleus. Assuming the nucleus is a sphere, a flat projection of a sphere is a circle, so we can use the equation for the area of a circle to determine the geometrical cross sectional area of a nucleus.
σ is the cross sectional area of the nucleus. In this case the functions we have only describe the geometrical cross section; this is a very simplified model. In actuality a cross section’s values vary wildly depending on the energy of the incoming neutron due to quantum mechanical effects (which, I promise, we will get to). This energy dependence is very important for the rest of our discussion however, so keep it in mind.
As there are many different types of reactions that could happen when a neutron interacts with a nucleus, it is often useful to think of specific cross sections for specific interactions. These can then be summed to find the total cross section at a given energy.
There are cross sections of elastic and inelastic scattering, neutron capture, fission and many others. If you sum them up you get the total.
Let’s give some examples using uranium-235. If we use the above nuclear radius function and assume an R0 of 1.2E-15 m, and a A value of 235 we find a σ value of 1.72E-28 m2. We’ll assume this is a total cross section. 1.72E-28 m2 is very small and to make these small numbers more manageable, physicists use the unit called the barn (bn) instead of meters-squared. One bn = 1E-28 m2 so our result is 1.72 barns. This is overly simple but I just wanted to use this as a way of giving an intuitive feel for the concept. When you hear cross sections, you’re just dealing with an area. The energy changes the area, as does the reaction you’re considering, but ultimately you’re dealing with an area.
For a little compare and contrast, pull up the JENDL ENDF graphing tool and take a look at the experimentally determined cross sections for uranium-235. Enter U235 into the Nuclides text box and 1 into the MT Numbers text box. Click the Draw Graph button and you should see something like this.
The y-axis is the cross section in barns and the x-axis is the energy in electron volts. Note that this graph has axes in a logarithmic scale so 10E-1 = 0.1, 10E0 = 1, 10E1 = 10 and so on. JENDL-5 is the data set, the 300 K means that the target was at roughly room temperature and the (n, total) refers to the reaction type, in this case total neutron cross section.
Now if we look at the graph we can see that 1.72 bn corresponds to nothing on the graph, being too small even at very high energies so there is clearly something missing in our very simplified cross section model outlined above. You’ll notice that the cross sections gradually get smaller as we increase our neutron energy, but around 1 eV things begin to get very wavy. That’s the resonance region, and at those energies quantum mechanical effects really come into play rapidly increasing and decreasing the cross sections. Beyond the resonance region, in the keV range and above, things broadly settle down again. Essentially R0 in the nuclear radius formula is not fixed and is energy dependent which is in turn dependent on nuclear quantum mechanical effects.
So let’s say we have a solid box, and we’re firing some neutrons at it. This box has a surface area S and a thickness dx. To find the total available target surface area for a neutron in this box you can’t just use S because the box is composed of atoms with nuclei and neutrons can slide into the box and hit nuclei further in. So you end up with a function that looks like this.
Where A is the effective area and σt is the total cross section at a given energy. n is the number density, or number of nuclei in a given volume. The number density is defined as the density ρ of the material multiplied by Avogadro’s number NA (which is equal to 6.022E23) and dived by the atomic weight A of the element in question.
Returning to our target area, we multiply the equation by the thickness dx. This is because between nuclei is hard vacuum so neutrons can penetrate into the material encountering nuclei throughout the entire volume. Surface area with depth, a bit of a weird concept.
So now, let’s blast our box with some neutrons, assuming a mono-energetic beam with a fixed velocity vn. Now of course in reality you have a spectrum of neutron energies in a beam, but to keep it simple we can assume mono-energetic neutrons and we’ll find that this assumption works well enough for calculations. So this beam of neutrons has a certain intensity I which means the number of neutrons in the beam passing by per second. To make this more useful we’ll divide this intensity by our surface area S to show the neutron flux, ϕ or the number of neutrons moving through a given area per second.
If we multiply our target area by the flux, we’ll find the rate at which neutrons interact with nuclei in the material; that is the reaction rate RR.
If we do a little algebra and divide the reaction rate by the intensity I we find the reaction probability. This means the likelihood or probability P that a given neutron in the beam will encounter a nucleus and undergo a reaction. The units here are simply reactions over number of neutrons as the seconds cancel out for you unit heads out there.
Now if we subtract this probability from one, we’ll find the probability of a neutron passing through a material without undergoing a reaction. They escape the material. This might seem like semantics but it is useful for the derivation we’re doing.
Now if we have some neutrons N0 and we multiply it by P we will know the number of neutrons that leave the material.
Now let’s imagine that our box’s thickness dx is very thin, becoming a plate and we have a collection of many of these plates placed back to back. The total number of plates is just the length x divided by the thickness of an individual plate dx. The entire length of the collection is x and the total number of plates is just the length x divided by the thickness of an individual plate dx.
One plate is N0P, and each added plate is means one more P is multiplied.
If we do this for the entire length of our layers of plates we get this.
Now if you know anything about calculus, you might notice we’re half way to the Euler limit. If we do a redefinition this will become more clear. Let’s say:
We can then rewrite the above function as:
Now take the limit as y →0 of our function and we get a classic exponential.
This illustrates the how the thickness of a block of material, the type of material as defined by the number density and the cross section all interplay. If the exponent goes to zero all the neutrons make it through the target, and the bigger the exponent value, the fewer neutrons make it through without interacting with nuclei.
From this equation, we can see how the total cross section of a given material can be found. You can do it using a relatively simple technique called a transmission measurement. To do it, a neutron detector is placed at one end of a tube under vacuum and a neutron source is placed at the other end. You need a vacuum because otherwise air molecules would absorb or scatter the neutrons interfering with your measurement. You do a run with an empty tube, no air, no sample and see what results you get on the detector. This is your N0 value. You then place a sample in the tube, in the path of the neutrons, and see what the detector reads out. Since your sample sits in the path of the beam, some of the neutrons interact with it while some sail on by to your detector. This detector reading is your N value. By knowing the ratio of N over N0 and by knowing the number density and thickness of your sample, you can solve for the total cross section.
Solving for σt gives us this relation.
The function is in absolute value brackets because as an area a cross section would not have a negative value. Now there is one more concept I want to touch on here, and I swear this is all going to pay off just bear with me. We know from earlier that the probability of escape P is:
This looks sort of like a relationship known in physics as the Beer-Lambert law.
It is an exponential law describing exactly the situation we’re considering here. If you have a beam of neutrons traveling a distance through some medium on average how far will it travel before interacting with a nucleus? λ here represents a distance in centimeters. By setting our 1/λ equal to our cross section and number density term we can find the distance traveled by a neutron of a given energy (remember the cross section varies with energy).
This is the mean free path (MFP); the average distance a given neutron with a given initial energy will travel in a medium. In our example the cross section value is for the total cross section but the mean free path for different types of reactions can be found using the same formula. Just replace σt with any other cross section value, for scattering or fission or whatever else. This is important for something called neutron diffusion which we will cover in more depth in a later post. It is a way to model the motion of neutrons in a medium, such as a block of uranium. We can use a simple function like this because neutrons are neutral. If you had an ion you would have to account for electromagnetic forces which makes the math more involved.
Let’s do a quick example to demonstrate the mean free path and consider fast fission in some uranium-235. A fast neutron means a neutron above a few keV of neutron energy. Typically We’ll find a fission cross section for u-235 assuming a neutron energy of 2 MeV; this is the typical energy of a neutron released in a fission reaction. Grabbing a chart from the JENDL site we can find the fission cross section of u-235 at 2 MeV.
Two MeV neutrons have a corresponding fission cross section σf of around 1.235 barns (1.1E-28 m2). The number density n of u-235 is 4.793E28 m2. Putting these into our MFP function gives:
So a neutron with an energy of 2 MeV will travel on average a distance of 16.9 cm in a block of uranium-235 before causing a nuclear fission reaction.
Anyway, this is a lot of background info, and I feel like that’s all I’ve been giving you for the past couple of entries but it will pay off. The next one of these posts will continue our discussion of the neutron and its interactions with the nucleus.
More things to read:
A Quantum Theory of the Scattering of X-rays by Light Elements, Compton, Physical Review- 1923 (A good overview of Compton scattering. The scenario is electron-X-ray interactions but the concepts and mathematics can be applied to protons and gamma-rays. The paper can be accessed here)
The Existence of a Neutron, Chadwick, Proccedings of the Royal Society- 1932 (Professor Chadwick’s paper on the discovery of the neutron. A useful and straightforward discussion it can be accessed here)
Handbook of Nuclear Engineering, Volume 1 Cacuci; Editor, Springer, 2010 (A refrence book, for our purposes the first chapter on neutron cross section measurements is useful)