4.4: Nuclear Cross Sections

Atomic nuclei are so small that they cannot be seen even through the most powerful existing microscopes. In order to investigate them, one has to use a number of other methods. One is to bombard them subatomic particles such as, e.g., protons, neutrons, or \(a\)-particles - and observe what happens.

In the available literature one can find the approximate formula for the radius \(R\) of an atomic nucleus of an atomic number \(A\) - assuming that it is a spherical object:

\[ R=r_{0} \mathrm{~A}^{\frac{1}{3}} \]

where \(r_{0}\) is usually given as \(1.2 \mathrm{fm}\) or \(1.25 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}=10^{-13} \mathrm{~cm}\right)\).

Suppose that a particle, e.g., a neutron, is flying towards the nucleus. For a moment, let's assume that the particle can see like a one-eyed creature. What would it see? Obviously, a sphere, of radius \(R\) watched from a distance by one eye looks like a disk of radius \(R\). The surface area of such disk is, of course, the same as the area of the cross section of the sphere.

The symbol used for the cross section area in nuclear physics is \(\sigma\) (Greek sigma). Consider a nucleus of an element from the middle of the Periodic Table - say, with \(\mathrm{A}\) - and find the value of \(\sigma\) for it:

\[ \sigma=\pi R^{2}=\pi \times\left(1.25 \times 10^{-13} \mathrm{~cm}\right)^{2} \times 100^{\frac{2}{3}}=1.058 \times 10^{-24} \mathrm{~cm}^{2} \]

For nuclei with other \(A\) values the results will be all of the order of \(10^{-24} \mathrm{~cm}^{2}\). Therefore nuclear physicists specializing in bombarding nuclei with different particles introduced, for convenience, a special unit of the name "barn": 1 barn \(=10^{-24} \mathrm{~cm}^{2}\). Also, physicists have discarded the word "area", now \(\sigma\) is referred to simply as the "cross section".

One may wonder: What is so special in the cross section that a special unit has been created for it? And what is the cross section good for in nuclear physics? We show an example below, and later on, it will become clear that understanding the notion of cross section is essential for understanding fully how a nuclear reactor works.

An example In lead metal \((\mathrm{Pb})\) the packing of atoms is such that there are \(3.296^{*} 1 \mathrm{O}^{22}\) atoms \(/ \mathrm{cm}^{3}\). If a low-energy neutron passing through the metal hits a nucleus, it's bounced off its path (in professional terminology, one says "scattered"). The cross section of a single nucleus of the metal for such process is \(11.2\) barns.

Suppose that a beam of neutrons is incident at a right angle on a lead plate of 2 cmthickness. Find what fraction of neutrons pass the plate without changing their flight direction, and what fraction of neutrons are bounced off their original flight direction?

Answer. Draw (mentally) a \(1 \mathrm{~cm} \times 1 \mathrm{~cm}\) square on the slab's surface. How many atoms there are behind the square's face? The volume of the material behind is \(1 \mathrm{~cm}^{2} 2 \mathrm{~cm}=2 \mathrm{~cm}^{3}\), so that behind the face there are \(6.592 \times 10^{22}\) atoms. What is the total cross section area of all nuclei of these atoms? The answer is: \(6.592 \times 10^{22} \times 11.2 \times 10^{-24} \mathrm{~cm}^{2}=0.751 \mathrm{~cm}^{2}\). So, an incoming neutron "sees" a \(1 \mathrm{~cm}^{2}\) square on the slab surface, and behind it it sees an enormous number of tiny discs, the total surface area of which obscure obscure \(0.751\) \(\mathrm{cm}^{2}\) of the square's surface area - it means, \(75.1 \%\) of it. Hence, a single neutron has \(75.1 \%\) chance of hitting one of the discs and being bounced out of its original flight path, while it has \(24.9 \%\) chance of passing through the plate without any collision. In other words, if there are many neutrons in the impinging beam, \(24.9 \%\) of them will emerge from the other side of the slab, as if there was no obstacle of any kind put across the beam's path.

So, as the example shows, working with cross sections maybe pretty straightforward - but now we have to reveal a secret concerning what has been said above: namely, the model in which the cross section is taken as \(\sigma=\) \(\pi R^{2}\), would be correct, yes, but if our "nuclei" were billiard balls, and our projectile-particles were, for instance, balls from a BB gun. In other words, in our "macro-world". But this model seldom describes correctly collisions in the micro-world, which is the realm of quantum phenomena.

The cross sections in the micro-world may be very different even for nuclei with the same A. So, the cross section \(\sigma\) is not the same as the cross section area of a simple sphere. For instance, the Xenon- 135 nucleus, a real neutron devourer - an accomplice of the Chernobyl disaster, and a source of strong headaches for nuclear reactor operators - has the geometric cross section of about \(1.5\) barn, and the cross-section for neutron capture (absorption) as incredibly high as 2,600,000 barns! It means that for an incoming neutron the Xe-125 nucleus "looks like" a disc with a radius one thousand times as large as the radius of the nucleus' geometrical cross section!

Another strange property of nuclear cross sections - especially, for neutron capture - is their dependence on the incident particle energy \(E_{i}\). For some nuclei, the \(\sigma\) vs. \(E_{i}\) plots exhibit sharp peaks, or even a large number of such peaks. Wewill return to discussing such characteristics shortly, because they are essential for understanding how nuclear reactors work.

One more example (we will need the result when discussing the design of a nuclear reactor): The total cross section of Uranium-238 for neutrons of \(1.5 \mathrm{MeV}\) energy ("total" means: for any kind of interaction) is \(\sigma=1.1\) barn. The atomic number density (i.e., number of atoms per 1 cubic \(\mathrm{cm}\) ) for Uranium metal is \(\rho_{\mathrm{A}}=4.8 \times 10^{22}\) atoms \(/ \mathrm{cm}^{3}\). What is the probability thata \(1.5 \mathrm{MeV}\) neutron passes "untouched" trough a \(1 \mathrm{~cm}\) thick plate of Uranium metal?

Answer: Consider a \(1 \mathrm{~cm} \times 1 \mathrm{~cm}\) square drawn on the surface of the plate. The plate is \(1 \mathrm{~cm}\) thick, so that the volume of the material behind the square is \(1 \mathrm{~cm}^{3}\), and there are \(4.8 \times 10^{22}\) atoms behind it. The total surface area of that many \(1.1\) barn discs "seen" by a neutron is \(4.8 \times 10^{22} \times 0.3 \times 10^{-24} \mathrm{~cm}^{2}=\) \(0.0528 \mathrm{~cm}^{2}\). Hence, the probability that a neutron interacts with a nucleus is \(5.28 \%\) - and it is \(94.72 \%\) that it will "fly untouched all the way through".