# Macroscopic Cross-section

The macroscopic cross-section represents the effective target area of all of the nuclei contained in the volume of the material (such as fuel pellet). The units are given in cm-1. It is the probability of neutron-nucleus interaction per centimeter of neutron travel. Codes commonly use these data for reactor core analyses and design. These codes are based on pre-computed assembly homogenized macroscopic cross-sections.

## Macroscopic Cross-section

The difference between the microscopic cross-section and macroscopic cross-section is very important and is restated for clarity. The microscopic cross-section represents the effective target area of a single target nucleus for an incident particle. The units are given in barns or cm2.

While the macroscopic cross-section represents the effective target area of all of the nuclei contained in the volume of the material, the units are given in cm-1.

A macroscopic cross-section is derived from microscopic cross-section and the atomic number density:

Σ=σ.N

Here σ, which has units of m2, is the microscopic cross-section. Since the units of N (nuclei density) are nuclei/m3, the macroscopic cross-section Σ has units of m-1. Thus, it is an incorrect name because it is not a correct unit of cross-sections. In terms of Σt (the total cross-section), the equation for the intensity of a neutron beam can be written as

-dI = N.σ.Σt.dx

Dividing this expression by I(x) gives

-dΙ(x)/I(x) = Σt.dx

Since dI(x) is the number of neutrons that collide in dx, the quantity –dΙ(x)/I(x) represents the probability that a neutron that has survived without colliding until x will collide in the next layer dx. It follows that the probability P(x) that a neutron will travel a distance x without any interaction in the material, which Σt characterizes, is:

P(x) = et.x

We can derive the probability that a neutron will make its first collision in dx from this equation. It will be the quantity P(x)dx. Suppose the probability of the first collision in dx is independent of its history. In that case, the required result will be equal to the probability that a neutron survives up to layer x without any interaction (~Σtdx) times the probability that the neutron will interact in the additional layer dx (i.e., ~et.x).

P(x)dx = Σtdx . et.x = Σt et.x dx

## Mean Free Path

From the equation for the probability of the first collision in dx, we can calculate the mean free path traveled by a neutron between two collisions. The symbol λ usually designates this quantity. It is equal to the average value of x, the distance traveled by a neutron without any interaction, over the interaction probability distribution. whereby one can distinguish λs, λa, λf, etc. This quantity is also known as the relaxation length because it is the distance in which the intensity of the neutrons that have not caused a reaction has decreased with a factor e.

For materials with a high absorption cross-section, the mean free path is very short, and neutron absorption occurs mostly on the material’s surface. This surface absorption is called self-shielding because the outer layers of atoms shield the inner layers.

## Macroscopic Cross-section of Mixtures and Molecules

Most materials are composed of several chemical elements and compounds. Most chemical elements contain several isotopes of these elements (e.g., gadolinium with its six stable isotopes). For this reason, most materials involve many cross-sections. Therefore, to include all the isotopes within a given material, it is necessary to determine the macroscopic cross-section for each isotope and then sum all the individual macroscopic cross-sections.

In this section, both factors (different atomic densities and cross-sections) will be considered in calculating the macroscopic cross-section of mixtures.

First, consider Avogadro’s number N0 = 6.022 x 1023, which is the number of particles (molecules, atoms) contained in the amount of substance given by one mole. Thus if M is the molecular weight, the ratio N0/M equals the number of molecules in 1g of the mixture. The number of molecules per cm3 in the material of density ρ and the macroscopic cross-section for mixtures is given by the following equations:

Ni = ρi.N0 / Mi Note that, in some cases, the cross-section of the molecule is not equal to the sum of cross-sections of its individual nuclei. For example, the cross-section of elastic neutron scattering of water exhibits anomalies for thermal neutrons. It occurs because the kinetic energy of an incident neutron is of the order or less than the chemical binding energy. Therefore, the scattering of slow neutrons by water (H2O) is greater than by free nuclei (2H + O).

Example - Macroscopic cross-section for boron carbide in control rods
A control rod usually contains solid boron carbide with natural boron. Natural boron consists primarily of two stable isotopes,11B (80.1%) and 10B (19.9%). Boron carbide has a density of 2.52 g/cm3.

Determine the total macroscopic cross-section and the mean free path.

Density:
MB = 10.8
MC = 12
MMixture = 4 x 10.8 + 1×12 g/mol
NB4C = ρ . Na / MMixture
= (2.52 g/cm3)x(6.02×1023 nuclei/mol)/ (4 x 10.8 + 1×12 g/mol)
= 2.75×1022 molecules of B4C/cm3

NB = 4 x 2.75×1022 atoms of boron/cm3
NC = 1 x 2.75×1022 atoms of carbon/cm3

NB10 = 0.199 x 4 x 2.75×1022 = 2.18×1022 atoms of 10B/cm3
NB11 = 0.801 x 4 x 2.75×1022 = 8.80×1022 atoms of 11B/cm3
NC = 2.75×1022 atoms of 12C/cm3

the microscopic cross-sections

σt10B = 3843 b of which σ(n,alpha)10B = 3840 b
σt11B = 5.07 b
σt12C = 5.01 b

the macroscopic cross-section

ΣtB4C = 3843×10-24 x 2.18×1022 + 5.07×10-24 x 8.80×1022 + 5.01×10-24 x 2.75×1022
= 83.7 + 0.45 + 0.14 = 84.3 cm-1

the mean free path

λt = 1/ΣtB4C = 0.012 cm = 0.12 mm (compare with B4C pellets diameter in control rods which may be around 7mm)
λa ≈ 0.12 mm

“Example
It was written the macroscopic cross-section is derived from microscopic cross-section and the atomic number density (N):

Σ=σ.N

In this equation, the atomic number density plays a crucial role as the microscopic cross-section. In the reactor core, the atomic number density of certain materials (e.g., water as the moderator) can be changed, leading to certain reactivity changes. To understand the nature of these reactivity changes, we must understand the term the atomic number density.

See theory: Atomic Number Density.

Most PWRs use uranium fuel, which is in the form of uranium dioxide (UO2). Typically, the fuel has enrichment of ω235 = 4% [grams of 235U per gram of uranium] of isotope 235U.

Calculate the atomic number density of 235U (N235U), when:

• the molecular weight of the enriched uranium MUO2 = 237.9 + 32 = 269.9 g/mol
• the uranium density UO2 = 10.5 g/cm3

NUO2 = UO2 . NA / MUO2

NUO2 = (10.5 g/cm3) x (6.02×1023 nuclei/mol)/ 269.9
NUO2 = 2.34 x 1022 molecules of UO2/cm3

NU = 1 x 2.34×1022 atoms of uranium/cm3
NO = 2 x 2.34×1022 atoms of oxide/cm3

N235U = ω235.NA.UO2/M235U x (MU/MUO2)

N235U = 0.04 x 6.02×1023 x 10.5 / 235 x 237.9 / 269.9 =9.48 x 1020 atoms of 235U/cm3

References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
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