## Macroscopic Cross-section

**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 cm**.

^{2}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 m^{2}, is the microscopic cross-section. Since the units of N (nuclei density) are nuclei/m^{3}, the macroscopic cross-section Σ have units of m^{-1}, thus in fact is an incorrect name, because it is not a correct unit of cross-sections. In terms of Σ_{t} (the total cross-section), the equation for the intensity of a neutron beam can be written as

**-dI = N.σ.Σ _{t}.dx**

Dividing this expression by I(x) gives

**-dΙ(x)/I(x) = Σ _{t}.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 is characterized by Σ_{t}, is:

**P(x) = e ^{-Σt.x}**

From this equation, we can derive the probability that a neutron will make its **first collision in dx**. It will be the quantity** P(x)dx**. If the probability of the first collision in dx is independent of its past history, the required result will be equal to the probability that a neutron survives up to layer x without any interaction (~Σ_{t}dx) times the probability that the neutron will interact in the additional layer dx (i.e. ~e^{-Σt.x}).

**P(x)dx = Σ _{t}dx . e^{-Σt.x} = Σ_{t} e^{-Σt.x} dx**

## Mean Free Path

**first collision in dx**we can calculate

**the mean free path**that is traveled by a neutron between two collisions. This quantity is usually designated by the symbol

**λ**and 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** λ _{s}, λ_{a}, λ_{f}**, 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 high absorption cross-section, the mean free path is **very short** and neutron absorption occurs mostly** on the surface** of the material. This surface absorption is called **self-shielding** because the outer layers of atoms shield the inner layers.

## Macroscopic Cross-section of Mixtures and Molecules

**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 different **cross-sections**) will be considered in the calculation of the **macroscopic cross-section of mixtures**.

First, consider the Avogadro’s number N_{0} = **6.022 x 10 ^{23}**, is the number of particles (molecules, atoms) that is contained in the amount of substance given by one mole. Thus if M is the

**molecular weight**, the ratio

**N**equals to the number of molecules in 1g of the mixture. The number of molecules per cm

_{0}/M^{3}in the material of density ρ and the macroscopic cross-section for mixtures are given by following equations:

**N _{i} = ρ_{i}.N_{0} / M_{i}**

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 neutron elastic 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** and therefore the scattering of slow neutrons by water (H_{2}O) is greater than by free nuclei (2H + O).

**control rod**usually contains solid

**boron carbide**with natural boron. Natural boron consists primarily of two stable isotopes,

**(80.1%) and**

^{11}B**(19.9%). Boron carbide has a density of**

^{10}B**2.52 g/cm**.

^{3}Determine the **total macroscopic cross-section** and the **mean free path**.

Density:

M_{B} = 10.8

M_{C} = 12

M_{Mixture} = 4 x 10.8 + 1×12 g/mol

N_{B4C} = ρ . N_{a} / M_{Mixture}

= (2.52 g/cm^{3})x(6.02×10^{23} nuclei/mol)/ (4 x 10.8 + 1×12 g/mol)

= **2.75×10 ^{22} molecules of B4C/cm^{3}**

N_{B} = 4 x 2.75×10^{22} atoms of boron/cm^{3}

N_{C} = 1 x 2.75×10^{22} atoms of carbon/cm^{3}

N_{B10} = 0.199 x 4 x 2.75×10^{22} = 2.18×10^{22} atoms of 10B/cm^{3}

N_{B11} = 0.801 x 4 x 2.75×10^{22} = 8.80×10^{22} atoms of 11B/cm^{3}

N_{C} = 2.75×10^{22} atoms of 12C/cm^{3}

**the microscopic cross-sections**

σ_{t}^{10B} = 3843 b of which σ_{(n,alpha)}^{10B} = 3840 b

σ_{t}^{11B} = 5.07 b

σ_{t}^{12C} = 5.01 b

**the macroscopic cross-section**

**Σ _{t}^{B4C} **= 3843×10

^{-24}x 2.18×10

^{22}+ 5.07×10

^{-24}x 8.80×10

^{22}+ 5.01×10

^{-24}x 2.75×10

^{22}

= 83.7 + 0.45 + 0.14 =

**84.3 cm**

^{-1}**the mean free path**

**λ _{t} **= 1/Σ

_{t}

^{B4C}= 0.012 cm =

**0.12 mm**(compare with B4C pellets diameter in control rods which may be around 7mm)

**λ**

_{a}≈ 0.12 mm**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 the crucial role as the microscopic cross-section, because in the reactor core the atomic number density of certain materials (e.g. water as the moderator) can be simply changed leading into certain **reactivity changes**. In order to understand the nature of these **reactivity changes**, we must understand the term the atomic number density.

See theory: Atomic Number Density

Most of PWRs use the **uranium fuel**, which is in the form of **uranium dioxide **(UO_{2}). Typically, the fuel have enrichment of ω_{235} = 4% [grams of ^{235}U per gram of uranium] of isotope ^{235}U.

**Calculate the atomic number density of ^{235}U**

**(N235U), when:**

- the molecular weight of the enriched uranium M
_{UO2}= 237.9 + 32 =**269.9 g/mol** - the uranium density ⍴
_{UO2}=**10.5 g/cm**^{3}

**N _{UO2} = ⍴_{UO2} . N_{A} / M_{UO2}**

**N _{UO2 }**= (10.5 g/cm

^{3}) x (6.02×10

^{23}nuclei/mol)/ 269.9

**N**= 2.34 x 10

_{UO2 }^{22}molecules of UO2/cm

^{3}

N_{U} = 1 x 2.34×10^{22} atoms of uranium/cm^{3}

N_{O} = 2 x 2.34×10^{22} atoms of oxide/cm^{3}

**N _{235U}** = ω

_{235}.N

_{A}.⍴

_{UO2}/M

_{235U}x (M

_{U}/M

_{UO2})

**N _{235U}** = 0.04 x 6.02×10

^{23}x 10.5 / 235 x 237.9 / 269.9 =

**9.48 x 10**

^{20}atoms of 235U/cm^{3}**Nuclear and Reactor Physics:**

- J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
- J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
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**Advanced Reactor Physics:**

- K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
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