## Mean Free Path

From the equation for the probability of the** first collision in dx** (**P(x)dx = Σ _{t}dx . e^{-Σt.x} = Σ_{t} e^{-Σt.x} 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.

## Transport Mean Free Path

**The transport mean free path (λ _{tr})** is an

**average distance**a neutron will move in its original direction

**after infinite number of scattering collisions**.

is average value of the cosine of the angle in the lab system at which neutrons are scattered in the medium. It can be calculated for most of the neutron energies as (A is the mass number of target nucleus):

## Example – Macroscopic Cross-section and Mean Free Path 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,** ^{11}B** (80.1%) and

**(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

**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