Question 9.1.Q2: Neutrons are classified, similarly to x rays and γ rays, as ......

Neutrons are classified, similarly to x rays and γ rays, as indirectly ionizing radiation that deposits energy in absorbing medium through an intermediate step involving release of secondary charged particles in the attenuating medium. Whereas photons interact with atomic electrons and release electrons and positrons in the attenuating medium, neutrons interact with nuclei of the attenuator and release nuclear particles, such as protons, deuterons, α particles, and heavier nuclear recoils.

(a) Define neutron fluence φ and neutron fluence rate (neutron flux density) \dot{\varphi}.
(b) Define total microscopic neutron cross section σ and total macroscopic cross section Σ.
(c) Describe attenuation of a collimated neutron beam in attenuating medium.
(d) Define the mean free path Λ of neutrons and reaction rate \dot{R} of neutrons.
(e) A lead attenuator (atomic number Z = 82, mass density ρ = 11.3 g/cm³, atomic weight A = 207.2 g/mol) of thickness z = 2 cm attenuates a collimated neutron beam of kinetic energy E_{\mathrm{K}}^{\mathrm{n}} = 10 MeV from neutron fluence rate \dot{\varphi}=2 \times 10^{12} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1} to a fluence rate of \dot{\varphi}=1.43 \times 10^{12} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1}. For this neutron beam interacting with lead attenuator calculate:

(1) Total macroscopic cross section Σ.
(2) Total microscopic cross section σ .
(3) Mean free path Λ.
(4) Neutron reaction rate \dot{R} at a depth of 1 cm in lead attenuator.

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(a) Neutron fluence φ and neutron fluence rate \dot{\varphi} are two basic physical quantities used for describing neutron beams and neutron fields. A neutron radiation field is established by neutron sources in conjunction with an attenuating medium that causes absorption as well as scattering of neutrons.

(1) Neutron fluence φ, according to the ICRU, is defined as the quotient of ΔN by Δa or φ = ΔN/Δa, with ΔN the number of particles that enter a sphere of cross sectional area Δa.

(2) Neutron fluence rate or neutron flux density \dot{\varphi}, in addition to neutron kinetic energy E_K, is a convenient parameter used for describing the presence of free neutrons in an attenuating medium. Several closely related definitions of neutron fluence rate (neutron flux) are in use:

(i) According to the ICRU, neutron fluence rate \dot{\varphi} is the quotient of Δφ by Δt where Δφ is the increment of neutron fluence in the time interval Δt;

(ii) Fluence rate \dot{\varphi} is also defined as the number of neutrons passing through a particular cross sectional area in any direction per unit time; and

(iii) Fluence rate \dot{\varphi} is the product of neutron velocity υ and neutron density (number of free neutrons per unit volume of the attenuator) n given as \dot{\varphi}=n v.

Typical unit of neutron fluence φ is cm^{−2}, expressing number of neutrons per cm². Typical unit of neutron fluence rate or flux density \dot{\varphi} \text { is } \mathrm{cm}^{-2} \cdot \mathrm{s}^{-1}, expressing number of neutrons per cm² per second.

(b) Neutrons being uncharged particles interact with nuclei of attenuating media through direct collisions rather than via Coulomb interactions. Many modes of interaction between incident neutron and nuclei of attenuating medium are available to a neutron propagating through an attenuating medium.

(1) Total microscopic cross sections. The probability of a given type of interaction i is expressed in terms of a microscopic cross section σ_i for the given target nucleus. For a given target nucleus and a given neutron kinetic energy E_{\mathrm{K}}^{\mathrm{n}} a set of partial microscopic cross sections σ_i is usually available for the various possible interaction modes i. At a given E_{\mathrm{K}}^{\mathrm{n}} and target nucleus, the sum \sum_i \sigma_i is referred to as the total microscopic cross section σ . Units of microscopic cross sections are cm²/atom, m²/atom, and barns per atom (b/atom).

(2) Total macroscopic cross section Σ for a given attenuating medium is obtained by multiplying the total microscopic cross section σ (sum of all relevant partial microscopic cross sections) with the atomic density n^{\square}.

\Sigma=n^{\square} \sigma \quad \text { (in units of } \mathrm{cm}^{-1} \text { or } \mathrm{m}^{-1} \text { ), }              (9.3)

with the atomic density n^{\square} defined as the number of atoms or nuclei N_{\mathrm{a}} per volume \mathcal{V} of the attenuator

n^{\square}=\frac{N_{\mathrm{a}}}{\mathcal{V}}=\rho \frac{N_{\mathrm{a}}}{m}=\rho \frac{N_{\mathrm{A}}}{A} \quad \text { (in units of } \mathrm{cm}^{-3} \text { or } \mathrm{m}^{-3} \text { ), }          (9.4)

where N_{\mathrm{A}} is the Avogadro number \left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right), \rho is mass density of the attenuator, and A is the atomic weight or atomic mass number in g/mol.
Like the microscopic cross section σ , the macroscopic cross section Σ depends on neutron kinetic energy E_{\mathrm{K}}^{\mathrm{n}} and physical properties of the attenuating medium. Relationship (9.4) is similar to the relationship in photon interactions with matter where the linear attenuation coefficient μ for a given photon interaction is a product of the atomic attenuation coefficient (also known as atomic cross section) {}_aμ and the atomic density n^{\square} \text { or } \mu=n^{\square}{}_a\mu.

(c) Just as the linear attenuation coefficient μ is used for description of photon beam attenuation in attenuating medium, so is the macroscopic cross section Σ used for describing attenuation of collimated neutron beams in attenuating medium. The decrease in beam intensity dI is proportional to the neutron beam intensity I , microscopic cross section σ of the attenuator, atomic density n^{\square} of attenuator, and thickness dx of the attenuating medium

\mathrm{d} I=-I \sigma n^{\square} \mathrm{d} x=-I \Sigma \mathrm{d} x \quad \text { or } \quad I=I_0 e^{-\Sigma x} \text {, }               (9.5)

where Σ is the macroscopic cross section of the attenuating medium for neutrons with kinetic energy E_{\mathrm{K}}^{\mathrm{n}}.

(d) Macroscopic quantities: cross section Σ, mean free path Λ, and reaction rate \dot{R} of neutrons traversing an attenuating medium depend upon atomic density n^{\square} of the attenuating medium and kinetic energy E_{\mathrm{K}}^{\mathrm{n}} of the incident neutron.

(1) Similarly to mean free path of photons, the mean free path Λ of neutrons is defined as that thickness Λ of attenuator that attenuates the neutron intensity I from original intensity I_0\ to\ I_0/e or, expressed mathematically, we can say

I(\Lambda)=\frac{I_0}{e}=0.368 I_0=I_0 e^{-\Sigma \Lambda} \quad \text { or } \quad e^{-1}=e^{-\Sigma \Lambda} \quad \text { or } \quad \Lambda=\frac{1}{\Sigma}             (9.6)

Mean free path Λ also is a measure of the mean distance that a neutron of given kinetic energy travels through a given attenuating medium before interacting with a nucleus. This definition is similar to the definition of mean free path \bar{x} of photons in attenuating medium, where \bar{x} = 1/μ with μ the linear attenuation coefficient.

(2) Reaction rate \dot{R} between neutrons and nuclei of attenuator is defined as the product of the macroscopic cross section Σ and neutron fluence rate \dot{ϕ} or

\dot{R}=\Sigma \dot{\varphi}=\left(n^{\square} \Sigma\right)(n v)            (9.7)

indicating that the reaction rate \dot{R} is linearly proportional to the atomic density n^{\square}, microscopic cross section σ , neutron density n, and velocity of neutrons v. From (9.7) we note that the unit of reaction rate \dot{R} \text { is } \mathrm{cm}^{-3} \cdot \mathrm{s}^{-1}.

(e) Attenuation of a collimated neutron beam in an attenuator is expressed as

I=I_0 e^{-\Sigma x}          (9.8)

where I_0 \text { and } I are the incident fluence rate \left(2 \times 10^{12} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1}\right) and transmitted fluence rate \left(1.43 \times 10^{12} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1}\right) , respectively, and Σ is the macroscopic cross section. Based on (9.8) and data provided on lead we now calculate:

(1) Total macroscopic cross section Σ.
(2) Total microscopic cross section σ .
(3) Mean free path Λ.
(4) Neutron reaction rate \dot{R} at a depth of 1 cm in lead attenuator.

(1) Solving (9.8) for Σ we get the following result for the macroscopic cross section Σ

\Sigma=\frac{1}{x} \ln \frac{I_0}{I}=\frac{1}{(2 \mathrm{~cm})} \ln \frac{2 \times 10^{22}}{1.43 \times 10^{22}}=0.168 \mathrm{~cm}^{-1}            (9.9)

(2) Next we calculate the microscopic cross section σ using (9.4) as follows

\begin{aligned} \sigma & =\frac{\Sigma}{n^{\square}}=\frac{A \Sigma}{\rho N_{\mathrm{A}}}=\frac{\left(207.2 \mathrm{~g} \cdot \mathrm{mol}^{-1}\right) \times\left(0.168 \mathrm{~cm}^{-1}\right)}{\left(11.3 \mathrm{~g} / \mathrm{cm}^3\right) \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)} \\ & =5.12 \times 10^{-24} \mathrm{~cm}^{-2} / \text { atom }=5.12 \mathrm{~b} / \text { atom. }\quad (9.10) \end{aligned}

(3) Mean free path Λ is calculated using (9.6) as follows

\Lambda=\frac{1}{\Sigma}=\frac{1}{0.168 \mathrm{~cm}^{-1}}=5.95 \mathrm{~cm}             (9.11)

This means that a 10 MeV neutron travels on average about 6 cm in lead before it experiences one of the possible nuclear interactions with a lead nucleus of the attenuator.
(4) Reaction rate \dot{R} of a neutron beam is, according to (9.7), expressed as \dot{R}=\Sigma \dot{\varphi}, \text { where } \Sigma is the total macroscopic cross section determined in (9.9) as \Sigma=0.168 \mathrm{~cm}^{-1} \text { and } \dot{\varphi} is the fluence rate at depth of x=1 cm in lead determined from (9.5) as follows

\begin{aligned} I(x=1 \mathrm{~cm}) & =I_0 e^{-\Sigma x}=\left(2 \times 10^{12} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1}\right) e^{-\left(0.168 \mathrm{~cm}^{-1}\right) \times(1 \mathrm{~cm})} \\ & =1.69 \times 10^{12} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1} . \end{aligned}      (9.12)

Reaction rate \dot{R} at a depth of x = 1 cm in lead is now calculated as

\dot{R}=\Sigma \dot{\varphi}=\left(0.168 \mathrm{~cm}^{-1}\right) \times\left(1.69 \times 10^{12} \mathrm{~cm}^{-2} \cdot \mathrm{s}^{-1}\right)=2.84 \times 10^{12} \mathrm{~cm}^{-3} \cdot \mathrm{s}^{-1}            (9.13)

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