Question 9.5.Q1: Neutrons, by virtue of their neutrality, similarly to photon......

(a) Depending on their kinetic energy E_{\mathrm{K}}^{\mathrm{n}}, neutrons are classified into many categories ranging from ultra-cold neutrons through thermal and intermediate neutrons to fast and relativistic neutrons. However, for dosimetric purposes, it is convenient to separate neutrons into only three categories: (1) thermal neutron region for E_{\mathrm{K}}^{\mathrm{n}}<1 \mathrm{eV},(2) \text { intermediate energy region for } 1 \mathrm{eV}<E_{\mathrm{K}}^{\mathrm{n}}<10 \mathrm{keV}, and (3) fast neutron region for E_{\mathrm{K}}^{\mathrm{n}}>10 \mathrm{keV}. A brief description of the three energy regions follows below and is summarized in Table 9.3.

(1) Dose deposition by neutrons of kinetic energy E_{\mathrm{K}}^{\mathrm{n}}<1 \mathrm{eV} (thermal neutron dosimetric region) in human tissue is governed by thermal neutron capture (absorption) reactions of neutron with { }_7^{14} \mathrm{~N} \text { and }{ }_1^1 \mathrm{H} atoms in tissue producing the following two reactions:

where Q_N\ and\ Q_H are reaction Q values that are both positive, making the reactions exothermic.

(2) Intermediate kinetic energy neutrons 1 \mathrm{eV}<E_{\mathrm{K}}^{\mathrm{n}}<10 \mathrm{keV} primarily interact with hydrogen nuclei of tissue through elastic scattering interaction and dose is deposited in tissue by the recoil energy picked up by hydrogen nuclei (proton) in the elastic scattering interaction. The mean energy transfer per one collision of neutron with hydrogen atom is 50 % of the kinetic energy of the neutron.

(3) Dose delivered to human tissue by interactions of fast neutrons \left(E_{\mathrm{K}}^{\mathrm{n}}>10 \mathrm{keV}\right) with atoms of human tissue is mainly due to elastic collisions of neutrons with atoms of tissue, most importantly with hydrogen atoms, because of the high efficiency of mean energy transfer from neutron to hydrogen atom (proton) in elastic collisions. Also of some importance are fast neutron interactions with carbon and oxygen atoms of tissue via inelastic collision processes that release α particles in reactions of the type \left(\mathrm{n}, \mathrm{n}^{\prime} 3 \alpha\right) and via non-elastic processes producing α particles and heavier ions. The α particles and heavier ions released in the fast neutron interactions are responsible for the dose deposition in tissue.

(b) The main constituent elements of human tissue are hydrogen H, carbon C, nitrogen N, and oxygen O and their fractions by weight f_X according to the ICRU are, respectively, 0.102, 0.123, 0.035, and 0.729. The number of atoms (nuclei) of a given element X per unit mass of tissue, N_X / m_{\text {tissue }}, is determined as follows:

(1) Number of atoms X per mole A_X of element X is constant and referred to as the Avogadro number N_A = 6.022\times 10^{23}\ mol^{−1}.

(2) Number of atoms X per gram of element X is given by N_A/A_X.

(3) Number of atoms X per gram of tissue, N_X / m_{\text {tissue }} \text {, is given as } N_X / m_{\text {tissue }}= ff_X N_{\mathrm{A}} / A_X.

Results of the calculation of N_X/m_{tissue} for H, C, N, and O are listed in row (5) of Table 9.4 and show that per unit mass of tissue hydrogen H atoms (nuclei) are the most abundant with 6.09\times 10^{22} atom/g, followed by oxygen O atoms with 2.74\times 10^{22} atom/g. Carbon C and nitrogen N make a much lower contribution to N_X/m_{tissue}, the number of atoms (nuclei) per unit mass of tissue with 0.62\times 10^{22} atom/g and 0.15\times 10^{22} atom/g, respectively.

N_X/m_{tissue} is of importance in determination of neutron kerma and absorbed dose, since both of these dosimetric quantities are linearly proportional to N_X/m_{tissue}.

(c) Mean energy transfer fraction \bar{f}_{\mathrm{tr}} for elastic scattering of neutrons is calculated using the following expression

\bar{f}_{\mathrm{tr}}=\frac{\overline{\Delta E_{\mathrm{K}}}}{E_{\mathrm{K}}^{\mathrm{n}}}=\frac{2 A}{(1+A)^2},              (9.77)

where

\overline{\Delta E_{\mathrm{K}}}   is the mean energy transferred from the incident neutron with kinetic energy E_{\mathrm{K}}^{\mathrm{n}} to the target (nucleus) of the attenuating medium.

A   is the atomic weight of the target atom (nucleus) in the attenuating medium.

Equation (9.77) is derived from the standard classical equation for energy transfer \Delta E_{\mathrm{K}} from projectile with mass m_1 to the stationary target with mass m_2 in elastic collision (T5.25)

\Delta E_{\mathrm{K}}=\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}\left(E_{\mathrm{K}}\right)_0 \cos ^2 \phi             (9.78)

where

\left(E_K\right)_0    is the kinetic energy of the projectile (incident particle) m_1.

φ    is the recoil angle of the target measured with respect to the direction of incident neutron.

Equation (9.77) can be obtained from (9.78) with the following assumptions:

(1) The incident particle (projectile) is the neutron: m_1 = m_n.

(2) Kinetic energy of the projectile is the kinetic energy of the incident neutron: \left(E_{\mathrm{K}}\right)_0=E_{\mathrm{K}}^{\mathrm{n}}.

(3) Mean energy transfer \overline{\Delta E_{\mathrm{K}}} \text { from projectile } m_1 \text { to stationary target } m_2 is given as

\begin{aligned} & \overline{\Delta E_{\mathrm{K}}}=\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}\left(E_{\mathrm{K}}\right)_0 \overline{\cos ^2 \phi}=\frac{2 m_1 m_2}{\left(m_1+m_2\right)^2}\left(E_{\mathrm{K}}\right)_0, \text { since } \\ & \overline{\cos ^2 \phi}=\frac{1}{2} . \end{aligned}              (9.79)

(4) The target is the nucleus of attenuator atom, i.e., m_2=Z m_{\mathrm{p}}+(A-Z) m_{\mathrm{n}} \approx  Am_{n}, where we assume that the proton mass m_p and the neutron mass m_n are approximately equal.

(5) Equation (9.78) for the mean energy transfer fraction \bar{f}_{tr} from neutron to nucleus of attenuating medium can now be written in the simplified form of (9.77) as follows

\bar{f}_{\mathrm{tr}}=\frac{\overline{\Delta E_{\mathrm{K}}}}{\left(E_{\mathrm{K}}\right)_0}=\frac{2 m_1 m_2}{\left(m_1+m_2\right)^2}=\frac{\overline{\Delta E_{\mathrm{K}}}}{E_{\mathrm{K}}^{\mathrm{n}}} \approx \frac{2 A}{(1+A)^2}                (9.80)

Mean energy transfer fractions \bar{f}_{tr} for hydrogen, carbon, nitrogen, and oxygen were calculated with (9.77) and results are displayed in row (6) of Table 9.4. Hydrogen is obviously the best nuclear target for energy transfer from neutron to target in elastic scattering, on the average receiving 50 % of neutron kinetic energy per elastic scattering collision. As shown in Table 9.4, \bar{f}_{tr} rapidly decreases with increasing number of nucleon A in the target, amounting to 0.142, 0.124, and 0.111 for carbon, nitrogen, and oxygen, respectively.