Question 9.4.Q1: Neutron kerma factor Fn plays an important role in neutron d......
Neutron kerma factor F_n plays an important role in neutron dosimetry similarly to the role that mass energy transfer coefficient μ_{tr}/ρ plays in photon dosimetry. However, F_n\ and\ μ_{tr}/ρ are defined differently to account for the differences in describing neutron fields in comparison to photon fields.
Photon fields are usually described by photon energy fluence ψ resulting in kerma expressed as K=\left(\mu_{\mathrm{tr}} / \rho\right) \psi, while neutron fields are described by neutron fluence φ resulting in kerma expressed as K=F_{\mathrm{n}} \varphi.
Total microscopic interaction cross section σ for neutrons with kinetic energy E_{\mathrm{K}}^{\mathrm{n}} of 10 MeV is 0.94 b in hydrogen and 1.25 b in oxygen. For E_{\mathrm{K}}^{\mathrm{n}} = 10 MeV neutrons determine:
(a) Total macroscopic interaction cross sections Σ of hydrogen and oxygen both at standard temperature and pressure (STP) of 0 °C and 101.3 kPa, respectively.
(b) Total macroscopic interaction cross section Σ of water.
(c) Neutron kerma factor F_n of hydrogen and oxygen at standard temperature and pressure.
(d) Neutron kerma factor F_n of water
(e) Verify your results calculated in (c) and (d) on the graph of Fig. 9.5A that plots F_{\mathrm{n}} \text { against } E_{\mathrm{K}}^{\mathrm{n}} for various attenuating media including hydrogen, oxygen, and water.
(a) The total macroscopic interaction cross sections of hydrogen \Sigma_{\mathrm{H}_2} and oxygen \Sigma_{\mathrm{O}_2} are calculated from the following expression linking the macroscopic and microscopic cross sections
\Sigma=n^{\square} \sigma, (9.60)
where σ is the total microscopic cross section for the given neutron interaction and n^{\square} is the atomic (nuclear) density, i.e., number of atoms (nuclei) per volume \mathcal{V} of the attenuating medium expressed as n^{\square}=N_{\mathrm{a}} / \mathcal{V}=\rho N_{\mathrm{a}} / m=\rho N_{\mathrm{A}} / A. Since hydrogen and oxygen are gases, we determine the appropriate nuclear densities n at STP (0 °C and 101.3 kPa) invoking the ideal gas law which states that 1 mol of ideal gas at STP occupies a volume of 22.4 liters, i.e., 22.4 \times 10^3 \mathrm{~cm}^3.
(1) Hydrogen gas in molecular form (H_2) has a molecular weight of M = 2\times 1.00794\ g/mol and since, according to ideal gas law, 1 mol of hydrogen gas at STP corresponds to a volume \mathcal{V} \text { of } 22.4 \times 10^2 \mathrm{~cm}^3, we obtain the following mass density of hydrogen gas
\rho_{\mathrm{H}_2}=\frac{2 \times(1.00794 \mathrm{~g} / \mathrm{mol})}{22.4 \times 10^3 \mathrm{~cm}^3 / \mathrm{mol}}=9 \times 10^{-5} \mathrm{~g} / \mathrm{cm}^3 (9.61)
(2) Oxygen gas in molecular form \left(O_2\right) has a molecular weight of M = 2\times 15.9994\ g/mol and since, according to ideal gas law, 1 mol of oxygen gas at STP corresponds to a volume V \text { of } 22.4 \times 10^2 \mathrm{~cm}^3, we obtain the following mass density of oxygen gas
\rho_{\mathrm{O}_2}=\frac{2 \times(15.9994 \mathrm{~g} / \mathrm{mol})}{22.4 \times 10^3 \mathrm{~cm}^3 / \mathrm{mol}}=1.429 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^3 (9.62)
We now use (9.60) to determine macroscopic interaction cross sections \Sigma_{\mathrm{H}_2} and \Sigma_{\mathrm{O}_2} of hydrogen at STP and oxygen at STP, respectively, for neutrons with E_{\mathrm{K}}^{\mathrm{n}} = 10 MeV.
(1) Total macroscopic cross section of hydrogen \Sigma_{\mathrm{H}_2} at STP for 10 MeV neutron
\begin{aligned} \Sigma_{\mathrm{H}_2} & =n_{\mathrm{H}_2}^{\square} \sigma_{\mathrm{H}_2}=\rho_{\mathrm{H}_2} \frac{N_{\mathrm{A}}}{A_{\mathrm{H}_2}} \sigma_{\mathrm{H}_2} \\ & =\frac{\left(9 \times 10^{-5} \mathrm{~g} / \mathrm{cm}^3\right) \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right) \times\left(0.94 \times 10^{-24} \mathrm{~cm}^2\right)}{2 \times 1.00794 \mathrm{~g} \cdot \mathrm{mol}^{-1}} \\ & =2.53 \times 10^{-5} \mathrm{~cm}^{-1} .\quad (9.63) \end{aligned}
(2) Total macroscopic cross section of oxygen \Sigma_{\mathrm{O}_2} at STP for 10 MeV neutron
\begin{aligned} \Sigma_{\mathrm{O}_2} & =n_{\mathrm{O}_2}^{\square} \sigma_{\mathrm{O}_2}=\rho_{\mathrm{O}_2} \frac{N_{\mathrm{A}}}{A_{\mathrm{O}_2}} \sigma_{\mathrm{O}_2} \\ & =\frac{\left(1.429 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^3\right) \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right) \times\left(1.25 \times 10^{-24} \mathrm{~cm}^2\right)}{2 \times 15.9994 \mathrm{~g} \cdot \mathrm{mol}^{-1}} \\ & =3.36 \times 10^{-5} \mathrm{~cm}^{-1} .\quad (9.64) \end{aligned}
(b) Total macroscopic interaction cross section of water \Sigma_{\mathrm{H}_2 \mathrm{O}} for 10 MeV neutrons is calculated from the sum of contributions from hydrogen and oxygen nuclei constituting the water molecule
\Sigma_{\mathrm{H}_2 \mathrm{O}}=\sum_i n_i^{\square} \sigma_i=n_{\mathrm{H}}^{\square} \sigma_{\mathrm{H}}+n_{\mathrm{O}}^{\square} \sigma_{\mathrm{O}} (9.65)
To find the atomic (nuclear) densities n_{\mathrm{H}}^{\square} \text { and } n_{\mathrm{O}}^{\square}, i.e., number of hydrogen and oxygen atoms (nuclei), respectively, per volume of water, we first determine the molecular density n_{\mathrm{H}_2 \mathrm{O}}^{\square} of water (i.e., number of water molecules per cm3 of water) as follows
\begin{aligned} n_{\mathrm{H}_2 \mathrm{O}}^{\square} & =\frac{N_{\mathrm{mol}}}{\mathcal{V}}=\rho_{\mathrm{H}_2 \mathrm{O}} \frac{N_{\mathrm{mol}}}{m_{\mathrm{H}_2 \mathrm{O}}} \\ & =\rho_{\mathrm{H}_2 \mathrm{O}} \frac{N_{\mathrm{A}}}{A_{\mathrm{H}_2 \mathrm{O}}}=\frac{\left(1 \mathrm{~g} \cdot \mathrm{cm}^{-3}\right) \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}{18.01528 \mathrm{~g} \cdot \mathrm{mol}^{-1}} \\ & =3.343 \times 10^{22} \mathrm{~cm}^{-3} .\quad (9.66) \end{aligned}
Since the molecular density of water is 3.343\times 10^{22} molecules per cm³ and each water molecule contains 2 atoms of hydrogen and 1 atom of oxygen, we conclude that the atomic density of hydrogen in water n_{\mathrm{H}}^{\square}\ \text { is } 2 \times 3.343 \times 10^{22}=6.686 \times 10^{22} hydrogen atoms per cm³ of water, while the atomic density of oxygen in water n_{\mathrm{O}}^{\square} is 3.343\times 10^{22} oxygen atoms per cm³ of water.
The macroscopic interaction cross section of water \Sigma_{\mathrm{H}_2 \mathrm{O}} for 10 MeV neutrons is now calculated as follows
\begin{aligned} \Sigma_{\mathrm{H}_2 \mathrm{O}}= & n_{\mathrm{H}}^{\square} \sigma_{\mathrm{H}}+n_{\mathrm{O}}^{\square} \sigma_{\mathrm{O}} \\ = & \left(6.686 \times 10^{22} \mathrm{~cm}^{-3}\right) \times\left(0.94 \times 10^{-24} \mathrm{~cm}^2\right) \\ & +\left(3.343 \times 10^{22} \mathrm{~cm}^{-3}\right) \times\left(1.25 \times 10^{-24} \mathrm{~cm}^2\right) \\ = & 0.0628 \mathrm{~cm}^{-1}+0.0418 \mathrm{~cm}^{-1}=0.1046 \mathrm{~cm}^{-1} .\quad (9.67) \end{aligned}
(c) Neutron kerma factor F_n of hydrogen and oxygen for 10 MeV neutrons is calculated from the following expression
F_{\mathrm{n}}=\frac{\Sigma}{\rho} \overline{\Delta E_{\mathrm{K}}}=\frac{n^{\square} \sigma}{\rho} \overline{\Delta E_{\mathrm{K}}}=\frac{N_{\mathrm{a}}}{\rho \mathcal{V}} \sigma \overline{\Delta E_{\mathrm{K}}}=\frac{N_{\mathrm{A}}}{A} \sigma \overline{\Delta E_{\mathrm{K}}}=\frac{N_{\mathrm{A}}}{A} \sigma \frac{2 A}{(1+A)^2} E_{\mathrm{K}}^{\mathrm{n}}, (9.68)
where
Σ is the total macroscopic interaction cross section of attenuating medium for neutrons.
ρ is the density of attenuating medium (hydrogen or oxygen).
n^{\square} atomic (nuclear) density of attenuating medium: n^{\square}=N_{\mathrm{a}} / \mathcal{V} \text { with } N_{\mathrm{a}} number of atoms and \mathcal{V} volume of attenuating medium.
σ is the total microscopic interaction cross section of attenuating medium for neutrons.
N_A is the Avogadro number: 6.022\times 10^{23} atoms (nuclei) per mol.
A is the atomic mass number (atomic weight) of the attenuating medium.
\overline{\Delta E_{\mathrm{K}}} is mean energy transferred from neutrons to charged particles at point-ofinterest in attenuating medium calculated for neutrons of kinetic energy E_{\mathrm{K}}^{\mathrm{n}} from the following expression
\overline{\Delta E_{\mathrm{K}}}=\frac{2 A}{(1+A)^2} E_{\mathrm{K}}^{\mathrm{n}} (9.69)
(1) Mean energy transfer \overline{\Delta E_{\mathrm{K}}} \text { from neutron with kinetic energy } E_{\mathrm{K}}^{\mathrm{n}}=10 \mathrm{MeV} interacting with hydrogen nucleus (A = 1) is from (9.69) given as
\overline{\Delta E_{\mathrm{K}}}=\frac{2 A}{(1+A)^2} E_{\mathrm{K}}^{\mathrm{n}}=\frac{2}{2^2} \times(10 \mathrm{MeV})=5 \mathrm{MeV} (9.70)
(2) Mean energy transfer \overline{\Delta E_{\mathrm{K}}} \text { from neutron with kinetic energy } E_{\mathrm{K}}^{\mathrm{n}}=10 \mathrm{MeV} interacting with oxygen nucleus (A = 16) is from (9.69) given as
\overline{\Delta E_{\mathrm{K}}}=\frac{2 A}{(1+A)^2} E_{\mathrm{K}}^{\mathrm{n}}=\frac{2 \times 16}{(1+16)^2} \times(10 \mathrm{MeV})=1.11 \mathrm{MeV} (9.71)
We now determine the neutron kerma factors F_n of hydrogen and oxygen using (9.68) in conjunction with (9.70) and (9.71), respectively
(1) Neutron kerma factor of hydrogen \left(F_n\right)_H for 10 MeV neutrons is given as
\begin{aligned} \left(F_{\mathrm{n}}\right)_{\mathrm{H}}=\frac{N_{\mathrm{A}}}{A_{\mathrm{H}}} \sigma_{\mathrm{H}} \overline{\left(\Delta E_{\mathrm{K}}\right)_{\mathrm{H}}} & =\frac{\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right) \times\left(0.94 \times 10^{-24} \mathrm{~cm}^2\right) \times(5 \mathrm{MeV})}{\left(1.00794 \mathrm{~g} \cdot \mathrm{mol}^{-1}\right)} \\ & \times \frac{\left(1.602 \times 10^{-13} \mathrm{~J} / \mathrm{MeV}\right) \times\left(10^2 \mathrm{cGy} / \mathrm{Gy}\right)}{\left(10^{-3} \mathrm{~kg} / \mathrm{g}\right)} \\ & =4.49 \times 10^{-8} \mathrm{cGy} \cdot \mathrm{cm}^2\quad (9.72) \end{aligned}
(2) Neutron kerma factor of oxygen \left(F_n\right)_O for 10 MeV neutrons is given as
\begin{aligned} \left(F_{\mathrm{n}}\right)_{\mathrm{O}}=\frac{N_{\mathrm{A}}}{A_{\mathrm{O}}} \sigma_{\mathrm{O}} \overline{\left(\Delta E_{\mathrm{K}}\right)_{\mathrm{O}}}= & \frac{\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right) \times\left(1.25 \times 10^{-24} \mathrm{~cm}^2\right) \times(1.11 \mathrm{MeV})}{\left(15.9994 \mathrm{~g} \cdot \mathrm{mol}^{-1}\right)} \\ & \times \frac{\left(1.602 \times 10^{-13} \mathrm{~J} / \mathrm{MeV}\right) \times\left(10^2 \mathrm{cGy} / \mathrm{Gy}\right)}{\left(10^{-3} \mathrm{~kg} / \mathrm{g}\right)} \\ = & 8.36 \times 10^{-10} \mathrm{cGy} \cdot \mathrm{cm}^2\quad (9.73) \end{aligned}
(d) Neutron kerma factor \left(F_{\mathrm{n}}\right)_{\mathrm{H}_2 \mathrm{O}} of water for 10 MeV neutrons is determined by summing up contributions to kerma factor from hydrogen and oxygen. Thus, 1 mol of water contains N_A molecules of water and, since each water molecule contains two hydrogen atoms (nuclei) as well as one oxygen atom (nucleus), one can say that 1 mol of water contains 2N_A atoms (nuclei) of hydrogen as well as N_A atoms of oxygen. \left(F_{\mathrm{n}}\right)_{\mathrm{H}_2 \mathrm{O}} is thus calculated as follows
\begin{aligned} \left(F_{\mathrm{n}}\right)_{\mathrm{H}_2 \mathrm{O}}= & \sum_i \frac{n_i^{\square}}{\rho_{\mathrm{H}_2 \mathrm{O}}} \sigma_i \overline{\left(\Delta E_{\mathrm{K}}\right)_i}=\sum_i \frac{N_{\mathrm{A}}}{A_{\mathrm{H}_2 \mathrm{O}}} \sigma_i \overline{\left(\Delta E_{\mathrm{K}}\right)_i} \\ = & \frac{2 N_{\mathrm{A}}}{A_{\mathrm{H}_2 \mathrm{O}}} \sigma_{\mathrm{H}} \overline{\left(\Delta E_{\mathrm{K}}\right)_{\mathrm{H}}}+\frac{N_{\mathrm{A}}}{A_{\mathrm{H}_2 \mathrm{O}}} \sigma_{\mathrm{O}} \overline{\left(\Delta E_{\mathrm{K}}\right)_{\mathrm{O}}} \\ = & \frac{2 \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right) \times\left(0.94 \times 10^{-24} \mathrm{~cm}^2\right) \times(5 \mathrm{MeV})}{18.0153 \mathrm{~g} \cdot \mathrm{mol}^{-1}} \\ & \times \frac{\left(1.6 \times 10^{-13} \mathrm{~J} / \mathrm{MeV}\right) \times\left(10^2 \mathrm{cGy} / \mathrm{Gy}\right)}{\left(10^{-3} \mathrm{~kg} / \mathrm{g}\right)} \\ = & \frac{\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right) \times\left(1.25 \times 10^{-24} \mathrm{~cm}^2\right) \times(1.11 \mathrm{MeV})}{\left(18.0153 \mathrm{~g} \cdot \mathrm{mol}^{-1}\right)} \\ & \times \frac{\left(1.6 \times 10^{-13} \mathrm{~J} / \mathrm{MeV}^{-3} \times\left(10^2 \mathrm{cGy} / \mathrm{Gy}\right)\right.}{\left(10^{-3} \mathrm{~kg} / \mathrm{g}\right)} \\ = & 0.503 \times 10^{-8} \mathrm{cGy} \cdot \mathrm{cm}^2+0.074 \times 10^{-8} \mathrm{cGy} \cdot \mathrm{cm}^2 \\ = & 5.77 \times 10^{-9} \mathrm{cGy} \cdot \mathrm{cm}^2 .\quad (9.74) \end{aligned}
Neutron kerma factor F_{\mathrm{n}} \text { is usually given in units of cGy } \cdot \mathrm{cm}^2 \text { meaning cGy } per neutron/cm². This follows from the definition of neutron kerma K=F_{\mathrm{n}} \varphi, so that the kerma factor F_{\mathrm{n}}=K / \varphi is sometimes referred to as fluence-to-kerma factor.
In neutron beam dosimetry kerma and dose in tissue are obtained by multiplying kerma and dose measured with an instrument that is generally not perfectly tissue equivalent by the ratio of kerma factors of the two attenuating media. This approach is similar to photon beam dosimetry where kerma and dose in tissue are obtained by multiplying kerma and dose in non-tissue equivalent medium by the ratio of mass energy transfer coefficients and ratio of mass energy absorption coefficients, respectively, of the two attenuating media. Note: unlike in photon beams, in neutron beams there is no energy loss to bremsstrahlung by charged particles released in the attenuating medium.
(e) Figure 9.5B provides the neutron kerma factor F_n against neutron kinetic energy E_{\mathrm{K}}^{\mathrm{n}} for various materials of interest in medical physics including hydrogen, oxygen, and water; the three materials used in this question. Data points at E_{\mathrm{K}}^{\mathrm{n}} = 10 MeV show results of our calculations of kerma factor in (c) and (d) in good agreement with the graphs published by the ICRU, Report 26.
