Question 9.5.Q2: Neutrons with kinetic energy below 10 eV fall into the therm......
Neutrons with kinetic energy below 10 eV fall into the thermal neutron dosimetric energy region that is categorized by thermal neutron capture interactions with elemental hydrogen { }_1^1 \mathrm{H} and elemental nitrogen { }_7^{14} \mathrm{~N} in tissue. Nitrogen { }_7^{14} \mathrm{~N} actually offers two neutron capture reactions, reaction (1){ }_7^{14} \mathrm{~N}(\mathrm{n}, \mathrm{p}){ }_6^{13} \mathrm{C} with cross section of \sigma\left({ }_7^{14} \mathrm{~N}\right) \approx 1.18 \mathrm{~b} / \text { atom } and reaction (2){ }_7^{14} \mathrm{~N}(\mathrm{n}, \gamma){ }_6^{15} \mathrm{~N} with a much smaller cross section of \sigma\left({ }_7^{14} \mathrm{~N}\right) \approx 0.06 \mathrm{~b} / \text { atom }. Hydrogen { }_1^1 \mathrm{H} contributes to neutron capture through reaction (3) { }_1^1 \mathrm{H}(\mathrm{n}, \gamma){ }_1^2 \mathrm{H} with a cross section of \sigma\left({ }_1^1 \mathrm{H}\right)=0.33 \mathrm{~b} / \text { atom }.
(a) Nuclear reactions are characterized with a Q value that can be positive, zero, or negative. Define the Q value for a nuclear reaction and describe the two methods that are used for its determination in a typical nuclear reaction.
(b) Calculate the Q values of the three neutron capture reactions that represent the most prominent interactions of thermal neutrons with tissue. In your calculations use the two methods described in (a) and show that both methods yield the same results for a given thermal neutron capture reaction. Appropriate rest energy and binding energy data are given in Appendix A.
(c) For each of the three reactions determine the kinetic energies of the recoil nucleus and particles released [proton in reaction (1) and γ photons in reactions (2) and (3)].
(a) Two methods are used for determination of reaction Q value:
(1) Rest energy method in which we subtract the sum of nuclear rest energies of reaction products after the reaction \sum_{i, \text { after }} M_i c^2 from the sum of nuclear rest energies of reactants (projectile and target) before the reaction \sum_{i, \text { before }} M_i c^2, or
Q=\sum_{i, \text { before }} M_i c^2-\sum_{i, \text { after }} M_i c^2 (9.81)
(2) Binding energy method in which we subtract the sum of nuclear binding energies of reactants before the reaction \sum_{i, \text { before }} E_{\mathrm{B}}(i) from the sum of nuclear binding energies of reaction products after the reaction \sum_{i, \text { before }} E_{\mathrm{B}}(i), or
Q=\sum_{i, \text { after }} E_{\mathrm{B}}(i)-\sum_{i, \text { before }} E_{\mathrm{B}}(i) (9.82)
(b) Two thermal neutron capture reactions are important in thermal neutron interactions with tissue: { }_1^1 \mathrm{H}(\mathrm{n}, \gamma){ }_1^2 \mathrm{H} \text { and }{ }_7^{14} \mathrm{~N}(\mathrm{n}, \mathrm{p}){ }_6^{14} \mathrm{C} \text {. } Another thermal neutron reaction { }_7^{14} \mathrm{~N}(\mathrm{n}, \gamma){ }_7^{15} \mathrm{~N} is possible for thermal neutrons interacting with tissue but it has a much smaller probability than the { }_7^{14} \mathrm{~N}(\mathrm{n}, \mathrm{p}){ }_6^{14} \mathrm{C} reaction.
(1) Thermal neutron capture reaction { }_7^{14} \mathrm{~N}(n, p){ }_6^{14} \mathrm{C} can be written as
{ }_7^{14} \mathrm{~N}+\mathrm{n}=\mathrm{p}+{ }_6^{14} \mathrm{C}+Q (9.83)
and its Q value is determined as follows:
Rest energy method
\begin{aligned} Q & =\sum_{i, \text { before }} M_i c^2-\sum_{i, \text { after }} M_i c^2=\left[M\left({ }_7^{14} \mathrm{~N}\right)+m_{\mathrm{n}} c^2\right]-\left[m_{\mathrm{p}} c^2+M\left({ }_6^{14} \mathrm{C}\right)\right] \\ & =[13040.2028 \mathrm{MeV}+939.5653 \mathrm{MeV}]-[938.2720 \mathrm{MeV}+13040.8703 \mathrm{MeV}] \\ & =[13979.7681 \mathrm{MeV}-13979.1423 \mathrm{MeV}]=0.626 \mathrm{MeV} .\quad (9.84) \end{aligned}Binding energy method
\begin{aligned} Q & =\sum_{i, \text { after }} E_{\mathrm{B}}(i)-\sum_{i, \text { before }} E_{\mathrm{B}}(i)=\left[0+E_{\mathrm{B}}\left({ }_6^{14} \mathrm{C}\right)\right]-\left[E_{\mathrm{B}}\left({ }_7^{14} \mathrm{~N}\right)+0\right] \\ & =105.2846 \mathrm{MeV}-104.6587 \mathrm{MeV}=0.626 \mathrm{MeV} .\quad (9.85) \end{aligned}(2) Thermal neutron capture reaction { }_7^{14} \mathrm{~N}(\mathrm{n}, \gamma){ }_7^{15} \mathrm{~N} can be written as
{ }_7^{14} \mathrm{~N}+\mathrm{n}=\gamma+{ }_7^{15} \mathrm{~N}+Q (9.86)
and its Q value is determined as follows:
Rest energy method
\begin{aligned} Q & =\sum_{i, \text { before }} M_i c^2-\sum_{i, \mathrm{after}} M_i c^2=\left[M\left({ }_7^{14} \mathrm{~N}\right)+m_{\mathrm{n}} c^2\right]-\left[0+M\left({ }_7^{15} \mathrm{~N}\right)\right] \\ & =[130402028 \mathrm{MeV}+939.5653 \mathrm{MeV}]-[13968.9350 \mathrm{MeV}] \\ & =[13979.7681 \mathrm{MeV}-13968.9350 \mathrm{MeV}]=10.83 \mathrm{MeV}\quad (9.87) \end{aligned}Binding energy method
\begin{aligned} Q & =\sum_{i, \text { after }} E_{\mathrm{B}}(i)-\sum_{i, \text { before }} E_{\mathrm{B}}(i)=\left[0+E_{\mathrm{B}}\left({ }_7^{15} \mathrm{~N}\right)\right]-\left[E_{\mathrm{B}}\left({ }_7^{14} \mathrm{~N}\right)+0\right] \\ & =115.4914 \mathrm{MeV}+104.6587 \mathrm{MeV}=10.83 \mathrm{MeV}\quad (9.88) \end{aligned}(3) Thermal neutron capture reaction { }_1^1 \mathrm{H}(\mathrm{n}, \gamma){ }_1^2 \mathrm{H} can be written as
{ }_1^1 \mathrm{H}+\mathrm{n}=\gamma+{ }_1^2 \mathrm{H}+Q (9.89)
and its Q value is determined as follows
Rest energy method
\begin{aligned} Q & =\sum_{i, \text { before }} M_i c^2-\sum_{i, \text { after }} M_i c^2=\left[M\left({ }_1^1 \mathrm{H}\right)+m_{\mathrm{n}} c^2\right]-\left[0+M\left({ }_1^2 \mathrm{H}\right)\right] \\ & =[938.2720 \mathrm{MeV}+939.5653 \mathrm{MeV}]-[0+1875.6128 \mathrm{MeV}] \\ & =[1877.8373 \mathrm{MeV}-1875.6128 \mathrm{MeV}]=2.225 \mathrm{MeV}\quad (9.90) \end{aligned}Binding energy method
Q=\sum_{i, \text { after }} E_{\mathrm{B}}(i)-\sum_{i, \text { before }} E_{\mathrm{B}}(i)=E_{\mathrm{B}}\left({ }_2^1 \mathrm{H}\right)-0=2.225 \mathrm{MeV} (9.91)
Q values for the three thermal neutron capture reactions have been calculated with the rest energy method and the binding energy method and, for a given nuclear reaction, the two techniques provided identical results.
(c) For a typical nuclear reaction triggered by projectile with mass m10 and kinetic energy \left(E_{\mathrm{K}}\right)_0 \text { striking a stationary target } m_{20} and resulting in two reaction products with masses m_{30}\ and\ m_{40}, the Q value is defined as
Q=\left(m_{10} c^2+m_{20} c^2\right)-\left(m_{30} c^2+m_{40} c^2\right) (9.92)
Conservation of total energy in nuclear reaction, on the other hand, results in the following expression
\left[m_{10} c^2+\left(E_{\mathrm{K}}\right)_0\right]+\left[m_{20} c^2+0\right]=\left[m_{30} c^2+\left(E_{\mathrm{K}}\right)_3\right]+\left[m_{40} c^2+\left(E_{\mathrm{K}}\right)_4\right], (9.93)
where \left(E_{\mathrm{K}}\right)_3 \text { and }\left(E_{\mathrm{K}}\right)_4 are kinetic energies of the reaction products m30 and m40 and the other quantities were defined above. Insertion of (9.92) into (9.93) results in the following simplification of (9.93)
\left(E_{\mathrm{K}}\right)_0+Q=\left(E_{\mathrm{K}}\right)_3+\left(E_{\mathrm{K}}\right)_4 . (9.94)
Equation (9.94) can now be further simplified for thermal neutron capture reaction since the incident thermal neutron energy \left(E_{\mathrm{K}}\right)_0=E_{\mathrm{K}}^{\mathrm{n}}=0.025 \mathrm{eV} \approx 0.
Thus, for thermal neutron capture reaction we can state that the sum of kinetic energies of the reaction products equals to Q value of the reaction
Q=\left(E_{\mathrm{K}}\right)_3+\left(E_{\mathrm{K}}\right)_4=\frac{\left|\mathbf{p}_{m_{30}}\right|^2}{2 m_{30}} \frac{\left|\mathbf{p}_{m_{40}}\right|^2}{2 m_{40}}=\frac{p^2}{2}\left[\frac{1}{m_{30}}+\frac{1}{m_{40}}\right], (9.95)
where \left|\mathbf{p}_{m_{30}}\right| \text { and }\left|\mathbf{p}_{m_{40}}\right| \text { are momenta of reaction products } m_{30} \text { and } m_{40} and we use the classical expression for kinetic energy of a particle. Total momentum before the capture reaction is zero, so that to fulfill the principle of momentum conservation the two momenta after the reaction must be equal in magnitude, i.e., \left|\mathbf{p}_{m_{30}}\right|=\left|\mathbf{p}_{m_{40}}\right|= p, and opposite in direction.
We now address the three neutron capture reactions of thermal neutrons with tissue atoms, two of the reactions produce a recoil atom and a γ ray and one reaction produces a recoil atom and a proton.
(1) Q value for reaction { }_7^{14} \mathrm{~N}(\mathrm{n}, \mathrm{p}){ }_6^{14} \mathrm{C} was determined in (b) as Q = 0.626 MeV. This energy is shared as kinetic energy between two reaction products: proton and { }_6^{14} \mathrm{C} nucleus in the inverse proportion of their masses, since both reaction products carry away the same momentum but in opposite directions. Equation (9.95) for this reaction is expressed as follows
Q=E_{\mathrm{K}}^{\mathrm{p}}+E_{\mathrm{K}}^{{ }^{14} \mathrm{C}}=\frac{p^2}{2}\left[\frac{1}{m_{\mathrm{p}}}+\frac{1}{M\left({ }_7^{14} \mathrm{C}\right)}\right]=\frac{p^2}{2}\left[\frac{\left[M\left({ }_7^{14} \mathrm{C}\right)+m_{\mathrm{p}}\right]}{m_{\mathrm{p}} M\left({ }_7^{14} \mathrm{C}\right)}\right] . (9.96)
After rearranging (9.96) we get the following expression for p²
p^2=2 Q \frac{m_{\mathrm{p}} M\left({ }_7^{14} \mathrm{C}\right)}{M\left({ }_7^{14} \mathrm{C}\right)+m_{\mathrm{p}}} (9.97)
so that we now express kinetic energy of the proton E_{\mathrm{K}}^{\mathrm{p}} as
and kinetic energy of the { }_6^{14} \mathrm{C} nucleus as
\begin{aligned} E_{\mathrm{K}}^{{ }_7^{14} \mathrm{C}} & =\frac{p^2}{2 M\left({ }_7^{14} \mathrm{C}\right)}=Q \frac{m_{\mathrm{p}}}{M\left({ }_7^{14} \mathrm{C}\right)+m_{\mathrm{p}}}=Q \frac{m_{\mathrm{p}} c^2}{M\left({ }_7^{14} \mathrm{C}\right) c^2+m_{\mathrm{p}} c^2} \\ & =(0.626 \mathrm{MeV}) \times \frac{938.2720}{13040.8703+938.2720}=0.042 \mathrm{MeV}\quad (9.99) \end{aligned}From (9.98) and (9.99) we note that in neutron capture reaction { }_7^{14} \mathrm{~N}(\mathrm{n}, \mathrm{p}){ }_6^{14} \mathrm{C} the proton and the { }_6^{14} \mathrm{C} nucleus share the Q = 0.626 MeV available energy in the inverse proportion of their rest masses, so that the proton carries away kinetic energy of E_{\mathrm{K}}^{\mathrm{p}}=0.584 \mathrm{MeV} \text { and the }{ }_6^{14} \mathrm{C} \text { nucleus carries away kinetic energy of } E_{\mathrm{K}}^{{ }^{14} \mathrm{C}}= 0.042 MeV.
Dose deposition in tissue from thermal neutron absorption is a two-step process. In the first step, energy is transferred from neutron to charged particles via thermal neutron capture contributing to tissue kerma. In the second step, energy is transferred from charged particles to tissue via Coulomb interactions with orbital electrons of tissue atoms, contributing to tissue dose.
In contrast to reaction (1) that in addition to recoil nucleus produces a proton, capture reactions (2) and (3) in addition to recoil nucleus produce a γ photon. The magnitude of the photon momentum p_ν is equal to the magnitude of the recoil nucleus momentum p\left(m_{30}\right) and the two momenta are opposite in direction to satisfy conservation of momentum in thermal neutron capture (note: total momentum before reaction is zero, since the neutron is of thermal energy and the target is stationary). The presence of γ photon after reaction requires a modification of (9.92) and (9.94) to read
Q=\left(m_{10} c^2+m_{20} c^2\right)-\left(m_{30} c^2+0\right) (9.100)
and
Q=\left(E_{\mathrm{K}}\right)_3+E_ν=\frac{p_ν^2}{2 m_{30}}+p_ν c . (9.101)
Solving (9.101) for p_ν results in the following quadratic equation
p_ν^2+2 m_{30} p_ν c-2 m_{30} Q=0 (9.102)
with the following physical solution for the magnitude of the photon and recoil nucleus momenta
p_ν=\frac{-2 m_{30} c+\sqrt{4 m_{30}^2+8 m_{30} Q}}{2}=\frac{m_{30} c^2}{c}\left[\sqrt{1+\frac{2 Q}{m_{30} c^2}}-1\right] . (9.103)
We now apply (9.103) to reactions (2) and (3) to determine p_ν .
(2) In neutron capture reaction { }_7^{14} \mathrm{~N}(\mathrm{n}, \gamma){ }_7^{13} \mathrm{~N} we determined the Q value as Q = 10.83 MeV and m_{30} \text { of }(9.103) \text { is } M\left({ }_7^{13} \mathrm{~N}\right), resulting in the following p_ν.
\begin{aligned} p_ν & =\frac{M\left({ }_6^{14} \mathrm{C}\right) c^2}{c}\left[\sqrt{1+\frac{2 Q}{M\left({ }_7^{13} \mathrm{~N}\right) c^2}}-1\right] \\ & =\frac{12111.1912 \mathrm{MeV}}{c}\left[\sqrt{1+\frac{2 \times 10.83}{12111.1912}}-1\right] \\ & =\frac{12111.1912 \mathrm{MeV}}{c} \times\left(8.94 \times 10^{-4}\right)=10.826 \mathrm{MeV} / c .\quad (9.104) \end{aligned}The photon momentum p_ν and recoil momentum p\left({ }_7^{13} \mathrm{~N}\right)=p_ν = 10.826 MeV/c give the following γ photon energy
E_\gamma=h ν=p_ν c=(10.826 \mathrm{MeV} / c) \times c=10.826 \mathrm{MeV} (9.105)
and kinetic energy of the recoil nucleus \text { s } E_{\mathrm{K}}{ }_7^{13} \mathrm{~N}.
E_{\mathrm{K}}{ }^{13} \mathrm{~N}=\frac{p_ν^2 c^2}{2 M\left({ }_7^{13} \mathrm{~N}\right) c^2}=\frac{10.826^2}{2 \times 12111.1912} \mathrm{MeV}=4.8 \times 10^{-3} \mathrm{MeV} (9.106)
(3) In neutron capture reaction { }_1^1 \mathrm{H}(\mathrm{n}, \gamma){ }_1^2 \mathrm{H} we determined the Q value as Q = 2.225 MeV and m_{30} of (9.103) is M\left({ }_1^2 \mathrm{H}\right) \text {, resulting in the following } p_ν.
\begin{aligned} p_ν & =\frac{M\left({ }_1^2 \mathrm{H}\right) c^2}{c}\left[\sqrt{1+\frac{2 Q}{M\left({ }_1^2 \mathrm{H}\right) c^2}}-1\right]=\frac{1875.6128 \mathrm{MeV}}{c}\left[\sqrt{1+\frac{2 \times 2.225}{1875.6128}}-1\right] \\ & =\frac{1875.6128 \mathrm{MeV}}{c} \times\left(1.19 \times 10^{-3}\right)=2.224 \mathrm{MeV} / c\quad (9.107) \end{aligned}The photon momentum p_ν\text { and recoil momentum } p\left({ }_1^2 \mathrm{H}\right)=p_ν=2.224 \mathrm{MeV} / c give the following γ photon energy
E_\gamma=hν=p_ν c=(2.224 \mathrm{MeV} / c) \times c=2.224 \mathrm{MeV} (9.108)
and kinetic energy of the recoil nucleus E_{\mathrm{K}}^{\mathrm{d}}, where d stands for the deuteron produced in the neutron capture reaction
E_{\mathrm{K}}^2 \mathrm{H}=E_{\mathrm{K}}^{\mathrm{d}}=\frac{p_ν^2 c^2}{2 M\left({ }_1^2 \mathrm{H}\right) c^2}=\frac{2.224^2}{2 \times 1875.6128} \mathrm{MeV}=1.3 \times 10^{-3} \mathrm{MeV} . (9.109)
From (9.105) and (9.106) as well as from (9.108) and (9.109) it is evident that the γ photon acquires most of the energy available from the Q value (of the order of 99.95 %) and the recoil nucleus, because of its large mass, receives only a small fraction of the available energy (0.05 %). In the first approximation, one can ignore the nuclear recoil energy and assume the full amount of the available Q value of the neutron thermal capture reaction is acquired by the photon when a photon is one of the reaction products.