Question 9.6.Q1: Boron neutron capture therapy (BNCT) is a binary radiotherap......

(a) As shown in Fig. 9.6, boron-10 exposed to thermal neutrons may undergo a neutron capture nuclear reaction that produces two reaction products ({ }_3^7 \mathrm{Li} ion and α particle) with two possible branches: \left({ }_3^7 \mathrm{Li}, \alpha\right) \text { and }\left({ }_3^7 \mathrm{Li}^*, \alpha\right) with a branching ratio of 6 % vs. 94 %. The lithium-7 excited state de-excites to ground state with emission of a γ ray of energy E_γ = 0.48 MeV.

Q value for both branches of the \left({ }_5^{10} \mathrm{~B}, \mathrm{n}\right) neutron capture reaction is calculated as follows [(T5.7) and (T5.8)]

(1) Rest energy method:

(2) Binding energy method:

Q value for both branches of the \left({ }_5^{10} \mathrm{~B}, \mathrm{n}\right) neutron capture reaction is 2.79 MeV; however, in the first branch \left({ }_3^7 \mathrm{Li}, \alpha\right) the two ions released share the full Q value in energy while in the second branch, which also produces a 0.48 MeV γ ray in the de-excitation process of { }_3^7 \mathrm{Li}^* \text {, } the two ions share the remaining energy Q − E_γ = 2.79 MeV − 0.48 MeV = 2.31 MeV.

(b) Kinetic energy of reaction products in the neutron capture reaction is determined from the known Q value for the reaction calculated in (a).

(1) Kinetic energies of the α particle and the { }_3^7 \mathrm{Li}{ }^{3+} ion produced in the \left({ }_3^7 \mathrm{Li}, \alpha\right) branch of the boron-10 neutron capture reaction are calculated as follows. The two reaction products share the Q value energy in the inverse proportion of their masses, since the momenta of the two reaction products \text { ( } \alpha particle and { }_3^7 \mathrm{Li} \text { ion) } are equal in magnitude but opposite in direction to satisfy the conservation of momentum principle in thermal neutron capture reaction. Note: Total momentum before reaction is zero, so total momentum after capture reaction must also be zero.
Q value can be expressed as follows

Q=E_{\mathrm{K}}^{\mathrm{Li}}+E_{\mathrm{K}}^\alpha=\frac{p_{\mathrm{Li}}^2}{2 M\left({ }_3^7 \mathrm{Li}\right)}+\frac{p_\alpha^2}{2 m_\alpha}=\frac{p^2}{2}\left[\frac{1}{M\left({ }_3^7 \mathrm{Li}\right)}+\frac{1}{m_\alpha}\right]=\frac{p^2}{2} \frac{m_\alpha+M\left({ }_3^7 \mathrm{Li}\right)}{M\left({ }_3^7 \mathrm{Li}\right) m_\alpha},           (9.112)

where E_{\mathrm{K}}^{\mathrm{Li}} \text { and } E_{\mathrm{K}}^\alpha are kinetic energy of the { }_3^7 \mathrm{Li} ion and α particle, respectively, and p stands for \left|\mathbf{p}_{\mathrm{Li}}\right|=p_{\mathrm{Li}} \text { as well as for }\left|\mathbf{p}_\alpha\right|=p_\alpha, since the magnitudes of the two momentum vectors are equal.

Solving (9.112) for p^2 allows us to express the kinetic energies E_{\mathrm{K}}^{\mathrm{Li}} \text { and } E_{\mathrm{K}}^\alpha \text {, } respectively, as follows

and

(2) Kinetic energies of the α particle and the { }_3^7 \mathrm{Li}^{3+} excited ion produced in the \left({ }_3^7 \mathrm{Li}^*, \alpha\right) branch of the boron-10 neutron capture reaction are calculated as follows. The excited nucleus { }_3^7 \mathrm{Li}^* attains the ground state of { }_3^7 \mathrm{Li} through emission of a γ ray of energy E_\gamma=0.48 \mathrm{~MeV} . Thus, the energy difference Q-E_\gamma = 2.31 MeV rather than the full energy of the Q value is available for sharing between the { }_3^7 \mathrm{Li} ion and α particle and the energy is again shared in the inverse proportion of the masses of the two ions. For this branch of the boron-10 neutron capture reaction kinetic energies E_{\mathrm{K}}^{\mathrm{Li}} \text { and } E_{\mathrm{K}}^\alpha, respectively, are given as follows

and

(c) Neutron capture reaction { }_5^{10} \mathrm{~B}(\mathrm{n}, \alpha){ }_3^7 \mathrm{Li} proceeds in two branches, the 6 % branch producing 1.78 MeV α particles and the 94 % branch producing 1.47 MeV α particles [see (b)]. To establish whether or not reaction { }_5^{10} \mathrm{~B}(\alpha, \mathrm{n}){ }_7^{13} \mathrm{~N} can run with these α particles we first evaluate the reaction Q value using the rest energy method and the binding energy method as follows

Since Q value of the reaction { }_5^{10} \mathrm{~B}(\alpha, \mathrm{n}){ }_7^{13} \mathrm{~N} is positive, the reaction is exothermic and can proceed without any threshold energy restrictions.