Boron-10 and cadmium-113 are used in control rods of nuclear reactors to maintain controlled chain reaction and to shut down the reactor. The respective nuclear reactions are: ^10 5B(n,α)^7 5Li and ^113 48Cd(n,γ)^114 48Cd. (a) Determine Q value for the ^10 5B(n,α)^7 5Li neutron capture nuclear

Question

Boron-10 and cadmium-113 are used in control rods of nuclear reactors to maintain controlled chain reaction and to shut down the reactor. The respective nuclear reactions are: [latex]{ }_5^{10} \mathrm{~B}(\mathrm{n}, \alpha){ }_5^7 \mathrm{Li} ext { and }{ }_{48}^{113} \mathrm{Cd}(\mathrm{n}, \gamma){ }_{48}^{114} \mathrm{Cd} ext {. }[/latex]

(a) Determine Q value for the [latex]{ }_5^{10} \mathrm{~B}(\mathrm{n}, \alpha){ }_5^7 \mathrm{Li}[/latex] neutron capture nuclear reaction. Use and compare three methods for Q value calculation: (1) Atomic rest energy method, (2) Nuclear rest energy method, and (3) Nuclear binding energy method. All required atomic and nuclear data are provided in Appendix A.

(b) For [latex]{ }_5^{10} \mathrm{~B}(\mathrm{n}, \alpha){ }_5^7 \mathrm{Li}[/latex] neutron capture nuclear reaction determine kinetic energy [latex]\left(E_{\mathrm{K}} ight)_\alpha ext { of the } \alpha[/latex] particle and recoil kinetic energy [latex]\left(E_{\mathrm{K}} ight)_{\mathrm{Li}-7}[/latex] of the Li-7 nucleus.

(c) Determine Q value for the [latex]{ }_{48}^{113} \mathrm{Cd}(\mathrm{n}, \gamma){ }_{48}^{114} \mathrm{Cd}[/latex] neutron capture nuclear reaction. Use and compare three methods for Q value calculation: (1) Atomic rest mass method, (2) Nuclear rest energy method, and (3) Nuclear binding energy method. All required atomic and nuclear data are provided in Appendix A.

(d) For [latex]{ }_{48}^{113} \mathrm{Cd}(\mathrm{n}, \gamma){ }_{48}^{114} \mathrm{Cd}[/latex] neutron capture nuclear reaction determine energy [latex]E_γ[/latex] of the γ ray photon and recoil kinetic energy [latex]\left(E_K ight)_{Cd-114}[/latex] of the Cd-114 nucleus.

Accepted Answer

Reaction energy or Q value of a nuclear reaction can be calculated with three methods and all three are expected to give the same result for a given nuclear reaction. The three methods are:

(1) Atomic rest energy method (T5.7).
(2) Nuclear rest energy method (T5.7).
(3) Nuclear binding energy method (T5.8).

We note the following general points related to the two neutron capture reactions:

(i) Neutrons in the two nuclear reactions of this problem are thermal neutrons.
In our calculations we can thus assume that their kinetic energy [latex]E_{\mathrm{K}}^{\mathrm{n}} \approx 0[/latex] is negligible and that the two target nuclei (boron-10 and cadmium-113) are initially at rest.
(ii) Total momentum before the nuclear reaction is zero causing the momenta of reaction products after the reaction to be opposite in direction and equal in magnitude.
(iii) Principles of total energy conservation and momentum conservation are used in the calculation of nuclear reaction Q value as well as energy and momentum of the reaction products.

(a) Q value of the [latex]{ }_5^{10} \mathrm{~B}(\mathrm{n}, \alpha){ }_3^7 \mathrm{Li}[/latex] neutron capture nuclear reaction is determined as follows:

(1) Atomic rest energy [latex]\mathcal{M}[/latex] method:

[latex]\begin{aligned} Q & =\left\{\mathcal{M}\left({ }_5^{10} \mathrm{~B} ight) c^2+m_{\mathrm{n}} c^2 ight\}-\left\{\mathcal{M}\left({ }_3^7 \mathrm{Li} ight) c^2+\mathcal{M}\left({ }_2^4 \mathrm{He} ight) c^2 ight\} \ & =\{(10.012937 \mathrm{u})+(1.008665 \mathrm{u})\} c^2-\{(7.016004 \mathrm{u})+(4.002603 \mathrm{u})\} c^2 \ & =\{11.021602-11.018607\} imes(931.494028 \mathrm{MeV})=2.79 \mathrm{MeV} .\quad (12.273) \end{aligned}[/latex]

(2) Nuclear rest energy M method:

[latex]\begin{aligned} Q= & \left\{M\left({ }_5^{10} \mathrm{~B} ight) c^2+m_{\mathrm{n}} c^2 ight\}-\left\{M\left({ }_3^7 \mathrm{Li} ight) c^2+m_\alpha c^2 ight\} \ = & \{9324.4360 \mathrm{MeV}+939.5654 \mathrm{MeV}\} \ & -\{6533.8328 \mathrm{MeV}+3727.3788 \mathrm{MeV}\} \ = & \{10264.0014 \mathrm{MeV}-10261.2116 \mathrm{MeV}\}=2.79 \mathrm{MeV} .\quad (12.274) \end{aligned}[/latex]

(3) Binding energy [latex]E_B[/latex] method:

[latex]\begin{aligned} Q & =\left\{E_{\mathrm{B}}\left({ }_3^7 \mathrm{Li} ight)+E_{\mathrm{B}}\left({ }_2^4 \mathrm{He} ight) ight\}-\left\{E_{\mathrm{B}}\left({ }_5^{10} \mathrm{~B} ight) c^2+0 ight\} \ & =\{39.2446 \mathrm{MeV}+28.2959 \mathrm{MeV}\}-\{64.7508 \mathrm{MeV}\}=2.79 \mathrm{MeV} .\quad (12.275) \end{aligned}[/latex]

The three methods used in determination of Q value provide identical results, as expected. Note that in the atomic rest energy method of (12.273) we use the atomic rest mass [latex]\mathcal{M}[/latex] of the helium-4 atom (α particle neutralized by two electrons) to account for all orbital electrons involved in the reaction, while in (12.274) we use the nuclear rest mass M of the α particle. In (12.275) the thermal neutron is free and thus has no binding energy

(b) To determine for the nuclear reaction [latex]{ }_5^{10} \mathrm{~B}(\mathrm{n}, \alpha){ }_3^7 \mathrm{Li}[/latex] kinetic energy [latex]\left(E_K ight)_α[/latex] of the α particle ejected in the nuclear reaction and recoil energy [latex]\left(E_{\mathrm{K}} ight)_{\mathrm{Pb}-218}\left(E_{\mathrm{K}} ight)_{\mathrm{Li}-7}[/latex] of the lithium-7 reaction product we use the principles of: (1) total energy conservation and (2) momentum conservation. The two conservation principles are for the nuclear reaction expressed as follows:

(1) For total energy conservation (T5.6) we recognize that kinetic energies in α decay are relatively small allowing us to use classical mechanics for expression of energy conservation

[latex]Q=\left(E_{\mathrm{K}} ight)_\alpha+\left(E_{\mathrm{K}} ight)_{\mathrm{Li}-7}=\frac{p_\alpha^2}{2 m_\alpha}+\frac{p_{\mathrm{Li}-7}^2}{2 \mathcal{M}_{\mathrm{Li}-7}}[/latex] (12.276)

(2) For momentum conservation we note that momentum before the nuclear reaction is zero causing the two momenta [latex]p_α\ and\ p_{Li-7}[/latex] after the nuclear reaction to be opposite in direction and equal in magnitude

[latex]0=\mathbf{p}_\alpha+\mathbf{p}_{\mathrm{Li}-7}[/latex] (12.277)

or

[latex]\left|\mathbf{p}_\alpha ight| \equiv p_\alpha=\sqrt{2 m_\alpha\left(E_{\mathrm{K}} ight)_\alpha}=\left|\mathbf{p}_{\mathrm{Li}-7} ight| \equiv p_{\mathrm{Li}-7}=\sqrt{2 \mathcal{M}_{\mathrm{Li}-7}\left(E_{\mathrm{K}} ight)_{\mathrm{Li}-7}},[/latex] (12.278)

where

[latex]p_α[/latex] is the magnitude of α particle momentum.
[latex]p_{Li-7}[/latex] is the magnitude of the lithium-7 atomic recoil momentum.
[latex]m_α[/latex] is the rest mass of the α particle.
[latex]\mathcal{M}_{Li-7}[/latex] is the rest mass of the lithium-7 atom

[latex]\mathcal{M}_{\mathrm{Li}-7} c^2=(7.016004 \mathrm{u}) imes(931.494024 \mathrm{MeV} / \mathrm{u})=6535.3658 \mathrm{MeV}[/latex]

Using (12.276) we can now express the conservation of energy as

[latex]Q=\left(E_{\mathrm{K}} ight)_\alpha+\left(E_{\mathrm{K}} ight)_{\mathrm{Li}-7}=\frac{p_\alpha^2}{2 m_\alpha}+\frac{p_{\mathrm{Li}-7}^2}{2 \mathcal{M}_{\mathrm{Li}-7}}=\frac{p_\alpha^2}{2}\left[\frac{1}{m_\alpha}+\frac{1}{\mathcal{M}_{\mathrm{Li}-7}} ight][/latex] (12.279)

from where it follows that

[latex]p_\alpha^2=p_{\mathrm{Li}-7}^2=2 Q\left[\frac{1}{m_\alpha}+\frac{1}{\mathcal{M}_{\mathrm{Li}-7}} ight]^{-1}[/latex] (12.280)

Inserting (12.280) into (12.279) we now get the following expressions for [latex]\left(E_K ight)_α\ and\ \left(E_K ight)_{Li-7}[/latex].

[latex]\left(E_{\mathrm{K}} ight)_\alpha=\frac{p_\alpha^2}{2 m_\alpha}=\frac{Q}{m_\alpha}\left[\frac{1}{m_\alpha}+\frac{1}{\mathcal{M}_{\mathrm{Li}-7}} ight]^{-1}=\frac{Q}{\left[1+\frac{m_\alpha}{M_{\mathrm{Li}-7}} ight]}[/latex] (12.281)

[latex]\begin{aligned} & =\frac{2.79 \mathrm{MeV}}{\left\{1+\frac{3727.3791 \mathrm{MeV}}{(7.016004 \mathrm{u}) imes(931.494028 \mathrm{MeV} / \mathrm{u})} ight\}}=\frac{2.79 \mathrm{MeV}}{1.5703} \ & =1.78 \mathrm{MeV}\quad (12.282) \end{aligned}[/latex]

and

[latex]\begin{aligned} \left(E_{\mathrm{K}} ight)_{\mathrm{Li}-7} & =\frac{p_{\mathrm{Li}-7}^2}{2 \mathcal{M}_{\mathrm{Li}-7}}=\frac{Q}{\mathcal{M}_{\mathrm{Li}-7}}\left[\frac{1}{m_\alpha}+\frac{1}{\mathcal{M}_{\mathrm{Li}-7}} ight]^{-1}=\frac{Q}{\left[1+\frac{\mathcal{M}_{\mathrm{Li}-7}}{m_\alpha} ight]} \ & =\frac{2.79 \mathrm{MeV}}{\left\{1+\frac{7.016004 \mathrm{u}) imes(931.494028 \mathrm{MeV} / \mathrm{u})}{3727.3791 \mathrm{MeV}} ight\}}=\frac{2.79 \mathrm{MeV}}{2.7533}=1.01 \mathrm{MeV}\quad (12.283) \end{aligned}[/latex]

(c) Q value of the [latex]{ }_{48}^{113} \mathrm{Cd}(\mathrm{n}, \gamma){ }_{48}^{114} \mathrm{Cd}[/latex] neutron capture nuclear reaction is determined as follows:

(1) Atomic rest energy [latex]\mathcal{M}[/latex] method:

[latex]\begin{aligned} Q & =\left\{\mathcal{M}\left({ }_{48}^{113} \mathrm{Cd} ight) c^2+m_{\mathrm{n}} c^2 ight\}-\left\{\mathcal{M}\left({ }_{48}^{114} \mathrm{Cd} ight) c^2+0 ight\} \ & =\{(112.904402 \mathrm{u})+(1.008665 \mathrm{u})\} c^2-\{(113.903359 \mathrm{u})\} c^2 \ & =\{113.913067-113.903359\} imes(931.494028 \mathrm{MeV})=9.04294 \mathrm{MeV}\quad (12.284) \end{aligned}[/latex]

(2) Nuclear rest energy M method:

[latex]\begin{aligned} Q & =\left\{M\left({ }_{48}^{113} \mathrm{Cd} ight) c^2+m_{\mathrm{n}} c^2 ight\}-\left\{M\left({ }_{48}^{114} \mathrm{Cd} ight) c^2+0 ight\} \ & =\{105145.2523 \mathrm{MeV}+939.5654 \mathrm{MeV}\}-\{106075.7748 \mathrm{MeV}\} \ & =\{106084.8177 \mathrm{MeV}-106075.7748 \mathrm{MeV}\}=9.0429 \mathrm{MeV}\quad (12.285) \end{aligned}[/latex]

(3) Binding energy [latex]E_B[/latex] method:

[latex]\begin{aligned} Q & =\left\{E_{\mathrm{B}}\left({ }_{48}^{114} \mathrm{Cd} ight)+0 ight\}-\left\{E_{\mathrm{B}}\left({ }_{48}^{113} \mathrm{Cd} ight) c^2+0 ight\} \ & =972.5984 \mathrm{MeV}-963.5555 \mathrm{MeV}=9.0429 \mathrm{MeV} .\quad (12.286) \end{aligned}[/latex]

(d) To determine for the nuclear reaction [latex]{ }_{48}^{113} \mathrm{Cd}(\mathrm{n}, \gamma){ }_{48}^{114} \mathrm{Cd} ext { energy } E_\gamma[/latex] of the γ ray ejected in the nuclear reaction and recoil energy [latex]\left(E_K ight)_{Cd-114}[/latex] of the cadmium114 reaction product we use the principles of:

(1) Conservation of total energy.
(2) Conservation of momentum.

The two conservation principles are for the nuclear reaction expressed as:

(1) Total energy conservation (T5.6) is expressed as follows

[latex]Q=E_\gamma+\left(E_{\mathrm{K}} ight)_{\mathrm{Cd}-114}=E_\gamma+\frac{p_{\mathrm{Cd}-114}^2}{2 M_{\mathrm{Cd}-114}}[/latex] (12.287)

(2) For momentum conservation we note that momentum before the nuclear reaction is zero causing the two momenta [latex]p_γ\ and\ p_{Cd-114}[/latex] after the nuclear reaction to be opposite in direction and equal in magnitude

[latex]0=\mathbf{p}_\gamma+\mathbf{p}_{\mathrm{Cd}-114}[/latex] (12.288)

or

[latex]\left|\mathbf{p}_\gamma ight| \equiv p_\gamma=\frac{E_\gamma}{c}=\left|\mathbf{p}_{\mathrm{Cd}-114} ight| \equiv p_{\mathrm{Cd}-114}=\sqrt{2 M_{\mathrm{Cd}-114}\left(E_{\mathrm{K}} ight)_{\mathrm{Cd}-114}}[/latex] (12.289)

where

[latex]p_γ[/latex] is the magnitude of γ photon momentum.
[latex]p_{Cd-114}[/latex] is the magnitude of the cadmium-114 atomic recoil momentum.
[latex]\mathcal{M}_{Cd-114}[/latex] is the rest mass of the cadmium-114 atom. Note: in the calculation we use atomic mass rather than nuclear mass, since the whole atom rather than just the nucleus recoils after the reaction. [latex]\mathcal{M}_{Cd-114c^2}[/latex] = (113.903359 u)×(931.494024 MeV/u) = 106100.2987 MeV.

Inserting (12.289) in the form of [latex]p_{\mathrm{Cd}-114}=E_\gamma / c[/latex] into (12.287) we obtain a quadratic equation for photon energy [latex]E_γ[/latex].

[latex]\frac{E_\gamma^2}{2 \mathcal{M}_{\mathrm{Cd}-114 c^2}}+E_\gamma-Q=0,[/latex] (12.290)

with the following physically relevant solution

[latex]\begin{aligned} E_\gamma & =\mathcal{M}_{\mathrm{Cd}-114 c^2}\left\{\sqrt{1+\frac{2 Q}{\mathcal{M}_{\mathrm{Cd}-114 c^2}}}-1 ight\} \ & =(106100.2987 \mathrm{MeV}) imes\left\{\sqrt{1+\frac{2 imes(9.0429 \mathrm{MeV})}{106100.2987 \mathrm{MeV}}}-1 ight\} \ & =9.0425 \mathrm{MeV} .\quad (12.291) \end{aligned}[/latex]

Recoil energy [latex]\left(E_K ight)_{Cd-114}[/latex] of the cadmium-114 atom is from (12.276) given as the difference between Q value of 9.0429 MeV and photon energy [latex]E_γ[/latex] = 9.0425 MeV of (12.291). The recoil energy of the cadmium-114 atom is thus given as

[latex]\left(E_{\mathrm{K}} ight)_{\mathrm{Cd}-114}=Q-E_\gamma=9.0429 \mathrm{MeV}-9.0425 \mathrm{MeV}=0.0004 \mathrm{MeV}=400 \mathrm{eV}[/latex] (12.292)

As shown in (12.291) and (12.292), the photon carries away most of the energy released by the nuclear reaction [latex]{ }_{48}^{113} \mathrm{Cd}(\mathrm{n}, \gamma){ }_{48}^{114} \mathrm{Cd}[/latex]. Only a minute, essentially negligible, fraction (0.004 %) is given to the cadmium-114 recoil atom.

Question 12.8.Q2: Boron-10 and cadmium-113 are used in control rods of nuclear......

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