Recoil Kinematics
Again, this material is really common, but problably comes from
(Lamarsh, et. al. 2001).
Given an \(\alpha\) of energy \(E\) incident on
\(\mathrm{\sideset{^{9}}{}{Be}}\), what is the identity and energy
of the recoil?
First, we can write up our reaction
\[\mathrm{\sideset{^{9}_{4}}{}{Be}} + \alpha \rightarrow \mathrm{\sideset{^{13}_{6}}{}C} + \gamma\]
, then we have to apply energy and momentum conservation and solve.
Energy Conservation
\[T_{\alpha} + m_{\alpha}c^{2} + m_{Be}c^{2} = T_{\gamma} + m_{C}c^{2} + T_{C}\]
Momentum Conservation
\[p_{\alpha} = p_{\gamma} + p_{C}\]
Solve
\[T_{\alpha} + \left[ m_{\alpha}c^{2} + m_{Be}c^{2} - m_{C}c^{2} = T_{\gamma} + T_{C}\right]\]
\[5.3\unit{MeV} + 931.5\unit{\nicefrac{MeV}{amu}} \left[9\unit{amu}+4\unit{amu}-13\unit{amu}\right] = T_{\gamma} + T_{C}\]
\[\frac{1}{c}\sqrt{\left(T_{\alpha}+2m_{\alpha}c^{2}\right)T_{\alpha}}=\frac{E_{\gamma}}{c}+\frac{1}{c}\sqrt{\left(T_{C}+2m_{C}c^{2}\right)T_{C}}\]
\[\sqrt{\left(T_{\alpha}+2m_{\alpha}c^{2}\right)T_{\alpha}}=E_{\gamma}+\sqrt{\left(T_{C}+2m_{C}c^{2}\right)T_{C}}\]
\[T_{\gamma}=15.55\unit{MeV}-T_{C}\]
\[\sqrt{\left(T_{\alpha}+2m_{\alpha}c^{2}\right)T_{\alpha}}=15.55\unit{MeV}-T_{C}+\sqrt{\left(T_{C}+2m_{C}c^{2}\right)T_{C}}\]
\[198.85\unit{MeV}-15.55\unit{MeV}=\sqrt{\left(T_{C}+2m_{C}c^{2}\right)T_{C}}-T_{C}\]
\[\left(T_{C}+183.3\unit{MeV}\right)^{2}=\left(T_{C}^{2}+T_{C}2m_{C}c^{2}\right)\]
\[\cancel{T_{c}^{2}}+\left(2\cdot183.3\unit{MeV}\right)T_{C}+\left(183.3\unit{MeV}\right)^{2}=\cancel{T_{C}^{2}}+T_{C}2m_{C}c^{2}\]
\[T_{C}=-\frac{\left(183.3\unit{MeV}\right)^{2}}{\left(2\cdot183.3\unit{MeV}\right)-2m_{C}c^{2}}\]
\[T_{C}=1.408\unit{MeV}\]
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