A 1-MeV proton is incident on \(^7\)Li at rest and causes the following reaction.
\begin{equation*}
^{7}\textrm{Li} + \ ^{1}\textrm{H} \longrightarrow\ ^{4}\textrm{He} + \ ^{4}\textrm{He}.
\end{equation*}
(a) Compute the \(Q\) of the reaction. (b) Assuming the two product particles move along the same line as the incident proton, find the kinetic energies of the product particles. The data needed: \(M_{Li7}\) = 7.016004 u, \(M_{He4}\) = 4.002 603 u, \(M_{H1}\) = 1.007825 u.
Hint.
(b) Assume nonrelativistic speeds.
Answer.
(a) \(17.34\:\textrm{MeV}\text{,}\) (b) \(10.7\:\textrm{ MeV}\text{,}\) \(7.64\:\textrm{ MeV}\text{.}\)
Solution.
(a) We compute \(Q\) from the rest energies of the particles in the reaction.
\begin{equation*}
Q = \left( M_{Li7} + M_{H1} - 2\times M_{He4} \right)\: c^2.
\end{equation*}
This gives
\begin{equation*}
Q = (0.018623\:\textrm{u}) c^2 \times 931.494\:\frac{\textrm{MeV}}{c^2} = 17.34\:\textrm{MeV}.
\end{equation*}
(b) Let \(K_{1}\) and \(K_{2}\) be the kinetic energies of the two product particles. Let us denote the kinetic energy of the incident proton by \(K_p\text{.}\) We assume that speeds of the particles are not in the relativistic domain so that
\begin{equation*}
p_1 = \sqrt{2M_{He4} K_1},\ \ \ p_2 = \sqrt{2M_{He4} K_2},
\end{equation*}
The conservation of energy gives
\begin{equation}
Q + K_p = K_1 + K_2. \tag{54.29}
\end{equation}
Let us label the helium particle with 1 that is in the same direction as the proton. The other one labeled 2 will be moving in the opposite direction. The conservation of momentum gives
\begin{equation}
\sqrt{2M_{H1} K_p} =\sqrt{2M_{He4} K_1} - \sqrt{2M_{He4} K_2}. \tag{54.30}
\end{equation}
Let \(x = \sqrt{K_1}\) and \(y = \sqrt{K_2}\text{.}\) Simplifying and putting the numerical values we get
\begin{equation*}
x^2 + y^2 = 18.34,\ \ x - y = 0.5.
\end{equation*}
Eliminating \(y\) from the first of these equations we get
\begin{equation*}
x^2 +(x-0.5)^2 = 18.35.
\end{equation*}
Solving for \(x\) we get
\begin{equation*}
x = 3.27\:\sqrt{\textrm{MeV}}.
\end{equation*}
Squaring we get
\begin{equation*}
K_1 = 10.7\:\textrm{ MeV}.
\end{equation*}
Use this \(K_1\) to get
\begin{equation*}
K_2 = 7.64\:\textrm{ MeV}.
\end{equation*}
Let us verify if the non-relativistic assumption was correct.
\begin{equation*}
v_1 = \sqrt{ 2 K_1/M_{He4}} = c\sqrt{ 2 K_1/M_{He4}c^2} = c\sqrt{2\times 10.7\:\textrm{ MeV}/3728.4\:\textrm{MeV}} = 0.076\:c.
\end{equation*}
This speed is only 7.6% of the speed of light, which can be safely treated as non-relativistic.
