Answer
See the detailed answer below.
Work Step by Step
We can use the nonrelativistic formulas of momentum and energy since we assume that the initial atoms were at rest.
This means that the resultant components must have a net momentum of zero.
Thus,
$$KE=\dfrac{1}{2}mv^2=\dfrac{1}{2}(mv) v\times \dfrac{m}{m}$$
$$KE=\dfrac{p^2}{2m}\tag 1$$
and hence,
$$p=\sqrt{2mKE}\tag 2$$
We know, from equation 31-8c, that the total energy released from this reaction is about 17.59 MeV.
Thus,
$$KE_{\rm ^4_2He}+KE_n+17.59=0$$
So that;
$$KE_{\rm ^4_2He}+KE_n=17.59 \tag 3$$
We ignored the negative sign here since we are dealing with magnitudes of energies.
The net momentum of the two of ${\rm ^4_2He} $ and $n$ must be zero which means that
$$p_n=p_{\rm ^4_2He} $$
Using (2);
$$\sqrt{2m_nKE_n}=\sqrt{2m_{\rm ^4_2He}KE_{\rm ^4_2He}} $$
Squaring both sides;
$$ \color{red}{\bf\not}2m_nKE_n = \color{red}{\bf\not}2m_{\rm ^4_2He}KE_{\rm ^4_2He} $$
$$ m_nKE_n = m_{\rm ^4_2He}KE_{\rm ^4_2He} $$
Plugging $KE_n$ from (3);
$$ m_n[17.59 -KE_{\rm ^4_2He}]= m_{\rm ^4_2He}KE_{\rm ^4_2He} $$
$$ 17.59 m_n -m_nKE_{\rm ^4_2He}= m_{\rm ^4_2He}KE_{\rm ^4_2He} $$
Solving for $KE_{\rm ^4_2He}$;
$$ 17.59 m_n = m_{\rm ^4_2He}KE_{\rm ^4_2He} +m_nKE_{\rm ^4_2He} $$
$$KE_{\rm ^4_2He}[ m_{\rm ^4_2He}+m_n]= 17.59 m_n$$
$$KE_{\rm ^4_2He}= \dfrac{17.59 m_n}{m_{\rm ^4_2He}+m_n}$$
Plugging the known;
$$KE_{\rm ^4_2He}= \dfrac{17.59 (1.008665)}{4.002603+1.008665}=\color{red}{\bf 3.54}\;\rm MeV$$
And hence,
$$KE_{n}= 17.59-3.54=\color{red}{\bf 14}\;\rm MeV$$
These results are not fixed values since increasing the temperature of the plasma increases the speed of the nucleus and neutrons.
