In the deuteron–triton fusion reaction of Eq. 43-15, what is the kinetic energy of (a) the alpha particle and (b) the neutron? Neglect the relatively small kinetic energies of the two combining particles.

Assume that the initial velocities is negligible, from the conservation of the energy,

$\mathbf{Q}\mathbf{=}{\mathbf{K}}_{\mathbf{\alpha }}\mathbf{+}{\mathbf{K}}_{\mathbf{n}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, Q is total energy,${\mathbf{K}}_{\mathbf{\alpha }}$ is kinetic energy of alpha particle, and${\mathbf{K}}_{\mathbf{n}}$ is kinetic energy of neutron.

From the conservation of the momentum,

$$\begin{gathered} \mathbf{0}\mathbf{=}{\mathbf{p}}_{\mathbf{\alpha }}\mathbf{+}{\mathbf{p}}_{\mathbf{n}} \\ {\mathbf{p}}_{\mathbf{\alpha }}^{\mathbf{2}}\mathbf{=}{\mathbf{p}}_{\mathbf{n}}^{\mathbf{2}} \end{gathered}$$

Divide both sides of the above equation by $\mathbf{2}{\mathbf{m}}_{\mathbf{n}}$.

role="math" localid="1661754581079" $\frac{{\mathbf{p}}_{\mathbf{\alpha }}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{n}}}=\frac{{\mathbf{p}}_{\mathbf{}\mathbf{n}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{n}}}$

Simplify further.

$\frac{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbf{m}}_{\mathbf{\alpha }}}\frac{{\mathbf{p}}_{\mathbf{\alpha }}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{n}}}=\frac{{\mathbf{p}}_{\mathbf{n}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{n}}}$

It is known that ${\mathit{K}}_{\mathit{\alpha }}\mathit{=}\frac{{\mathit{p}}_{\mathit{\alpha }}^{\mathit{2}}}{\mathit{2}{\mathit{m}}_{\mathit{\alpha }}}\mathit{,}\mathit{a}\mathit{n}\mathit{d}\mathit{}{\mathit{K}}_{\mathit{n}}\frac{{\mathit{p}}_{\mathit{n}}^{\mathit{2}}}{\mathit{2}{\mathit{m}}_{\mathit{n}}}$.

From the equation,$\frac{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbf{m}}_{\mathbf{\alpha }}}\frac{{\mathbf{p}}_{\mathbf{\alpha }}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{n}}}\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{n}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{n}}}$

$\frac{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbf{m}}_{\mathbf{\alpha }}}{\mathit{K}}_{\mathbf{\alpha }}\mathbf{=}{\mathit{K}}_{\mathbf{n}}$

From equation (1),

$$\begin{gathered} \mathit{Q}\mathbf{=}{\mathit{K}}_{\mathbf{\alpha }}\frac{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbf{m}}_{\mathbf{\alpha }}}{\mathit{K}}_{\mathbf{\alpha }} \\ \mathit{Q}\mathbf{=}{\mathit{K}}_{\mathbf{\alpha }}\left(1+\frac{m_{\alpha }}{m_{\alpha }}\right) \\ {\mathit{K}}_{\mathbf{\alpha }}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{(}\mathbf{1}\mathbf{+}\frac{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbf{m}}_{\mathbf{\alpha }}}\mathbf{)}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

The mass of the alpha particle is$m_{\alpha }=4.0015u$ and the mass of the neutron is $m_n=1.008665u$, where $\mathrm{Q}=17.59\mathrm{MeV}$.

Substitute all the known values in equation (2).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{\alpha }}=\frac{\left(17.59\mathrm{MeV}\right)}{\left(1+{\displaystyle \frac{4.0015\mathrm{u}}{1.008665\mathrm{u}}}\right)} \\ =3.541\mathrm{MeV} \end{gathered}$$

Therefore, the kinetic energy of alpha particle is 3.541MeV.

Substitute all the known values in equation (1).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{n}}=\mathrm{Q}-{\mathrm{K}}_{\mathrm{\alpha }} \\ =17.59\mathrm{MeV}-3.541\mathrm{MeV} \\ =14.05\mathrm{MeV} \end{gathered}$$

Therefore, the kinetic energy of neutron is 14.05 MeV.