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Found in: Page 1333

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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.

1. The kinetic energy of alpha particle is 3.541 MeV.
2. The kinetic energy of neutron is 14.05 MeV.
See the step by step solution

## Step 1: Describe the expression for kinetic energy of alpha particle

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,

${\mathbf{0}}{\mathbf{=}}{{\mathbf{p}}}_{{\mathbf{\alpha }}}{\mathbf{+}}{{\mathbf{p}}}_{{\mathbf{n}}}\phantom{\rule{0ex}{0ex}}{{\mathbf{p}}}_{{\mathbf{\alpha }}}^{{\mathbf{2}}}{\mathbf{=}}{{\mathbf{p}}}_{{\mathbf{n}}}^{{\mathbf{2}}}$

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 ${{\mathbit{K}}}_{{\mathbit{\alpha }}}{\mathbit{=}}\frac{{\mathbit{p}}_{\mathbit{\alpha }}^{\mathbit{2}}}{\mathbit{2}{\mathbit{m}}_{\mathbit{\alpha }}}{\mathbit{,}}{\mathbit{a}}{\mathbit{n}}{\mathbit{d}}{\mathbit{}}{{\mathbit{K}}}_{{\mathbit{n}}}\frac{{\mathbit{p}}_{\mathbit{n}}^{\mathbit{2}}}{\mathbit{2}{\mathbit{m}}_{\mathbit{n}}}$.

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

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

From equation (1),

${\mathbit{Q}}{\mathbf{=}}{{\mathbit{K}}}_{{\mathbf{\alpha }}}\frac{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbit{K}}}_{{\mathbf{\alpha }}}\phantom{\rule{0ex}{0ex}}{\mathbit{Q}}{\mathbf{=}}{{\mathbit{K}}}_{{\mathbf{\alpha }}}\left(1+\frac{{m}_{\alpha }}{{m}_{\alpha }}\right)\phantom{\rule{0ex}{0ex}}{{\mathbit{K}}}_{{\mathbf{\alpha }}}{\mathbf{=}}\frac{\mathbf{Q}}{\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{{\mathbf{m}}_{\mathbf{\alpha }}}{{\mathbf{m}}_{\mathbf{\alpha }}}\mathbf{\right)}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{2}}{\mathbf{\right)}}$

## Step 2(a): Find the kinetic energy of alpha particle

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).

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

## Step 3(b): Find the kinetic energy of neutron

Substitute all the known values in equation (1).

${\mathrm{K}}_{\mathrm{n}}=\mathrm{Q}-{\mathrm{K}}_{\mathrm{\alpha }}\phantom{\rule{0ex}{0ex}}=17.59\mathrm{MeV}-3.541\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=14.05\mathrm{MeV}$

Therefore, the kinetic energy of neutron is 14.05 MeV.