Chapter 31, Problem 32P

To determine

To show:The energy nucleus and neutron is 3.5 MeV and 14MeV.Andthese value dependent of temperature.

Explanation of Solution

Given data:

. As a result, the momenta of the two products are equal in magnitude. The available energy of $17.57\text{MeV}$ is much smaller than the masses involved, so we use

Formula used:

The nonrelativistic relationship between momentum and kinetic energy,

  $\begin{array}{c}\text{KE}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}\\ \mathrm{p}=\sqrt{2\mathrm{m}\text{KE}}\end{array}$

Calculation:

Total energy of nucleus and reaction $17.57\text{MeV}$

  ${\text{KE}}_{\text{H}\text{e}}+{\text{KE}}_{\text{n}}={\text{KE}}_{\text{total }}=17.57\text{MeV}$

We assume that the reactants are at rest when they react, so the total momentum of the system is 0 $\begin{array}{c}{\mathrm{p}}_{\text{H}\text{e}}={\mathrm{p}}_{\text{n}}\\ \sqrt{2{\mathrm{m}}_{\text{H}\text{e}}{\text{KE}}_{\text{H}\text{e}}}=\sqrt{2{\mathrm{m}}_{\text{n}}{\text{KE}}_{\text{n}}}\\ {\mathrm{m}}_{\text{H}\text{e}}{\text{KE}}_{\text{H}\text{e}}={\mathrm{m}}_{\text{n}}{\text{KE}}_{\text{n}}\\ {\mathrm{m}}_{{}_2^4\text{He}}{\text{KE}}_{\text{H}\text{e}}={\mathrm{m}}_{\text{n}}\left({\text{KE}}_{\text{total }}-{\text{KE}}_{\text{H}\text{e}}\right)\end{array}$

The kinetic energy of $\text{H}\text{e}$ nucleus

  $\begin{array}{c}{\text{KE}}_{\text{H}\text{e}}=\frac{{\mathrm{m}}_{\text{n}}}{{\mathrm{m}}_{{}_2^4\text{He}}+{\mathrm{m}}_{\text{n}}}{\text{KE}}_{\text{total }}\\ =\left(\frac{1.008665}{4.002603+1.008665}\right)17.57\text{MeV}\\ =3.536\text{MeV}\\ \approx 3.5\text{MeV}\end{array}$

The kinetic energy ofneutron

  $\begin{array}{c}{\text{KE}}_{\text{n}}={\text{KE}}_{\text{total }}-{\text{KE}}_{\text{H}\text{e}}\\ =17.57\text{MeV}-3.54\text{MeV}\\ =14.03\text{MeV}\\ \approx 14\text{MeV}\end{array}$

A higher plasma temperature would result in higher values for the energies.

Conclusion:

The kinetic energy of $\text{H}\text{e}$ nucleus is $3.5\text{MeV}$

The kinetic energy of neutron is $14\text{MeV}$

A higher plasma temperature would result in higher values for the energies.