Chapter 32, Problem 58GP

To determine

To show:

The threshold energy is $6m_{\text{p}}{c}^2,$ equal to three times the magnitude of the $Q$ -value of the reaction.

Explanation of Solution

Given data:

The given reaction is

  $p+p\to 3p+\overline{p}$

All final product have same speed

Calculation:

The mass of the products is

  $M=3m_p+m_p=4m_p$

The $Q$ value of the given reaction is the mass energy of the reactants minus the mass energy of the products.

  $\begin{array}{c}Q=\left(m_p{c}^2+m_p{c}^2\right)-M{c}^2\\ =2m_p{c}^2-4m_p{c}^2\\ =-2m_{\text{p}}{c}^2\end{array}$

Energy conservation gives the following,

  $\left({\text{KE}}_{\text{th}}+m_{\text{p}}{c}^2\right)+m_{\text{p}}{c}^2={\text{KE}}_M+M{c}^2$

  $K{E}_{bh}+2m_p{c}^2=K{E}_M+M{c}^2\mathrm{...}(1)$

From the conservation of momentum

  $\begin{array}{l}{p}_{\text{p}}=p_{\text{M}}\\ {\left(p_{\text{p}}c\right)}^2={\left(p_{\text{M}}c\right)}^2\end{array}$

Using the relativistic relationship $E^2=p^2{c}^2+m^2{c}^4$

  ${\left({\text{KE}}_{\text{th}}+m_{\text{p}}{c}^2\right)}^2-m_{\text{p}}^2{c}^4={\left({\text{KE}}_M+M{c}^2\right)}^2-M^2{c}^4$

Replace $\left({\text{KE}}_M+M{c}^2\right)$ term using equation (1)

  ${\text{KE}}_{\text{th}}^2+2{\text{KE}}_{\text{th}}{m}_{\text{p}}{c}^2+m_{\text{p}}^2{c}^4-m_{\text{p}}^2{c}^4={\text{KE}}_{\text{th}}^2+4{\text{KE}}_{\text{th}}{m}_{\text{p}}{c}^2+4m_{\text{p}}^2{c}^4-{\left(4m_{\text{p}}\right)}^2{c}^4$

  $\begin{array}{l}2{\text{KE}}_{\text{th}}{m}_{\text{p}}{c}^2=4{\text{KE}}_{\text{th}}{m}_{\text{p}}{c}^2+4m_{\text{p}}^2{c}^4-16m_{\text{p}}^2{c}^4\\ 2{\text{KE}}_{\text{th}}{m}_{\text{p}}{c}^2=12m_{\text{p}}^2{c}^4\end{array}$

  ${\text{KE}}_{\text{th}}=6m_{\text{p}}{c}^2=3\left|Q\right|$

Conclusion:

The threshold energy $6m_{\text{p}}{c}^2$ is equal to three times the magnitude of the $Q$ -value of the reaction. ${\text{KE}}_{\text{th}}=6m_{\text{p}}{c}^2=3\left|Q\right|$