Chapter 41, Problem 39P

(a)

To determine

The unknown particle in the reaction.

Explanation of Solution

Given: The reaction is $\text{p}+\text{p}\to {\Lambda }^0+{{\rm K}}^0+\text{p}+(?)$ .

Introduction:

The reaction process for given particle is balanced if charge, baryon number, lepton numbers and spin are conserved.

The total charge before and after the reaction must be same; it means, the charge must be conserved in the reaction.

Conserve the total charge before and after the reaction.

  $\begin{array}{c}+1+1=0+0+1+Q\\ +2=+1+Q\\ Q=+1\end{array}$

Here, $Q$ is the charge number of unknown particles.

The Baryon number before and after the reaction must be same; it means, the Baryon number must be conserved in the reaction.

Conserve the Baryon number before and after the reaction.

  $\begin{array}{c}+1+1=+1+0+1+B\\ +2=+2+B\\ B=0\end{array}$

Here, $B$ is the baryon number of unknown particles.

The strangeness before and after the reaction must be same; it means, the strangeness must be conserved in the reaction.

Conserve the strangeness before and after the reaction.

  $\begin{array}{c}0+0=-1+1+0+S\\ S=0\end{array}$

Here, $S$ is the strangeness of unknown particles.

The spin before and after the reaction must be same; it means, the spin must be conserved in the reaction.

Conserve the spin before and after the reaction.

  $\begin{array}{c}+\frac{1}{2}+\frac{1}{2}=+\frac{1}{2}+0+\frac{1}{2}+s\\ s=0\end{array}$

Here, $s$ is the strangeness of unknown particles.

All the above properties illustrate that the unknown particle is pion particle.

Conclusion:

Thus, the unknown particle in the reaction is $\text{pion}\left({\pi }^{+}\right)$ particle.

(b)

To determine

The $Q$ value for the reaction.

Explanation of Solution

Given:

The reaction is $\text{p}+\text{p}\to {\Lambda }^0+{{\rm K}}^0+\text{p}+{\pi }^{+}$ .

Formula used:

Write the expression for change in mass.

  $\Delta m=\left(m_{\text{p}}+m_{\text{p}}\right)-\left(m_{{\Lambda }^0}+m_{{{\rm K}}^0}+m_{\text{p}}+m_{{\pi }^{+}}\right)$

Here, $m_{{\Lambda }^0}$ is mass of baryon, $m_{\text{p}}$ is mass of proton, $m_{{\pi }^{+}}$ is mass of pion, $m_{{{\rm K}}^0}$ is mass of meson and $\Delta m$ is change in mass.

Write the expression for $Q$ value.

  $Q=\Delta m{c}^2$

Substitute $\left(m_{\text{p}}+m_{\text{p}}\right)-\left(m_{{\Lambda }^0}+m_{{{\rm K}}^0}+m_{\text{p}}+m_{{\pi }^{+}}\right)$ for $\Delta m$ in above expression.

  $Q=\left[\left(m_{\text{p}}+m_{\text{p}}\right)-\left(m_{{\Lambda }^0}+m_{{{\rm K}}^0}+m_{\text{p}}+m_{{\pi }^{+}}\right)\right]c^2$ …… (1)

Here, $Q$ is the energy released and $c$ is the speed of light.

Calculation:

Substitute $1116\text{MeV}/c^2$ for $m_{{\Lambda }^0}$ , $938.3\text{MeV}/c^2$ for $m_{\text{p}}$ , $497.7\text{MeV}/c^2$ for $m_{{{\rm K}}^0}$ and $139.6\text{MeV}/c^2$ for $m_{{\pi }^{+}}$ in equation (1).

  $\begin{array}{c}Q=\left[\left(\begin{array}{l}938.3\text{MeV}/c^2\\ +938.3\text{MeV}/c^2\end{array}\right)-\left(\begin{array}{l}1116\text{MeV}/c^2+497.7\text{MeV}/c^2\\ +938.3\text{MeV}/c^2+139.6\text{MeV}/c^2\end{array}\right)\right]c^2\\ =-815\text{MeV}\end{array}$

Conclusion:

Thus, the $Q$ value for the reaction is $-815\text{MeV}$ .

(c)

To determine

The thresholdenergy for this reaction.

Explanation of Solution

Given:

The reaction is $\text{p}+\text{p}\to {\Lambda }^0+{{\rm K}}^0+\text{p}+{\pi }^{+}$ .

Formula used:

Write the expression for change in mass.

  $\Delta m=\left(m_{\text{p}}+m_{\text{p}}\right)-\left(m_{{\Lambda }^0}+m_{{{\rm K}}^0}+m_{\text{p}}+m_{{\pi }^{+}}\right)$

Here, $m_{{\Lambda }^0}$ is mass of baryon, $m_{\text{p}}$ is mass of proton, $m_{{\pi }^{+}}$ is mass of pion, $m_{{{\rm K}}^0}$ is mass of meson and $\Delta m$ is change in mass.

Write the expression for $Q$ value.

  $Q=\Delta m{c}^2$

Substitute $\left(m_{\text{p}}+m_{\text{p}}\right)-\left(m_{{\Lambda }^0}+m_{{{\rm K}}^0}+m_{\text{p}}+m_{{\pi }^{+}}\right)$ for $\Delta m$ in above expression.

  $Q=\left[\left(m_{\text{p}}+m_{\text{p}}\right)-\left(m_{{\Lambda }^0}+m_{{{\rm K}}^0}+m_{\text{p}}+m_{{\pi }^{+}}\right)\right]c^2$

Here, $Q$ is the energy released and $c$ is the speed of light.

Write the expression for threshold energy.

  $K_{\text{th}}=-\frac{Q}{2m_{\text{p}}}\left(m_{\text{p}}+m_{\text{p}}+m_{{\Lambda }^0}+m_{{{\rm K}}^0}+m_{\text{p}}+m_{{\pi }^{+}}\right)$ …… (2)

Here, $K_{\text{th}}$ is the threshold energy for the reaction.

Calculation:

Substitute $1116\text{MeV}/c^2$ for $m_{{\Lambda }^0}$ , $938.3\text{MeV}/c^2$ for $m_{\text{p}}$ , $497.7\text{MeV}/c^2$ for $m_{{{\rm K}}^0}$ , $139.6\text{MeV}/c^2$ for $m_{{\pi }^{+}}$ and $-815\text{MeV}$ for $Q$ in equation (2).

  $\begin{array}{c}{K}_{\text{th}}=\frac{-815\text{MeV}}{2\left(938.3\text{MeV}/c^2\right)}\left[\begin{array}{l}938.3\text{MeV}/c^2+938.3\text{MeV}/c^2+1116\text{MeV}/c^2+\\ 497.7\text{MeV}/c^2+938.3\text{MeV}/c^2+139.6\text{MeV}/c^2\end{array}\right]c^2\\ =1984\text{MeV}\left(\frac{1\text{GeV}}{1000\text{MeV}}\right)\\ =1.984\text{GeV}\end{array}$

Conclusion:

Thus, the threshold energy for this reaction is $1.984\text{GeV}$ .