Chapter 15, Problem 14P

(a)

To determine

The threshold for production of strange particles in the reaction $\text{p}+\text{p}\to \text{n}+{\Sigma }^{+}+{\text{K}}^0+{\pi }^{+}$.

Answer to Problem 14P

The threshold for production of strange particles in the reaction $\text{p}+\text{p}\to \text{n}+{\Sigma }^{+}+{\text{K}}^0+{\pi }^{+}$ is $\underset{\_}{2.201\text{GeV}}$.

Explanation of Solution

Consider the reaction,

    $\text{p}+\text{p}\to \text{n}+{\Sigma }^{+}+{\text{K}}^0+{\pi }^{+}$        (I)

Write the expression for the threshold for production of strange particles in the reaction.

    $K_{th}=\frac{{\left(m_n+m_{{\Sigma }^{+}}+m_{K^0}+m_{{\pi }^{+}}\right)}^2{c}^2-4m_p^2{c}^2}{2m_p}$        (II)

Here, $K_{th}$ is the threshold energy.

Conclusion:

Substitute $939.6\text{MeV}/{\text{c}}^2$ for $m_n$, $1189.4\text{MeV}/{\text{c}}^2$ for $m_{{\Sigma }^{+}}$, $497.7\text{MeV}/{\text{c}}^2$ for $m_{K^0}$, $139.6\text{MeV}/c^2$ for $m_{{\pi }^{+}}$, $938.3\text{MeV}/c^2$ for $m_p$ in equation (II) to find $K_{th}$.

    $\begin{array}{c}{K}_{th}=\frac{{\left(939.6+1189.4+497.7+139.6\right)}^2{\left(\text{MeV}\right)}^2{c}^2-\left(4\right){\left(938.3\text{ MeV}\right)}^2{c}^2}{\left(2\right)\left(938.3\text{ MeV}\right)}\\ =2.201\text{ GeV}\end{array}$

Therefore, the threshold for production of strange particles in the reaction $\text{p}+\text{p}\to \text{n}+{\Sigma }^{+}+{\text{K}}^0+{\pi }^{+}$ is $\underset{\_}{2.201\text{GeV}}$.

(b)

To determine

The threshold for production of strange particles in the reaction ${\pi }^{-}+\text{p}\to {\Sigma }^0+{\text{K}}^0$.

Answer to Problem 14P

The threshold for production of strange particles in the reaction ${\pi }^{-}+\text{p}\to {\Sigma }^0+{\text{K}}^0$ is $\underset{\_}{903.2\text{MeV}}$.

Explanation of Solution

Consider the reaction,

    ${\pi }^{-}+\text{p}\to {\Sigma }^0+{\text{K}}^0$        (III)

Write the expression for the threshold for production of strange particles in the reaction ${\pi }^{-}+\text{p}\to {\Sigma }^0+{\text{K}}^0$.

    $K_{th}=\frac{{\left(m_{{\Sigma }^0}+m_{K^0}\right)}^2{c}^2-{\left(m_{{\pi }^{-}}+m_p\right)}^2{c}^2}{2m_p}$        (IV)

Conclusion:

Substitute $1192.5\text{MeV}/{\text{c}}^2$ for $m_{{\Sigma }^0}$, $497.7\text{MeV}/{\text{c}}^2$ for $m_{K^0}$, $139.6\text{MeV}/c^2$ for $m_{{\pi }^{-}}$, $938.3\text{MeV}/c^2$ for $m_p$ in equation (IV) to find $K_{th}$.

    $\begin{array}{c}{K}_{th}=\frac{{\left(1192.5+497.7\right)}^2{\left(\text{MeV}\right)}^2{c}^2-{\left(139.6+938.3\right)}^2{\left(\text{MeV}\right)}^2{c}^2}{\left(2\right)\left(938.3\text{ MeV}\right)}\\ =\frac{\left(2856776{\left(\text{MeV}\right)}^2/c^4\right)c^2-\left(1\text{161}\text{868 }{\left(\text{MeV}\right)}^2/c^4\right)c^2}{1876\text{MeV}/c^2}\\ =903.2\text{ MeV}\end{array}$

Therefore, the threshold for production of strange particles in the reaction ${\pi }^{-}+\text{p}\to {\Sigma }^0+{\text{K}}^0$ is $\underset{\_}{903.2\text{MeV}}$.