Chapter 46, Problem 68CP

(a)

To determine

Show that the threshold kinetic energy is $K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$.

Answer to Problem 68CP

The threshold kinetic energy is $K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$.

Explanation of Solution

Write the equation showing the conservation of energy.

    $E_{\min }+m_2{c}^2=\sqrt{{\left(m_3{c}^2\right)}^2+{\left(p_{3}c\right)}^2}$                                                                         (I)

Here, $E_{\min }$ is the minimum energy required for the bombarding particle to induce the reaction, $m_2$ is the mass of the stationary particle, $m_3$ is the mass of the product, $p_3$ is the momentum of the product and $c$ is the speed of light.

Write the relativistic energy equation and substitute $p_1$ for $p_3$ and $m_1$ for in equation (I) because $p_1=p_3$ due to conservation of momentum.

    $\begin{array}{c}{\left(p_{3}c\right)}^2={\left(p_{1}c\right)}^2\\ =E_{\min }^2-\left(m_1{c}^2\right)\end{array}$                                                                                    (II)

Substitute equation (II) in (I).

    $E_{\min }+m_2{c}^2=\sqrt{{\left(m_3{c}^2\right)}^2+E_{\min }^2-\left(m_1{c}^2\right)}$

Conclusion:

Take square on both sides.

    $\begin{array}{c}{E}_{\min }^2+2E_{\min }{m}_2{c}^2+{\left(m_2{c}^2\right)}^2={\left(m_3{c}^2\right)}^2+E_{\min }^2-\left(m_1{c}^2\right)\\ E_{\min }=\frac{\left(m_3^2-m_1^2-m_2^2\right)c^2}{2m_2}\end{array}$                                   (III)

Write the equation for minimum kinetic energy.

    $K_{\min }=E_{\min }-m_1{c}^2$

Substitute equation (III) in above equation to find $K_{\min }$.

    $\begin{array}{c}{K}_{\min }=\frac{\left(m_3^2-m_1^2-m_2^2\right)c^2}{2m_2}-m_1{c}^2\\ =\frac{\left(m_3^2-m_1^2-m_2^2\right)c^2-2m_1{m}_2{c}^2}{2m_2}\\ =\frac{\left(m_3^2-m_1^2-m_2^2-2m_1{m}_2\right)c^2}{2m_2}\\ =\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}\end{array}$

Thus, the threshold kinetic energy is $K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$.

(b)

To determine

The threshold energy for the reaction $p+p=p+p+p+\overline{p}$.

Answer to Problem 68CP

The threshold energy for the reaction $p+p=p+p+p+\overline{p}$ is $5.63\text{GeV}$.

Explanation of Solution

Write the equation for the threshold energy.

    $K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$

Substitute $m_p+m_p+m_p+m_{\overline{p}}$ for $m_3$, $m_p$ for $m_1$ and $m_p$ for $m_2$ in the above equation.

    $K_{\min }=\frac{\left[{\left(m_p+m_p+m_p+m_{\overline{p}}\right)}^2-{\left(m_p+m_p\right)}^2\right]c^2}{2m_p}$

Conclusion:

Substitute $938.3{\text{MeV/c}}^2$ for $m_p$ and $938.3{\text{MeV/c}}^2$ for $m_{\overline{p}}$ to find $K_{\min }$.

    $\begin{array}{c}{K}_{\min }=\frac{\left[\begin{array}{l}{\left(938.3{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2\right)}^2\\ -{\left(938.3{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2\right)}^2\end{array}\right]c^2}{2\left(938.3{\text{MeV/c}}^2\right)}\\ =\frac{\left[4{\left(938.3{\text{MeV/c}}^2\right)}^2-2{\left(938.3{\text{MeV/c}}^2\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^2\right)}\\ =\left(5630\text{MeV}\right)\left(\frac{{10}^{-3}\text{GeV}}{1\text{MeV}}\right)\\ =5.63\text{GeV}\end{array}$

Thus, the threshold energy for the reaction $p+p=p+p+p+\overline{p}$ is $5.63\text{GeV}$.

(c)

To determine

The threshold energy for the reaction ${\pi }^{-}+p={\text{K}}^0+{\Lambda }^0$.

Answer to Problem 68CP

The threshold energy for the reaction ${\pi }^{-}+p={\text{K}}^0+{\Lambda }^0$ is $768\text{MeV}$.

Explanation of Solution

Write the equation for the threshold energy.

    $K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$

Substitute $m_{{\text{K}}^0}+m_{{\Lambda }^0}$ for $m_3$, $m_{{\pi }^{-}}$ for $m_1$ and $m_p$ for $m_2$ in the above equation.

    $K_{\min }=\frac{\left[{\left(m_{{\text{K}}^0}+m_{{\Lambda }^0}\right)}^2-{\left(m_{{\pi }^{-}}+m_p\right)}^2\right]c^2}{2m_p}$

Conclusion:

Substitute $497.7{\text{MeV/c}}^2$ for $m_{{\text{K}}^0}$, $1115.6{\text{MeV/c}}^2$ for $m_{{\Lambda }^0}$, $139.6{\text{MeV/c}}^2$ for $m_{{\pi }^{-}}$ and $938.3{\text{MeV/c}}^2$ for $m_{\overline{p}}$ to find $K_{\min }$.

    $\begin{array}{c}{K}_{\min }=\frac{\left[{\left(497.7{\text{MeV/c}}^2+1115.6{\text{MeV/c}}^2\right)}^2-{\left(139.6{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^2\right)}\\ =768\text{MeV}\end{array}$

Thus, the threshold energy for the reaction ${\pi }^{-}+p={\text{K}}^0+{\Lambda }^0$ is $768\text{MeV}$.

(d)

To determine

The threshold energy for the reaction $p+p=p+p+{\pi }^0$.

Answer to Problem 68CP

The threshold energy for the reaction $p+p=p+p+{\pi }^0$ is $280\text{MeV}$.

Explanation of Solution

Write the equation for the threshold energy.

    $K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$

Substitute $m_p+m_p+m_{{\pi }^0}$ for $m_3$, $m_p$ for $m_1$ and $m_p$ for $m_2$ in the above equation.

    $K_{\min }=\frac{\left[{\left(m_p+m_p+m_{{\pi }^0}\right)}^2-{\left(m_p+m_p\right)}^2\right]c^2}{2m_p}$

Conclusion:

Substitute $135{\text{MeV/c}}^2$ for $m_{{\pi }^0}$ and $938.3{\text{MeV/c}}^2$ for $m_{\overline{p}}$ to find $K_{\min }$.

    $\begin{array}{c}{K}_{\min }=\frac{\left[\begin{array}{l}{\left(938.3{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2+135{\text{MeV/c}}^2\right)}^2\\ -{\left(938.3{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2\right)}^2\end{array}\right]c^2}{2\left(938.3{\text{MeV/c}}^2\right)}\\ =\frac{\left[{\left(2\left(938.3{\text{MeV/c}}^2\right)+135{\text{MeV/c}}^2\right)}^2-2{\left(938.3{\text{MeV/c}}^2\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^2\right)}\\ =280\text{MeV}\end{array}$

Thus, the threshold energy for the reaction $p+p=p+p+{\pi }^0$ is $280\text{MeV}$.

(e)

To determine

The threshold energy for the reaction $p+\overline{p}={\text{Z}}^0$.

Answer to Problem 68CP

The threshold energy for the reaction $p+\overline{p}={\text{Z}}^0$ is $4.43\text{TeV}$.

Explanation of Solution

Write the equation for the threshold energy.

    $K_{\min }=\frac{\left[m_3^2-{\left(m_1+m_2\right)}^2\right]c^2}{2m_2}$

Substitute $m_{{\text{Z}}^0}$ for $m_3$, $m_p$ for $m_1$ and $m_{\overline{p}}$ for $m_2$ in the above equation.

    $K_{\min }=\frac{\left[{\left(m_{{\text{Z}}^0}\right)}^2-{\left(m_p+m_{\overline{p}}\right)}^2\right]c^2}{2m_{\overline{p}}}$

Conclusion:

Substitute $91.2\times {10}^3{\text{MeV/c}}^2$ for $m_{{\text{Z}}^0}$ and $938.3{\text{MeV/c}}^2$ for $m_{\overline{p}}$ and $m_{\overline{p}}$ to find $K_{\min }$.

    $\begin{array}{c}{K}_{\min }=\frac{\left[{\left(91.2\times {10}^3{\text{MeV/c}}^2\right)}^2-{\left(938.3{\text{MeV/c}}^2+938.3{\text{MeV/c}}^2\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^2\right)}\\ =\frac{\left[{\left(91.2\times {10}^3{\text{MeV/c}}^2\right)}^2-2{\left(938.3{\text{MeV/c}}^2\right)}^2\right]c^2}{2\left(938.3{\text{MeV/c}}^2\right)}\\ =\left(4.43\times {10}^6\text{MeV}\right)\left(\frac{{10}^{-6}\text{TeV}}{1\text{MeV}}\right)\\ =4.43\text{TeV}\end{array}$

Thus, the threshold energy for the reaction $p+\overline{p}={\text{Z}}^0$ is $4.43\text{TeV}$.