Chapter 26, Problem 92P

To determine

The kinetic energies (in the lab frame) of the neutron and pion after decay.

Answer to Problem 92P

The kinetic energies (in the lab frame) of the neutron and pion after decay is $5.7\text{MeV}$ and kinetic energy pion after decay is $35.4\text{MeV}$

Explanation of Solution

Write the equation for the relativistic energy of a particle.

  ${\left(pc\right)}^2=K^2+2K{E}_0$        (I)

Here, $p$ is the momentum of the particle, $K$ is the kinetic energy of the particle, and  $E_0$ is the rest energy of the particle.

Write the expression for the relativistic energy of a neutron.

  ${\left(p_{n}c\right)}^2=K_n{}^2+2K_n{E}_{0n}$        (II)

Here, $p_n$ is the momentum of the neutron, $K_n$ is the kinetic energy of the neutron, and  $E_{0n}$ is the rest energy of the neutron.

Write the expression for the relativistic energy of a neutron.

  ${\left(p_{\pi }c\right)}^2=K_{\pi }{}^2+2K_{\pi }{E}_{0\pi }$        (III)

Here, $p_{\pi }$ is the momentum of the pion, $K_{\pi }$ is the kinetic energy of the pion, and  $E_{0\pi }$ is the rest energy of the pion.

Since lambda hyperon is at rest in the lab frame, by conservation of momentum $p_n=p_{\pi }$ .

Equate equation (II) and (III)

  $K_n{}^2+2K_n{E}_{0n}=K_{\pi }{}^2+2K_{\pi }{E}_{0\pi }$        (IV)

Thus, total energy of the system is conserved. Apply conservation of energy for the decay process.

$m_{\Lambda }{c}^2=m_n{c}^2+m_{\pi }{c}^2+K_n+K_{\pi }$        (V)

Here, $m_{\Lambda }$ is the mass of the lambda hyperon, $m_n$ is the mass of neutron, $m_{\pi }$ is the mass of the pion and $c$ is the speed of light in vacuum.                   

Rearrange above equation to get $K_n+K_{\pi }$

  $\begin{array}{r}\left(m_{\Lambda }-m_n-m_{\pi }\right)c^2=K_n+K_{\pi }\\ =K\end{array}$        (VI)

Here, $K$  is the total kinetic energy.

Rearrange the equation $K=K_n+K_{\pi }$ to get $K_n$

  $K_n=K-K_{\pi }$        (VII)

Substitute $\left(m_{\Lambda }-m_n-m_{\pi }\right)c^2$ for $K$ in expression (VII)                            $K_n=\left(m_{\Lambda }-m_n-m_{\pi }\right)c^2-K_{\pi }$        (VIII)

Substitute expression (VII) in expression (IV) and solve for $K_{\pi }$.

    $\begin{array}{c}{\left(K-K_{\pi }\right)}^2+2\left(K-K_{\pi }\right)E_{0n}=K_{\pi }{}^2+2K_{\pi }{E}_{0\pi }\\ K^2+2K{K}_{\pi }+K_{\pi }{}^2+2K{E}_{0n}=K_{\pi }{}^2+2K_{\pi }{E}_{0\pi }\\ 2K_{\pi }\left(K+E_{0n}+E_{0\pi }\right)=K\left(K+2E_{0n}\right)\\ K_{\pi }=\frac{K\left(K+2E_{0n}\right)}{2\left(K+E_{0n}+E_{0\pi }\right)}\end{array}$        (IX)

Substitute $\left(m_{\Lambda }-m_n-m_{\pi }\right)c^2$ for $K$, $m_n{c}^2$ for $E_{0n}$, and $m_{\pi }{c}^2$ for $E_{0\pi }$ in expression (IX),

  $K_{\pi }=\frac{\left(m_{\Lambda }-m_n-m_{\pi }\right)c^2\left(\left(m_{\Lambda }-m_n-m_{\pi }\right)c^2+2m_n{c}^2\right)}{2\left(\left(m_{\Lambda }-m_n-m_{\pi }\right)c^2+m_n{c}^2+m_{\pi }{c}^2\right)}$        (X)

Conclusion:

Substitute the $1115.7{\text{MeV/c}}^{\text{2}}$ for $m_{\Lambda }$,  $939.6{\text{MeV/c}}^{\text{2}}$ for $m_n$, $135.0{\text{MeV/c}}^{\text{2}}$ for $m_{\pi }$ in expression (x)

  $\begin{array}{c}{K}_{\pi }=\frac{\left(1115.7{\text{MeV/c}}^{\text{2}}-939.6{\text{MeV/c}}^{\text{2}}-135.0{\text{MeV/c}}^{\text{2}}\right)c^2\left(\begin{array}{l}\left(\begin{array}{l}1115.7{\text{MeV/c}}^{\text{2}}\\ -939.6{\text{MeV/c}}^{\text{2}}\\ -135.0{\text{MeV/c}}^{\text{2}}\end{array}\right)c^2\\ +2\left(939.6{\text{MeV/c}}^{\text{2}}\right)c^2\end{array}\right)}{2\left(\begin{array}{l}\left(\begin{array}{l}1115.7{\text{MeV/c}}^{\text{2}}\\ -939.6{\text{MeV/c}}^{\text{2}}\\ -135.0{\text{MeV/c}}^{\text{2}}\end{array}\right)c^2\\ +\left(939.6{\text{MeV/c}}^{\text{2}}\right)c^2+\left(135.0{\text{MeV/c}}^{\text{2}}\right)c^2\end{array}\right)}\\ =35.4\text{MeV}\end{array}$

Substitute the $1115.7{\text{MeV/c}}^{\text{2}}$ for $m_{\Lambda }$,  $939.6{\text{MeV/c}}^{\text{2}}$ for $m_n$, $135.0{\text{MeV/c}}^{\text{2}}$ for $m_{\pi }$ , and $35.4\text{MeV}$ for $K_{\pi }$ in expression (VII)

  $\begin{array}{c}{K}_n=\left(1115.7{\text{MeV/c}}^{\text{2}}-939.6{\text{MeV/c}}^{\text{2}}-135.0{\text{MeV/c}}^{\text{2}}\right)c^2-35.4\text{MeV}\\ =5.7\text{MeV}\end{array}$

Therefore the kinetic energies (in the lab frame) of the neutron and pion after decay is $5.7\text{MeV}$ and kinetic energy pion after decay is $35.4\text{MeV}$