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

Explanation of Solution

Write the expression for the relativistic energy of a particle.

  ${\left(pc\right)}^2=K^2+2K{E}_0$

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.

  ${\left(p_{\text{n}}c\right)}^2=K_{\text{n}}^2+2K_{\text{n}}{E}_{\text{0n}}$                                                                                         (I)

Here, $p_{\text{n}}$ is the momentum of neutron, $K_{\text{n}}$ is the kinetic energy of neutron and $E_0{}_{\text{n}}$ is the rest energy of neutron.

Write the expression for the relativistic energy of a pion.

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

Here, $p_{\pi }$ is the momentum of pion, $K_{\pi }$ is the kinetic energy of pion and $E_0{}_{\pi }$ is the rest energy of pion.

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

Equate equation (I) and (II).

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

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

  $m_{\Lambda }{c}^2=m_{\text{n}}{c}^2+m_{\pi }{c}^2+K_{\text{n}}+K_{\pi }$

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

Rearrange above equation to get $K_{\text{n}}+K_{\pi }$.

  $\begin{array}{c}\left(m_{\Lambda }-m_{\text{n}}-m_{\pi }\right)c^2=K_{\text{n}}+K_{\pi }\\ =K\end{array}$                                                                   (IV)

Here, $K$ is the total kinetic energy.

Rearrange the equation $K=K_{\text{n}}+K_{\pi }$ to get $K_{\text{n}}$.

  $K_{\text{n}}=K-K_{\pi }$

Substitute $\left(m_{\Lambda }-m_{\text{n}}-m_{\pi }\right)c^2$ for $K$ in above equation to get $K_{\text{n}}$.

  $K_{\text{n}}=\left(m_{\Lambda }-m_{\text{n}}-m_{\pi }\right)c^2-K_{\pi }$                                                                             (V)

Substitute $K-K_{\pi }$ for $K_{\text{n}}$ in equation (III) to get $K_{\pi }$.

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

Solve above equation to get $K_{\pi }$.

  $K_{\pi }=\frac{K\left(K+2E_{\text{0n}}\right)}{2\left(K+E_{\text{0n}}+E_{\text{0}\pi }\right)}$

Substitute $\left(m_{\Lambda }-m_{\text{n}}-m_{\pi }\right)c^2$ for $K$ , $m_{\text{n}}{c}^2$ for $E_{\text{0n}}$ and $m_{\pi }{c}^2$ for $E_{0\pi }$ in above equation to get expression of $K_{\pi }$ in terms of masses of particles.

  $K_{\pi }=\frac{\left(m_{\Lambda }-m_{\text{n}}-m_{\pi }\right)c^2\left(\left(m_{\Lambda }-m_{\text{n}}-m_{\pi }\right)c^2+2m_{\text{n}}{c}^2\right)}{2\left(\left(m_{\Lambda }-m_{\text{n}}-m_{\pi }\right)c^2+m_{\text{n}}{c}^2+m_{\pi }{c}^2\right)}$                                     (VI)

Conclusion:

Substitute $1115.7\text{MeV}/c^2$ for $m_{\Lambda }$ , $939.6\text{MeV}/c^2$ for $m_{\text{n}}$, $135.0\text{MeV}/c^2$ for $m_{\pi }$ in above equation to get $K_{\pi }$.

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

Substitute $1115.7\text{MeV}/c^2$ for $m_{\Lambda }$ , $939.6\text{MeV}/c^2$ for $m_{\text{n}}$, $135.0\text{MeV}/c^2$ for $m_{\pi }$ and $35.4\text{MeV}$ for $K_{\pi }$ in equation (V) to get $K_{\text{n}}$.

  $\begin{array}{c}{K}_{\text{n}}=\left(1115.7\text{MeV}/c^2-939.6\text{MeV}/c^2-135.0\text{MeV}/c^2\right)c^2-35.4\text{MeV}\\ =\text{5}\text{.7}\text{MeV}\end{array}$

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