Chapter 32, Problem 52GP

To determine

To Calculate: The mass of ${\pi }^0$ .

Answer to Problem 52GP

The mass of the ${\pi }^0$ is 133.5 MeV.

Explanation of Solution

Given:

  ${\pi }^{-}+p\to {\pi }^0+n$

Initial kinetic energy is zero and the emitted neutron energy is 0.60Mev.

Formula used:

The mass of the ${\pi }^0$ can be calculated using the formula:

  $m_{{\pi }^0}{c}^2=\sqrt{E_{{\pi }^0}{}^2-E_{{}_n}^2+m_n^2{c}^4}$

Where, $E_{{\pi }^0}$ is the kinetic energy for ${\pi }^0$ , $E_{{}_n}^2$ is kinetic energy of neutron and $m_n$ is the mass of the neutron and c is the light velocity.

Calculation:

Initially the interaction must happen at rest which means that the initial kinetic energy and momentum is zero. The momentum of neutron and that of ${\pi }^0$ have the same magnitude. According to the energy conservation law, total energy of ${\pi }^0$ is

  $\begin{array}{l}{M}_{{\pi }^0}{c}^2+m_p{c}^2=E_{{\pi }^0}+m_n{c}^2+K{E}_n\hfill \\ E_{{\pi }^0}=M_{{\pi }^0}{c}^2+m_p{c}^2-m_n{c}^2-K{E}_n\hfill \\ \text{ =139}\text{.6}\text{MeV+938}\text{.3}\text{MeV-939}\text{.6}\text{MeV-0}\text{.60MeV}\hfill \\ \text{ = 137}\text{.7}\text{MeV}\hfill \end{array}$

The mass energy of ${\pi }^0$ on the bases of momentum conservation law is given as,

  $\begin{array}{l}{P}_n=P_{{\pi }^0}\\ \Rightarrow {(p_{n}c)}^2={(p_{{\pi }^0}c)}^2\\ \Rightarrow E_{{}_n}^2-m_n^2{c}^4=E_{{\pi }^0}{}^2-m_{{\pi }^0}^2{c}^4\\ \Rightarrow m_{{\pi }^0}^2{c}^4=-E_{{}_n}^2+m_n^2{c}^4+E_{{\pi }^0}{}^2\\ \Rightarrow m_{{\pi }^0}{c}^2=\sqrt{E_{{\pi }^0}{}^2-E_{{}_n}^2+m_n^2{c}^4}\\ =\sqrt{{\text{(137}\text{.7}\text{MeV)}}^2\text{-(939}\text{.6}\text{MeV+0}\text{.60}{\text{MeV)}}^2\text{+(939}\text{.6}{\text{MeV)}}^{\text{2}}}\\ \text{=133}\text{.5}\text{MeV}\\ m_{{\pi }^0}=133.5{\text{MeV/c}}^{\text{2}}\end{array}$

Conclusion:

The mass is 133.5 MeV.