Chapter 30, Problem 25P

To determine

Find the kinetic energy of muon.

Answer to Problem 25P

The kinetic energy of muon is $\text{4}\text{.2 MeV}$ .

Explanation of Solution

The given decay reaction is shown below,

${\pi }^{-}\to {\mu }^{-}+{\overline{\nu }}_{\mu }$

Write the equation for conservation of momentum.

$p_{\mu }=p_{\overline{\nu }}$ (I)

Here, $p_{\mu }$ is the momentum of muon and $p_{\overline{\nu }}$ is the momentum of antineutrino.

The kinetic energy of muon is given below,

$\begin{array}{l}K=\frac{p_{\mu }^2}{2m}\\ p_{\mu }=\sqrt{2mK}\end{array}$ (II)

Here, $K$ is the kinetic energy of muon and $m$ is the mass of muon.

The energy of antineutrino is given below,

$\begin{array}{l}{E}_{\overline{\nu }}=p_{\overline{\nu }}c\\ p_{\overline{\nu }}=\frac{E_{\overline{\nu }}}{c}\end{array}$ (III)

Here, $E_{\overline{\nu }}$ is the energy of antineutrino and $c$ is speed of light.

Substitute equation II and III in equation I.

$\begin{array}{c}\sqrt{2mK}=\frac{E_{\overline{\nu }}}{c}\\ K=\frac{E_{\overline{\nu }}^2}{2m{c}^2}=\frac{E_{\overline{\nu }}^2}{2E_{0\mu }^2}\end{array}$ (IV)

Here, $E_{0\mu }$ is the rest energy of muon.

Write the equation for energy of antineutron using conservation energy.

$E_{0\pi }=E_{0\mu }+K+E_{\overline{\nu }}$ (V)

Here, $E_{0\pi }$ is rest energy of pion.

Substitute equation IV in V.

$0=\frac{1}{2E_{0\mu }}{E}_{\overline{\nu }}^2+E_{\overline{\nu }}-\left(E_{0\pi }-E_{0\mu }\right)$

Solve the above quadratic equation for $E_{\overline{\nu }}$.

$E_{\overline{\nu }}=\frac{-1\pm \sqrt{1^2+4\left(\frac{1}{2E_{0\mu }}\right)\left(E_{0\pi }-E_{0\mu }\right)}}{E_{0\mu }{}^{-1}}$ (VI)

Conclusion:

Substitute $0.106\text{ GeV}$ for $E_{0\mu }$ and $0.140\text{ GeV}$  for $E_{0\pi }$ in equation VI

$\begin{array}{c}{E}_{\overline{\nu }}=\frac{-1\pm \sqrt{1+2\left(\frac{0.140\text{ GeV}}{0.106\text{ GeV}}-1\right)}}{{\left(0.106\text{ GeV}\right)}^{-1}}\\ =0.106\text{ GeV}\left(-1\pm \sqrt{1+2\left(\frac{0.140}{0.106}-1\right)}\right)\end{array}$

Solution of the above equation will give the following two values.

$0.030\text{ GeV}$ and $-0.242\text{ GeV}$

In the above two values $-0.242\text{ GeV}$ is not valid. Thus substitute $0.030\text{ GeV}$ for $E_{\overline{\nu }}$ and $0.106\text{ GeV}$ for $E_{0\mu }$ in equation IV.

$\begin{array}{c}K=\frac{{\left(0.030\text{ GeV}\right)}^2}{2\left(0.106\text{ GeV}\right)}\\ =0.0042\text{ GeV}\\ =4.2\text{ MeV}\end{array}$

Therefore, the kinetic energy of muon is $\text{4}\text{.2 MeV}$