Chapter 39, Problem 61P

(a)

To determine

The kinetic energy of the muon.

Answer to Problem 61P

The kinetic energy of the muon is $\underset{\_}{4.08\text{MeV}}$.

Explanation of Solution

A pion at rest decays to a muon and an antineutrino. The reaction for the decay is shown below.

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

Here, ${\pi }^{-}$ is the pion, ${\mu }^{-}$ is the muon, and $\overline{\nu }$ is the antineutrino.

The decay reaction obeys both energy conservation law and momentum conservation law. That is energy and momentum of the particles before and after the decay remains the same.

Write the expression for the energy of pion at rest.

    $E_p=m_{\pi }{c}^2$                                                                                                                (I)

Here, $E_p$ is the energy of pion at rest, $m_{\pi }$ is the mass of pion, and $c$ is the speed of light.

After decay of pion, muon and antineutrino are formed.

Write the expression for the relativistic energy of muon.

    $E_{\mu }=\gamma m_{\mu }{c}^2$                                                                                                             (II)

Here, $E_{\mu }$ is the energy of muon, $\gamma$ is the Lorentz factor, and $m_{\mu }$ is the mass of muon.

Write the expression of the energy of antineutrino.

    $E_{\overline{\nu }}=\left|p_{\overline{\nu }}\right|c$                                                                                                               (III)

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

Pion is at rest. So its momentum is zero. Thus the final momentum also should be zero. Thus sum of momentum of muon and antineutrino is equal to zero.

    $\begin{array}{c}{p}_{{\mu }^{-}}+p_{\overline{\nu }}=0\\ p_{\overline{\nu }}=-p_{{\mu }^{-}}\end{array}$                                                                                                     (IV)

Here, $p_{{\mu }^{-}}$ is the momentum of muon.

Write the expression for the momentum of muon.

    $p_{\mu }=\gamma m_{\mu }u$                                                                                                               (V)

Here, $u$ is the speed of muon.

Use expression (V) in (IV).

    $p_{\overline{\nu }}=-\gamma m_{\mu }u$                                                                                                           (VI)

Use expression (VI) in (III) to find $E_n$.

    $E_n=\gamma m_{\mu }uc$                                                                                                          (VII)

By conservation of energy, energy of pion is equal to the sum of energy of muon and antineutrino.

    $E_p=E_m+E_n$                                                                                                        (VIII)

Use expressions (VII), (II), and (I) in expression (VIII).

    $m_{\pi }{c}^2=\gamma m_{\mu }{c}^2+\gamma m_{\mu }cu$                                                                                          (IX)

Solve expression (IX) for $m_{\pi }$.

    $m_{\pi }=m_{\mu }\left(\gamma +\frac{\gamma u}{c}\right)$                                                                                                   (X)

Write the expression for Lorentz factor.

    $\gamma =\frac{1}{\sqrt{1-{\left(\frac{u}{c}\right)}^2}}$                                                                                                        (XI)

Use expression (XI) in (X).

    $\begin{array}{c}{m}_{\pi }=m_{\mu }\left(\frac{1}{\sqrt{1-{\left(u/c\right)}^2}}+\frac{\frac{u}{\sqrt{1-{\left(u/c\right)}^2}}}{c}\right)\\ \frac{m_{\pi }}{m_{\mu }}=\left(\frac{1}{\sqrt{1-{\left(u/c\right)}^2}}+\frac{u}{c\sqrt{1-{\left(u/c\right)}^2}}\right)\\ =\frac{c\sqrt{1-{\left(u/c\right)}^2}+u\sqrt{1-{\left(u/c\right)}^2}}{\left[1-{\left(u/c\right)}^2\right]c}\end{array}$                                                               (XII)

Reduce expression (XII).

    $\begin{array}{c}\frac{m_{\pi }}{m_{\mu }}=\frac{\sqrt{1-{\left(u/c\right)}^2}\left[c+u\right]}{c\left(1-{\left(u/c\right)}^2\right)}\\ =\frac{c\left(1+\frac{u}{c}\right)}{c\sqrt{1-{\left(\frac{u}{c}\right)}^2}}\\ =\frac{1+\frac{u}{c}}{\sqrt{1-{\left(\frac{u}{c}\right)}^2}}\end{array}$                                                                                   (XIII)

Solve the right hand side of expression (XIII).

    $\begin{array}{c}\frac{1+\frac{u}{c}}{\sqrt{1-{\left(\frac{u}{c}\right)}^2}}=\frac{\sqrt{1+\frac{u}{c}}\times \sqrt{1+\frac{u}{c}}}{\sqrt{\left(1-\left(\frac{u}{c}\right)\right)\left(1+\left(\frac{u}{c}\right)\right)}}\\ =\sqrt{\frac{1+\frac{u}{c}\left(1+\frac{u}{c}\right)}{\left(1-\left(\frac{u}{c}\right)\right)\left(1+\left(\frac{u}{c}\right)\right)}}\\ =\sqrt{\frac{1+\frac{u}{c}}{\left(1-\left(\frac{u}{c}\right)\right)}}\end{array}$                                                                       (XIV)

Use expression (XIV) in (XIII).

    $\frac{m_{\pi }}{m_{\mu }}=\sqrt{\frac{1+\frac{u}{c}}{\left(1-\left(\frac{u}{c}\right)\right)}}$                                                                                               (XV)

Write the expression for kinetic energy of muon.

    $K_{\mu }=\left(\gamma -1\right)m_{\mu }{c}^2$                                                                                                     (XVI)

Here, $K_{\mu }$ is the kinetic energy of muon.

Conclusion:

Substitute $273m_e$ for $m_{\pi }$, and $207m_e$ for $m_{\mu }$ in equation (XV) and solve for $\frac{u}{c}$.

    $\begin{array}{c}\frac{273m_e}{207m_e}=\sqrt{\frac{1+\frac{u}{c}}{\left(1-\left(\frac{u}{c}\right)\right)}}\\ \frac{{\left(273\right)}^2}{{\left(207\right)}^2}=\frac{1+\frac{u}{c}}{\left(1-\left(\frac{u}{c}\right)\right)}\\ \frac{u}{c}=\frac{{\left(273\right)}^2-{\left(207\right)}^2}{{\left(273\right)}^2+{\left(207\right)}^2}\\ =0.270\end{array}$

Substitute $0.270$ for $\gamma$ in expression (XI) to find $\gamma$.

    $\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\left(0.270\right)}^2}}\\ =1.0385\end{array}$

Substitute $207m_e$ for $m_{\mu }$, and $1.0385$ for $\gamma$ in equation (XVI) to find $K_{\mu }$.

    $\begin{array}{c}{K}_{\mu }=\left(1.0385-1\right)\left(207m_e\right)c^2\\ =0.0385\left(207\times 0.511\text{MeV}\right)\\ =4.08\text{MeV}\end{array}$

Therefore, the kinetic energy of the muon is $\underset{\_}{4.08\text{MeV}}$.

(b)

To determine

The energy of the antineutrino in electron volts.

Answer to Problem 61P

The energy of the antineutrino is $\underset{\_}{29.6\text{MeV}}$.

Explanation of Solution

Write the expression for the kinetic energy of antineutrino.

    $K_{\overline{\nu }}=E_p-\left(E_m+K_{\mu }\right)$                                                                                              (XVII)

Here, $K_{\overline{\nu }}$ is the kinetic energy of antineutrino.

Use expression (I), and (II) in expression (XVII) to find $K_{\overline{\nu }}$.

    $K_{\overline{\nu }}=m_{\pi }{c}^2-\left(\gamma m_{\mu }{c}^2+K_{\mu }\right)$                                                                                   (XVIII)

Substitute $207m_e$ for $m_{\mu }$, and $273m_e$ for $m_{\pi }$, and $4.08\text{MeV}$ for $K_{\mu }$ in equation (XVIII) to find $K_{\overline{\nu }}$.

    $K_{\overline{\nu }}=273m_e{c}^2-\left(207m_e{c}^2+K_{\mu }\right)$                                                                            (XIX)

Conclusion:

Substitute $0.511\text{MeV}$ for $m_e{c}^2$, and $4.08\text{MeV}$ for $K_{\mu }$ in expression (XIX) to find $K_{\overline{\nu }}$.

    $\begin{array}{c}{K}_{\overline{\nu }}=\left(273\times 0.511\text{MeV}\right)-\left(207\times 0.511\text{MeV}+4.08\text{MeV}\right)\\ =29.6\text{MeV}\end{array}$

Therefore, the energy of the antineutrino is $\underset{\_}{29.6\text{MeV}}$.