Chapter 39, Problem 39.61P

A pion at rest (m_π = 273m_e) decays to a muon (m_µ = 207m_e) and an antineutrino (m_p ≈ 0). The reaction is written π^– → µ^– + $\bar{v}$. Find (a) the kinetic energy of the muon and (b) the energy of the antineutrino in electron volts.

To determine

The kinetic energy of the muon.

Answer to Problem 39.61P

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

Explanation of Solution

Given info: The mass of pion at rest $m_{\text{π}}$ is $273m_{\text{e}}$, where $m_{\text{e}}$ is the mass of an electron, the mass of muon $m_{\text{μ}}$ is $207m_{\text{e}}$, the mass of antineutrino $m_{\overline{\text{v}}}$ is $0$ and the reaction is ${\pi }^{-}\to {\mu }^{-}+\overline{v}$.

Write the equation of kinetic energy of the muon.

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

Here,

$\gamma$ is a constant.

$m_{\text{μ}}$ is the mass of muon.

$c$ is the speed of light.

The value of $\gamma$ is $\frac{1}{\sqrt{1-u^2/c^2}}$.

Substitute $1/\sqrt{1-u^2/c^2}$ for $\gamma$ and $207m_{\text{e}}$ for $m_{\text{μ}}$ in above equation.

$K_{\text{μ}}=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)\left(207m_{\text{e}}\right)c^2$

The rest mass energy of an electron $m_{\text{e}}{c}^2$ is $0.511\text{MeV}$.

Substitute $0.511\text{MeV}$ for $m_{\text{e}}{c}^2$ in above equation to find $K_{\text{μ}}$.

$\begin{array}{l}{K}_{\text{μ}}=\left(\frac{1}{\sqrt{1-u^2/c^2}}-1\right)\left\{207\left(0.511\text{MeV}\right)\right\}\\ K_{\text{μ}}=\left(\frac{1}{\sqrt{1-{\left(u/c\right)}^2}}-1\right)\left(105.77\text{MeV}\right)\end{array}$ (1)

Write the equation of conservation of energy.

$m_{\text{π}}{c}^2=\gamma m_{\text{μ}}{c}^2+\left|p_{\overline{v}}\right|c$ (2)

Here,

$m_{\text{π}}$ is the mass of the pion.

$\left|p_{\overline{v}}\right|$ is the momentum.

Write the equation of conservation of momentum.

$p_{\overline{\text{v}}}=-p_{\text{μ}}$

Here,

$p_{\text{μ}}$ is the momentum of muon.

Write the equation of momentum of muon.

$p_{\text{μ}}=\gamma m_{\text{μ}}u$

Here,

$u$ is the speed of the muon.

Substitute $\gamma m_{\text{μ}}u$ for $p_{\text{μ}}$ in above equation to find $p_{\overline{\text{v}}}$.

$p_{\overline{\text{v}}}=-\gamma m_{\text{μ}}u$

Substitute $-\gamma m_{\text{μ}}u$ for $p_{\overline{\text{v}}}$ in equation (2) to find $m_{\text{π}}$.

$\begin{array}{c}{m}_{\text{π}}{c}^2=\gamma m_{\text{μ}}{c}^2+\left|-\gamma m_{\text{μ}}u\right|c\\ m_{\text{π}}=\gamma m_{\text{μ}}+\gamma m_{\text{μ}}u\left(1/c\right)\\ m_{\text{π}}=\gamma m_{\text{μ}}\left(1+u/c\right)\end{array}$

Substitute $273m_{\text{e}}$ for $m_{\text{π}}$ and $207m_{\text{e}}$ for $m_{\text{μ}}$ in above equation.

$\begin{array}{c}\left(273m_{\text{e}}\right)=\gamma \left(207m_{\text{e}}\right)\left(1+u/c\right)\\ \frac{273}{207}=\gamma \left(1+u/c\right)\end{array}$

Substitute $1/\sqrt{1-u^2/c^2}$ for $\gamma$ in above equation.

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

Substitute $0.270$ for $u/c$ in equation (1) to find $K_{\text{μ}}$.

$\begin{array}{c}{K}_{\text{μ}}=\left(\frac{1}{\sqrt{1-{\left(0.270\right)}^2}}-1\right)\left(105.77\text{MeV}\right)\\ =4.0797\text{MeV}\\ \approx \text{4}\text{.08}\text{MeV}\end{array}$

Conclusion:

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

To determine

The energy of the antineutrino in electron volts.

Answer to Problem 39.61P

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

Explanation of Solution

Given info: The mass of pion at rest $m_{\text{π}}$ is $273m_{\text{e}}$, where $m_{\text{e}}$ is the mass of an electron, the mass of muon $m_{\text{μ}}$ is $207m_{\text{e}}$, the mass of antineutrino $m_{\overline{\text{v}}}$ is $0$ and the reaction is ${\pi }^{-}\to {\mu }^{-}+\overline{v}$.

Write the equation of energy of the antineutrino.

$K_{\overline{\text{v}}}=m_{\text{π}}{c}^2-\gamma m_{\text{μ}}{c}^2$

Substitute $273m_{\text{e}}$ for $m_{\text{π}}$, $1/\sqrt{1-u^2/c^2}$ for $\gamma$ and $207m_{\text{e}}$ for $m_{\text{μ}}$ in above equation to find $K_{\overline{\text{v}}}$.

$K_{\overline{\text{v}}}=273m_{\text{e}}{c}^2-\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)\left(207m_{\text{e}}\right)c^2$

Substitute $0.511\text{MeV}$ for $m_{\text{e}}{c}^2$ and $0.270$ for $u/c$ in above equation to find $K_{\overline{\text{v}}}$.

$\begin{array}{c}{K}_{\overline{\text{v}}}=273\left(0.511\text{MeV}\right)-\left(\frac{1}{\sqrt{1-{\left(0.270\right)}^2}}\right)\left\{207\left(0.511\text{MeV}\right)\right\}\\ =29.6\text{MeV}\end{array}$

Conclusion:

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