Question

A charged pion ( ${\mathit{m}}_{\mathbf{p}}{\mathit{c}}^{\mathbf{2}}$= 139.6 MeV) at rest decays into a muon (${\mathit{m}}_{\mathbf{\mu }}$= 105.7 MeV) and a neutrino (whose mass is very nearly zero). Find the kinetic energy of the muon and the kinetic energy of the neutrino, in MeV (1 MeV = $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$eV).

Short answer

The kinetic energy of neutrino is 59.57 MeV and the kinetic energy of the muon is 4.12 MeV.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The energy of charged pion is, $m_{\mathrm{\pi }}{c}^2=139.6\mathrm{MeV}$.
• The mass of a muon is, $m_{\mu }=105.7\mathrm{MeV}$.
• The value of 1 MeV is, $1\mathrm{MeV}=1\times {10}^6\mathrm{eV}$.

Concept/Significance of the nuclear decay

The nuclear decay discovered Two of the natural four fundamental forces. The unstable nucleus releases surplus energy for its nucleon configuration by generating radiation.

Determination of the kinetic energy of the muon and the kinetic energy of the neutrino

From energy conservation relation the energy of the pion is given by,

$m_{\mathrm{\pi }}{c}^2=\sqrt{c^2{p}^2+m_{\mu }^2}+pc$

Here, pcis the energy of neutrino and $m_{\mu }$is the mass of muon and c is the speed of light.

Substitute values in the above,

$$\begin{gathered} pc=\frac{(m_{\mathrm{\pi }}^2-m_{\mu }^2)c^2}{2m_{\mathrm{\pi }}} \\ =\left[\frac{{(139.6\mathrm{MeV}/{\mathrm{c}}^2)}^2-(105.7\mathrm{MeV}/{\mathrm{c}}^2)}{139.6\mathrm{meV}/{\mathrm{c}}^2}\right]{\mathrm{c}}^2 \\ =59.57\mathrm{MeV} \end{gathered}$$

Thus, the kinetic energy of neutrino is 59.57 MeV.

The kinetic energy of the muon is given by,

$$\begin{gathered} T_{\mu }=m_{\mathrm{\pi }}{c}^2-\frac{(m_{\mathrm{\pi }}^2-m_{\mu }^2)c^2}{2m_{\mu }}-m_{\mu }{c}^2 \\ =\frac{(m_{\mathrm{\pi }}^2-m_{\mu }^2)c^2}{2m_{\mathrm{\pi }}} \end{gathered}$$

Substitute all the values in the above,

$$\begin{gathered} T_{\mu }=\frac{\left(139.6{\displaystyle \frac{\mathrm{MeV}}{{\mathrm{c}}^2}}-105.7\frac{\mathrm{MeV}}{{\mathrm{c}}^2}\right){\mathrm{c}}^2}{2\left(139.6\frac{\mathrm{MeV}}{{\mathrm{c}}^2}\right)} \\ =4.12\mathrm{MeV} \end{gathered}$$

Thus, the kinetic energy of the muon is 4.12 MeV.