The primary decay mode for the negative pion \({\pi ^{\rm{ - }}} \to {{\rm{\mu }}^{\rm{ - }}}{\rm{ + }}{{\rm{\bar \upsilon }}_{\rm{\mu }}}\).

(a) What is the energy release in \({\rm{MeV}}\) in this decay?

(b) Using conservation of momentum, how much energy does each of the decay products receive, given the \({\pi ^{\rm{ - }}}\) is at rest when it decays? You may assume the muon antineutrino is massless and has momentum \(p = \frac{{{E_\nu }}}{c}\), just like a photon.

The relation between mass and energy is given by the energy expression,

\(E = \Delta m{c^2}\)

Here\(E\)is the energy of the physical system,\(m\)is the mass of the system and\(c\)is the speed of the light in vacuum.

(a)

As the negative pion\({\pi ^{\rm{ - }}}\)was at rest before the decay, energy released is calculated as the change in rest mass –

\(\Delta m = {m_{{\pi ^ - }}} - {m_{{\mu ^ - }}} - {m_{{{\bar \nu }_\mu }}}\)…… (i)

Inputting the known rest masses

\(\begin{aligned}{}\Delta E &= {c^2}\Delta m\\ &= {c^2}\left( {{\rm{139}}{\rm{.6 MeV/}}{{\rm{c}}^{\rm{2}}}} \right) - {c^2}\left( {{\rm{105}}{\rm{.7 MeV/}}{{\rm{c}}^{\rm{2}}}} \right) - {c^2}\left( {{\rm{0 MeV/}}{{\rm{c}}^{\rm{2}}}} \right)\\ &= {\rm{33}}{\rm{.9 Me}}{{\rm{V}}^{\rm{2}}}\end{aligned}\)

Therefore, the value for energy released is obtained as \(\Delta E = 33.9{\rm{ Me}}{{\rm{V}}^{\rm{2}}}\).

(b)

Since energy and momentum are conserved in the process, use the momentum-energy relation for the muon –

\(E_\mu ^2 = {m^2}{c^4} + {p^2}{c^2}\) …… (ii)

And it's massless reduction for the neutrino –

\(p = \frac{{{E_\nu }}}{c}\) …… (iii)

To solve for energy and momentum, the momentum is conserved, so input equation (ii) values into the equation (i)

\(E_\mu ^2 = m_\mu ^2{c^4} + E_\nu ^2\) …… (iv)

Since energy is conserved, it can be written as –

\({E_\mu } + {E_\nu } = {m_\pi }{c^2}\)

Where \({m_\pi }{c^2}\) is the rest mass energy of the pion. Plugging (iii) into (iv)

\(\begin{aligned}{c}\sqrt {m_\mu ^2{c^4} + E_\nu ^2} + {E_\nu } &= {m_\pi }{c^2}\\\sqrt {m_\mu ^2{c^4} + E_\nu ^2} &= {m_\pi }{c^2} - {E_\nu }\\m_\mu ^2{c^4} + E_\nu ^2 &= m_\pi ^2{c^4} - 2{m_\pi }{c^2}{E_\nu } + E_\nu ^2\\2{m_\pi }{c^2}{E_\nu } &= m_\pi ^2{c^4} - m_\mu ^2{c^4}\\{E_\nu } &= \frac{{m_\pi ^2{c^4} - m_\mu ^2{c^4}}}{{2{m_\pi }{c^2}}}\\ &= \frac{{m_\pi ^2{c^2} - m_\mu ^2{c^2}}}{{2{m_\pi }}}\end{aligned}\)

Inputting the known masses

\(\begin{aligned}{}{E_\nu } &= \frac{{{{\left( {{\rm{139}}{\rm{.6 MeV/}}{{\rm{c}}^{\rm{2}}}} \right)}^{\rm{2}}}{{\rm{c}}^{\rm{2}}}{\rm{ - }}{{\left( {{\rm{105}}{\rm{.7 MeV/}}{{\rm{c}}^{\rm{2}}}} \right)}^{\rm{2}}}{{\rm{c}}^{\rm{2}}}}}{{{\rm{2}}\left( {{\rm{139}}{\rm{.6 MeV/}}{{\rm{c}}^{\rm{2}}}} \right)}}\\ &= 29.8{\rm{ MeV}}\end{aligned}\)

To obtain the energy given to the muon use (4) and further subtract the muon rest mass energy –

\(\begin{aligned}{}\Delta {E_\mu } &= {m_\pi }{c^2} - {E_\nu } - {m_\mu }{c^2}\\ &= \left( {{\rm{139}}{\rm{.6 MeV/}}{{\rm{c}}^{\rm{2}}}} \right){{\rm{c}}^{\rm{2}}}{\rm{ - (29}}{\rm{.8 MeV) - }}\left( {{\rm{105}}{\rm{.7 MeV/}}{{\rm{c}}^{\rm{2}}}} \right){{\rm{c}}^{\rm{2}}}\\ &= {\rm{4}}{\rm{.1 MeV}}\end{aligned}\)

Therefore, the value for energy received by each decay product is obtained as\({E_\nu } = 29.8{\rm{ MeV}}\) and\(\Delta {E_\mu } = 4.1{\rm{ MeV}}\).