91Ó°ÊÓ

We'll start by applying the conservation of energy. In this decay process, the total energy must be conserved. This means that the initial energy of the pion at rest is equal to the final combined energy of the muon and neutrino: \(E_{initial} = E_{final}\) Since the pion is initially at rest, its energy is just its rest mass energy, given by \(E_{\pi} = m_{\pi}c^2\). The energies of the muon and neutrino after the decay are denoted as \(E_{\mu}\) and \(E_{\nu}\), respectively. So, we can write the energy conservation equation as: \(m_{\pi}c^2 = E_{\mu} + E_{\nu}\)

Since the pion is initially at rest, its momentum is zero. After the decay, the total momentum of the decay products, muon and neutrino, must also be zero to conserve momentum: \(p_{\pi} = p_{\mu} + p_{\nu} = 0\)

We can use the relativistic energy-momentum relation to express the energies and momenta of the decay products in terms of their masses and velocities. The relation for each particle is given by: \(E^2 = (mc^2)^2 + (pc)^2\) Applying this relation for the muon and neutrino, we get: \(E_{\mu}^2 = (m_{\mu}c^2)^2 + (p_{\mu}c)^2\) \(E_{\nu}^2 = (p_{\nu}c)^2\) Note that we didn't include a mass term for the neutrino, since the mass of the neutrino \((v)\) is negligible.

We want to find the total kinetic energy of the decay products. This is given by the sum of the kinetic energies of the muon and neutrino: \(K_{total} = K_{\mu} + K_{\nu}\) Remember that the kinetic energy of a particle is given by \(K = E - mc^2\). Therefore, we can write: \(K_{total} = (E_{\mu} - m_{\mu}c^2) + (E_{\nu})\) We can substitute the expressions for \(E_{\mu}\) and \(E_{\nu}\) from the energy conservation equation: \(K_{total} = (m_{\pi}c^2 - E_{\nu}) - m_{\mu}c^2 + E_{\nu} = m_{\pi}c^2 - m_{\mu}c^2\) Now, we can plug in the given values for the masses of the pion and muon: \(K_{total} = (139.6 \mathrm{MeV} - 105.7 \mathrm{MeV})\) Calculating the difference, we get: \(K_{total} = 33.9 \text{ MeV}\) Therefore, the total kinetic energy of the decay products (muon and neutrino) is \(\boxed{33.9\text{ MeV}}\).