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Chapter 30: Problem 14

A muon decay is described by $\mu^{-} \rightarrow \mathrm{e}^{-}+v_{\mu}+\bar{v}_{c} .$ What is the maximum kinetic energy of the electron, if the muon was at rest? Assume that the electron is extremely relativistic and ignore the small masses of the neutrinos.

Short Answer

Expert verified
Answer: The maximum kinetic energy of the electron produced in the muon decay is approximately 105.189 MeV.

Step by step solution

01

Identify the decay process and initial conditions

The decay process is given by \(\mu^{-} \rightarrow \mathrm{e}^{-} + v_{\mu} + \bar{v}_{c}\). We are told to assume that the muon is initially at rest.
02

Apply conservation of energy

The total energy before the decay is equal to the muon's rest energy, which is given by \(E_0 = m_{\mu} c^2\). After the decay, the total energy is shared among the electron, muon neutrino, and electron antineutrino. Since we are asked to find the maximum kinetic energy of the electron, we will assume that the neutrinos have negligible kinetic energy. Therefore, the total energy after the decay is given by the electron's kinetic energy \(K_e\) plus its rest energy, or \(E_{total} = K_e + m_e c^2\). By conservation of energy, we have: \(m_{\mu} c^2 = K_e + m_e c^2\)
03

Solve for the maximum kinetic energy of the electron

We will now solve for \(K_e\): \(K_e = m_{\mu} c^2 - m_e c^2\) The masses of the muon and electron are \(m_{\mu} = 105.7 \mathrm{MeV}/c^2\) and \(m_e = 0.511 \mathrm{MeV}/c^2\), respectively. Plugging in these values, we get: \(K_e = (105.7 \, \mathrm{MeV} - 0.511 \, \mathrm{MeV}) c^2 = 105.189 \, \mathrm{MeV} \, c^2\)
04

Present the answer

The maximum kinetic energy of the electron in the muon decay is \(\approx 105.189 \, \mathrm{MeV}\).

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