Chemistry Labs

Problem 25

Muon. The muon (μ\mu) is a subatomic particle of the lepton family with the same charge and magnetic behaviour as the electron, but a different mass; it is unstable and disintegrates within microseconds. Determine the mass of the muon by two different approaches. (a) The most common spontaneous disintegration is μX−→eX−+νˉXe+νXμ\ce{\mu- -> e- + \bar{\nu}_e + \nu_{\mu}}. In an experiment with a stationary muon, the two neutrinos carried away a total energy of 2.000×10−122.000\times10^{-12} J while the electron had a kinetic energy of 1.4846×10−111.4846\times10^{-11} J. Determine the mass of the muon. (me=9.109×10−31m_e = 9.109\times10^{-31} kg, c=2.998×108c = 2.998\times10^{8} m s−1^{-1}.) (b) Exotic atoms in which a muon replaces the electron are formed in various excited states. For an atom consisting of a X1X221H\ce{^1H} nucleus with an attached muon, the transition from the third excited state to the first excited state is observed at a wavelength of 2.615 nm. Determine the mass of the muon using the Bohr model with reduced-mass correction. (En=−109700 cm−1×hc×μ/me×1/n2E_n = -109700\ \mathrm{cm}^{-1}\times hc\times\mu/m_e \times 1/n^2 with μ=mμmH/(mμ+mH)\mu = m_\mu m_H/(m_\mu + m_H); mH=1.673×10−27m_H = 1.673\times10^{-27} kg.)
Step 3 of 3: Solve for the muon mass
mμ=μ mHmH−μ=1.693×10−28×1.673×10−271.673×10−27−1.693×10−28=1.884×10−28 kgm_\mu = \frac{\mu\, m_H}{m_H - \mu} = \frac{1.693\times10^{-28}\times1.673\times10^{-27}}{1.673\times10^{-27} - 1.693\times10^{-28}} = 1.884\times10^{-28}\ \text{kg}
Analysis

Inverting the reduced-mass relation μ=mμmH/(mμ+mH)\mu = m_\mu m_H/(m_\mu + m_H) gives mμ=μmH/(mH−μ)m_\mu = \mu m_H/(m_H - \mu); inserting μ=1.693×10−28\mu = 1.693\times10^{-28} kg and mH=1.673×10−27m_H = 1.673\times10^{-27} kg yields mμ=1.884×10−28m_\mu = 1.884\times10^{-28} kg — in excellent agreement with the decay-energy value.