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 2 of 3: Reduced mass from the spectrum
1λ=109700 cm−1×μme(122−142) ⇒ μ=185.9 me=1.693×10−28 kg\frac{1}{\lambda} = 109700\ \mathrm{cm}^{-1}\times\frac{\mu}{m_e}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\ \Rightarrow\ \mu = 185.9\,m_e = 1.693\times10^{-28}\ \text{kg}
Analysis

The transition from the third excited state (n=4n = 4) to the first excited state (n=2n = 2) obeys 1/λ=RH(μ/me)(1/4−1/16)1/\lambda = R_H (\mu/m_e)(1/4 - 1/16). With 1/λ=3.824×1061/\lambda = 3.824\times10^{6} cm−1^{-1}, μ/me=185.9\mu/m_e = 185.9, giving μ=1.693×10−28\mu = 1.693\times10^{-28} kg.

Common pitfall. “Third excited state” means n=4n = 4, not n=3n = 3: the ground state is n=1n = 1 and the first excited state is n=2n = 2.