The charged kaon has two leptonic decay channels

PHYSICS · Multi-part · Fermi University Physics · Particle Physics & Quantum Fields · Electroweak theory and the Standard Model · Flavour, neutrinos and CP violation

Symmetry and Flavour, Ch Quark mixing: the Cabibbo angle, the GIM mechanism and the CKM matrix

Problem

The charged kaon has two leptonic decay channels, K+→e+νeK^+ \to e^+\nu_e and K+→μ+νμK^+ \to \mu^+\nu_\mu, both mediated by the same WW-exchange diagram with the same ∣Vus∣|V_{us}|. Data: B(K+→e+νe)=1.582×10−5\mathcal{B}(K^+ \to e^+\nu_e) = 1.582\times10^{-5}, B(K+→μ+νμ)=63.56%\mathcal{B}(K^+ \to \mu^+\nu_\mu) = 63.56\%, τK+=1.238×10−8 s\tau_{K^+} = 1.238\times10^{-8}\,\mathrm{s}, mK=493.68 MeVm_K = 493.68\,\mathrm{MeV}, mμ=105.66 MeVm_\mu = 105.66\,\mathrm{MeV}, me=0.511 MeVm_e = 0.511\,\mathrm{MeV}, fK=155.7 MeVf_K = 155.7\,\mathrm{MeV} (the leptonic-decay normalisation), GF=1.1664×10−5 GeV−2G_F = 1.1664\times10^{-5}\,\mathrm{GeV^{-2}}, ℏ=6.582×10−25 GeV s\hbar = 6.582\times10^{-25}\,\mathrm{GeV\,s}. ((a)) Using the two-body decay-rate formula Γ(P→ℓν)=GF28π∣Vus∣2fK2mℓ2mK(1−mℓ2mK2)2\Gamma(P\to\ell\nu) = \dfrac{G_F^2}{8\pi}|V_{us}|^2f_K^2m_\ell^2m_K\left(1-\dfrac{m_\ell^2}{m_K^2}\right)^2, with ∣Vus∣|V_{us}| common to both channels, predict the ratio Γ(K+→e+ν)/Γ(K+→μ+ν)\Gamma(K^+\to e^+\nu)/\Gamma(K^+\to\mu^+\nu) and compare it with the measured value built from the branching fractions above. State what this comparison tests. ((b)) Using the measured Γ(K+→μ+ν)\Gamma(K^+\to\mu^+\nu) alone, extract ∣Vus∣|V_{us}| from the same formula and compare with the accepted value 0.22430.2243.

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