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If two semileptonic decays differ by $s\to u$ versus $d\to u$, an equal-hadronic-strength approximation and a fifth-power phase space ansatz give $r_s=K{\sin }^2{\theta }_C{\Delta }_s^5$ and $r_d=K{\cos }^2{\theta }_C{\Delta }_d^5$. Their partial decay rates are $r_i=B_i/{\tau }_i$, where $B_i$ is the appropriate branching fraction and ${\tau }_i$ the total mean lifetime. Taking the ratio gives the displayed expression. Different hadronic hadronic form factors, finite Electron mass and recoil corrections prevent this simplified extraction from being an exact determination.

A leptonic decay has only leptons in its non-radiative final state. For example, a charged pion can decay as ${\pi }^{+}\to {\mu }^{+}{\nu }_{\mu }$. A semileptonic decay instead retains at least one hadron alongside the leptons. This classification concerns the final particles, not whether the initial particle is a hadron or a lepton.

In both semileptonic decays, an active quark emits a virtual W boson, which produces an Electron and an electron antineutrino:The unchanged $u,d$ spectator quarks combine with the produced $u$ into the final proton. The relevant Cabibbo-Kobayashi-Maskawa matrix elements are $V_{ud}\simeq \cos {\theta }_C$ and $V_{us}\simeq \sin {\theta }_C$.

Use the quoted masses in energy units, equivalently natural units $c=1$, and define the energy releases before subtracting the electron rest energy asThe prescribed density of states ansatz uses precisely these differences to the fifth power. A Cabibbo-angle extraction from semileptonic decay rates also needs a common reduced hadronic strength. In that approximation the partial decay rates, measured in ${\mathrm{s}}^{-1}$, areThe supplied mean lifetimes describe all decays, so multiply their inverses by the relevant branching fractions:Taking their ratio cancels $K$ and givesChoose the conventional first-quadrant Cabibbo angle:The Lambda baryon partial rate is much larger despite its smaller weak flavour coupling because the fifth-power phase space factor is enormous.

There is an assumption beyond the stated density of states ansatz: the neutron and Lambda reduced hadronic strengths have been set equal. If their ratio is $H=K_{\Lambda }/K_n$, the same data instead imply ${\tan }^2{\theta }_C=0.0588300/H$. Without specifying $H$, the angle is not uniquely determined. Unequal hadronic form factors, finite electron mass and recoil make the displayed common-strength result an exercise-level estimate, rather than a precision measurement.

A semileptonic decay contains both final-state hadrons and leptons, typically through a weak charged current. This distinguishes it from a purely leptonic decay.