 Solution

The weak charged current creates a negative charged lepton in a left-chiral state. For a massless particle, that would have negative helicity; part (e) instead requires positive helicity. A nonzero mass supplies the necessary wrong-helicity component. This helicity suppression is extremely strong for the fast Electron, but much weaker for the slow muon.

For a positive-helicity charged lepton with spinors normalized to unit probability, the left-chiral weight is $(1-{\beta }_{\ell })/2$. With the box-state normalization of part (c), this gives the useful consistent rate ratioIndeed $1-{\beta }_{\ell }=2m_{\ell }^2/(m_{\pi }^2+m_{\ell }^2)$ cancels the extra energy-dependent factor in that density ratio, leaving the standard pion electron-to-muon decay ratio in inverted form:Using the rounded masses in this paper gives $R\simeq 8185$, close to the stated order of $8000$. Precision predictions require more accurate masses and electromagnetic radiative corrections. The essential mechanism is the charged-lepton-mass dependence of the decay amplitude; simply counting available final momenta misses it.