= Solution
For the charged <lepton>, $p_\ell=\beta_\ell\gamma_\ell m_\ell$ and $E_\ell=\gamma_\ell m_\ell$. Their ratio is therefore its speed in <natural units>. Using part (a),
$$
\boxed{\beta_\ell=\frac{p_\ell}{E_\ell}
=\frac{m_\pi^2-m_\ell^2}{m_\pi^2+m_\ell^2}.}
$$
Numerically,
$$
\boxed{\beta_e=\frac{19600-0.25}{19600+0.25}=0.99997449,\qquad
\beta_\mu=\frac{19600-11236}{19600+11236}=0.271241.}
$$
The <Electron> is almost ultrarelativistic, while the <muon> is relatively slow. In ordinary units these are approximately $0.999974c$ and $0.271c$.
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