 Solution

ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2021/section-b/5/f/solution

For the charged lepton,

p_{ℓ} = β_{ℓ} γ_{ℓ} m_{ℓ}

and

E_{ℓ} = γ_{ℓ} m_{ℓ}

. Their ratio is therefore its speed in natural units. Using part (a),

β_{ℓ} = \frac{p _{ℓ}}{E _{ℓ}} = \frac{m _{π}^{2} - m _{ℓ}^{2}}{m _{π}^{2} + m _{ℓ}^{2}} .

Numerically,

β_{e} = \frac{19600 - 0.25}{19600 + 0.25} = 0.99997449, β_{μ} = \frac{19600 - 11236}{19600 + 11236} = 0.271241.

The Electron is almost ultrarelativistic, while the muon is relatively slow. In ordinary units these are approximately

0.999974 c

and

0.271 c

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