Past exam of the particle and nuclear physics course of the University of Cambridge / 2022 / Section B / 4 / d / iii

Table of contents
- Solution iii

Solution

Using the observed counts,

N_{1} + N_{4} + N_{6} = 1275240, 1 - Γ_{0} = 0.33.

Thus

Γ_{1} Γ_{2} Γ_{3} = 0.33 \frac{1 230 470}{1 275 240} = 0.318415 ≃ 31.84%, = 0.33 \frac{42 269}{1 275 240} = 0.0109382 ≃ 1.094%, = 0.33 \frac{2501}{1 275 240} = 0.000647196 ≃ 0.06472%.

Together with the

67%

direct ground state branch, these sum to unity. No measurement uncertainties were supplied for these ratios, so the extra digits only document the arithmetic.

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